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Help with this difference equation

  1. Jul 21, 2006 #1
    Let be the equation:

    [tex] A(2n)-A(2n-1)=3 [/tex]

    i,m a bit stuck..i don,t know how to solve it :frown: :frown: it it weren't for the 3 term i would try [tex] A(n)=r^n [/tex] where r is an unknown number..however the 3 factor spoils all..also we could try the identity:

    [tex] \sum_{n=1}^{k}A(n)-A(n-1)=3k=A(k)-A(1) [/tex] but it seems not to work..i,m really messed up with this nasty equation..:grumpy: :grumpy:
     
  2. jcsd
  3. Jul 21, 2006 #2

    arildno

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    Try
    [tex]A(n)=an^{2}+bn+c[/tex]
    Determine what a,b,c must be
    EDIT:
    See latter post.
     
    Last edited: Jul 21, 2006
  4. Jul 21, 2006 #3
    Is n required to be an integer? If so, I don't think you have enough information. Well, for starters you'd need to know what A(0) is, or at least what A(n) is for some n, otherwise the solution will have a constant at the end.

    But if n is required to be an integer, you only know the difference between an even and the odd directly below it. You'd need another equation to find the difference between an odd and the even below it, or between adjacent odds or evens.
     
  5. Jul 21, 2006 #4

    arildno

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    "between adjacent odds or evens" Eeh? :confused:

    eljose:
    Your system of equations is DECOUPLED; in each equation, only two unknowns appear, neither of which appears anywhere else.

    What follows from this?
     
    Last edited: Jul 21, 2006
  6. Jul 22, 2006 #5

    HallsofIvy

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    This is a "non-homogeous, linear difference equation with constant coefficients". It can be treated exactly like you would a similar differential equation: find the general solution to the associated homogeneous equation, A(2n)- A(2n-1)= 0, then add any one solution to the entire equation. The general solution is A(n)= C+ 3n where C is any constant.
     
  7. Jul 22, 2006 #6

    arildno

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    No, HallsofIvy!
    You have too few equations here, each unknown member of the sequence appears (in pairs) in only one equation.

    Thus, the general solution is:
    [tex]C_{2n-1}=b_{n}, C_{2n}=3+C_{2n-1}, n=1,2\cdots,[/tex]
    where [itex]b_{n}[/itex] is an arbitrary sequence.
     
  8. Jul 22, 2006 #7

    HallsofIvy

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    So you are saying that my solution, An= C+ 3n is does satisfy the equation but is not the general solution?

    If we are given bn= 1, 3, 7 , 5, 9, ... then
    A1= 1, A2= 3+ 1= 4, A3= 3, A4= 3+ 3= 6, A5= 7, A6= 3+ 7= 10, ...
    Your point, then, is that we don't need to worry about the fact that
    A5- A4= 7- 6 is not 3, because the equation
    A2n- A2n-1 requires that the first index always be even. Okay, that's clear. Thanks.
     
  9. Jul 22, 2006 #8

    arildno

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    It was BoTemp who was the first to point this out; I stepped into the same "trap" of indicating a linear solution as the gen. solution.
     
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