Gamma(s+1) = s*Gamma(s).

log Gamma(s+1) – log Gamma(s) = log s.

Gamma'(s+1)/Gamma(s+1) – Gamma'(s)/Gamma(s) = 1/s.

[Gamma”(s+1)*Gamma(s+1)-Gamma'(s+1)^2]/Gamma(s+1)^2 – [Gamma”(s)*Gamma(s)-Gamma'(s)^2]/Gamma(s)^2 = -1/s^2.

[Gamma”'(s+1)/Gamma(s+1) – 3*Gamma”(s+1)*Gamma'(s+1)/Gamma(s+1)^2 + 2*Gamma'(s+1)^3/Gamma(s+1)^3] – [Gamma”'(s)/Gamma(s) – 3*Gamma”(s)*Gamma'(s)/Gamma(s)^2 + 2*Gamma'(s)^3/Gamma(s)^3] = 2/s^3.

\lim_{N\to\infty} [Gamma”'(N)/Gamma(N) – 3*Gamma”(N)*Gamma'(N)/Gamma(N)^2 + 2*Gamma'(N)^3/Gamma(N)^3] – [Gamma”'(1)/Gamma(1) – 3*Gamma”(1)*Gamma'(1)/Gamma(1)^2 + 2*Gamma'(1)^3/Gamma(1)^3] = 2*zeta(3).

Gamma(1)=1.

Gamma'(1)=-gamma.

Gamma”(1)=gamma^2+pi^2/6.

Gamma”'(1)=?????

It seems the []ed term with Gamma(N), Gamma'(N), Gamma”(N), Gamma”'(N) goes to 0, but I have only numerical evidence to support this.

Gamma”'(1)=-5.44487…

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## About Simon

Hi. I'm Simon Rubinstein-Salzedo. I'm a mathematics postdoc at Dartmouth College. I'm also a musician; I play piano and cello, and I also sometimes compose music and study musicology. I also like to play chess and write calligraphy. This blog is a catalogue of some of my thoughts. I write them down so that I understand them better. But sometimes other people find them interesting as well, so I happily share them with my small corner of the world.

Oops. I forgot to mention that I might be able to prove that the []ed part with the Ns goes to 0 with Stirling’s Formula, but I haven’t tried yet.

Can you help me with my problem?

What’s your problem?

To find a polynomial f(x) with integral coefficients such that f(2^(1/2) + 2^(1/3)) = 0.

I have solved it but with tremendous effort and what seems to be a rather specific method. Is there a technique? Is there something here I don’t understand?

x^6-6x^4-4x^3+12x^2-24x-4. What I did was to construct a polynomial with roots the sum of the roots of two other polynomials with Viète’s formula. I got x^6-6x^4-4x^3+36x^2+48x-4 because I made some arithmetic errors. then tried it and didn’t make arithmetic errors and got the above (correct) answer.

Thanks, I find this method more morally satisfying.

i think ill go hide in a hole now