## anonymous 5 years ago Find the Taylor series of 1/(1-x)^2 about a=0

1. anonymous

1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+O(x^6)

2. anonymous

saubhik has it without finding derivatives, evaluating etc

3. anonymous

just by noting that $\frac{1}{(1-x)^2}$ is the derivative of $\frac{1}{1-x$

4. anonymous

why dont you get alternating signs? when you differentiate 1/(1-x)^2 you get - sign

5. anonymous

oop i meant $\frac{1}{1-x}$

6. anonymous

yeah I get that part

7. anonymous

the well known formula for summing a geometric sequence $\frac{1}{1-x}=1+x+x^2+x^3+...$

8. anonymous

differentiate term by term to get the answer

9. anonymous

oh hell, I see... there are 2 minus signs

10. anonymous

oh in the chain rule? yes

11. anonymous

tripped me up the first time i saw it.

12. anonymous

:-) Now I get it, I just could not see why it is not 1-2 x+3 x^2-4 x^3+ and so on. but I see it now

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