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Math Help - need help with a/(1-r) form

  1. #1
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    need help with a/(1-r) form

    f(x)=\frac{4}{4+x^2}, c=0

    f(x)=\frac{4}{4(1+(\frac{x^2}{4}))} = \frac{1}{1+\frac{x^2}{4}} = \frac{1}{1-(\frac{-x^2}{4}))}

    \sum^ \infty_ {n=0} (\frac{-x^2}{4})^n = \frac{-1}{4}\sum^ \infty_ {n=0} (x^2)^n

    If I did this right, where do I go from here?? Thank you.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by saiyanmx89 View Post
    f(x)=\frac{4}{4+x^2}, c=0

    f(x)=\frac{4}{4(1+(\frac{x^2}{4}))} = \frac{1}{1+\frac{x^2}{4}} = \frac{1}{1-(\frac{-x^2}{4}))}

    \sum^ \infty_ {n=0} (\frac{-x^2}{4})^n = \frac{-1}{4}\sum^ \infty_ {n=0} (x^2)^n

    If I did this right, where do I go from here?? Thank you.
    Almost

    \frac{4}{4+x^2}=\frac{1}{1+\left( \frac{x}{2}\right)^2}=...
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  3. #3
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    Quote Originally Posted by saiyanmx89 View Post
    f(x)=\frac{4}{4+x^2}, c=0

    f(x)=\frac{4}{4(1+(\frac{x^2}{4}))} = \frac{1}{1+\frac{x^2}{4}} = \frac{1}{1-(\frac{-x^2}{4}))}

    \sum^ \infty_ {n=0} (\frac{-x^2}{4})^n = \frac{-1}{4}\sum^ \infty_ {n=0} (x^2)^n
    You need to stop with \color{blue}\sum^ \infty_ {n=0} (\frac{-x^2}{4})^n.
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  4. #4
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    you mean I didn't need to go any further with it?
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  5. #5
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    Quote Originally Posted by saiyanmx89 View Post
    you mean I didn't need to go any further with it?
    No, I meant that you cannot factor the \frac{-1}{4} out of the sum!
    Do you see why?
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  6. #6
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    oh yea. the x^2, which is why I must use= (x/2)^2

    But, where do I go from here:
    \sum^ \infty_ {n=0} (\frac{-x}{2})^{2n} = ??
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by saiyanmx89 View Post
    oh yea. the x^2, which is why I must use= (x/2)^2
    well, you could distribute the power. what do you think it would look like then?
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  8. #8
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    could I use the Ratio Test on \frac{-x^{2n}}{4^n}
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    Quote Originally Posted by saiyanmx89 View Post
    could I use the Ratio Test on \frac{-x^{2n}}{4^n}
    review how to distribute powers.

    Spoiler:
    \sum_{n = 0}^\infty (-1)^n \frac {x^{2n}}{4^n}

    this is to highlight the fact it is an alternating series
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  10. #10
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    so, by using the Alternating Series, I can take the absolute value of the function leaving me with (x^2/4) or (x/4) ??? I'm not sure which...
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  11. #11
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    Quote Originally Posted by saiyanmx89 View Post
    so, by using the Alternating Series, I can take the absolute value of the function leaving me with (x^2/4) or (x/4) ??? I'm not sure which...
    Are you not making too much of this? Over think it?
    The common ratio is \frac{-x^2}{4}.
    So what is the absolute value?
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  12. #12
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    x^2/4
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