How can we say that $\exp(0)=1$ when $\exp(0)=\sum_{n=0}^\infty \frac{0^n}{n!}=0^0+0+0+...$ which doesn't seem to have a defined value. Obviously $\lim_{x\to 0}\exp(x)=1$ and $e^0=1$ but I don't think the former proves that $\exp$ doesn't have a gap and the latter seems to me like an example of the tail wagging the dog.
Asked
Active
Viewed 60 times
-1
-
1We define $0^0=1$ in this case. – pshmath0 Feb 26 '20 at 09:08
-
Also: https://math.stackexchange.com/q/1585105, https://math.stackexchange.com/q/1805745 – Martin R Feb 26 '20 at 09:10
2 Answers
4
When we write $e^{x}$ as $ \sum\limits_{k=0}^{\infty} \frac {x^k} {k!}$ we make the convention that the starting term is always interpreted as $1$, even when $x=0$.
Enrico M.
- 26,114
Kavi Rama Murthy
- 311,013
4
Actually,$$\exp(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$and therefore$$\exp(0)=1+0+0+0+\cdots=1.$$
Enrico M.
- 26,114
José Carlos Santos
- 427,504