Part by part:

a. (True) We have $$ 0.9^2 = {9 \over 10}\cdot{9 \over 10} = {81 \over 100} = 0.81 $$ and $0.81 < 0.9$.

b. (True) We have $$ 0.1^2 = {1 \over 10} \cdot {1 \over 10} = {1 \over 100} = 0.01, $$

and $0.1$ is nonnegative, so $\sqrt{0.01} = 0.1$.

c. (True) In fact, $\sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}}$ and $\sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}}$ are both equal to $\sqrt[6]{\rule{0pt}{0.6em}2}$. To convince yourself, note that $$ \begin{align} &\,\, (\sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}}\hspace{0.1em})^6 \\ =&\,\, \rule{0pt}{1.3em} \sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}} \times \sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}} \times \sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}} \times \sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}} \times \sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}} \times \sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}}\qquad\\ =&\,\, \rule{0pt}{1.3em} (\!\hspace{0.15em}\sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}} \times \sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}}\hspace{0.11em}) \times (\!\hspace{0.15em}\sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}} \times \sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}}\hspace{0.11em}) \times (\!\hspace{0.15em}\sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}} \times \sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}}\hspace{0.11em}) \\ =& \,\, \rule{0pt}{1.3em} (\sqrt[3]{\rule{0pt}{0.64em}2}\hspace{0.1em}) \times (\sqrt[3]{\rule{0pt}{0.64em}2}\hspace{0.1em}) \times (\sqrt[3]{\rule{0pt}{0.64em}2}\hspace{0.1em})\\ =& \,\, \rule{0pt}{1.4em} 2 \end{align} $$

$$ \sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}} \times \sqrt[2]{\rule{0pt}{0.75em}\sqrt[3]{2}}\, =\, \sqrt[3]{\rule{0pt}{0.64em}2} $$

and $$ \begin{align} &\,\, (\sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}}\hspace{0.1em})^6 \\ =&\,\, \rule{0pt}{1.3em} \sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}} \times \sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}} \times \sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}} \times \sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}} \times \sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}} \times \sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}}\\ =& \,\, \rule{0pt}{1.3em} (\!\hspace{0.15em}\sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}} \times \sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}} \times \sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}}\hspace{0.11em}) \times (\!\hspace{0.15em}\sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}} \times \sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}} \times \sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}}\hspace{0.11em})\\ =&\,\, \rule{0pt}{1.3em} \sqrt[2]{\rule{0pt}{0.65em}2} \times \sqrt[2]{\rule{0pt}{0.65em}2}\\ =&\,\, \rule{0pt}{1.4em} 2 \end{align} $$

$$ \sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}} \times \sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}} \times \sqrt[3]{\rule{0pt}{0.75em}\sqrt[2]{2}} \,=\, \sqrt[2]{\rule{0pt}{0.65em}2} $$

so $(\sqrt[2]{\rule{0pt}{0.76em}\sqrt[3]{2}}\hspace{0.1em})^6 = (\sqrt[3]{\rule{0pt}{0.76em}\sqrt[2]{2}}\hspace{0.1em})^6 = 2$.

Technically, however, a number $x$ such that $$ x^6 = 2 $$

is not necessarily $\sqrt[6]{\rule{0pt}{0.6em}2}$, because $x = -\sqrt[6]{\rule{0pt}{0.6em}2}$ satisfies this equation as well!

The last step, therefore, is to note that $\sqrt[2]{\rule{0pt}{0.76em}\sqrt[3]{2}}$ and $\sqrt[3]{\rule{0pt}{0.76em}\sqrt[2]{2}}$ are both nonnegative numbers (taken as obvious), and which implies that they are the unique nonnegative solution to $x^6 = 2$.

d. (True) In general, $$ {\sqrt{x} \over \sqrt{y}} = \sqrt{\rule{0pt}{0.7em}x \over y} $$ for all $x \geq 0$, $y > 0$ (you each root to be defined), so $$ {\sqrt{2} \over 2} = {\sqrt{2} \over \sqrt{4}} = \sqrt{\rule{0pt}{0.8em}2 \over 4} = \sqrt{0.5} $$ ...ta-daa!

Note 1. One can also proceed by “direct verification”: $$ \left({\sqrt{2} \over 2}\right)^{\!2} = {\sqrt{2} \over 2}\cdot{\sqrt{2} \over 2} = {\sqrt{2}\cdot\sqrt{2} \over 4} = {2 \over 4} = 0.5. $$ (This, together with the fact that ${\sqrt{2} \over 2}$ is not negative, establishes that ${\sqrt{2} \over 2} = \sqrt{0.5}$.)

e. (True) Using the “${\sqrt{x} \over \sqrt{y}} = \sqrt{\rule{0pt}{0.7em}x \over y}$” identity: $$ {1 \over \sqrt{2}} = {\sqrt{1} \over \sqrt{2}} = \sqrt{\rule{0pt}{0.8em}1 \over 2} = \sqrt{0.5}. $$ Or by direct verification: $$ \left({1 \over \sqrt{2}}\right)^{\!2} = {1 \over \sqrt{2}}\cdot{1 \over \sqrt{2}} = {1 \over \sqrt{2}\cdot\sqrt{2}} = {1 \over 2} = 0.5. $$ (And $1 \over \sqrt{2}$ is nonnegative.) Or by reducing to part d: $$ {1 \over \sqrt{2}} = {\sqrt{2} \over \sqrt{2} \cdot \sqrt{2}} = {\sqrt{2} \over 2}. $$ (The point being: we already know that ${\sqrt{2} \over 2} = \sqrt{0.5}$ by part d.)

f. (True) We have $$ 2^{30} = 2^{10} \times 2^{10} \times 2^{10} = (2^{10})^3 $$ and $$ (2^{10})^3 = (1024)^3 > 1000^3. $$

Note 2. The first ten or so powers of $2$ are worth knowing by heart (here's eleven powers, mind you): $$ \begin{array}{c|c} \,\,\,\,n\,\,\,\, & 2^n\Rule{0pt}{0em}{0.3em} \\ \hline 0 & 1 \rule{0pt}{1.1em}\\ 1 & 2 \\ 2 & 4 \\ 3 & 8 \\ 4 & 16 \\ 5 & 32 \\ 6 & 64 \\ 7 & 128 \\ 8 & 256 \\ 9 & 512 \\ 10 & 1024 \end{array} $$ Among which, the fact that $$ 2^{10} \approx 10^3 $$

can be particularly useful to know! For example, if a 1-millimeter-thick napkin is folded $50$ times over, doubling the width each time, one obtains something of thickness $$ 2^{50}\hspace{0.1em}\text{mm} = (2^{10})^5\hspace{0.1em}\text{mm} \approx (10^3)^5\hspace{0.1em}\text{mm} = 10^{15}\hspace{0.1em}\text{mm}. $$ As $$ 1\hspace{0.1em}\text{mm} = 10^{-6}\hspace{0.1em}\text{km} $$ this is

$$ 10^{15}\hspace{0.1em}\text{mm} = 10^{15}\hspace{0.1em}(10^{-6}\hspace{0.1em}\text{km}) = \dots $$

$$ 10^{9}\hspace{0.1em}\text{km} $$

or one billion kilometers. By comparison, the distance from the Earth to the Sun is a mere $150$ million kilometers.

(The point being: that we could go from the relatively mysterious $$ \text{“}2^{50}\hspace{0.1em}\text{mm}\text{”} $$ to the relatively less mysterious $$ \text{“}\hspace{0.1em}10^{15}\text{mm}\text{”} $$ by the approximation $2^{10} \approx 10^3$.)

g. (True) As an inequality can be multiplied on both sides by a positive number while preserving the inequality, one has $$ \begin{align} & {1 \over 0.95} > 1.05\\ \iff & 1 > 1.05 \cdot 0.95\rule{0pt}{1.4em}\\ \iff & 1 > (1 + 0.05)(1 - 0.05)\rule{0pt}{1.4em}\\ \iff & 1 > 1 - 0.05^2\rule{0pt}{1.4em} \end{align} $$ (using the fact that $(1+x)(1-x) = 1-x^2$, of $$ \text{“}\,(a+b)(a-b) = a^2-b^2\,\text{”} $$ fame), and since the last inequality is true, the first inequality is true! (Recall that “$\!\iff\!$” means “$\hspace{0.1em}$if and only if”.)

Note 3. More generally, even though $$ {1 \over 1 - \epsilon} > 1 + \epsilon $$ for any small $\epsilon > 0$, the number $1 + \epsilon$ remains a good approximation to ${1 \over 1 - \epsilon}$. For example, $$ %{1 \over 0.99} 1.01 $$ is a good approximation to $$ %1.01 {1 \over 0.99} $$ while $$ 1.001 %{1 \over 0.999} $$ is a good approximation to $$ %1.001 \,{1 \over 0.999}, $$ etc.

h. (True) Here are the first few powers of $-1$ (note how each additional multiplication by $-1$ simply changes the sign of the previous result):

(Etc.) Obviously, even powers of $(-1)$ are equal to $1$, while odd powers of $(-1)$ are equal to $-1$. As $101$ is odd, $(-1)^{101}$ is $-1$.

i. (False) We have $$ {100 \over 99} = {99 + 1 \over 99} = 1 + {1 \over 99} $$ and $$ {101 \over 100} = {100 + 1 \over 100} = 1 + {1 \over 100} $$ so the smaller of the two fractions is ${101 \over 100}$, since ${1 \over 100} < {1 \over 99}$.

Note 4. The difference $$ {1 \over 99} - {1 \over 100} $$ is interesting in its own right, being connected to a famous infinite sum.

To visualize this sum, picture a hare poised at $x = 0$ on the number line. This hare runs forward by one unit and backwards by half a unit, stopping at the number $$ 1 - \frac{1}{2} $$

by virtue of this back-and-forth movement. The hare then proceeds to run forward by half a unit and back by a third of a unit, stopping at $$ \begin{align} &\, \left(1 - {1 \over 2}\right)\\ + \,&\, \left({1 \over 2} - {1 \over 3}\right)_{\color{white} a_{a_a}\!\!\!\!\!\!\!\!\!\!}\\ \hline = \,&\, \left(1 - {1 \over 3}\right)^{\color{white} a^{a^a}} \end{align} $$

for another break. Keeping with this pattern, the hare then stops at $$ \begin{align} &\, \left(1 - {1 \over 2}\right)\\ + \,&\, \left({1 \over 2} - {1 \over 3}\right)\\ + \,&\, \left({1 \over 3} - {1 \over 4}\right)_{\color{white} a_{a_a}\!\!\!\!\!\!\!\!\!\!}\\ \hline = \,&\, \left(1 - {1 \over 4}\right)^{\color{white} a^{a^a}} \end{align} $$ and then at $$ \begin{align} &\, \left(1 - {1 \over 2}\right)\\ + \,&\, \left({1 \over 2} - {1 \over 3}\right)\\ + \,&\, \left({1 \over 3} - {1 \over 4}\right)\\ + \,&\, \left({1 \over 4} - {1 \over 5}\right)_{\color{white} a_{a_a}\!\!\!\!\!\!\!\!\!\!}\\ \hline = \,&\, \left(1 - {1 \over 5}\right)^{\color{white} a^{a^a}} \end{align} $$ and so on. Clearly, the successive positions at which the hare stops are approaching the number $1$ from the left, pointing to the fact that the infinite sum $$ \begin{align} &\, \left(1 - {1 \over 2}\right)\\ + \,&\, \left({1 \over 2} - {1 \over 3}\right)\\ + \,&\, \left({1 \over 3} - {1 \over 4}\right)\\ + \,&\, \left({1 \over 4} - {1 \over 5}\right)\\ + \,&\, \left({1 \over 5} - {1 \over 6}\right)\\ + \,&\, \left({1 \over 6} - {1 \over 7}\right)\\ + \,&\, \,\,\,\,\,\,\,\,\dots\rule{0pt}{1.3em} \end{align} $$ is “equal” (in some sense) to $1$. But how much, exactly, is the $n$-th term $$ {1 \over n} - {1 \over n+1} $$ of the sum? (By the way, this $n$-th term is the difference ${1 \over 99} - {1 \over 100}$ for $n = 99$, which is how we came to be reminded of this infinite sum in the first place.) Well... $$ \begin{align} {1 \over n} - {1 \over n+1} &= {1 \over n}\cdot{n+1 \over n+1}\, - \, {1 \over n+1}\cdot{n \over n}\rule{0pt}{1.5em}\\ &= {n+1 \over n(n+1)} - {n \over n(n+1)}\rule{0pt}{1.5em}\\ &= {1 \over n(n+1)}\rule{0pt}{1.5em} \end{align} $$

$$ \begin{align} {1 \over 99} - {1 \over 100} &= {1 \over 99}\cdot{100 \over 100}\, - \,{1 \over 100}\cdot{99 \over 99}\rule{0pt}{1.5em}\\ &= {100 \over 99\cdot 100} - {99 \over 99\cdot 100}\rule{0pt}{1.5em}\\ &= {1 \over 99\cdot 100}\rule{0pt}{1.5em} \end{align} $$

...it's that much. (For example, $$ {1 \over 1} - {1 \over 2} = {1 \over 1 \cdot 2} = {1 \over 2} $$ and $$ {1 \over 2} - {1 \over 3} = {1 \over 2 \cdot 3} = {1 \over 6} $$ and so on.) So the infinite sum $$ \begin{align} &\, \left(1 - {1 \over 2}\right)\\ + \,&\, \left({1 \over 2} - {1 \over 3}\right)\\ + \,&\, \left({1 \over 3} - {1 \over 4}\right)\\ + \,&\, \left({1 \over 4} - {1 \over 5}\right)\\ + \,&\, \left({1 \over 5} - {1 \over 6}\right)\\ + \,&\, \left({1 \over 6} - {1 \over 7}\right)\\ + \,&\, \,\,\,\,\,\,\,\,\dots\rule{0pt}{1.3em}\Rule{0pt}{0em}{1em}\\ \hline = \,&\, 1\rule{0pt}{1.5em} \end{align} $$ can also be written $$ {1 \over 1 \cdot 2} + {1 \over 2 \cdot 3} + {1 \over 3 \cdot 4} + {1 \over 4 \cdot 5} + {1 \over 5 \cdot 6} + \dots \,=\, 1 $$ (or $$ {1 \over 2} + {1 \over 6} + {1 \over 12} + {1 \over 20} + {1 \over 30} + \dots \,=\, 1 $$ equivalently) which is not obvious at first glance, and kind of interesting!

Note 5. The fact that $$ {1 \over n} - {1 \over n+1} = {1 \over n(n+1)} $$ means, in particular, that ${1 \over n} - {1 \over n+1}$ is roughly ${1 \over n^2}$ for large $n$, which is sometimes handy to know. For example, $$ {1 \over 10} - {1 \over 11} $$ is approximately $1/10^2 = 0.01$, while $$ {1 \over 100} - {1 \over 101} $$ is approximately $1/100^2 = 0.01^2 = 0.0001$, etc.

Since $$ {1 \over n} - {1 \over n+1} = {1 \over n(n+1)} $$ (as previously computed) is smaller than

$${1 \over 99\cdot 100}$$

$${1 \over 99^2}$$

$$ {1 \over n^2} $$ we will compute the difference $$ {1 \over n^2} - {1 \over n(n+1)} $$ as opposed to the “other” difference $$ {1 \over n(n+1)} - {1 \over n^2} $$

to avoid a minus sign in the result. (Computing the second difference and having a minus sign does not constitute a mistake, however.) Having said this, the difference is:

$$ \begin{align} {1 \over 99^2} - {1 \over 99\cdot 100} &= {1 \over 99^2}\cdot{100 \over 100}\, - \, {1 \over 99\cdot 100}\cdot{99 \over 99}\\ &= {100 \over 99^2 \cdot 100} - {99 \over 99^2 \cdot 100}\rule{0pt}{1.5em}\\ &= {1 \over 99^2\cdot 100}\rule{0pt}{1.5em} \end{align} $$

$$ \begin{align} {1 \over n^2} - {1 \over n(n+1)} &= {1 \over n^2} \cdot {n+1 \over n+1} - {1 \over n(n+1)} \cdot {n \over n}\\ &= {n+1 \over n^2(n+1)} - {n \over n^2(n+1)}\rule{0pt}{1.5em}\\ &= {1 \over n^2(n+1)}\rule{0pt}{1.5em} \end{align} $$ which, for large $n$, is roughly $$ {1 \over n^3} $$

since $n^2(n+1) \approx n^3$ for large $n$.

Example 1. Above, we estimated $$ {1 \over 10} - {1 \over 11} $$ to be roughly $1/100 = 0.01$, but $1/100$ is bigger than the actual value of ${1 \over 10\cdot 11} = {1 \over 110}$ by $$ {1 \over 10^2\cdot 11} $$

or close to $1/10^3 = 0.001$. (So $$ {1 \over 10} - {1 \over 11} $$ is about $0.01$, while making an error of about $0.001$.) (In fact, $$ {1 \over 10^2\cdot 11} $$ is less than $1/10^3$, so the error is less than $0.001$.)

This is true, as illustrated by the following pair of lines:

In more detail, the two triangles are related by a $90^\circ$ rotation and so, likewise, are the lines defined by their hypotenuses!

Note 1. More generally, a line of slope $p$ is perpendicular to a line of slope $-1/p$, for all $p \ne 0$. By a similar drawing:

(If the axes are oriented the usual way then the above drawing covers all the cases $p > 0$, whereas if we flip the two number axes to point down/left the above drawing covers all the cases $p < 0$—magic!)

A point $(x,y) \ne (x_0, y_0)$ is on the line of slope $p$ if and only if

$$p = {y - y_0 \over x - x_0}$$ because $${y - y_0 \over x - x_0}$$ is the slope of the line segment from $(x_0,y_0)$ to $(x,y)$, and it is necessary and sufficient for this segment to have slope $p$ in order for the point $(x,y)$ to be on the line!

Unfortunately, the equation $$ p = {y - y_0 \over x - x_0} $$ is not an entirely satisfactory answer, because the point $(x,y) = (x_0,y_0)$ itself does not satisfy the equation. (We find $$ p = {0 \over 0} $$ if we plug in $x = x_0$, $y = y_0$, which is not a valid equality because the right-hand side is an undefined quantity.)

Instead, multiplying $$ p = {y - y_0 \over x - x_0} $$

on both sides by $x-x_0$, we find the fraction-less equation $$ p(x-x_0) = y-y_0 $$ which is satisfied by the point $(x,y) = (x_0,y_0)$ as well as by every other point on the line. This can be a final answer, and, pleasingly, has the form $$ \text{“}\text{slope} \times \text{run} = \text{rise}\text{”} $$ which can also make it easy to remember!

Note 1. The answer we gave is more often written $$ y - y_0 = p(x - x_0) $$ with the two sides of the equation swapped, or $$ y = p(x - x_0) + y_0 $$ with $y$ isolated on the left-hand side. From there one can also distribute $p(x-x_0)$, obtaining (after putting “$-px_0$” last) $$ y = px + y_0 - px_0 $$ which has the form $$ y = ax + b $$ with $a = p$, $b = y_0 - px_0$.

Here is the “official” graph of the (vertical) velocity:

On each interval, the velocity is rate of change of the height, i.e., the slope of the height. For example, the rate of change of the height is $$ {1\text{m} \over 1\text{s}} = 1\text{m}/\text{s} $$ between $-4$s and $-3$s, where the mosquito goes up by one meter during a one second period, so the vertical velocity is 1m$/$s for that time interval, etc.

