Posted by: holdenlee | June 3, 2011

## Number Theory and Pi

In this post we give a number theory proof of the following identity

$\frac{\pi}{4}=1-\frac13+\frac15-\frac17+\cdots.$

(Prerequisite: knowing how factoring works in $\mathbb Z[i]$.)

Step 1: Approximate the sum with a finite sum.

We claim that

$1-\frac13+\frac 15-\cdots=\lim_{N\to \infty} \frac 1N\left(\left\lfloor \frac N1\right\rfloor-\left\lfloor \frac N3\right\rfloor+\left\lfloor \frac N5\right\rfloor-\cdots\right).$

(Bear with me… this is the most technical part of the proof.) Indeed, we have

(Note we used the trivial bound 2 on each term in the first sum and telescoped the last sum.)

Step 2:

$\frac 1N\left(\left\lfloor \frac N1\right\rfloor-\left\lfloor \frac N3\right\rfloor+\left\lfloor \frac N5\right\rfloor-\cdots\right)=\sum_{n=1}^N (d_1(n)-d_3(n))$

where $d_k(n)$ denotes the number of divisors of $n$ that are congruent to $k$ modulo 4.

Indeed, the LHS equals

$\sum_{n\equiv 1\pmod 4} \left\lfloor \frac Nn\right\rfloor -\sum_{n\equiv 3\pmod 4}\left\lfloor \frac N{n}\right\rfloor.$

The term $\left\lfloor\frac Nn\right\rfloor$ counts the number of multiples of $n$ less than $N$, so $\sum_{n\equiv 1\pmod 4} \left\lfloor \frac Nn\right\rfloor$ counts the number of pairs $(n,n')$ where $n\equiv 1\pmod 4$ and $n'\le N$ is a multiple of $n$. Summing the number of such pairs over $n'$ instead we get $\sum_{n'\le N} d_1(n')$. Similarly, $\sum_{n\equiv 3\pmod 4}\left\lfloor \frac N{n}\right\rfloor=\sum_{n'\le N} d_3(n')$.

Step 3: $4(d_1(n)-d_3(n))$ is the number of integer solutions to $x^2+y^2=n$.

One proof of this uses Jacobi’s triple product identity (see http://www.jstor.org/pss/2323169). We give a proof using Gaussian integers.

Each solution to $x^2+y^2=n$ corresponds to a factoring $(x+yi)(x-yi)=n$ over the Gaussian integers $\mathbb Z[i]$. Thus the number of solutions is the number of $z$ such that $z\bar{z}=n$, or 4 times the number of nonassociated $z\in \mathbb Z[i]$ such that $z\bar{z}=n$. (Two Gaussian numbers are associated if they differ by a unit $\pm 1, \pm i$, so $x+yi, -y+xi, -x-yi, y-xi$ are considered the same.)

Now factor $n=2^ap_1^{b_1}\cdots p_k^{b_k}q_1^{c_1}\cdots q_m^{c_m}$ where $p_j$ and $q_j$ are primes congruent to 1, 3 modulo 4, respectively. From knowledge of factoring in $\mathbb{Z}[i]$ we know that

1. 2 ramifies in $\mathbb Z[i]$, that is, it is the product of two associated primes $1+i,1-i$.
2. The $p_j\equiv 1\pmod 4$ split, that is, $p_j=z_j\bar{z_j}$ where $z$ is prime in $\mathbb{Z}[i]$ and not associated to $\bar z$.
3. The $q_j\equiv 3\pmod 4$ remain prime.

