Solving Power Series Equations if we introduce Logarithmic Terms

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If we have a complex power series equation like $$ \sum_{n=0}^{\infty} a_n z^n = \sum_{n=0}^{\infty} b_n z^n $$ then we can conclude $a_n = b_n$. We can see this by viewing $z^n$ as basis elements, or by generating a collection of equations by taking successive derivatives and then setting $z = 0$ for each.

I'd like to know whether we can still deal with equations like this (even if new approaches are required) if we have logarithmic terms appearing.

Suppose we have an equation like $$ \sum_{n=0}^{\infty} a_n z^n = \sum_{n=0}^{\infty} b_n z^n + \sum_{n=0}^{\infty} c_n \log \left( z \right) z^n $$ or perhaps like $$ \sum_{n=0}^{\infty} a_n z^n = \sum_{n=0}^{\infty} b_n z^n + \sum_{n=0}^{\infty} c_n \log \left( z \right) z^n + \sum_{n=0}^{\infty} d_n \left( \log \left( z \right) \right)^2 z^n. $$ In these cases, are there any techniques we can use to deduce relationships between the coefficients $a_n, b_n, c_n, d_n$?

Clearly we can't expand our logarithmic terms as power series around $z=0$. It also seems like our usual basis argument or usual method for using derivatives and taking $z=0$ both fail here. Can we deduce anything from equations like this?

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There are 2 best solutions below

2
On

$$ \sum_{k=0}^{\infty} a_n z^n = \sum_{k=0}^{\infty} b_n z^n + \sum_{k=0}^{\infty} c_n \log \left( z \right) z^n $$

First, it is easy to verify $c_0=0$, define $t_n=a_n-b_n$, then we write it as

$$ \ln z=\frac{\displaystyle\sum_{n=0}^{\infty} (a_n - b_n) z^n}{\displaystyle\sum_{n=1}^{\infty} c_n z^n}=\frac{\displaystyle\sum_{n=0}^{\infty} t_n z^n}{\displaystyle\sum_{n=1}^{\infty} c_n z^n} $$

Don't know the exact form of your coefficients, but in principle, we can compute the inversion of the series $$\left(\displaystyle\sum_{n=1}^{\infty} c_n z^n \right)^{-1}=\sum_{n\in\mathbb{Z}} s_n z^n$$

So we get:

$$ \ln z=\displaystyle\sum_{n=0}^{\infty} t_n z^n\sum_{n\in\mathbb{Z}} s_n z^n $$

Finally, take derivative on each sides and compare coefficients.

0
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In general

Let $$F(z) = \sum_{n=0}^{\infty}\sum_{m=0}^M c_{n,m} z^n \log(z)^{-m}$$ If for some $r>0$, $$\forall z \in (0, r),\, F(z) = 0\;\implies \;\forall n\in \mathbb N,\, \forall m,\, c_{n, m} = 0$$

Using Leibniz rule and induction you easly can prove that

$$F^{(k)} (z) = \sum_{n=0}^\infty \sum_{m=0}^M c_{n,m}\left(\frac{n!}{(n-k) !} + m \sum_{j=1}^{k} a_{m,n,k} \log(z)^{-k}\right)\log(z)^{-m}z^{n-k} $$

Now take the limit $z\to 0^+$ and you will have, $c_{k, 0} = 0$.

Can you complete the proof?