Q-find-d-dx-x-

Question Number 80613 by M±th+et£s last updated on 04/Feb/20

Q . f i n d \frac{d}{d x} (x!)

Commented by Rio Michael last updated on 04/Feb/20

i {don}^{'} t think this exist, from my understanding

x! = x (x - 1)! has a domain of D = x \in N^{*}

tbat means we have no specific interval

for which x! is not continous, A non - continous function

is non - defferentiable hence i think

\frac{d}{d x} (x!) cannot be calculated .

please give me your oppinion

Commented by MJS last updated on 04/Feb/20

\frac{d}{d x} [x!] does not exist because x! is only

defined on N \Rightarrow {it}^{'} s not continuous

Γ (x) is the extension of x! but both are not

the same

so if you ask for \frac{d}{d x} [x!] the correct answer

is “ does not {exist}^{″}

mathematical language is precise language

x! = Γ (x + 1) is true \forall x \in N

but

Γ (x) \neq (x - 1)! \forall x \in R ∖ N

Answered by M±th+et£s last updated on 04/Feb/20

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

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

l n (\frac{1}{Γ (x)}) = - l n (Γ (x)) = l n x + γ x + \underset{k = 1}{\sum^{\infty}} [l n (1 + \frac{x}{k}) - \frac{x}{k}]

\frac{d}{d x} (- l n (Γ (x))) = \frac{Γ^{'} (x)}{Γ (x)} = \frac{1}{x} + γ + \underset{k = 1}{\sum^{\infty}} [\frac{1}{1 + \frac{x}{k}} . \frac{1}{k} - \frac{1}{k}]

\Rightarrow \frac{Γ^{'} (x)}{Γ (x)} = - \frac{1}{x} - γ + \underset{k = 1}{\sum^{\infty}} [\frac{1}{k} - \frac{1}{k + x}] = Ψ (x)

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

Ψ (x + 1) = - γ + \underset{k = 1}{\sum^{\infty}} [\frac{1}{k} - \frac{1}{k + x}] = \frac{Γ^{'} (x + 1)}{Γ (x + 1)}

Γ^{'} (x + 1) = Ψ (x + 1) Γ (x + 1)

\frac{d}{d x} = Γ (x + 1) Ψ (x + 1)

Commented by M±th+et£s last updated on 04/Feb/20

Ψ^{(m)} (z) = {(- 1)}^{m + 1} \int_{0}^{\infty} \frac{t^{m} e^{- z t}}{1 - e^{- t}} d t : m > 0 a n d R e (z) > 0

Commented by mathmax by abdo last updated on 04/Feb/20

Γ (x) = (x - 1)! i s o n l y a n o t a t i o n \dots!