Now if $z\bar z=n$ and a Gaussian prime divides $z$, then its conjugate must divide $\bar z$. Thus, since we have unique factorization in $\mathbb{Z}[i]$, each such $z$, up to multiplication by associates, corresponds to a way of splitting the prime factors of $n$ into complex conjugate pairs. We note the following:

1. The factors $q_j$ are their own conjugates, so $z$ and $\bar z$ must each get $q_j^{c_j/2}$. If one of the $c_j$ is odd there is no solution. So we suppose they are all even.
2. It doesn’t matter how the prime factors of $2^a$ are split since they are all associates.
3. There are $b_j+1$ ways to split the factors of $q_j^{b_j}$, since we can have either $z_j^{b_j}$, or $z_j^{b_j-1}\bar{z_j}$,… or $\bar z_j^{b_j}$ divide $z$. Thus there are $(b_1+1)\cdots (b_k+1)$ solutions to $z\bar z=n$ up to associates.

Now if $c_k$ is odd, then the number of factors that are congruent to 1 or 3 modulo 4 are the same: Indeed, for every factor $d$ having an even power of $q_k$, we can pair it up with the factor $dq_k$, and these two factors are different modulo 4.

If all the $c_k$ are even, then we claim $(b_1+1)\cdots (b_k+1)=d_1(n)-d_3(n)$. We induct on the number of 3-mod-4 factors $m$. For the case $m=0$, all odd factors of $N=2^ap_1^{b_1}\cdots p_m^{b_m}$ are 1 mod 4, and $N$ has $(b_1+1)\cdots (b_k+1)$ odd factors. For the induction step, note that each 1-mod-4 factor of $N$ is obtained by multiplying an even power of $q_m$ (there are $\frac{c_j}{2}+1$ choices since $c_j$ is even) by a 1-mod-4 factor of $\frac n{q_m}$, or multiplying an odd power of $q_m$ (there are $\frac{c_j}2$ choices) by a 3-mod-4 factor. Hence $d_1(n)=\left(\frac{c_j}{2}+1\right)d_1\left(\frac n{q_m^{c_m}}\right)+\frac{c_j}{2}d_3\left(\frac n{q_m^{c_m}}\right)$. We get a similar formula for $d_3(n)$. Thus

by the induction hypothesis.

Thus in either case, the number of solutions to $x^2+y^2=n$ equals the $4(d_1(n)-d_3(n))$.

Step 4: Counting lattice points in circles

From step 3, we now know

$\sum_{n=1}^N 4(d_1(n)-d_3(n))$

is the number of integer solutions to $1\le x^2+y^2\le N$, i.e. the number of lattice points in the circle centered at the origin with radius $\sqrt N$ excluding the origin. The area formula for the circle gives that there are around $\pi(\sqrt N)^2$ lattice points inside, so $\lim_{N\to \infty}\frac 1N \sum_{n=1}^N 4(d_1(n)-d_3(n))=\pi$, which together with steps 1-2, complete the proof.

Note to make the above geometric argument rigorous, for each lattice point $(x,y)$ in the circle, shade the square with opposite corners $(x,y)$ and $(x+1,y+1)$. Then every point in the circle with radius $\sqrt N-\sqrt 2$ is shaded (since the square that contains the point is entirely within the original circle) and no point outside the circle with radius $\sqrt N+\sqrt 2$ is shaded (since the square that contains the point is entirely outside the original circle). The area of the shaded region is $f(n)=1+\sum_{n=1}^N 4(d_1(n)-d_3(n))$. Hence $\pi (\sqrt N-\sqrt 2)^2-1 \le \sum_{n=1}^N 4(d_1(n)-d_3(n))\le \pi (\sqrt N+\sqrt 2)^2-1$ which gives the desired after dividing by $N$ and taking the limit.

(Note: The regular calculus proof can be found at http://en.wikipedia.org/wiki/Leibniz_formula_for_pi.)

## Responses

1. On a deeper level we can see this as a special case of the analytic class number formula. See lecture 3 of http://math.uchicago.edu/~chonoles/expository-notes/promys/promys2012-analyticclassnumberformulanotes.pdf and http://math.stackexchange.com/questions/292104/how-to-derive-the-class-number-formula

2. And step 3 is the identity zeta(s)L(s,(*/4))=zeta_{Q(i)}(s).

3. […] proved this identity in a post several years ago on Number Theory and Pi. It turns out that this is a special case of a more general identity called the analytic class […]

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