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Let $f :(-1,1) \to \mathbb{R};\ \ f(x)=\sum_{n=0}^\infty a_n x^n$ be an analytical function expressible as a power series.

Also, let

$$g : (-1,1) \to \mathbb{R}; \ \ g(x)=\frac{d}{dx} \log{f(x)} = \frac{\sum_{n=1}^\infty n a_n x^{n-1}}{\sum_{n=0}^\infty a_n x^n} =\sum_{n=0}^\infty d_n x^n$$

Assume that $\log f(x)$ is defined for all $\lvert x\rvert<1$. Is it possible to obtain a non-recursive expression for the coefficients $d_n$?

I came across some other answers related to quotients of power series (like this one or this other one), but they all rely on recursive expressions. I wonder whether there is a closed form for expressing those coefficients.

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You start with Faà de Bruno formula (iterated chain rule): We have for $I, J$ open subsets of $\mathbb R$, $f: I\rightarrow J$, $g: J\rightarrow \mathbb R$, smooth functions, $k\in \mathbb N^{*}$, $$ \frac{(g\circ f)^{(k)}}{k!}=\sum_{1\le r\le k}\frac{g^{(r)}\circ f}{r!} \sum_{\substack{(k_{1},\dots, k_{r})\in {(\mathbb N^{*})}^{r}\\k_1+\dots+k_r=k}}\prod_{1\le j\le r}\frac{f^{(k_{j})}}{k_{j}!}. \tag{$\ast$}$$ It is enough to take $g=\log$, note that for $r\ge 1$, $g^{(r)}(y)=(-1)^{r-1}(r-1)!y^{-r}$, so that you get (assuming $f(0)\not=0$ or something giving sense to the composition) $$ d_k=\sum_{1\le r\le k} (-1)^{r-1}\frac1rf(0)^{-r} \sum_{\substack{(k_{1},\dots, k_{r})\in {(\mathbb N^{*})}^{r}\\k_1+\dots+k_r=k}}\prod_{1\le j\le r}a_{k_j}. $$

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  • $\begingroup$ We know that $f(0)=a_0$. $\endgroup$ – Max Alekseyev Feb 20 '18 at 16:11
  • $\begingroup$ I had never seen before a formulation of Faà di Bruno's formula involving that factorial in the denominator, and I seem to be having some trouble working with it. Take, for example, $f(x)=\sum_{n=0}^\infty x^n$. Then, $d_n$ should always equal $1$, but I am not getting that result. Is the latest sum over all partitions of $k$ in exactly $r$ summands? Does it allow to repeat the same summands but in different order? (i.e. taking first $k_1=2, \ k_2=1$ and then $k_1=1, \ k_2=2$) $\endgroup$ – user3141592 Feb 22 '18 at 8:56
  • $\begingroup$ I suppose that what it is meant to be said in this answer is a variation of the formula (a3) from encyclopediaofmath.org/index.php/Bell_polynomial . Thank you. $\endgroup$ – user3141592 Feb 27 '18 at 10:36

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