Series Identity: Showing f_(a+b) is Equivalent to f_(a)f_(b)

[Easty]

1. Homework Statement [/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)

Homework Equations

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, I am thinking there is a better way to approach this question but i can't see it.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[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 ;)

 

[HallsofIvy]

Or look at the derivatives: f^a(0)= 1, f^a'(0)= a, f^a"(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:
$$\frac{d^n fg}{d x^n}= \sum_{i=0}^n \left(\begin{array}{c}n \\ i\end{array}\right)\frac{d^{n-i}f}{dx}\frac{d^ig}{dx}$$

 

[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.