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Thread: Solving Differential Equation - series solution

  1. #1
    Sep 2009

    Solving Differential Equation - series solution

    Hi all;
    Let $\displaystyle a_{1}=a_{3}=0 $ and $\displaystyle a_{0} $\neq$0 $and $\displaystyle a_{2} $\neq$0$

    and $\displaystyle a_{n} =\frac{a_{n-4}}{n(n-2)} ; n\geq4$

    I need the general formula for the following summation in terms of $\displaystyle a_{0}$ and $\displaystyle a_{2}$:
    $\displaystyle y=\sum_{i=1}^{\infty}a_{n}x^n$

    Last edited by mr fantastic; Jul 2nd 2010 at 01:26 AM. Reason: Added to title.
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  2. #2
    MHF Contributor

    Apr 2005
    Start by writing out some values and looking for a pattern:
    $\displaystyle a_0$
    $\displaystyle a_1= 0$
    $\displaystyle a_2$
    $\displaystyle a_3= 0$
    $\displaystyle a_4= \frac{a_0}{4(2)}$
    $\displaystyle a_5= \frac{a_1}{5(3)}= 0$
    $\displaystyle a_6= \frac{a_2}{6(4)}$
    $\displaystyle a_7= \frac{a_3}{7(5)}= 0$
    $\displaystyle a_8= \frac{a_4}{8(6)}= \frac{a_0}{8(6)(4)(2)}$
    $\displaystyle a_9= \frac{a_5}{9(8)}= 0$
    $\displaystyle a_{10}= \frac{a_6}{10(8)}= \frac{a_2}{10(8)(6)(4)}$
    $\displaystyle a_{11}= \frac{a_7}{11(10)}= 0$.

    It should be clear that if n is odd, then $\displaystyle a_n= 0$.

    It should also be clear that if n is a multiple of 4, we have $\displaystyle a_n$ equal to $\displaystyle a_0$ divided by a product of even numbers from n down to 2: if n= 4i, then $\displaystyle a_{4i}= \frac{a_0}{(4i)(4i- 2)(4i- 4)\cdot\cdot\cdot (6)(4)(2)}$ and we can treat that product as a factorial by factoring out "2":$\displaystyle a_{4i}= \frac{a_0}{[2(2i)][2(2i-1)][2(2i-2)]\cdot\cdot\cdot[2(3)][2(2)][2(1)]}= \frac{a_0}{2^{2i} (2i)!}$

    If n is even but not a multiple of 4, n= 4i+ 2, then it is almost exactly the same except that the numerator is $\displaystyle a_2$, of course, and there is no "2" in the product in the denominator. We can fix that by taking a "2" out of the denominator: by multplying by 2.
    $\displaystyle a_{4n+2}= \frac{a_2}{2^{2i-1}(2i)!}$

    The whole sum can then be written as two separate sums:
    $\displaystyle y= \sum_{i=0}^\infty \frac{a_0}{2^{2i}(2i)!} x^{4i}+ a_2\sum_{i=1}^\infty \frac{1}{2^{2i-1}(2i)!} x^{4i+2}$

    For future reference, a product of odd numbers can be written by multiplying and dividing by the even numbers:
    $\displaystyle (2n+1)(2n-1)\cdot\cdot\cdot(5)(3)(1)= \frac{(2n+1)(2n)(2n-1)(2n-2)\cdot\cdot\cdot(5)(4)(3)(2)(1)}{(2n)(2n-2)\cdot\cdot\cdot(4)(2)}$
    and then treating the product of even numbers in the denominator as above:
    $\displaystyle (2n+1)(2n-1)\cdot\cdot\cdot(5)(3)(1)= \frac}(2n+1)!}{2^n(n!)}$
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