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Thread: Solve The Difference Equation

  1. #1
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    Solve The Difference Equation

    $\displaystyle f_{n+2} - 2f_{n+1}+ 5f_n = 0$

    given that $\displaystyle f_0=1$ and $\displaystyle f_1=3$

    Auxiliary equation:

    $\displaystyle m^2 - 2m + 5 = 0$

    Therefore:

    $\displaystyle m = 1 + 2i$ or $\displaystyle m = 1 - 2i$

    What would the complementry fuction be and paticular solution to work towards the solution of:

    $\displaystyle f_n = [(1-i)/2][1+2i]^n + [(1-i)/2][1-2i]^n$
    Last edited by CaptainBlack; Apr 29th 2009 at 08:43 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by louboutinlover View Post
    $\displaystyle f_{n+2} - 2f_{n+1}+ 5f_n = 0$

    given that $\displaystyle f_0=1$ and $\displaystyle f_1=3$

    Auxiliary equation:

    $\displaystyle m^2 - 2m + 5 = 0$
    Which is derived from an assumed solution of the form $\displaystyle f_n=m^n$

    Therefore:

    $\displaystyle m = 1 + 2i$ or $\displaystyle m = 1 - 2i$
    OK, so you have a general solution of the form:

    $\displaystyle f_n=A (1+2\text{i})^n+B(1-2\text{i})^n$

    and the initial conditions will allow you to determine $\displaystyle A$ and $\displaystyle B$, which you appear to have done.

    What would the complementry fuction be and paticular solution to work towards the solution of:

    $\displaystyle f_n = [(1-i)/2][1+2i]^n + [(1-i)/2][1-2i]^n$
    What are you asking here?

    CB
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  3. #3
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    So using the initial conditions would give:

    $\displaystyle A+B=1$

    $\displaystyle A(1+2i)+B(1-2i)=3$

    But then how do you solve these equations ?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by louboutinlover View Post
    So using the initial conditions would give:

    $\displaystyle A+B=1$

    $\displaystyle A(1+2i)+B(1-2i)=3$

    But then how do you solve these equations ?
    $\displaystyle A=1-B$ from 1st,

    $\displaystyle (1-B)(1+2\text{i})+B(1-2\text{i})=3$ by substituting from the above into the second.

    So:

    $\displaystyle
    B[-(1+2\text{i})+(1-2\text{i})]=3-(1+2\text{i})
    $

    $\displaystyle
    B=\frac{-1+1\text{i}}{2\text{i}}=\frac{1+i}{2}
    $

    Assuming my algebra is right (check it, it appears to differ from what you are give by a sign here or there).

    CB
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