KindiakMath

What is the Exponential Unit?

You might have seen the number e \approx 2.718 in various mathematical discussions. What makes e such a crucial number? It turns out to be the key number in calculus, because

\displaystyle \frac{\mathrm d}{\mathrm dx} e^x = e^x.

Our goal is to define e rigorously, then prove this result, which is a highly nontrivial result.

Theorem 1. The sequence \{e_n\} defined by e_n := (1+1/n)^n converges. Thus, the exponential unit is defined by

\displaystyle e := \lim_{n \to \infty} e^n \equiv \lim_{n \to \infty} \left(1 + \frac 1n\right)^n.

Proof of Theorem 1. By the monotone convergence theorem, we need to show that \{e_n\} is bounded above and non-decreasing, or bounded below and non-increasing.

We need to compute e_{n+1}/e_n and show that this ratio satisfies either \leq 1 or \geq 1. By definition,

\displaystyle \begin{aligned} \frac{e_{n+1}}{e_n} = \frac{ (1+\frac{1}{n+1})^{n+1} }{(1+\frac{1}{n})^n} &= \frac{ (1+\frac{1}{n+1})^{n+1} }{(1+\frac{1}{n})^{n+1}} \cdot \left(1+\frac{1}{n} \right) \\ &= \left(\frac{ 1+\frac{1}{n+1}}{1+\frac{1}{n}}\right)^{n+1} \cdot \left(1+\frac{1}{n} \right) \\ &= \left( \frac{n(n+2)}{(n+1)^2} \right)^{n+1} \cdot \left(1+\frac{1}{n} \right) \\ &= \left( 1 - \frac{1}{(n+1)^2} \right)^{n+1} \cdot \left(1+\frac{1}{n} \right). \end{aligned}

Heuristically, we would think that 1/(n+1)^2 \to 0 more quickly than 1/(n+1) \to 0, so that the whole limit \to 1^+. But how do we formulate this idea rigorously?

With a stroke of genius, we will need Bernoulli’s inequality, which states that for real numbers x > -1 and integers r \geq 0,

(1+x)^r \geq 1 + rx.

We will prove this result later. For now, since n+1 \geq 0 and x = -1/(n+1)^2 > -1, we can apply Bernoulli’s inequality to obtain

\displaystyle \begin{aligned} \frac{ e_{n+1} }{ e_n } &= \left( 1 - \frac{1}{(n+1)^2} \right)^{n+1} \cdot \left(1+\frac{1}{n} \right) \\ &\geq \left( 1 + (n+1) \cdot \left(-\frac{1}{(n+1)^2} \right) \right) \cdot \left(1+\frac{1}{n} \right) \\ &= \left( 1 - \frac{1}{n+1} \right) \cdot \left(1+\frac{1}{n} \right) = \frac{n}{n+1} \cdot \frac{n+1}{n} = 1, \end{aligned}

which implies that e_{n+1} \geq e_n, which establishes that \{e_n\}. This hints that we need to establish that \{e_n\} is bounded above. Experimentally, we notice that e_n \leq 3, so we aim to prove this result rigorously, via induction.

Now for n = 1, we get

\displaystyle e_1 = \left( 1 + \frac 11 \right)^1 = 2 \leq 3.

Now inductively, we first suppose that e_k \leq 3, and aim to prove that e_{k+1} \leq 3. To that end,

\begin{aligned} e_{k+1} &= \left(1 + \frac 1{k+1} \right)^{k+1} \\ &= \left(1 + \frac 1{k+1} \right)^{k} \cdot \left(1 + \frac 1{k+1}\right) \\ &= \left(1 + \frac 1{k} \right)^{k} \cdot \frac{ \left(1 + \frac 1{k+1} \right)^{k} }{ \left(1 + \frac 1{k} \right)^{k} } \cdot \left(1 + \frac 1{k+1}\right) \\ &= \left(1 + \frac 1{k} \right)^{k} \cdot \left( 1 - \frac{1}{(k+1)^2} \right)^k \cdot \left(1 + \frac 1{k+1}\right) \\ &\leq 3 \cdot \left( 1 - \frac{1}{k+1} \right)^k \cdot \left(1 + \frac 1{k+1}\right)^k \\ &\leq 3 \cdot \left( 1 - \frac{1}{(k+1)^2} \right)^k \leq 3 \cdot (1-0)^k = 3, \end{aligned}

as required. Thus, \{e_n\} is bounded above and non-decreasing. By the monotone convergence theorem, \{e_n\} converges, as required.

Now, this proof hinges on Bernoulli’s inequality, which we will now prove using induction.

Lemma 1 (Bernoulli’s Inequality). For real numbers x > -1 and integers r \geq 0,

(1+x)^r \geq 1 + rx.

Proof of Bernoulli’s Inequality. We will prove by induction on r. The result for r = 0 is obvious. Suppose now that

(1+x)^k \geq 1+kx.

We need to prove that (1+x)^{k+1} \geq 1 + (k + 1)x. Then

\begin{aligned} (1+x)^{k+1} &= (1+x)^k \cdot (1+x) \\ &\geq (1+kx)\cdot (1+x) \\ &= 1 + (k+1)x + kx^2 \\ &\geq 1 + (k+1)x + 0 \\ &= 1 + (k+1)x.\end{aligned}

Now we aim to prove the real result of interest.

