# Functions to power series Find power series representations centered at 0 for the following functions using known power series. Give the interval of convergence for the resulting series. f(x)=lnsqrt{4-x}

Functions to power series Find power series representations centered at 0 for the following functions using known power series. Give the interval of convergence for the resulting series.
$f\left(x\right)=\mathrm{ln}\sqrt{4-x}$
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Given, $f\left(x\right)=\mathrm{ln}\sqrt{4-x}$
Find power series representation, the radius of converges, the interval of converges.
Use the power series formula,
$f\left(x\right)=\mathrm{ln}\left(1-{x}^{n}\right)=-\sum _{k=1}^{\mathrm{\infty }}\frac{{x}^{k}}{k},-1\le x<1$

Consider
$f\left(x\right)=\mathrm{ln}\sqrt{4-{x}^{2}}$
$f\left(x\right)=\mathrm{ln}\left(4-{x}^{2}{\right)}^{\frac{1}{2}}$
Use the power rule of logarithmic,
$\mathrm{ln}{m}^{2}=n\mathrm{ln}m$
$=\frac{1}{2}\mathrm{ln}\left(4-{x}^{2}\right)$
Now multiple and divide "4" on the logarithmic function we get,
$=\frac{1}{2}\mathrm{ln}4\left(\frac{4-{x}^{2}}{4}\right)$
$=\frac{1}{2}\mathrm{ln}4\left(\frac{4}{4}-\frac{{x}^{2}}{4}\right)$
$=\frac{1}{2}\mathrm{ln}4\left(1-\frac{{x}^{2}}{4}\right)$
Now use the product rule of logarithmic we get,
$\mathrm{ln}\left(mn\right)=\mathrm{ln}\left(m\right)+\mathrm{ln}\left(n\right)$
$=\frac{1}{2}\mathrm{ln}4+\mathrm{ln}\left(1-\frac{{x}^{2}}{4}\right)$
Now again use the power rule of logarithmic we get,
$=\mathrm{ln}{4}^{\frac{1}{2}}+\mathrm{ln}\left(1-\frac{{x}^{2}}{4}\right)$
$=\mathrm{ln}{2}^{\frac{2}{2}}+\mathrm{ln}\left(1-\frac{{x}^{2}}{4}\right)$
$=\mathrm{ln}2+\mathrm{ln}\left(1-\frac{{x}^{2}}{4}\right)$
Now use the power series formula we get,
$=\mathrm{ln}2-\sum _{k=1}^{\mathrm{\infty }}\frac{\left(\frac{{x}^{2}}{4}{\right)}^{k}}{k}$
$=\mathrm{ln}2-\sum _{k=1}^{\mathrm{\infty }}\frac{\left(\frac{{x}^{2k}}{{4}^{k}}\right)}{k}$
For this we get interval, $-2\le x\le 2$
Use Ratio test to check the series converges or diverges:
Ratio test definition:
Suppose we have series $\sum {a}_{n}$ defined,
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{{a}_{n+1}}{{a}_{n}}|$
If $L>1$ then the series diverges,
If $L<1$ then the series converges,
If $L=1$ then the test fails.
Consider,
${a}_{k}=\frac{{x}^{2k}}{{4}^{k}k}$
${a}_{k+1}=\frac{{x}^{2\left(k+1\right)}}{{4}^{\left(k+1\right)}\left(k+1\right)}$
$\underset{k\to \mathrm{\infty }}{lim}|\frac{\frac{{x}^{2k}{x}^{2}}{{4}^{k}4\left(k+1\right)}}{\frac{{x}^{2k}}{{4}^{k}k}}|=\underset{k\to \mathrm{\infty }}{lim}|\frac{{x}^{2k}{x}^{2}}{{4}^{k}4\left(k+1\right)}×\frac{{4}^{k}k}{{x}^{2k}}|$
$=\underset{k\to \mathrm{\infty }}{lim}|\frac{{x}^{2}}{4}|\frac{k}{k+1}|$
$=|\frac{{x}^{2}}{4}|\underset{k\to \mathrm{\infty }}{lim}|\frac{k}{k\left(1+\frac{1}{k}\right)}|$
$=|\frac{{x}^{2}}{4}|\underset{k\to \mathrm{\infty }}{lim}|\frac{1}{1+\frac{1}{k}}|$
$=|\frac{{x}^{2}}{4}$

Jeffrey Jordon