When I was working in my next post I've found the problem to compute the kth derivative of $f(x)=\log^2 x$, for $x>0$,
Fact. If $f(x)=\log^2 x$, for $x>0$, then the kth derivative, $k\geq 1$, is given by $$f^{k)}(x)=\frac{2(-1)^k}{x^k}(a_k-(k-1)!\cdot\log x)$$ where $a_1=0$ and $a_k=(k-1)a_{k-1}+(k-2)!$ when $k\geq 2$.
My question,
Question a) I am looking for a proof verification of my proof of previous Fact. b) Suggestions to compute a closed solution for the cited sequence $a_k=(k-1)a_{k-1}+(k-2)!$ for $k>1$ and $a_1=0$. c)(Optional and reissued) Can you provide us a proof or disproof that the sequence $(a_k)_{k\geq 1}$ is $A000254$, corresponding to unsigned Stirling numbers of first kind (see [1])?
My attempts were,
Proof. By mathematical induction.
Step 1 (We prove that k=1 holds). In a hand by the chain rule we obtain $f'(x)=(2\log x)\cdot 1/x$, and in the other hand $(2/x)\log x =(2(-1)^1/x^1)\cdot (0-0!\log x)$, we assume that $0!=1$.
Step 2 (Induction statement). We take as induction hypothesis that for every an integer $k>1$ the identity involving $f^{k)}(x)$ in Fact is true.
Step 3 (We prove that $k\Rightarrow k+1$). Deriving one more time using induction statement and the rule for the derivative of a product we obtain
$$f^{k+1)}=\frac{d}{dx} \left( f^{k}(x)\right)= 2(-1)^{k+1}x^{-k-1}k(a_k-(k-1)!\log x)+2(-1)^{k+1}x^{-k-1}(k-1)!$$ this is RHS is equal, when we use the recursion formula for $a_k$, to
$$\frac{2(-1)^{k+1}}{x^{k+1}}(ka_k+(k-1)!-k!\log x)=\frac{2(-1)^{k+1}}{x^{k+1}}(a_{k+1}-k!\log x),$$ and thus we've finished the proof.
For b) my attempt was work in substitutions of these kind:
$$a_{k+1}=k(k-1)a_{k-1}+(k-2)!\cdot k+(k-1)!$$
since $ka_k=k(k-1)a_{k-1}+(k-2)!\cdot k$, in the same way
$$a_{k+1}=k(k-1)(k-2)a_{k-2}+k(k-1)\cdot (k-3)!+(k-2)!\cdot k+(k-1)!.$$
I don't know if this is a good way to sum the general term. Neither if this sequence $a_k$ is known, see
For c) I search in The On-Line Encyclopedia of Integer Sequences, the first six terms of the sequence $A000254$ match with us sequence since $(a_k)$ starts as
$$0, 1, 3, 11, 50, 274, 1764, 13068,\cdots$$
seems the same sequence. I don't know if this is obvious, today.
In my next post I want to essay a summation wth Euler Maclaurin formula, and I need know, previously derivatives of cited function $(\log (x))^2$ to try compute my error term. Thanks in advance, I apprecite your help.
References:
[1] https://oeis.org, The On-Line Encyclopedia of Integer Sequences, sequence $A000254$, Unsigned Stirling numbers of first kind.
a) It looks like your proof is appropriate.
b) It can be simply shown by induction that $a_{n}=(n-1)!\left(a_1+H_{n-1}\right)$ where $H_n$ is $n$th Harmonic number, $H_n=\sum\limits_{k=1}^{n}\frac{1}{k}$. Here $a_1=0$ so $a_n=(n-1)!H_{n-1}$.