We have discussed various interesting properties of elliptic functions and related theta functions in detail in previous posts. In particular we discussed that these elliptic functions are doubly periodic. It is only natural then to seek the Fourier series corresponding to these functions. However in this case we will use only the real periods to expand these functions in a Fourier series. It turns out that the Fourier expansions provide us many important identities which can be used in surprisingly many ways to connect to number theory.
Let z=πu/2K so that the functions sn(u,k),cn(u,k),dn(u,k) become periodic with periods 2π,2π,π respectively when considered as functions of z. Also these are analytic in the strip |ℑ(z)|<πK′/2K=(π/2)ℑ(τ). We can then consider the case for dn(u) which is an even function and periodic with period π and therefore can be expressed as dn(u)=∞∑n=−∞ane2inz=d02+∞∑n=1dncos(2nz) where a−n=an so that dn=2an=1π∫π−πdn(u)e−2inzdz Calculating the above integral requires the use of residue calculus. We can first look at the integral along the parallelogram (in counterclockwise direction) joining the points −π,π,πτ,−2π+πτ. Clearly the function at hand has poles at πτ/2 and −π+πτ/2 and at these points the residue is given by the expression −i(π2K)e−niπτ=−i(π2K)q−n By residues theorem the integral is given by ∫Cdn(u)e−2inzdz=2πi(sum of residues in the parallelogram)=2πi⋅−i(πK)q−n=(2π2K)q−n It is now easy to observe that while evaluating the integral along boundary C of the parallelogram the integrals on the left and right boundaries cancel each other due to periodicity and the integral on the upper boundary is given by ∫−2π+πτπτdn(u)e−2inzdz=−∫πτ−2π+πτdn(u)e−2inzdz Putting z=t−π+πτ the integral reduces to e−2niπτ∫π−πe−2intdn(2Kπt)dt=q−2n∫π−πdn(u)e−2inzdz It then follows that ∫Cdn(u)e−2inzdz=(1+q−2n)∫π−πdn(u)e−2inzdz and thus finally ∫π−πdn(u)e−2inzdz=11+q−2n(2π2K)q−n The Fourier coefficient is now given by dn=2πKqn1+q2n We finally have the Fourier series expansion as dn(u,k)=π2K+2πK∞∑n=1qncos(2nz)1+q2n Similarly we can establish sn(u,k)=2πKk∞∑n=0qn+1/2sin(2n+1)z1−q2n+1cn(u,k)=2πKk∞∑n=0qn+1/2cos(2n+1)z1+q2n+1
The number of ways in which a positive integer can be expressed as the sum of two squares is four times the difference between the number of its divisors which are of the form 4i+1 and the number of its divisors which are of the form 4i+3.
Two corollaries immediately follow from this theorem:
A prime number p of the form p=4i+1 can be expressed as the sum of two squares in essentially one way.
A prime number p of the form p=4i+3 can not be expressed as the sum of two squares.
The above two corollaries are more properly known as Fermat's theorem on two squares or sometimes as Thue's Lemma. The same technique can be used to prove similar theorems about expressing a number as sum of four squares but the proof is a bit complicated.
