Question

How do you find the Maclaurin series for `ln((1+x)/(1-x))`?

Answer 1

`ln( (1+x)/(1-x) ) = 2x+2/3x^3 +2/5x^5 + ...`

Explanation:

We want to find the Maclaurin Series for

`ln( (1+x)/(1-x) )`

We can start with a well known standard series

`ln(1+x) = x-x^2/2+x^3/3-x^4/4 + ...`

If we replace `x` by -`x` we get the following:

`ln(1-x) = -x-x^2/2-x^3/3-x^4/4 - ...`

Using the property of logs we can write the original function as:

`ln( (1+x)/(1-x) ) = ln(1+x) - ln(1-x)`

Then substituting the above two series we get:

`ln( (1+x)/(1-x) ) = { x-x^2/2+x^3/3-x^4/4 + ... } -`
`" " { -x-x^2/2-x^3/3-x^4/4 - ... }`

`" " = x-x^2/2+x^3/3-x^4/4 + ...`
`" " +x+x^2/2+x^3/3+x^4/4 - ...`

`" " = 2x+2x^3/3 +2x^5/5 + ...`