It's actually quite simple: Take the Gamma reflection formula Γ(z)Γ(1-z)=\frac{π}{sin(πz)} Take natural log on both sides ln(Γ(z))+ln(Γ(1-z))=ln(π)-ln(sin(πz)) Take derivative on both sides ψ(z)-ψ(1-z)=-π cot(πz) Therefore ψ(1-z)-ψ(z)=π cot(πz) Hope this was helpful
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Amazingly clear exposition. Please don't stop making videos.
Please keep uploading the videos and complete the series ,please please
Hello Dear QN3.
Thank you for Topology playlist.
Hello Dear *QN^3* .
I have a request. Please if it's possible, make a video about Digamma functions and its reflection formula.
Thank you so much
It's actually quite simple:
Take the Gamma reflection formula
Γ(z)Γ(1-z)=\frac{π}{sin(πz)}
Take natural log on both sides
ln(Γ(z))+ln(Γ(1-z))=ln(π)-ln(sin(πz))
Take derivative on both sides
ψ(z)-ψ(1-z)=-π cot(πz)
Therefore
ψ(1-z)-ψ(z)=π cot(πz)
Hope this was helpful
Can you please suggest me some good book to follow for this series ,so that I can get more practice problems and keep up with the series ?
Munkres Topology 2nd edition