Note 1. As explained in Chapter 3, an empty circle of this type

indicates a “missing” value. Specifically, in our case, the vertical velocity is undefined wherever the graph of the height has a sharp corner. (Because the slope of the graph is not well-defined at such corners.)

Note 2. For the time interval from $2$s to $2.5$s, the slope is $$ {-2\hspace{0.05em}\text{m} \over 0.5\hspace{0.05em}\text{s}} = -\hspace{0.07em}4\hspace{0.1em}\text{m}/\text{s} $$ and similarly for the time interval from $2.5$s to $3$s the slope is $$ {2\hspace{0.05em}\text{m} \over 0.5\hspace{0.05em}\text{s}} = 4\hspace{0.1em}\text{m}/\text{s} $$ because $2/0.5 = 4$. (Think: how many times does $0.5$ go into $2$?)

E.g.:

The rise from the $y$-intercept to $(x_0, y_0)$ is $px_0$, as found by “rise equals slope times run”, implying that $$ y_0 - px_0 $$ is the $y$-intercept.

Note 1. Our drawing makes some implicit assumptions, such as $p > 0$ and $x_0 > 0$. But $$ px_0 $$ is the rise from the $y$-intercept to $(x_0, y_0)$ regardless of the sign of $p$ or $x_0$ (because “$x_0$” is the run in all cases), making $$ y_0 - px_0 $$ the $y$-intercept in all cases.

The absolute value function is $$ x \rightarrow \begin{cases} x & \text{if $x \geq 0$,}\\ -x\!\!\rule{0pt}{1.2em} & \text{if $x < 0$}\end{cases} $$ because $-(-1) = 1$, $-(-5) = 5$, etc.

We have $$ |x| = \sqrt{x^2} $$ because $\sqrt{(-1)^2} = 1$, $\sqrt{(-5)^2} = 5$, etc.

Note 1. This definition is less ad-hoc than might seem, being a 1-dimensional form of the Pythagorean theorem.

The answers are:

As $x$ grows, so does $\lfloor{x}\rfloor$, but $\lfloor{x}\rfloor$ only “levels up” each time $x$ reaches a new integer, and “flatlines” otherwise; this gives rise to the following staircase-shaped graph:

(For example, $\lfloor{1}\rfloor = 1$ because the greatest integer less than or equal to $1$ is $1$, $\lfloor{-0.5}\rfloor = -1$ because the greatest integer less than or equal to $-0.5$ is $-1$, and so on.)

For the second part note that the following two displacements, excerpted from the “factory roof” graph in the statement, are equal:

The red dot to the left of $x$ has $x$-coordinate $\lfloor{x}\rfloor$, so the horizontal displacement is $$ x - \lfloor{x}\rfloor $$ so the equation of the graph is $$ y = x - \lfloor{x}\rfloor $$ because the $y$-coordinate is the vertical displacement, given that the vertical displacement starts at $y = 0$, and because the vertical and horizontal displacements are equal.

We would like to argue the correctness of the following two-step process (divide the input by $2$, apply the function from Exercise 4):

Indeed, the two graphs featured above differ only by a horizontal dilation; dividing the input by $2$ “undoes” the dilation, at which point it suffices to apply the function pictured in the second graph; having declared our method correct, the answer is thus... $$ {x/2 - \lfloor x/2 \rfloor} $$ ...as obtained by “sticking” $x/2$ (the halved input) in place of “$x$” in “$\,x - \lfloor x \rfloor$”, the formula for the function from Exercise 4.

Note 1. One can check the answer by typing “x/2 - floor(x/2)” in DESMOS. Viz:

Note 2. Alternately, enter “f(x) = x - floor(x)” and then “f(x/2)”, viz:

Or we can be even fancier:

What you see above (the graph in orange) is the so-called

composition

of the functions $\Rule{0.12em}{0.8pt}{-0.8pt}f$ and $g$; in more detail, if we switch the “input tube” and “output tube” sides of a function...

...(compared to the drawing at the top of the chapter), then the composition of $\Rule{0.12em}{0.8pt}{-0.8pt}f$ and $g$, written $$ {\Rule{0.12em}{0.8pt}{-0.8pt}f \circ g} $$ and read

“$f$ of $\!\hspace{0.1em}{}g$”

(mathematicians have to invent a notation for everything—that little circle “$\circ$” is called the composition operator, by the way) is the function that you get by gluing $g$'s box to the right of $\Rule{0.12em}{0.8pt}{-0.8pt}f$'s box, like so:

In other words, $g$'s output is passed on to $\Rule{0.12em}{0.8pt}{-0.8pt}f$ for further processing. (A certain movie called “The Human Centipede” comes to mind.)

(To be perfectly clear,

$f \circ g$ is a function, defined as the above assemblage of “$g$ first, $f$ second”.)

Note 3. For a formal definition of “$f \circ g$”—something not based on pictures—one need only specify what $f \circ g$ does to inputs. Specifically: $$ \,{(f \circ g)(x) = f(g(x))}. $$ (So that equation is a formal definition.) One can also clarify that $$ {\text{dom}\, f \circ g = \{x\, \in\, \text{dom}\, g:\, g(x)\, \in\, \text{dom}\, f\}} $$ which is to say that the domain of $f \circ g$ consists of all $x$ such that: (i) $g(x)$ exists (a.k.a, “$x \in \text{dom}\, g$”) and, (ii) $f(g(x))$ exists (a.k.a., “$g(x) \in \text{dom}\, f$”).

Note 4. Amusingly—or not—both sides of $$ {(f \circ g)(x) = f(g(x))} $$ are read

“$f$ of $\!\hspace{0.1em}{}g$ of $x\hspace{0.1em}$”

since “$f \circ g$” is read “$f$ of $g\hspace{0.1em}$”, and “$f(\dots)$” is read “$f$ of ...”.

For the first graph, the an answer is $$ 2 \cdot(x/2 - \lfloor x/2 \rfloor) $$ which simplifies to $$ x - 2\lfloor x/2 \rfloor $$ because all we have to do is to multiply Exercise 5's formula by $2$.

For the second graph, the an answer is $$ x/3 - \lfloor x/3 \rfloor $$ because the problem is similar to Exercise 5 except with a factor $3$ horizontal dilation.

For the third graph, we will first stop to find a formula for the function depicted here:

And that formula is...

...iiiiiiiS... $$ (x-1)/3 - \lfloor (x-1)/3 \rfloor $$ as obtained by substituting “$x - 1$” (the input, minus $1$) in place of “$x$” in “$\,x/3 - \lfloor x/3 \rfloor$”, the formula for the second graph. Then we multiply that by $3$ (to go from “” to “”, namely), meaning that the final answer is $$ 3 \cdot ((x-1)/3 - \lfloor (x-1)/3 \rfloor) $$ or $$ (x - 1) - 3\lfloor (x-1)/3 \rfloor $$ after simplification. (Or just $$ x - 1 - 3\lfloor (x-1)/3 \rfloor $$

though we personally prefer the previous form, it being more “talkative”.)

Consider how to “read off” a value of $y = \cos(1000x)$ from the graph $y = \cos(x)$:

By the first step, a

horizontal dilation by a factor 1000

maps the first graph onto the second graph—i.e., a point $$ (x, y) $$

is on the first graph if and only the dilated point $$ (1000x, y) $$ is on the second graph. The first graph is therefore some very compressed thing, full of scrunched bumps!

Note 1. One can also reason that a small change in $x$ results in a large change in ${1000x}$, so that ${\cos(1000x)}$ must “cycle” much faster through values than ${\cos(x)}$ does.

Setting “$r$” to “$f$” and “$s$” to “$(g \circ h)$” in (**) yields

...which already constitutes progress towards our goal, since only one copy of “$\circ$” exists on the right-hand side! But

by the “definitional equation” again, so

...and this completes the computation!

Note 1. We can collect both steps of the computation into a single string of equalities:

We will again evaluate the “outer” composition operator first and the “inner” composition operator second, where the “outer” composition operator is the one that is fewer pairs of parentheses away from the outside world:

So the first step is...

...by setting $r = f \circ g$, $s = h$ in the definitional equation, and the second step is...

...by setting $r = f$, $s = g$, and while replacing “$x$” by “$h(x)$”.

Note 1. The fact that $$(f \circ (g \circ h))(x)$$ and $$((f \circ g) \circ h)(x)$$ both evaluate to $$f(g(h(x)))$$ actually implies that $$f \circ (g \circ h)$$ and $$(f \circ g) \circ h$$ are the same function; this function is namely the function that maps $x$ to $f(g(h(x)))$ for all $x$ (or $$x \rightarrow f(g(h(x)))$$ in lambda notation).

Note 2. Because of this, we can write $$f \circ g \circ h$$ without any parentheses. (The point is: either way you parenthesize it you obtain the same function, so why bother?)

Note 3. The fact that

for all numbers $a$, $b$, $c$ is known as the

associativity

of addition; likewise, the fact that

for all numbers $a$, $b$, $c$ is known as the

associativity

of multiplication; and again likewise, the fact that

for all functions $f$, $g$, $h$ is known as the

associativity

of function composition.

Note 4. One of the best ways to explain & understand the associativity of function composition uses this picture:

In the above $A$, $B$, $C$, $D$ are sets while the arrows encode functions $f$, $g$ and $h$ that, respectively in reverse order, go from $D$ to $C$, $C$ to $B$, and $B$ to $A$. For example,

${\Large h(}$${\Large{}) =\hspace{0.2em}}$

because the arrow that originates at in set $D$ lands at in set $C$, and

${\Large g(h(}$${\Large{})){}=\hspace{0.2em}}$

because, pursuing that path onwards, the arrow that originates at in set $C$ lands at in set $B$, etc.

Under this representation one can “compute” $f \circ g \circ h$ by gluing arrows end-to-end. First, say, obliviate set $C$ in the middle right, then do the same with set $B$ in the middle left:

We can also get rid of $B$ first, $C$ second:

The first order of computation corresponds to the parenthetization “$f \circ (g \circ h)$” while the second corresponds to the parenthetization “$(f \circ g) \circ h$”. Intuitively, the reason they come out the same (in “step 6”, bottom left) is because each final arrow in the last diagram comes from a path-of-arrows in the original diagram, and the order in which the waypoints along a path are “straightened” (or “collapsed”) does not affect the origin point or destination point of the final arrow.

Note 5. The last series of diagrams might leave one with the impression that the composition of two or more functions can be “precomputed” by looking ahead along the path of yellow arrows. Just so you know, computers do not generically do this. For reason, computers are not given functions as tables of input-output values to know by heart but rather as “recipes” (synonyms: algorithms, code, programs) that allow them to compute an output for any given input. Moreover, there is no general way of flattening two recipes into a single, shorter one—when composing two functions the computer has, in general, no choice but to diligently apply each recipe in order—the first function first, the second function second.

Note 6. We have taken for granted the fact that two functions $f$ and $g$ are “equal” if and only if they produce the same outupt for every input but this is a actually subtle thing that has to do with how functions are defined “under the hood”. Specifically, mathematicians view functions as long lists of sets of ordered pairs; for example—conceptual cold water shock ahead—

because VX-11/78A maps $0$ to $3.5$ and maps $1$ to $2.5$. (The presence of an ordered pair

means that input $a$ produces output $b$.) So two functions are equal if and only if they are equal

as sets of ordered pairs

because the set of ordered pairs is the underlying “thing” that the function is. In particular, there is no notion of a “formula” or of a “procedure” being attached to a function, that might cause two functions to be considered unequal even if they produce the same output on every input—producing the same output on every input implies that the list of set of ordered pairs is equal, and, perforce, that the two functions are equal!!

The answer is $$ (x + 5)^2 + 5 $$ because $$ \begin{align} \,\,\,\,(x + 5)^2 &= x^2 + (2\cdot 5)x + 5^2 \\ &= \rule{0pt}{1.4em} x^2 + 10x + 25 \end{align} $$ and adding $5$ gives $x^2 + 10x + 30$.

Put an unknown “$U$” for the first set of dots and an unknown “$V$” for the second set of dots. Then $$ (x + U)^2 + V = x^2 + 10x + 30 $$ becomes the equation to satisfy. Expanding the left-hand side, we get: $$ x^2 + 2Ux + U^2 + V = x^2 + 10x + 30. $$ In order for this equation to hold as an equality between polynomials (i.e., for all $x$) the coefficients of $x^2$ on both sides of the equation must be equal, the coefficients of $x$ on both sides of the equation must be equal, and the constant terms on boths of the equation must be equal—this gives us $$ 1 = 1 $$ (equating the coefficients of $x^2$), and $$ 2U = 10 $$ (equating the coefficients of $x$), and $$ U^2 + V = 30 $$ (equating the constant terms). Only the latter two equations contain information. In particular, $$ 2U = 10 $$ implies $U = 5$, so $U^2 + V = 30$ becomes $25 + V = 30$, and $V = 30 - 25 = 5$. So $U = V = 5$, as previously found. (But now we know that the solution is unique, because the only number $U$ that satisfies $$ 2U = 10 $$ is $U = 5$, and the only number $V$ that satisfies $$ 25 + V = 30 $$ is $V = 5$.)

The equation is equivalent to $$ (x + 5)^2 + 5 = 0 $$ by Exercise 21, but this implies $$ (x + 5)^2 = -5 $$ which is an equation with no solution over the reals because the square of a real number is nonnegative.

The equation can be written $$ (x + 5)^2 - 55 = 0 $$ because $(x + 5)^2 = x^2 + 10x + 25$ and $25 - 55 = -30$. Passing $55$ to the other side, we find $$ (x + 5)^2 = 55 $$ which holds if and only if $$ \,x + 5 = \pm\sqrt{55} $$ or $$ \,x = -5 \pm\sqrt{55} $$ constituting two distinct solutions.

Write $$ {Ax^2 + Bx} $$ as $$ {A\Big(x^2 + {B \over A}x\Big)} $$ and then write $$ {x^2 + {B \over A}x} $$ as $$ {\Big(x + {B \over 2A}\Big)^2 - {B^2 \over 4A^2}} $$ so that, altogether, $$ {Ax^2 + Bx} $$ is rewritten $$ {A\left[\Big(x + {B \over 2A}\Big)^2 - {B^2 \over 4A^2}\right]} $$ that can be seen as descending from $y = x^2$ in three steps: $$ {y = x^2} $$ $$ {\downarrow} $$ $$ {y = \,\Big(x + {B \over 2A}\Big)^2} $$ $$ {\downarrow} $$ $$ {y = \,\Big(x + {B \over 2A}\Big)^2 - {B^2 \over 4A^2}} $$ $$ {\downarrow} $$ $$ {y = A\left[\Big(x + {B \over 2A}\Big)^2 - {B^2 \over 4A^2}\right]}. $$

Three steps, three geometric transformations! The third step effects a

vertical scaling by $A$

i.e., vertically stretches the graph by a factor $A$, because we multiply the value of $y$ by $A$. The second step effects a

vertical translation by ${-{B^2 \over 4A^2}}$

i.e., lowers the height of the entire graph by ${B^2 \over 4A^2}$, because we add $-{B^2 \over 4A^2}$ to the value of $y$. The first step, on the other hand, is entirely different: it is a

preprocessing

step, in that we mess with the input (i.e., $x$), instead of adding on (or “multiplying on”) to the current value of $y$.

To understand how a preprocessing step affects the shape of a graph, note that, more generally, a graph of the form $$ {y = f(x + a)} $$ (for some constant $a$) “fetches” values on the graph $$ {y = f(x)} $$ by going $a$ units to the right. The larger $a$ is, thus, the further $$ {y = f(x + a)} $$ drifts off to the left. For example, $$ {y = f(x + 20)} $$ has value $f(0)$ at $x = -20$, and if you replace $20$ with something larger, that position (i.e., $x = -20$) drifts even further off to the left! In any case, the graph $y = f(x + a)$ is the

leftward

translate by $a$ units of $y = f(x)$ and, as a consequence, the first step effects a

leftward translation by ${B\over 2A}$

of the curve $y = x^2$, or

horizontal translation by $-{B\over 2A}$

more elegantly put. (The second formulation doesn't assume a particular orientation of the $x$-axis, that's why it's “more elegantly put”, in our opinion.)

To recapitulate, the three transformations are, in order:

1. horizontal translation by $-{B \over 2A}$

2. vertical translation by ${-{B^2 \over 4A^2}}$

3. vertical scaling by $A$

* * * *

Note 1. You could do the vertical translation before the horizontal translation, geometrically it comes out the same. That order of geometric transformations would correspond to the following sequence of algebraic transformations: $$ {y = x^2} $$ $$ {\downarrow} $$ $$ {y = x^2 - {B^2 \over 4A^2}} $$ $$ {\downarrow} $$ $$ {y = \,\Big(x + {B \over 2A}\Big)^2 - {B^2 \over 4A^2}} $$ $$ {\downarrow} $$ $$ {y = A\left[\Big(x + {B \over 2A}\Big)^2 - {B^2 \over 4A^2}\right]} $$ ...in which the second step is a preprocessing step. (I.e., a step that replaces “$x$” with something else.)

Start by noting that the point $(\sqrt{y_0}, y_0)$ is on the curve $y = x^2$, as well as the point $(-\sqrt{y_0}, y_0)$, because $(\sqrt{y_0})^2 = (-\sqrt{y_0})^2 =$ $y_0;$ here is a sketch of the situation before anything happens:

After vertically translating by $-y_0$ the roots will therefore be at $x = \pm\sqrt{y_0}$:

Then after horizontally translating by $x_0$ the roots mosey over to $x = x_0\pm\sqrt{y_0}$:

Lastly, vertical scaling does not affect the position of the roots, because it stretches the graph about the $x$ axis (here $a \approx 1.7$):

So the roots are at: $x = x_0 \pm \sqrt{y_0}$. (Like we found them after the second step.)

Well, $$ Ax^2 + Bx = 0 $$ obviously has solution $x = 0$ to start with, so we don't need the previous exercises for one of the roots at least—actually, $$ Ax^2 + Bx = x(Ax + B) $$ so the equation is equivalent to $$ x(Ax + B) = 0 $$ and so one of the roots is $$ x = 0 $$ and the other root is the value of $x$ such that $$ Ax + B = 0 $$ which is $x = -B/A$. (In order for the product $$ x(Ax + B) $$ to be $0$ you either need the first term to be $0$, leading us to $x = 0$, or the second term to be $0$, leading us to $Ax + B = 0$—the product of two things is $0$ if and only if one of the two things is $0$.)

So the roots are $x = 0$ and $x = -B/A$.

To complete the problem as we were asked, however, we will use the fact that $y = Ax^2 + Bx$ is obtained from $y = x^2$ by the following sequence of transformations (cf. Exercise 14):

1. vertical translation by ${-{B^2 \over 4A^2}}$

2. horizontal translation by $-{B \over 2A}$

3. vertical scaling by $A$

(We put the vertical translation first.) By Exercise 15, the roots of $y = Ax^2 + Bx$ are thus at

which looks a little different than our previous result of $x = 0$ and $x = -B/A$ until you realize that $$ \pm \sqrt{B^2 \over 4A^2} = \pm {B \over 2A} $$ (because $$ \left({B \over 2A}\right)^{\!2} = {B^2 \over 4A^2} $$ and even though ${B\over 2A}$ could be negative, the “$\pm$” on either side of the equation means that the set of values on either side of the equation is the same), so that (*) becomes $$ x = -{B\over 2A} \pm {B \over 2A} $$ and, on the one hand, $$ -{B\over 2A} + {B \over 2A} = 0 $$ while, on the other hand, $$ -{B\over 2A} - {B \over 2A} = -{2B\over 2A} = -{B\over A} $$ so here too we find that the roots are $x = 0$ and $x = -B/A$. (It must be the right answer!)

