Results 1 to 2 of 2

Thread: Understanding Series stuff

  1. #1
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591

    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 $\displaystyle a_{n}=t^n$ then I solve the equation $\displaystyle t^n = 3t^{n-1}-t^{n-2}$. then I was told that $\displaystyle a_{n}$ 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
    Attached Thumbnails Attached Thumbnails Understanding Series stuff-picture-2.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Given a sequence:$\displaystyle a_n$

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

    Note that $\displaystyle
    A'\left( x \right)
    $ is the exponential generating function of $\displaystyle
    a_{n + 1}

    $ (try differentiating)

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


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

    Therefore: $\displaystyle A(x)=
    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 }}
    {{n!}} \cdot x^n } \right)} + B \cdot \sum\limits_{n = 0}^\infty {\left( {\frac{{t_2 ^n }}
    {{n!}} \cdot x^n } \right)}
    $

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

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

    Read here to find out more about this topic: http://www.mathhelpforum.com/math-he...tionology.html
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Mar 30th 2010, 01:44 PM
  2. stuff
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: Jul 18th 2009, 02:36 PM
  3. Need help understanding some probability stuff..
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Mar 6th 2009, 09:07 AM
  4. Understanding Taylor Series.
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: Aug 6th 2008, 01:55 AM
  5. More log stuff
    Posted in the Math Topics Forum
    Replies: 18
    Last Post: May 12th 2008, 07:24 PM

Search Tags


/mathhelpforum @mathhelpforum