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I am trying to find the power series representation for $\frac{1}{2-x}$.

I realize that I can write it as $\frac{1}{2}\cdot \frac{1}{1-\frac{x}{2}}$ and then use the geometric series.

My question is, what is wrong with writing it as $\frac{1}{1-(x-1)}$ and then saying the power series is $\sum(x-1)^n$?

Does the term $t$ in the expression $\frac{1}{1-t}$ always need to be a monomial to apply this trick? Why?

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    $\begingroup$ do you want the power series about $x=0$ or $x=1$? $\endgroup$ – J. W. Tanner Dec 1 '19 at 20:57
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It all depends on where you want to center your power series.

The power series $\sum(x-1)^n$ is centered in $x_0=1$, meaning its truncates give good results (i.e. approximate well your function) near $1$.

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  • $\begingroup$ Don't $\sum \frac{x^n}{2^{n+1}}$ and $\sum (x-1)^n$ have intervals of convergence that overlap? I don't see how these two expressions can both approximate the original function well $\endgroup$ – user162520 Dec 1 '19 at 21:06
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    $\begingroup$ As noted in the answer below, where the two intervals overlap, the series give the same results. The "approximation" result is in the sense of polinomials, i.e. truncated series. The truncated series are different and the approximations they give are not equal. $\endgroup$ – Alberto Saracco Dec 1 '19 at 21:12
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As pointed out, these two power series are both valid, just centered at different points. It should be noted that $\sum (x-1)^n$ has a radius of convergence of $1$, so this is valid on $(0,2)$ while $\frac{1}{2}\sum (x/2)^n$ has a radius of convergence of 2, so this is valid on $(-2,2)$. The two power series agree where they are both defined

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