# From power series to differential equations

Let $$f$$ be define by the power series \begin{align} f(x)=\sum_{n=0}^\infty a_n x^n. \end{align} where $$f(1)=1$$.

Moreover, let $$f$$ coefficient of power series satisfy \begin{align} \sum_{n=0}^\infty (n+1) a_{n+1} x^n = \sum_{n=0}^\infty ( c_k n^k+c_{k-1} n^{k-1}+ \cdots +c_0 ) a_n x^n \end{align} for some fixed and given $$c_k,...,c_0$$. We have to solve for $$f$$ or find $$a_k$$'s.

My approach is to solve it by trying to re-write the above as differential equation in terms of $$f$$. For example, the first term can be written as \begin{align} \sum_{n=0}^\infty (n+1) a_{n+1} x^n= \sum_{n=1}^\infty n a_{n} x^{n-1}=f'(x) \end{align}

So we have that \begin{align} f'(x)= \sum_{n=0}^\infty ( c_k n^k+c_{k-1} n^{k-1}+\cdots+c_0 ) a_n x^n \end{align}

Now how to re-write $$\sum_{n=0}^\infty ( c_k n^k+c_{k-1} n^{k-1}+\cdots+c_0 ) a_n x^n$$ in terms of derivatives of $$f$$?

• To clarify, you given the coefficients $c_i$ along with their relationship to the coefficients $a_j$ and asked to determine the coefficients $a_j$? – parsiad Feb 26 at 1:14
• @parsiad Yes, that is correct. – Lisa Feb 26 at 1:15
• You have $a_{n+1}/a_n=P(n)/Q(n)$ with $P(n)=c_k n^k+c_{k-1} n^{k-1}+ \cdots +c_0$ and $Q(n)=n+1$. The function $f$ is hypergeometric. – AccidentalFourierTransform Feb 26 at 1:16
• @AccidentalFourierTransform Can you please clarify these steps. I am not sure what hypergeometric function is. It be great if you could put the details as an answer. – Lisa Feb 26 at 1:18
• It is hard to be more explicit than that without knowing more about the $\{c_i\}$. Do you have any explicit sequence in mind? I don't think you will be able to get a closed-form expression for arbitrary $\{c_i\}$. If you pick a specific example, people might be able to help. – AccidentalFourierTransform Feb 26 at 1:21

Note that $$D_x^j(x^jf(x))=\sum_{n\ge0}a_n n^{(j+1)}x^n$$ with $$n^{(\,\cdot\,)}$$ the rising factorial. Thus, \begin{aligned} \sum_{n\ge0} c_0 a_n x^n&=c_0f(x)\\ \sum_{n\ge0} c_1 n a_nx^n&=c_1\left[(xf(x))'-f(x)\right]\\ \sum_{n\ge0} c_2 n^2a_nx^n&=c_2\left[(x^2f(x))''-3\left[(xf(x))'-f(x)\right]-2f(x)\right]\\ &\cdots \end{aligned}
Therefore, your function indeed satisfies a linear differential equation of order $$k$$. Writing it down explicitly for arbitrary $$\{c_i\}$$ is left as an exercise to the reader. Hint: think of the Stirling numbers of the second kind:
$$\sum_{n\ge0}c_kn^ka_nx^n=c_k\sum_{j=1}^k S^{(j)}_k x^j f^{(j)}(x)$$
Note also that one may write $$f(x)$$ as a $$_kF_0$$ hypergeometric function whose parameters are essentially the roots of $$c_kn^n+\cdots+c_0=0$$. Again, for arbitrary $$\{c_i\}$$ this is as explicit as it gets.
• @Lisa I wrote down the formula for arbitrary $k$, but you'll have to prove it yourself. – AccidentalFourierTransform Feb 26 at 2:01
• Great. Thanks a lot. I will try to do this. I will put the derivation. Do you mind checking it once I write it? Also, are $S_k^{(j)}$ the Stirling numbers that you mentioned? – Lisa Feb 26 at 2:07