Exponential of Zero

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Theorem

Let $\exp x$ be the exponential of $x$.


Then:

$\exp 0 = 1$


Proof 1

We have that the exponential function is the inverse of the natural logarithm function:

$\ln 1 = 0$

Hence the result.

$\blacksquare$


Proof 2

Using the definition of the exponential as a limit of a sequence:

\(\displaystyle \exp 0\) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \left({1 + \frac 0 n}\right)^n\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)

$\blacksquare$


Proof 3

This proof assumes the series definition of $\exp$.

That is, let:

$\displaystyle \exp x = \sum_{k \mathop = 0}^{\infty} \frac {x^k} {k!}$


Then:

\(\displaystyle \exp 0\) \(=\) \(\displaystyle \sum_{k \mathop = 0}^{\infty} \frac {0^k} {k!}\)
\(\displaystyle \) \(=\) \(\displaystyle 1\) Definition of $0^k$

$\blacksquare$


Proof 4

This proof assumes the Definition of $\exp x$ as the unique continuous extension of $e^x$.

\(\displaystyle \exp 0\) \(=\) \(\displaystyle e^0\)
\(\displaystyle \) \(=\) \(\displaystyle 1\) Definition of $x^0$, where $x \ne 0$

$\blacksquare$


Proof 5

\(\displaystyle \map \exp {z + \paren {-z} }\) \(=\) \(\displaystyle \exp z \, \map \exp {-z}\) Exponential of Sum
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \exp {z - z}\) \(=\) \(\displaystyle \dfrac {\exp z} {\exp z}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \exp 0\) \(=\) \(\displaystyle 1\)

$\blacksquare$


Sources