User:Scm

DE implies sum of series and vice versa
Differentiating . $e^x := \displaystyle \sum_{n = 0}^\infty \frac {x^n} {n!}$ using Differentation of uniformly convergent series and Powerseries are uniformly convergent (and maybe Weierstrass M-Test) we get
 * $\displaystyle \sum_{n = 1}^\infty \frac {x^{n-1}} {(n-1)!}$

which after adjusting the index, by replacing $n$ by $n+1$ is the same. Evaluating in $0$ yields $1$, hence the series solves the DE and by uniqueness has to be equal to $\exp(x)$.

For the converse again oberve that the series solves the DE and hence must be equal to the unique solution. Use Uniquess of solutions for Differential Equation, to see that the equation has only one solution.


 * I considered doing that, but I don't know how to adjust indices. Wanna teach me? --GFauxPas 17:06, 9 February 2012 (EST)
 * Adjustment of Indices and its related result Permutation of Indices are results still missing from this site. I keep meaning to get to them, but formulating them neatly and concisely takes work and thought (both of which I like to postpone). --prime mover 17:34, 9 February 2012 (EST)