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Finding the Natural exponential, One of Euler's Ways

Motivation for this pure math is to understand the background of the anti-derivative of the function $1/x=x^{-1}$, which one needs in his toolbox calculating work against a gravitational field,
and also in using the multipole expansion,
where one needs it to understand convergence of the Legendre polynomials,
and for that one needs the
integral test of the harmonic series, which is briefly $\int (1/x)dx$.

The logarithm is the inverse of the exponential function,
keeping the same base,
and a natural logarithm is the inverse of the natural base exponent.
The natural exponential function
(as in $e^x$, here as a function of $x$) is encountered in Statistical mechanics,
where it is found with an exponent of energy, as opposed to $x$, the identity function.
The natural exponential is also found in all wave mechanics,
due to use of the complex numbers.

The
irrational number
$e$ (2.718...) can be calculated many different ways, one of which will be demonstrated here with graphical plots.
It was posited by L. Euler that the natural base can be found by
solving the equation $b^x=1+x$,
where one orients with the equation first by noting that both left and right are easily satisfied for $x=0$
[Dunham]
.
For points very close to $x=0$, regardless of the base $b$, the curve is approximately a line, of varying slope, all passing through $(0,1)$:
$$
b^x=1+kx
$$
Because we know $b^0=1$ (link to-be-created),
the curves of different bases would go through the point $(0,1)$, and it turns out they have different slopes.
Below are plots of different bases, with lines drawn at slopes $+0.1$ greater slope, and $-0.1$ less slope than the slope of the curve at the point $(0,1)$.

By looking at the plot of $b^x$, for some $b \in \{2, e, 3\}$,
and drawing a line in the right place, we can find one way.
First, inspect a plot of the exponential function with base $2$, below.

This plot of the base of $2$ from $-2$ to $2$ (light pistachio color)
has a red line drawn through point (0,1) at a slope of 0.6 and a similar
blue line at slope 0.8 (indicating $k=0.7$, minimal precision).

The pistachio curve, and the red, and blue lines are the same width, so looking at
the triple crossing, drawn at the exponent of zero, point (0,1),
one can see that the
red, and blue lines have slopes just a little less
and a little more than the slope of the exponential curve at
that point.

This plot of the "natural base" of $2.718$ ,
from $-2$ to $2$ (pistachio)
has a red line drawn through point (0,1) at a slope of 0.9 and a similar
blue line at slope 1.1, indicating the "natural", ideal, slope of one.

This plot of the base of $3$ from $-2$ to $2$ (pistachio)
has a red line drawn through point (0,1) at a slope of 1.0 and a similar
blue line at slope 1.2, indicating $k=1.1$.

This plot of the base of $3$ from $(-0.0001)$ to $0.0001$ (pistachio)
has a red line drawn through point (0,1) at a slope of 1.0 and a similar
blue line at slope 1.2, indicating a slope of 1.1 for the exponent, very close to zero.

Next step:
plots of $ln(x)$ (abreviation for logarithm with natural base) and $1/x$ (not at zero),
so plots showing the slope of $ln(x)$ at interval 0.0005 +/- 0.0001 ($x$ close to zero), and around $x=1$ (something negative, because $1\lt e$),
and around $e$ (ordinate is equal to one).
Also, in a subsequent, post-calculus article, the lines drawn will have more terms added to the approximation of the exponent around zero.