# Taylor Polynomial for ln(1-x)

by Schrodinger's Dog
Tags: ln1x, polynomial, taylor
 P: 1,136 1. The problem statement, all variables and given/known data $ln(1+x)=x-\frac{1}{2}x^2+ \frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5-.... -1<\ x\ <1$ Is there a Taylor polynomial for ln(1-x) for -1< x <1, if so how would I go about working it out from the above? This is not really a homework question just a thought I had, as they do it for other changes of sign. I just seem to be having a hard time figuring out how it would work for some reason? Any help appreciated. Are they all minuses for ln(1-x)? Or something else? I'm sure I could just look it up on the web, but I'd like to see how it's done from (1+x) I tried multiplying each x term by a - and got: $-x+\frac{1}{2}x^2 -\frac{1}{3}x^3+\frac{1}{4}x^4-\frac{1}{5}x^5+....$ Is that right? Be gentle with me I've only just started on this
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,571 You are NOT "multiplying each x by -1", you are multiplying the entire series by -1 so you really have -ln(1+x). If you multiply x by -1- that is, replace x by -x, the even powers of x do not change sign- you would have instead $ln(1+x)=-x-\frac{1}{2}x^2- \frac{1}{3}x^3-\frac{1}{4}x^4-\frac{1}{5}x^5-....$ Also, ln(x) is not defined for $x\le 0$ so the Taylor's series you give for ln(1+x) converges only for (-1, 1). (Since it is a power series it converges in some radius of convergence. The center is at x= 0, since it cannot converge for x= -1 (where ln(1+x)= ln(1-1)= 0) that radius is 1 and so it cannot converge for x> 1.) Since 1- x= 0 when x= 1 and ln(1-x) not defined for $x\ge 1$, The radius of convergence is still 1: the series converges for -1< x< 1 still.