Could someone explain how I would find the derivative of the following function? I am completely lost: $$f(x) = e^{i(x!^{\log x})}$$
Finding the derivative of a function?
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calculus
derivatives
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0Do you really mean "$x!$" ? – 2012-06-10
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0@ArturoMagidin No, $x!^{\log x}$ – 2012-06-10
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1$\,x!=\Gamma(x+1)\,\,,\,\,x+1\neq\,$ non-positive integer? – 2012-06-10
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0@DonAntonio Sorry, I don't understand. What are you asking? I am not asking a derivative for only the factorial- I am asking for a derivative for the whole function... – 2012-06-10
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1@Riddler Yes, it is, but then I ask just what, apparently, Arturo meant to ask: how would you define $\,x!\,$ for a positive non-integer? – 2012-06-10
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0@DonAntonio I see. $x!$ for a positive non-integer would just be the associated value on the gamma function. For example Γ(3/2) = $\frac{\sqrt \pi}{2}$ – 2012-06-10
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1Well, *then* is what I asked before! You're going to need the derivative of the gamma function, the polygamma: – 2012-06-10
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1@Riddler: You can't restrict $x$ to positive integers and ask about derivatives. The computation of the derivative requires you to evaluate the function arbtirarily close to the point you want. Even if the "point you want" is an integer, points arbitrarily close to it will **not** be. So, no matter what, you'll end up having to deal with the derivative of the Gamma function. – 2012-06-10
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0@ArturoMagidin I guess I don't really have an choice then. If you are aware of one, could you please provide a solution with the derivative of the Gamma function? – 2012-06-10
1 Answers
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For complex $\,z\,$ with $\,\operatorname{Re}(z)>0\,$ , $$\Gamma'(z)=\int_0^\infty t^{z-1}e^{-t}\log t\,dt$$ so $$\left(e^{ix!^{\log x}}\right)'=e^{ix!^{\log x}}\left(ix!^{\log x}\right)'$$ Now, since $\,\displaystyle{x!^{\log x}=e^{\log x\log(x!)}}\,\,and\,\,\log x!=\log\Gamma(x+1)$ , we get$$\left(ie^{\log x\log x!}\right)'=i\left[\frac{\log x!}{x}+\log x\frac{\Gamma'(x+1)}{\Gamma(x+1)}\right]e^{\log x\log x!} $$Now put together all the above and have sweet mathematical nightmares.