Geometric intuition for $\pi /4 = 1 - 1/3 + 1/5 - \cdots$?

Question

Following reading this great post : Interesting and unexpected applications of $\pi$, Vadim's answer reminded me of something an analysis professor had told me when I was an undergrad - that no one had ever given him a satisfactory intuitive explaination for why this series definition should be true (the implied assumption being that it is so simple, there should be some way to look at this to make it intuitive).

Now it follows simply from putting $x=1$ in the series expansion $$ \tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots $$

but I don't see anything geometrically intuitive about this formula either!

Answer 1

I doubt that there is a simple geometric proof that can be visualized directly, like some circle area computation using some inclusion-exclusion procedure.

Here's a nice article.

Edit : see also here.

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We could split the proof in three steps:

1. The area of a quarter of unit circle equals the area below $f(x)=\frac{1}{1+x^2}$ in the interval $[0,1]$

2. $\frac{1}{1+x^2}=1 - {x^2} + {x^4} - \cdots$ for $|x|<1$

3. $\int_0^1 x^{2n} dx= \frac{1}{2n+1}$

Step 1 admits several visual-geometric proofs (below I insert a diagram I made for the linked question, that shows the area equivalence for the full graph: $\int_{-\infty}^{\infty}1/(1+x^2) dx =\pi$; just trim it to the $[0,1]$ range).

But steps 2 and 3, though elementary from a analysis point of view, do not seem easy to visualize.

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Explanation added by comment request: the graph (edited) intends to show that the two areas in yellow are (asymptotically) equal.

First, we have a circular sector with radius 1 and arc $s$, which is approximated by a triangle with the same height and base ; hence its area is $s/2$.

The other area is below the function $\frac{1}{2(1+x^2)}$, so its area is $ \overline{CE} \times y = a \frac{1}{2(1+x^2)}$. We then compute $a$:

Because $B0C$ is rectangular $\ell=\overline{BC}=\sqrt{1+\overline{0C}^2}=\sqrt{1+x^2}$

Because triangles $CDB$ and $C'D'B$ are similar, $\ell=\frac{\overline{BC}}{\overline{B C^{'}}}=\frac{\overline{CD}}{\overline{C^{'}D^{'}}}=\frac{t}{s}$

Because $CED$ and $BOC$ are similar $\frac{a}{t}=\frac{\overline{CE}}{\overline{CD}}=\frac{\overline{BC}}{\overline{0B}}=\ell$

Hence $a= s \ell^2 = s (1+x^2)$ and the area is also $s/2$.

Applying this to the range $[0,1]$ we deduce that the area of an octave of a unit circle equals $\int_0^1 \frac{1}{2(1+x^2)}dx$

Answer 2

By elementary means:

In the figure below, we have drawn a quarter circle inside a unit square and divided the top side in $n$ equal parts. Then we have chosen the height of the blue rectangles in such a way that the area of a rectangle equals the area of the corresponding sector (the aperture angles go decreasing).

If you look at the enhanced triangle, its long sides have length $\sqrt{1+x^2}$, by Pythagoras, so that its short side is $\sqrt{1+x^2}$ times longer than the arc on the circle ($x$ horizontal in $[0,1]$). Then by similarity of triangles, the ratio of the width of a rectangle to this short side is also $\sqrt{1+x^2}$, and we conclude that the equation of the green curve is

$$y(x)=\frac1{2(1+x^2)}.$$

Furthermore, the area under the curve is the same as the area of the eighth of a circle, i.e. $\dfrac\pi8$.

Now the (doubled) area is obtained as the sum of the areas of the rectangles,

$$\frac\pi4=2\sum_{k=1}^n\frac1ny\left(\frac kn\right)=\frac1n\sum_{k=1}^n\frac1{1+\dfrac{k^2}{n^2}}.$$

To evaluate it $^{(1)}$, we use the identity

$$\frac1{1+\dfrac{k^2}{n^2}}=1-\frac{k^2}{n^2}+\frac{k^4}{n^4}-\frac{k^6}{n^5}+\cdots$$

and by the Faulhaber formula $^{(2)}$, we know that the sum of the $n$ first $p^{th}$ powers is $\dfrac{n^{p+1}}{p+1}$.

This gives

$$\frac\pi4=\frac1n\left(n-\frac{n^3}{3n^2}+\frac{n^5}{5n^4}-\frac{n^7}{7n^6}+\cdots\right)$$ which is the claimed formula.

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$(1)$ This is "justified" by

$$(1+x)(1-x+x^2-x^3+x^4-\cdots)=1+x-x-x^2+x^2+x^3-x^3-x^4+x^4+x^5\cdots=1\pm x^\infty$$ where the last term vanishes when $x<1$.

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$(2)$ This is "justified", using the binomial theorem, by

$$n^p=\sum_{k=1}^n k^p-\sum_{k=1}^{n-1} k^p\approx\frac{n^{p+1}}{p+1}-\frac{(n-1)^{p+1}}{p+1}=\frac{n^{p+1}-(n^{p+1}-(p+1)n^p+\binom{p+1}2n^{p-1}-\binom{p+1}3n^{p-2}+\cdots)}{p+1}=n^p+\cdots.$$

The neglected terms are of lower degree in $n$ and can be ignored for large $n$.