The true statements are i, iv, because $f \circ g$ is the function that maps an input $x$ to $f(g(x))$, and symmetrically for $g \circ f$.

The first equality is false because the right-hand side is actually $$ (f \circ g) + (f \circ h) $$ not $f \circ (g + h)$; the second equality is true; the third equality is true even though you would expect the right-hand side to be written $$ (x \rightarrow g(f(x)) + h(f(x))) $$ with “$g$” and “$h$” switched (but addition is commutative, so it doesn't matter); the fourth equality is true: it is the definition of “$\circ$”.

Write $$ { Ax^2 + Bx + C} $$ as $$ { A\Big(x^2 + {B \over A}x + {C\over A}}\Big) $$ and, similarly to Exercise 14, write $$ { x^2 + {B \over A}x} $$ as $$ { \Big(x + {B \over 2A}\Big)^2 - {B^2 \over 4A^2}} $$ so that, altogether, $Ax^2 + Bx + C$ becomes $$ { A\left[\Big(x + {B \over 2A}\Big)^2 - {B^2 \over 4A^2} + {C\over A}\right]} $$ or $$ { A\left[\Big(x + {B \over 2A}\Big)^2 - {B^2 - 4AC\over 4A^2}\right]} $$ by putting things on a common denominator. (We have endeavored to keep the minus sign out front of the common denominator fraction in order to maintain the most similarity with the term “$-{B^2\over 4A}$” of Exercise 14, that also has a minus sign out front.)

By direct analogy with Exercise 14, the three transformations are thus...

1. horizontal translation by $-{B \over 2A}$

2. vertical translation by ${-{B^2 - 4AC \over 4A^2}}$

3. vertical scaling by $A$

...or...

1. vertical translation by ${-{B^2 - 4AC \over 4A^2}}$

2. horizontal translation by $-{B \over 2A}$

3. vertical scaling by $A$

...if we put the vertical translation first.

The curve $$ y = Ax^2 + Bx + C $$ is obtained from the curve $y = x^2$ by the following sequence of transformations (cf. Exercise 19):

1. vertical translation by ${-{B^2 - 4AC \over 4A^2}}$

2. horizontal translation by $-{B \over 2A}$

3. vertical scaling by $A$

~

On the one hand, if $$ {B^2 - 4AC \over 4A^2} < 0 $$ then $$ -{B^2 - 4AC \over 4A^2} > 0 $$ and the vertical translation is upward, the curve detaches from the $x$ axis never to see it again, and there are no roots!

On the other hand, if $$ {B^2 - 4AC \over 4A^2} \geq 0 $$ then the roots are given by $$ x = -{B \over 2A} \pm \sqrt{B^2 - 4AC \over 4A^2} $$ by Exercise 15. $\rightarrow$ ~The End~ $\leftarrow$

Note 1. In fact, $$ \pm\sqrt{B^2 - 4AC \over 4A^2} = \pm {\sqrt{B^2 - 4AC} \over 2A} $$ (square both sides of the equation—in general, $$ \pm E = \pm F $$ as one set of two values equalling another set of two values, if and only if $$ |E| = |F| $$ or $$ E^2 = F^2 $$ —so that's why we say “square both sides”), so the formula for the roots can also be written $$ x = -{B \over 2A} \pm {\sqrt{B^2 - 4AC} \over 2A} $$ or $$ x = {{-B \pm \sqrt{B^2 - 4AC}} \over 2A} $$ as briefly flashed by, e.g., in Chapter 1.

Note 2. If $$ {B^2 - 4AC \over 4A^2} < 0 $$ then $$ \sqrt{B^2 - 4AC \over 4A^2} $$ does not exist, alerting you to the absence of roots, if you try to use the first formula we gave. Also $$ {B^2 - 4AC \over 4A^2} < 0\iff B^2 - 4AC < 0 $$ because $4A^2 > 0$ for all $A \ne 0$, so the second set of formulas would alert you to the absence of roots in that case, as well.

...in an expression such as... $$ (x \rightarrow x^3)(6) $$ ...we pair the $x$ with $6$...

...and $6$ becomes the value to use for $x$ in “$x^3$”:

...; in an expression such as... $$ (x \rightarrow y \rightarrow x^3y)(6) $$ ...we also pair the $x$ with $6$...

...and $6$ becomes the value to use for $x$ in “$y \rightarrow x^3y$”:

...(in this case the result is not a number, but a function—a function is a mathematical object like another, after all); in an expression such as... $$ (x \rightarrow y \rightarrow x^3y)(6)(8) $$ ...we pair the $x$ with $6$ and the $y$ with $8$...

...and $6$ and $8$ become respectively the values to use for $x$ and $y$ in “$x^3y$”:

...; now in an expression such as... $$ (f \rightarrow x \rightarrow h \rightarrow {f(x+h) - f(x)\over h})(x \rightarrow x^2)(5)(0.1) $$ ...we pair the $f$ with $x \rightarrow x^2$, the $x$ with $5$, and the $h$ with $0.1$...

...and $x \rightarrow x^2$, $5$ and $0.1$ become respectively the values to use for $f$, $x$ and $h$ in “${f(x + h) - f(x)\over h}$”:

...; evaluating... $$ {(x \rightarrow x^2)(5 + 0.1) - (x \rightarrow x^2)(5) \over 0.1} $$ ...we... $$ {(x \rightarrow x^2)(5.1) - (x \rightarrow x^2)(5) \over 0.1} $$ ...get... $$ {5.1^2 - 5^2 \over 0.1} $$ ...this... $$ {26.01 - 25 \over 0.1} $$ ...thiiis... $$ {1.01 \over 0.1} = 1.01 \times 10 = 10.1 $$ ...result! (The answer is: ten point one.)

That would be:

(The derivative is $1/2$ when the slope is $1/2$, is $-1/2$ when the slope is $-1/2$, and is undefined at the corners.)

The graph of $y = {1 \over x}$ looks like so:

As one can see, the slope is very negative near $x = 0$, on either side. So the answer is: “very large negative”.

The second derivative is zero wherever the first derivative is flat, and is undefined wherever the first derivative is undefined; this gives the second derivative the following pockmarked appearance:

Note 1. Taking even further derivatives produces the same graph back, over and over again.

Note 2. “first derivative” is a synonym of “derivative”.

The units on the $y$ axis become kilometers, kilometers per hour, and kilometers per hours squared (or “kilometers per hour, per hour”), including the first graph. (Each time another derivative is taken, divide the units of the $y$ axis by the units of the $x$ axis.) These are the position, velocity, and acceleration of the car as a function of time:

The equation is true! Syntactically, $$ (x \rightarrow x + 1) \,+\, (u \rightarrow 2u + 1) $$ is a

sum of functions

because $x \rightarrow x + 1$ and $u \rightarrow 2u + 1$ are both functions. Now by definition, the sum $$ f + g $$ of functions $f$ and $g$ is the function $$ x \rightarrow f(x) + g(x) $$ that maps a number to the sum of the individual values of the functions. So—for example— $$ \begin{align} & \,\,\,((x \rightarrow x + 1) \,+\, (u \rightarrow 2u + 1))(5) \\ =& \,\,\,(x \rightarrow x + 1)(5) + (u \rightarrow 2u + 1)(5) \rule{0pt}{1.5em} \\ =& \,\,\,(5 + 1) + (2\cdot 5 + 1) \rule{0pt}{1.5em} \\ =& \,\,\,3\cdot 5 + 2 = 17 \rule{0pt}{1.5em} \end{align} $$ and—with a general input $t$— $$ \begin{align} & \,\,\,((x \rightarrow x + 1) \,+\, (u \rightarrow 2u + 1))(t) \\ =& \,\,\,(x \rightarrow x + 1)(t) + (u \rightarrow 2u + 1)(t) \rule{0pt}{1.5em} \\ =& \,\,\,(t + 1) + (2t + 1) \rule{0pt}{1.5em} \\ =& \,\,\,3t + 2 \rule{0pt}{1.5em} \end{align} $$ which implies that, indeed, $$ (x \rightarrow x + 1) \,+\, (u \rightarrow 2u + 1) $$ is the function that maps each real number $t$ to $3t + 2$, i.e., is equal to the function $t \rightarrow 3t + 2$. (!!)

Note 1. One can also do the main computation with $x$ in place of $t$: $$ \begin{align} & \,\,\,((x \rightarrow x + 1) \,+\, (u \rightarrow 2u + 1))(x) \\ =& \,\,\,(x \rightarrow x + 1)(x) + (u \rightarrow 2u + 1)(x) \rule{0pt}{1.5em} \\ =& \,\,\,(x + 1) + (2x + 1) \rule{0pt}{1.5em} \\ =& \,\,\,3x + 2 \rule{0pt}{1.5em} \end{align} $$ Here we have two different $x$'s: the $x$ that denotes the input, and the $x$ that is used as a placeholder to describe how the first function acts.

The pattern to respect is that, each time you take a derivative, the units on the $x$ axis stay the same, while the units on the $y$ axis become divided by those on the $x$ axis. This gives the unique solutions:

Note 1. A unit of “$1$” is a

dimensionless

unit. Dimensionless units arise when quantities are divided by like quantities. Think of dimensionless quantities as “pure fractions” or “pure ratios”. (Percentages are dimensionless—in fact the term

percentage

is synonymous with

dimensionless ratio

though if you spoke to people about “dimensionless ratios” they would look at you funny. Also percentages are a system of notation, whereby the symbol “%” means “divide the preceding number by 100, in order to discover the numerical value of the ratio I'm talking about”.) (To drive it home: In Chinese, the written expressions “$23\%$” and “$23/100$” are indistinguishable when read out loud; they are both read “$23$ over $100$”; that is the simple & correct way!)

The graphs are already in the right order: if “$f$” is the original function then $f$ is on the left, $f'$ is in the middle, and $f''$ is on the right:

For example, the graph on the left has a slope that starts at $\sim\!-1$ and ends at $\sim\!1$, while those are the values at which the graph in the middle starts and ends (and not coincidentally, since the graph in the middle is the derivative of the graph on the left!):

Moreover the middle graph has slope close to $0$ at either end, and some slope near $1.5$ or $2$ towards the middle, matching the values of the graph on the right:

(Taking one more derivative would produce a zigzag, by the way.)

To start with, the slope of $g$ seems to be about $-1.5$, $0$ and (a bit greater than) $2$ at $x = -2$, $x = 0$ and $x = 2$ respectively:

This already gives us three points from which to interpolate a basic approximation to the graph $y = g'(x)$:

But the graph of $g''(x)$ indicates more, namely that $g'(x)$ has a slope that rises from $\approx 0.2$ near the left edge of the graph up to $1.3$ at $x = 0.5$, before falling again to $0.6$ past $x = 2$:

As a second step, we thus “bend into shape” our previous sketch to produce these slopes...

...achieving our final answer.

Note 1. For reference, the actual derivative looks like so:

Firstly, the graph of $h(x)$ seems to have slope $0$ around $x = 0.6$...

...which gives us one data point on the curve $y = h'(x)$ to start with...

...moreover, by the graph of $h''(x)$, the slope of $h'(x)$ is near $-1/3$ on an interval that is approximately (say) $[-0.85,0.7]$....

...so, as a second step, we can extend the graph of $h'(x)$ by a segment of slope $-1/3$ on this interval:

(To achieve a passable slope of $-1/3$ we modeled ourselves on a nearby grid segment.) Next, $h(x)$ has slope $\approx 1.2$ at $x = -2$, and slope $\approx -0.9$ (?) at $x = 2$:

This gives us two more points on the graph $y = h'(x)$:

Then, because the second derivative has value $\approx -1/3$ for $x \leq -1.6$ (about) and for $x \geq 1.5$ (about)...

...we extend these two new data points by segments of slope $-1/3$...

...on the relevant intervals. (I.e., for $x \leq -1.6$ and for $x \geq 1.5$.) The last step is to join the existing segments by some kind of “connector curves” of yet-to-be-determined shape:

Since $h''(x)$ shows that the two connectors have slopes of about $-1/3$ at their edges and slopes of about $-1.4$ and $-1.6$ (respectively) near their middles...

...our final answer, given by the following sketch, is obtained by “bending into shape” the connector curves...

...to give them a slope of $-1/3$ at their endpoints, and slopes of $-1.4$, $-1.6$, respectively, in their middles.

Note 1. Here is the actual graph of $h'$:

Yes, this is the case. For a joke way of seeing it, here is a graph of a putative function $f$, before and after scaling:

The second graph truly is the first graph vertically scaled by a factor $2$, because the scale on the $y$ axis has been doubled. This means that the ratio $$ {\text{rise}\over \text{run}} $$ has doubled in the second graph, because “rise” has doubled (each $y$-coordinate is twice as large!), whereas “run” stays the same. (So the slope of the tangent has doubled, so the derivative is doubled.)

The rate of change is the slope, which is greatest along the right-hand portion of the curve:

On the other hand,

the rate of change of the rate of change

[a.k.a., second derivative] is the rate of change of the slope, and that will be greatest at the first bend of the curve, where the slope is changing at the fastest rate:

(Well, believe us or not, but we're right!)

As it happens—and by the exact method that we used to generate these curves—the blue is the derivative of the red:

Likewise, the derivative of the blue is the yellow, the derivative of the yellow is the green, and the derivative of the green is the red, at which point it starts all over again! (For example, the fifth derivative of the red curve is the blue, because the fourth derivative of the red curve is itself.)

Note 1. After all, the slope of these curves keeps oscillating between two fixed values—the “most slanted up” and the “most slanted down”—so their derivatives were always going to have an oscillatory pattern, as well.

Note 2. Because “most slanted up” occurs when the curve has not yet crested, but when the derivative is already in the process of cresting (that's why it's “most slanted up”), the derivative is ahead of the original curve by half a bump, not the other way around:

Note 3. When we examine the velocity of a particle moving in the plane, we examine the velocities of its shadow on the $x$- and $y$-axes:

The velocities of the two shadows encode the overall “two-dimensional” velocity of the particle. (No need for quotes, really: the velocity is two-dimensional.)

Here's another point of view: just like

position

is encoded by a pair of numbers—sometimes known as the position vector by the way, where “vector” is a term of art for “pair of numbers”—so the

velocity

is encoded by a pair of numbers—equally known as the velocity vector—which is no coincidence, because the first coordinate of the

velocity vector

is the derivative of the first coordinate of the

position vector

and likewise for the second coordinate—two coordinates, two rates of change!

Geometrically, if we use the $x$- and $y$-components $v_x$ and $v_y$ of the velocity to draw an arrow emanating from a point on the curve, this arrow is tangent to the curve, and the

length

of the arrow is the

speed

of the particle at that moment in time. More precisely, if you let the particle drift at the exact same $x$- and $y$-velocities $v_x$ and $v_y$ that you measured at the root of the arrow for one unit of time, the particle would cover exactly the length of the arrow in that one unit of time, no more no less, because the particle would cover $v_x$ units in $x$ and $v_y$ units in $y$. And speed being

distance per unit time

the length of the arrow is, therefore, the speed!

Now consider not one but four particles, going around a unit circle in clockwise fashion, 90° apart in phase, at unit speed (“unit speed” = speed 1, “unit circle” = radius 1) (ps: We center the circle at the origin):

The

position vectors

of the particles are as follows:

(You can't really see it so well, but each arrow originates at $(0, 0)$.) While the

velocity vectors

are as follows:

(Like the position vectors, the velocity vectors keep changing instant by instant—this is the subtlety of calculus!) The velocity vectors have length $1$ because the speed is $1$, & are brushed in the direction of travel.

(Nb: When we draw a vector as an arrow we mean that the first coordinate of the vector is equal to the horizontal displacement from the tail of the arrow to the head of the arrow, and likewise that the second coordinate is equal to the vertical displacement from the tail of the arrow to the head of the arrow.)

Due to the 90° rotations and uniform lengths of $1$, one particle's velocity vector is another particle's position vector; for example, the red particle's velocity vector is the blue particle's position vector:

Focusing on the $x$-coordinates, for example,

at any given moment in time. This also means:

the rate of change of the $x$-coordinate of the red particle is the $x$-coordinate of the blue particle

...because “velocity in $x$” is the same as “rate of change of the $x$-coordinate”.

Concretely, if you graph the $x$-coordinates of the red and blue particles on the same graph, the rate of change of the red particle's $x$-coordinate will equal the value of the blue particle's $x$-coordinate. These are the reds and blue curves from the problem statement, if we start the red particle at position $$ (1, 0) $$ at time $t = 0$:

If we add the $x$-coordinates of the green and yellow particles, we find the graph from the problem statement!

Note 4. If needed, here is an illustration of one $360^\circ$ rotation of the particles of Note 3, with each curve being an $x$-coordinate:

(If the above just looks like a confusing mess then don't sweat it—it's not that important.)

Note 5. Take a look at this figure again:

The derivative is

ahead

of the function—the values that you see now on the derivative are values you will see a little later on the function—because the blue particle is, likewise,

ahead

of the red particle, so that $x$-coordinates you see now on the blue particle will be seen a little later on the red particle! (In particular, if you wanted to generate the same curves with a counterclockwise rotation, you could do that, but you would have to reverse the order of the particles around the circle to keep the blue particle ahead of the red particle, the yellow particle ahead of the blue particle, etc.)

The values oscillate between $+2$ and $-2$, so we need a circle of radius $2$ to generate these curves. Also the values go through one cycle in the same amount of time as the particles of Exercise 12, but the circle has twice the circumference (having twice the radius), so the particles are going twice as fast! (I.e.: speed 2, since the particles of Exercise 12 have unit speed.)

Note 1. In this and in the previous exercise the units of time and distance are “anonymous”: distance could be meters, kilometers, or anything, and time could be seconds, hours, etc—it doesn't matter. However, one should be aware that what amounts to

unit speed

under one set of units is no longer “unit speed” under a different set of units—this is not a “physical” property of the particles, but, rather, a “mathematical” property that holds only for one specific “tweaking” of the units.

We can naïvely try to imitiate how the curves of Exercise 16 are generated by placing five equally spaced particles around the unit circle (“the” unit circle is the one centered at $(0, 0)$, by convention), instead of 4:

The idea would be that the

velocity vector

of the red particle is the

position vector

of the blue particle, likewise for the blue and yellow particles, and so on. (Position vectors shown above.) For example, at the instant above, the velocity vectors would be as follows:

The velocity vectors are

NOT

tangent to the unit circle, and so the particles will leave the circle! (But that's OK.) In one-tenth a unit of time, for example, the particles would travel approximately one-tenth their velocity vectors, that would bring them to approximately these new positions:

In the next one-tenth unit of time we can apply a similar approximation again, advancing the particles by ${1\over 10}$th of [the current approximation to] their velocity vectors. Skipping the construction lines:

Applying the same process for $8$ more steps:

To be clear, in the above figure, the position of the red particle at, say, the fifth step...

...is obtained by starting from the red particle's position at the fourth step...

...and adding one-tenth of the approximation that we have to the red particle's velocity vector at that moment, that approximation being namely the blue particle's position vector at the fourth step ($t = {4\over 10}$)...

...and we do the same for each particle, to advance to the next step.

