Hyperbolic Cotangents And Sums Of Squares

Calculus Level 5

It can be shown that n = 1 coth 2 π n n 2 = P Q G + R S π T \sum_{n=1}^\infty \frac{\coth^2 \pi n}{n^2} \; = \; \frac{P}{Q}G + \frac{R}{S}\pi^T where G = n = 0 ( 1 ) n ( 2 n + 1 ) 2 = 0.915965594 G \;=\; \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} \;=\; 0.915965594\ldots is Catalan's constant , and where P , Q , R , S P,Q,R,S and T T are all positive integers with P , Q P,Q and R , S R,S coprime pairs. Find P + Q + R + S + T P+Q+R+S+T .


Inspiration .


The answer is 206.

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2 solutions

Mark Hennings
Jul 27, 2016

Let C N C_N be the positively oriented square contour with vertices at ξ ( N + 1 2 ) + i η ( N + 1 2 ) \xi(N+\tfrac12) + i\eta(N + \tfrac12) for ξ , η = 1 , 1 \xi,\eta = 1,-1 .

For any b N b \in \mathbb{N} the function f b ( z ) = π cot π z ( z 2 + b 2 ) 2 f_b(z) \,=\, \frac{\pi \cot \pi z}{(z^2 + b^2)^2} is meromorphic with simple poles at every integer, and with double poles at ± i b \pm ib . Thus 1 2 π i C N f b ( z ) d z = R e s z = i b f b ( z ) + R e s z = i b f b ( z ) + a = N N R e s z = a f b ( z ) = π 2 2 b 2 π coth π b 2 b 3 π 2 coth 2 π b 2 b 2 + 1 b 4 + 2 a = 1 N 1 ( a 2 + b 2 ) 2 \begin{array}{rcl} \displaystyle \frac{1}{2\pi i}\int_{C_N} f_b(z)\,dz & = & \displaystyle \mathrm{Res}_{z=ib}f_b(z) + \mathrm{Res}_{z=-ib}f_b(z) + \sum_{a=-N}^N \mathrm{Res}_{z=a}f_b(z) \\ & = & \displaystyle \frac{\pi^2}{2b^2} - \frac{\pi \coth\pi b}{2b^3} - \frac{\pi^2 \coth^2\pi b}{2b^2} + \frac{1}{b^4} + 2\sum_{a=1}^N \frac{1}{(a^2+b^2)^2} \end{array} Since f b ( z ) = O ( N 4 ) |f_b(z)| \,=\, O(N^{-4}) uniformly on C N C_N , it follows, letting N N \to \infty , that a = 1 1 ( a 2 + b 2 ) 2 = π coth π b 4 b 3 + π 2 coth 2 π b 4 b 2 1 2 b 4 π 2 4 b 2 \sum_{a=1}^\infty \frac{1}{(a^2+b^2)^2} \; = \; \frac{\pi \coth\pi b}{4b^3} + \frac{\pi^2 \coth^2\pi b}{4b^2} - \frac{1}{2b^4} - \frac{\pi^2}{4b^2} for any b N b \in \mathbf{N} , and hence it follows that a , b = 1 1 ( a 2 + b 2 ) 2 = 1 4 π b = 1 coth π b b 3 + 1 4 π 2 b = 1 coth 2 π b b 2 1 2 ζ ( 4 ) 1 4 π 2 ζ ( 2 ) \sum_{a,b=1}^\infty \frac{1}{(a^2+b^2)^2} \; = \; \tfrac14\pi\sum_{b=1}^\infty \frac{\coth\pi b}{b^3} + \tfrac14\pi^2\sum_{b=1}^\infty \frac{\coth^2\pi b}{b^2} - \tfrac12\zeta(4) - \tfrac14\pi^2\zeta(2)

