# Definition of $\exp(x)$

I've been thought at school that the definition of $\exp(x)$ is the unique function satisfying $\exp'(x)=\exp(x)$ and $\exp(0)=1$. But I've never found this definition somewhere else. On Wikipedia there are multiple definitions which are : $$\exp(x)=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$ $$\text{The solution y to the equation }x=\int_1^y\frac{1}tdt$$ $$\text{and } \exp(x)=\lim_{n\to+\infty}\left(1+\frac{x}{n}\right)^n$$ As I know, something can have only one definition, but can have multiply ways to define it and properties. So which one is the real definition of the exponential function?

• They are all completely equivalent definitions, and they are all equally "real" or "correct". May 4, 2013 at 17:29
• @moray95 add "the inverse of $\log$" May 4, 2013 at 17:30
• The one you use to find most properties is the one with the series. But they are equivalent anyway. May 4, 2013 at 17:30
• When definitions are equivalent, mathematicians stop carrying which is "real". They will just use the one they like. May 4, 2013 at 17:32
• I don't get "equivalent" definitions, a definition should be unique no? May 4, 2013 at 17:33

The following are equivalent definitions for $\exp(x)$. \begin{align} 1. & f(x) = \sum_{k=0}^{\infty} \dfrac{x^k}{k!}\\ 2. & \dfrac{d f(x)}{dx} = f(x) \text{ with } f(0) = 1\\ 3. & f(x) = \lim_{n \to \infty} \left(1+\dfrac{x}n\right)^n\\ 4. & f(x+y) = f(x) \cdot f(y) \text{ with }f(x) >0 \text{ being continuous at one point and } f(1) = e \end{align} If you start with any one, you can derive/prove the others.

EDIT

The important thing is that you can start with anyone and derive the others as property. If a statement $A$ implies a statement $B$ and vice-versa, both are equivalent statements. We may, hence, use any one of them as a definition.

For instance, if you choose $(1)$ to define $\exp(x)$ as $\exp(x) = \displaystyle \sum_{k=0}^{\infty} \dfrac{x^k}{k!}$, the rest from $(2)$ to $(4)$ become properties.

• Then the one you start will be the definition and the other ones properties... May 4, 2013 at 17:35
• @moray95 Yes. But the important thing you can start with anyone and derive the other as properties. If a statement $A$ implies a statement $B$ and vice-versa, both are equivalent statements. We may, hence, use any one of them as a definition.
– user17762
May 4, 2013 at 17:38
• @moray95 For instance, if you choose to define $\exp(x) = \sum_{k=0}^{\infty} \dfrac{x^k}{k!}$, the rest become properties.
– user17762
May 4, 2013 at 17:40
• Thanks much more clear now! May 4, 2013 at 17:43
• The fourth one misses an assumption of continuity. Aug 26, 2017 at 17:39

To address the comment below your question, and your assertion that something must have "one real" definition:

"Equivalent definitions" simply mean that there are many ways to define/describe $e^x$, all of which define/describe the unique function: $f(x) = e^x$. When definitions are equivalent, we can choose any one of them to derive the others. So we are free to choose which, among equivalent defintions, will serve our purpose best, depending on when and how we need to use it.

That is true of many mathematical entities: e.g., there is no *ONE* true definition of $\pi$: there are many ways to define the unique number $\pi$.

• Agreed, context matters and sometimes there are even slightly different definitions of things (just like the English language, which makes it really hard with potentially dozens of meanings for a single word). +1 May 5, 2013 at 0:36

Another Wikipedia article of relevance is Characterizations of the exponential function.

Which characterization is most appropriate to take to be the definition depends on the context.