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Problem

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

Note by Brilliant Member
1 year, 3 months ago

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Gamma function \(\Gamma (1+x) = x!\). \(\implies \dfrac d{dx}x! = \dfrac d{dx} \Gamma (1+x) = \Gamma' (1+x) \). As digamma function is defined as: \(\psi (z) = \dfrac d{dz} \ln \Gamma (z) = \dfrac {\Gamma' (z)}{\Gamma (z)} \). \(\implies \dfrac d{dx}x! = \Gamma' (1+x) = \Gamma (1+x) \psi (1+x)\).

Chew-Seong Cheong - 1 year, 3 months ago

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okay thanks sir

Brilliant Member - 1 year, 3 months ago

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

Chew-Seong Cheong - 1 year, 3 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 !

Brilliant Member - 1 year, 3 months ago

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

Chew-Seong Cheong - 1 year, 3 months ago

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

Sabhrant Sachan - 1 year, 3 months ago

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

Chew-Seong Cheong - 1 year, 3 months ago

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@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.

Brilliant Member - 1 year, 3 months ago

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

Brilliant Member - 1 year, 3 months ago

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