What is ∂∂xi(xi!) where xi is a discrete variable? Do you consider (xi!)=(xi)(xi−1)⋯1 and do product rule on each term, or something else?

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What is ∂∂xi(xi!) where xi is a discrete variable?  Do you consider (xi!)=(xi)(xi−1)⋯1 and do product rule on each term, or something else?

Mary Herrera · Accepted Answer

Step 1The derivative of a function of a discrete variable doesn&#039;t really make sense in the typical calculus setting. However, there is a continuous variant of the factorial function called the Γ function, for which you can take derivatives and evaluate the derivative at integer values.In particular, since n≠Γ(n+1), there is a nice formula for Γ′ at integer values:Γ′(n+1)=n!(−γ+∑k=1n1k)where γ is the Euler-Mascheroni constant.Step 2As has been mentioned, the Gamma function Γ(x) is the way to go.Integration by parts yieldsΓ(x)=∫0∞e−ttx−1dt=(x−1)∫0∞e−ttx−2dt=(x−1)Γ(x−1)Taking the derivative of the logarithm of Γ(x) givesΓ′(x)Γ(x)=1x−1+Γ′(x−1)Γ(x−1)Because Γ(x) is log-connvex andlimx⇒∞Γ′(x)Γ(x)−log⁡(x)=0we get thatΓ′(x)Γ(x)=−γ+∑k=1∞(1k−1k+x−1)For integer n, n≠Γ(n+1), so the derivative isΓ(n+1)=Γ(n+1)(−γ+∑k=1∞nk(k+n))=n!(−γ+Hn)where Hn is the nth Harmonic Number (with the convention that H0=0)

Debbie Moore · Accepted Answer

Step 1Start fromx≠x(x−1)!x!′=x(x−1)!′+(x−1)!So we are looking for a function that satisfiesf(x)=xf(x−1)+(x−1)!Replacing we havef(x)=x((x−1)f(x−2)+(x−2)!)+(x−1)!Againf(x)=x((x−1)((x−2)f(x−3)+(x−3)!)+(x−2)!)+(x−1)!Notice that we have at this stagef(x)=x(x−1)(x−2)f(x−3)+x(x−1)(x−3)!+x(x−2)!+(x−1)!So we can extendf(x)=x(x−1)(x−2)⋯(x−(k−1))f(x−k)+x(x−1)⋯(x−(k−2))(x−k)!+⋯+x(x−1)(x−3)!+x(x−2)!+(x−1)!orf(x)=x(x−1)(x−2)⋯(x−(k−1))f(x−k)+∑m=xx−(k−1)x!mStep 2Taking k=x and x integer we havef(x)=x!f(0)+∑m=x1x!mNotice that this must be completely valid no matter what extension of factorial we take. And we could essentially stop here.Still, since we can, it all now comes to defining f(0) which is 0!′, first derivative of factorial at 0.What is the value of 0!′?Well f(0) is a constant so there is no harm of replacing it with f(0)=−γ+c (We use γ so we could argue about the asymptotic evaluation as it is obviosly needed to reach ln⁡(x))

nick1337 · Accepted Answer

Step 1 It&#039;s probably best to use an analytic continuation of the factorial function, rather than the factorial itself. Consider the gamma function: Γ(x)=∫0∞tx−1e−tdt Obviously, Γ(1)=1, and we also have: Γ(x+1)=∫0∞txe−tdt =[−txe−t]0∞+x∫0∞tx−1e−tdt So, Γ(x)=(x−1)!. So, just freely take derivatives now.

What is \frac{\partial}{\partial x_{i}}(x_{i}!) where x_{i} is a discrete variable? Do

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