# Prove that $e>2$ geometrically.

Q: Prove that $e>2$ geometrically.

Attempt: I only know one formal definition of $e$ that is $\lim_\limits{n\to\infty} (1+\frac{1}{n})^n=e$. I could somehow understand that this is somehow related to rotation in the complex plane. $$e^{i\theta}=\cos \theta + i \sin \theta$$ Hence we have $$e^{i\pi}=-1$$ But how can I bring out the value of $e$ when I am showing this rotation in a geometrical figure?

Any hints are appreciated.

EDIT: As per the comments, I am making a small addition to the question which will not affect the existing answers. It is that, as a definition of $e$, one can use any definition which does not use the fact $2<e<3$.

• Does this help? : Look at the area under $\frac 1x$ from $[1,2]$. That is clearly less than $1$. Thus, defining $e$ as the solution to $ln(x)=1$, we must have $e>2$.
– lulu
Dec 19 '15 at 17:42
• A nice approach would be to somehow show that $(1 + \frac 1n)^n$ is increasing as $n \to \infty$, then simply observe that $$\left(1 + \frac 11 \right)^1 = 2$$ there are nice proofs of this fact, but I don't know if any of them are particularly "geometrical". Dec 19 '15 at 17:49
• @Aniket Well, you need some definition of $e$ in order to prove anything at all...and every definition that leaps to mind involves some "calculus-like" operation (integration, as in my area, power series, limit, as in your definition, differential equation, and so on). To me, my construction is as geometric as you're going to get! But, then, there are so many ways to look at $e$ that I wouldn't rule anything out.
– lulu
Dec 19 '15 at 17:56
• @Omnomnomnom: perhaps this answer qualifies.
– robjohn
Dec 19 '15 at 23:10

In this image we see that $$\color{#00A000}{1}+\color{#C000C0}{x}\le\left(1+\frac x2\right)^2$$ Therefore, \begin{align} 1+1 &\le\left(1+\frac12\right)^2\\ &\le\left(1+\frac14\right)^4\\ &\le\left(1+\frac18\right)^8\\ &\dots\\ &\le\lim_{n\to\infty}\left(1+\frac1{2^n}\right)^{\large2^n}\\[9pt] &=e \end{align}

• @jkabrg: With Intaglio.
– robjohn
Dec 19 '15 at 23:39
• Nice moustache, robjohn! :-)
– mvw
Dec 19 '15 at 23:42
• @mvw: Thanks! The hats just don't seem to fit a mean square.
– robjohn
Dec 19 '15 at 23:52
• @robjohn why does $\left(1+\frac12\right)^2 \le\left(1+\frac14\right)^4$ ? Jan 8 '16 at 11:03
• Square the inequality for $x=\frac12$.
– robjohn
Jan 8 '16 at 11:16

A better (or at least alternative) definition of $e$ is this:

Let $$L(x) = \int_1^x \frac{1}{t} dt$$ $L$ is well-defined for positive $x$ by the fundamental theorem of calculus.

With a little work, you can show that $L$ is surjective onto $R$, and since it's clearly increasing and continuous, it's also injective. So it has an inverse, $E$. $L$ is usually known as $\ln$ and $E$ is known as $\exp$.

Then $e = E(1)$ defines a new constant, called Euler's constant.

To show $e > 2$, you need only show that $L(2) < 1$. You can do this by computing an upper bound for the integral that is $L(2)$, i.e $\int_1^2 \frac{1}{t} dt$, using the partition $1, 1.5, 2$; and the left-hand ends as sample points (because $y = 1/x$ is a decreasing functions. The upper integral is then $$\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{2}{3} = \frac{5}{6} < 1,$$ and you are done, because the integral is no larger than any of its upper integrals.

• You're right; I was picturing log in my head as I wrote that, rather than $y = 1/x$. Thanks! Dec 19 '15 at 19:26

A simple geometrical representation can be given noting that $e=f(1)$ for a function $f(x)$ such that $f'(x)=f(x)$ and $f(0)=1$ (This can be a definition of $e$ derived from the definition of the exponential function as the function that represents an exponential growth).

So, representing a graphical approximation of the function (a representation of the Euler Method), as in the figure, We can see that $f(1)=e>2$ • how did you produce this image? Dec 19 '15 at 23:37
• Using LaTex with :pgfplots.sourceforge.net. Dec 20 '15 at 9:13

You can give a (hyper)geometical interpretation to the inequality

$$(1+x)^n\gt 1+nx\quad\text{when }n\ge2$$

by viewing the left hand side as the volume of an $n$-dimensional hypercube with sides of length $1+x$ and the right hand side as the sum of the volumes of the unit hypercube and the $n$ hyperrectangles of size $1\times1\times\cdots\times1\times x$. Plugging in $x=1/n$ leads to the inequality $e\gt2$.

• I tried to carry off a similar idea in two-dimensions that could be iterated.
– robjohn
Dec 19 '15 at 23:56

If $f(x)$ is a convex function, then:

$$f(x + h) \geq f(x) + h f'(x)~\forall x, \forall h > 0. ~~~~~~~~~~~~~~~~(1)$$

We know that the function $f(x) = e^x$ has an important geometrical property:

the slope in any point is equal to the function itself

Also, we know that $f(x)$ is positive because of exponentiation. This implies that it is convex, since $f(x)'' = f'(x) = f(x) > 0$ (thanks to A.S. for this point).

Then, (1) holds for $f(x) = e^x$. That is:

$$f(x + h) \geq f(x) + h f'(x)~\forall x, \forall h > 0 \Rightarrow \\ f(x + h) \geq (1+h)f(x)~\forall x, \forall h > 0.$$

If we choose $x=h=1$, then:

$$f(2) \geq 2f(1) \Rightarrow \\ e^2 \geq 2e \Rightarrow \\ e \geq 2.$$

• @A.S. yeah, you are right. I'm going to fix this. Dec 19 '15 at 20:24
• @A.S. thanks again, I have fixed it. Dec 19 '15 at 20:27

You could use the fact that $e^x$ is its derivative: $$\int\limits_a^b e^x \, dx = e^b - e^a$$ So $$e - 1 = \int\limits_0^1 e^x dx = \int\limits_0^1 \left(1 + (\underbrace{e^x - 1}_{\ge 0}) \right) dx > \int\limits_0^1 1\, dx = 1$$ So $e - 1$ can be interpreted as the area beneath the curve $e^x$ from $x = 0$ to $x = 1$. This area can be divided into the area below the constant function $f(x) = 1$ (green square) and some non-zero area which corresponds to the area below the function $g(x) = e^x - 1 \ge 0$ (the area having the points $A$, $B$, $D$).

• how did you produce this image? Dec 19 '15 at 23:38
• @jkabrg I used GeoGebra. A very nice software. I use version 5 with 3D support on Win 7 and iOS 9 and version 4 on Linux.
– mvw
Dec 19 '15 at 23:40
• cheers, i'll check that out Dec 19 '15 at 23:43