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One common way to introduce Euler's number $\mathrm e$ is $$\mathrm e = \lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n,$$ where the right-hand expression has an "interest rate interpretation" (so that a high-school student can see why this expression might be of interest and worth attaching a name to).

Separately, we can show that $$\mathrm e = \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots.$$ Is there, similarly, any real-world meaning/"interpretation" we can attach to the expression $\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots$?

(An ideal answer would target a secondary/high-school student who has no prior knowledge about calculus or the properties of $\mathrm e$ and $\exp$, as is the case with the "interest rate interpretation".)

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    $\begingroup$ Discussion (Raugh, Probst, 2019) of an unpublished manuscript of Leibniz, who used the sum to approximate (what is now called) $e$ and construct a catenary: sciencedirect.com/science/article/pii/S0315086018301290 $\endgroup$
    – user1815
    Commented Jun 2, 2023 at 16:20
  • $\begingroup$ @Raciquel: No, they may have already been introduced to limits and infinite series, though perhaps not rigorously. For example, they may have encountered asymptotes and $1+1/2+1/4+\dots=2$ (but may not be able to rigorously define/prove these). $\endgroup$
    – user18187
    Commented Jun 3, 2023 at 4:23
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    $\begingroup$ I think a better (or at least, also interesting )question is whether there is a real-world setting where both $\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n,$ and $\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots$ have some sort of seemingly different meaning (within the same real-world scenario), but then turn out to be the same thing. $\endgroup$ Commented Jun 9, 2023 at 10:16
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    $\begingroup$ @TonyMathew: Yes that will be fine. But ideally I want it to be independent of the "interest-rate interpretation". $\endgroup$
    – user18187
    Commented Jun 10, 2023 at 1:59
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    $\begingroup$ @AdamRubinson If you're willing to settle for a setting where $\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n$ and $\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\ldots\right)^{-1}$ have a seemingly different meaning but turn out to be the same thing then David E Speyer's answer provides an example. (Set $\lambda=1$.) $\endgroup$ Commented Jun 13, 2023 at 19:50

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You could ask this series of questions; I don't know how "real world" it is but it does indicate the specific contributions of each individual term of the series.

Suppose we have a car, and it is 1 meter away from us. How far away will it be after 1 second?

1 meter.

Suppose we have a car, and it is 1 meter away from us, but it is also moving away from us at 1 meter per second. How far away will it be after 1 second?

2 meters.

Suppose we have a car, and it is 1 meter away from us, but it is also moving away from us at 1 meter per second, and it is also accelerating away from us at 1 meter per second per second. How far away will it be after 1 second?

$2 \frac12$ meters...

Suppose we have a car, and it is 1 meter away from us, but it is also moving away from us at 1 meter per second, and it is also accelerating away from us at 1 meter per second per second, AND it is also getting jerked away from us at 1 meter per second per second per second. How far away will it be after 1 second?

$2 \frac23$ meters...

We don't seem to be gaining much with each of these. If you keep adding these 1s forever, how far away can the car get in 1 second?

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I think the one with useful interpretation is $$ \frac{1}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots = \exp(x) $$ You can check $\exp(x+y) = \exp(x)\exp(y)$, so it does behave like you would expect a power $e^x$ to behave. More advanced students can check that it satisfies the differential equation $\exp'(x) = \exp(x)$.

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A possible motivation is the solution to the so-called "hat check problem."

There are $N$ people who throw their hats into a box and then randomly take one out. The probability that each person ends up with a hat that is not their own is the alternating exponential series:

$$P(N)=\sum_{n=0}^N (-1)^n\frac{1}{n!}\approx e^{-1}$$

If you search for the word "derangements" you will find more details. The formula follows directly from the inclusion–exclusion principle and the binomial formula.

