$$$$

Project

This project illustrates just how far we have come in a semester of real analysis: we can now prove the identity that made Euler famous:

$$\sum _{n\ge 1}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$

And, the identity that bears his name today: $$e^{i\pi }=-1$$

This project is a collaboration between you and the text: the full proof takes place over as sequence of smaller results, some of which are proven here, and some of which are left as exercises.

Trigonometry

Both of these identities involve $\pi$, a number intimately connected to trigonometry. So, one would (correctly) assume that the trigonometric functions will play a crucial role in their proofs. But how can we suitably define sine and cosine (and $\pi$ itself) in a way that is both fully rigorous and helpful for our purposes?

One such method is to define them via a functional equation - below we give the same functional equation presented in class as the definition.

Definition 1 (Angle Identities) A pair of two functions $(c,s)$ are trigonometric if they are a continuous & differentiable nonconstant solution to the angle identities $$s(x-y)=s(x)c(y)-c(x)s(y)$$ $$c(x-y)=c(x)c(y)+s(x)s(y)$$

Remark 1. In fact one can drop the hypothesis that they are differentiable, and actually prove this from the others. But such a proof takes us a bit astray from the main goal of the project, so we instead opt for what initially appears to be a stronger characterization.

Identities

A good warm-up to functional equations is using them to prove some identities! I’ll do the first one for you

Lemma 1 (Values at Zero) If $s,c$ are trigonometric, then we can calculate their values at $0$: $$s(0)=0c(0)=1$$

Proof. Setting $x=y$ in the first immediately gives the first claim $$s(0)=s(x-x)=s(x)c(x)-c(x)s(x)=0$$

Evaluating the second functional equation also at $x=y$ $$c(0)=c(x-x)=c(x)c(x)+s(x)s(x)=c(x{)}^2+s(x{)}^2$$

From this we can see that $c(0)\ne 0$, as if it were, we would have $c(x{)}^2+s(x{)}^2=0$: since both $c(x{)}^2$ and $s(x{)}^2$ are nonnegative this implies each are zero, and so we would have $c(x)=s(x)=0$ are constant, contradicting the definition. Now, plug in $0$ to what we’ve derived, and use that we know $s(0)=0$

$$c(0)=c(0{)}^2+s(0{)}^2=c(0{)}^2$$

Finally, since $c(0)$ is nonzero we may divide by it, which gives $c(0)=1$ as claimed.

An important corollary showed up during the proof here, when we observed that $c(0)=c(x{)}^2+s(x{)}^2$: now that we know $c(0)=1$, we see that $(c,s)$ satisfy the Pythagorean identity!

Corollary 1 (Pythagorean Identity) If $s,c$ are trigonometric, then for every $x\in \mathbb{R}$ $$s(x{)}^2+c(x{)}^2=1$$

Continuing this way, we can prove many other trigonometric identities: for instance, the double angle identity (which will be useful to us later)

Lemma 2 (Evenness and Oddness) If $s,$ are trigonometric, then $s$ is odd and $c$ is even:

$$s(-x)=-s(x)c(-x)=c(x)$$

Lemma 3 (Angle Sums) If $s,c$ are trigonometric, then for every $x\in \mathbb{R}$ $$s(x+y)=c(x)s(y)+s(x)c(y)$$ $$c(x+y)=c(x)c(y)-s(x)s(y)$$

Corollary 2 (Double Angles) If $s,c$ satisfy the angle sum identities, then for any $x\in \mathbb{R}$, $$s(2x)=2s(x)c(x)$$

Exercise 1 Prove Lemma Evenness and Oddness, Lemma Angle Sums and Corollary Double Angles.

Another useful identity that I will need at the very end (but you will not, for any of the exercises) is the ‘Half Angle Identity’ for $c(x)$:

Lemma 4 If $s,c$ are trigonometric functions, then $$c(x{)}^2=\frac{1+c(2x)}{2}$$

Proof. Using the angle sum identity we see $$c(2x)=c(x)c(x)-s(x)s(x)=c(x{)}^2-s(x{)}^2$$ Then applying the pythagorean identity $$\begin{array}{rl}c(2x)& =c(x{)}^2-s(x{)}^2\\ & =c(x{)}^2-(1-c(x{)}^2)\\ & =2c(x{)}^2-1\end{array}$$

Re-arranging yields the claimed identity.

