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 $$\frac{\sum_1^{\infty} na_nx^{n-1}}{\sum_0^{\infty} a_nx^n}$$ Then do the term by term power series division and get a new power series $$ \sum_0^{\infty} b_nx^n=\frac{\sum_1^{\infty} na_nx^{n-1}}{ \sum_0^{\infty} a_nx^n}$$ In the end, can I integrate the power series $$ \sum_0^{\infty} b_nx^n$$ to get $$ \sum_0^{\infty} \frac{1}{n+1}b_nx^{n+1} + \ln a_0$$ 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$?
Best Answer
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 $$b_0=\log(a_0)$$ $$b_1=\frac{a_1}{a_0}$$ $$b_2=-\frac{a_1^2}{2 a_0^2}+\frac{a_2}{a_0}$$ $$b_3=\frac{a_1^3}{3 a_0^3}-\frac{a_2 a_1}{a_0^2}+\frac{a_3}{a_0}$$ $$b_4=-\frac{a_1^4}{4 a_0^4}+\frac{a_2 a_1^2}{a_0^3}-\frac{a_3 a_1}{a_0^2}-\frac{a_2^2}{2 a_0^2}+\frac{a_4}{a_0}$$ $$b_5=\frac{a_1^5}{5 a_0^5}-\frac{a_2 a_1^3}{a_0^4}+\frac{a_3 a_1^2}{a_0^3}+\frac{a_2^2 a_1}{a_0^3}-\frac{a_4 a_1}{a_0^2}-\frac{a_2 a_3}{a_0^2}+\frac{a_5}{a_0}$$