?Use the formula\(\tanh ^{-1} x=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)

Chapter 10, Problem 50

(choose chapter or problem)

Use the formula

\(\tanh ^{-1} x=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) \quad-1<x<1\)

and the Maclaurin series for \(\ln (1+x)\)to show that

\(\tanh ^{-1} x=\sum_{n=0}^{\infty} \frac{x^{2 n+1}}{2 n+1}\)

Equation Transcription:

 

Text Transcription:

tanh^-1 x = frac{1}{2} ln (frac{1+x}{1-x})    -1 < x < 1

ln(1 + x)

tanh^-1 x = the sum from n=0 infinity frac{x^2n+1}{2n+1}