anonymous
  • anonymous
Find the Taylor series of 1/(1-x)^2 about a=0
Mathematics
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anonymous
  • anonymous
Find the Taylor series of 1/(1-x)^2 about a=0
Mathematics
schrodinger
  • schrodinger
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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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