This question is probably very basic, but I've been away from school for a while and the answer eludes me.


I was tempted to prove that d/dx(e^x) = (e^x) for old times sake and that was easy enough. I just expressed e^x as a power series where n goes from 0 to infinity for ((x^n)/n!).


During the derivation I started to wonder, how did they know that a power series where n goes from 0 to infinity for A(subscript n)\*X^n would converge to the form of (someconstant)^x.


Is there some theorem that proves this?


This question is pretty probably trivial for the hardcore math types but it's been bothering me, so I thought I'd ask :-) ....



Why does the power series expressing e^x have the form of a constant raised to x ?
Some people would say that your question is trivial because we *define* the power function by a^x = exp(x log a).


However, that's not a very satisfying answer.


Clearly one wants the power series for exp(x) to satisfy exp(z+w) = exp(z) exp(w), and exp(0) = 1, from what we know the power function should be if z and w are integers. (I'm writing exp(x), not e^x, because I'm assuming exp(x) hasn't yet been shown to have this property.)


So say exp(x) = a0 + a1 x + a2 x^2 + a3 x^3 ... is a formal power series satisfying exp(z+w) = exp(z) exp(w).


Then since exp(0) = 1, we must have a0 = 0.


So exp(x) = 1 + a1 x + a2 x^2 + a3 x^3 + ...; therefore


exp(2x) = exp(x) exp(x) = (1 + a1 x + a2 x^2 + a3 x^3 + ...) (1 + a1 x + a2 x^2 + a3 x^3 + ...)


and expanding the rightmost member of this equation as a formal power series,


exp(2x) = 1 + 2a1 x + (2a2 + a1^2) + (2a3 + 2 a2 a1) x^3 + ...


However, exp(2x) = 1 + 2a1 x + 4a2 x^3 + 8a3 x^3 + ... by substituting 2x into the formal power series for exp(x).


By equating the coefficients of x, x^2, and x^3, you get


2 a1 = 2 a1
2 a2 + a1^2 = 4 a2
2 a3 + 2 a2 a1 = 8 a3


and so on. The first equation tells you nothing. The second tells you a1^2 = 2a2, so a2 = a1^2/2. The third becomes


2 a3 + 2 (a1^2 / 2) a1 = 8 a3


from which you get a1^3 = 6 a3, and a3 = a1^3/6. The pattern here continues, with an = a1^n/n!, as can be proven by induction.


This gives the series


exp(x) = 1 + a1 x + a1^2/2! x^2 + a1^3/3! x^3 + ...


and now we just have to choose a1. We pick 1 just because it's simple to do so.


