text: 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 […]
  
  4. I am now not sure the place you’re getting your information, however great topic.
  I must spend some time learning much more or understanding more.
  Thank you for fantastic info I was on the lookout for this information for
  my mission.
  
  5. pls i nid more explaination on it
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distilabel_metadata.content: 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 […]
  
  4. I am now not sure the place you’re getting your information, however great topic.
  I must spend some time learning much more or understanding more.
  Thank you for fantastic info I was on the lookout for this information for
  my mission.
  
  5. pls i nid more explaination on it
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a detailed mathematical proof that demonstrates the identity \(\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\). The proof is structured into several steps, each building on the previous one to arrive at the final result. The proof uses concepts from number theory, particularly the properties of Gaussian integers (\(\mathbb{Z}[i]\)) and the distribution of divisors modulo 4.
  
  ### Summary of the Proof Steps:
  1. **Approximate the Sum with a Finite Sum:**
     - The infinite series is approximated by a finite sum involving floor functions.
     - The approximation is justified by bounding the error terms.
  
  2. **Relate the Finite Sum to Divisors:**
     - The finite sum is expressed in terms of the number of divisors of \(n\) that are congruent to 1 or 3 modulo 4.
     - This step uses the properties of floor functions and the counting of multiples.
  
  3. **Count Solutions to \(x^2 + y^2 = n\):**
     - The number of integer solutions to \(x^2 + y^2 = n\) is