Re: Proof of the existence of a function f such the f(x)=f'(x)

The definition of $\displaystyle e^{x}$ as the solution of the 'initial value problem'...

$\displaystyle y^{'}= y\ ,\ y(0)=1$ (1)

... is [in principle...] perfecly acceptable. The 'initial condition' guarantees that $\displaystyle y(x)$ is analytic in $\displaystyle x=0$ , so that we can write...

$\displaystyle y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n}$ (2)

From (1) we can easily derive that $\displaystyle \forall n\ y^{(n)} (0)=1$ , so that is...

$\displaystyle y(x)= e^{x}= \sum_{n=0}^{\infty} \frac{x^{n}}{n!}$ (3)

The 'more traditional' alternative is to define...

$\displaystyle e^{x}= \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n}$ (4)

... and from (4) is possible to derive all the properties of $\displaystyle e^{x}$ , including the fundamental identity...

$\displaystyle \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n} = \lim_{n \rightarrow \infty}\sum_{k=0}^{n} \frac{x^{k}}{k!}$ (5)

... so that defintions (3) and (4) are fully equivalent...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$