Dirichlet’s Prime Number Theorem – Part 2

The next step in this series of blog posts, which started with the first part, ist to show why the Riemann Zeta function does not diverge at $s=2$ and why its value at this point is $\frac{\pi^2}{6}$ .

I recently found 2 ways to show that the infinite sum of all reciprocals of perfect squares is $\frac{\pi^2}{6}$ :

The first way is via constructing a representation of the $\sin$ function from its roots and then comparing it to its well-known Taylor series. I first found this proof in the mathologer video https://www.youtube.com/watch?v=yPl64xi_ZZA. [Sullivan2013] elaborates in more detail on that solution.
The second way is via the total amount of apparent brightness received from an array of light sources arranged equidistantly around an infinitely large circle. ( https://www.youtube.com/watch?v=d-o3eB9sfls)

I will only discuss the second proof in this post whose proof is based on the paper at [Wästlund2010]. The proof presented here is identical to that one, I have even copied the figures from that paper. However, I have tried to put in some more details and make it easier to read for someone who encounters this for the first time.

The star light proof

This proof is based on the idea that the perceived brightness of a star is proportional to the inverse square of the distance between an observer and the star. Based on this observation it is possible to construct a circular (that’s where $\pi$ comes in) arrangement of equidistant stars that has a known perceived brightness for an observer on the circle. In particular, the circle can be scaled up arbitrarily (with the amount of stars scaled accordingly) without changing the total perceived brightness. In the limiting case of an infinitely large circle we have an infinite amounf of stars looking in both directions along the circle which effectively then gives us our infinite sum of inverse squares while still summing up to the same original perceived brightness.

In order to break that proof down, we first need a lemma. This lemma is called the Inverse Pythagorean Theorem. In any right-angled triangle like the following we have that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{h^2}$ . So instead of the squared legs adding up to the squared hypotenuse (as in the Pythagorean Theorem), their inverses add up to the inverse of the squared height!

Figure 1. Right triangle. Figure taken from [Wästlund2010]

There’s an easily understandable proof of this that I’ve taken from [Wästlund2010]. It involves only the Pythagorean Theorem and the formula for the are of a triangle. The latter gives us the relation $\frac{ab}{2} = \frac{ch}{2} \Rightarrow ab = ch$ . This is due to the fact that we assumed a right angle at C. From the Pythagorean Theorem we obviously know that $a^2 + b^2 = c^2$ . Now, we can finish the proof by just manipulating the following sum and by applying those 2 facts: $\frac{1}{a^2} + \frac{1}{b^2} = \frac{a^2 + b^2}{a^2b^2} = \frac{c^2}{c^2h^2} = \frac{1}{h^2}$ . And we’re done.

There’s also another proof of this lemma which is explained in the 3blue1brown video mentioned above but I did not take the time to fully understand it so I’ll leave it out for now.

The case N=1

Back to the original problem. As mentioned before, we want to model the perceived brightness of stars on a circle at a point P on that same circle. See Figure 2. We define the relative placement of the closest star and P as the distance along the circle and call it x. The circumference of this circle we set to 1 and thus obtain a radius of $r=\frac{1}{2\pi}$ . The angle between the two radii connecting the center to both points is therefore $\alpha = 2\pi x$ . To show that the perceived brightness of the red star at point P is $f_1(x) = \left(\frac{\pi}{\sin\pi x}\right)^2$ we need to know the straight distance between both points, not the distance x along the circle’s circumference. This distance, call it d, can be evaluated by the following equation, together with the right-hand side of Figure 2: $\sin\frac{\alpha}{2} = \frac{d}{2} \ \frac{1}{2\pi} = d\pi$ . This gives us $d=\frac{\sin\pi x}{\pi}$ , the squared inverse of which is exatcly $f_1(x)$ .

The case N=2

Next, we show that we can scale the circle’s diameter up by a factor of 2 and double the amount of light sources on the circle while still keeping the total perceived brightness the same. In other words, $f_2(x)=f_1(x)$ . To understand that part, it is helpful to look at Figure 3 and observe that we have the original circle in their from our previous case and the larger circle. We have constructed the points $B_1, B_2$ by putting a line through R and the new circle’s center Q and marked the points where this line crosses the new, greater circle. Next, observe that the triangle $B_2PB_1$ is a right triangle (Thales’ theorem). Therefore, we can apply the Inverse Pythagorean Theorem and see that the sum of the inverse squared distances to the blue points (from P) is the same as the inverse squared distance to the red point. This shows us that the total perceived brightness did not change while doubling the number of light sources in such a way.

Figure 3: Circles of circumference 1 and 2 with 1 and 2 stars, respectively. Figure taken from [Wästlund2010]

Next, we still need to show that the relative displacement of P with respect to R on the inner circle’s circumference is the same as that with respect to $B_2$ on the outer circle. Because otherwise, we would have proven $f_2(x') = f_1(x)$ . For that part of the proof, we need the Central Angle Theorem which states that the central angle, subtended by two points on a circle is twice the inscribed andle subtended by those points. In our case of Figure 3 that implies the following relation of angles: $2 \angle PQR = \angle POR$ . And since the length of the radius QP is twice that of the inner circle’s radius OP, the arc lengths from P to R and from P to $B_2$ must be equal in length.