Answer 3

There is a proof of this formula using the geometry and arithmetic of the Gaussian integer ring $R=\mathbf Z[i]$. It turns out that this formula can be boiled down to the claim that the unit circle has area $\pi$!

For $\Re s>1$, the Dedekind zeta function of $R$ is defined as the convergent series

$$\zeta_R(s) = \sum_{I \subseteq R} \|I\|^{-s}$$

where the sum is taken over the nonzero ideals of $R$. Since $R$ is a PID, and since $R^\times = \{\pm 1, \pm i\}$, we can also write this as

$$\zeta_R(s) = \frac{1}{4}\sum_{(m,n)}(m^2+n^2)^{-s}$$

the sum being taken over pairs $(m,n) \in \mathbf Z^2$ with $m^2+n^2 \neq 0$.

The function $\zeta_R(s)$ has a simple pole at $s=1$, and we can calculate its residue there in two different ways: using arithmetic, and using geometry.

1. Using geometry: According to the Wiener-Ikehara theorem applied to the Dirichlet series $\zeta_R(s)$, we can calculate the residue as $$\mathrm{Res}_{s=1}\zeta_R(s) = \lim_{N\to \infty} \frac{1}{4N}\sum_{m^2+n^2\leq N} 1$$ The sum simply counts the number of lattice points inside the circle of radius $\sqrt{N}$, which is approximately the area of this circle, namely $\pi N$ (up to an error which is killed when dividing by $N$ and taking the limit).
2. Using arithmetic: By Quadratic Reciprocity, the function $\zeta_R(s)$ factors as $$\zeta_R(s) = \zeta(s) L(\chi, s)$$ where $\zeta(s)$ is the Riemann zeta function and $\chi : (\mathbf Z/4\mathbf Z)^\times \to \{\pm 1\}$ is the primitive quadratic Dirichlet character of conductor $4$, $\chi(n) = (-1)^{(n-1)/2}$, which encodes the splitting of primes in $R$. We have $$L(\chi, 1) = 1 - \frac13 + \frac15 - \dots,$$ and the residue of $\zeta(s)$ at $s=1$ is $1$; therefore $\mathrm{Res}_{s=1}\zeta_R(s) = 1 - \frac13 + \frac15 - \dots$.

Thus, $$\mathrm{Res}_{s=1}\zeta_R(s) = \frac\pi 4 = 1 - \frac13 + \frac15 - \dots.$$

Answer 4

I would say that

$\int_0^1 \frac{dx}{1+x^2}=\tan^{-1}(1)-\tan^{-1}(0)=\frac{\pi}{4}$

so this is the area under the curve $\frac{1}{1+x^2},\,\, x\,\in [ 0,1]$.

$\int_0^1 \frac{dx}{1+x^2}=\int_0^1(1-x^2+x^4\cdots)dx=[x-\frac{x^3}{3}+\frac{x^5}{5}\cdots]_0^1=1-\frac{1}{3}+\frac{1}{5}\cdots$

Answer 5

Let $\mathbf T = S^1 \times S^1$ be the torus with coordinates $(\theta, \psi)$, with the Riemannian metric $d\theta \otimes d\theta+ d\psi \otimes d\psi$. The Laplace operator on $\mathbf T$ is $\Delta = \frac{­\partial^2}{\partial \theta^2}+\frac{­\partial^2}{\partial \psi^2}$. Eigenfunctions of the Laplacian are the functions $e^{2\pi i (n\theta+m\psi)}$ for $(m,n) \in \mathbf Z^2$, with eigenvalues $(2\pi i)^2(m^2+n^2)$.

The spectral zeta function of $\mathbf T$ is therefore

$$\zeta_{\mathbf T}(s) = \mathrm{tr}(\Delta^{-s}) = (2\pi)^{-2s}\sum_{(m,n)} (m^2+n^2)^{-s}.$$

By a theorem of Minakshisundaram-Pleijel, it has a simple pole at $s=\frac{\dim \mathbf T}{2}=1$ with residue given by

$$\frac{1}{(2\sqrt \pi)^{\dim \mathbf T}\Gamma(\dim \mathbf T/2)} = \frac{1}{(2\sqrt \pi)^{2}\Gamma(1)} = 1/4\pi.$$

On the other hand, using part 2. of my other answer, $$\mathrm{Res}_{s=1}{\zeta_{\mathbf T}(s)} = 4(2\pi)^{-2}\left(1-\frac13+\frac15 - \dots\right).$$ Therefore,

$$1-\frac13+\frac15 - \dots = \frac{(2\pi)^2}{4 \times 4\pi} = \frac{\pi}{4}.$$

(It would be nice to make this self-contained, and give an explanation for the factorization $\zeta_{\mathbf T}(s) = 4(2\pi)^{-2s}\zeta(s)L(\chi,s)$ using only the interpretation of $\zeta_{\mathbf T}$ as a spectral zeta function...)