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A nonlinear difference equation

  1. Jun 2, 2012 #1
    Consider the equation
    [tex]\frac{a_n-a_{n-1}}{1+a_na_{n-1}}=\frac{1}{2n^2}[/tex]
    I know one special solution is
    [tex]a_n=\frac{n}{n+1}[/tex]
    But how to solve and find the general solution?

    Thanks in advance.
     
  2. jcsd
  3. Jun 3, 2012 #2

    tiny-tim

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    hi asmani! :smile:

    hint: trig substitution :wink:
     
  4. Jun 3, 2012 #3
    Thank you tiny-tim. I don't think that works, since the the equation is actually coming from trigonometry!

    Mathematica gives the following solution:

    [tex]a_n=\frac{a_0+\left (1+a_0 \right )n}{1+\left ( 1-a_0 \right )n}[/tex]

    But I don't know how to derive this solution analytically.

    P.S. The original problem was to show that:

    [tex]\sum_{n=1}^{\infty}\tan^{-1}\left (\frac{1}{2n^2} \right )=\frac{\pi}{4}[/tex]
     
  5. Jun 3, 2012 #4

    tiny-tim

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    yes it does!!

    try it! :smile:
     
  6. Jun 3, 2012 #5
    [itex]a_n=\tan\theta_n [/itex]? Another hint please!
     
  7. Jun 3, 2012 #6

    tiny-tim

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    yes!! :biggrin:

    so the LHS of the original equation is … ? :smile:
     
  8. Jun 3, 2012 #7
    I think I get it, Thanks a lot.

    [tex]\tan \left (\theta_n-\theta_{n-1} \right )=\frac{1}{2n^2}[/tex]
    and then
    [tex]\theta_n =\tan^{-1}\left (\frac{1}{2n^2} \right )+\theta_{n-1}[/tex]
    and then
    [tex]\theta_n =\theta_{0}+\sum_{k=1}^{n}\tan^{-1}\left (\frac{1}{2k^2} \right )[/tex]
    By knowing that special solution mentioned in post #1, I can derive the formula in post #3. What If I didn't know that special solution?
     
    Last edited: Jun 3, 2012
  9. Jun 3, 2012 #8

    tiny-tim

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    hmm :confused:

    putting ao = 0 or 1 gives an = 1 + 2n or -1/(1 + 2n)

    and putting ao = ±i looks interesting :rolleyes: (i haven't followed it through :redface:) …

    do either of those help? :smile:
     
  10. Jun 3, 2012 #9
    Putting a0=0 in which equation?
     
  11. Jun 3, 2012 #10

    tiny-tim

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    Mathematica's …
    [tex]a_n=\frac{a_0+\left (1+a_0 \right )n}{1+\left ( 1-a_0 \right )n}[/tex]
     
  12. Jun 3, 2012 #11
    OK. So far, first I 'guessed' a special solution, then I derived the general solution by using that special solution. Is there any analytic way to find (not guess) a special solution?
     
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