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Math Help - Generating Functions

  1. #1
    Junior Member
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    Generating Functions

    I had a test I took and my work or steps are right, but final answer is wrong and I can't figure out the error. I was given the generating function

    g(x)=\frac{5+2x}{(1-5x)(1+4x)}

    My function or formula is

    h_{n}=2(5^{n})+3(-4)^{n}

    And my recurrence I got was
    h_{n}=h_{n-1}+20h_{n-2}; h_{0}=5, h_{1}=7

    The only problem is my recurrence and function don't spit the same values out. Can anyone help me and tell me which one is wrong.
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  2. #2
    Super Member PaulRS's Avatar
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    This is what I get:
    \frac{1}{(1-5x)\cdot{(1+4x)}}=\sum_{n=0}^{\infty}{\left(\sum_{  k=0}^n{5^k\cdot{(-4)^{n-k}}}\right)\cdot{x^n}}

    \sum_{k=0}^n{5^k\cdot{(-4)^{n-k}}}=<br />
\left( { - 4} \right)^n  \cdot \sum\limits_{k = 0}^n {\left( { - \tfrac{5}<br />
{4}} \right)^k }  = \left( { - 4} \right)^n  \cdot \frac{{1 - \left( { - \tfrac{5}<br />
{4}} \right)^{n + 1} }}<br />
{{1 + \tfrac{5}<br />
{4}}}<br />
 = \left( { - 4} \right)^n  \cdot 4 \cdot \frac{{1 - \left( { - \tfrac{5}<br />
{4}} \right)^{n + 1} }}<br />
{9}<br /> <br /> <br />

    Thus: <br />
\frac{1}<br />
{{\left( {1 - 5x} \right) \cdot \left( {1 + 4 \cdot x} \right)}} = \sum\limits_{n = 0}^\infty  {\left( {\tfrac{{\left( { - 4} \right)^n  \cdot 4 + 5^{n + 1} }}<br />
{9}} \right) \cdot } x^n <br /> <br />

    Therefore: <br />
\frac{{2x + 5}}<br />
{{\left( {1 - 5x} \right) \cdot \left( {1 + 4 \cdot x} \right)}} = 5 + \sum\limits_{n = 1}^\infty  {\left( {2 \cdot \tfrac{{\left( { - 4} \right)^{n - 1}  \cdot 4 + 5^n }}<br />
{9} + 5 \cdot \tfrac{{\left( { - 4} \right)^n  \cdot 4 + 5^{n + 1} }}<br />
{9}} \right) \cdot } x^n <br /> <br /> <br />

    I've got to go now
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  3. #3
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    \frac{5+2x}{(1-5x)(1+4x)}=\frac{2}{1-5x}+\frac{3}{1+4x}

    \frac{2}{1-5x} = 2*[1+ 5x +5^{2}x^{2} +5^{3}x^{3}+..+5^{n}x^{n}] = 2*5^{n}


    \frac{3}{1+4x}=3*[1 + (-4x) + (-4x)^{2} +...+(-4)^{n}x^{n}]= 3*(-4)^{n}

    So the formula is

    h_{n}=2*5^{n} + 3*(-4)^{n}

    I think that's the formula, not what you posted.
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  4. #4
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Jrb599 View Post
    \frac{5+2x}{(1-5x)(1+4x)}=\frac{2}{1-5x}+\frac{3}{1+4x}
    Actually it is: \frac{5+2x}{(1-5x)(1+4x)}=\frac{3}{1-5x}+\frac{2}{1+4x}

    So the formula is: h_n=3\cdot{5^n}+2\cdot{(-4)^n}

    The formula I posted is ok (it just doesn't look nice because it wasn't simplified)
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  5. #5
    Super Member PaulRS's Avatar
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    Quote Originally Posted by PaulRS View Post
    the formula is: h_n=3\cdot{5^n}+2\cdot{(-4)^n}
    Thus the corresponding equation is: 0=(x-5)\cdot{(x+4)}=x^2-x-20 => x^2=20+x

    And h_{n+2}=20\cdot{h_n}+h_{n+1} with h_0=5 and h_1=7 so the recurrence you got was ok
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