Differentiability

We have assumed as part of the definition that the trigonometric functions $(s,c)$ are differentiable: here we calculate their derivatives and use them to provide some important limits and estimates.

Lemma 5 Show that the derivatives of the trigonometric functions are completely determined by their derivatives at zero:

$$s^{\prime }(x)=s^{\prime }(0)c(x)+c^{\prime }(0)s(x)$$ $$c^{\prime }(x)=c^{\prime }(0)c(x)-s^{\prime }(0)s(x)$$

Further, what we already know about $s$ and $c$ (together with the fact that they are nonconstant) implies a bit about their values at zero:

Lemma 6 If $(s,c)$ are trigonometric, then $c^{\prime }(0)=0$ and $s^{\prime }(0)\ne 0$.

Exercise 2 Prove Lemma lem-trig-derivatives-determined and Lemma lem-trig-derivatives-zero

As a rather direct corollary of this, we have complete formulas for the derivatives of trigonometric functions, at every point along the real line!

Corollary 3 (Derivatives of $s,c$) If $s,c$ are trigonometric, then for a fixed nonzero $\lambda \in \mathbb{R}$ $$s^{\prime }(x)=\lambda c(x)c^{\prime }(x)=-\lambda s(x)$$

Definition 2 The sine and cosine functions are the trigonometric pair where ${\sin }^{\prime }(0)=1$.

Thus, we see ${\sin }^{\prime }(x)=\cos (x)$ and ${\cos }^{\prime }(x)=-\sin (x)$. Recalling the limit definition of the derivative, this gives us the ability to compute one particularly useful limit:

Proposition 1 For any $x\in \mathbb{R}$ $$limn\sin \left(\frac{x}{n}\right)=x$$

Exercise 3 Prove Proposition prp-sinc-limit-zero

Periodicity

We’ve already learned a lot about the trigonometric functions $\sin$ and $\cos$. But you might notice the fundamental constant $\pi$ is nowhere to be found! How do we rigorously define this number in the context of trigonometry? You may recall from previous classes that the trig functions have periods $2\pi$. This provides a possible path: we may hope to define $\pi$ a half the period of the trigonometric functions!

But to do so, we first need to show the functions even have a period. Why must solutions to the angle identities be periodic?

Lemma 7 The cosine function has a root.

Proof. Since the cosine is not constant and $\cos (0)=1$, there must be some $t_0$ for which $\cos (t_0)\ne 0$. Since ${\cos }^2+{\sin }^2=1$ we see that $-1\le \cos \le 1$ so $\cos (t_0)<1$.

If $\cos (t_0)$ is negative, we are done by the intermediate value theorem - there is a zero between 0 and $t_0$. So, we may assume $0<\cos (t_0)<1$, and define the sequence

$$c_n=\cos (2^n{t}_0)$$

To show $\cos$ is eventually negative (and thus, has a root by the intermediate value theorem argument) it suffices to see that $c_n$ is eventually negative; and thus that $L=inf\{c_n\}$ is negative (note the infimum exists as the set $\{c_n\}$ is bounded below by $-1$).

First, notice that the half angle identity implies $2c_0^2-1=c_1$. For $x\in (0,1)$, we see $2x^2-x-1$ is negative: plugging in $c_0$ yields $2c_0^2-c_0-1<0$, or $c_1=2c_0^2-1<c_0$. Thus, $c_0$ is not the smallest term in our sequence, and we can truncate it without changing the infimum: $$\underset{n\ge 0}{inf}\{c_n\}=\underset{n\ge 0}{inf}\{c_{n+1}\}$$

Using again the half angle identity, $2c_n^2-1=c_{n+1}$, so

$$L=inf\{c_n\}=inf\{c_{n+1}\}=inf\{2c_n^2-1\}=2inf\{c_n^2\}-1$$

If our sequence were never negative, then $inf\{c_n\}=L\ge 0$ and $inf\{c_n^2\}=L^2$. Combining with the above, this implies $L=2L^2-1$ whose only positive solution is $L=1$ (which we know is not the infimum, as $c_0<1$). Thus, this is impossible, so it must be that $L<0$, and our sequence eventually reaches a negative term.

