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Use differentiation to find a power series of $f(x) = \frac{1}{(8+x)^2}$

$ f'(x) = \frac{-2}{(8+x)^3} $

how do I find the power series of this? I can not go next step.

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$$ g(x) = \frac{1}{(8+x)} = \sum_{n=0}^{\infty} \frac{(-1)^nx^n}{8^{n+1}} $$ Taking the derivative in both sides: (n=0 is constant) $$ g'(x) = \frac{-1}{(8+x)^2} = \sum_{n=1}^{\infty} \frac{n(-1)^nx^{n-1}}{8^{n+1}} $$ $$ -f(x) = \frac{-1}{(8+x)^2} = \sum_{n=0}^{\infty} \frac{(n+1)(-1)^{n+1}x^{n}}{8^{n+2}} $$ $$ f(x) = \sum_{n=0}^{\infty} \frac{(n+1)(-1)^{n}x^{n}}{8^{n+2}} $$

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  • $\begingroup$ Thank you so much. I'll try figure it out through your answer! $\endgroup$ – devDNA Apr 3 '16 at 0:20
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Hint: To find a series expression for $\frac{2}{(8+x)^2}$, differentiate the power series of (more or less) $\frac{1}{8+x}$. Note that $\frac{1}{8+x}$ has derivative $-\frac{1}{(8+x)^2}$.

To find the series for $\frac{1}{8+x}$, rewrite as $\frac{1}{8}\cdot \frac{1}{1+x/8}$, and use the familiar series for $\frac{1}{1-t}$.

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  • $\begingroup$ I found a power series of $\frac{1}{8+x}$ as $\sum_{n=0}^{\infty} \frac{(-1)^n*x^n}{8^{n+1}}$ How do I combine it with the derivative? $\endgroup$ – devDNA Apr 3 '16 at 0:03
  • $\begingroup$ The derivative of this is $-\frac{1}{(8+x)^2}$. Differentiate the series you wrote down above term by term. That gives you the series for $-\frac{1}{(8+x)^2}$. Then multiply each term by $-2$ to get the series for your function $\frac{2}{(8+x)^2}$. $\endgroup$ – André Nicolas Apr 3 '16 at 0:09
  • $\begingroup$ I multiplied both sides by -2 and then I got $\frac{2}{(8+x)^2} = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}nx^{n-1}2}{8^{n+1}}$ In order to get the left side as same as differentiation of f(x) I multiplied both sides by $\frac{-1}{2}$ again. so I got $\frac{-2}{(8+x)^3} = \sum_{n=0}^{\infty} \frac{(-1)^n n(n-1)x^{n-2}}{8^{n+1}}$ My answer is wrong. Where did I make a mistake? $\endgroup$ – devDNA Apr 3 '16 at 0:19
  • $\begingroup$ You should have differentiated only once. You are not interested in $\frac{1}{(8+x)^3}$. By the way, in your sum the first term is $0$. Omit it, and change the variable of summation, changing $n-1$ to $k$. $\endgroup$ – André Nicolas Apr 3 '16 at 0:22
  • $\begingroup$ Oh Cool. Now I understand what I have to do! Thank you a lot. $\endgroup$ – devDNA Apr 3 '16 at 0:35

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