# Math Maps The Island of Utopia

Under the category of “Somebody Had To Figure It Out,” Dr. Andrew Simoson of King University (Bristol, Tennessee) used calculus to determine the shape of the island of Utopia in the 500-year-old book by Sir Thomas More based on the description of island given in the book’s introduction.

Paper by Dr. Simoson: http://archive.bridgesmathart.org/2016/bridges2016-65.html

# A natural function with discontinuities (Part 1)

The following tidbit that was published on the American Mathematical Monthly’s Facebook page caught my attention:

Here’s the relationship between $r$, $R$, and $\theta$ in case it isn’t clear from the description. The gray sector is determined by $r$ and $\theta$, and then the blue circle with radius $r$ is chosen to enclose the sector.

Unfortunately, there was typo for the third case; it should have been $r = R \sin \frac{1}{2} \theta$ if $90^\circ \le \theta \le 180^\circ$. Here’s the graph if $R = 1$, using radians instead of degrees:

As indicated in the article, there’s a discontinuity at $t=0$. However, the rest of the graph looks nice and smooth.

Here’s the graph of the first derivative:

The first derivative is continuous (and so the original graph is smooth). However, there are obvious corners in the graph of the first derivative, which betray discontinuities in the graph of the second derivative:

# An Alternative Proof of the Product Rule

I saw this and immediately groaned, wondering why I hadn’t thought of this myself.

# Stump the Prof: An Activity for Calculus I

After finishing the Product, Quotient, and Chain Rules in my calculus class, I’d tell my class the following: “Next time, we’re going to play Stump the Prof. Anything that you can write on the board in 15 seconds, I will differentiate. Anything. I don’t care how hard it looks, I’ll differentiate it (if it has a derivative). So do your best to stump me.”

At the next lecture, I would devote the last 15-20 minutes of class time to Stump the Prof. Students absolutely loved it… their competitive juices got flowing as they tried to think of the nastiest, hairiest functions that they could write on the board in 15 seconds. And I’d differentiate them all using the rules we’d just covered.. though I never promised that I would simplify the derivatives!

Sometimes the results were quite funny. Every once in a while, a student would write some amazingly awful expression but forgot to include an $x$ anywhere. Since the given function was a constant, the derivative of course was zero.

The worst one I ever got was something like this:

$y = \csc^4(\sec^5(\csc^8(\sec^7(\csc^4(\sec^5(x)))))$

Differentiating this took a good 3-4 minutes and took maybe 5 lines across the entire length of the chalkboard; I remember that my arm was sore after writing down the derivative. Naturally, some wise guy used his 15 seconds to write $y =$ in front of my answer, asking me to find the second derivative. At that, I waved my white handkerchief and  surrendered.

The point of this exercise is to illustrate to students that differentiation is a science; there are rules to follow, and by carefully following the rules, one can find the derivative of any “standard” function.

Later on, when we hit integration, I’ll draw a contrast: differentiation is a science, but integration is a combination of both science and art.

# Different Ways of Computing a Limit: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on different ways of computing the limit

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$

Part 1: Algebra

Part 2: L’Hopital’s Rule

Part 3: Trigonometric substitution

Part 4: Geometry

Part 5: Geometry again

# How I Impressed My Wife: Part 6g

This series was inspired by a question that my wife asked me: calculate

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Originally, I multiplied the top and bottom of the integrand by $\tan^2 x$ and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.
Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals.

Since $Q$ is independent of $a$, I can substitute any convenient value of $a$ that I want without changing the value of $Q$. As shown in previous posts, substituting $a =0$ yields the following simplification:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

$= \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}$

$= 2\pi i \left[\displaystyle \frac{r_1}{r_1^2-1} + \displaystyle \frac{r_2}{r_2^2-1} \right]$,

where I’ve made the assumption that $|b| < 1$. In the above derivation, $C_R$ is the contour in the complex plane shown below (graphic courtesy of Mathworld).

Also,

$r_1 = \sqrt{1-b^2} + |b|i$

and

$r_2 = -\sqrt{1-b^2} + |b|i$

are the two poles of the final integrand that lie within this contour.

