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Expand the function $f(x)=\ln(1+x+x^2+x^3)$ into a power series.

This is my solution: with $|x| <1$ we have $$\ln(1+x+x^2+x^3)=\ln\left(\dfrac{1-x^4}{1-x}\right)=\ln(1-x^4)-\ln(1-x) \\=-\sum_{n=1}^{\infty} \dfrac{x^{4n}}{n}+ \sum_{n=1}^{\infty} \dfrac{x^{n}}{n}=(1-x^4)\sum_{n=1}^{\infty} \dfrac{x^n}{n}$$

Is the last series called a power series? Anyone help me please?

Winther
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liverpool29
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    Yes it is. If you want you can go the last step and try to derive an explicit expression for $a_n$ in $f(x) = \sum a_n x^n$ (consider the two cases 1) $n$ on the form $4m$ and 2) $n$ not on the that form). btw the last equality is not correct since $x^{4n} \not= x^4 \cdot x^n$ – Winther Nov 18 '16 at 04:03
  • One more question: if I write: $\sum_{n=1}^{\infty} \dfrac{x^n-x^{4n}}{n}$ then is this a power series? – liverpool29 Nov 18 '16 at 04:17
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    Yes, however the canonical way to present a power-series is on the form $\sum a_n x^n$ so if this was a book problem / homework etc. then I would expect that this is the form you are supposed to deliver the answer on. For example $a_1 = 1$, $a_2 = \frac{1}{2}$, $a_3 = \frac{1}{3}$, $a_4 = \frac{1}{4} - 1$, $\ldots$. Thus if I were you I would try to give a formula / description for the value of the $a_n$'s. – Winther Nov 18 '16 at 04:22
  • Yes, thank you again. I understand now. – liverpool29 Nov 18 '16 at 04:25

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