Exponential of Zero

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:

\(\ds \exp 0\)

\(=\)

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

\(\ds \)

\(=\)

\(\ds 1\)

$\blacksquare$

Proof 3

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

That is, let:

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

Then:

\(\ds \exp 0\)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \frac {0^k} {k!}\)

\(\ds \)

\(=\)

\(\ds 1\)

Definition of Power of Zero

$\blacksquare$

Proof 4

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

\(\ds \exp 0\)

\(=\)

\(\ds e^0\)

\(\ds \)

\(=\)

\(\ds 1\)

Definition of $x^0$, where $x \ne 0$

$\blacksquare$

Proof 5

\(\ds \map \exp {z + \paren {-z} }\)

\(=\)

\(\ds \exp z \, \map \exp {-z}\)

Exponential of Sum

\(\ds \leadsto \ \ \)

\(\ds \map \exp {z - z}\)

\(=\)

\(\ds \dfrac {\exp z} {\exp z}\)

\(\ds \leadsto \ \ \)

\(\ds \exp 0\)

\(=\)

\(\ds 1\)

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 14.4$