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Math Help - Understanding Series stuff

  1. #1
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    Understanding Series stuff

    Now i have managed to solve the attached problem using some instructions given to me. But i really do not understand how this works.

    Under what premise can I assume that a_{n}=t^n then I solve the equation t^n = 3t^{n-1}-t^{n-2}. then I was told that a_{n} is a linear combination of the roots.

    I solve this question fine, I just really do not understand any of the theory.

    Could someone show me some reading material or a proof of this method.

    Many Thanks

    Bobak
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  2. #2
    Super Member PaulRS's Avatar
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    Given a sequence: a_n

    Let: <br />
 A\left( x \right) = \sum\limits_{n = 0}^\infty  {\left( {\frac{{a_n }}<br />
{{n!}} \cdot x^n } \right)} <br />
    (this is called the exponential generating function of the sequence)

    Note that <br />
A'\left( x \right)<br />
is the exponential generating function of <br />
a_{n + 1} <br /> <br />
(try differentiating)

    So if we have the recurrence equation: <br />
a_{n + 2}  = c_1  \cdot a_{n + 1}  + c_2  \cdot a_n <br />
we must have: <br />
A''\left( x \right) = c_1  \cdot A'\left( x \right) + c_2  \cdot A\left( x \right)<br />


    Suppose there are two different solutions to this equation: t^2-c_1\cdot{t}-c_2=0(1) (maybe complex)
    Then the general solution to this differential equation is: <br />
A\left( x \right) = A \cdot e^{t_1  \cdot x}  + B \cdot e^{t_1  \cdot x} <br />
(for some constants A and B that depend on the initial conditions) where t_1 and t_2 are the roots of the equation (1)

    Therefore: A(x)=<br />
A \cdot e^{t_1  \cdot x}  + B \cdot e^{t_1  \cdot x}  = A \cdot \sum\limits_{n = 0}^\infty  {\left( {\frac{{t_1 ^n }}<br />
{{n!}} \cdot x^n } \right)}  + B \cdot \sum\limits_{n = 0}^\infty  {\left( {\frac{{t_2 ^n }}<br />
{{n!}} \cdot x^n } \right)} <br />

    Thus: <br />
A\left( x \right) = \sum\limits_{n = 0}^\infty  {\left( {\frac{{A \cdot t_1 ^n  + B \cdot t_2 ^n }}<br />
{{n!}} \cdot x^n } \right)} <br />

    And we finally get <br />
A \cdot t_1 ^n  + B \cdot t_2 ^n  = a_n <br />
(because of the uniqueness of the power series expansion of a function)

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