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February 1st, 2019, 09:57 PM   #1
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Derivatives of a function

Hello,

Calculate the derivatives of the function:

1) $\displaystyle f(x)=x!$

2) $\displaystyle f(x)=(2-\sqrt{2})\cdot (2-\sqrt[3]{2})\cdot \dots \cdot (2-\sqrt[x]
{2})$.

All the best,

Integrator
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February 1st, 2019, 10:59 PM   #2
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Neither function is continuous, so neither function is differentiable.
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February 2nd, 2019, 12:54 PM   #3
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$x!=\Gamma(x+1)=\int_0^\infty u^xe^{-u}du$. You can get the derivative from this. The second expression is defined only for integers, so you can't get a derivative.
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February 2nd, 2019, 03:31 PM   #4
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Originally Posted by mathman View Post
$x!=\Gamma(x+1)=\int_0^\infty u^xe^{-u}du$. You can get the derivative from this. The second expression is defined only for integers, so you can't get a derivative.
This is essentially 100% wrong. Its true that $\Gamma$ is an analytic extension of the factorial function, but it isn't the only one. There are arbitrarily many such extensions and you can give them whatever derivative you like. It makes no sense to assume that THE derivative of $x!$ must be the one obtained by differentiation of $\Gamma$.

On the other hand, the second function, which also can be analytically extended from the positive integers, has a natural choice for this extension which makes its derivative unique. This choice comes from the continuous extension of $f(x) = a^x$ from the integers to the rationals. This automatically determines a unique extension to the positive reals and this extension happens to be analytic.

In a nutshell, the first expression has no "correct" answer without more assumptions and the second one does.
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February 2nd, 2019, 09:45 PM   #5
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Quote:
Originally Posted by mathman View Post
$x!=\Gamma(x+1)=\int_0^\infty u^xe^{-u}du$. You can get the derivative from this. The second expression is defined only for integers, so you can't get a derivative.
Hello,

1) From "WolframAlpha" reading:
https://www.wolframalpha.com/input/?i=x!%27.
Is it correct what WolframAlpha says?
2) I do not know how to use "WolframAlpha" for the first derivative of the function $\displaystyle f(x)=(2-\sqrt{2})\cdot (2-\sqrt[3]{2})\cdot \dots \cdot (2-\sqrt[x]{2})$.
Thank you very much!

All the best,

Integrator
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February 3rd, 2019, 04:46 AM   #6
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Quote:
Originally Posted by Integrator View Post
Hello,

1) From "WolframAlpha" reading:
https://www.wolframalpha.com/input/?i=x!%27.
Is it correct what WolframAlpha says?
2) I do not know how to use "WolframAlpha" for the first derivative of the function $\displaystyle f(x)=(2-\sqrt{2})\cdot (2-\sqrt[3]{2})\cdot \dots \cdot (2-\sqrt[x]{2})$.
Thank you very much!

All the best,

Integrator
Honestly I don't really see that your instructor wants you to use the Gamma function for this problem. I've never taken Analysis but I doubt they cover Gamma functions? And making x a continuous variable in the second problem also falls into that category. Ask your instructor, but don't borrow trouble and keep it simple.

-Dan
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