It now remains to simplify the final algebraic expression. To begin, I note

$\displaystyle \frac{r_1}{r_1^2-1} = \displaystyle \frac{\sqrt{1-b^2} + |b|i}{[\sqrt{1-b^2} + |b|i]^2 - 1}$

$= \displaystyle \frac{\sqrt{1-b^2} + |b|i}{1-b^2 + 2|b|i\sqrt{1-b^2} - |b|^2 - 1}$

$= \displaystyle \frac{\sqrt{1-b^2} + |b|i}{-2|b|^2 + 2|b|i\sqrt{1-b^2}}$

$= \displaystyle \frac{\sqrt{1-b^2} + |b|i}{2|b|i(|b|i +\sqrt{1-b^2})}$

$= \displaystyle \frac{1}{2|b|i}$.

Similarly,

$\displaystyle \frac{r_2}{r_2^2-1} = \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{[-\sqrt{1-b^2} + |b|i]^2 - 1}$

$= \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{1-b^2 - 2|b|i\sqrt{1-b^2} - |b|^2 - 1}$

$= \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{-2|b|^2 - 2|b|i\sqrt{1-b^2}}$

$= \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{2|b|i(|b|i -\sqrt{1-b^2})}$

$= \displaystyle \frac{1}{2|b|i}$.

Therefore,

$Q = 2\pi i \left[\displaystyle \frac{r_1}{r_1^2-1} + \displaystyle \frac{r_2}{r_2^2-1} \right] = 2\pi i \left[ \displaystyle \frac{1}{2|b|i} + \frac{1}{2|b| i} \right] = 2\pi i \displaystyle \frac{2}{2|b|i} = \displaystyle \frac{2\pi}{|b|}$.

And so, at long last, I’ve completed a fifth different evaluation of $Q$.

# How I Impressed My Wife: Part 6f

This series was inspired by a question that my wife asked me: calculate

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Originally, I multiplied the top and bottom of the integrand by $\tan^2 x$ and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.
Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals.

Since $Q$ is independent of $a$, I can substitute any convenient value of $a$ that I want without changing the value of $Q$. As shown in previous posts, substituting $a =0$ yields the following simplification:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

$= \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}$,

where $C_R$ is the contour in the complex plane shown below (graphic courtesy of Mathworld).

Amazingly, contour integrals can be simply computed by evaluating the residues at every pole located inside of the contour. (See Wikipedia and Mathworld for more details.) I have already handled the case of $|b| = 1$ and $|b| > 1$. Today, I begin the final case of $|b| < 1$.

Earlier in this series, I showed that

$z^4 + (4b^2 - 2) z^2 + 1 = (z^2 + 2z \sqrt{1-b^2} + 1)(z^2 - 2z \sqrt{1-b^2} + 1)$

if $|b| < 1$, and so the quadratic formula can be used to find the four poles of the integrand:

$r_1 = \sqrt{1-b^2} + |b|i$,

$r_2 = -\sqrt{1-b^2} + |b|i$,

$r_3 = \sqrt{1-b^2} - |b|i$,

$r_4 = -\sqrt{1-b^2} - |b|i$.

Of these, only two lie ($r_1$ and $r_2$) within the contour for sufficiently large $R$ (actually, for $R > 1$ since all four poles lie on the unit circle in the complex plane).

As shown earlier in this series, the residue at each pole is given by

$\displaystyle \frac{r^2 + 1}{2r^3 + (4b^2-2)r}$

I’ll now simplify this considerably by using the fact that $r^4 + (4b^2-2)r^2 + 1 = 0$ at each pole:

$\displaystyle \frac{r^2 + 1}{2r^3 + (4b^2-2)r} = \displaystyle \frac{r(r^2+1)}{2r^4+(4b^2)-r^2}$

$= \displaystyle \frac{r(r^2+1)}{r^4+r^4 + (4b^2)-r^2}$

$= \displaystyle \frac{r(r^2+1)}{r^4-1}$

$= \displaystyle \frac{r(r^2+1)}{(r^2+1)(r^2-1)}$

$= \displaystyle \frac{r}{r^2-1}$.

Therefore, to evaluate the contour integral, I simply the sum of the residues within the contour and multiply the sum by $2\pi i$:

$Q = \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1} = 2\pi i \left[\displaystyle \frac{r_1}{r_1^2-1} + \displaystyle \frac{r_2}{r_2^2-1} \right]$.

So, to complete the evaluation of $Q$, I need to simplify the right-hand side. I’ll complete this in tomorrow’s post.