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  • $\begingroup$ Also look for "rencontres numbers": $\frac{D_{N,k}}{N!}$ is the probability that exactly $k$ people get their own hats. When $N$ is much larger than $k$, this is approximately $\frac{1}{e}\frac{1}{k!}$, which is the Poisson probability for a process with mean $1$. Summing over all $k$ from $0$ to infinity must, of course, give $1$, $$\frac{1}{e}\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\ldots\right)=1.$$ This relates this answer to Raciquel's answer. $\endgroup$ Commented Jun 2, 2023 at 19:31
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    $\begingroup$ Note that the interest rate expression given in the original post is also closely related to the Poisson process. Roll a fair $N$-sided die $N$ times. You expect to roll a $1$ one time. By the Bernoulli formula, the probability that you roll exactly $k$ $1$s is $$\binom{N}{k}\left(\frac{1}{N}\right)^k\left(1-\frac{1}{N}\right)^{N-k}.$$ You will recognize the probability of rolling no $1$s ($k=0$) as the compound interest formula with a rate $-1$, which approaches $\frac{1}{e}$. $\endgroup$ Commented Jun 2, 2023 at 19:44
  • $\begingroup$ For fixed $k$ much less than $N$ the Bernoulli expression is approximately $\frac{1}{e}\frac{1}{k!}$, again the Poisson probability for a process with mean $1$. $\endgroup$ Commented Jun 2, 2023 at 19:58
  • $\begingroup$ I just noticed that the calculation just above is a specialization to the case where the mean $\lambda$ equals $1$ of the calculation David E Speyer does in his answer. Instead of getting the power series for $e$ divided by the constant $e$, David Speyer gets the power series for $e^\lambda$ divided by the constant $e^\lambda$. In regard to this, it's interesting to compute the expected value--we know from the outset what it is going to be--and see how this relates to the familiar derivative property of $e^\lambda$. $\endgroup$ Commented Jun 3, 2023 at 4:28
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For what it's worth: The distinction between $e = \lim_{n\to\infty} (1+\frac{1}{n})^n$ and $e = \sum_{n=0}^\infty \frac{1}{n!}$ has, to me, become very blurry in recent years. In fact, just assume $n$ is "very big", and use the binomial formula to multiply out:

$$\left(1+\frac{1}{n}\right)^n = 1 + \underbrace{n\cdot \frac{1}{n}}_{=1} + \underbrace{\dfrac{n(n-1)}{2!}\cdot\frac{1}{n^2}}_{\approx \frac{1}{2!}}+\underbrace{\dfrac{n(n-1)(n-2)}{3!} \cdot \frac{1}{n^3}}_{\approx \frac{1}{3!}} + \dots$$

and the approximations become better and better for $n \to \infty$.

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  • $\begingroup$ Agreed. Though I think what the OP is trying to address is that even if we agree that two expressions are mathematically saying the same thing, one may feel more intuitive and comfortable just based on how it is expressed. $\endgroup$ Commented Jun 12, 2023 at 7:52
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    $\begingroup$ I believe this is how Euler works out the series for $e$, taking for $n$ an infinite number. $\endgroup$
    – KCd
    Commented Jun 14, 2023 at 8:39
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If we have a random event that occurs $\lambda=1$ time per day on average, then over time, the proportion of days in which the event occurs $k$ times approaches $P(k) = e^{-1}/k!$, the probability that the event occurs $k$ times in a day (Poisson distribution). So $$\sum_{k=0}^\infty {e^{-1} \over k!} = e^{-1} \sum_{k=0}^\infty {1 \over k!} = 1$$ is the sum of the probabilities that the number of events in a day is 0, or 1, or 2,..., which of course is $1$. Consequently, for $k=0$, the probability of $0$ events is $${e^{-1} \over 0!} = e^{-1} = 1 \Big/ \sum_{k=0}^\infty {1 \over k!} \,.$$ Alternatively, this is the proportion of days over time in which the event does not occur.

