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I'm having some trouble finding the Taylor series for the following function at zero (Maclaurin series). \begin{equation} \frac{1}{(1+x)^t} \end{equation} Where $t$ is a constant that is greater than zero.

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An approach is to obtain successive derivatives of $\displaystyle f(x)=\frac{1}{(1+x)^t}$ with respect to $x$ and then plug it in the Taylor series expansion near $0$. You easily find by induction that $$ f^{(n)}(x)=(-1)^n\frac{t(t+1) \cdots (t+n-1)}{(1+x)^{t+n}}, \quad n=1,2,\ldots, $$ leading to

$$ \frac{1}{(1+x)^t}=1+\sum_{n=1}^{\infty}(-1)^n\frac{t(t+1) \cdots (t+n-1)}{n!}x^n $$

for $x$ near $0$.

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  • $\begingroup$ Thank you so much, could you maybe also clarify for me what it means if I am told to write, for example, the third degree Taylor polynomial of the function. $\endgroup$ – friedmanfanboy Feb 25 '15 at 13:00
  • $\begingroup$ You are welcome! You may write, for the third degree Taylor polynomial : $$\frac{1}{(1+x)^t}=1-t x+\frac{1}{2} t(t+1) x^2-\frac{1}{6} t(t+1)(t+2) x^3+\mathcal{O}(x^4) $$ $\endgroup$ – Olivier Oloa Feb 25 '15 at 13:03
  • $\begingroup$ I'm having some trouble figuring out how you calculate those terms. $\endgroup$ – friedmanfanboy Feb 25 '15 at 13:09
  • $\begingroup$ Please, if you mean those in my preceding comment, you just have to put $n=1,\,2,\,3$ in the formula for $f^{(n)}(x)$ given in my answer. $\endgroup$ – Olivier Oloa Feb 25 '15 at 13:13
  • $\begingroup$ I'm not sure how you get the first term ($-tx$) because I do not get that when I plug in the value for $n$. $\endgroup$ – friedmanfanboy Feb 25 '15 at 13:17

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