Applying the intermediate value theorem to on the interval between $c(t_0)>0$ and $c(2^n{t}_0)<0$ furnishes a zero.

This shows that cosine has a zero somewhere. Because it will be convenient below, we carry this reasoning a little farther and show that cosine actually has a first positive zero.

Lemma 8 There is a $z>0$ such that $\cos z=0$, but the cosine is positive on the interval $[0,z)$: that is, $z$ is the first zero of the cosine.

Proof. Let $x$ be a zero of the cosine function. Since the cosine is even we know $-x$ is also a zero: and, since $\cos (0)=1$ we know neither $x\ne 0$ so at least one of $\pm x$ is positive. Thus, the cosine has at least one positive real root.

Let $R=\{x>0\mid \cos (x)=0\}$ be the set of all positive roots of the cosine function. We prove this set has a minimum element, which is the first zero. Since $R$ is nonempty (our first observation) and bounded below by zero (by definition) completeness implies $r=infR$ exists. For every $n\in \mathbb{N}$, since $r+1/n$ is not an upper bound we may choose some $x_n\in \mathbb{R}$ with $r\le x_n\le r+1/n$. By the squeeze theorem $x_n\to r$, and by continuity of the cosine this implies $$lim\cos (x_n)=\cos (lim{x}_n)=\cos (r)$$

However each $x_n$ is a zero of cosine by definition! Thus this is the constant sequence $0,0,\dots ,$ which converges to $0$. All together this means $\cos (r)=0$, and so $r\in R$. But if the infimum is an element of the set then that set has a minimum element, so $r$ is the smallest positive zero of the cosine!

Remark 2. The argument above works generally for continuous functions: we did not use special properties of the cosine. Indeed, being familiar with the cosine function itself this argument might seem a little strange: the zeroes of cosine are evenly spaced out (at intervals of size $\pi \approx 3$) so what are the zeroes we are finding between $r$ and $r+1/n$? In fact - these are all just $r$, and the sequence constructed in the above argument is constant (but we don’t need to know that, for the argument to go through).

It turns out that simply knowing the existence of a single zero of the cosine function is enough to resolve everything.

Proposition 2 (Periodicity of $s,c$) The functions $\sin (x)$ and $\cos (x)$ are periodic, with the same period $P>0$.

Exercise 4 Prove Proposition Periodicity of s,c. Hint This period is four times the first zero of cosine.

Definition 3 $\pi$ is the half-period of sine and cosine. Equivalently, $\pi$ is the first positive zero of the sine function.

This brings us to the final major result we need about the trigonometric functions:

Proposition 3 For all $x\in \mathbb{R}$, the cosine is just a shifted version of the sine function., and that sine is symmetric about $\pi /2$: $$\cos x=\sin (x+\frac{\pi }{2})$$ $$\sin (\frac{\pi }{2}+x)=\sin (\frac{\pi }{2}-x)$$

Exercise 5 Prove Proposition prp-cosine-shifted-sine.

Finally, we will need one more fact about the sine function (which will be crucial to us extracting $\pi$ in the Basel problem), which relates its values before its first maximum to a secant line connecting the origin to that point:

Proposition 4 For all $x\in [0,\pi /2]$, $$\frac{2x}{\pi }<\sin (x)$$

Exercise 6 Prove Proposition prp-sine-below-line

Euler’s Identity

Starting from the functional equations, we have done a lot of work to understand the properties that trigonometric functions must have, but we have not yet proven that there are such functions! We are going to use the functional properties of trigonometry to solve the Basel problem, so to be fully rigorous we better make sure the tools we use actually exist!

There are many routes to doing so, but the path we follow here meanders through some particularly beautiful mathematics in its own right. We will – as a corollary of proving that trigonometry exists – discover the famous identity of Euler $e^{i\pi }=-1$.

