# Connecting the two definitions of $e$

There are two different ways I've been taught to understand the meaning of $$e$$.

1. $$e$$ can be defined as the total growth from continuously compounding interest in a single period. To make the growth rate the same as the dollar amount, consider I start with $$\1$$. Then, $$a_n$$ represents the amount I have after compounding $$n$$ times. \begin{align} a_n&=1\cdot\left(1+\frac{1}{n}\right)^n\\[5pt] e=a_{\max}&=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n \end{align} As we increase the number of times we compound the interest, the final amount (in the case that we start with $$\1$$), approaches $$e$$.
2. All exponential functions increase at a rate proportional to their current value. For example, consider the derivative of $$2^x$$: \begin{align} \frac{d\left(2^x\right)}{dx}&=\lim_{h \to 0}\frac{2^{x+h}-2^x}{h}\\[5pt] &=\lim_{h \to 0}\frac{2^x2^h-2^x}{h}\\[5pt] &=\lim_{h \to 0}\frac{2^x\left(2^h-1\right)}{h}\\[5pt] &=2^x\lim_{h \to 0}\frac{\left(2^h-1\right)}{h} \end{align} From this we can see that the rate at which $$2^x$$ changes is proportional to its current value, with a proportionality constant of $$\lim_{h \to 0}\frac{\left(2^h-1\right)}{h}.$$ For $$2^x$$, this constant happens to be $$\ln(2)$$ (why is this the case?). $$e$$ can be defined as the base where this proportionality constant is $$1$$; therefore, the rate $$e^x$$ grows at is exactly its current value.

My question is this: how can the same constant have two (apparently) separate definitions? Is there a way of looking at $$e$$ that will combine these two definitions, and make it obvious as to why they are true? Also, why is the proportionality constant of any exponential functions growth rate the natural log of its base? Is there a way to understand why this is the case intuitively?

• $\exp(1)=\sum\limits_{n=0}^\infty \dfrac{1}{n!}$ is another possible definition – Henry Jan 20 at 16:04
• @Henry Yes, this makes sense as it's the Taylor Series expansion of $e$ – John Hippisley Jan 20 at 16:09
• "how can the same constant have two (apparently) separate definitions?" Happens all the time. Definition 1: $k$ is the smallest even prime. Definition 2: $m$ is the sum of the multiplicative identity with itself. Definition 3: $j$ is the square root of the smallest composite natural number. We can calculate that $k= 2$ and $m=2$ and $j=2$. And we could analysis and figure out why those must be equal... or we could accept they are. – fleablood Jan 20 at 16:36
• From an intuitive perspective, the two characterisations of the exponential function should definitely be related, if not equivalent. The second characterisation of $e^x$ that you gave shows that the exponential growth of a quantity is proportional to the quantity itself. This is the same as in the compound interest example, where the more money you have in the bank, the faster the rate at which it grows. Then taking things to the limit means that you are dealing with continuous (as opposed to discrete) growth. – Joe Jan 22 at 16:06

Let's start with your second definition. We want to find the derivative of $$f(x) = a^x$$, for some $$a>0.$$ Using the definition of derivative, we get

\begin{align} f'(x) &= \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}\\ &= \lim_{h\to 0}\frac{a^{x+h} - a^{x}}{h}\\ &=a^{x}\lim_{h\to 0}\frac{a^{h} - 1}{h}. \end{align}

Now, as you said, one way to define $$e$$ is to define it such that $$\lim_{h\to 0}\frac{e^{h} - 1}{h} = 1.\tag{*}$$ Now, I'm going to be somewhat informal here. Suppose we have $$\frac{e^{h} - 1}{h} = 1.$$ Then, doing some manipulation: \begin{align} \frac{e^{h} - 1}{h} &= 1\\ e^{h} - 1 &= h\\ e^{h} &= h + 1\\ e &= (h+1)^{1/h} \end{align}

Now, by taking limits we have that $$\begin{equation*} \lim_{h\to 0}\frac{e^{h} - 1}{h} = 1 \iff e = \lim_{h\to 0}(h+1)^{1/h}. \end{equation*}$$

Taking the new limit, $$e = \lim_{h\to 0}(h+1)^{1/h}$$, and setting $$n = 1/h$$, we note that $$h\to 0 \implies n\to \infty$$, and we can change variables to get $$e = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^{n},$$ which matches your first definition.