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Fourier Series Expansion of Elliptic Functions
Let f(z) be a periodic function with period 2π and let us also assume that it is analytic in some strip a<ℑ(z)<b. Then we can clearly expand it in a Fourier series of the form f(z)=∞∑n=−∞cneinz,cn=12π∫z0+2πz0f(z)e−inzdz where the integration has to be done along a path connecting z0 and z0+2π which lies entirely in the strip a<ℑ(z)<b.Let z=πu/2K so that the functions sn(u,k),cn(u,k),dn(u,k) become periodic with periods 2π,2π,π respectively when considered as functions of z. Also these are analytic in the strip |ℑ(z)|<πK′/2K=(π/2)ℑ(τ). We can then consider the case for dn(u) which is an even function and periodic with period π and therefore can be expressed as dn(u)=∞∑n=−∞ane2inz=d02+∞∑n=1dncos(2nz) where a−n=an so that dn=2an=1π∫π−πdn(u)e−2inzdz Calculating the above integral requires the use of residue calculus. We can first look at the integral along the parallelogram (in counterclockwise direction) joining the points −π,π,πτ,−2π+πτ. Clearly the function at hand has poles at πτ/2 and −π+πτ/2 and at these points the residue is given by the expression −i(π2K)e−niπτ=−i(π2K)q−n By residues theorem the integral is given by ∫Cdn(u)e−2inzdz=2πi(sum of residues in the parallelogram)=2πi⋅−i(πK)q−n=(2π2K)q−n It is now easy to observe that while evaluating the integral along boundary C of the parallelogram the integrals on the left and right boundaries cancel each other due to periodicity and the integral on the upper boundary is given by ∫−2π+πτπτdn(u)e−2inzdz=−∫πτ−2π+πτdn(u)e−2inzdz Putting z=t−π+πτ the integral reduces to e−2niπτ∫π−πe−2intdn(2Kπt)dt=q−2n∫π−πdn(u)e−2inzdz It then follows that ∫Cdn(u)e−2inzdz=(1+q−2n)∫π−πdn(u)e−2inzdz and thus finally ∫π−πdn(u)e−2inzdz=11+q−2n(2π2K)q−n The Fourier coefficient is now given by dn=2πKqn1+q2n We finally have the Fourier series expansion as dn(u,k)=π2K+2πK∞∑n=1qncos(2nz)1+q2n Similarly we can establish sn(u,k)=2πKk∞∑n=0qn+1/2sin(2n+1)z1−q2n+1cn(u,k)=2πKk∞∑n=0qn+1/2cos(2n+1)z1+q2n+1
Some Number Theory
We can use the above results to establish some theorems of number theory. To proceed if we put u=0 in the expansion for dn(u,k) we get 2Kπ=1+4∞∑n=1qn1+q2n=1+4∞∑n=1qn∞∑m=0(−1)mq2mn=1+4∞∑n=1∞∑m=0(−1)mq(2m+1)n We can now put i=(2m+1)n,d=2m+1,m=(d−1)/2 so that 2Kπ=1+4∞∑i=1(∑d∣i,dodd(−1)(d−1)/2)qi=1+4∞∑n=1(∑d∣n,dodd(−1)(d−1)/2)qn The terms in the inner sum are of the form (−1)(d−1)/2 so that the term is +1 if d≡1(mod4) and −1 if d≡3(mod4). Hence we can write the inner sum as a function δ(n) which can be expressed as: δ(n)=#(divisors ofnof the form4i+1)−#(divisors ofnof the form4i+3) where # represents the count of things under its argument. Therefore we have 2Kπ=1+4∞∑n=1δ(n)qn The magic happens when we consider the fact that 2K/π can be expressed in terms of theta functions: 2Kπ=θ23(q)=(∞∑n=−∞qn2)2=∞∑n=0r2(n)qn=1+∞∑n=1r2(n)qn where r2(n) is number of ways in which n can be written as sum of two squares (counting order as well as sign of the integers) and r2(0)=1. Upon comparing the coeffcients it follows that r2(n)=4δ(n) for all positive integers n. Thus we have the theorem (first established by Jacobi using elliptic function theory):The number of ways in which a positive integer can be expressed as the sum of two squares is four times the difference between the number of its divisors which are of the form 4i+1 and the number of its divisors which are of the form 4i+3.
Two corollaries immediately follow from this theorem:
A prime number p of the form p=4i+1 can be expressed as the sum of two squares in essentially one way.
A prime number p of the form p=4i+3 can not be expressed as the sum of two squares.
The above two corollaries are more properly known as Fermat's theorem on two squares or sometimes as Thue's Lemma. The same technique can be used to prove similar theorems about expressing a number as sum of four squares but the proof is a bit complicated.
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Wow,this is so exciting !! Thank-you for writing!
mathter2357
January 20, 2017 at 9:42 AMDear Mr. Paramanand Singh,
Your calculation of the Fourier coefficient for the elliptic function dn(u,k) is very nice and direct. Could you, please, send me the corresponding calculation of the Fourier coefficient for the function sn(u,k)? Or even a reference (it may already have been developed by some textbook author)?
Thank you in advance,
Fabio M. S. Lima (fabio_msl@yahoo.com)
Institute of Physics, University of Brasilia
Fabio
December 22, 2017 at 1:26 AM@Fabio,
Thanks for your positive feedback. You can see the development of these Fourier Series in the classic text "A Course of Modern Analysis" by Whittaker and Watson. It should be available in any university library. Do refer pages 510 and later.
Regards
Paramanand
Paramanand
December 24, 2017 at 8:46 PM