First, we use what we learned about differentiation to produce candidate functions as infinite series

Definition 4 (The Series $C(x)$ and $S(x)$) Define the following two series $$C(x)=\sum _{n\ge 0}\frac{(-1{)}^n}{(2n)!}{x}^{2n}S(x)=\sum _{n\ge 0}\frac{(-1{)}^n}{(2n+1)!}{x}^{2n+1}$$

Exercise 7 Prove that $C(x)$ and $S(x)$ are absolutely convergent on the entire real line, and that they satisfy the differential equation required of sine and cosine: $$S^{\prime }(x)=C(x)C^{\prime }(x)=-S(x)$$

Now, we show these candidate functions actually satisfy the angle sum identites.

There are many possible arguments here. The one we’ll take uses some complex numbers, but does not require any complex analysis. The only fact needed is that $i$ is a number where $i^2=-1$, which allows complex multiplication to be defined by the distributive property:

$$(a+bi)(c+di)=ac+adi+bci+bd{i}^2=(ac-bd)+(ad+bc)i$$

We can use the fact that a complex number $a+bi$ has a real component ($a$) and an imaginary component $b$ to define convergence: a complex series converges if both its real and imaginary parts converge.

Definition 5 (The Function $\mathrm{CIS}(x)$) Define the function $\mathrm{CIS}(x)$ as $$\mathrm{CIS}(x)=C(x)+iS(x)$$

Using the series for $C(x)$ and $S(x)$ we can produce a series for $\mathrm{CIS}(x)$.

Proposition 5 $$\mathrm{CIS}(x)=\sum _{n\ge 0}\frac{1}{n!}(ix{)}^n$$

Proof. First, we recall the definition as a limit of finite sums, and work with the limit laws to get to a single partial sum: $$\begin{array}{rl}\mathrm{CIS}(x)& =\sum _{n\ge 0}\frac{(-1{)}^n}{(2n)!}{x}^{2n}+i\sum _{n\ge 0}\frac{(-1{)}^n}{(2n+1)!}{x}^{2n+1}\\ & =\underset{N}{lim}\sum _{0\le n\le N}\frac{(-1{)}^n}{(2n)!}{x}^{2n}+i\underset{N}{lim}\sum _{0\le n\le N}\frac{(-1{)}^n}{(2n+1)!}{x}^{2n+1}\\ & =\underset{N}{lim}(\sum _{0\le n\le N}\frac{(-1{)}^n}{(2n)!}{x}^{2n}+i\sum _{0\le n\le N}\frac{(-1{)}^n}{(2n+1)!}{x}^{2n+1})\\ & =\underset{N}{lim}\sum _{0\le n\le N}(\frac{(-1{)}^n}{(2n)!}{x}^{2n}+i\frac{(-1{)}^n}{(2n+1)}{x}^{2n+1})\end{array}$$

Where in the last line here we re-arranged the terms of this finite sum (between $0$ and $N$) Writing out the $N=3$ case for concreteness we see

$$(1+ix)+(\frac{-1}{2!}{x}^2+i\frac{-1}{3!}{x}^3)+(\frac{1}{4!}{x}^4+i\frac{1}{5!}{x}^5)+(\frac{-1}{6!}{x}^6+\frac{-1}{7!}{x}^7)$$

Using that $i^2=-1$ we see that we can re-write this summation as

$$(1+ix)+(\frac{1}{2!}(i{x}^2)+\frac{1}{3!}(ix{)}^3)+(\frac{1}{4!}(ix{)}^4+\frac{1}{5!}(ix{)}^5)+(\frac{1}{6!}(ix{)}^6+\frac{1}{7!}(ix{)}^7)$$

This has a nice enough pattern that we can re-package it into summation notation $\sum _{0\le n\le 7}\frac{1}{n!}(ix{)}^n$, or more generally

$$\sum _{0\le n\le 2N+1}\frac{1}{n!}(ix{)}^n$$

Because this is exactly equal to the partial sums defining $\mathrm{CIS}(x)$ (all we did was algebra to finite arithmetic!), taking the limit as $N\to \mathrm{\infty }$ gives the claim.

But this looks awfully familiar: we are acquainted with the power series $E(x)=\sum _{n\ge 0}{x}^n/n!$: this defines the exponential function!