As to understanding where the natural log comes from, i.e. why we have $$\lim_{h\to 0}\frac{a^{h}-1}{h} = \ln(a),$$ this is a straightforward consequence of $$(*)$$: \begin{align} \lim_{h\to 0}\frac{a^{h} - 1}{h} &= \lim_{h\to 0}\frac{e^{\ln(a^{h})}-1}{h}\\ &=\lim_{h\to 0}\frac{e^{h\ln(a)}-1}{h}\\ &=\frac{\ln(a)}{\ln(a)}\lim_{h\to 0}\frac{e^{h\ln(a)}-1}{h}\\ &=\ln(a)\lim_{h\to 0}\frac{e^{h\ln(a)}-1}{h\ln(a)}.\\ \end{align} Now, set $$k = h\ln(a)$$, and note that because $$a > 0$$, we have that $$h\to 0\implies k \to 0$$ as well, and so we have by $$(*)$$ that: \begin{align} \lim_{h\to 0}\frac{a^{h} - 1}{h} &=\ln(a)\lim_{h\to 0}\frac{e^{h\ln(a)}-1}{h\ln(a)}\\ &=\ln(a)\lim_{k\to 0}\frac{e^{k} - 1}{k}\\ &=\ln(a)\cdot 1\\ &=\ln(a). \end{align}

This Wikipedia article: Characterizations of the exponential function, shows six ways to define the exponential function $$e^x$$ (thus for the constant $$e$$) and a detailed discussion of equivalency of all the six definitions.

This happens a lot in mathematics. The same mathematical object may have many properties that can characterize the object. Once they are proved to be equivalent to the definition, they can all be regarded as a definition. More generally, there are equivalent definitions of mathematical structures.

For an intuitive understanding of the constant $$e$$ and exponential growth, you may read this article https://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/ from the "BetterExplained" website.

Take the second definition:

Consider: $$\lim_{h\to 0}\frac {b^h -1}h = \ln b = 1$$. And solve for $$b$$. That should be $$e$$ and we could define, supposedly, $$e$$ as the $$b$$ where that limit is $$1$$.

....

$$\lim_{h\to 0}\frac {b^h -1}h =$$

$$\lim_{\frac 1n\to 0}\frac {b^{\frac 1n} -1}{\frac 1n} =$$

$$\lim_{n\to \infty} n(b^{\frac 1n} - 1)=1$$

Now for any $$k$$ if $$k(b_k^{\frac 1k} - 1)=1$$ then $$b_k= (1+ \frac 1k)^k$$ and so if we wave our hands really fast and point out that "Look, it's Halley's comet" it should be reasonable that

If $$\lim_{n\to \infty}n(b^{\frac 1n} -1) = 1$$ then

$$b = \lim_{n \to \infty}(1 + \frac 1n)^n$$.

So the definitions are consistent.

Here is an alternative approach: if we first define the natural logarithm as $$\log x = \int_{1}^{x}\frac{1}{t} \, dt$$ and the exponential function as the inverse of $$\log$$, then it is not much work to show that $$e^x$$ is its own derivative. (I go into more detail about this here.) Then, the limit $$\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n$$ can be rewritten as $$\lim_{n \to \infty}e^{n\log(1+1/n)} \, .$$ If we make a change of variables $$h=1/n$$, then this limit is equal to $$\lim_{h \to 0^+}e^{1/h\log(1+h)}$$ For small $$h$$, $$\log(1+h) \approx h$$. This can be easily justified by considering the graph of $$\log$$:

$$\log x$$ denotes the area of the region bounded by the hyperbola $$y=1/t$$, the $$t$$-axis, and the vertical lines $$t=1$$ and $$t=x$$. When $$x = 1+h$$ (and $$h$$ is small), this region is approximately rectangular, with a width of $$h$$ and a height of $$1$$. Hence, $$\log(1+h) \approx h$$, with $$h$$ being a slight overestimate of the region's area.

This provides us with a nice heuristic argument that $$\lim_{h \to 0^+}e^{1/h\log(1+h)} = e^{1/h \cdot h} = e \, .$$