How do you prove only a Taylor series converges to a function without knowing the original function?

Question

Suppose you only had the Taylor series for $e^{x}$ but you didn't know what function it was derived from, $$ \sum_{k=0}^{ \infty} \frac{x^{k}}{k!}.$$ How do you prove this series converges to $e^{x}$ without relying on the preconceived process of taking derivatives of $ e^{x}$?

Answer 1

I believe knowing the taylor expansions of $e^x$ and other functions such as $sin(x)$ an $cos(x)$ are usually used when trying to identify a taylor series as reference. But in reference to this particular one I think the simplest way to show that $$\sum^\infty_{k=0}\frac{x^k}{k!} = e^x$$ is by showing that the derivative of the sum is the original, like so $$\frac{d}{dx}\sum^\infty_{k=0}\frac{x^k}{k!} = \sum^\infty_{k=1}\frac{x^{k-1}}{(k-1)!} = \sum^\infty_{k=0}\frac{x^k}{k!}$$ And so then it must be $Ce^x$ for some $C$, from here it could be shown that $C=1$ by showing that no fractional term can be simplified further.

Answer 2

It depends on what you mean by $e^x$ in the first place (some people define $e^x$ by as its Taylor series, and prove its properties later. I dislike this approach).

If you define $e$ as $\lim_{n\to \infty} \left( 1 + \frac{1}{n}\right)^n$ and the exponential through some continuity argument, then you can show that any function with the property $$ f(x+y) = f(x) f(y) $$ must be an exponential function. To find the base you need only find $f(1)$, or $\sum \frac{1}{k!}$ in this case. Then, show this is equal to the limit definition of $e$.

On the other hand, I like to define $e^x$ as the fixed point of the derivative, that is $$\frac{d}{dx} e^x = e^x.$$ In fact, $e^x$ is the unique function which satisfies this (well, technically $c e^x$ for some constant $c$, but again evaluate at $1$ and you get $ce$ so). If you know(/prove) this, then all you need to do is differentiate the power series and notice it's the same function. This is more complicated than it might seem, you aren't allowed to differentiate power series term by term: $$ \frac{d}{dx} \sum f_k(x) = \sum \frac{d}{dx} f_k(x) $$ generally only works for a finite sum. Fortunately, this series is uniformly convergent everywhere and for such series you can differentiate term by term (actually there is another sublety here, the series must converge somewhere, but this series rather obviously converges when $x$ is $0$).

Answer 3

Let us define $e^x$ to be the unique solution to $f' = f$ with $f(0)=1.$ Then it's easy to show that $\sum_{n=0}^{\infty} \frac{1}{n!} x^n = e^x$:

Let $f(x) = \sum_{n=0}^{\infty} \frac{1}{n!} x^n.$ Then, $$ f'(x) = \sum_{n=0}^{\infty} \frac{1}{n!} n x^{n-1} = \{ \text{ first term vanishes so we can start at $n=1$ } \} \\ = \sum_{n=1}^{\infty} \frac{1}{n!} n x^{n-1} = \sum_{n=1}^{\infty} \frac{1}{(n-1)!} x^{n-1} = \{ \text{ set $m=n-1$ } \} = \sum_{m=0}^{\infty} \frac{1}{m!} x^{m} = f(x) $$ Also, $$f(0) = \sum_{n=0}^{\infty} \frac{1}{n!} 0^n = \{ \text{$0^0 = 1$ and $0^n=0$ when $n>0$ } \} = \frac{1}{0!} = 1.$$

Thus $f'=f$ and $f(0) = 1$ so $f(x) = e^x.$