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Taylor Polynomial for ln(1x) 
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#1
Aug407, 01:47 PM

P: 1,142

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(1x) 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(1x)? 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 


#2
Aug407, 02:03 PM

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P: 39,293

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(11)= 0) that radius is 1 and so it cannot converge for x> 1.) Since 1 x= 0 when x= 1 and ln(1x) not defined for [itex]x\ge 1[/itex], The radius of convergence is still 1: the series converges for 1< x< 1 still. 


#3
Aug407, 02:10 PM

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 1x are all negative. 


#4
Jun609, 08:41 AM

P: 8

Taylor Polynomial for ln(1x)
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 


#5
Jun609, 08:58 AM

HW Helper
P: 1,495

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



#6
Jun609, 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



#7
Jun609, 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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