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Taylor Polynomial for ln(1-x)

by Schrodinger's Dog
Tags: ln1x, polynomial, taylor
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Schrodinger's Dog
#1
Aug4-07, 01:47 PM
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1. The problem statement, all variables and given/known data

[itex]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[/itex]

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

Is that right?

Be gentle with me I've only just started on this
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HallsofIvy
#2
Aug4-07, 02:03 PM
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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
[itex]ln(1+x)=-x-\frac{1}{2}x^2- \frac{1}{3}x^3-\frac{1}{4}x^4-\frac{1}{5}x^5-....[/itex]

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
#3
Aug4-07, 02:10 PM
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P: 1,142
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
#4
Jun6-09, 08:41 AM
P: 8
Taylor Polynomial for ln(1-x)

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
#5
Jun6-09, 08:58 AM
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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
#6
Jun6-09, 09:06 AM
P: 8
by normal do you mean real? and if you mean 0 then i thought that 0^0 still equals one
Cyosis
#7
Jun6-09, 09:08 AM
HW Helper
P: 1,495
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.


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