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El_Arrow

  • one year ago

Help! I need to find a power series representation for the function

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  1. El_Arrow
    • one year ago
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    f(x)=x^2/(1-7x)^2

  2. El_Arrow
    • one year ago
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    @Michele_Laino @Concentrationalizing @jim_thompson5910

  3. El_Arrow
    • one year ago
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    this is what i have so far |dw:1437181408215:dw|

  4. El_Arrow
    • one year ago
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    dont know if its right or not

  5. anonymous
    • one year ago
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    That's not correct. You cannot use the geometric series on that since you still have the denominator squared. You would have to eliminate that squared denominator before you go to that.

  6. El_Arrow
    • one year ago
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    how do i do that?

  7. anonymous
    • one year ago
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    You can ignore the \(x^{2}\) and integrate \(\frac{1}{(1-7x)^{2}}\)

  8. El_Arrow
    • one year ago
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    so have |dw:1437182143406:dw|

  9. El_Arrow
    • one year ago
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    what do i do after that?

  10. anonymous
    • one year ago
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    Thats what you got for the power series representation of the antiderivative?

  11. El_Arrow
    • one year ago
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    oh no

  12. anonymous
    • one year ago
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    Yeah, basically what I'm having you do is: \[\int\limits_{}^{}\frac{ 1 }{ (1-7x)^{2} }dx = \sum_{n=0}^{\infty}??\]

  13. El_Arrow
    • one year ago
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    i got 1/7 ln(1-7x)

  14. anonymous
    • one year ago
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    Shouldn't have ln in your answer.

  15. El_Arrow
    • one year ago
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    |dw:1437182482351:dw|

  16. anonymous
    • one year ago
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    Right, there ya go. Now of course you need + C, but that won't be an issue with what we are doing. So what is the power series representation of that function?

  17. El_Arrow
    • one year ago
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    would it be |dw:1437182579940:dw|

  18. anonymous
    • one year ago
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    Yep. I like to separate everything with power series, though, makes it easier to see simplification a lot of the time. So I would have it written like: \[\sum_{n=0}^{\infty}7^{n-1}x^{n}\] Okay, so this series is equal to the antiderivative of what we started with. SO we want to go backwards and get back to where we started, which means I need to differentiate my result. You've seen how to differentiate and integrate series?

  19. El_Arrow
    • one year ago
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    no

  20. El_Arrow
    • one year ago
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    so this is the power series representation |dw:1437182937373:dw|

  21. El_Arrow
    • one year ago
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    so we need to find the radius now

  22. anonymous
    • one year ago
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    Well, you essentially take the derivative as normal, you worry about the variable terms while constants just stay as is. So kind of informally, the derivative would be something like this: \[\sum_{n=0}^{\infty}7^{n-1}\frac{ d }{ dx }(x^{n})\] Where n is a constant. So what's rhe derivative of \(x^{n}\)?

  23. El_Arrow
    • one year ago
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    nx^n

  24. El_Arrow
    • one year ago
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    right?

  25. anonymous
    • one year ago
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    \(nx^{n-1}\)

  26. El_Arrow
    • one year ago
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    why minus 1?

  27. El_Arrow
    • one year ago
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    oh nvm

  28. anonymous
    • one year ago
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    But one thing we must also do is raise the index from 0 to 1. This is because, as is, letting the first n be 0 would make the series 0, so we fix that by raising the index by one. Thus taking the derivative would give us: \[\sum_{n=1}^{\infty}7^{n-1}nx^{n-1}\] Now just multiply in the \(x^{2}\)

  29. El_Arrow
    • one year ago
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    like this |dw:1437183623856:dw|

  30. El_Arrow
    • one year ago
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    cause the x^-1 and x^2 add each other right?

  31. anonymous
    • one year ago
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    Yep. Just have to make the index 1 on your series.

  32. El_Arrow
    • one year ago
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    so using the power series representation how do i determine its radius of convergence?

  33. El_Arrow
    • one year ago
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    @dumbcow is the radius 1/7?

  34. dumbcow
    • one year ago
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    yes the radius of convergence is 1/7

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