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    $\begingroup$ I think there is some circularity here: How do we know $P(k)=\mathrm e^{-1}/k!$ (first sentence)? $\endgroup$
    – user18187
    Commented Jun 3, 2023 at 2:45
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    $\begingroup$ @user24096 You asked about "interpretations" not "proofs." I took it as given that the mathematics has been proved. $\endgroup$
    – user1815
    Commented Jun 3, 2023 at 2:49
  • $\begingroup$ $$\begin{array}{l} \tt\text{import math} \\ \tt\text{import numpy as np} \\ \tt\text{import random} \\ \\ \tt\text{n = 1000000} \\ \tt\text{x = np.ones(n)} \\ \tt\text{for i in range(n):} \\ \quad \tt\text{x[math.floor(random.uniform(0, n))] = 0} \\ \tt\text{print(n / np.sum(x))} \end{array}$$ $\endgroup$
    – user1815
    Commented Jun 3, 2023 at 5:27
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Here's a problem I designed, and my solution 'discovering' the summation expression as an expected value of a triangular number of objects.

A builder is tasked to make stairs from a set of wooden planks. The owner has already bought the planks, and told the builder to use the first plank for the first step of the stairs, the next two planks for the second step, the next three for the third step, and so on. The builder does so, but then the owner realises they forget to tell the builder how to arrange the planks within each step, as there was a pattern on the side of the stairs they wanted to have seen. If there are many steps in the stairs, estimate how many are correctly arranged.

We do not have enough information to determine the exact number of correctly arranged steps. But assuming the builder chose a random arrangement of the planks within each step, we can calculate the expected number of correct steps using probability.

There is only 1 plank in the first step. So there is only one possible arrangement, and it must be the correct one. Hence, the first step is guaranteed correct.

The second step has 2 planks; so $2!=2$ possible arrangements, out of which only 1 is correct. This arrangement has a correctness value of "1", while the other arrangement has a correctness value of "0". And assuming uniform distribution, each arrangement has an equal likelihood of $\frac{1}{2!}$ of being chosen. Hence, the expected correctness of the second step is $$1\times \frac{1}{2!} + 0\times \frac{1}{2!} = \frac{1}{2!}$$

Repeating this logic, the expected correctness of the third step (which has 3 planks with $3!=6$ possible arrangements) is, $$1\times \frac{1}{3!} + 0\times \frac{1}{3!} + 0\times \frac{1}{3!} + 0\times \frac{1}{3!} + 0\times \frac{1}{3!} + 0\times \frac{1}{3!} = \frac{1}{3!}$$

Continuing, step $n$ has an expected correctness of $\frac{1}{n!}$. Then the expected total 'number' of correct steps is, $$\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} ... \frac{1}{n!}$$

If we calculate this sum for increasing $n$, we see it tends towards a constant value of roughly 1.72. |n |1 |2 |3 |4 |5 |6| |----|----|----|----|----|----|-| |$\sum_{i=1}^n \frac{1}{i!}$|1|1.5|1.67|1.71|1.717|1.718

This constant comes up often in mathematics, and so it has a special symbol, $\mathrm e$. But with one difference of starting from $\frac{1}{0!} = 1$, as this form is more useful. $$\mathrm e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} ... $$

Therefore, $\mathrm e - 1$ can be physically interpreted as the expected number of correctly arranged steps in the above staircase problem.

