Problem

What is ddx(x!)\dfrac d{dx}(x!)?

#Calculus

Note by A Former Brilliant Member
4 years, 9 months ago

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Comments

Gamma function Γ(1+x)=x!\Gamma (1+x) = x!.   ⟹  ddxx!=ddxΓ(1+x)=Γ′(1+x)\implies \dfrac d{dx}x! = \dfrac d{dx} \Gamma (1+x) = \Gamma' (1+x) . As digamma function is defined as: ψ(z)=ddzln⁡Γ(z)=Γ′(z)Γ(z)\psi (z) = \dfrac d{dz} \ln \Gamma (z) = \dfrac {\Gamma' (z)}{\Gamma (z)} .   ⟹  ddxx!=Γ′(1+x)=Γ(1+x)ψ(1+x)\implies \dfrac d{dx}x! = \Gamma' (1+x) = \Gamma (1+x) \psi (1+x).

Chew-Seong Cheong - 4 years, 9 months ago

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okay thanks sir

A Former Brilliant Member - 4 years, 9 months ago

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I think there is no closed form for ddxx!\dfrac d{dx}x!, that is why we use digamma function.

Chew-Seong Cheong - 4 years, 9 months ago

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@Chew-Seong Cheong – okay... i was thinking why there was a need of new function just to define d/dx of gamma function. now i understand it !

A Former Brilliant Member - 4 years, 9 months ago

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@A Former Brilliant Member – I actually used Wolfram Alpha to check ddxx!\dfrac d{dx}x!, it gave the same result involving ψ(x)\psi(x). Check here.

Chew-Seong Cheong - 4 years, 9 months ago

@A Former Brilliant Member –  Gamma function is not in the syllabus of IIT Advanced  The function is defined as Γ(1+x)=∫0∞e−ttxdtFor integer values of x the value of integral is Γ(1+x)=x! \text{ Gamma function is not in the syllabus of IIT Advanced } \\ \text{ The function is defined as } \Gamma{(1+x)}=\displaystyle \int_{0}^{\infty} e^{-t}t^{x} dt \\ \text{For integer values of x the value of integral is } \Gamma{(1+x)}=x!

Sabhrant Sachan - 4 years, 9 months ago

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@Sabhrant Sachan – The motivation of the invention by Euler of Γ(z)\Gamma (z) is to extend factorial beyond the integers. It connects the integer factorials in a smooth curve.

Chew-Seong Cheong - 4 years, 9 months ago

@Sabhrant Sachan – yes you are right * but * some questions can be easily solved using gamma and beta functions also i did'nt knew the thing i asked was related to gamma or digamma function , i asked purely out of curiosity as i thought of it in my leisure time.

A Former Brilliant Member - 4 years, 9 months ago

Thanks @Chew-Seong Cheong and @Sambhrant Sachan for stopping by and helping me with my problem , plz keep supporting.

A Former Brilliant Member - 4 years, 9 months ago
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