If we stop $10$ times as often, advancing the clock by ${1\over 100}$th of a unit of time at each step, the same figure becomes just a blur (still going from $t = 0$ to $t = 1$):

To visualize such a fine-grained approximation we need to revert to drawing the particles as points. In the following figure the colored paths are points that come from a “${1\over 100}$th” approximation, while the orange dots are the old positions obtained from a “${1\over 10}$th” approximation:

Zooming in a bit (or else we still can't see anything):

Fig. 1

In any case, even the “${1\over 100}$th” approximation is just an approximation, but the point is that such approximations do converge to a set of “true” particle paths, as pictured in Fig$.$ 1, that can be computed by some wizards; as time can be played forward or backward, these paths form doubly-infinite spirals—in to infinity, out to infinity.

In any case [take two] the point is that whether or not you are one of the wizards, you can

guess

the existence of these five paths—sort of “feel” that they exist! (This is a moral consolation prize, at least.)

We can also convert the paths into a function $$ f $$ that satisfies the problem requirements.

For example let $f$ be the function that, given a time $t$, outputs the $x$-coordinate of the red particle at $t$; then, to spell it all out, since

the rate of change of the $x$-coordinate of the red particle is the $x$-coordinate of the blue particle

$f'$ is the $x$-coordinate of the blue particle; and since

the rate of change of the $x$-coordinate of the blue particle is the $x$-coordinate of the yellow particle

$f''$ is the $x$-coordinate of the yellow particle; and since

the rate of change of the $x$-coordinate of the yellow particle is the $x$-coordinate of the green particle

$f'''$ is the $x$-coordinate of the green particle; and since

the rate of change of the $x$-coordinate of the green particle is the $x$-coordinate of the purple particle

$f''''$ is the $x$-coordinate of the purple particle; and since

the rate of change of the $x$-coordinate of the purple particle is the $x$-coordinate of the red particle

$f''''' = f^{(5)}$ equals $f$.

Note 1. If you graph the $x$-coordinates of the 5 particles over time, each in their color, you get a graph like so, in which blue is the derivative of red, yellow is the derivative of blue, etc; the function $f$ can be taken to be any one of these curves:

Note 2. There is nothing special about $x$-coordinates vis-à-vis $y$-coordinates. You can also define $f(t)$ to be, e.g., the $y$-coordinate of the red particle at time $t$.

Note 3. It is worth noting that, in fact, the $x$- and $y$-coordinates live separate lives. The rate of change of each $x$-coordinate is some other $x$-coordinate, and the rate of change of each $y$-coordinate is some other $y$-coordinate.

Note 4. Adding to this observation, we don't need to start the particles in a symmetric configuration. Symmetry only helps to picture how the positions of the particles will evolve without making any computations. We also don't need to work in two dimensions. We can place the particles in a one-dimensional world, e.g., ...

...(the initial positions really don't matter much, as long as you don't give all the particles the same initial position, or else you won't have $f \ne f'$ etc) and stipulate the same rules, namely that the

velocity

(now $1$-dimensional) of the red particle be the

position

(now $1$-dimensional) of the blue particle and so on—you can “release” the particles from their initial configuration and simulate (or compute exactly, if you have the know-how) their motion by the same methods as above. The five position functions obtained are each a solution $f$ to the problem. (But this solution will typically look more chaotic than the curves from Note 1.)

Note 5. In fact, our symmetric two-dimensional solution is an instance in which you can say that

the whole is simpler than the parts

in that you would never spot the symmetry at play, or have a chance of eyeballing the long-term evolution of the system, if you were shown just the $x$-coordinates, or just the $y$-coordinates, on their own!

Take $17$ particles equally spaced out along the unit circle, such as these (shown here with position vectors):

Set the velocity of particle $$ {\Large 1} $$ equal to the position of particle $$ {\Large 5} $$ and keep going by this pattern, making the velocity of each particle equal to the position of the particle that is $4$ later; in the configuration above, the velocity vectors end up looking like so, for example:

Maintaining this relationship at all points in time, and given that the velocity vectors point very slightly outward from the unit circle, and because all the symmetry and all the angles are maintained as we play time forward or backward, the particles spiral gently outward/inward from the circle for time forward/backward, respectively. Taking $f(t)$ to be the $x$- or $y$-coordinate of any one of the particles (e.g., particle $1$) at time $t$ gives an oscillating function whose $17$th derivative is itself (because the rate of change of the $x$-coordinate of particle $1$ is the $x$-coordinate of particle $5$, etc, until we make it all the way back to particle $1$), and that grows comparatively slowly over time. ~The End~

Note 1. In case you're curious, the actual spiral paths of the particles look like so:

...and if you take the $x$-coordinates of the particles over time, with time $t = 0$ corresponding to the original configuration depicted where particle 1 is at $(1, 0)$, you find paths functions that look like so:

For example, the derivative of curve , highlighted below in red, is curve , highlighted in blue:

...and taking sixteen more derivatives starting from curve we would go through curves , , , , ..., , before finally coming back to curve !

Note 2. It can be interesting to examine what goes wrong if we attempt to make the velocity vectors even more tangent to the unit circle. For example, if we start the particles so that particle 5 is at $90^\circ$ exactly from particle 1, particle 9 is at $90^\circ$ exactly from particle 5, and so on, until we reach particle 14, the last particle in this order; then we have the following starting configuration:

To parse the above figure, understand that:

(Note that the red arrows have to form a cycle of length 17 in order for us to later extract a function $f$ such that $$ f^{(17)} = f $$ but this is the case: the red arrows only “close the loop” after going through all 17 particles!)

In this starting configuration, all velocity vectors are exactly tangent to the unit circle

EXCEPT

for particle 14, whose velocity vector, being the position of particle 1, is straight out from the circle! So, as we “start time”, particle 14 will push out from the circle, that will in turn affect particle 10, and so on, until all particles end up being “peeled off” from the circle, in due time; if you are so curious, the particle trajectories end up like so (shown only for $t \geq 0$):

The particles shoot of to infinity in short order—the solution is much worse—for fun we have also highlighted two particle trajectories in this figure:

(The point is: if your velocity vector is tracing a circle centered at $(0, 0)$—at a uniform rate—then you, too, are going around in a circle—it's just that your circle could be centered anywhere, not necessarily at $(0, 0)$!)

This version pictures all the elements that appear in the fraction:

The point is: the fraction $$ {f(x + h) - f(x)\over h} $$ is seen to have the form rise over run, and is more precisely equal to the slope of the pale brown line going through the point

$$ (x, f(x)) $$

at one end, and

$$ (x + h, f(x + h)) $$

at the other end. (This is also the case if $h$ is negative, by the way.)

Note 1. A fraction of this form is called a Newton quotient.

Note 2. The pale brown line is sometimes known as the secant [through $(x, f(x))$, $(x+h, f(x+h))$]. “Secant” is a general term for “line passing through two specified points on another curve”.

Note 3. If we let $h$ drop to $0$, and if $f$ is differentiable at $x$, the Newton-quotient-a.k.a.-slope-of-the-secant approaches $$ f'(x) $$ because the secant approaches the tangent, in that case, and the slope of the secant is also, perforce, approaching the slope of the tangent, which is $f'(x)$. (But you cannot directly set $h = 0$, because $$ {f(x+0)-f(x)\over 0} = {0 \over 0} $$ is undefined.)

Dividing by $\Delta{}t$:

As $\Delta{}t$ approaches $0$, the term on the left-hand side approaches $$ (AB)'(t_0) $$ where we view $A$ and $B$ as functions of time with, therefore, the product $AB$ also becoming a function of time. (By definition, $AB$ is the function $$ t \rightarrow A(t)B(t) $$ where $A(t)$ is the value of $A$ at time $t$, $B(t)$ is the value of $B$ at time $t$.) Indeed, a ratio of the form $$ {f(t_0 + \Delta{}t) - f(t_0)\over \Delta{}t} $$ is a Newton quotient (cf. Exercise 16), that approaches $$ f'(t_0) $$ as $\Delta{}t$ approaches $0$, assuming $f$ is differentiable at $t_0$ (cf. Exercise 16 Note 3), and

has the form $$ {f(t_0 + \Delta{}t) - f(t_0)\over \Delta{}t} $$

for $f = AB$.

The first term on the right-hand side, for its part, approaches

$$ B(t_0)A'(t_0) $$ as $\Delta{}t$ approaches $0$. Indeed, when you write it out, that term becomes the algebraic expression

where every term on top contains a “$B(t_0)$”, that can therefore be factored out, giving us the equivalent expression $$ B(t_0)\cdot{A(t_0 + \Delta{}t) - A(t_0) \over \Delta{}t} $$ that, you will notice, has the form $$ B(t_0)\cdot{f(t_0 + \Delta{}t) - f(t_0) \over \Delta{}t} $$ for $f = A$, and thus approaches $$ B(t_0) \cdot A'(t_0) $$ as $\Delta t$ approaches $0$, by the property of the Newton quotient.

Lastly the most interesting term is the second term on the right-hand side! Symmetrically to the first term on the right-hand side, the second term approaches

$$ A(t_0)B'(t_0) $$ as $\Delta{}t$ approaches $0$, but the reasons are slightly different! (Slightly.) Indeed, this term, written out, is $$ {A(t_0 + \Delta{}t)B(t_0+\Delta{}t) - A(t_0+ \Delta{}t)B(t_0) \over \Delta{}t} $$ which is equal to $$ A(t_0 + \Delta{}t)\cdot{B(t_0+\Delta{}t) - B(t_0) \over \Delta{}t} $$ by factoring out the common term $A(t_0 + \Delta{}t)$; and $$ {B(t_0+\Delta{}t) - B(t_0) \over \Delta{}t} $$ approaches $$ B'(t_0) $$ as $\Delta{}t$ approaches $0$, like before (when we had $AB$ or $A$ instead of $B$) whereas $$ A(t_0 + \Delta{}t) $$ —which is a bit different from before—approaches $$ A(t_0) $$ as $\Delta{}t$ approaches $0$—so that makes up $A(t_0)B'(t_0)$. (The

differentiability

of $A$ and $B$ at $t_0$—that we are tacitly assuming—implies

continuity

as well, which implies that $A(t_0 + \Delta{}t)$ approaches $A(t_0)$ as $\Delta t$ approaches $0$.)

Summarizing, the three terms separately approach $$ (AB)'(t_0) $$ $$ B(t_0)A'(t_0) $$ $$ A(t_0)B'(t_0) $$ as $\Delta{}t$ approaches $0$ and, in fact, because the equation holds no matter how close we make each term to its respective limit above, one can conclude that $$ (AB)'(t_0) = B(t_0)A'(t_0) + A(t_0)B'(t_0) $$ for functions $A$, $B$ differentiable at a point $t_0$.

Nb: This result is known as the product rule.

Note 1. Keeping things alphabetical everywhere, the same equation is more often written $$ (AB)'(t_0) = A'(t_0)B(t_0) + A(t_0)B'(t_0) $$ with “$A'(t_0)B(t_0)$” in the middle. (But which is the same, of course, as $B(t_0)A'(t_0)$.)

One can say

the derivative of the sum is the sum of the derivatives

or

the rate of change of the sum is the sum of the rates of change

or (we made this one up)

the rate of change of the aggregate is the sum of the rates of change of the components

(etc).

The form of... $$ (fg)'(t_0) = f'(t_0)g(t_0) + f(t_0)g'(t_0) $$ ...that follows the style of... $$ (f + g)' = f' + g' $$ ...is... $$ (fg)' = f'g + fg' $$ ...this. (Valid for differentiable functions $f$, $g:$ $\mathbb{R} \rightarrow \mathbb{R}$.)

Note 1. Whereas $$ (fg)'(t_0) = f'(t_0)g(t_0) + f(t_0)g'(t_0) $$ is an equality between real numbers, $$ (fg)' = f'g + fg' $$ is an equality between functions. So there is a more-than-skin-deep difference between the two forms. Also note that each form has its own “qualitatively distinct” qualifying conditions.

(To wit, $$ (fg)'(t_0) = f'(t_0)g(t_0) + f(t_0)g'(t_0) $$ holds “for $t_0$ at which $f$ and $g$ are differentiable”, while $$ (fg)' = f'g + fg' $$ holds “for differentiable functions $f$, $g$”.)

That would be the fact that $$ (cf)' = cf' $$ for all differentiable functions $f : \mathbb{R} \rightarrow \mathbb{R}$, for all $c \in \mathbb{R}$, mentioned in Exercise 10 for $c = 2$.

Note 1. You can also write $$ (cf)' = c \cdot f' $$ if it helps clarify the difference between the left- and right-hand sides. (The difference being namely “($c$ times $f$) prime” on the left vs. “c times ($f$ prime)” on the right.)

The “logical” definition is: $$ cf = (x \rightarrow cf(x)) $$ where the product “$cf(x)$” is an ordinary multiplication between two real numbers, because $c$ is a real number and $f(x)$ is a real number! (In this way, the product of a function by a real number “bootstraps” off of the ordinary product of real numbers—this is already similar to what happens for the definition... $$ fg = (x \rightarrow f(x)g(x)) $$ of the product of two functions from $\mathbb{R}$ to $\mathbb{R}$, or with the case of function addition, that relies on real number addition.) BUT. There is a MORE CLEVER way of doing the definition. Which is to define $$ cf = (x \rightarrow c)f $$ where the right-hand-side is one function times another, i.e., a product of functions, which is something that has ITSELF ALREADY BEEN DEFINED. (!) (To wit, the definition of function multiplication is that $$ fg = (x \rightarrow f(x)g(x)) $$ of course.) (Wait we just mentioned that already.) Mathematicians LOVE to bootstrap off an intermediate step, instead of going back to the beginning, so the second way is clearly the superior definition!!

E.g.: $$ (f \circ g)(x) = f(g(x)) $$ $$ (fg)(u) = f(u)g(u) $$ $$ (f/g)(z) = f(z)/g(z) $$ $$ (f - g)(t) = f(t) - g(t) $$ (Looking at these definitions we must really admit that we prefer the first form, with the arrow, found at the end of the chapter—it's more explicit!)

We can start with the fraction in the middle of the product:

This is seen to be a Newton quotient (cf. Exercise 16) $$ {f(x + h) - f(x)\over h} $$ with $f = A_i$, $x = t_0$, per which (Exercise 16 Note 3), the fraction approaches $$ A_i'(t_0) $$ as $h$ approaches $0$, given also the assumption that each of the functions $A_1$, ..., $A_n$ (including $A_i$) is differentiable at $t_0$.

Next down in order of interesting-ness we presumably have the terms $A_1(t_0 + h)$ through $A_{i-1}(t_0 + h)$ at the beginning of the product...

...; here the

differentiability

of $A_1$ at $t_0$ implies the

continuity

of $A_1$ at $t_0$, which implies that $$ A_1(t_0 + h) $$ approaches $$ A_1(t_0) $$ as $h$ approaches $0$. (These various technicalities concerning a generic function $f : \mathbb{R} \rightarrow \mathbb{R}$ are mentioned in the solution to Exercise 17.) Similarly for $A_2(t_0 + h)$, etc, up to $A_{i-1}(t_0 + h)$, so $$ A_1(t_0 + h) \,\times\, \dots \,\times\, A_{i-1}(t_0 + h) $$ approaches $$ A_1(t_0) \,\times\, \dots \,\times\, A_{i-1}(t_0) $$ as $h$ approaches $0$. (If some quantities are each approaching a different value, then the product-of-the-quantities will approach the product-of-the-values—something not mentioned in the solution to Exercise 17, but that might have been.)

Lastly one has the tail end of the product, where $h$ does not even appear:

Because $h$ does not appear here, the tail end

stays put

where it is, irrespective of the value of $h$. So that was easy! Altogether, the answer is therefore:

...with a lone “$A_i'$” in the middle.

Because the particle is going at unit speed the circumference of the unit circle is equal to the amount of time it takes the particle to complete one revolution of the circle. That is, for example, the length of this yellow interval:

One revolution around the circle is also made up of four quarter-revolutions, where each quarter-revolution of the circle is “half a bump”, on the graph:

Going a bit further, seven of these quarter-revolutions appear to take up exactly $t = 11$ units of time (!!!!!!!!!!!) (or maybe just a little less than $11$ units, if you zoom in):

Therefore $$ \Large {11 \over 7} $$ is an approximation to the quarter-circumference of the circle, and $$ \Large 4 \cdot {11 \over 7} = {44 \over 7} $$ is an approximation to the circumference of a unit circle.

Note 1. This approximation ends up being about half-a-part-in-a-thousand too large ($0.040249943...\%$ too large) (or just: “$0.00040249943...$ too large”), which is strikingly good, all things considered.

Note 2. Numerically, note that $$ \Large {44 \over 7} = 6.285714\dots $$ is a bit larger than $6$, which agrees with what we see here for the length of a full revolution...

...whereas $$ \Large {11 \over 7} = 1.571428\dots $$ is about $1.6$, which also appears to agree with what we can see on the graph about the length of a quarter-revolution:

(So, we have some secondary “visual confirmation” of our approximations.)

Let's start by examining Pacman's first six displacements, appearing here in blue (positive displacements, going to the right or up) and red (negative displacements, going to the left or down):

We can estimate the duration of each displacement to the closest 10th of a second (mistakes of estimation can be made, we shall recover):

We can also estimate the velocity to be $$ \pm 6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s} $$ when it is nonzero (for displacement the velocity might seem more like $-6.8\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}$ at the least, but we've already made more significant errors while eyeballing the durations, so nevermind). Using

$$ (\text{velocity}) \times (\text{amount of time}) = (\text{displacement}) $$

then gives us the following estimates for the amount of travel during the six displacements:

$-6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}\,\,\times\,\,0.5\text{s}\,\,=\,\,-3.375\,\text{cells}$

Given the horizontal/vertical alternation of displacements, this would nominally imply the following set of initial motions:

But these are approximate numbers and the true values must be integers, except for . (Because we don't know where Pacman started out. For the next displacement, if you look back at the graphs, is horizontal, so yes.) In fact, if you look at the maze, $$ 3 $$ cells is the smallest amount that Pacman can travel vertically when changing $y$-coordinate, between two moments of horizontal motion. The next smallest possible amounts are

$$ 4 $$

and

$$ 6 $$

and

$$ 7 $$

cells, with $5$ not being a possibility. In the horizontal direction, the smallest amounts are $$ 3, 6, \text{ and } 9 $$ (and $12$ and ...) which is even more restrictive. Now if each of our duration measurements carries an error of no more than $$ \pm{}0.2\text{s} $$ each computed displacement is at most $$ 6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}\,\times\,\pm{}0.2\text{s}\,=\,\pm1.35\text{c} $$ from the truth, give or take the small difference between $6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}$ and the actual velocity. So $$ -2.7\text{c} $$ must be either $$ -3\text{c} $$ or $$ -4\text{c} $$ these being the only two posssible integer vertical displacements within $\pm1.35$c of $-2.7$c. Then, applying similar logic to each measurement, the initial motions must be:

The maze fits these constraints in only two places (note that and equal $-3\text{c}$ and $3\text{c}$, in each case):

Looking into the future, the next three displacements are right/down/right and last ~$3.2$s/~$0.4$s/~$1.3$s respectively:

(Nb: Imagine translating these intervals to the left or right until the start of the interval is at an integer value: this is a good way to estimate the length.)

Because displacement is horizontal to the right the only possible remaining solution is the right-hand one, or else Pacman would collide with the ghost cage, with displacement equal to 6c not 7c, or else Pacman would collide with a wall:

Since $$ 6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}\,\times\,3.2\text{s}\,=\,21.6\text{c} $$ it seems that displacement brings Pacman all the way around the maze to the left edge of the ghost cage, like so...

