Taylor Polynomial for ln(1-x)

Schrodinger's Dog

1. The problem statement, all variables and given/known data

[latex]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[/latex]

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: [latex]-x+\frac{1}{2}x^2 -\frac{1}{3}x^3+\frac{1}{4}x^4-\frac{1}{5}x^5+....[/latex]

Is that right?

Be gentle with me I've only just started on this

HallsofIvy

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
[latex]ln(1+x)=-x-\frac{1}{2}x^2- \frac{1}{3}x^3-\frac{1}{4}x^4-\frac{1}{5}x^5-....[/latex]

Also, ln(x) is not defined for [itex]x\le 0[/itex] 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 [itex]x\ge 1[/itex], The radius of convergence is still 1: the series converges for -1< x< 1 still.

Schrodinger's Dog

Ah I see, I figured it might go something like that. But I also figured the limits were different because logs of negative numbers are not calculable, well not by calculator anyway.

Much appreciated.

The reason I thought they might be all minus is because the values of x between 0 and 1 in 1-x are all negative.

3hlang

can you help me with a problem my dad set me?
he asked what to the power of 0 does not equal one? im guessing its to do with expansions of logs. help would be greatly appreciated. thankyou

Cyosis

Just think of normal numbers. There is one number that when raised to the power 0 yields an indeterminate expression, any idea which one?

3hlang

by normal do you mean real? and if you mean 0 then i thought that 0^0 still equals one

Cyosis

With normal I meant the numbers you know. But yes 0^0 is an indeterminate form. Check http://en.wikipedia.org/wiki/0^0#Zero_to_the_zero_power out.