Next, the function g ( z ) = π cot π z coth π z z 3 g(z) = \frac{\pi \cot \pi z \coth\pi z}{z^3} is meromorphic with simple poles at n n and i n in for any nonzero integer n n , and a quintuple pole at 0 0 . Thus 1 2 π i C N g ( z ) d z = R e s z = 0 g ( z ) + b = 1 N ( R e s z = b + R e s z = b + R e s z = i b + R e s z = i b ) g ( z ) = 4 b = 1 N coth π b b 3 7 45 π 3 \begin{array}{rcl} \displaystyle \frac{1}{2\pi i} \int_{C_N} g(z)\,dz & = & \displaystyle \mathrm{Res}_{z=0}g(z) + \sum_{b=1}^N \big(\mathrm{Res}_{z=b} + \mathrm{Res}_{z=-b} + \mathrm{Res}_{z=ib} + \mathrm{Res}_{z=-ib}\big) g(z) \\ & = & \displaystyle 4\sum_{b=1}^N \frac{\coth\pi b}{b^3} - \tfrac{7}{45}\pi^3 \end{array} Since g ( z ) = O ( N 3 ) |g(z)| = O(N^{-3}) uniformly on C N C_N , letting N N \to \infty gives us that b = 1 coth π b b 3 = 7 180 π 3 \sum_{b=1}^\infty \frac{\coth \pi b}{b^3} \; = \; \tfrac{7}{180}\pi^3 and hence a , b = 1 1 ( a 2 + b 2 ) 2 = 1 4 π 2 b = 1 coth 2 π b b 2 3 80 π 4 \sum_{a,b=1}^\infty \frac{1}{(a^2+b^2)^2} \; = \; \tfrac14\pi^2 \sum_{b=1}^\infty \frac{\coth^2\pi b}{b^2} - \tfrac{3}{80}\pi^4

Finally, using results from Number Theory we know that a , b = 1 1 ( a 2 + b 2 ) 2 = ζ ( 2 ) G ζ ( 4 ) = 1 6 π 2 G 1 90 π 4 \sum_{a,b=1}^\infty \frac{1}{(a^2+b^2)^2} \; = \; \zeta(2)G - \zeta(4) \; = \; \tfrac16\pi^2 G - \tfrac{1}{90}\pi^4 where G G is Catalan's constant. Putting this all together gives b = 1 coth 2 π b b 2 = 2 3 G + 19 180 π 2 \sum_{b=1}^\infty \frac{\coth^2\pi b}{b^2} \; =\; \tfrac23G + \tfrac{19}{180}\pi^2 making the answer 2 + 3 + 19 + 180 + 2 = 206 2 + 3 + 19 + 180 + 2 = 206 .

Well done! If the function can be differentiated as a function of z z , except at some at most countable collection of isolated points, where the function takes the value \infty , it is meromorphic. Meromorphic equals entire with some isolated singularities.

Mark Hennings - 4 years, 10 months ago

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Ohh I got it, Thanks !

Aditya Narayan Sharma - 4 years, 10 months ago

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Stick with my final sentence: meromorphic equals entire except at some isolated singularities. The value of a function at an isolated essential singularity (infinite pole) is not \infty ; the function takes values arbitrarily close to any complex number arbitrarily close to any essential singularity. Only at (finite) poles does the function tend to \infty near the singularity.

Mark Hennings - 4 years, 10 months ago

It's wonderful, I used the same steps but I'm confused that mathematically how to determine the function is meromorphic or not so that we can apply residue theorem

Aditya Narayan Sharma - 4 years, 10 months ago

I do not know contour yet, so I do not know how to generalise this sum.

Generalising this sum can give interesting results regarding the generating function of the divisor function. For instance:

n = 1 coth 2 ( ln x 2 n ) n 2 = π 2 6 + 4 [ n = 1 σ 3 ( n ) n 2 x n ] \sum _{n=1}^{\infty }\frac{\coth ^2\left(\frac{\ln x}{2}n\right)}{n^2}=\frac{\pi ^2}{6}+4\left[\sum _{n=1}^{\infty }\frac{σ_3\left(n\right)}{n^2}x^{-n}\right]

n = 1 coth ( ln x 2 n ) n 2 = π 2 6 + 2 p = 1 σ 2 ( p ) p 2 x p \sum _{n=1}^{\infty }\frac{\coth \left(\frac{\ln x}{2}n\right)}{n^2}=\frac{\pi ^2}{6}+2\sum _{p=1}^{ \infty}\frac{σ_2\left(p\right)}{p^2} x^{-p}

Julian Poon - 4 years, 10 months ago
Aman Rajput
Jul 30, 2016

Consider

C f ( z ) d z = C π cot ( π z ) coth 2 ( π z ) z 2 d z \displaystyle \int_C f(z)dz = \int_C \frac{\pi \cot(\pi z)\coth^2(\pi z)}{z^2} dz

