# Represent the function 3ln(8-x) as a power series (Maclaurin series) What is the radius of convergence?

Represent the function $3\mathrm{ln}\left(8-x\right)$ as a power series (Maclaurin series)
What is the radius of convergence?
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To write the function $f\left(x\right)=3\mathrm{ln}\left(8-x\right)$ as a power series (Maclaurin series). Also to calculate the radius of convergence.
From the standard power series, the Maclaurin series of $\mathrm{ln}\left(1+x\right)$ is given by,

Here,
|x|<1 is its radius of converhence
Rewrite the function as shown below
$f\left(x\right)=3\mathrm{ln}\left(8-x\right)$
$f\left(x\right)=3\mathrm{ln}\left(8\left(1-\frac{x}{8}\right)\right)$
$f\left(x\right)=3\mathrm{ln}\left(8\right)+3\mathrm{ln}\left(1-\frac{x}{8}\right)$
Now, $\mathrm{ln}\left(1-\frac{x}{8}\right)=\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n-1}\frac{\left(-\frac{x}{8}{\right)}^{n}}{n},\phantom{\rule{1em}{0ex}}|-\frac{x}{8}|<1$
$\mathrm{ln}\left(1-\frac{x}{8}\right)=\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n-1}\frac{\left(-1{\right)}^{n}{x}^{n}}{n{8}^{n}}\phantom{\rule{1em}{0ex}}\frac{|x|}{8}<1$
$\mathrm{ln}\left(1-\frac{x}{8}\right)=\sum _{n=1}^{\mathrm{\infty }}-\left(-1{\right)}^{2n}\frac{{x}^{n}}{n{8}^{n}}\phantom{\rule{1em}{0ex}}|x|<8$
$\mathrm{ln}\left(1-\frac{x}{8}\right)=\sum _{n=1}^{\mathrm{\infty }}-\frac{{x}^{n}}{n{8}^{n}}\phantom{\rule{1em}{0ex}}|x|<8$
Use it above
$f\left(x\right)=3\mathrm{ln}\left(8\right)-3\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{n{8}^{n}}$
This is the required power series and the radius of convergence is 8 (from $|x|<8$)
Power series: $f\left(x\right)=3\mathrm{ln}\left(8\right)-3\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{n{8}^{n}}$
The radius of convergence is 8.

Jeffrey Jordon