...though it is hard to measure that distance; but this is confirmed by the fact that the next two displacements are “down by $3$ and to the right”; specifically, since $$ -6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}\,\times\,0.4\text{s}\,=\,-2.7\text{c} $$ displacement must be $-3$c or $-4$c; must actually be $-3$c since displacement is to the right; so, notwithstanding the exact length of displacement , there is only one possibility for displacements through :

So at $t = 49$s, between displacements and , Pacman is immediately to the left of the ghost cage.

It seems well-advised to start by heuristically verifying that Pacman's speed remains $$ \approx 6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s} $$ no matter the direction that Pacman is headed, as long as Pacman is in motion.

For example, take the instant $t \approx 23.7$s, when the $x$- and $y$-velocities are both about (?) $4.8\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}$:

The velocity vector (cf. Exercise 12) is therefore about $$ (4.8, 4.8) $$ in units of $\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}$ at $t \approx 23.7$s, and the speed, being the length of the velocity vector (cf. Exercise 12), is about

$$ \sqrt{4.8^2 + 4.8^2} = \sqrt{2} \times 4.8 = 6.788... $$

(Pythagoras!) in units of $\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}$ as well, and $$ 6.788... \approx 6.75 $$ which supports, in this case, the hypothesis that Pacman's speed is $\approx 6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}$ regardless of the direction of travel.

For more verification, take $t = 26\text{s}$, at which point the velocity vector is roughly $$ (6.5, -2) $$ cells per second:

This gives a speed of

cells per second, Again close to $6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}$. (!)

For two more verifications take $t = 27\text{s}$ and $t = 30\text{s}$:

The speed at $t = 27\text{s}$ is approximately $$ \sqrt{3^2 + 6.2^2} = \sqrt{47.44} = 6.888 $$ cells per second, while the speed at $t = 30\text{s}$ is approximately $$ \sqrt{5.6^2 + 3.6^2} = \sqrt{44.32} = 6.657 $$ cells per second. Both close-ish to $6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}$! For one last verification (truly the last, we promise) consider $t = 31\text{s}$:

This yields a speed of $$ \sqrt{2.6^2 + 6.2^2} = \sqrt{45.2} = 6.723... $$ cells per second, again close to $6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}$! (Closest so far, in fact.)

We now admit, after this “heuristic verification”, that Pacman goes approximately the same speed regardless of direction, namely something in the vicinity of $6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s}$.

Now consider the time interval from $t = 22.7\text{s}$ to $t = 24.7\text{s}$:

Both the $x$- and $y$-velocities are nonzero during this interval, which indicates the presence of a curve. The curve starts with vertical motion and ends with horizontal motion:

Thus Pacman starts the curve going up, and ends the curve going right. Moreover, it takes Pacman $$ 24.7\text{s} - 22.7\text{s} = 3\text{s} %24.7 - 22.7 = 3 $$ to complete the curve (we know the curve is fully completed from the purely horizontal motion at either end), from which the curve must be approximately $$ 6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s} \times 3\text{s} = 20.25\text{c} $$ in length! In turn, using the approximation $$ \approx {11 \over 7} $$ for the quarter-circumference of a unit circle (cf$.$ Exercise 24), this would indicate that the curve (which is a quarter-circle, as all curves in this maze) has radius $$ \approx {20.25\text{c} \over 11/7} = {7 \times 20.25\text{c} \over 11} = {141.75\text{c} \over 11} = 12.886...\text{c} $$ where we give up and use a calculator at the last step. But the possible radii are $3$, $6$, $9$ and $12$. This all but rules out all of the maze curves except the one that has radius $12$, and that allows a traversal that starts upward and ends rightward; we mean the upper left curve of the maze:

From there, Pacman goes right for a bit, then takes another curve 3 seconds long, that starts rightward and ends downward:

...this second curve must, of course, be the upper right-hand corner of the maze, that has the appropriate length, position, and orientation:

What is extremely strange, however, is that Pacman immediately follows the end of this curve with rightward motion:

In fact, it is also strange that Pacman preceded the first curve with rightward motion (when that curve starts at the leftmost edge of the maze):

Looking back over our work, we find that we made a mistake when we wrote $$ 24.7\text{s} - 22.7\text{s} = 3\text{s} $$ the corrected version of that being of course $$ 24.7\text{s} - 22.7\text{s} = 2\text{s} $$ (the second curve likewise lasted $2\text{s}$, not $3\text{s}$) making the length and radius of the first curve two-thirds of whatever we previously computed (because $2\text{s}$ is two-thirds of $3\text{s}$), i.e., $$ {2 \over 3} \times 12.886...\text{c} $$ for the radius of the first (and second) curve, which means that the first and second curves actually had radii $9$, undoubtedly, and that Pacman's initial motion followed the one-inside track (the two rightward motions are easily seen to be ~$3\text{c}$ each):

Next, after some downward motion we are faced with a long, juicy, down-and-then-left curve, which must surely be the bottom-right curve of radius $12$:

Indeed, the curve lasts ~$2.7$s, and $$ 6.75\text{c}\!\hspace{0.1em}/\!\hspace{0.1em}\text{s} \times 2.7\text{s} = 18.225\text{c} $$ is approximately the same as $$ {11 \over 7} \cdot 12\text{c} = 18.85...\text{c} $$ confirming the radius of $12\text{c}$ and the location of the curve. Pacman's trajectory so far is then:

Next Pacman seems to reverse course, and briefly re-enters the curve (going right and up a tiny bit):

But then changes again, and re-exits the curve (going left and down a tiny bit):

Then Pacman goes left-and-then-right-again by some small amount:

At this point—and in particular at $t = 34\text{s}$—Pacman is between a moment of purely horizontal motion and purely vertical motion; since the left-and-then-right-again motion did obviously not bring Pacman $3$ cells over to the left (which is the next place after the curve exit that is connected to both horizontal and vertical paths), Pacman must be at the bottom-left exit of the bottom-right maze corner, still.

Note 1. Feel free to follow Pacman all the way to the end of the timeseries. He ends up somewhere near...

As the second derivative is the

rate of change of the slope

places where the second derivative is zero are places where the slope of the function is constant. So the function will have a constant slope over each of these purple intervals (we're going to assume that what looks like $0$ is $0$, and eyeball where that starts and ends, the statement does say “broadly” anyway):

As

constant slope

means

line segment

the function will therefore be a line segment, over each of the purple intervals!

Between these line segments, however, things happen, and there is a change in slope! The change in slope is given by the “amount of bump” in the second derivative between the intervals. As it turns out, the area enclosed by the bump gives the total change in slope:

(We won't argue this right now, but it's sort-of-intuitive.) For bumps lying below the $x$-axis the area counts as negative; that negative area is, again, the total change in slope from one end of the bump to the other:

In any case the areas are all the same in absolute value, meaning that whatever slope is gained as we pass over a positive bump, the same amount is lost again as we pass over a negative bump! Thus, the line segments of the original function will alternate between “low” and “high” slopes—as we pass over a positive bump we switch from a “low slope” line segment to a “high slope” line segment, and vice-versa when we pass over a negative bump.

Put $$ \Large a $$

for the area of a positive bump (per appearances, $a \approx 1$), and

$$ \Large c $$ for the slope of a “low slope” line segment. Then a “high slope” line segment has slope $$ \Large c + a $$ since we add $a$ to the slope each time we go over a positive bump. (And the slope goes back down to $$ \Large (c + a) + (-a) = c $$ when we pass over a negative bump, with $-a$ being the (negative) area of a negative bump.)

With these variables in place, here is a generic illustration of a graph (in black) whose second derivative is the one from the statement (faded in the background):

In this example $c \approx 0.2$, but $c$ can be any value—this is not constrained by the second derivative. Moreover any amount of vertical translation can also be introduced to the graph. (Vertical translation does not affect the derivative, much less the second derivative.)

For another example, if $c = -a/2$, meaning $c \approx -1/2$, the graph ends up a perfectly balanced see-saw that stays confined to a bounded range of $y$-values:

Moreover, like the previous graph, this solution can also be vertically translated by any amount! (And same for any solution.)

For yet another example, here is a graph in which $c + a = 0$, $c = -a \approx -1$:

Again, any of these graphs are

equally valid

solutions, and, for the last time,

any amount of vertical translation can be introduced

(you can move the graphs up and down). So in other words we have a “two-parameter family of solutions”: one parameter of the solution—free to choose—is $c$—while another parameter—independently free to choose—is the amount of vertical translation.

To claim a truly good “theoretical” understanding of the solution, however, we should also determine this rise here, if we can, as a function of $c$ and $a$, i.e., the amount of rise between the end of one line segment and the start of the next:

In fact, is not entirely clear that there aren't possibly two different values of this rise, for the two different kinds of “connector curves” that exist (the concave ones and the convex ones):

(It will turn out that the rises are all the same but we're just pointing out.) Focusing on the case of a convex connector curve, note that the rise is lower bounded by $$ \Large 1.6c $$ where $1.6 = 0.8 + 0.8$ is the length (run) of the connector curve, because $c$ is the lowest slope found anywhere inside the connector curve:

Symmetrically, $$ \Large 1.6(c + a) $$ is an upper bound on the rise, because $c + a$ is the greatest slope anywhere inside the connector curve:

To go any further we must add the first derivative to this sketch—the first derivative has value $$ \Large c $$ where the function has slope $c$, has value $$ \Large c + a $$ where the function has slope $c + a$, and climbs up/down along an S-shaped curve outside of those intervals, adhering to a slope that is given by the value of the second derivative:

The afore-mentioned lower bound of $$ \Large 1.6c $$ coincides with the area of a rectangle that lies below the graph of the derivative:

Whereas the afore-mentioned upper bound of $$ \Large 1.6(a + c) $$ coincides with the area of a rectangle that lies above the graph of the derivative:

In other words, the rise of the convex connector curve is lower and upper bounded by these two areas. It will be helpful to write this as a pictorial inequality:

But we can tighten the inequality by dividing the areas halfway (we'll let you think about this one—if you don't get it, don't worry, because we'll revisit the same topic in detail at some point):

Or even:

If we take this logic to its bitter conclusion, we find the equality:

And because the S-curve is centrally symmetric (the slopes at equal distance from the center are the same because those slopes can be read off the second derivative, and the second derivative bump is left-right symmetric) we can compute the area that the curve encloses exactly, by a geometric surgery:

Long story short, the area enclosed, which is also the rise of the connector curve, is $$ \Large 1.6\cdot (c + {a\over 2}) $$ ...that can be read as “run times average slope” (because $$ \Large 1.6 $$ is the run while the slope (first derivative) spends equal amounts of time, in equal measure, above and below the value $$ \Large c + {a \over 2} $$ that is, indeed, the average of $c$ and $c + a$). For concave connector curves the S-curve of the derivative is...

...flipped around from before, going from high to low, but the area enclosed by the S-curve is the same. This area is also the rise of the connector curve. Hence, long story short—for the second time—all connector curves have rise $$ \Large 1.6\cdot (c + {a\over 2}) $$ and we can annotate our sketch of the “generic” solution with this additional piece of information, if we want. (Well...

...there, no one can accuse us of not doing the homework ourselves.)

It is necessary and sufficient to show that $$ ((fg)h)(u) $$ is the same as $$ (f(gh))(u) $$ for an arbitrary input $u \in \mathbb{R}$, in order to show that $$ (fg)h $$ and $$ f(gh) $$ are the same function. (Function equality is based on input-output behavior: two functions are equal if and only if every input is mapped to the same output under either function. See Note 6, Exercise 9, Chapter 3.)

Starting up, $$ ((fg)h)(u) = (fg)(u) \cdot h(u) $$ by the definition of function multiplication, and $$ (f(gh))(u) = f(u) \cdot (gh)(u) $$ likewise. Moreover, $$ (fg)(u) = f(u) \cdot g(u) $$ and $$ (gh)(u) = g(u) \cdot h(u) $$ by the same definition again. Therefore, $$ ((fg)h)(u) = (f(u) \cdot g(u)) \cdot h(u) $$ on the one hand, and $$ (f(gh))(u) = f(u) \cdot (g(u) \cdot h(u)) $$ on the other hand. But $$ (f(u) \cdot g(u)) \cdot h(u) = f(u) \cdot (g(u) \cdot h(u)) $$ by the associativity of ordinary real number multiplication. (Not function multiplication: real number multiplication.) So $$ ((fg)h)(u) $$ equals $$ (f(gh))(u) $$ for arbitrary $u$, which completes the proof.

Note 1. In words, we have just established the

associativity of function multiplication

while we had previously established the

associativity of function composition

(if you recall that one) in Exercise 9 of Chapter 3.

Note 2. By this result, we can write $$ fgh $$ without any parentheses at all: it doesn't matter whether we think of this product as $(fg)h$ or $f(gh)$, the result is the same.

Given an arbitrary $u \in \mathbb{R}$ we have $$ (f + g)(u) = f(u) + g(u) $$ and $$ (g + f)(u) = g(u) + f(u) $$ by the definition of function addition. But $$ f(u) + g(u) = g(u) + f(u) $$ by the commutativity of real number addition [$f(u)$ and $g(u)$ are both real numbers—the

commutativity

of real number addition is the fact that $$ a + b = b + a $$ for all real numbers $a$, $b$, mentioned in the statement—so we can use this here]; thus $$ (f + g)(u) = (g + f)(u) $$ for all $u \in \mathbb{R}$, which implies $$ f + g = g + f $$ by definition of function equality.

For the second half we have, similarly, $$ \begin{align} (fg)(u) &= f(u) \cdot g(u) \\ &= g(u) \cdot f(u) \rule{0pt}{1.5em}\\ &= (gf)(u) \rule{0pt}{1.5em} \end{align} $$ for arbitrary $u\in \mathbb{R}$, where the first and last equality are by the definition of a product of functions and where the middle equality is by commutativity of real number multiplication. [That would be the fact that $$ ab = ba $$ for all $a, b \in \mathbb{R}$, as mentioned in the statement.] Hence $$ fg $$ and $$ gf $$ agree on an arbitrary input, hence $fg = gf$ by definition of function equality.

It computes the amount earned by the rat as a function of time. In more detail, $ (f \circ g)(t) $ is the number of $\text{\$}$'s earned by the rat at $t$ seconds after the start of the race.

Note 1. In even more detail, $$ g(t) $$ is the position in meters of the rat $t$ seconds after start, by definition of $g$, at which position the rat has earned $$ f(g(t)) $$ $\text{\$}$'s in total, by definition of $f$. And $$ f(g(t)) $$ is $$ (f \circ g)(t) $$ by definition of “$\circ$”.

Note 2. If it helps, here is a pictorialization of the “units transformation pipeline” that occurs inside $f \circ g$:

Note 3. To emphasize, $f(x)$ is the

~ total ~

amount earned when position $x$ is reached. In real life $f$'s graph might therefore look something like this, while inventing some numbers:

In the above the rat earns $\text{\$}$3 for the first 50m, after which the dollars-per-meter rate is reduced.

It computes the dollars-per-second earnings rate as a function of time. In more detail, $$ (f \circ g)'(t) $$ is the dollars-per-second rate which the rat is fundraising at $t$ seconds after the start of the race.

Note 1. You don't need to know anything about “$f$” or “$g$” to answer this question. You only need to know what “$f \circ g$” is.

At $t = 2$s the rat is running at a velocity of $$ {5\over 3\rule{0pt}{1em}}[{\text{m/s}}] $$ by the slope of this line segment on $y = g(t)$:

Moreover at $t = 2$s the rat has reached $2 \cdot (5/3) = 10/3 = 3.\overline{33}$m, where the dollars-per-meter earnings rate is one-tenth of a dollar per meter, by the slope of this segment on the graph $y = f(x)$:

Multiplying the $5/3$ meters-per-second velocity by the $1/10$ dollars-per-meter rate gives us the dollars-per-second rate at $t = 2$s (our final answer—recall that $(f \circ g)'(2)$ is the dollars-per-second rate at $t = 2$s, by Exercise 35):

$$ \left({5\over 3\rule{0pt}{1em}}\left[{\text{m} \over \text{s}}\right]\right) \times \left({1 \over 10\rule{0pt}{1em}}\left[{\text{\$} \over \text{m}}\right]\right) = {1\over 6\rule{0pt}{1em}}\left[{\text{\$} \over \text{s}}\right] $$

...or

$$ {0.1666...}[\text{\$/s}] $$

in decimal. (I.e., ~sixteen~ point $666...$ cents per second!)

The sought-for formula is $$ g'(t)\cdot f'(g(t)) $$ because—to come back to the example of the fundraising race—one must multiply the meters-per-second velocity at time $t$ (that is, $g'(t)$) by the dollars-per-meter earnings rate at position $g(t)$ (that is, $f'(g(t))$) to obtain the dollars-per-second earning rate at time $t$ (that is, $(f \circ g)'(t)$).

(For example, the solution to Exercise 36 can actually be written $$ g'(2) \cdot f'(g(2)) $$ since, indeed, $g(2) = 3.333...$. [Remember that we ended up multiplying $g'(2) = {5\over 3}\text{m/s}$ by $f'(3.333...) = {1\over 10}\text{\$/s}$—the “$3.333...$” is $g(2)$.])

Nb: This result is known as the chain rule.

Note 1. Said chain rule is more commonly written $$ (f \circ g)'(t) = f'(g(t))g'(t) $$ with “$g'(t)$” last, or

with “$x$” instead of “$t$”.

Note 2. In “function form” (instead of “value form”), the chain rule can be written: $$ \,(f \circ g)' = (f' \circ g)g'. $$ (Which is very succinct.)

The velocity vector is an arrow of length $3$ tangent to a circle of radius $2$, brushed clockwise:

If we bring the tail of the vector back to $(0, 0)$ we find an arrow of length $3$ tracing a circle of radius $3$:

Lastly, the velocity vector does a full revolution of the red circle in the same amount of time that the particle does a full revolution of the blue circle, which is $$ {2\cdot (\text{circumference of a unit circle}) \over \text{3}} $$ because the circumference of the blue circle is twice the circumference of a unit circle, and the particle is going at speed $3$.

Note 1. In such diagrams we recycle the axes to plot quantities of several different dimensions: positions (in blue, in this case) have dimensions of length ([L]) while velocities (in red, in this case) have dimensions of length over time ([L/T]).

The velocity vector of Exercise 39 travels in a circle of radius $3$ in the same amount of time that the position vector travels around a circle of radius $2$. The speed of the velocity vector is therefore $1.5$ times the speed of the position vector, or $1.5 \times 3 = 4.5$.

As the derivative of the velocity vector, the acceleration vector is therefore a vector of length $4.5$ (= the speed of the velocity vector) brushed clockwise along the path of the velocity vector:

Or, if we translate the acceleration vector back to the origin and trace out its path over time (either way is fine):

Note 1. You truly have to think of the acceleration vector as “the velocity of the velocity vector”—if the velocity vector is changing, the acceleration vector is nonzero!

