A cute proof that makes$e$natural

For the full article covering many properties of$e$, including history and comparison with existing methods of teaching: PDF from arXiv. A video explanation will be posted here shortly.

This webpage pulls out the part of the article which uses Pre-Calculus language to explain what is so natural about$e$, while intuitively connecting the following two important properties:

• The slope of the tangent line to$y=e^x$at the point$(x,e^x)$is just$e^x$. (In Calculus language:$e^x$is its own derivative.)
• The expression$(1+\frac{1}{n}{)}^n$approaches$e$as$n$grows.

Key conceptual starting point

Geometrically, there really is only one exponential function curve shape, because all exponential function curves$y=a^x$(with positive real bases$a$) are just horizontal stretches of each other. This is exactly like how all ellipses are just stretches of each other (and for the same reason).

For example,$y=8^x$, stretched horizontally by a factor of$6$, is$y=8^{x/6}=(\sqrt{2}{)}^x$.

Geometrically, since stretching is a continuous process, exactly one of these horizontally stretched exponential curves has the property that its tangent line at its$y$-intercept has the particularly nice and natural slope of$1$.

We define$e$to be the unique positive real base corresponding to that curve.

Easy$e$approximation

Let's find a number whose exponential curve has tangent slope$1$at the$y$-axis. For this, we take the curve$y=3^x$and estimate what factor to horizontally stretch it. To start, we must estimate the slope of the tangent line to$y=3^x$at its$y$-intercept$A(0,3^0)$. But how? Does that need Calculus? No! Algebra is enough!

Consider a very-nearby point on the curve:$B(h,3^h)$, where$h$is tiny but not zero. The slope of line$AB$is$$\frac{3^h-3^0}{h-0}$$Use$h=0.0001$to approximate that tangent slope:$$\frac{3^{0.0001}-1}{0.0001-0}\approx 1.09867$$Thus a horizontal stretch by a factor of$\approx 1.09867$will make the tangent slope$\approx 1$. So$3^{x/1.09867}=(3^{1/1.09867}{)}^x$has a tangent slope of$\approx 1$.

Therefore,$3^{1/1.09867}\approx 2.71814$is close to$e$. This is pretty good, because actually$e\approx 2.718281828459045$.

Beautiful tangent slopes everywhere

The same method derives the slope of the tangent line to$y=e^x$at any point$P(x,e^x)$. Consider a very-nearby point on the curve:$Q(x+h,e^{x+h})$, where$h$is tiny but not zero. The slope of line$PQ$is$$\frac{e^{x+h}-e^x}{(x+h)-x}=e^x\cdot \left(\frac{e^h-e^0}{h-0}\right).$$

The bracket is the slope of the line through$(0,e^0)$and$(h,e^h)$, so as$h$shrinks, the bracket becomes the slope of the tangent to$y=e^x$at the$y$-intercept. That miraculously cleans up to just$1$by our definition of$e$. (And that is precisely why we built the definition this way.)

So, the slope of the tangent at$P(x,e^x)$is just$e^x$.

Rephrased in Calculus language:$e^x$is its own derivative. This is perhaps the single most important property of$e$, because all of the Calculus facts stemming from$e$can be deduced from this fact.

Compound interest limit

Pre-Calculus usually teaches a different definition of$e$, as the limit of the expression$(1+\frac{1}{n}{)}^n$which arises from continuously compounded interest. To reconcile the approaches, we now visually prove that$(1+\frac{1}{n}{)}^n$approaches the same number we defined.

Since${\log }_{b}x$is the inverse function of$b^x$for any base$b$, using our base$e$we get$$\begin{array}{rl}{(1+\frac{1}{n})}^n& ={\left(e^{{\log }_e(1+\frac{1}{n})}\right)}^n\\ & =e^{n{\log }_e(1+\frac{1}{n})}\end{array}$$We used base$b=e$(instead of, say,$10$) because it now conveniently suffices to show that the expression in the exponent tends to$1$as$n$grows. That expression rearranges into a slope calculation!$$n{\log }_e(1+\frac{1}{n})=\frac{{\log }_e(1+\frac{1}{n})}{\frac{1}{n}}$$$$=\frac{{\log }_e(1+\frac{1}{n})-{\log }_e(1)}{(1+\frac{1}{n})-(1)}$$That's the slope of the line through the point$(1,0)$on the curve$y={\log }_{e}x$and another point very nearby on the curve. As$n$grows, that tends to the slope of the tangent line at$(1,0)$. We are done as soon as we prove that slope is$1$(which is also a natural objective to seek).

To that end, since${\log }_{e}x$is the inverse function of$e^x$, their graphs are reflections over the line$y=x$.

Both of the following lines have slope$1$:

• the tangent line to$y=e^x$through$(0,1)$by definition of$e$; and
• the line$y=x$.

So, they are parallel, making this nice reflection:

Therefore, the slope of the tangent line to$y={\log }_{e}x$at$(1,0)$is indeed$1$, completing the proof!