- #1
Schrodinger's Dog
- 835
- 7
Homework Statement
[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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