 # Find the power series representation for g centered at 0 by differentiating or integrating the power series for f(perhaps more than once). ka1leE 2021-03-07 Answered

Find the power series representation for g centered at 0 by differentiating or integrating the power series for f(perhaps more than once). Give the interval of convergence for the resulting series.
$g\left(x\right)=\mathrm{ln}\left(1-2x\right)$ using $f\left(x\right)=\frac{1}{1}-2x$

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To deduce the power series of g(x) from the power series for f(x) and identify its radius of convergence
The power series for f(x) is just the geometric series derived from $\frac{1}{1-y}$, setting $y=2x$. Its radius of convergence is 0.5
Let
$f\left(x\right)=\frac{1}{1-2x}=1+\left(2x\right)+\left(2x{\right)}^{2}+...+\left(2x{\right)}^{n}+...$
the power series expansion (geometric series),
valid for
so, radius of convergence $=0.5$
The power series for g(x) is found by integrating term by term the power series of f(x) (upto a constant). The radius of converngence of g(d) is the same as that of f(x) (from general theory) $=0.5$
Now, $g\left(x\right)=\mathrm{ln}\left(1-2x\right)$
$=-2\int \frac{dx}{\left(1-2x\right)}=-2\int f\left(x\right)dx$
$=-2\int \left[1+\left(2x\right)+\left(2x{\right)}^{2}+...+\left(2x{\right)}^{n}+...\right]dx$
$=-2x-\frac{\left(2x{\right)}^{2}}{2}-\frac{\left(2x{\right)}^{3}}{3}-\frac{\left(2x{\right)}^{4}}{4}-...-\frac{\left(2x{\right)}^{n}}{n}-...$
is the power series expansion for g(x),
radius of convergence $=0.5$

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