# How do you find the first coefficients of a power series?

So, I've found them, but I don't understand the first few. Let me explain.

The problem I was working on was:

Suppose that

$$\frac{10 x}{12 + x} = \sum_{n=0}^{\infty}c_nx^n.$$

Find the first few coefficients : $c_0,c_1,c_2,c_3,c_4,\dots$ Now, I figured out (through a bit of odd luck) that:

$c_0 = 0$
$c_1 = 10/12$
$c_2 = -10/144$

and you continue to multiply by $-1/12$ to get further ones.

Anyways, I don't understand why $c_0$ is $0$ and $c_1$ is $10/12$

See, I transformed the left side $\frac{10 x}{12 + x}$ into:

$$\frac{10}{12}\sum_{n=0}^{\infty}(-1/12)^n x^{n+1} )$$

Now, when I substitute in $0$ for $n$ (for $c_0$), the coefficient I get is $(10/12) \times 1$, or $10/12$. So why isn't $c_0=10/12$?

Any help is greatly appreciated!

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If, as you say, you plug in $n=0$, you get the coeffiecient leading $x^{n+1}=x^{0+1}=x$, not the coefficient leading the constant term, so you are in effect calculating $c_1$. – Olivier Bégassat Jul 15 '11 at 23:30
Learn to format the \LaTeX so things look better. – ncmathsadist Jul 15 '11 at 23:34
I'm confused Olivier, but I think your comment is the answer I've been looking for if only I could understand it. See, I thought the coefficient leading x was the constant term multiplying x, or for c0: $\frac{5}{6} \frac{-1}{12}^0 )$ which = $\frac{5}{6} (1) )$ right? – user13327 Jul 15 '11 at 23:47
I understand what you mean now Olivier, thanks! – user13327 Jul 15 '11 at 23:56
The generic method for such problems is of course to repeatedly differentiate your function and then evaluate at the expansion point... of course, if the function is simple enough, answers like Bill's or ncms's methods are applicable. – J. M. Jul 16 '11 at 7:26

HINT $\rm\displaystyle\quad \frac{1}{c+x}\ =\ \frac{1}c\ \frac{1}{1+x/c}\:.\:$ Now apply the formula for the geometric series to the latter.

Thus $\rm\displaystyle\quad\ \frac{10\:x}{12+x}\ =\ \frac{5\:x}6\ \frac{1}{1-(-x/12)}\ =\ \frac{5\:x}6\ (1 - \frac{x}{12} + \frac{x^2}{144} - \:\cdots\:)$

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Thanks man, but I can get it past this point to the point where I actually have the series (see above comment), I just don't understand why the first coefficient is 0 and the second is 5/6. I thought that the first (c0) should be (5/6) – user13327 Jul 15 '11 at 23:37
@Silver The first coef is $0$ because the geometric series is multiplied by $\rm\:x\:,\:$ so the resulting product has constant term $= 0\:.$ – Bill Dubuque Jul 15 '11 at 23:47
Alright, so what you're saying is that there is no constant term (x^0), and so the constant term is 0. OOOOOOOOOOHHHH... so the coefficients are not found by plugging in 0,1,2,3 for n, but by finding the coefficients of x^0, x^1, x^2 etc? ooooooooooooooooohhhhhhhhhhh! Thank you so much! – user13327 Jul 15 '11 at 23:55
@Silver Yes $c_i$ is by definition the coef of $x^i$. Thus $c_0$ is the coef of $x^0$, i.e. the constant term. If the constant term is $0$ then then it is not the leading coef (except if the series $= 0$). – Bill Dubuque Jul 16 '11 at 0:09

You have $${10 x\over 12 + x} = {5\over 6}{x\over 1 + x/12}= {5\over 6} x\sum_{n=0}^\infty (-x/12)^n = {5\over 6} \sum_{n=0}^\infty {(-1)^nx^{n+1}\over 12^n}.$$ You can separate the rest out.

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Right, I got that, except I got (5/6) * sum of n=0 to infinity of x^(n+1)/(-12)^n, because when we substitute, u should equal -x/12 right? Anyways, I can get to this point, but I still don't see why the first coefficient is 0 and the second is 5/6. – user13327 Jul 15 '11 at 23:36
You have a sign error in your series expansion. – Bill Dubuque Jul 15 '11 at 23:37
Thanks, Bill for seeing that. – ncmathsadist Jul 15 '11 at 23:50
Silver, re-index the sum. – ncmathsadist Jul 16 '11 at 0:44