# How I Impressed My Wife: Part 6d

This series was inspired by a question that my wife asked me: calculate

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Originally, I multiplied the top and bottom of the integrand by $\tan^2 x$ and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.
Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals.

nvenient value of $a$ that I want without changing the value of $Q$. As shown in previous posts, substituting $a =0$ yields the following simplification:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

$= \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}$

$= 2\pi \displaystyle \left[ \frac{1-r_1^2}{-2r_1^3 + (4b^2-2) r_1} +\frac{1-r_2^2}{-2r_2^3 + (4b^2-2) r_2} \right]$

where I’ve assumed $|b| > 1$, the contour $C_R$ in the complex plane is shown below (graphic courtesy of Mathworld),

and the positive constants $r_1$ and $r_2$ are given by

$r_1 = \sqrt{2b^2 - 1 + 2|b| \sqrt{b^2 - 1}}$,

$r_2 = \sqrt{2b^2 - 1 - 2|b| \sqrt{b^2 - 1}}$.

Now we have the small matter of simplifying our expression for $Q$. Actually, this isn’t a small matter because Mathematica 10.1 is not able to simplify this expression much at all:

Fortunately, humans can still do some things that computers can’t. As observed yesterday, The numbers $r_1$ and $r_2$ are chosen so that $\pm ir_1$ and $\pm ir_2$ are the roots of the denominator $z^4 + (4 b^2 - 2) z^2 + 1$, so that

$r_1^2 + r_2^2 = 4b^2 - 2$,

$r_1 r_2 = 1$.

These relationships will be very handy for simplifying our expression for $Q$:

$Q = 2\pi \displaystyle \left[ \frac{1-r_1^2}{-2r_1^3 + (4b^2-2) r_1} +\frac{1-r_2^2}{-2r_2^3 + (4b^2-2) r_2} \right]$

$= 2\pi \left[ \displaystyle \frac{1-r_1^2}{r_1 (-2r_1^2 + 4b^2-2)} + \frac{1-r_2^2}{r_2(-2r_2^2 + 4b^2-2)} \right]$

$= 2\pi \left[ \displaystyle \frac{1-r_1^2}{r_1 (-2r_1^2 +r_1^2 + r_2^2)} + \frac{1-r_2^2}{r_2(-2r_2^2 + r_1^2 + r_2^2)} \right]$

$= 2\pi \left[ \displaystyle \frac{1-r_1^2}{r_1 (-r_1^2 +r_2^2)} + \frac{1-r_2^2}{r_2(r_1^2 -r_2^2)} \right]$

$= 2\pi \left[ \displaystyle \frac{r_1^2-1}{r_1 (r_1^2 -r_2^2)} + \frac{1-r_2^2}{r_2(r_1^2 -r_2^2)} \right]$

$= 2\pi \displaystyle \frac{(r_1^2-1)r_2 + r_1(1-r_2^2)}{r_1 r_2 (r_1^2 -r_2^2)}$

$= 2\pi \displaystyle \frac{r_1^2 r_2- r_2 + r_1- r_1 r_2^2)}{r_1 r_2 (r_1^2 -r_2^2)}$

$= 2\pi \displaystyle \frac{r_1 - r_2 + r_1 r_2 (r_1 - r_2)}{r_1 r_2 (r_1-r_2)(r_1 + r_2)}$

$= 2\pi \displaystyle \frac{(r_1 - r_2)(1 + r_1 r_2)}{r_1 r_2 (r_1-r_2)(r_1 + r_2)}$

$= 2\pi \displaystyle \frac{1 + r_1 r_2}{r_1 r_2 (r_1 + r_2)}$

$= 2\pi \displaystyle \frac{1 + 1}{1 \cdot (r_1 + r_2)}$

$= \displaystyle \frac{4\pi}{r_1 + r_2}$

To complete the calculation, I observe that

$(r_1 + r_2)^2 = r_1^2 + 2r_1 r_2 + r_2^2 = 4b^2 -2 + 2 = 4b^2$,

so that

$r_1 + r_2 = 2|b|$.

Therefore,

$Q = \displaystyle \frac{4\pi}{r_1 + r_2} = \displaystyle \frac{4\pi}{2|b|} = \displaystyle \frac{2\pi}{|b|}$.

In tomorrow’s post, I’ll present another way to simplify this nasty algebraic expression.

# How I Impressed My Wife: Part 6c

This series was inspired by a question that my wife asked me: calculate

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Originally, I multiplied the top and bottom of the integrand by $\tan^2 x$ and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.
Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals.

