Question

What is the power series representation of $\ln \left( \dfrac{1+x}{1-x} \right)$?

Answer 1

Hint: Assume the given expression as E. Use the property of log given as $\log \left( \dfrac{m}{n} \right)=\log m-\log n$ to simplify the expression. Now, write the expansion formula of $\ln \left( 1+x \right)$ given by the Maclaurin series as $\ln \left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+\dfrac{{{x}^{5}}}{5}-....$. Now, to find the expansion of $\ln \left( 1-x \right)$ substitute $-x$ in place of $x$ in the expansion formula of $\ln \left( 1+x \right)$. Consider the difference between these series to get the answer.

Complete step-by-step solution:
Here we have been provided with the expression $\ln \left( \dfrac{1+x}{1-x} \right)$ and we are asked to write its power series expression. Let us assume the expression as E, so we have,
$\Rightarrow E=\ln \left( \dfrac{1+x}{1-x} \right)$
Using the property of log given as $\log \left( \dfrac{m}{n} \right)=\log m-\log n$ we get,
$\Rightarrow E=\ln \left( 1+x \right)-\ln \left( 1-x \right)$
Now, power series representation of a function is obtained by the Maclaurin series which is a special case of the Taylor series. According to this the expansion of the logarithmic function is given as $\ln \left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+\dfrac{{{x}^{5}}}{5}-....$. Now, substituting $-x$ in place of $x$ in this relation we get,
\[\begin{align}
  & \Rightarrow \ln \left( 1-x \right)=\left( -x \right)-\dfrac{{{\left( -x \right)}^{2}}}{2}+\dfrac{{{\left( -x \right)}^{3}}}{3}-\dfrac{{{\left( -x \right)}^{4}}}{4}+\dfrac{{{\left( -x \right)}^{5}}}{5}-.... \\
 & \Rightarrow \ln \left( 1-x \right)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-\dfrac{{{x}^{5}}}{5}-.... \\
 & \Rightarrow \ln \left( 1-x \right)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-\dfrac{{{x}^{5}}}{5}-.... \\
\end{align}\]
Subtracting $\ln \left( 1-x \right)$ from $\ln \left( 1+x \right)$ we get,
\[\Rightarrow \ln \left( 1+x \right)-\ln \left( 1-x \right)=\left( x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+\dfrac{{{x}^{5}}}{5}-.... \right)-\left( -x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-\dfrac{{{x}^{5}}}{5}-.... \right)\]
Clearly we can see that the terms containing even power of x will get cancelled while the terms containing the off powers of x will add up, so we get,
\[\therefore E=2\left( x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}+\dfrac{{{x}^{7}}}{7}+\dfrac{{{x}^{9}}}{9}.... \right)\]
Hence, the above relation is our answer.

Note: Note that you need to remember the expansion formula of the functions $\ln \left( 1+x \right),\sin x,\cos x,\tan x,{{e}^{x}}$ because they are used in the chapter limit and differentiation. The proof of the expansion of these functions is given in higher mathematics. Note that the above expansion series of log function is only for the natural log and not of log with any other base. In case we need to determine the expansion of log function with a different base then first we need to convert it into the natural log using the base change rule.