# Series identity

1. Sep 9, 2009

### Easty

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

f $$_{a}$$ (z) is defined as

f(z) = 1 + az + $$\frac{a(a-1)}{2!}$$z$$^{2}$$+....+$$\frac{a(a-1)(a-2)...(a-n+1)}{n!}$$z$$^{n}$$ + ......

where a is constant

Show that for any a,b

f $$_{a+b}$$ (z)= f $$_{a}$$(z)f $$_{b}$$(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. Sep 9, 2009

### CompuChip

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 ;)

3. Sep 9, 2009

### HallsofIvy

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

Try using the product rule, extended to higher derivatives:
$$\frac{d^n fg}{d x^n}= \sum_{i=0}^n \left(\begin{array}{c}n \\ i\end{array}\right)\frac{d^{n-i}f}{dx}\frac{d^ig}{dx}$$

Last edited: Sep 10, 2009
4. Sep 10, 2009

### Easty

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.