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I was thinking about the factorial function today and I wondered:

Is there a function $f$ that satisfies $g(s)=f^{(s)}(x)^s|_{x=1/e}=s!$ where the notation is the $s$th derivative of $f$ to the $s$th power evaluated at $x=1/e$? Here $s>0$.

I hope not. However, I'm still curious. I feel like I did something egregiously wrong somewhere.

J. Zimmerman
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1 Answers1

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Define $$f(x) = \sum_{k=1}^\infty \frac{(k!)^{1/k}}{k!}\left(x-\frac{1}{e}\right)^k$$

It follows from the Taylor formula for the series coefficients that $$f^{(k)}(1/e)=(k!)^{1/k}$$

jjagmath
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