e^x = x⁰/0! + x¹/1! + x²/2! + x³/3! + x⁴/4! + x⁵/5! + ...

     That's the assertion, but why would we assert that? Where does it come from?

     It all comes from something called the Taylor series, which might be worth going over by itself, but to summarize, we assume that,

     f(x) = e^x = c₀ + c₁(x-a) + c₂(x-a)² + c₃(x-a)³ + ...

     Where c_n represents the n^th element in a sequence of constants and "a" is a constant.

     If we set x=a then we have,

     f(a) = e^a = c₀

     Likewise we can derive the following:

     f'(x) = c₁ + 2c₂(x-a) + 3c₃(x-a)² + 4c₄(x-a)³ + ...
     f'(a) = e^a = c₁

     f''(x) = 2c₂ + (2 * 3)c₃(x-a) + (3 * 4)c₄(x-a)² + ...
     f''(a)/2 = e^a/2 = c₂

     f'''(x) = (2 * 3)c₃ + (2 * 3 * 4)c₄(x-a) + ...
     f'''(a)/(2 * 3) = e^a/(2 * 3) = c₃

     From this we can deduce,

     c_n = f⁽ⁿ⁾(a)/(n!) where f⁽ⁿ⁾ refers to the n^th derivative of the function "f".
     c_n = e^a/(n!) because f(x) = e^x and e^x is it's own derivative.

     Using this definition for c_n we can say the following:

     f(x) = e^x = c₀ + c₁(x-a) + c₂(x-a)² + c₃(x-a)³ + ...
     f(x) = e^x = e^a + e^a(x-a) + (e^a/2)(x-a)² + (e^a/(3!))(x-a)³ + ...

     This equation, of course is easiest to deal with if we use a = 0, because then x - a = x, but even more importantly e^a = e⁰ = 1. Thus we simplify and obtain:

     f(x) = e^x = 1 + x + (1/2)x² + (1/(3!))x³ + ...
     f(x) = e^x = 1 + x + x²/2 + x³/(3!) + ...

What's It Good For?

ex	Calculating the Trig Functions
	Calculating xi
	Calculating ln(i)