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d-dx-of-x-




Question Number 145197 by Gbenga last updated on 03/Jul/21
d/dx of x!=?
d/dxofx!=?
Commented by Dwaipayan Shikari last updated on 03/Jul/21
x!=Γ(x+1)  ((dx!)/dx)=Γ′(x+1)=Γ(x+1)ψ(x+1)      As ((Γ′(x+1))/(Γ(x+1)))=ψ(x+1)
x!=Γ(x+1)dx!dx=Γ(x+1)=Γ(x+1)ψ(x+1)AsΓ(x+1)Γ(x+1)=ψ(x+1)
Answered by puissant last updated on 03/Jul/21
Γ(x)=(x−1)!=∫_0 ^(+∞) t^(x−1) e^(−t) dt  (1/(Γ(x)))= xe^(γx) Π_(k=1) ^∞ (1+(x/k))e^(−(x/k))   ln((1/(Γ(x))))=−ln(Γ(x))                     =ln(x)+γx+Σ_(k=1) ^∞ (1+(x/k))−(x/k)  (d/dx)(ln((1/(Γ(x)))))=−(ln(Γ(x)))′ =− ((Γ′(x))/(Γ(x)))  =(1/x)+γ+Σ_(k=1) ^∞ (1/(1+(x/k)))×(1/k)−(1/k)  =−(1/x)−γ+Σ_(k=1) ^∞ (1/k)−(1/(x+k))  λ(x)=−γ+Σ_(k=1) ^∞ (1/k)−(1/(k−1+x))  ⇒λ(x+1)=−γ+Σ_(k=1) ^∞ (1/k)−(1/(k+x))  ⇒(d/dx)(x!)=((Γ′(x+1))/(Γ(x+1)))....
Γ(x)=(x1)!=0+tx1etdt1Γ(x)=xeγxk=1(1+xk)exkln(1Γ(x))=ln(Γ(x))=ln(x)+γx+k=1(1+xk)xkddx(ln(1Γ(x)))=(ln(Γ(x)))=Γ(x)Γ(x)=1x+γ+k=111+xk×1k1k=1xγ+k=11k1x+kλ(x)=γ+k=11k1k1+xλ(x+1)=γ+k=11k1k+xddx(x!)=Γ(x+1)Γ(x+1).

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