Back to e

Above, we defined $ e$ as the particular real number satisfying

$\displaystyle \lim_{\delta\to 0} \frac{e^\delta-1}{\delta} \isdef 1
$

which gave us $ (a^x)^\prime = a^x$ when $ a=e$. From this expression, we have, as $ \delta\to 0$,

\begin{eqnarray*}
e^\delta - 1 & \rightarrow & \delta \\
\Rightarrow \qquad e^...
... \\
\Rightarrow \qquad e & \rightarrow & (1+\delta)^{1/\delta},
\end{eqnarray*}

or

$\displaystyle \zbox {e \isdef \lim_{\delta\to0} (1+\delta)^{1/\delta}.}
$

This is one way to define $ e$. Another way to arrive at the same definition is to ask what logarithmic base $ e$ gives that the derivative of $ \log_e(x)$ is $ 1/x$. We denote $ \log_e(x)$ by $ \ln(x)$.

Numerically, $ e$ is a transcendental number (a type of irrational number3.5), so its decimal expansion never repeats. The initial decimal expansion of $ e$ is given by3.6

$\displaystyle e = 2.7182818284590452353602874713526624977572470937\ldots\,.
$

Any number of digits can be computed from the formula $ (1+\delta)^{1/\delta}$ by making $ \delta$ sufficiently small.


Next Section:
e^(j theta)
Previous Section:
Derivatives of f(x)=a^x