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I'm asked to represent the function $\displaystyle \frac{2 x}{10 + x}$ as a power series $f(x) = \displaystyle \sum_{n=0}^\infty c_n x^n$

I found this to be $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{5 \cdot 10^n} x^{n+1}$

The question is asking for $\ c_0, c_1, c_2, c_3,$ and $\ c_4$

I entered $\displaystyle c_0 = \frac{1}{5}, c_1 = \frac{-1}{50}, c_2 = \frac{1}{500}, c_3 = \frac{-1}{5000},$ and $\displaystyle c_4 = \frac{1}{50000}$ but I am told this is incorrect.

Can someone please explain what I'm doing wrong? Thanks!

UPDATE: using another formula I was able to find $\displaystyle c_n = \frac{f^{n}(a)}{n!}$ so $\displaystyle c_0 = 0, c_1 = \frac{1}{5}, c_2 = \frac{-1}{50},$ etc.. which are the correct answers. So I'm still confused.. Assuming I calculated the power series representation correctly to get $\ c_0$ don't you just let n=0 and take the coefficient of the resulting term?

UPDATE: $\ c_n$ is actually the coefficient for the $\ x^n$th term. So $\ c_0$ is the constant term. This helped me for another question I had where the terms were $\ c_0 + c_2x^2 + c_4x^4 \cdots $ and I didn't understand why the coefficient $\ c_1$ was 0

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You were asked to find a series in the form $$\displaystyle \sum_{n=0}^\infty c_n x^n$$ but your answer is $$\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{5 \cdot 10^n} x^{n+1}$$ which is in the form $$\displaystyle \sum_{n=0}^\infty c_n x^{n+1}$$ Do you see the difference? The power of $x$ is off by one.

If you correct for that off-by-one error, you will get your new $c_0$ is $0$, your new $c_1$ is your old $c_0$, and so on. That is exactly the correct answer that you got the other way.

ADDED:

If you need a general formula for your new $c_n$ you could use

$$ c_n = \begin{cases} 0, & n=0 \\[2ex] \displaystyle \frac{(-1)^{n-1}}{5 \cdot 10^{n-1}}, & n>0 \end{cases} $$

Note that I changed the $n$'s to $n-1$'s in the formula for $c_n$ for this to work. It would be difficult to write that into an explicit series, as you did for $\sum_{n=0}^\infty \frac{(-1)^n}{5 \cdot 10^n} x^{n+1}$.

But you do not need such a formula, since you were only asked for $c_0$ through $c_4$.

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  • $\begingroup$ Thanks I see what you mean.. Can you explain how I change the series to write it in the correct form? $\endgroup$ – Devin Crossman Dec 2 '14 at 23:18
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Hint: $\dfrac{2x}{10+x} = \dfrac{x}{5}\cdot \dfrac{1}{1-\left(-\frac{x}{10}\right)}$, and

use the well-known geometric series: $\dfrac{1}{1-a} = 1 + a + a^2+....$

for $|a| < 1$.

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  • $\begingroup$ Thanks but I already did this much to determine the power series representation (did I do it wrong?).. $\endgroup$ – Devin Crossman Dec 2 '14 at 22:03
  • $\begingroup$ I think you got it right. $\endgroup$ – DeepSea Dec 2 '14 at 22:04
  • $\begingroup$ What about the coefficients of the first 5 series terms? I don't understand why I'm being told they're wrong $\endgroup$ – Devin Crossman Dec 2 '14 at 22:07

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