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I have the following power series:

$f(T) = \sum_{n\geq 0}(n+1)T^n \in \mathbb{C} [T]$

and I have to compute the multiplicate inverse.

I've tryed to use the formula to calculate the coefficients of the inverse one by one in order to find if there were any rule but it wasnt usefull.

$b_0= a_0^{-1}$

$b_n = -a_0^{-1}\sum_{i=1}^{n}a_ib_{n-i}$

then $b_0= 1, b_1 = -2, b_2 = 1$ and $b_n=0 \space \forall n \geq 3$

Thank you in advance.

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  • $\begingroup$ note that this is the formal derivative of the power series $\sum_{k \ge 1}T^k$ and one can then argue using complex power series $\endgroup$
    – Conrad
    Jul 1 '20 at 12:28
  • $\begingroup$ @Conrad can you develope it a little more please? $\endgroup$
    – Hojas
    Jul 1 '20 at 18:19
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    $\begingroup$ $G(T)=T/(1-T)=-1+1/(1-T), F(T)=1/(1-T)^2, 1/F=(T-1)^2=1-2T+T^2$ all of his is "formal" but can be justified riguorusly by using $T \to z \in \mathbb C, |z| <1$ $\endgroup$
    – Conrad
    Jul 1 '20 at 19:17
  • $\begingroup$ You have obtained a correct result. (Now what's the question?..) $\endgroup$
    – metamorphy
    Jul 2 '20 at 11:13
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I rewrite the prove of Conrad.

Let $g(T)=\sum_{n\geq1}T^n$ then $g'(T)=f(T)$

It is known that the multiplicative inverse of $\sum_{n\geq0}T^n$ is $(1-T)$ then

$(g(T)+1)(1-T)=1;g(T)+1=\frac{1}{1-T}$ and taking the derivative

$g'(T)=f(T)=\frac{1}{(1-T)^2}$

then the multiplicative inverse of $f$ is $(1-T)^2$

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