Poles

At z = 0 z=0 order 5 5

At z = n , n Z { 0 } z=n , n \in Z-\{0\} order 1 1

At z = k i , k Z { 0 } z=ki , k \in Z-\{0\} order 2 2

Evaluation

At z = 0 z=0 , Laurent series of f ( z ) = 1 π 2 z 5 + 1 3 z 3 8 π 2 45 z + f(z)=\frac{1}{\pi^2 z^5}+\frac{1}{3z^3}-\frac{8\pi^2}{45z}+\cdots

Residue is 8 π 2 45 -\frac{8\pi^2}{45}

At z = n z=n

lim z n ( z n ) π cot ( π z ) coth 2 ( π z ) z 2 = coth 2 ( π z ) n 2 \displaystyle \lim_{z \to n} \frac{(z-n)\pi\cot(\pi z)\coth^2(\pi z)}{z^2}=\frac{\coth^2(\pi z)}{n^2}

At z = k i z=ki

lim z k i d d z ( ( z k i ) 2 π cot ( π z ) coth 2 ( π z ) z 2 ) = ( k π + sinh ( 2 k π ) ) \cosech 2 ( k π ) π k 3 \displaystyle \lim_{z \to ki}\frac{d}{dz}\left( \frac{(z-ki)^2\pi\cot(\pi z)\coth^2(\pi z)}{z^2} \right)=-\frac{(k\pi+\sinh(2k\pi))\cosech^2(k\pi)}{\pi k^3}


C π cot ( π z ) coth 2 ( π z ) z 2 d z = 0 \displaystyle \int_C \frac{\pi \cot(\pi z)\coth^2(\pi z)}{z^2} dz=0 C π cot ( π z ) coth 2 ( π z ) z 2 d z = 8 π 2 45 + 2 n 1 coth 2 ( π n ) n 2 2 k 1 ( k π + sinh ( 2 k π ) ) \cosech 2 ( k π ) π k 3 \displaystyle \int_C \frac{\pi \cot(\pi z)\coth^2(\pi z)}{z^2} dz=-\frac{8\pi^2}{45} + 2\sum_{n\geq 1}\frac{\coth^2(\pi n)}{n^2}-2\sum_{k\geq 1}\frac{(k\pi+\sinh(2k\pi))\cosech^2(k\pi)}{\pi k^3}

But k 1 ( k π + sinh ( 2 k π ) ) \cosech 2 ( k π ) π k 3 = 2 3 G + 1 10 ζ ( 2 ) = 2 3 G + π 2 60 \displaystyle \sum_{k\geq 1} \frac{(k\pi+\sinh(2k\pi))\cosech^2(k\pi)}{\pi k^3}=\frac23G+\frac{1}{10}\zeta(2)=\frac23G+\frac{\pi^2}{60}

Thus, n 1 coth 2 ( π n ) n 2 = 4 π 2 45 + 2 3 G + π 2 60 \displaystyle \sum_{n\geq 1}\frac{\coth^2(\pi n)}{n^2} = \frac{4\pi^2}{45}+\frac23G+\frac{\pi^2}{60}

n 1 coth 2 ( π n ) n 2 = 2 3 G + 19 π 2 180 \displaystyle \sum_{n\geq 1}\frac{\coth^2(\pi n)}{n^2}=\frac23G+\frac{19\pi^2}{180}

You just state a formula for k 1 ( k π + sinh 2 k π ) c o s e c h 2 k π π k 3 \sum_{k\ge1}\frac{(k\pi +\sinh2k\pi)\mathrm{cosech}^2k\pi}{\pi k^3} Where does that come from? If you combine it with the result for k 1 coth k π k 3 \sum_{k\ge1}\frac{\coth k\pi}{k^3} That I gave, and the identity c o s e c h 2 x = coth 2 x 1 \mathrm{cosech}^2x = \coth^2x -1 , then the result is immediate...

Mark Hennings - 4 years, 10 months ago

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You mean , you want proof for the formula ?

Aman Rajput - 4 years, 10 months ago

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Yes. If you have a direct proof of it, it might yield a much simpler derivation of the result, putting your and my approaches together. As it stands, it is a rather strong statement to be given without proof or reference.

Mark Hennings - 4 years, 10 months ago

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