Theorem 2. \displaystyle \frac{\mathrm d}{\mathrm dx} e^x = e^x.

Proof of Theorem 2. Fix x \in \mathbb R. By the definition of the derivative, we need to take h \to 0 of the difference quotient

\displaystyle \frac{e^{x+h} - e^x}{h} = \frac{e^x e^h - e^x}{h} = e^x \cdot \frac{e^h - 1}{h}.

If we can prove that (e^h - 1)/h \to 1 as h \to 0^+, then we will obtain

\displaystyle \begin{aligned} \lim_{h \to 0^-} \frac{e^h - 1}{h} &= \lim_{h \to 0^+} \frac{e^{-h} - 1}{-h} \\ &= \lim_{h \to 0^+} \frac{1 - e^{-h} }{h} \\ &= \lim_{h \to 0^+} \left( e^{-h} \cdot \frac{e^h - 1}{h} \right) \\ &= \lim_{h \to 0^+} e^{-h} \cdot \lim_{h \to 0^+} \frac{e^h - 1}{h} \\ &= 1 \cdot 1 = 1, \end{aligned}

so that (e^h - 1)/h \to 1 as h \to 0 in either direction and the result follows. To that end, let’s use the limit definition of e to our advantage. Our first observation is that by a useful re-indexing,

\displaystyle \begin{aligned} e^h &= \left(\lim_{n \to \infty} \left( 1 + \frac 1n \right)^n \right)^h \\ &= \lim_{n \to \infty} \left( 1 + \frac 1n \right)^{nh} \\ &= \lim_{n \to \infty} \left( 1 + \frac h{nh} \right)^{nh} \\ &= \lim_{n \to \infty} \left( 1 + \frac hn \right)^n, \end{aligned}

since h > 0. The expression (1+h/n)^n suggests the use of the binomial theorem:

\displaystyle \begin{aligned} \left( 1 + \frac hn \right)^n &= \sum_{k=0}^n {n \choose k} \frac{h^k}{n^k} = 1 + h + \sum_{k=2}^n {n \choose k} \frac{h^k}{n^k} \end{aligned}

With the help of algebra and the triangle inequality,

\displaystyle \begin{aligned} \left| \frac{\left( 1 + h/n \right)^n - 1}{h} - 1 \right| &\leq h \cdot \sum_{k=0}^{n-2} {n \choose k+2} \frac{ h^k }{n^{k+2}}.\end{aligned}

We will need to take h \to 0, so we can assume |h| \leq 1/2. Let’s analyse each term inside the sum (perhaps that is where real analysis gets its name from). We observe that

\begin{aligned} {n \choose k+2} \frac{ h^k }{n^{k+2}} &= \frac{n(n-1)\cdot \cdots \cdot (n-k-1)}{n^{k+2} } \cdot \frac{h^k}{(k+2)!} \\ &= \frac{n}{n} \cdot \frac{n-1}{n} \cdot \cdots \cdot \frac{n-k-1}{n} \cdot \frac{h^k}{(k+2)!} \\ &\leq 1^n \cdot (1/2)^k = (1/2)^k. \end{aligned}

Using the formula for a geometric series that we will prove in just a moment,

\displaystyle \begin{aligned} \sum_{k=0}^{n-2} {n \choose k+2} \frac{ h^k }{n^{k+2}} &\leq \sum_{k=0}^{n-2} (1/2)^k = \frac{1 - (1/2)^{n-1}}{1-1/2} \leq \frac{1 - 0}{1-1/2} = 2. \end{aligned}

Therefore,

\displaystyle \begin{aligned} \left| \frac{\left( 1 + h/n \right)^n - 1}{h} - 1 \right| &\leq h \cdot \sum_{k=0}^{n-2} {n \choose k+2} \frac{ h^k }{n^{k+2}} \leq 2 h.\end{aligned}

Taking n \to \infty,

\displaystyle \left| \frac{e^h - 1}{h} - 1 \right| = \lim_{n \to \infty} \left| \frac{\left( 1 + h/n \right)^n - 1}{h} - 1 \right| \leq \lim_{n \to \infty} (2h) = 2 h.

Taking h \to 0^+, the squeeze theorem yields

\displaystyle \lim_{h \to 0^+} \left| \frac{e^h - 1}{h} - 1 \right| = 0.

Therefore,

\displaystyle \lim_{h \to 0^+} \frac{e^h - 1}{h} = 1,

as required.

Lemma 2 (Geometric Series). For any r \neq 0,

\displaystyle \sum_{k=0}^{n-1} r^k = \frac{1-r^n}{1-r}.

Proof of Lemma 2. The proof is immediate by the observation

\displaystyle \begin{aligned} (1-r) \sum_{k=0}^{n-1} r^k &= \sum_{k=0}^{n-1} r^k - \sum_{k=0}^{n-1} r^{k+1} \\ &= 1 + \sum_{k=1}^{n-1} r^k - \sum_{k=0}^{n-2} r^{k+1} - r^n \\ &= 1 + \sum_{k=1}^{n-1} r^k - \sum_{k=1}^{n-1} r^{k} - r^n \\ &= 1 - r^n.\end{aligned}

Dividing yields the desired result.

—Joel Kindiak, 26 Dec 24, 1845H

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