MHB Can the Exponential Power Series Be Defined Without a Function?

[evinda]

Hello!

I am looking at the exponential power series:

$$\sum_{n=0}^{\infty} \frac{x^n}{n!}$$

It is $R=\displaystyle{\frac{1}{\lim_{n \to \infty} \sup \sqrt[n]{|a_n|}}}=\frac{1}{\lim_{n \to \infty} \sup \sqrt[n]{n!}}=+\infty$

So,the power series converges at $(-\infty,+\infty)$,so $\exists s(x): \mathbb{R} \to \mathbb{R} \text{ such that } \displaystyle{s(x)=\sum_{n=0}^{\infty} \frac{x^n}{n!}}, x \in \mathbb{R}$

$(*)\displaystyle{ \sum_{n=1}^{\infty} n \frac{1}{n^!}x^{n-1}=\sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}=\sum_{n=0}^{\infty} \frac{x^n}{n!}}$

According to a theorem: $$s'(x)=\sum_{n=1}^{\infty} n \frac{1}{n^!}x^{n-1}=\sum_{n=0}^{\infty} \frac{x^n}{n!}, x \in \mathbb{R}$$

So, $s(x)=s'(x), x \in \mathbb{R}$.In my notes,they continue,defining $f(x)=e^{-x} s(x),x \in \mathbb{R}$ and they conclude that $f(x)=c, \forall x \in \mathbb{R}$

Then,using the fact that $s(1)=e$,they conclude that $s(x)=e^x, \forall x \in \mathbb{R}$.

Instead of defining a function, could I do it like that: $$s'(x)=s(x) \Rightarrow \frac{s'(x)}{s(x)}=1 \Rightarrow ln|s(x)|=x+c \Rightarrow s(x)=Ce^x$$ ??

Or isn't it possible,because we don't know if it is $s(x)=0$ ?

 

[I like Serena]

Hello!

I am looking at the exponential power series:

$$\sum_{n=0}^{\infty} \frac{x^n}{n!}$$

It is $R=\displaystyle{\frac{1}{\lim_{n \to \infty} \sup \sqrt[n]{|a_n|}}}=\frac{1}{\lim_{n \to \infty} \sup \sqrt[n]{n!}}=+\infty$

So,the power series converges at $(-\infty,+\infty)$,so $\exists s(x): \mathbb{R} \to \mathbb{R} \text{ such that } \displaystyle{s(x)=\sum_{n=0}^{\infty} \frac{x^n}{n!}}, x \in \mathbb{R}$

$(*)\displaystyle{ \sum_{n=1}^{\infty} n \frac{1}{n^!}x^{n-1}=\sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}=\sum_{n=0}^{\infty} \frac{x^n}{n!}}$

According to a theorem: $$s'(x)=\sum_{n=1}^{\infty} n \frac{1}{n^!}x^{n-1}=\sum_{n=0}^{\infty} \frac{x^n}{n!}, x \in \mathbb{R}$$

So, $s(x)=s'(x), x \in \mathbb{R}$.In my notes,they continue,defining $f(x)=e^{-x} s(x),x \in \mathbb{R}$ and they conclude that $f(x)=c, \forall x \in \mathbb{R}$

Then,using the fact that $s(1)=e$,they conclude that $s(x)=e^x, \forall x \in \mathbb{R}$.

Instead of defining a function, could I do it like that: $$s'(x)=s(x) \Rightarrow \frac{s'(x)}{s(x)}=1 \Rightarrow ln|s(x)|=x+c \Rightarrow s(x)=Ce^x$$ ??

Or isn't it possible,because we don't know if it is $s(x)=0$ ?

Hi! :D

Yes, that is also possible.
And well... it is not the case that $s(x)=0$.

But we do have that $s(0)=1$! (Angel)
That follows because $\lim\limits_{x \to 0} x^0 = 1$.

 

[evinda]

Hi! :D

Yes, that is also possible.
And well... it is not the case that $s(x)=0$.

But we do have that $s(0)=1$! (Angel)
That follows because $\lim\limits_{x \to 0} x^0 = 1$.

I understand...Thank you very much! :)