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Math Help - Finding a series ?

  1. #1
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    Finding a series ?

    Consider the series for 1/1+x = 1+x^2+.....+x^n (1)

    now use that to find the series for 1/1+x^2 (which is basically replace x with x^2

    which produces

    1/1+x^2 = 1+x^2+x^4+....+x^2n (2)


    and now find a series with 2x/1+x^2

    basically multiply (2) with 2x ....

    no my question is how do i find the series for

    ln(1+x^2) using those facts above me


    i used equation (2)

    1+x^2 = 1/(1+x^2+x^4+....+x^2n)
    now i log both sides

    ln(1+x^2) = ln(1/(1+x^2+x^4+....+x^2n)

    which gives me

    ln(1+x^2) = -ln(1+x^2+x^4+....+x^2n)

    is this correct ??

    or am i going no where with this proof ?

    Tanks
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  2. #2
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    Quote Originally Posted by Khonics89 View Post
    Consider the series for 1/1+x = 1+x^2+.....+x^n (1)

    now use that to find the series for 1/1+x^2 (which is basically replace x with x^2

    which produces

    1/1+x^2 = 1+x^2+x^4+....+x^2n (2)


    and now find a series with 2x/1+x^2

    basically multiply (2) with 2x ....

    no my question is how do i find the series for

    ln(1+x^2) using those facts above me


    i used equation (2)

    1+x^2 = 1/(1+x^2+x^4+....+x^2n)
    now i log both sides

    ln(1+x^2) = ln(1/(1+x^2+x^4+....+x^2n)

    which gives me

    ln(1+x^2) = -ln(1+x^2+x^4+....+x^2n)

    is this correct ??

    or am i going no where with this proof ?

    Tanks
    Note that \int \frac{2x}{1+x^2} \, dx = \ln (1 + x^2) + C ....
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  3. #3
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    Hmm a little confused... why do i need to integrate ?
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    Note that \int \frac{2x}{1+x^2} \, dx = \ln (1 + x^2) + C ....
    I had hoped from my first reply that you would realise that you need to integrate the series for \frac{2x}{1+x^2}. For reasons already explained the result is equal to \ln (1 + x^2) + C. To get the value of C, substitute x = 0 into the result (unsurprisingly you find C = 0).
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  5. #5
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    mr fantastic.. you are hopeless at explaining stuff you know that!
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  6. #6
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    Quote Originally Posted by Khonics89 View Post
    mr fantastic.. you are hopeless at explaining stuff you know that!
    My replies are not intended to be explanations. They are hints (and blunt hints at that). You're meant to think about what I posted and see how it relates to the question. Read the replies again and then, after you have applied appropriate thought, say what you don't understand if you're still stuck.
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  7. #7
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    Quote Originally Posted by Khonics89 View Post
    Consider the series for 1/1+x = 1+x^2+.....+x^n (1)

    now use that to find the series for 1/1+x^2 (which is basically replace x with x^2

    which produces

    1/1+x^2 = 1+x^2+x^4+....+x^2n (2)


    and now find a series with 2x/1+x^2

    basically multiply (2) with 2x ....

    no my question is how do i find the series for

    ln(1+x^2) using those facts above me


    i used equation (2)

    1+x^2 = 1/(1+x^2+x^4+....+x^2n)
    now i log both sides

    ln(1+x^2) = ln(1/(1+x^2+x^4+....+x^2n)

    which gives me

    ln(1+x^2) = -ln(1+x^2+x^4+....+x^2n)

    is this correct ??

    or am i going no where with this proof ?

    Tanks
    First the series are infinite series not finite series.

    Alternative method (not compliant with some of the restrictions you imply on method):

    \frac{1}{1+x}=\sum_{r=0}^{\infty}x^r,\ \ \ |x|<1

    Now:

    \frac{d}{dt}\ln(1+x)=\frac{1}{1+x}

    so:

    \ln(1+x)=\int_{\zeta=0}^x \frac{1}{1+\zeta}\; d\zeta=\int_{\zeta=0}^x \sum_{r=0}^{\infty}\zeta^r\; d\zeta

    Now interchange the integral and summations, do the integrals to get a series representation for \ln(1+x), ...

    CB
    Last edited by CaptainBlack; October 17th 2009 at 03:01 AM.
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