Corollary 4 $$CIS(x)=e^{ix}$$

From here, it is easy work to show that the the functions $C$ and $S$ satisfy the angle identities: this is just the law of exponents, which was the defining property of any exponential $E(x+y)=E(x)E(y)$

Proposition 6 The functions $C$ and $S$ satisfy the angle identities: $$S(x-y)=S(x)C(y)-C(x)S(y)$$ $$C(x-y)=C(x)C(y)+S(x)S(y)$$

Thus, trigonometric functions exist! This is all we need to move on, and use trigonometry in our solution of the Basel problem.

But…let’s not rush so quickly. From our work with the functional equations, we know that $C(x)=\cos (x)$ and $S(x)=\sin (x)$ are periodic, with half period $\pi$, which is also the first zero of the sine function. Using this immediately yields something beautiful:

Corollary 5 $$e^{i\pi }+1=0$$

Exercise 8 Prove Proposition prp-cis-angle-identities, and Corollary cor-euler-identity

The Basel Problem

Having now essentially developed the theory of trigonometry from scratch, we now aim to put what we’ve learned to use, in a rigorous solution to the Basel problem. This identity is so surprising because on one side there are only the natural numbers (as reciprocals, squared) and on the other side is the circle constant $\pi$ (squared, over six)!

The outline of our approach is as follows:

• Find a trigonometric identity that involves the sum of squares of reciprocals (of something).
• Use this to get an expression that has both $\pi$ in it, and sums of reciprocals of stuff.
• Take a limit to get an infinite sum of reciprocals, and do some algebra turn this into the sum we want.

A Trigonometric Identity

Step one is to find a trigonometric identity that allows us to expand something as a sum of reciprocal squares. This partial fractions decomposition of $1/{\sin }^2$ is a good start:

Proposition 7 The following trigonometric identity holds for $1/{\sin }^2$, whenever $x$ is not a multiple of $\pi$. $$\begin{array}{rl}\frac{1}{{\sin }^{2}x}& =\frac{1}{4{\sin }^2\left(\frac{x}{2}\right){\cos }^2\left(\frac{x}{2}\right)}\\ & =\frac{1}{4}[\frac{1}{{\sin }^2\left(\frac{x}{2}\right)}+\frac{1}{{\cos }^2\left(\frac{x}{2}\right)}]\\ & =\frac{1}{4}[\frac{1}{{\sin }^2\left(\frac{x}{2}\right)}+\frac{1}{{\sin }^2\left(\frac{x+\pi }{2}\right)}]\end{array}$$

Exercise 9 Prove the equalities claimed in Proposition prp-useful-identity hold.

Let’s see what we can do with this: applying it twice recursively,

$$\begin{array}{rl}\frac{1}{{\sin }^{2}x}& =\frac{1}{4}[\frac{1}{{\sin }^2\left(\frac{x}{2}\right)}+\frac{1}{{\sin }^2\left(\frac{x+\pi }{2}\right)}]\\ & =\frac{1}{4}[\frac{1}{4}[\frac{1}{{\sin }^2\left(\frac{x/2}{2}\right)}+\frac{1}{{\sin }^2\left(\frac{x/2+\pi }{2}\right)}]+\frac{1}{4}[\frac{1}{{\sin }^2\left(\frac{(x+\pi )/2}{2}\right)}+\frac{1}{{\sin }^2\left(\frac{(x+\pi )/2+\pi }{2}\right)}]]\\ & =\frac{1}{16}[(\frac{1}{{\sin }^2\left(\frac{x}{4}\right)}+\frac{1}{{\sin }^2(\frac{x}{4}+\frac{\pi }{2})})+(\frac{1}{{\sin }^2(\frac{x}{4}+\frac{\pi }{4})}+\frac{1}{{\sin }^2(\frac{x}{4}+\frac{3\pi }{4})})]\\ & =\frac{1}{16}[\frac{1}{{\sin }^2\left(\frac{x}{4}\right)}+\frac{1}{{\sin }^2(\frac{x}{4}+\frac{\pi }{4})}+\frac{1}{{\sin }^2(\frac{x}{4}+\frac{2\pi }{4})}+\frac{1}{{\sin }^2(\frac{x}{4}+\frac{3\pi }{4})}]\end{array}$$

https://homepage.univie.ac.at/josef.hofbauer/02amm.pdf

Applying once more to each term of the sum (and skipping the algebraic simplifications) yields

