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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+O(x^6)

  2. anonymous
    • 5 years ago
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    saubhik has it without finding derivatives, evaluating etc

  3. anonymous
    • 5 years ago
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    just by noting that \[\frac{1}{(1-x)^2}\] is the derivative of \[\frac{1}{1-x\]

  4. anonymous
    • 5 years ago
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    why dont you get alternating signs? when you differentiate 1/(1-x)^2 you get - sign

  5. anonymous
    • 5 years ago
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    oop i meant \[\frac{1}{1-x}\]

  6. anonymous
    • 5 years ago
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    yeah I get that part

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

  8. anonymous
    • 5 years ago
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    differentiate term by term to get the answer

  9. anonymous
    • 5 years ago
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    oh hell, I see... there are 2 minus signs

  10. anonymous
    • 5 years ago
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    oh in the chain rule? yes

  11. anonymous
    • 5 years ago
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    tripped me up the first time i saw it.

  12. anonymous
    • 5 years ago
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    :-) 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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