Iterating

Now we show that in general $f_{2N}(x) = f_N(x)$ . We basically use the same line of reasoning as in the previous section but add the fact that the blue points on the outer circle (with double radius) are still equally spaced along that circle. See Figure 4.

We’ll focus on the smaller circle with red dots. Remember that we have so far an original circle having N equally spaced points on it and a total circumference of N. So from the point of view of the circle’s center, each “piece of the pie” has an equal angle of $\frac{2\pi}{N}$ . So from the point of view of the black dot in Figure 4 (the larger circle’s center) the angles between to adjacent arcs to red dots a this black dot are also euqal and evaluate to $\frac{\pi}{N}$ by the central angle theorem. Therefore all the angles between adjacent diameters between blue dots on the outer circle must also have euqal angles.

Figure 4: Circles of circumference 5 and 10 with 5 and 10 stars, respectively. Figure taken from [Wästlund2010]

Wrapping up

So far, we have only shown that the total amount of light emanating from equally spaced light sources on a circle is equal to $\left(\frac{\pi}{\sin\pi x}\right)^2$ . Now, we want to show that this is equivalent to the equation $\sum_{n = -\infty}^{\infty} \frac{1}{(n-x)^2} = \left(\frac{\pi}{\sin\pi x}\right)^2$ . And from there we can deduce Euler’s identity. Let me show you how.

If we plug into the previous equation the value $x=1/2$ , then we get $\sum_{n = -\infty}^{\infty} \frac{1}{(n-1/2)^2} = \pi^2$ . Dividing that by 4 gives $\sum_{n = -\infty}^{\infty} \frac{1}{(2n-1)^2} = \frac{\pi^2}{4}$ . And this is almost what we want because now we are summing the inverse squares of all odd integers. If we restrict this to the positive half of the summation (the positive odd integers) we get $\frac{\pi^2}{8}$ . Now we only need to observe that this is actually already three fourth of the sum of all inverse integer squares and therefore $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{8}\frac{4}{3} = \frac{\pi^2}{6}$ . To see the last step, just multiply the sum on the left-hand side by one fourth and observe that these are just the missing terms in the sum of the odd inverse squares: $\frac{1}{4} \left(1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \ldots\right) = \frac{1}{4} + \frac{1}{16} + \frac{1}{36} + \frac{1}{64} + \ldots$ .

Now we have almost established the connection between the light sources on a circle and Euler’s identity. The missing link is to show that the light sources actually also make up for the left-hand side of the first equation in this paragraph. The trick here is to play the game of the previous sections ad infinitum and construct ever larger circles but deleting some of them in each step so that we end up with a line of light sources tangential to all circles through P which has infinitely many light sources in both directions. In detail, this goes as follows.

Imagine that in each step we first delete all the light sources on the upper half of the current circle before applying the “doubling procedure”. In Figure 5, we have first deleted the most distant (from P) red light sources and then projected the blue light sources onto the larger circle via diameters through the center and each red source. This leaves us then with the closest and farthest quarter of the blue dots. The error in the total perceived brightness after deleting the red sources is at most $N\frac{1}{(N/\pi)^2} = \frac{\pi^2}{N}$ because all light sources are at least one radius’ length away (in fact even more than $\sqrt{2} r$ but a precise bound is not important here). Now for the blue sources on the outer circle, the deletion process gives again an error of at most $N\frac{1}{(2N/\pi)^2} = \frac{\pi^2}{4N}$ and in general for every step k, this error will be bounded by $\frac{\pi^2}{4^k N}$ .

Figure 5: Scaling up the circle. Figure taken from [Wästlund2010]

As $k\rightarrow\infty$ , the set of N closest stars will approach the points at coordinates $n-x$ measured from P along a straight line tangent to all the circles at P. Therefore, in the limit $k\rightarrow\infty$ , the total radiation at P from the N remaining stars will approach $\sum_{\vert n-x\vert<N/2}\frac{1}{(n-x)^2}$ . The total error from all the deletions of stars will be at most $\frac{\pi^2}{N} + \frac{\pi^2}{4N}+\frac{\pi^2}{16N}+\dots = \frac{\pi^2}{N}\frac{1}{1-1/4} = \frac{4\pi^2/3}{N}$ (Geometric Series).

Since we already know that $f_N(x) =f_{2N}(x)$ , we conclude that for every N, $f_N(x) = \sum_{\vert n-x\vert < N/2}\frac{1}{(n-x)^2}+\frac{\theta}{N}$ , where $\theta$ may depend on N but satisfies $0\leq \theta \leq \frac{4\pi^2}{3}$ . In the final twist we apply the previous equation with N replaced by $2^kN$ , although we aready know that $f_{2^kN}(x) =f_N(x)$ . We conclude that for every N, $f_N(x) = \lim_{k\rightarrow\infty} f_{2^kN}(x) = \lim_{k\rightarrow\infty}\sum_{\vert n-x\vert < 2^kN/2}\frac{1}{(n-x)^2}+\frac{\theta}{2^kN}=\sum_{n =-\infty}^{\infty}=\frac{1}{(n-x)^2}$ . And since we have constructed our circles such that $f_1(x) = f_N(x)$ , we have finally proven both sides of the equation and with the above-mentioned transformation arrive at Euler’s identity.