# Problem

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

Note by Brilliant Member
1 year, 11 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: $$\psi (z) = \dfrac d{dz} \ln \Gamma (z) = \dfrac {\Gamma' (z)}{\Gamma (z)}$$. $$\implies \dfrac d{dx}x! = \Gamma' (1+x) = \Gamma (1+x) \psi (1+x)$$.

- 1 year, 11 months ago

okay thanks sir

- 1 year, 11 months ago

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

- 1 year, 11 months ago

okay... i was thinking why there was a need of new function just to define d/dx of gamma function. now i understand it !

- 1 year, 11 months ago

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

- 1 year, 11 months ago

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

- 1 year, 11 months ago

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.

- 1 year, 11 months ago

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.

- 1 year, 11 months ago

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

- 1 year, 11 months ago