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

Question

Ezekiel Alexander · Accepted Answer

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

The Maclaurin Series is what we're looking for.

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

We may begin with a widely recognized 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 - ... #

Because of the logs' property, we can rewrite the original function as follows:

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

Substituting the two series mentioned above, we obtain:

# 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 + ... #

Paisley Johnson · Answer

undefined To find the Maclaurin series for ln((1+x)/(1-x)), you can start by expanding ln(1+x) and ln(1-x) using their respective Maclaurin series.

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

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

Now, subtract the series for ln(1-x) from ln(1+x):

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^2/2) + (x^3/3 + x^3/3) + (x^4/4 + x^4/4) + ...

Simplify the expression:

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

This is the Maclaurin series for ln((1+x)/(1-x)).