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Math Help - Showing this equals that

  1. #1
    Senior Member chella182's Avatar
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    Showing this equals that

    Sorry for the vague title, it's not on a specific topic, it's a section in a question and differential equations and I've been told to show that...

    \frac{1}{n(1-\lambda n^{\alpha})}=\frac{1}{n}+\frac{\lambda n^{\alpha -1}}{1-\lambda n^{\alpha}}

    ...then I've got to integrate the RHS. I know how to integrate that, but I'm stuck on how to show LHS = RHS. Any ideas?
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  2. #2
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    Hello,
    Quote Originally Posted by chella182 View Post
    Sorry for the vague title, it's not on a specific topic, it's a section in a question and differential equations and I've been told to show that...

    \frac{1}{n(1-\lambda n^{\alpha})}=\frac{1}{n}+\frac{\lambda n^{\alpha -1}}{1-\lambda n^{\alpha}}

    ...then I've got to integrate the RHS. I know how to integrate that, but I'm stuck on how to show LHS = RHS. Any ideas?
    It's like a partial fractions decomposition :
    Solve for a and b in \frac{1}{n(1-\lambda n^\alpha)}=\frac an+\frac{b}{1-\lambda n^\alpha}
    (by putting on the same fraction)


    Otherwise, you can start from the RHS to prove the LHS.


    Another method is to note that 1=(1-\lambda n^\alpha)+\lambda n^\alpha
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  3. #3
    Senior Member chella182's Avatar
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    I tried partial fractions and got in a right mess with it :\.
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  4. #4
    Senior Member chella182's Avatar
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    Okay, proved the thing... I can't figure how to integrate

    \frac{\lambda n^{\alpha-1}}{1-\lambda n^{\alpha}}

    though :\.
    Last edited by chella182; February 28th 2009 at 12:05 PM.
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