Intuitive Understanding of the constant "$e$"

Question

Potentially-related questions, shown before posting, didn't have anything like this, so I apologize in advance if this is a duplicate.

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I know there are many ways of calculating (or should I say "ending up at") the constant e. How would you explain e concisely?

It's a rather beautiful number, but when friends have asked me "what is e?" I'm usually at a loss for words -- I always figured the math explains it, but I would really like to know how others conceptualize it, especially in common-language (say, "English").

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related but not the same: https://math.stackexchange.com/questions/3006/could-you-explain-why-dex-dx-ex-intuitively

Answer 1

One way of understanding what is $e$, is to see it as a rate of growth.

This article explains it very well.

Answer 2

For the somewhat-calculus-literate, your "related but not the same" question is what I'd go for: $e$ is the number for which the exponential function with that base is its own derivative.

Without calculus, I'd go for the notion of compound interest: With a rate $r$ per period, compounded $n$ times per period, $A$ grows to $A(1+\frac{r}{n})^n$ after 1 period; as $n\to\infty$, $A(1+\frac{r}{n})^n\to Ae^r$.

Answer 3

It is my opinion there is no "intuitive understanding of the number $e$".

Presumably, what you want to explain to your friends is not some mythical intuitive content of the number but some actual, concrete property it has which makes you appreciate it. Explain that.

Answer 4

Suppose you have a particle with the following property: starting at time zero (in seconds, s), the magnitude of its velocity $v = |\mathbf{v}|$ (in meters per second, m/s) at time $t$ is exactly equal to its distance $d$ (in meters, m) from its starting place at the same time.

At $1$ m its speed is $1$ m/s, at $2$ m, $2$ m/s and so on.

What is the distance function $d = d(t)$ of this particle as a function of time, you might ask? Since $v(t) = d^{\prime}(t) = d(t)$, then $d(t) = d_{0} \ e^{t}$, where $d_{0}$ is some multiplicative constant. Since we haven't yet specified any initial data, we can (without loss of generality) simply take $d_0 = 1 \ \text{m}$ and $v_{0} = 1 \ \text{m/s}$ both at $t = 0 \ \text{s}$.

The constant $e$ is the magnitude of this particle's distance or velocity at time $t = 1$ s.

Answer 5

I also believe this to be equivalent to the cited question about e^x, but will answer the explicit question for the sake of argument.

To be concise, I remark to friends and students that e is the most important number in calculus, just as pi is the most important constant in geometry.

Certainly these claims are arguable, but my friends/students don't argue!

Answer 6

Preface: The best first definition of e is one that nobody has yet given in this thread, namely, that it is the base such that the corresponding exponential function has a slope of unity at 0. If it is not immediately clear what this means, then sketch two or three exponential functions with different bases (say, 2, 3, and 1/2), note that they all pass through the point (0,1), and then note that they can be distinguished by what slope they have at that point (ie, what slope their tangent line has at that point). The “most convenient” one is the one that has a slope of unity. This base can be taken as the definition of e. However, once the edifice of one-variable calculus has been erected, it turns out to be convenient to define e in another manner, and this “first definition” is obtained as a theorem.

Now, on to the answer: There is a joke, from the old West, about how to sell a covered wagon. As best as I recall, it goes something like this: “Say that the price is one hundred dollars. If the buyer doesn’t wince, add “just for the wheels”. If the buyer still doesn’t wince, add “for each one”…” You get the idea.

In a similar vein, the answer to this question ought to be layered. The first best answer is that given by The Chaz (whose answer I have therefore up-voted), that is, just state that it is the most important number in calculus. Now, the audience mentioned by The Chaz is very limited, namely, his friends and students, however, the OP, naturally, wants us to consider the public at large, such as a casual acquantance encountered in an elevator who, stuck for something to say to fill the silence, asks you, whom they know to be an adept of mathematics, about this mysterious number e. The answer that The Chaz gives his friends and students is the best first answer for anyone, the best, as they say, elevator pitch (ie, something intelligible that you can say in the time that it takes an elevator to go between floors).

If the questioner does not wince, that is, does not inquire further, then just leave it at that. If the questioner is not satisfied, then the next answer to give is the “exponential base whose tangent at 0 is 1” one (that I mentioned above). If the questioner does not inquire further, then just leave it at that. If the questioner is not satisfied, then the next answer to give is the compound-interst one. If the questioner does not wince, then leave it at that. If the questioner is still not satisfied, then give the calculus rate-of-change explanation – the velocity version by user02138 is especially good (and so I have up-voted it too).

If the questioner is still not satisfied, then say, “You have reached the limit, no pun intended, of my anyone’s ability to explain this to you, short of you yourself learning calculus.”

