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Math Help - how to repeated evalution of the recurrence relation

  1. #1
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    Question how to repeated evalution of the recurrence relation

    u_{r}=4u_{r}+3\\u_{1}=6
    Test the answer by finding the value of u_{4} in 2 different ways:
    by substitution in your solution and by repeated evaluation of the equation.

    by substitution I have found u2=27,u3=111 and u4=447
    I'm wondering how to find these values by repeated evaluation of the equation?
    Cheers


    Moderator edit: Please use tex tags NOT math tags. See Questions etc. subforum.
    Last edited by mr fantastic; May 2nd 2011 at 01:06 PM. Reason: Changed math to tex.
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  2. #2
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    Find the solution to the equation Ur+1= 4Ur + 3?
    U1 = 6

    Test your answer by finding the value of u4 in 2 different ways: by substitution in your solution and by repeated evaluation of the equation.

    sorry about the equation problem
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  3. #3
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    Quote Originally Posted by hazeleyes View Post
    u_{r}=4u_{r}+3\\u_{1}=6
    Test the answer by finding the value of u_{4} in 2 different ways:
    by substitution in your solution and by repeated evaluation of the equation.

    by substitution I have found u2=27,u3=111 and u4=447
    I'm wondering how to find these values by repeated evaluation of the equation?
    Cheers
    I think you are trying to solve the difference equation

    u_{r+1}-4u_{r}=3 I think there is a typo and you are missing the +1

    So to solve we do it in two parts:

    First we solve the homogenous equation

    u_{r+1}-4u_{r}=0

    we use the "guess"

    u_{r}=n^{r} \implies u_{r+1}=n^{r+1}

    Plugging this in gives

    n^{r+1}-4n^{r}=0 \iff n^{r}(n-4)=0

    this gives n=4

    so the solution has the form

    u_{r}=C4^{r}

    Since the right hand side is a constant we guess that the particular solution is also a constant

    u_r=C4^r+A

    C4^{r+1}+A-4(C4^r+A)=3 \iff A-4A=3 \iff A=-1

    So the solution is

    u_{r}=C4^r-1

    Now using the initial condition gives

    u_{1}=4C-1=6 \iff C=\frac{7}{4}

    So the solution is

    u_{r}=7(4)^{r-1}-1
    Now if we plug in

    r=4

    we get

    u_{4}=7\cdot 4^3-1=447
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  4. #4
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    Quote Originally Posted by hazeleyes View Post
    Find the solution to the equation Ur+1= 4Ur + 3?
    U1 = 6
    Look at this.
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  5. #5
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    thanks so much for your super clear explanation!!!
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