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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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    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.2011-07-16

2 Answers 2

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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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    @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 $= 0$).2011-07-16
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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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    Silver, re-index the sum.2011-07-16