On the Summation of Hypergeometric series

Since, 2F1(a,b,c,z)=1+abcz+a(a+1)b(b+1)c(c+1)z22!+a(a+1)(a+2)b(b+1)(b+2)c(c+1)(c+2)z33!+..._2F_1(a,b,c,z)=1+\frac{ab}{c}z+\frac{a(a+1)b(b+1)}{c(c+1)}\frac{z^2}{2!}+\frac{a(a+1)(a+2)b(b+1)(b+2)}{c(c+1)(c+2)}\frac{z^3}{3!}+... Then using the following result by Sir Leonard Euler, B(b,cb)2F1(a,b,c,z)=01xb1(1x)cb1(1zx)adx\Beta(b,c-b)_2F_1(a,b,c,z)=\int_0^1x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}dx We will convert it as follows by sayinga=aa=-a , dividing 1-1 and aa to both sides, B(b,cb)2F1(a,b,c,z)a=(1)a1a01xb1(1x)cb1(zx1)adx\frac{-\Beta(b,c-b)_2F_1(-a,b,c,z)}{a}=\frac{(-1)^{a-1}}{a}\int_0^1x^{b-1}(1-x)^{c-b-1}(zx-1)^{a}dx B(b,cb)a=12F1(a,b,c,z)a=01xb1(1x)cb1a=1(1)a1(zx1)aadx-\Beta(b,c-b)\sum_{a=1}^{\infty}\frac{_2F_1(-a,b,c,z)}{a}=\int_0^1x^{b-1}(1-x)^{c-b-1}\sum_{a=1}^{\infty}\frac{(-1)^{a-1}(zx-1)^a}{a}dx Applying the Taylor series of the logarithm,we get, B(b,cb)a=12F1(a,b,c,z)a=01xb1(1x)cb1ln(zx)dx-\Beta(b,c-b)\sum_{a=1}^{\infty}\frac{_2F_1(-a,b,c,z)}{a}=\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (zx)dx Splitting the integral part we yield, B(b,cb)a=12F1(a,b,c,z)a=01xb1(1x)cb1ln(x)dx+ln(z)01xb1(1x)cb1dx-\Beta(b,c-b)\sum_{a=1}^{\infty}\frac{_2F_1(-a,b,c,z)}{a}=\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (x)dx+\ln (z)\int_0^1x^{b-1}(1-x)^{c-b-1}dx Then realising that the integral with ln(z)\ln (z) is simply B(b,cb)\Beta(b,c-b) but, 01xb1(1x)cb1ln(x)dx=B(b,cb)(ψ(b)ψ(c))\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (x)dx=\Beta(b,c-b)(\psi(b)-\psi(c)) I'll show this result in the next discussion ,right now let's take it for granted.Inputting it to the integral and dividing both sides by B(b,cb)-\Beta(b,c-b) we get this surprising result , a=12F1(a,b,c,z)a=ψ(c)ψ(b)+ln(1/z)\sum_{a=1}^{\infty}\frac{_2F_1(-a,b,c,z)}{a}=\psi(c)-\psi(b)+\ln (1/z) Corollary c=1-b and z=1 we get the reflection equation of diagamma function i.e.ψ(1s)ψ(s)=πcot(πs)\psi(1-s)-\psi(s)=\pi\cot(\pi s) a=12F1(a,b,1b,1)a=πcot(πb)\sum_{a=1}^{\infty}\frac{_2F_1(-a,b,1-b,1)}{a}=\pi\cot(\pi b) So it also provides an alternative representation of the Cotangent function and as always another wonderful equation is born.(Please report any problem in this discussion)

Note by Aruna Yumlembam
3 months, 1 week ago

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Really good! You know you should team up with @Neeraj Anand Badgujar, @Naren Bhandari and @Gandoff Tan, @Aruna Yumlembam?

Yajat Shamji - 3 months, 1 week ago

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Hey they are the best people the Brilliant community has ,I am just 15 and I have much more to learn.Thank you.

Aruna Yumlembam - 3 months, 1 week ago

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No! I am not talking about that!

I am talking about calculus - they post some amazing calculus stuff, just like you!

You four should get together and create a group!

How's the sound of Calculus Geeks?

Yajat Shamji - 3 months, 1 week ago

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@Yajat Shamji Sorry got you wrong . I'll try.

Aruna Yumlembam - 3 months, 1 week ago

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