$$\begin{array}{rl}\frac{1}{{\sin }^{2}x}=\frac{1}{64}& [\frac{1}{{\sin }^2\left(\frac{x}{8}\right)}+\frac{1}{{\sin }^2(\frac{x}{8}+\frac{\pi }{8})}+\frac{1}{{\sin }^2(\frac{x}{8}+\frac{2\pi }{8})}+\frac{1}{{\sin }^2(\frac{x}{8}+\frac{3\pi }{8})}+\\ & +\frac{1}{{\sin }^2(\frac{x}{8}+\frac{4\pi }{8})}+\frac{1}{{\sin }^2(\frac{x}{8}+\frac{5\pi }{8})}+\frac{1}{{\sin }^2(\frac{x}{8}+\frac{6\pi }{8})}+\frac{1}{{\sin }^2(\frac{x}{8}+\frac{7\pi }{8})}]\end{array}$$

Putting this all together and looking at the three rounds of expansion, we see that

$$\begin{array}{rl}\frac{1}{{\sin }^{2}x}& =\frac{1}{4}\sum _{k=0}^1\frac{1}{{\sin }^2(\frac{x}{2}+\frac{k\pi }{2})}\\ & =\frac{1}{4^2}\sum _{k=0}^3\frac{1}{{\sin }^2(\frac{x}{4}+\frac{k\pi }{4})}\\ & =\frac{1}{4^3}\sum _{k=0}^7\frac{1}{{\sin }^2(\frac{x}{8}+\frac{k\pi }{8})}\end{array}$$

Carrying this out inductively $n$ times straightforwardly yields

Proposition 8 For any $n\in \mathbb{N}$, the function $1/{\sin }^2(x)$ can be expressed as the following finite sum: $$\frac{1}{{\sin }^{2}x}=\frac{1}{4^n}\sum _{0\le k<2^n}\frac{1}{{\sin }^2\left(\frac{x+k\pi }{2^n}\right)}$$

While this trigonometric identity is interesting in its own right, we will only require a special case for our application: evaluating at $x=\pi /2$ we find

Corollary 6 $$1=\frac{1}{4^n}\sum _{0\le k<2^n}\frac{1}{{\sin }^2\left(\frac{{\mathfrak{o}}_k\pi }{2^{n+1}}\right)}$$ Where ${\mathfrak{o}}_k:=2k+1=1,3,5,7,\dots$ is the sequence of odd numbers.

This gives, for every $n$, a large finite sum (its $2^n$ terms long!) of reciprocals of squares. There are just two obstacles in our way:

• The sum is finite!
• The reciprocals are complicated trigonometric functions

We can solve both by (carefully) taking the limit as $n\to \mathrm{\infty }$.

Taking the Limit

Because for all $n$ this finite sum exactly equals 1 we know in the limit as $n\to \mathrm{\infty }$ it equals $1$ as well:

$$1=\underset{n\to \mathrm{\infty }}{lim}\sum _{0\le k<2^n}\frac{1}{4^n{\sin }^2\left(\frac{{\mathfrak{o}}_k\pi }{2^{n+1}}\right)}$$

Now we have the delicate issue of taking the limit as $n\to \mathrm{\infty }$. It is tempting to take the limit of the terms individually but this is not always justified: as the simple example below shows

Example 1 $$\begin{array}{rl}1& =\frac{1}{2}+\frac{1}{2}\\ & =\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\\ & =\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\end{array}$$

Taking the termwise limit and adding them up gives $$1=0+0+0+\cdots +0=0$$

What’s going on here is that we are implicitly exchanging two limits and we haven’t justified that such an exchange is possible: in the toy example above, one may define for each $n$ the series $$a_n(k)=\{\begin{array}{ll}1/2^n& 0\le k<2^n\\ 0& \mathrm{else}\end{array}$$

Then each of the rows above is the sum $1=\sum _{k\ge 0}{a}_n(k)$ for $n=2,3,4$. Since this is constant it is true that the limit is $1$, but it is not true that the limit of the sums is the sum of the limits, which is zero.

$$1=\underset{n}{lim}\sum _{k\ge 0}{a}_n(k)\ne \sum _{k\ge 0}lim{a}_n(k)=0$$

The reason this fails is that our sum does not satisfy the hypotheses of dominated convergence. Recall that requires that all of the $a_n(k)$ (for all $n$) must be less than some convergent series. Here we can see that $a_n(k)<\frac{1}{k}$ (and we can’t do better than this) but the harmonic series diverges! So the hypotheses of dominated convergence are violated, and switching limits leads to disaster.