The velocity vector has length $2$, because the particle has speed $2$. So the velocity vector looks like so, while attached to the particle path (top) or brought back to the origin (bottom):

Moreover (!) the speed of the velocity vector is $2/3$ the speed of the particle, because the velocity vector goes around a circle of $2/3$ the radius in the same amount of time. So the velocity vector has speed $$ %\Large {2\over 3}\cdot 2 = {4\over 3} \Large 2 \cdot {2\over 3} = {4\over 3} $$ from which the acceleration vector—that can be described as “the velocity vector of the velocity vector”—has length ${4\over 3}$ (the speed of the velocity vector), and looks like so (in either representation):

Lastly the acceleration vector has speed $$ \Large {4\over 3}\cdot {2\over 3} = {8 \over 9} $$ by virtue of circling a circle of radius $2/3$ that of the velocity vector, that has speed $4/3$, in the same amount of time. Since the jerk is the derivative of the acceleration, this becomes the length of the jerk vector, that is exactly opposite to the velocity vector, being twice $90^\circ$ away:

Note 1. If the particle's original path is centered at $(0, 0)$ then that path constitutes a fourth circle obeying the same pattern of $2/3$-ratios between the successive radii:

While the particle goes around a circle of radius $r$, the velocity vector goes around a circle of radius $v$. (Indeed $v$, being the speed, is the length of the velocity vector, and the length of the velocity vector is the radius of the circle traced by the velocity vector.) So the circle traced by the velocity vector is $$ \Large {v \over r} $$ times as large as the circle traced by the position vector. Therefore, the velocity vector goes $$ \Large {v \over r} $$ times as fast as the position vector! (The two vectors trace their respective circles in the same amount of time, so the only difference in speed is caused by differences in the radii—and this is the ratio of the radii.) Therefore, the velocity vector has speed $$ \Large v \cdot {v \over r} = {v^2 \over r} $$ ...as obtained by multiplying the speed of the position vector ($v$) by the ratio of the speeds ($v/r$). This is also the length of the acceleration vector. (Speed of velocity vector = length of acceleration vector.)

Next, the ratio $$ \Large {\text{speed}\over \text{radius}} $$ is the same for the velocity vector as it is for the position vector, because both “speed” and “radius” are scaled up by a factor $$ \Large {v \over r} $$ compared to the position vector. So $$ \Large {\text{speed}\over \text{radius}} = {v \over r} $$ for the velocity vector as well as for the position vector. But we can also write this ratio as $$ \Large {\text{length of acceleration vector}\over \text{radius}} $$ since the speed of the velocity vector is the length of the acceleration vector, or as $$ \Large {\text{length of acceleration vector}\over \text{radius of velocity vector circle}} $$ to be more exact, or as $$ {\Large {\text{radius of acceleration vector circle}\over \text{radius of velocity vector circle}}} $$ in yet another way! Therefore, the circle traced by the acceleration vector is $$ {\Large {v \over r}} $$ times as large as the circle traced by the velocity vector, and the same pattern starts all over again!

(In other words, each time we take a derivative we find that the vector whose derivative we are taking has speed $$ {\Large {v \over r}} $$ times the speed of the previous vector whose derivative we took, resulting in a circle $$ {\Large {v \over r}} $$ times as large as the current circle, resulting in a future speed $v/r$ times as large for the next derivative, etc, etc.)

Concretely, the length of the jerk vector will be $$ {\Large {v^2 \over r} \cdot {v \over r} = {v^3 \over r^2}} $$ because the length of the acceleration is $v^2/r$, and the length of the derivative of the jerk would be $$ {\Large {v^3 \over r^2} \cdot {v \over r} = {v^4 \over r^3}} $$ because the length of the jerk is $v^3/r^2$, etc. (Not that we needed to go beyond the jerk.)

Coming back to a sketch of all this, if $$ {\Large {v \over r} \approx 1.2} $$ the sketch will involve concentric circles of successive ratio $\approx 1.2$ with the successive vectors being off by $90^\circ$. The position circle might not be centered at $(0, 0)$, so we didn't include it in this sketch (this sketch presumes clockwise motion, but it's unimportant):

But if the position circle is centered at $(0, 0)$, it becomes the first circle in the sequence:

Note 1. The ratio $$ {\Large {v \over r}} $$ is known as the

angular velocity

of the particle. You can think of the angular velocity as

$$ {\Large {\text{speed}\over \text{radius}}} $$

directly per the expression above, or as

$$ {\Large {\text{distance per unit time}\over \text{radius}}} $$

since that is just the definition of “speed”, but which also means that you can think of the angular velocity as $$ \Large {\text{number of radii per unit time}} $$ or, say, just as $$ \Large {\text{radii per unit time}} $$ in other words. (The “number or radii” covered by an arc is also known as the radian measure of the arc—an alternate measure of angle—so this can also be phrased radians per unit time, in that sense.) What is noteworthy is that the angular velocity of the position vector is the same as the angular velocity of the velocity vector, of the acceleration vector, etc, and it also constitutes the ratio between the successive lengths of all these vectors!

Note 2. A common notation for the angular velocity of a particle is $$ {\Large \omega} $$ which means that the velocity vector, acceleration vector, and jerk vector have lengths have lengths $$ {\Large \omega{}^1r} $$ $$ {\Large \omega{}^2r} $$ $$ {\Large \omega{}^3r} $$ respectively, where $r$ is the radius of the circle, as the angular velocity is the ratio of the lengths of the successive vectors, as noted. (PS: As the length of the velocity vector is also known as the speed, $\omega^1r = \omega{}r$ is also the speed, by another name.) (PPS: We couldn't resist writing “$\omega^1r$” instead of “$\omega{}r$”, to keep things extra symmetric & typographically aligned.)

Starting with the second part of the question, because the particles are going around a circle of radius $3$ at speed $3$ the angular velocity (cf. Exercise 37) is $$ {\Large {3\over 3} = 1} $$ which means that the ratio of the lengths of all the vectors will be $1$, i.e., all vectors (velocity, acceleration, jerk) will have the same length as the radius, which is $3$.

Keeping in mind that the jerk is $90^\circ$ ahead of the acceleration is $90^\circ$ ahead of the velocity, etc, in the direction of rotation, the sixteen vectors—four for each particle—are therefore as follows:

(In particular, the purple particle's position ends up being the velocity of the red particle, and many other identities of the sort.)

Solution 1. Angular velocity is

speed over radius

which has dimensions $$ {\Large {\text{L/T} \over \text{L}}} $$ because speed has dimensions of length over time, L/T, while the radius has dimensions of length, L; this simplifies...

...down to dimensions of “one over time”.

Solution 2. Angular velocity is

radians per unit time

or

number of radii per unit time

(if you prefer), which is a “one over time” quantity, because radians are dimensionless.

Indeed, “radian” is short for “number of radii that fit inside the arc length”, which is one length divided by another length, which is, therefore, dimensionless.

The angular velocity, being speed over radius, is

or some $3$ and a third million radii per hour! (That's $925.\overline{925}$ radii per second, to bring it down to more “human” dimensions.) (You can also say $925.\overline{925}$ radians per second.)

Going back in time, examine how long it would take the particle to cross each of the intervals defined by the following geometric progression* (*see Note 1):

The interval from $0.5$ to $1$ takes at least

$$ {0.5 \over 1} = 0.5 $$

time to cross, because the maximum speed of the particle inside of that interval is $1$. Similarly, the interval from $$ x = 0.25 $$ to $$ x = 0.5 $$ takes at least $$ {0.25 \over 0.5} = 0.5 $$ time to cross, because the maximum speed of the particle inside of that interval is $0.5$! And, again, the interval from $$ x = 0.125 $$ to $$ x = 0.25 $$ takes at least $$ {0.125 \over 0.25} = 0.5 $$ time to cross, because the maximum speed of the particle inside of that interval is $0.25$. Etc—each interval takes at least $$ 0.5 $$ units of time to cross, because the length of each interval is half of the maximum speed within the interval! But there are infinitely many intervals, and, therefore, it takes

at least

infinitely much time to make it to $x = 0$, where the “infinitely” comes from adding infinitely many $0.5$'s together! (In other words, the particle never makes it to $x = 0$, no matter how far back in time you look.)

Note 1. A so-called

geometric progression

is a sequence of numbers in which each number is a fixed multiple of the previous number. For example, $$ 100,\, 300,\, 900,\, 2700 $$ is a (finite) geometric progression, because each number is the previous number multiplied by $3$, and $$ 1,\, 0.5,\, 0.25,\, 0.125,\, 0.0625,\, \ldots $$ is an (infinite) geometric progression, because each number is the previous multiplied by $0.5$.

For $t > 0$ the yellow and blue particles approach $0$ in a kind of “radioactive decay” pattern; for $t < 0$, they spin off to $-\infty$ and $\infty$ respectively at an accelerating rate:

If we add the graph of the red particle to the mix, it is simply the mirror image of the blue particle's position through the $y$ axis ($y$ axis that is ironically labeled “$x$”):

Indeed, for the red graph,

the slope equals the $y$-value

(velocity = position), while

the slope equals minus the $y$-value

for the blue graph (velocity = position of yellow = minus own position). (We forgot to mention that the blue and yellow graphs are mirror images of one another through the horizontal axis—this is one of the “symmetries” that the problem statement asks about, though.) As taking a mirror image through the $y$ axis negates slopes without affecting $y$-coordinates, while both the blue and red graphs have the same value at $t = 0$, this explains why the mirror image of one graph fits the constraints of the other and vice-versa:

Also note that all graphs have slopes of $ \pm 1 $ at $t = 0$, as we tried to reflect in the sketches, because each corresponding particle position is either $1$ or $-1$.

At $t = 0$ the particles have position vectors that are up and the right, so the velocity vectors will be up and to the right, so the particles will move more up and to the right, and the velocity vectors will become more “up and to the right”, and so on—broadly speaking, for $t > 0$ there will ensue a kind of four-particle explosion that goes up and to the right, off to $(+\infty, +\infty)$.

For $t < 0$ motion will be down and to the left, at least initially—it is hard to forecast off the top of one's head (unless you have a sudden flash of insight) what will happen for larger negative values of $t$.

HOWEVER.

It is possible to say much more.

To go deeper, we introduce eight new particles, comprising the original colors but in white and black flavors:

, , , , , , , .

At $t = 0$ the white particles are just a translate of the original particles, such that the circle on which they lie is centered at $(0, 0)$:

The black particles, for their part, are piled on top of one another at $(x, y) = (2, 4)$ at $t = 0$, that we draw as four quarter-pies of different colors, like a UNO card:

Within each group we set the velocity of the purple particle to the position of the yellow particle, the velocity of the yellow particle to the position of the green particle, and so on, like for the original set of particles.

In this case the white particles will rotate at unit speed around their circle of radius $1$ centered at the origin, just like the particles discussed in the solution to Exercise 12, that obey a similar set of constraints (albeit with a different set of colors).

The black particles, for their part, behave as a single fused-together particle whose velocity is equal to its position, and will see their motion confined to an infinite half-line through $(0, 0)$ and $(2, 4)$, as their velocity—being equal to their position—stays parallel to the line between them and the origin, meaning they are “stuck” to that line.

Also note that the

speed

of the black particles, being equal to the

length of the velocity vector

of said particles, is equal to the

length of the position vector

of said particles, is equal to the

distance to the origin

of said particles, since the length of the position vector is the distance to the origin.

This means that if we introduce gradations to the afore-mentioned half-line through $(0, 0)$ and $(2, 4)$...

...indicating the distance to the origin, the black particles behave like a one-dimensional system comprising a single particle on a half-line (or entire line, it doesn't hurt) whose velocity is equal to its position on this line:

The behavior of such a particle is identical to the behavior of the red particle from Exercise 41, except that the current “UNO particle” has a slight head-start over the red particle from Exercise 41, being at position $x = 2\sqrt{5}$ instead of at position $x = 1$ at $t = 0$. (!!)

This describes an “understandable” behavior of the black and white particles.

Next we write

$_x$

for the function that gives the $x$-coordinate of the purple-white particle as a function of time (in more detail,

$_x : \mathbb{R} \rightarrow \mathbb{R}$

to emphasize that WE ARE TALKING ABOUT A FUNCTION, e.g.,

$_x(2)$

is the $x$-coordinate of the purple-white particle at $t = 2$, etc), and write

$_y$

for the function that gives the $y$-coordinate of the purple-white particle as a function of time, and so on for all the other particles.

For example,

$_x\,\!\!\!' = $ $_x$

because the rate of change of the $x$-coordinate of the purple-white particle is the value of the $x$-coordinate of the yellow-white particle; we also have

$_x\,\!\!\!' + $ $_x\,\!\!\!' = $ $_x + $ $_x$

by adding two such equations together; this can also be written

$($$_x + $ $_x)' = $ $_x + $ $_x$

by the sum rule; but this gives us an idea!; we can try to define the original particles , , , by setting...

$_x =$ $_x + $ $_x$

$_{y\hspace{0.05em}} =$ $_{y\hspace{0.05em}} + $ $_{y\hspace{0.05em}}$

$_x = $ $_x + $ $_x$

$_{y\hspace{0.05em}} = $ $_{y\hspace{0.05em}} + $ $_{y\hspace{0.05em}}$

$_x =\ $ $_x + $ $_x$

$_{y\hspace{0.05em}} =\ $ $_{y\hspace{0.05em}} + $ $_{y\hspace{0.05em}}$

$_x =\ $ $_x + $ $_x$

$_{y\hspace{0.05em}} =\ $ $_{y\hspace{0.05em}} + $ $_{y\hspace{0.05em}}$

...and see if these definitions satisfy the constraints of the problem! (We momentarily have two different purple particles: the one from the problem statement, and the one that we have just defined; but that's ok, as long as we are aware of this small semantic transgression, it is not such a big deal, and we shall soon prove that these two particles are one and the same.) For starters...

$_x\,\!\!\!' =\ $ $($$_x + $ $_x\!\hspace{0.05em})' =\ $ $_x + $ $_x =$ $_x$

$_{y\hspace{0.05em}}\,\!\!\!' =\ $ $($$_{y\hspace{0.05em}} + $ $_{y\hspace{0.05em}}\!\hspace{0.05em})' =\ $ $_{y\hspace{0.05em}} + $ $_{y\hspace{0.05em}} =$ $_{y\hspace{0.05em}}$

$_x\,\!\!\!' =\ $ $($$_x + $ $_x\!\hspace{0.05em})' =\ $ $_x + $ $_x =$ $_x$

$_{y\hspace{0.05em}}\,\!\!\!' =\ $ $($$_{y\hspace{0.05em}} + $ $_{y\hspace{0.05em}}\!\hspace{0.05em})' =\ $ $_{y\hspace{0.05em}} + $ $_{y\hspace{0.05em}} =$ $_{y\hspace{0.05em}}$

$_x\,\!\!\!' =\ $ $($$_x + $ $_x\!\hspace{0.05em})' =\ $ $_x + $ $_x =$ $_x$

$_{y\hspace{0.05em}}\,\!\!\!' =\ $ $($$_{y\hspace{0.05em}} + $ $_{y\hspace{0.05em}}\!\hspace{0.05em})' =\ $ $_{y\hspace{0.05em}} + $ $_{y\hspace{0.05em}} =$ $_{y\hspace{0.05em}}$

$_x\,\!\!\!' =\ $ $($$_x + $ $_x\!\hspace{0.05em})' =\ $ $_x + $ $_x =$ $_x$

$_{y\hspace{0.05em}}\,\!\!\!' =\ $ $($$_{y\hspace{0.05em}} + $ $_{y\hspace{0.05em}}\!\hspace{0.05em})' =\ $ $_{y\hspace{0.05em}} + $ $_{y\hspace{0.05em}} =$ $_{y\hspace{0.05em}}$

...or...

$_x\,\!\!\!' =\ $ $_x$

$_{y\hspace{0.05em}}\,\!\!\!' =\ $ $_{y\hspace{0.05em}}$

$_x\,\!\!\!' =\ $ $_x$

$_{y\hspace{0.05em}}\,\!\!\!' =\ $ $_{y\hspace{0.05em}}$

$_x\,\!\!\!' =\ $ $_x$

$_{y\hspace{0.05em}}\,\!\!\!' =\ $ $_{y\hspace{0.05em}}$

$_x\,\!\!\!' =\ $ $_x$

$_{y\hspace{0.05em}}\,\!\!\!' =\ $ $_{y\hspace{0.05em}}$

...cutting out the middle computation, so the constraints relating particle velocities to particle positions are satisfied (e.g., the velocity vector of the purple particle is the position vector of the yellow particle); for seconders, evaluating these definitions at $t = 0$ gives...

$_x(0) =\ $ $_x(0) + $ $_x(0) = 2\,\,+$ $_x(0)$

$_{y\hspace{0.05em}}(0) =\ $ $_{y\hspace{0.05em}}(0) + $ $_{y\hspace{0.05em}}(0) = 4\,\,+$ $_{y\hspace{0.05em}}(0)$

$_x(0) =\ $ $_x(0) + $ $_x(0) = 2\,\,+$ $_x(0)$

$_{y\hspace{0.05em}}(0) =\ $ $_{y\hspace{0.05em}}(0) + $ $_{y\hspace{0.05em}}(0) = 4\,\,+$ $_{y\hspace{0.05em}}(0)$

$_x(0) =\ $ $_x(0) + $ $_x(0) = 2\,\,+$ $_x(0)$

$_{y\hspace{0.05em}}(0) =\ $ $_{y\hspace{0.05em}}(0) + $ $_{y\hspace{0.05em}}(0) = 4\,\,+$ $_{y\hspace{0.05em}}(0)$

$_x(0) =\ $ $_x(0) + $ $_x(0) = 2\,\,+$ $_x(0)$

$_{y\hspace{0.05em}}(0) =\ $ $_{y\hspace{0.05em}}(0) + $ $_{y\hspace{0.05em}}(0) = 4\,\,+$ $_{y\hspace{0.05em}}(0)$

...or...

$_x(0) = 2\,\,+$ $_x(0)$

$_{y\hspace{0.05em}}(0) = 4\,\,+$ $_{y\hspace{0.05em}}(0)$

$_x(0) = 2\,\,+$ $_x(0)$

$_{y\hspace{0.05em}}(0) = 4\,\,+$ $_{y\hspace{0.05em}}(0)$

$_x(0) = 2\,\,+$ $_x(0)$

$_{y\hspace{0.05em}}(0) = 4\,\,+$ $_{y\hspace{0.05em}}(0)$

$_x(0) = 2\,\,+$ $_x(0)$

$_{y\hspace{0.05em}}(0) = 4\,\,+$ $_{y\hspace{0.05em}}(0)$

...cutting out the middle computation, which is to say that the positions at time $t = 0$ of our newly-defined particles , , and are the translate of the white particle positions at $t = 0$ back up and to the right by the vector $(2, 4)$, which brings those positions back to the original positions of , , and as they appear in the problem statement! I.e., our newly-defined particles , , and are in the desired place at $t = 0$!

In other words, the proposed definitions of , , and “work” in the sense of satisfying all the conditions of the problem statement, and are, indeed, the solution we seek.

Qualitatively, this implies that the particles can be understood as four particles rotating at unit speed around a circle of radius $1$ (the white particles) where the center of circle (the UNO particle) is moving at exponential rate along a half-line. In particular, the particles remain at constant distance from one another for all $t$, whether that seems intuitive or not.

Concretely, the particle trajectories end up like so, locally around $t = 0$:

The above plot goes from $t = -5$ to $t \approx 1$—winding further back in time would produce near-perfect counterclockwise circular motion, as the black particles rush up to $(0, 0)$ and come to a near-halt rather fast, leaving only the residual motion of the white particles!

Note 1. When we said that, for $t > 0$, there ensues

“a [kind of] four-particle explosion”

in the first paragraph of the solution, the word “explosion” might be misleading, implying increased distances between the particles over time. This is not the case! (But we didn't know any better, back then.)

