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Series identity

  1. Sep 9, 2009 #1
    1. The problem statement, all variables and given/known data[/b]

    f [tex]_{a}[/tex] (z) is defined as

    f(z) = 1 + az + [tex]\frac{a(a-1)}{2!}[/tex]z[tex]^{2}[/tex]+....+[tex]\frac{a(a-1)(a-2)...(a-n+1)}{n!}[/tex]z[tex]^{n}[/tex] + ......

    where a is constant

    Show that for any a,b

    f [tex]_{a+b}[/tex] (z)= f [tex]_{a}[/tex](z)f [tex]_{b}[/tex](z)

    2. Relevant equations

    3. The attempt at a solution

    I've tried starting directly from f_a+f_b and trying to show it is equivalent to f_ab and vice versa but i keep getting stuck with the last general term, im thinking there is a better way to approach this question but i cant see it.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Sep 9, 2009
  2. jcsd
  3. Sep 9, 2009 #2


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    Homework Helper

    Is n a fixed number, or is it an infinite series?
    Perhaps it helps if you write the numerators in terms of factorials as well, perhaps you will even recognize some binomial coefficients ;)
  4. Sep 9, 2009 #3


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    Or look at the derivatives: fa(0)= 1, fa'(0)= a, fa"(0)= a(a-1) and, in general
    [tex]\frac{d^n f^a}{dx^n} (0)= a(a-1)\cdot\cdot\cdot (a-n+1)[/tex]
    and of course,
    [tex]\frac{d^n f^b}{dx^n} (0)= b(b-1)\cdot\cdot\cdot (b-n+1)[/tex]

    Try using the product rule, extended to higher derivatives:
    [tex]\frac{d^n fg}{d x^n}= \sum_{i=0}^n \left(\begin{array}{c}n \\ i\end{array}\right)\frac{d^{n-i}f}{dx}\frac{d^ig}{dx}[/tex]
    Last edited by a moderator: Sep 10, 2009
  5. Sep 10, 2009 #4
    Thank you Hallsofivy. Once i took your advice the answer was quite simple to obtain, it was a nice way to approach the problem that i would have never seen.
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