Question

Find the Taylor series of 1/(1-x)^2 about a=0

Answer 1

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

Answer 2

saubhik has it without finding derivatives, evaluating etc

Answer 3

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

Answer 4

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

Answer 5

oop i meant \[\frac{1}{1-x}\]

Answer 6

yeah I get that part

Answer 7

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

Answer 8

differentiate term by term to get the answer

Answer 9

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

Answer 10

oh in the chain rule? yes

Answer 11

tripped me up the first time i saw it.

Answer 12

:-) 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