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anonymous

  • 5 years ago

Find a power series representation for the function f(x)= 2/(1-2x). For what values of x does it equal the function?

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  1. anonymous
    • 5 years ago
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    You can take the Taylor Series Expansion, but I am not so sure what you mean by for what values of x does it equal the function. In theory all values of x should equal the function so long as the series converge.

  2. anonymous
    • 5 years ago
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    Taylor Expansion at center x=0 would be 2+4 x+8 x^2+16 x^3+32 x^4+64 x^5+..

  3. anonymous
    • 5 years ago
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    Hmm. Well the geometric progression a+ax^2+ax^3... converges to a/(1-x) when -1<x<1. So let x=ax. Then a+a(ax^2)+a(ax^3)... converges to a/(1-ax) for (-1/a)<x<(1/a). So i guess 2/(1-2x) would be 2+4x^2+8x^3... for (-1/2)<x<(1/2).

  4. anonymous
    • 5 years ago
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    Hmm...i think i get what you mean

  5. anonymous
    • 5 years ago
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    Should be a+ax+a(ax)^2+a(ax)^3... Missed the x term all over lol

  6. anonymous
    • 5 years ago
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    Sigh. Well the geometric progression a+ax+ax^2+ax^3... converges to a/(1-x) when -1<x<1. So let x=ax. Then a+a(ax)+a(ax)^2+a(ax)^3... converges to a/(1-ax) for (-1/a)<x<(1/a). So i guess 2/(1-2x) would be 2+4x+8x^2+16x^3... for (-1/2)<x<(1/2).

  7. anonymous
    • 5 years ago
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    so how did you come up w/ it converging to a/(1-x)

  8. anonymous
    • 5 years ago
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    Its a geometric series. It's derived without calculus. It goes: Let a=1+x+x^2+x^3... So ax=x+x^2+x^3... a-ax=1. So a=1/(1-x) I said it only converges for -1<x<1 because of the ratio test

  9. anonymous
    • 5 years ago
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    ok thanks so much! :)

  10. anonymous
    • 5 years ago
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    Np

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