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Can someone help me to confirm this identity that I have established, I really have no idea how I would go about proving that this is true.

By the way, this is not a homework assignment, I am just genuinely curious. Thank you!

If $f(x)=x^x$, then

$$f^{(n+1)}(x)=f^{(n)}(x)(\ln(x)+1)+\sum_{k=1}^n (-1)^{k+1}\frac{n!}{k(n-k)!}f^{(n-k)}(x)$$ $$\text{ for } n\in\mathbb{N}$$

where $f^{(n)}(x)$ is the $n^{\text{th}}$ derivative of $f(x)$

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For $n = 1$,

$$\begin{align*} f(x) &= x^x\\ & = e^{x\ln x}\\ f'(x) &= \frac{d}{dx} e^{x\ln x}\\ &= e^{x\ln x} \left(\ln x + 1\right)\\ &= x^x\left(\ln x + 1\right)\\ \end{align*}$$

Let $g(x) = \ln x + 1$. Then for $k \ge 1$, $$g^{(k)}(x) = (-1)^{k-1}(k-1)!\ x^{-k}$$

From the second derivative onwards, i.e. $n \ge 1$, using the general Leibniz rule,

$$\begin{align*} f^{(n+1)}(x) &= \frac{d^n}{dx^n}f'(x)\\ &= \frac{d^n}{dx^n}\left[f(x)\ g(x)\right]\\ &=\sum_{k=0}^{n}\binom{n}{k}\ f^{(n-k)}(x)\ g^{(n)}(x)\\ &= f^{(n)}(x)\ g^{(0)}(x) + \sum_{k=1}^{n}\binom{n}{k}\ f^{(n-k)}(x)\ g^{(n)}(x)\\ &= f^{(n)}(x) \left(\ln x + 1\right) + \sum_{k=1}^{n}\binom{n}{k}\ f^{(n-k)}(x)\left[(-1)^{k-1}(k-1)!\ x^{-k}\right]\\ &= f^{(n)}(x) \left(\ln x + 1\right) + \sum_{k=1}^{n}(-1)^{k+1}\frac{n!}{k(n-k)!}\ f^{(n-k)}(x)\ \color{red}{x^{-k}}\\ \end{align*}$$

The red $x^{-k}$ is what I think the question may be missing, but I could be wrong.

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  • $\begingroup$ You are right! When I originally did this on paper, I did include an x^k in the denominator. I simply forgot to insert it into my question when I typed it. $\endgroup$ – Brothersquid Oct 6 '18 at 21:57
  • $\begingroup$ Oh yeah there we go! I like it. $\endgroup$ – clathratus Oct 6 '18 at 21:58
  • $\begingroup$ Also in the last two lines, your n seems to have changed to a Zero $\endgroup$ – Brothersquid Oct 6 '18 at 21:58
  • $\begingroup$ @Brothersquid fixed, thanks. $\endgroup$ – peterwhy Oct 6 '18 at 22:01
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I recently found something related. Given that $f(x)$ and $g(x)$ are continuous and $n$-times differentiable on some interval $I\subseteq \Bbb R$, and $u(x)=f(x)g(x)$, then $$u^{(n)}(x)=\sum_{k=0}^{n}{n\choose k}f^{(n-k)}(x)g^{(k)}(x).$$

If you can find $n$-times differentiable $f(x)$ and $g(x)$ such that $x^x=f(x)g(x)$, then you have your theorem.

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    $\begingroup$ I thought of that, though be careful that the fraction inside the summation has denominator $k$, not $k!$. Wikipedia call this the general Leibniz rule. $\endgroup$ – peterwhy Oct 6 '18 at 21:24
  • $\begingroup$ @peterwhy what fraction in the summation? $\endgroup$ – clathratus Oct 6 '18 at 21:59
  • $\begingroup$ That $k$ in the $\frac{n!}{k(n-k)!}$ in the question. But after I worked out my answer, that $k$ instead of $k!$ actually makes sense. $\endgroup$ – peterwhy Oct 6 '18 at 22:03
  • $\begingroup$ @peterwhy I didn't catch that. But I'm sure that the method that I suggested would yield the same result. $\endgroup$ – clathratus Oct 6 '18 at 22:06

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