I was also perturbed by the jump from the definition of e to the taylor expansion. I know how to get there the long way (define e first, derive properties of the exponential derivative, then construct the Taylor series), does anyone know a shortcut?
it's a little ugly, because you have to have some strong conditions to use associativity on infinite series (and I forget what they are off the top of my head). Of course, this is true for splitting up the e^x into cos(x) and sin(x) as well.
The series has to be absolutely convergent. That is, you can rearrange the terms of \sum_{i=0}^\infty a_n freely if and only if \sum_{i=0}^\infty |a_n| converges. See http://en.wikipedia.org/wiki/Riemann_series_theorem
The Taylor series is actually the expansion of the limit in the line above. There's some trickery in proving that the limit converges, but you can derive one line from the other with some straightforward combinatorics.
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I was also perturbed by the jump from the definition of e to the taylor expansion. I know how to get there the long way (define e first, derive properties of the exponential derivative, then construct the Taylor series), does anyone know a shortcut?
Here is another way (a bit informal):
e = lim_{n->infinity} (1 + 1/n)^n
Now, apply the binomial theorem:
1 + n * 1/n + n! / (2 (n-2)! n^2) + ... + n! / (m! (n - m!) n^m) + ...
Now, for each m, we have this sequence:
a_n = n! / (m! (n - m)! n^m)
Which converges on 1/m!, so we are left with this:
1 + 1 + 1/2! + 1/3! + 1/4! + ...
it's a little ugly, because you have to have some strong conditions to use associativity on infinite series (and I forget what they are off the top of my head). Of course, this is true for splitting up the e^x into cos(x) and sin(x) as well.
The series has to be absolutely convergent. That is, you can rearrange the terms of \sum_{i=0}^\infty a_n freely if and only if \sum_{i=0}^\infty |a_n| converges. See http://en.wikipedia.org/wiki/Riemann_series_theorem
The Taylor series is actually the expansion of the limit in the line above. There's some trickery in proving that the limit converges, but you can derive one line from the other with some straightforward combinatorics.