Note 2. As you might already have caught on, but is maybe worth emphasizing,

speed

is not the same thing as

velocity

because, specifically, speed is

distance per unit time

—a nonnegative number—whereas velocity is

displacement per unit time

—a vector-valued quantity, or $\pm$-valued quantity, in 1 dimension!

Recall the curves from Exercise 12:

The blue curve is the derivative of the red curve but is also the horizontal translate of the red curve by $a$ units to the left, where

$$ a $$

is the distance between adjacent bumps. Thus if $$ f $$ is the function that generates the red curve, then $$ %\Large f'(x) = f(x + a) $$ using the fact that $$ %\Large y = f(x + a) $$ is the horizontal translate of $y = f(x)$ by $a$ units to the left, in general for any function $f$ and constant $a \in \mathbb{R}$, as discussed in Exercise 14 of Chapter 3. (Well, this shows one solution, at least.)

For the functions:

For the real numbers:

Note 1. We never took the time to prove the associativity of function addition, but it is easy to prove! (For other proofs see Exercise 32, Exercise 33, as well as Exercise 9, Chapter 3.)

The function $x \rightarrow \cos(x + 0.1)$ fetches its values

in the future

by $0.1$, compared to $\cos(x)$. It is therefore

ahead

or its graph

to the left

of $y = \cos(x)$, by $0.1$.

It appears that the function is “a line plus something”, in the sense of the following diagram:

(Or in the technical sense of taking the sum of two functions, to be more exact—that is what the sum of two functions looks like, pictorialized!)

The line appears to be $$ y = {1\over 4}x $$ making it a true linear function (as opposed to affine, cf. Chapter 3).

As for the “something”—the bumpy one—it appears to just be a “sped up cosine”, and note that the graph goes through approximately $10$ cycles between $x = 0$ and $x = 6.28 \approx 4\eta$, as we can count on the graph of the original function:

As cos goes through one cycle from $x = 0$ to $x = 4\eta$, the “bumpy function” is therefore (roughly, from what we can see) a “$10$​$\times$” sped-up version of cosine, i.e., $$ y = \cos(10x) $$ from which the guesstimate for the initial function would be $$ y = \cos(10x) + {1\over 4} x $$ adding our two separate guesstimates together.

Having intimated that the answer is the sum of a parabola and of some cosine deviant, let us focus on the parabolic portion first, that would namely be roughly this purple curve:

[Nb:

parabolic

is a synonym of

quadratic

or

degree $\mathit{2}$ polynomial

i.e., a function of the form $$ x \rightarrow a_2x^2 + a_1x + a_0 $$ for constants $a_2$, $a_1$, $a_0 \in \mathbb{R}$, cf. Chapter 3.]

As the parabola is symmetric about the $y$ axis it will be of the form $$ y = Ax^2 + C $$ for some constants $A$, $C \in \mathbb{R}$. (And specifically without a “$Bx$” term, that would break symmetry.)

The value $$ C $$ is easy because it is the value of the parabola at $x = 0$, which in this case appears to be $y = -3$:

...so... $$ C = -3 $$

(we say). For $A$, note that the parabola appears to have value $y \approx 3$ at $x = \pm 12$, resulting in an increase of $\approx 6$ between $x = 0$ and $x = \pm 12$:

That increase being entirely due to the term $Ax^2$, we get $$ A \cdot 12^2 \approx 6 $$ (in more detail, $$ %(A\cdot 12^2) - (A\cdot 0^2) \approx 6 A\cdot 12^2 - A\cdot 0^2 \approx 6 $$ but $A\cdot 0^2$ goes away), meaning $$ A \approx {6\over 12^2} = {1\over 24} $$ meaning that the quadratic portion of the function is $$ {x^2\over 24} - 3 $$ per this estimate.

The other portion of the answer—what is left after the parabola is subtracted—is a cosine-like function (or sinusoid) whose amplitude (the height of a bump) is roughly $0.5$, which is half the amplitude of sine/cosine:

Moreover at $x = 0$ we find more or less exactly the bottom of a bump, so altogether we can use a function of the form $$ -0.5\cos(Bx) $$ to model this sinusoid, where the multiplication by $0.5$ gives us the desired amplitude and where the ‘$-$’ gives us an anti-bump (“trough”?) instead of a bump at $x = 0$; on the other hand the value $B$ will control the amount of horizontal compression inside the curve; specifically, $$ B = 1 $$ will give a curve that goes through one full cycle per interval of length $4\eta$, while, in general, an arbitrary value of $B$ will give a curve that goes $B$ full cycles per interval of length $4\eta$ (the larger $B$ is, the more “frenzied” the curve); in our case, it seems that $B \approx 20$:

So an estimate for the second function would be... $$ -0.5\cos(20x) $$ ...giving us... $$ -0.5\cos(20x) + {x^2\over 24} - 3 $$ ...for our final answer, after adding the parabola back.

First we estimate a function for the large-scale curve in purple:

It appears to be a linear function (line through $(0, 0)$ sloping down) plus a sinusoid. To estimate the slope of the linear function we can take two points in like relation to the sinusoid, draw a line between them and estimate the slope:

This makes the linear function portion of the purple curve $$ y = -{1\over 4}x $$ or thereabouts.

To model the sinusoid portion of the large-scale purple curve we need more measurements, such as the total height of the sinusoid from top of bump to bottom of trough; we can add a third “bottom of trough” dot, in same relation to the top of bumps (but we won't actually draw this dot or else we won't be able to see where we're measuring):

So it appears that the sinusoid has a total height of $\approx 4$ from top of bump to bottom of trough.

(And in case you're confused by what we're trying to do, let us re-explain that we are trying to measure the vertical width of this blue band...

...that, indeed, seems near $4$.)

What this means is that if we remove the linear portion $$ y = -{1\over 4}x $$ from the purple curve, what we will find is a sinusoid whose individual bumps have height $\approx 2$; something like this (we switch the color to aquamarine, so that “purple curve” retains its unique meaning):

This graph has the form $$ y = -2\sin(Bx) $$ for some value of $B \in \mathbb{R}$ that, chosen correctly, will give us the desired “wavelength”. (Note that $$ y = -\sin(x) $$ has graph...

...and that $$ y = -2\sin(x) $$ has graph...

...and, from there, all that remains is to “slow down” the oscillation to match the aquamarine graph—the “slowing down” is what $B$ is for.)

To know how much $B$ must be, we must measure the cycle length (it is, admittedly, hard to accurately determine the position of the top of each large-scale bump, but we do our best by basing ourselves off of what appear to be identical patterns in the small-scale oscillations at the top of each large-scale bump):

As $$ 12.6 \approx 12.56 = 2\times 6.28 \approx 2\cdot 4\eta $$ the period of the large-scale sinusoid is near twice the period of sin or cos; i.e., we need to “slow down” $$ -2\sin(x) $$ by a factor $2$, i.e., put $$ B = 0.5 $$ i.e., use $$ y = -2\sin(0.5x) $$ for the large-scale sinusoid. (Aquamarine graph.)

(So far we have $$ -{1\over 4}x - 2\sin(0.5x) $$ for our approximation to the purple curve, putting the linear and sinusoidal parts together. Now we move on from the purple curve.)

It remains to add in the small-scale oscillation from the original curve; we can do the tedious part first, and count the number cycles in an interval of length $4\eta \approx 6.28$:

So the small-scale oscillation is running at $\approx 19$ times the frequency of an ordinary sine or cosine, and we can model the small-scale oscillation by $$ x \rightarrow -0.5\sin(19x) $$ since, like the large-scale oscillation from the purple curve, it shares the same phase as $-\sin(x)$, and since, like the small-scale oscillation from Exercise 3, it has an amplitude of $\approx 0.5$.

Altogether, we get $$ y = -{1\over 4}x - 2\sin(0.5x) - 0.5\sin(19x) $$ as our “guesstimate”, while adding the linear function, the large-scale sinusoid, and the small-scale sinusoid together.

The culprits are:

For reference (if you need help checking), the graphs of $\sin$ and $\cos$ are as follows:

(Then imagine summing together, squaring, etc.)

Note 1. It is, indeed, intriguing that all of these functions appear to be sinusoids. (Formally defined as a function of the form $x \rightarrow A\cdot \sin(Bx + C) + D$ for some constants $A$, $B$, $C$, $D \in \mathbb{R}$.)

Yes. Imagine two wooden cutouts made from the graphs of $y = \sin^2(x)$, $y = \cos^2(x)$:

After vertically flipping the cutout of $y = \cos^2(x)$, the cutouts fit together to make the constant function $y = 1$ (like a parquet):

And this occurs because $$ \cos^2(x) + \sin^2(x) = 1 $$ for all $x \in \mathbb{R}$, which is the pythagorean identity.

What we want is the multiplicative ratio that would take us from the bottom to the top side of this quadrilateral, so that we can multiply by that ratio over and over again:

But the two triangles involved are

NOT RIGHT TRIANGLES

and we must break them into smaller parts that are right triangles in order to use trigonometric functions. Specifically, as per this drawing:

The multiplicative ratios that correspond to the first and third arrows (in arrow-order from bottom to top) are $$ \sin(68^\circ\!\hspace{0.1em}) $$ $$ \sin(59^\circ\!\hspace{0.1em}) $$ because these are “hypotenuse-to-opposite” arrows, while the multiplicative ratios that correspond to the second and fourth arrows are $$ \csc(71^\circ\!\hspace{0.1em}) $$ $$ \csc(60^\circ\!\hspace{0.1em}) $$ because these are the opposite (no pun intended), i.e.,“opposite-to-hypotenuse” arrows. The “big grey arrow” ratio from two diagrams ago is obtained by multiplying these four small-arrow ratios together, or $$ \sin(68^\circ\!\hspace{0.1em})\times\csc(71^\circ\!\hspace{0.1em})\times\sin(59^\circ\!\hspace{0.1em})\times\csc(60^\circ\!\hspace{0.1em}) $$ (that can also be written $$ {\sin(68^\circ\!\hspace{0.1em})\times\sin(59^\circ\!\hspace{0.1em}) \over \sin(71^\circ\!\hspace{0.1em})\times\sin(60^\circ\!\hspace{0.1em})} $$ because $\csc = {1\over \sin}$) which, numerically, comes out to $$ 0.97057870529467... $$ meaning that the top side of the quadrilateral tile is $$ 97\% $$ and some the length of the bottom side; taking the $36$-th power of $0.9705\dots$, because $36$ is the number of times that the quadrilateral repeats within the spiral, we find $$ 0.34127722635785... $$ which is the desired ratio $A/B$, and which agrees with the drawing, as $A$ seems plausibly to be about one-third of $B$, from the drawing!

Note 1. While the final answer can be written $$ (\sin(68^\circ\!\hspace{0.1em})\cdot\csc(71^\circ\!\hspace{0.1em})\cdot\sin(59^\circ\!\hspace{0.1em})\cdot\csc(60^\circ\!\hspace{0.1em}))^{36} $$ or $$ \left({\sin(68^\circ\!\hspace{0.1em})\cdot\sin(59^\circ\!\hspace{0.1em}) \over \sin(71^\circ\!\hspace{0.1em})\cdot\sin(60^\circ\!\hspace{0.1em})}\right)^{\!36} $$ teachers typically want to see such expressions evualuated out, to make sure that you and your calculator form a good team. (And, to be fair, catching one's own calculator mistakes by virtue of spotting a nonsensical number is a skill in itself.)

Note 2. Instead of counting the number of times that the quadrilateral appears in the double spiral by hand one can calculate the amount of rotation between one quadrilateral and the next, which is this purple angle:

The key to measuring this angle is the concept of an “alternating angle”, whereby $68^\circ$ reappears to the left of $71^\circ$, here:

Then we can calculate the purple angle as

meaning that each quadrilateral is rotated by $20^\circ$ from the previous, and in one turn of the spiral there are $$ {360^\circ\over 20^\circ} = 18 $$ quadrilaterals, or $$ 2 \times 18 = 36 $$ quadrilaterals for two turns!

The $x$- and $y$-coordinates are respectively $$ x_0 + r\cdot \cos(\phi) $$ and $$ y_0 + r\cdot \sin(\phi) $$ because

$$ r\cdot\cos(\phi) $$

is the difference from the center of the circle to $P$ in $x$ and

$$ r\cdot\sin(\phi) $$

is the difference from the center of the circle to $P$ in $y$, as per scaling a unit circle to radius $r$.

The standard definition is that the sum $$ \vec{u} + \vec{v} $$ of a vector $$ \vec{u} = (u_x, u_y) $$ and of a vector $$ \vec{v} = (v_x, v_y) $$ is the vector $$ (u_x + v_x, u_y + v_y) $$ whose first coordinate is the sum of the first coordinates of $\vec{u}$ and $\vec{v}$ and whose second coordinate is the sum of the second coordinates of $\vec{u}$ and $\vec{v}$.

Example 1. If $$ \vec{u} = (100, 100) $$ and $$ \vec{v} = (1, -1) $$ then

$$ \,\vec{u} + \vec{v} = (101, 99) $$

because $$ 100 + 1 = 101 $$ on the one hand, and $$ %100 + (-1) = 99 100 - 1 = 99 $$ on the other hand.

Note 1. The little arrow “$\vec{{\color \white{x}}}$” is a notation used to denote vectors. (Feel free to choose your own notation.)

Note 2. Represent the vectors $\vec{u}$, $\vec{v}$ by arrows whose components are displacements $u_x$, $u_y$, $v_x$, $v_y$ in $x$, $y$, $x$ and $y$ again respectively:

Then $$ u_x + v_x $$ may be geometrically realized as the concatenation of the $u_x$ and $v_x$...

...component displacements, while $$ u_y + v_y $$ may be geometrically realized as the concatenation of the $u_y$ and $v_y$...

...component displacements; moreover, both concatenations may be simultaneously obtained by concatenating the original $\vec{u}$ and $\vec{v}$ arrows...

...which actually implies that $\vec{u} + \vec{v}$ is the vector going from the head of $\vec{u}$ to the tail of $\vec{v}$ in the afore-mentioned concatenation, because of how we defined $\vec{u} + \vec{v}$:

This makes a mess, but the point is that this gives us a

geometric interpretation

or

geometric representation

or

geometric method of evaluation

for the sum of two vectors: concatenate the arrows of the vectors you're summing, and take the final displacement from the head of the first arrow to the tail of the last arrow.

Note 1. The solution to Exercise 7 can be cast in terms of vector addition, with, specifically, the position vector of the point being equal to the sum of the position vector of the circle's center with the “radial vector” from the center to the point:

The simplest motion that would produce such an $x$-coordinate (according to subjective human standards of simplicity, admittedly) is circular motion at uniform speed. In this case:

And either:

Or:

In more detail, every time $$ t $$ increases by $1$, $$ Bt + C $$ increases by $B$, but $$ Bt + C $$ is an amount of radians, because anything fed to ‘cos’ is an amount of radians; and so $B$ ends up being the

radian increase per unit time,

or angular speed, of the particle.

However, said “increase” in radians can be associated to either clockwise or a counterclockwise motion, there is no telling. (Quite aside from the fact that $B$ might be negative.) Indeed, while we usually think of $\cos(x)$ as

the $x$-coordinate of a point $x$ units
counterclockwise from $(1, 0)$ on the unit circle

$\cos(x)$ is also

the $x$-coordinate of a point $x$ units
clockwise from $(1, 0)$ on the unit circle

[“counterclockwise” $\rightarrow$ “clockwise”]. Per the one interpretation of ‘cos’, $$ Bt + C $$ is an amount of counterclockwise radians; per the other, $$ Bt + C $$ is an amount of clockwise radians. The following diagram illustrates the two possibilities:

(Or... ~ ~ ~ ~ ~ ~ ~ ~ ~

...to put everything in terms of counterclockwise-ness.) This accounts for the two solutions listed above. (But it can only be one of those two solutions, having made the “Occam's razor” assumption that the particle is traveling at uniform speed around a circle.) (Nb: In particular, “uniform speed” precludes sudden reversals of direction at either end of the circle, if you were at all thinking of that, for speed would be undefined at those points where direction is reversed.)

Note 1. If you harbor any doubts about there being no more than two solutions, picture this diagram...

...and imagine the vertical line scanning to the left
and to the right again as it tracks the $x$-coordinate of a particle going around the circle, unseen. Then there is one clockwise particle that tracks with the line, and one counterclockwise particle that tracks with the line, but no more, insofar as non-direction-reversing particles are concerned!

Let $$ x $$ be the number of cycles per $4\eta$ units of time.

Since one cycle is $4\eta$ radians, we can, instead of saying that the particle travels

$x$ cycles per $4\eta$ units of time

say that the particle travels

$x\cdot 4\eta{}$ radians per $4\eta$ units of time

or that the particle travels

$x$ radians per $1$ unit of time

dividing by $4\eta$. In other words, $x$ is the so-called

angular speed

or

angular frequency

of the particle. (That's the answer: “angular speed”, or, equivalently, “angular frequency”.)

An angle is $$ 0.2 $$ radians if the length of the subtended arc is $$ 20\% $$ the length of the radius; proceeding by elimination—many things are obviously not $20\%$ of the radius—that's this one:

If you zoom in a little bit you can actually see “$100\%$”, “$20\%$” written in fine print:

( Joking.)

While $\eta$ was defined as the quarter-circumference of a unit circle, more generally, $$ \eta $$ is the quarter-circumference of a circle (any circle) divided by its radius, and, correspondingly, $$ 4\eta $$ is the circumference of a circle (any circle) divided by its radius.

Note 1. As one consequence, it follows that the circumference of a circle is $$ 4\eta\cdot r $$ where $r$ is the radius.

Note 2. These “ratio descriptions” of $\eta$ and $4\eta$ also follow by viewing $\eta$ and $4\eta$ as the radian values of a right angle and a full angle, respectively.

Here is the division in American notation overlayed on top of a “Plaza” wallpaper to help demarcate the different columns of digits (in American notation each column of digits is associated to a power of $10$, with both the numerator and the quotient [the result] living inside the same set of columns, and only the denominator living outside, in a time-space porthole of its own):

The division stops when we see the same remainder twice—here ‘$40$’ reappears, which means that the next digit of the quotient will be $5$ (like the second digit of the quotient, that we obtained back when we had a remainder of $40$), the next one $7$ (the third digit of the quotient), etc—digits will repeat and the “final” quotient when we let the division unravel infinitely far to the right will be

...where the decimal point is after the first ‘$1$’ because the first ‘$1$’ is in the ‘$10^0$’ column.

Note 1. If you've never done this kind of thing before, the division starts in this blank state:

We take the first digit of the numerator, which is ‘$1$’, ask “how many times does $7$ go into $1$?”, we will write the answer here:

The answer is ‘$0$’ ($7$ goes $0$ times into $1$):

We next add a digit from the numerator, giving us $11$, we ask “how many times does $7$ go into $11$?”, we will write the answer here:

The answer is ‘$1$’ ($7$ goes $1$ times into $11$):

We subtract

from $11$, giving us a new remainder of $4$ (the very first “remainder” is actually $11$, before anything starts) (before anything started we had $11 = 7 \times 0 + 11$, and now we have $11 = 7 \times 1 + 4$):

Because $7$ does not fit into $4$ (and if it did, we would have done something wrong) we “bring down a $0$” that is actually part of a hidden sequence of $0$'s sitting to the right of $11$:

We ask “how many times does $7$ go into $40$?”, we will write the answer here:

The answer is ‘$5$’ ($7$ goes $5$ times into $40$):

We subtract

from $40$, giving us a new remainder of $5$ (well, to be technical, the remainder is actually $5 \times 10^{-1}$, not $5$, but the teacher at the board will often say “$5$”):

Because $7$ does not fit into $5$, we bring another $0$ down from our infinite reserve of $0$'s:

(Etc.)

