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What is the power series representation of #ln((1+x)/(1-x))#?

Answer 1

# ln((1+x)/(1-x)) =2x^3/3+2x^5/5+2x^7/7 ... = 2sum_(n=1)^oox^(2n+1)/(2n+1)#

I would use the following

So: # \ \ \ \ \ ln((1+x)/(1-x)) =ln(1+x)-ln(1-x) # # :. ln((1+x)/(1-x)) ={1-x^2/2+x^3/3-x^4 + ...} - {1-(-x)^2/2+(-x)^3/3-(-x)^4 + ...} # # :. ln((1+x)/(1-x)) ={1-x^2/2+x^3/3-x^4 + ...} - {1-x^2/2-x^3/3-x^4 + ...} # # :. ln((1+x)/(1-x)) =1-x^2/2+x^3/3-x^4 + ... - 1+x^2/2+x^3/3+x^4 + ... # # :. ln((1+x)/(1-x)) =2x^3/3+2x^5/5+2x^7/7 ... = 2sum_(n=1)^oox^(2n+1)/(2n+1)#

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Answer 2

Scarlett Allison

The power series representation of ( \ln\left(\frac{1+x}{1-x}\right) ) is given by:

[ \ln\left(\frac{1+x}{1-x}\right) = 2\left(x + \frac{x^3}{3} + \frac{x^5}{5} + \frac{x^7}{7} + \frac{x^9}{9} + \cdots\right) ]

This series is valid for ( -1 < x \leq 1 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.