I'm trying to find a power series, centered at $x=-1$
What do I have to do for next step?
$f(x)=\frac{1}{2x-3} = \frac{1}{2(x+1)-2-3} = \frac{1}{5}[\frac{1}{\frac{2}{5}(x+1)-1}]$
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Do you know the power series for $\frac{1}{1-x}$ ? – Teddy Apr 04 '16 at 08:17
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@Teddy Yes. $\sum_{n=0}^{\infty} x^n$ – devDNA Apr 04 '16 at 08:33
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If you substitute $t = \frac{2}{5}(x+1)$, you get the form I mentioned, in $t$. – Teddy Apr 04 '16 at 08:40
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@Teddy how do I substitute $t$ into $\frac{1}{1-x}$? I have tried, but i can't. so I have asked this question. $ – devDNA Apr 04 '16 at 08:44
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@Teddy Oh, wait! Never mind! I solved. Thank you. – devDNA Apr 04 '16 at 08:47
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Substituting $t = \frac{2}{5}(x+1)$, your expression equals $ \frac{1}{5}[\frac{1}{t-1}] = -\frac{1}{5}[\frac{1}{1-t}] = -\frac{1}{5}(\sum_{n=0}^{\infty} t^n) = \sum_{n=0}^{\infty} -\frac{1}{5} t^n =...$ – Teddy Apr 04 '16 at 08:47
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@Teddy Thank you for help, Teddy!:-) – devDNA Apr 04 '16 at 19:19