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El_Arrow

  • one year ago

Help! find the taylor series for f(x) center at the given value

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  1. El_Arrow
    • one year ago
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    f(x)=9x^-2 and a=1

  2. El_Arrow
    • one year ago
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    i already got the series i just need the sum |dw:1437185617267:dw|

  3. El_Arrow
    • one year ago
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    @triciaal @Michele_Laino

  4. El_Arrow
    • one year ago
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    i got 9 - 18/1! (x-1) + 52/2! (x-1)^2 - 216/3! (x-1)^3 + 1080/4! (x-1)^4 - 6480/5! (x-1)^5

  5. El_Arrow
    • one year ago
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    @dumbcow

  6. El_Arrow
    • one year ago
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    @Concentrationalizing help with one more problem

  7. anonymous
    • one year ago
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    What do you mean? You have the series but need the sum? Saying you need a sum implies that there's a numerical answer when there isnt.

  8. anonymous
    • one year ago
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    You mean you need the nth term?

  9. El_Arrow
    • one year ago
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    yeah thats what i meant

  10. anonymous
    • one year ago
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    I would suggest doing what we did before, actually. Consider integrating the function and then applying the new center. \[\int\limits_{}^{}\frac{ 9 }{ x^{2} }dx = -\frac{ 9 }{ x }+C\] Now I want to shift this antiderivative. It would have been too difficult to do so while it was squared, though. So what we can do is write this in the form of a geo-series by doing this: \[-\frac{ 9 }{ x } + C = \frac{ 9 }{ 1-x-1 } = \frac{ 9 }{ 1-(x+1) } = 9\sum_{n=0}^{\infty} (x+1)^{n} + C\] Now proceed exactly like we did in the last problem. Since we'll be differentiating, we don't need to worry about the +C.

  11. El_Arrow
    • one year ago
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    okay so that would be (9x+1)^n right

  12. anonymous
    • one year ago
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    No, the 9 is just stuck outside. Not much you can do with it. Gotta head out, though. Good luck :)

  13. El_Arrow
    • one year ago
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    @iambatman could you please help?

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