The form of this example is exactly replicated in our question: if we define for each $n$ an infinite series with terms $a_n(k)$ as

$$a_n(k)=\{\begin{array}{ll}1/4^n{\sin }^2\left(\frac{{\mathfrak{o}}_k\pi }{2^{n+1}}\right)& 0\le k<2^n\\ 0& k\ge 2^n\end{array}$$

we see that $1=\underset{n}{lim}\sum _{k\ge 0}{a}_n(k)$ (Where again, ${\mathfrak{o}}_k=(2k-1)$ is just shorthand for the sequence of odd numbers) but to make progress computing this limit, we need to be able to exchange it with the order of summation, so we need a means of applying dominated convergence.

Unfortunately, this is tricker than one might hope: our series fails the hypotheses of dominated convergence as written!

Remark 3. Try to see this yourself, by drawing a graph of $a_n(k)$ as a function of $k$, for various $n$. For any given $n$, can you find where the maximal value of $a_n(k)$ occurs (at which $k$), and what it’s value is?

Why does this prevent you from building a dominating sequence?

Luckily, there is a way to proceed: for a fixed $n$, the coefficients we are summing up involve the sine function evaluated at odd multiples of $\pi /2^{n+1}$. This list of numbers is symmetric about $\pi /2$, and since $\sin (x)$ is symmetric about $\pi /2$, so are our coefficients.

Completing the first half of the sum and doubling the result gives the same value: illustrating with the $n=2$ iteration,

$$\begin{array}{rl}1& =\frac{1}{16}[\frac{1}{{\sin }^2\left(\frac{\pi }{8}\right)}+\frac{1}{{\sin }^2\left(\frac{3\pi }{8}\right)}+\frac{1}{{\sin }^2\left(\frac{5\pi }{8}\right)}+\frac{1}{{\sin }^2\left(\frac{7\pi }{8}\right)}]\\ & =\frac{1}{16}[(\frac{1}{{\sin }^2\left(\frac{\pi }{8}\right)}+\frac{1}{{\sin }^2\left(\frac{7\pi }{8}\right)})+(\frac{1}{{\sin }^2\left(\frac{3\pi }{8}\right)}+\frac{1}{{\sin }^2\left(\frac{5\pi }{8}\right)})]\\ & =\frac{1}{16}[\frac{2}{{\sin }^2\left(\frac{\pi }{8}\right)}+\frac{2}{{\sin }^2\left(\frac{3\pi }{8}\right)}]\end{array}$$

Carrying out this finite re-arrangement for the $n^{th}$ iteration yields

Corollary 7 $$1=\frac{1}{4^n}\sum _{0\le k<2^n}\frac{1}{{\sin }^2\left(\frac{{\mathfrak{o}}_k\pi }{2^{n+1}}\right)}=\frac{1}{4^n}\sum _{0\le k<\frac{1}{2}{2}^n}\frac{2}{{\sin }^2\left(\frac{{\mathfrak{o}}_k\pi }{2^{n+1}}\right)}$$ Where ${\mathfrak{o}}_k=1,3,5,\dots$ is the sequence of odd numbers.

Since our new upper summation limit is below $\frac{1}{2}{2}^n=2^{n-1}$, the values we are plugging into the sine function are all less than $\pi /2$, and so Proposition prp-sine-below-line applies, allowing us to bound $\sin$ below, and hence $1/\sin$ above.

Proposition 9 The series $$\sum _{0\le k<2^{n-1}}\frac{1}{4^n{\sin }^2\left(\frac{{\mathfrak{o}}_k\pi }{2^{n+1}}\right)}$$ Satisfies the hypotheses of dominated convergence: precisely, define $$a_n(k)=\{\begin{array}{ll}1/4^n{\sin }^2\left(\frac{{\mathfrak{o}}_k\pi }{2^{n+1}}\right)& 0\le k<2^{n-1}\\ 0& k\ge 2^n\end{array}$$

• Independently of $n$ the $k^{th}$ term $a_n(k)$ is bounded above by $2/{\mathfrak{o}}_k^2$; that is $2/(2k-1{)}^2$
• The sum of these bounds converges

Exercise 10 Prove Proposition prp-bounding-coefs

This justifies the exchange of limits, which will be crucial to our the main step.