The divisions, pursued up to the point where remainders repeat, look as follows:

From which...

...because the digits of the quotient are, in each case, about to restart from the first digit after the decimal point.

Note 1. Because...

...we thus have...

...though none of these estimates are actually accurate to more than two decimal places. (While all are accurate up to at least two decimal places.)

Note 2. As mentioned in Note 1 of Exercise 24, Chapter 3, these approximations are about half-a-part-in-a-thousand too large, or to be more exact, $$ \approx 0.0004 $$ too large in relative terms. I.e., you can subtract $$ \approx 1.\overline{571428} \cdot 0.0004 \approx 0.0006 $$ from $$ \approx 1.\overline{571428} $$ to get a better approximation for $\eta$, subtract $$ \approx 3.\overline{142857} \cdot 0.0004 \approx 0.0012 $$ from $$ \approx 3.\overline{142857} $$ to get a better approximation for $2\eta$, subtract $$ \approx 4.\overline{714285} \cdot 0.0004 \approx 0.0019 $$ from $$ \approx 4.\overline{714285} $$ to get a better approximation for $3\eta$, subtract $$ \approx 6.\overline{285714} \cdot 0.0004 \approx 0.0025 $$ from $$ \approx 6.\overline{285714} $$ to get a better approximation for $4\eta$. The resulting approximations end up being... $$ \Large \eta \approx 1.5708\\ \Large \rule{0pt}{1.7em}2\eta \approx 3.1416\\ \Large \rule{0pt}{1.7em}3\eta \approx 4.7123\\ \Large \rule{0pt}{1.7em}4\eta \approx 6.2832 $$ ...that are correct approximations up to the fourth digit, it turns out, modulo rounding off of the fifth digit. (But these are not worth learning by heart, by any means.)

Here is the long division, pursued up to the “$10^{-3}$” column of the quotient, including one last remainder computation that occurs pursuant to adding the digit in the “$10^{-3}$” column of the quotient (this is what the problem statement refers to as the “last remainder”):

Said quotient is $$ 20003200099.033 $$ while the new (last) remainder is $$ 0.359 $$ and the equation linking the two is $$ 15542486476949 = 777 \times 20003200099.033 + 0.359 $$ per properties of the long division algorithm.

Note 1. Lest anyone get left behind, we can go over the division “on the board” for a bit.

We start by asking “how many times does $777$ go into $1$?”, the answer will go here:

The answer is $0$ (that we need not write down, but we can), we move to asking “how many times does $777$ go into $15$?”, the answer will go here:

The answer is $0$, we move to asking “how many times does $777$ go into $155$?”, the answer will go here:

The answer is $0$, we move to asking “how many times does $777$ go into $1554$?”, the answer will go here:

The answer is $2$, we subtract $2 \times 777$ from $1554$, giving us a new “prefix” for the remainder:

(FYI, the remainder now consists of these yellow digits:)

Continuing, we bring down a $2$, ask “how many times does $777$ go into $2$?”, the answer will go here:

The answer is $0$; we bring down a $4$, ask “how many times does $777$ go into $24$?”, the answer will go here:

The answer is $0$; we bring down an $8$, ask “how many times does $777$ go into $248$?”, the answer will go here:

The answer is $0$; we bring down a $6$, ask “how many times does $777$ go into $2486$?”, the answer will go here:

The answer is $3$, we subtract $3 \times 777$ from $2648$, giving us a new remainder:

(And to be specific, the remainder is now formed by...

...these yellow digits.) Etc.

Note 2. If you need help brushing up on your long-form subtraction, say your subtraction is this:

You can either treat the top or bottom number as an odometer, and count how much a car with this odometer must be driven back/forth to reach the other number.

Viewing the top number as an odometer, the odometer would be as follows:

Starting from the right end of the subtraction, we ask “by how much (or how little) does a car need to be driven backwards, to turn the ‘$2$’ into an ‘$8$’?”, we will write the answer below:

The answer is: $4$ [miles*] (*say):

But the odometer will pass from ‘$0$’ to ‘$9$’ on its way down to $8$, so we also turn the ‘$6$’ into a ‘$5$’ (or “borrow a $10$” to do $12 - 8 = 4$, as some teachers put it):

Next we ask “by how many [$10$s of miles] does the car need to be driven back, to turn the ‘$5$’ into a ‘$9$’?”, we will write the answer below:

The answer is: $6$ [$10$s of miles]:

But here too the odometer will pass from ‘$0$’ to ‘$9$’ as we reduce it, so the wheel to its left must be turned back a notch as well; and because that wheel is a ‘$0$’ already, the wheel to its left must be turned back a notch; and so on, resulting in a small cascade effect:

Next we ask “by how many [$100$s of miles] does the car need to be driven back, to turn the ‘$9$’ into a ‘$0$’?”, we will write the answer below:

The answer is: $9$ [$100$s of miles]:

Next we ask “by how many [$1000$s of miles] does the car need to be driven back, to turn the ‘$9$’ into a ‘$5$’?”:

The answer is: $4$ [$1000$s of miles]:

Etc—we finally obtain:

(And like we alluded to above, one can also consider the bottom number to be the odometer, and “count up” towards the top number, leading to a symmetric algorithm, but one method is not better than the other.)

Note 3. Generally,

at any point in a long division after the remainder is updated to account for a new term added to the quotient. That's what the remainder is for: to satisfy this equation!

It is $\eta - \theta$. (Since... $$ \theta + (\eta - \theta) + \eta = 2\eta $$ ...you know that's the right answer!) (Nb: $2\eta = 180^\circ$.)

The sum $$ x + y $$ where $$ (x, y) $$ is a point in $\mathbb{R}^2$ can be understood as a certain vertical displacement added to a certain horizontal displacement, but where the sum is numerical. (Not vectorial.) To realize the sum geometrically we must either align the vertical displacement to be horizontal, or else align the horizontal displacement to be vertical:

Either way, the upshot is that the sum $$ x + y $$ can be found as the intersection of a line of angle $-45^\circ$ through the point $(x, y)$ with either the $x$- or $y$-axis:

(By the way: when we say “line of angle $-45^\circ$” we refer to the standard position of $-45^\circ$ on the unit circle, and, more specifically, to a line that is parallel to a line going through $(0, 0)$ and that standard position.)

To add a little imagery, if we make a heatmap of $\mathbb{R}^2$ according to the value of the coordinate sum... $$ x + y $$ ...over all points $(x, y)$ we will obtain diagonal bands of slope $-1$:

If we are confined to some region of the plane and we need to find a point that maximizes the coordinate sum we must go as far up and to the right as possible, towards brown—whereas to minimize the sum we must go as far down and to the left as possible, towards purple!

Having said this, $$ \cos(x) + \sin(x) $$ can be interpreted as the sum of the $x$- and $y$-coordinates of the point $$ (\cos(x), \sin(x)) $$ that is a point on the unit circle. In other words, the unit circle is “the region of the plane” (cf. previous paragraph) to which we are confined—we must choose a value of $x$ that puts us as far “up and to the right” as possible on the circle. That value is... $x = \eta/2$ (!!!!):

...or with any multiple of $4\eta$ added, making the set of solutions $x$ actually equal to $$ \Large \{\eta/2 + 4\eta{}k : k \in \mathbb{Z}\} $$ (to be read “$\eta/2$ plus any multiple of $4\eta$”) (*or: “the set of values of the form $\eta/2$, plus any multiple of $4\eta$”).

The actual value of $\cos(x) + \sin(x)$ achieved at this point is

using the fact that

and that $$ \Large \sqrt{0.5} = {\sqrt{2} \over 2} $$ (cf. Exercise 1, Chapter 1).

Note 1. The fact that $$ \Large \cos(\eta/2) + \sin(\eta/2) = \sqrt{2} $$ can also be seen from this diagram...

...which is an application of the Pythagorean theorem. (In the above, $x_0 = \cos(\eta/2)$, $y_0 = \sin(\eta/2)$, and the sum is seen to be $\sqrt{1^2 + 1^2} = \sqrt{2}$.)

Note 2. On the original graph from the statement, the $y$-value of the maximum is therefore $y = \sqrt{2}$, attained at $x = \eta/2$, $x = 9\eta/2$, $x = -7\eta/2$, etc:

(Note that $$ \Large {\eta\over 2} + 4\eta = {\eta\over 2} + {8\eta\over 2} = {9\eta\over 2} $$ $$ \Large {\eta\over 2} - 4\eta = {\eta\over 2} - {8\eta\over 2} = -{7\eta\over 2} $$ and, in general, the numerators of these fractions will be some multiple of $8\eta$ apart.)

It is $\theta$, as well. One method of deduction uses the fact that “the complement of my complement is myself”:

Another method of deduction uses the fact that, together with the angle immediately to its left (which happens to be $\eta - \theta$, because it is the complement of $\theta$ via the smallest right triangle present), the sought-for angle makes up $90^\circ$:

(In one case we use the fact that the medium-sized triangle is a right triangle, in another case that the smallest-size triangle is a right triangle—and in both cases that the original, largest triangle is a right triangle.)

The ending and starting half-lines of the angle marked ‘?’ are both $90^\circ$ counterclockwise from the ending and starting half-lines, respectively, of the central angle:

The angle marked ‘?’ is therefore obtained by a $90^\circ$ rotation (and then translation) of the central angle, and is, therefore, equal to the central angle $\theta$.

Note 1. This holds no matter which quadrant we push $\theta$ to:

Verbalized: the counterclockwise angle from the positive $x$ axis to the radial vector equals the counterclockwise angle from the positive $y$ axis to the counterclockwise tangent.

(Nb: When we say the “radial vector” and “counterclockwise tangent” we mean those objects that are illustrated here:)

The isosets of $$ (x, y) \rightarrow xy $$ (lambda-notation for a two-variable function) have this general appearance (it depends on the window and on the exact isosets that you choose to draw—we chose a few different random ones):

Note that each isoset consists of the union of TWO disjoint curves, except for the isoset $$ xy = 0 $$ (or: “the isoset $$ \{(x, y) \in \mathbb{R}^2 : xy = 0 \} $$ ...” to pedantically indicate that we are talking about a set of points in the plane), that, for its part, cannot be said to consist of two _~_~_disjoint_~_~_ curves, because it is the union of the $x$- and $y$- axes, that intersect.

(For a throwback, the solution of Exercise 16 of Chapter 3 mentions that the product of two numbers is $0$ if and only if one of the numbers is $0$. In our case, $$ xy = 0 $$ if and only $$ \,x = 0\, $$ or $$ \,y = 0 $$ where $$ x = 0 $$ happens to be the equation of the $y$ axis, and $$ y = 0 $$ happens to be the equation of the $x$ axis, which explains the shape of the isoset.)

If we draw a “heatmap” of $$xy$$ in some region of the plane, similarly to Exercise 18, the larger (more positive) values show up in the first and third quadrants, while the smaller (more negative) values show up in the second and fourth quadrants:

Of particular interest to us: at a given distance from the origin, the line $$ x = y $$ can be seen seen to hold the largest of values of $xy$:

In particular, $$ \cos\theta\,\sin\theta $$ will reach its maximum at those values of $\theta$ that put the point $(\cos \theta, \sin \theta)$ at either $(\sqrt{0.5}, \sqrt{0.5})$ or $(-\sqrt{0.5}, -\sqrt{0.5})$ on the unit circle; these values of $\theta$ are $$ \Large \{0.5\eta + 4\eta{}k : k \in \mathbb{Z}\} \\ \Large \cup \{2.5\eta + 4\eta{}k : k \in \mathbb{Z}\}\rule{0pt}{1.5em} $$ as per this illustration...

...and the maximum value of $$ \Large \cos\theta\,\sin\theta $$ itself will be $$ \Large \sqrt{0.5} \cdot \sqrt{0.5} = {1\over 2} $$ or $$ \Large (-\sqrt{0.5}) \cdot (-\sqrt{0.5}) = {1\over 2} $$ equivalently; though one should also note that

which is the “clever” way of writing the set of $\theta$'s for which the maximum is achieved.

Note 1. As a consequence, the function $$ %y = \sin x{}\cos x x \rightarrow \sin x{}\cos x $$ discussed in Exercise 4 has maximum value $$ {1\over 2} $$ achieved for inputs in the set

comprising the sequence of values... $$ \large \dots,\,\, -{7\eta\over 2},\,\, -{3\eta\over 2},\,\, {\Rule{0pt}{0em}{0.25em}\eta \over 2},\,\, {5\eta \over 2},\,\, {9\eta \over 2},\,\, \dots $$ ...or... $$ \large \dots,\,\, {-3.5\eta},\,\, {-1.5\eta},\,\, {0.5\eta},\,\, {2.5\eta},\,\, {4.5\eta},\,\, \dots $$ ...(maybe more legibly); annotating the graph given in Exercise 4:

Note 2. For completeness, here is a closer look at the isoset $xy = 1$, including some labeled points:

Note 3. Because a point $(x, y)$ satisfies $$ xy = 1 $$ if and only if the point $(2x, y)$ satisfies $$ xy = 2 $$ (one has $$ x_0y_0 = 1 $$ if and only if $$ (2x_0)y_0 = 2 $$ surprise or not) the curve $$ xy = 2 $$ is the horizontal dilation of the curve $$ xy = 1 $$ by a factor $2$; likewise, it is also the

vertical dilation

of the curve $$ xy = 1 $$ by a factor $2$; the two dilations are illustrated here:

More generally, the curve $$ xy = C $$ for $C \ne 0$ is the

$(a, b)$-dilation

[meaning: a horizontal dilation by a factor $a$ followed by a vertical dilation by a factor $b$, or vice-versa, the order doesn't matter] of the curve $$ xy = 1 $$ for all pairs $(a, b)$ such that $ab = C$; for example, $$ xy = 3 $$ is the

$(3, 1)$-dilation

[horizontal dilation by factor $3$] of $xy = 1$, as it is the

$(1, 3)$-dilation

[vertical dilation by factor $3$] of $xy = 1$, but is also the

$(\sqrt{3}, \sqrt{3})$-dilation

of $xy = 1$, since $\sqrt{3}\cdot\sqrt{3} = 3$, and the

$(12, {1\over 4})$-dilation

of $xy = 1$, since $12 \cdot {1\over 4} = 3$, etc.

For another specifically noteworthy instance of this phenomenon, $$ xy = 1 $$ is the

$(-1, -1)$-dilation

of itself, since $(-1)\cdot(-1) = 1$, a fact that is also known as the “central symmetry” of $xy = 1$. (You can take this last statement as the definition of “centrally symmetric”. I.e., a set $S \subseteq \mathbb{R}^2$ is centrally symmetric if and only if $S$ is equal to the $(-1, -1)$-dilation of itself.)

(Indeed, since $$ xy = C $$ if and only if $$ (-x)(-y) = C $$ each of the isosets is centrally symmetric, not only $xy = 1$.)

Note 4. Among other additional symmetries, the line $$ x = y $$ is an axis of symmetry of each isoset, meaning that each isoset equals its mirror reflection about that line:

Note that, technically, a set $S \subseteq \mathbb{R}^2$ [meaning: $S$ is a set of points in the plane] is symmetric about $x = y$ if and only if $$ \Large (x_0, y_0) \in S \iff (y_0, x_0) \in S $$ [read “$(x_0, y_0)$ is in $S$ if and only if $(y_0, x_0)$ is in $S$”] for all $(x_0, y_0)$. Illustrated:

In our case, a point $(x_0, y_0)$ is on the curve $$ xy = C $$ if and only if the point $(y_0, x_0)$ is on the curve, because $x_0y_0 = y_0x_0$, by commutativity of multiplication. This observation constitutes the “proof” that each isoset is mirror symmetric through $x = y$.

(Or... $$ \large \begin{align} \large & (x_0, y_0) \in \{(x,y)\in \mathbb{R}^2: xy = C\} \\ \large\rule{0pt}{1.4em} \iff & \,x_0y_0 = C \\ \large\rule{0pt}{1.4em} \iff & \,y_0x_0 = C \\ \large\rule{0pt}{1.4em} \iff & (y_0, x_0) \in \{(x,y)\in \mathbb{R}^2: xy = C\} \end{align} $$ ...to put it over-the-top formally.)

Note 5. Lastly, each isoset is symmetric through the line $x = -y$:

Indeed, this symmetry can be obtained as the

composition

of a symmetry through $x = y$ and a central symmetry:

In other words, any set that is symmetric through $x = y$ and that is centrally symmetric is also symmetric through $x = -y$, so there is nothing “new” to prove here, except to make this observation about composition!

We shall use the height $C$ of the triangle as a stopover between $A$ and $B$:

On the one hand, $C/B = \tan(\theta)$:

On the other hand, $A/C = \tan(\theta)$ also, by the result of Exercise 19 (whereby $\theta$ reappears as the top left angle of the middle-sized right triangle):

The answer is therefore: $$ %{A\over B} = {C\over B}\cdot {A\over C} = \tan(\theta)\cdot\tan(\theta) = \tan^2(\theta). $$ (As per the fact that $(A/B) = (C/B)\cdot(A/C)$.)

Here are two solutions:

Solution 1. We use the small leg $D$ of the triangle as a stopover between $A$ and $B$:

On the one hand, $D/B = \sec(\theta)\,\, (= 1/\cos(\theta))$:

On the other hand, $A/D = \sec(\theta)$, also (!?):

Thus: $$ {A\over B} = {D\over B}\times {A\over D} = \sec(\theta)\cdot \sec(\theta) = \sec^2(\theta). $$

Solution 2. We decompose $A$ as $B + A'$ where $A' = B - A$ is the “old $A$” from Exercise 22:

We find: $$ {A\over B} = {{B + A'}\over B} = {B\over B} + {A'\over B} = 1 + \tan^2(\theta) $$ ...since $$ {A'\over B} = \tan^2(\theta) $$ by Exercise 22. (The End.)

Note 1. Since the two solutions compute answers to the same question, one can in particular deduce that

for all $0 < \theta < \eta$, which is the range of $\theta$ covered by these diagrams. (The same identity holds more generally than just those $\theta$'s, however.)

Let $\theta \in \text{dom}\, \tan = \text{dom}\, \sec$. Then $$ \cos(\theta) \ne 0 $$ and $$ %1 = {\cos(\theta)\over \cos(\theta)} = {\cos^2(\theta)\over \cos^2(\theta)} 1 = {\cos^2(\theta)\over \cos^2(\theta)} $$ and, by the definitions, $$ \begin{align} 1 + \tan^2(\theta) \,\,&=\,\, 1 + \left({\sin(\theta)\over \cos(\theta)}\right)^{\!2} \\ &=\,\, \rule{0pt}{2em} {\cos^2(\theta)\over \cos^2(\theta)} + {\sin^2(\theta)\over \cos^2(\theta)} \\ &=\,\, \rule{0pt}{2em} {\cos^2(\theta) + \sin^2(\theta)\over \cos^2(\theta)} \\ &=\,\, \rule{0pt}{2em} {1\over \cos^2(\theta)} \\ &=\,\, \rule{0pt}{2em} \left({1\over \cos(\theta)}\right)^{\!2} \\ &=\,\, \rule{0pt}{2em} \sec^2(\theta) \end{align} $$ while using the pythagorean identity in the fourth step.