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!
$\endgroup$ 52 Answers
$\begingroup$Hint $\rm\displaystyle\,\ \ \frac{1}{c-x}\ =\ \frac{1}c\ \frac{1}{1-x/c}\,.\, $ Now apply the formula for the geometric series to the latter.
Thus $\rm\displaystyle\,\ \frac{10\:x}{12+x}\ =\ \frac{10\,x}{12}\ \frac{1}{1-(-x/12)}\ =\ \frac{5\,x}6\ (1 - \frac{x}{12} + \frac{x^2}{144} - \:\cdots\:)$
$\endgroup$ 4 $\begingroup$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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