Exponential of Zero
Jump to navigation
Jump to search
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): $\S 14.4$