Theorem 1 $$\underset{n}{lim}\sum _{0\le k<2^{n-1}}\frac{2}{4^n{\sin }^2\left(\frac{{\mathfrak{o}}_k\pi }{2^{n+1}}\right)}=\sum _{k\ge 0}\underset{n}{lim}\frac{2}{4^n{\sin }^2\left(\frac{{\mathfrak{o}}_k\pi }{2^{n+1}}\right)}$$

The Termwise Limit

Taking the limit term by term allows us to start by finding the limit of the denominator:

Proposition 10 $$\underset{n}{lim}{4}^n{\sin }^2\left(\frac{{\mathfrak{o}}_k\pi }{2^{n+1}}\right)={\left(\frac{{\mathfrak{o}}_k\pi }{2}\right)}^2$$

Exercise 11 Prove this

Putting this back into our original trigonometric identity gives

$$1=\sum _{k\ge 0}\frac{2}{{\left(\frac{{\mathfrak{o}}_k\pi }{2}\right)}^2}$$

This is the key to a rigorous derivation of Euler’s incredible identity.

Exercise 12 Re-arrange the sum above to show $$\frac{{\pi }^2}{8}=\sum _{k\ge 0}\frac{1}{{\mathfrak{o}}_k^2}=1+\frac{1}{3^2}+{\frac{1}{5}}^2+\cdots$$ And show that this implies $$\sum _{n\ge 1}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$

Exercise 13 Before taking the termwise limit, we had to be careful and replace our sum with different sum (having the same value) so that we could apply Dominated Convergence. Show that this is necessary: take the termwise limit of the original sum and show you get the wrong answer for $\sum _{k\ge 0}\frac{1}{{\mathfrak{o}}_k^2}$.

Hints

Below are hints to the various exercises in this document: I encourage you to try them without the hints! But feel free to get inspiration here, and to ask for further hints in office hours if a particular problem has you stuck.

Exercise 1: Look at $s(0-x)$ or $c(0-x)$ now that we know the values of $s$ and $c$ at zero…

Exercise 2: For the first part: use the definition of the derivative, and the angle sum identites. For the follow-up: we can know $c^{\prime }(0)=0$ because we can prove $0$ is the location of a max and apply theorems. If both $s^{\prime }$ and $c^{\prime }$ were zero at $x=0$, can you show $s(x)$ and $c(x)$ must be constant?

Exercise 3: The definition of the derivative is a limit, and limits of functions are defined in terms of sequences…

Exercise 4: If $z$ is the first zero of cosine, show that $c(4z)=1$ and $s(4z)=0$, and that this implies $c(x+4z)=c(x)$, $s(x+4z)=s(x)$.

Exercise 5: Use the definition of $\pi$ and the angle sum/difference identities.

Exercise 6: Use a theorem relating the second derivative to convexity. A convex function always has its secant line above the graph. What about the negative of a convex function?

Exercise 7: Show convergence with a test for power series, then use a theorem on differentiating power series within their radius of convergence.

Exercise 8: Use the definition of $\mathrm{CIS}(x)$, the properties of complex multiplication, and the definition of $\pi$.

Exercise 9: Use the trigonometric identities derived from the original functional equation.

Exercise 10: Use Exercise 6 to get the bound, and comparison to prove convergence.

Exercise 11: Exercise 3 tells us about $limn\sin (x/n)$. Can you rewrite what you have here into something like $lim{2}^{n+1}\sin (x/2^{n+1})$? Then use facts about subsequences and limits.

Exercise 12: The first part is just algebra. For the second, can you break the sum of $1/n^2$ into the sum of even and odd terms? Can you do something about the even terms, to get an equation relating the sum of all reciprocals to just the sum of the odd ones?

Exercise 13: Just repeat the calculation you worked through above, but for the other sum.