The questioner may then say, “I appreciate all you have said, but what I want to know is why e has the value it does, rathter than some other value.” You should then say, “That’s a very good question, and no one has ever answered that, but for that matter, no one has ever answered the corresponding question for π. For example, why is the third decimal digit of π 1, rather than, say, 2? But remember that mathematics is not unique in this regard. In Physics, there is a number, approximately equal to 1/137, the “intuitive understanding” of which is a genuine mystery. This number is called the fine structure constant. The big mystery is exactly what you are asking regarding e, namely, why it has the value that it does. There is a long-standing brouhaha about it, nothing like the placid acceptance of e in mathematics. After reading up on it, you just might come back to mathematics and look upon e as an old friend.”

Answer 7

If someone asks me, "what is $e$?" I sketch the graph of $y=1/x$, draw a line segment from $(1,1)$ on the curve down to $(1,0)$ on the $x$-axis, and ask, how far to the right do I have to draw another vertical segment to rope off an area of 1? Anyone who is familiar with the idea of graphing a function can appreciate that definition, and it's not surprising that something with such a down-to-earth definition is going to turn up in lots of other places in Mathematics. And anyone who knows Calculus can be shown that all the other properties of $e$ and of $e^x$ and of $\log x$ can be derived from this one property of $e$.

The yellow area equals the red one:

Answer 8

Geometric interpretations help with the intuition, and I liked Gerry Myerson's explanation for that reason. Here's another geometric explanation you might give, using exponential decay. Start by imagining a process, such as radioactive decay, where at the end of every hour you have half the amount of material you started with at the begining of the hour. So if you start with 1 unit of material, then the amount of material remaining at hours 0, 1, 2, 3 is 1, 1/2, 1/4, 1/8. This process is described by $(1/2)^t$ or $2^{-t}$. Or, you can imagine faster decay where you have only 1/3 of the material left at the end of each hour, so that the amounts are 1, 1/3, 1/9, 1/27 and the process is described by $(1/3)^t$ or $3^{-t}$.

You can then sketch these two functions; both asymptotically approach 0, with $3^{-t}$ getting there faster than $2^{-t}$. You can then mention the amazing fact that the area bounded by either curve and the horizontal and vertical axes, although infinite in extent, has a definite finite area. This is highly plausible since the curve is approaching the horizontal so quickly. Tell them that if you calculate this area, you find that for $(1/2)^t$ the area is bigger than 1, while for $(1/3)^t$, the area is smaller than 1. Then ask how can you adjust the decay rate, or equivalently, the fraction remaining after each hour, so that the area exactly equals 1? The answer turns out to be that $1/e\approx1/2.718$ of the material should remain after each hour - a process described by $e^{-t}$. Not surprisingly, this is a number between 2 and 3.

For the very curious and dedicated listener who knows about geometric series, you can justify the assertions that the areas are finite, and that the area is greater than 1 for $2^{-t}$ and less than 1 for $3^{-t}$. For example, for $2^{-t}$ you can get an overestimate of the area using rectangles: $1+1/2+1/4+\ldots=2$. So the area under the curve is finite. To show that for $2^{-t}$ the area is bigger than 1, you can do an underestimate using rectangles: $1/2+1/4+1/8+\ldots=1$.
To show that $3^{-t}$ is smaller than 1, you can do an overestimate using trapezoids. If you break each trapezoid into a rectangle and a triangle, you get the overestimate $$ \left(\frac{1}{3}+\frac{1}{2}\cdot\frac{2}{3}\right)+\left(\frac{1}{9}+\frac{1}{2}\cdot\frac{2}{9}\right)+\ldots=2\left(\frac{1}{3}+\frac{1}{9}+\ldots\right)=1. $$

This provides some explanation for the magnitude of $e$.

Answer 9

It's easy to show that $\dfrac{d}{dx} 2^x = (2^x\cdot\text{constant})$. And $\left.\dfrac{d}{dx} 2^x\right|_{x=0} = \text{that constant}$.

Since the graph of $y=2^x$ gets steeper as $x$ grows, the slope at $x=0$ must be less than the slope of the secant line involving $x=0$ and $x=1$. That latter slope is 1. Therefore the "constant" is less than 1.

By thinking about $y=4^x$ and considering the secant line involving $x=-1/2$ and $x=0$, one sees that that "constant" is more than 1.

Therefore 2 is too small, and 4 is too big, to be $e$.

For $y=e^x$, the "constant" is exactly 1.

(One can show that 3 is too big via the secant line at $x=-1/6$ and $x=0$, but the arithmetic is a bit messy.) Similarly $2.5$ is too small, via $x=0$ and $x={}$ . . . . I don't remember which number I used here. A positive number, obviously, and less than 1. Messy arithmetic again.

Answer 10

If $a > 0$, draw the curve $y = a^x$. You will notice it has a tangent line when it strikes the $y$-axis. If $a$ is much larger than 1, the tangent line is steep. If $a$ is small the slope is small. This slope depends continuously on $a$. The unique value making it 1 is $e$. This unlocks all of the magic if you think carefully.

