d-dx-of-x-

Question Number 145197 by Gbenga last updated on 03/Jul/21

d / d x o f x! = ?

Commented by Dwaipayan Shikari last updated on 03/Jul/21

x! = Γ (x + 1)

\frac{d x!}{d x} = Γ^{'} (x + 1) = Γ (x + 1) ψ (x + 1) A s \frac{Γ^{'} (x + 1)}{Γ (x + 1)} = ψ (x + 1)

Answered by puissant last updated on 03/Jul/21

Γ (x) = (x - 1)! = \int_{0}^{+ \infty} t^{x - 1} e^{- t} dt

\frac{1}{Γ (x)} = {xe}^{γ x} \underset{k = 1}{\prod^{\infty}} (1 + \frac{x}{k}) e^{- \frac{x}{k}}

\ln (\frac{1}{Γ (x)}) = - \ln (Γ (x))

= \ln (x) + γ x + \underset{k = 1}{\sum^{\infty}} (1 + \frac{x}{k}) - \frac{x}{k}

\frac{d}{dx} (\ln (\frac{1}{Γ (x)})) = - {(\ln (Γ (x)))}^{'} = - \frac{Γ^{'} (x)}{Γ (x)}

= \frac{1}{x} + γ + \underset{k = 1}{\sum^{\infty}} \frac{1}{1 + \frac{x}{k}} \times \frac{1}{k} - \frac{1}{k}

= - \frac{1}{x} - γ + \underset{k = 1}{\sum^{\infty}} \frac{1}{k} - \frac{1}{x + k}

λ (x) = - γ + \underset{k = 1}{\sum^{\infty}} \frac{1}{k} - \frac{1}{k - 1 + x}

\Rightarrow λ (x + 1) = - γ + \underset{k = 1}{\sum^{\infty}} \frac{1}{k} - \frac{1}{k + x}

\Rightarrow \frac{d}{dx} (x!) = \frac{Γ^{'} (x + 1)}{Γ (x + 1)} \dots .

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