# Similar Power Series

• Mar 31st 2013, 08:19 PM
mzez
Similar Power Series
so ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 .... i know for ln(1+x^2) i just need to sub x^2 where x is so = x^2 - x^4/2 + x^6/3 .... what about ln(1+x^2/4) ? can i do the same thing and just sub so x^2/4 - x^4/32 + x^6/192 ... or do i need to use integration to find this power series?
• Apr 1st 2013, 01:14 AM
chiro
Re: Similar Power Series
Hey mzez.

You can do what you are thinking of and just for some general u, then replace ln(1 + u) = f(u) with some general u = g(x).

The only thing though that you need to be careful of is that for a general u = g(x) is that the function is defined for that particular g(x). If it isn't defined and isn't analytic around that point, then you can't use that function.

Otherwise if it's analytic and converges around that particular point, then it should be OK.
• Apr 1st 2013, 04:54 AM
Prove It
Re: Similar Power Series
Quote:

Originally Posted by mzez
so ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 .... i know for ln(1+x^2) i just need to sub x^2 where x is so = x^2 - x^4/2 + x^6/3 .... what about ln(1+x^2/4) ? can i do the same thing and just sub so x^2/4 - x^4/32 + x^6/192 ... or do i need to use integration to find this power series?

Since you have the series for $\displaystyle \displaystyle \ln{ \left( 1 + x^2 \right) }$, if you want $\displaystyle \displaystyle \ln{ \left( 1 + \frac{x^2}{4} \right) }$, rewrite it as $\displaystyle \displaystyle \ln {\left[ 1 + \left( \frac{x}{2} \right) ^2 \right] }$ and replace the x with x/2.