Answer 11

There is a phenomenon called "exponential growth". We all have a feeling for it. Exponential growth of some quantity $Q$ with time is characterized by the property that in equal time intervals the quantity $Q$ increases by the same factor. Exponential growth can be slow or fast; therefore we have to have a "unit" in order to measure and compare the speed of growth in individual cases. Mathematicians tell us that the natural unit for this speed is encoded in a certain number $e\doteq 2.718$, in the same way that the unit of length was encoded in a rod of platinum and stored 1799 in a vault somewhere in Paris.

Answer 12

Nobody has yet given a combinatorial way of visualizing $e,$ so I thought I'd add one.

The constant $\pi$ is, of course, a ratio, that of the circumference of a circle to its diameter. If the corresponding ratios for regular polygons are used as a starting point, then $\pi$ is defined by taking the limit as the number of sides of the polygon goes to infinity. In a similar way, the constant $e$ is the limit of a ratio as a certain parameter is taken to infinity.

Imagine a lottery in which, to enter, you must submit a sequence of $20$ numbers in the range $1$ to $20.$ Order matters, and numbers can be used multiple times. So, for example, you can submit all $1\text{s}$ if you like. The winning number could be any of $20^{20}$ possibilities.

Now imagine that a similar lottery has been set up in a different locale, except that, because of local superstition, use of the number $13$ has been banned. Lottery entries still consist of $20$ numbers in the range $1$ to $20$, but only $19$ number choices are available, giving $19^{20}$ possible winning numbers. Clearly it's easier to win the second lottery, but by how much? The answer is the ratio $(1/19^{20})/(1/20^{20})=20^{20}/19^{20}\approx2.79.$ You are $2.79$ times as likely to win the second lottery as the first.

To generalize, imagine comparing a lottery in which submissions consist of $n$ numbers in the range $1$ to $n$ with a lottery in which one of the numbers in the range $1$ to $n$ is unlucky and may not be used. The probability of winning the second lottery is higher by a factor of $n^n/(n-1)^n.$ The limiting value of this ratio as $n$ goes to infinity is $e\approx2.718281828$.

What does this have to do with rates of growth of exponential processes? Imagine a quantity that goes up by a fixed ratio every year, say $72\%.$ Suppose we preferred to use months as units rather than years. It would be incorrect to simply divide the yearly rate by $12$ to get $72\%/12=6\%$ monthly growth, or, if we did so, we'd be describing a different process. The reason it's different is compounding. Growth of $6\%$ per month means growth by a factor of $1.06$ each month. In two months this translates to growth by a factor of $1.06\times1.06=1.1236$ or an increase of $12.36\%,$ which is bigger than $2\times6\%=12\%$ since the $6\%$ of quantity gained in the first month itself grows by $6\%$ in the second month. Similarly, over the course of a year, the factor of increase is $1.06^{12}\approx2.0122,$ which translates to $101.22\%$ growth per year instead of $72\%.$

More generally, if we prefer to use time units of $1/n$ years, an increase of $72\%/n$ every $1/n$ years is bigger than an increase of $72\%$ every year because of compounding. Specifically, it corresponds to growth by a factor of $(1+0.72/n)^n$ every year, which, if you compute it for particular $n,$ is bigger than $72\%$ per year. Moreover, if we make the time unit smaller by making $n$ bigger, the annual growth gets bigger because the compounding effect gets enhanced.

In calculus we like talking about instantaneous rates of change, which involves make the time interval over which the change is measured arbitrarily small. If we do this in our problem, by taking the limit as $n$ goes to infinity of $(1+0.72/n)^n,$ we maximize the compounding effect, obtaining a growth factor of $2.05443,$ or $105.443\%$ per year. The limiting growth factor of $2.05443$ turns out to equal to $e^{0.72}.$

To relate this to the combinatorics problem, imagine a quantity that doubles every year, that is, that increases by $100\%$ per year. If we instead split this $100\%$ over $n$ equal time intervals, we get a growth factor of $(1+1/n)^n=(n+1)^n/n^n.$ As $n$ goes to infinity, this has the same limiting value as $n^n/(n-1)^n,$ namely $e.$ Hence, because we are compounding on arbitrarily short time intervals, we grow by a factor of approximately $2.718281828$ per year rather than by a factor of $2.$

Interestingly, there's a related combinatorics problem in which $e$ appears. Imagine, once again, a lottery in which $20$ numbers in the range $1$ to $20$ must be chosen. But in this lottery all $20$ numbers must be used. So our number selection boils down to choosing a permutation of the sequence $1,2,3,\ldots,20.$ Now imagine a similar lottery with the extra stipulation that the first number may not be $1,$ the second number may not be $2,$ the third number may not be $3,$ and so on. Such a permutation is called a derangement. Clearly there are fewer derangements than permutations, so the second lottery is easier to win. Once again, the probability of winning the second lottery is approximately $e$ times bigger than the probability of winning the first.