# An easy way to define $\exp(x)$ - does it work?

$\exp(x)$ is usually defined in three different ways:

1) By its Taylor series: $\exp(x)=\sum_{k=0}^{\infty} \frac{x^k}{k!}$

2) By its derivative: $\exp(x)'=\exp(x)$

3) By the limit $\exp(x)=\lim_{N \rightarrow \infty} \left(1+\frac{x}{N} \right)^N$

In textbooks they mostly use the 3rd way, because $\exp(x)$ is introduced before the derivative.

But it just occured to me - we don't need this much information to define this function. There is a very simple definition which allows to recover all the other properties:

4)

$$\lim_{x \rightarrow 0} \exp(x) = \lim_{x \rightarrow 0} (1+x)$$

Now, it's just two terms of Taylor expansion - it doesn't seem like much. Yet we can prove the following properties:

a) It follows that $\lim_{x \rightarrow 0} \exp(a x) = \lim_{x \rightarrow 0} \exp^a(x)$

$$\lim_{x \rightarrow 0} \exp^a(x) = \lim_{x \rightarrow 0} (1+x)^a = \lim_{x \rightarrow 0} (1+a x) = \lim_{x \rightarrow 0} \exp(a x)$$

b) Using this property we recover the 3rd definition

$$\lim_{N \rightarrow \infty} \left( 1+\frac{x}{N} \right)^N=\lim_{N \rightarrow \infty} \left( \exp \left(\frac{x}{N}\right) \right)^N=\lim_{N \rightarrow \infty} \left( \exp^{\frac{1}{N}}(x) \right)^N=\exp(x)$$

Does this work, or did I make a mistake somewhere?

Edit

I was obviously wrong - there is an infinite number of functions with the property 4. So the definition needs to be:

4) For $x \rightarrow 0$ $$\exp(x) \approx 1+x$$

For any $x$

$$\exp^a(x) = \exp(a x)$$

• Your definition (4) is satisfied by any function $f(x)$ such that $\lim_{x\to 0} f(x) = 1$, so there is no way that you can prove $f(x)=\exp(x)$ from it... – Hans Lundmark Feb 11 '16 at 11:34
• If $a$ is real, how do you define $\text{thing}^a$? – Martín-Blas Pérez Pinilla Feb 11 '16 at 11:51
• As $(\exp(x))^a$ – Yuriy S Feb 11 '16 at 11:52
• See the edit of my previous comment. – Martín-Blas Pérez Pinilla Feb 11 '16 at 11:53
• $\approx$ doesn't have a standard meaning. However, if we define $f(x)\approx g(x)$ to mean $\lim\limits_{x\to0}\frac{f(x)-g(x)}x=0$, then $\exp(x)\approx 1+x$ is actually equivalent to $\exp'(0)=\exp(0)$. Along with the rule $\exp(a)\exp(b)=\exp(a+b)$, this would imply $\exp'(x)=\exp(x)$ (why?), which is the second definition at the top. – Akiva Weinberger Feb 11 '16 at 11:55

$$\lim_{x\to 0} (1+x)$$ is simply equal to $1$, so the equation you wrote is equal to the equation $$\lim_{x\to0} \exp(x) = 1$$
which is most certainly not enough to characterise the exponential function, for example $\cos(x)$.
• I wanted to write $\exp(x) \approx 1+x$ for $x \rightarrow 0$ at first, would it be better? – Yuriy S Feb 11 '16 at 11:39
• @YuriyS Not really, since the function $1+x$ also satisfies that condition. – 5xum Feb 11 '16 at 11:40
• And together with the property $\exp^a(x)=\exp(ax)$? – Yuriy S Feb 11 '16 at 11:42