Proof that $e^x = 1 + x +\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots$

Question

I'm very new to Taylor and/or Maclaurin Series, and the main thing that I understand is actually not rigorous at all, and is that:

Taylor series involves expressing a function as the sum of an infinite series of terms in the form $c_k(x-a)^k$, where $k$ is the term number starting at $k=0$. Hence, the general rule for a Taylor series expansion is

$f(x) = \displaystyle c_0+c_1(x-a)+c_2(x-a)^2+\ldots+c_k(x-a)^k$

From here, I've been shown how each term can be found as follows:

$f(a)=c_0$

$f'(x)=c_1 + 2c_2(x-a) + 3c_3(x-a)^2+\ldots$

$f'(a)=c_1$

$f''(x) = 2c_2+6c_3(x-a)+12c_4(x-a)^2$

$f''(a) = 2c_2$

$\displaystyle \frac{f''(a)}{2}=c_2$

So actually, over the course of the terms, there is a pattern emerging, which is that each term is multiplied by $(x-a)^k$ and has a derivative of one higher than the prior term, and is divided by $k!$, so:

$\displaystyle f(x)=f(a)+f'(a)+\frac{f''(a)}{2!}(x-a)^2+\ldots+\frac{f^k(a)}{k!}$

So what proof is there to show that

$\displaystyle e^x = f(a)+ f'(a)+\frac{f''(a)}{2!}(x-a)^2+\ldots+\frac{f^k(a)}{k!}=1+x+\frac{x^2}{2!}+\ldots+\frac{x^k}{k!}$

Like I said, I am very new to Taylor series, and if there are any corrections or references that can be provided I would greatly appreciate them. Thank you.

One thing to note is that $e^x$ is sometimes defined by this series expansion. So, in order to prove what you intend to prove, you need to use a different definition for $e$, and thus the question arises - Which definition of $e$ are you using here? — Prasun Biswas, Sep 02 '17 at 07:10
Setting $\exp(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}$ you can easily see that $\exp'(x) = \exp(x)$ and that $\exp(0) = 1$. Since $e^x$ is the only function with these properties, clearly $\exp(x) = e^x$. — md2perpe, Sep 02 '17 at 07:58
@YvesDaoust Oh! Thank you for the patience, I understand what you mean now, but another question, how come term $1$ is not multiplied by $(x-a)$? Or is it and I've simply not done enough research? — joshuaheckroodt, Sep 02 '17 at 08:53

score 1 · Accepted Answer · answered Sep 02 '17 at 08:03

1

First of all, the formula you have written for $f(x)$ is completely wrong. When the summation $\sum$ is opened the value of $k$ is changed by each integer in the given interval. Also you have changed $(x-a)^n$ to $(a-x)^n$ which is wrong. So,

$$\begin{align}f(x)&=\frac{f(a)}{0!}(x-a)^0+\frac{f'(a)}{1!}(x-a)^1+\frac{f''(a)}{2!}(x-a)^2+\dots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\dots\\ &=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\dots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\dots \end{align}$$

Let $f(x)=e^x$. Then

$$f'(x)=f''(x)=f'''(x)=f^{(4)}(x)=\dots=f^{(n)}(x)=\dots\displaystyle{\dots}=e^x$$ and $$f'(0)=f''(0)=f'''(0)=f^{(4)}(0)=\dots=f^{(n)}(0)=\dots\displaystyle{\dots}=e^0=1$$ Now in the expansion

$$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\dots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\dots$$ put $a=0$, $$\begin{align}f(x)&=f(0)+f'(0)(x-0)+\frac{f''(0)}{2!}(x-0)^2+\dots+\frac{f^{(n)}(0)}{n!}(x-0)^n+\dots\\ &=1+1(x)+\frac1{2!}(x^2)+\dots+\frac1{n!}(x^n)+\dots \end{align}$$

Hence, $$e^x=1+x+\frac{x^2}{2!}+\dots$$ and the proof is complete.

answered Sep 02 '17 at 08:03

Faiq Irfan

1,313

Are you hence trying to say that the use of summation in this particular context is not necessary or that it shouldn't be used? – joshuaheckroodt Sep 02 '17 at 08:12
I didn't mean that. However the last and second last equation in your question are algebraically wrong because each of your terms includes $(x-a)^k$ which means that the only power of $x$ in your expansion is $k$. – Faiq Irfan Sep 02 '17 at 08:20
But it shouldn't matter, should it? At $0, (x-a)^k = 1$, and at $1, (x-a)^k = x-a$, as it should. How come this is algebraically incorrect? – joshuaheckroodt Sep 02 '17 at 08:25
But the values of $k$ aren't determined by the situation. It is rather determined by the sum $\sum_{k=0}^\infty$ – Faiq Irfan Sep 02 '17 at 08:51
Also by rules of summation, the letter $k$ must be removed when you open the summation $\sum$ on $k$. – Faiq Irfan Sep 02 '17 at 08:53
I understand, thanks for the help :) – joshuaheckroodt Sep 02 '17 at 08:55
1

Don't mention it. It's what this site is for. ;) – Faiq Irfan Sep 02 '17 at 08:57

score 0 · Answer 2 · 2017-09-02T08:27:56.383

In you last identity, compare the terms and conclude that to make them identical (with the fix $a-x\leftrightarrow x-a$ and correct denominators and exponents), you need

$$a=0$$ and

$$f^{(k)}(0)=1.$$

So you need a way to evaluate the derivatives of the exponential function, and this depends on the definition you have learnt, which you didn't tell us.

(It turns out that $(e^x)^{(k)}=e^x$ for all $x$, but this need to be proven.)

Proof that $e^x = 1 + x +\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots$

2 Answers2