Since $Q$ is independent of $a$, I can substitute any convenient value of $a$ that I want without changing the value of $Q$. As shown in previous posts, substituting $a =0$ yields the following simplification:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

$= \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}$,

where $C_R$ is the contour in the complex plane shown below (graphic courtesy of Mathworld).

Amazingly, contour integrals can be simply computed by evaluating the residues at every pole located inside of the contour. (See Wikipedia and Mathworld for more details.) I handled the case of $|b| = 1$ in yesterday’s post. Today, I’ll begin the case of $|b| > 1$.

To find the poles of the integrand, I use the quadratic formula to set the denominator equal to zero:

$z^4 + (4 b^2 - 2) z^2 + 1 = 0$

$z^2 = \displaystyle \frac{2-4b^2 \pm \sqrt{(4b^2-2)^2 - 4}}{2}$

$z^2 = \displaystyle \frac{2-4b^2 \pm \sqrt{16b^4 - 16b^2 + 4 - 4}}{2}$

$z^2 = \displaystyle \frac{2-4b^2 \pm \sqrt{16b^4 - 16b^2}}{2}$

$z^2 = \displaystyle \frac{2-4b^2 \pm 4|b| \sqrt{b^2 - 1}}{2}$

$z^2 = 1-2b^2 \pm 2|b| \sqrt{b^2 - 1}$

As shown earlier in this series, the right-hand side is negative if $|b| > 1$. So, for the sake of simplicity, I’ll define

$r_1 = \sqrt{2b^2 - 1 + 2|b| \sqrt{b^2 - 1}}$,

$r_2 = \sqrt{2b^2 - 1 - 2|b| \sqrt{b^2 - 1}}$,

so that the four poles of the integrand are $ir_1$, $ir_2$, $-ir_1$, and $-ir_2$. Of these, only two ($ir_1$ and $ir_2$) lie within the contour for sufficiently large $R$, and so I’ll need to compute the residues for these two poles.

Before starting that task, I notice that

$z^4 + (4 b^2 - 2) z^2 + 1 = (z - ir_1)(z + ir_1)(z - ir_2)(z + ir_2)$,

or

$z^4 + (4b^2 - 2)z^2 + 1 = (z^2 + r_1^2)(z^2 + r_2^2)$,

or

$z^4 + (4b^2 -2)z^2 + 1 = z^4 + (r_1^2 + r_2^2) z^2 + r_1^2 r_2^2$.

Matching coefficients, I see that

$r_1^2 + r_2^2 = 4b^2 - 2$,

$r_1^2 r_2^2 = 1$.

These will become very handy later in the calculation.

The integrand has the form $\displaystyle g(z)/h(z)$, and each pole has order one. As shown earlier in this series, the residue at such pole is equal to

$\displaystyle \frac{g(r)}{h'(r)}$.

In this case, $g(z) = 2(1+z^2)$ and $h(z) = z^4 + (4b^2-2)z^2 + 1$ so that $h'(z) = 4z^3 + 2(4b^2-2)z$, and so the residue at $r_1$ and $r_2$ are given by

$\displaystyle \frac{2(1+[ir_1]^2)}{4 [ir_1]^3 + 2(4b^2-2) [ir_1]} = \displaystyle \frac{-i(1-r_1^2)}{-2r_1^3 + (4b^2-2) r_1}$

and

$\displaystyle \frac{2(1+[ir_2]^2)}{4 [ir_2]^3 + 2(4b^2-2) [ir_2]} = \displaystyle \frac{-i(1-r_2^2)}{-2r_2^3 + (4b^2-2) r_2}$

Finally, to evaluate the contour integral, I simply the sum of the residues within the contour and then multiply the sum by $2\pi i$:

$Q = \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1} = 2\pi i \left[\frac{-i(1-r_1^2)}{-2r_1^3 + (4b^2-2) r_1} +\frac{-i(1-r_2^2)}{-2r_2^3 + (4b^2-2) r_2} \right]$

$= 2\pi \displaystyle \left[ \frac{1-r_1^2}{-2r_1^3 + (4b^2-2) r_1} +\frac{1-r_2^2}{-2r_2^3 + (4b^2-2) r_2} \right]$

So, to complete the evaluation of $Q$, I’m left with the small matter of simplifying the right-hand side. I’ll tackle this in tomorrow’s post.