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Let R be a real positive number, $$f(x)=\sum_{n=0}^\infty a_n x^n \\ g(x)=\sum_{n=0}^\infty b_n x^n $$ where $|x|<R$.

I have to give the first 4 addends depending on $\{a_n\}$ and $\{b_n\}$.


I know that $$f(x)= a_0 + a_1 x + a_2 x^2 + ... \\ g(x)= b_0 + b_1 x + b_2 x^2 + ... $$ I've multiplied some elements and I think that I've got the result: $$f(x)g(x)= a_0b_0+(a_0b_1 + a_1b_0)x+(a_0b_2 + a_1b_1 + a_2b_0)x^2+(a_0b_3 + a_1b_2 + a_2b_1 + a_3b_0)x^3+...$$ But I think that's not the good way to do it, I suppose there is an other better way to do it. Anyway, I don't know if my answer is ok...

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  • $\begingroup$ Further to the answer posted by @nala, if power series converges absolutely and so does the product of two power series. Therefore, rearrangement of terms doesn't change the sum of the series. $\endgroup$
    – Koro
    May 11 '20 at 12:10
  • $\begingroup$ An other question, now how can I find the first 4 addends of the Taylor serie of f(x)=(e^x)/(3+2x) in the point x_0=0? $\endgroup$
    – User160
    May 11 '20 at 13:43
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    $\begingroup$ Write series for $e^x$, which is $1+x +x^2/2!+ x^3/3!+...$ and then for $\frac{1}{1+(2x/3)}$. Multiply them and collect the terms of x with power upto 3 to get first four terms. $\endgroup$
    – Koro
    May 11 '20 at 15:22
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this kind of product is called the Cauchy product of two series :) You got the hang of it for the first few terms, now, by induction, it generalizes to:

$\left(\displaystyle\sum_{n=0}^{+\infty} a_n x^n\right) \cdot \left(\displaystyle\sum_{n=0}^{+\infty} b_n x^n\right) = \displaystyle\sum_{n=0}^{+\infty} c_n x_n$, where each of the $c_n$ is defined as $c_n = \displaystyle\sum_{j = 0}^{n} a_j b_{n-j}$.

This coincides with the first few terms you computed :). You can see more on the wikipedia page on Cauchy product.

PS: try to LaTeX your question next time, it makes it clearer to read :)

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  • $\begingroup$ An other question, now how can I find the first 4 addends of the Taylor serie of f(x)=(e^x)/(3+2x) in the point x_0=0? $\endgroup$
    – User160
    May 11 '20 at 13:26
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    $\begingroup$ Are you supposed to use a Cauchy product? :) Cause if so, try and see $\frac{1}{3+2x}$ as $\frac{1}{3}\cdot\frac{1}{1 + 2x/3}$, and then expend it as a geometric series :) $\endgroup$
    – Azur
    May 11 '20 at 19:39

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