anonymous
  • anonymous
Find the Taylor series of 1/(1-x)^2 about a=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+O(x^6)
anonymous
  • anonymous
saubhik has it without finding derivatives, evaluating etc
anonymous
  • anonymous
just by noting that \[\frac{1}{(1-x)^2}\] is the derivative of \[\frac{1}{1-x\]

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anonymous
  • anonymous
why dont you get alternating signs? when you differentiate 1/(1-x)^2 you get - sign
anonymous
  • anonymous
oop i meant \[\frac{1}{1-x}\]
anonymous
  • anonymous
yeah I get that part
anonymous
  • anonymous
the well known formula for summing a geometric sequence \[\frac{1}{1-x}=1+x+x^2+x^3+...\]
anonymous
  • anonymous
differentiate term by term to get the answer
anonymous
  • anonymous
oh hell, I see... there are 2 minus signs
anonymous
  • anonymous
oh in the chain rule? yes
anonymous
  • anonymous
tripped me up the first time i saw it.
anonymous
  • 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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