# How can I express the natural logarithm of a power series in terms of another power series?

Can I express the expression $$\ln\left( \sum_0^{\infty} a_nx^n\right)$$ in terms of another power series?

Can I differentiate the term $$\ln\left( \sum_0^{\infty} a_nx^n\right)$$ and get $$\frac{\sum_0^{\infty} na_nx^{n-1}}{\sum_0^{\infty} a_nx^n}$$ Then do the term by term power series division and get a new power series $$\sum_0^{\infty} b_nx^n=\frac{\sum_0^{\infty} na_nx^{n-1}}{ \sum_0^{\infty} a_nx^n}$$ In the end, can I integrate the power series $$\sum_0^{\infty} b_nx^n$$ to get $$\sum_0^{\infty} \frac{1}{n+1}b_nx^{n+1}$$ and say it is equal to the original expression $\ln\left( \sum_0^{\infty} a_nx^n\right)$ ?

I saw some one set $\ln\left( \sum_0^{\infty} a_nx^n\right)$ equal to another power series in terms of another power series with coefficents related to $a_1,a_2,a_3,\ldots$. But when I try to do what I just shown to confirm what is in the book, the result did not match.

If it is possible to express $\ln\left( \sum_0^{\infty} a_nx^n\right)$ in terms of another power seize,s can you show me the first two terms in the new power series and their relation to $a_1,a_2,a_3,\ldots$?

You can do both and get the same results using for sure the integration constant leading to $b_0=\log(a_0)$.

What is the simplest method ? Probably the long division if you need only few terms.

Using both methods, you should get things like $$b_0=\log(a_0)$$ $$b_1=\frac{a_1}{a_0}$$ $$b_2=-\frac{a_1^2}{2 a_0^2}+\frac{a_2}{a_0}$$ $$b_3=\frac{a_1^3}{3 a_0^3}-\frac{a_2 a_1}{a_0^2}+\frac{a_3}{a_0}$$ $$b_4=-\frac{a_1^4}{4 a_0^4}+\frac{a_2 a_1^2}{a_0^3}-\frac{a_3 a_1}{a_0^2}-\frac{a_2^2}{2 a_0^2}+\frac{a_4}{a_0}$$ $$b_5=\frac{a_1^5}{5 a_0^5}-\frac{a_2 a_1^3}{a_0^4}+\frac{a_3 a_1^2}{a_0^3}+\frac{a_2^2 a_1}{a_0^3}-\frac{a_4 a_1}{a_0^2}-\frac{a_2 a_3}{a_0^2}+\frac{a_5}{a_0}$$

This is not a full answer, but too long to put in a comment.

You can try

$$\ln\left(a_0 + \sum_{n=1}^{\infty}a_nx^n \right) = \ln(a_0) + \ln\left(1 +\sum_{n=1}^{\infty}c_n x^n \right)$$

where $a_n = a_0c_n$, then apply the power series for $$\ln(1+t) = -\sum_{k=1}^{\infty}\frac{(-t)^k}{k} = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$$

of course this only converges if the inner series $$\left| \sum_{n=1}^\infty a_nx^n \right| < a_0$$