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  • $\begingroup$ The specific problem I designed may not be as appealing and feel artificial compared to the popular interest-rate problem. But if you understand the mathematics, perhaps you might think of a situation more relevant or interesting to students. $\endgroup$ Commented Jun 12, 2023 at 8:30
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This post is just an elaboration of my comment to user52817's answer on the hat check problem. The rencontres numbers $D_{n,m}$ enumerate the permutations of $n$ items with exactly $m$ fixed points. Their number can be computed by multiplying the number of ways of choosing which items are the fixed ones by the number of derangements of the remaining items, that is, by multiplying the number of ways of choosing which $m$ people get their own hats by the number of ways of assigning hats not their own to the remaining $n-m$ people, $$ D_{n,m}=\binom{n}{m}D_{n-m,0}=\frac{n!}{m!\,(n-m)!}\sum_{k=0}^{n-m}(-1)^k\frac{(n-m)!}{k!}. $$ A standard expression for the number of derangements has been used on the right. (A derivation is at the end of this post.) As a consequence, the probability that a random permutation of $n$ items has exactly $m$ fixed points is $$ \frac{D_{n,m}}{n!}=\frac{1}{m!}\sum_{k=0}^{n-m}(-1)^k\frac{1}{k!}. $$ Fixing $m$ and letting $n$ go to infinity we get $$ \lim_{n\to\infty}\frac{D_{n,m}}{n!}=\mathcal{N}\frac{1}{m!}, $$ where we can think of $\mathcal{N}=\sum_{k=0}^\infty(-1)^k\frac{1}{k!}$ as a normalization constant for the probability distribution. You will recognize that $\mathcal{N}=\frac{1}{e}$, but we can pretend we don't know that yet. Probabilities for all possible numbers of fixed points must sum to $1$, which gives $$ 1=\mathcal{N}\sum_{m=0}^\infty\frac{1}{m!}. $$ So in addition to the expression for $\mathcal{N}$ already given, $\mathcal{N}$ is also the reciprocal of $\sum_{m=0}^\infty\frac{1}{m!}$. The consistency of these two characterizations can be verified by multiplying out the two sums and using an identity for the alternating sum of binomial coefficients to show that the product is $1$. (A version of this calculation for finite $n$ is given at the end of this post.) In addition to its role as normalization constant, $\mathcal{N}$ is the asymptotic probability that a random permutation has no fixed points, and also the asymptotic probability that it has exactly one fixed point. It should be noted that the probability distribution we have obtained by taking the $n\to\infty$ limit is the Poisson distribution with $\lambda=1$.

Whether students will see rencontres numbers as a compelling enough problem to motivate the definition of a new mathematical constant is a good question. My feeling is that the Poisson distribution is the fundamental thing since it shows up over and over again, with the rencontres numbers being but one application.

One potential objection that seems to have been raised is that Poisson probability is almost always defined as a limit of Bernoulli probability and that taking this limit requires the use of (a generalization of) the limit that shows up in the interest rate application. Which, I guess makes the Poison interpretation somehow "not independent" of the interest rate formula. All of this is discussed in David E Speyer's answer and the comments below it.

I'm actually not sure why that's so bad. Two points I would make are (1) that the rencontres numbers application shows that limits not involving the interest rate formula also arrive at the Poisson distribution, and (2) that it is just as interesting or more so to show interconnections between ideas as it is to show completely independent derivations. If the Poisson distribution is too complicated for students at this level to derive from scratch, it is still useful to them to teach some of the bare facts about it, such as that the probability of exactly $k$ hits is proportional to $\frac{\lambda^k}{k!}$.

With regard to the last point, it may be that students at the introductory calculus level will see the Poisson distribution as intimidating because it involves a transcendental expression $e^\lambda$. A key point is that the ratio of any two Poisson probabilities is just a power of $\lambda$ times a rational number--so the kind of thing that is very familiar from algebra. The only role the transcendental expression plays is as an overall normalization constant that divides every probability equally and that cancels out if you take ratios of probabilities.

So my view is that the Poisson distribution is very down-to-earth motivation for introducing the transcendental function $e^\lambda$ and the transcendental number $e$, the latter being relevant to the very important special case $\lambda=1$. (The importance of $\lambda=1$ is that, for a Poisson process unfolding in continuous time or continuous space, the inverse of the hit rate provides a natural time or distance scale for the problem. Hence one of the bald facts that I think students ought to know is that in a Poisson process the probability of zero hits within the natural time scale for the problem is $\frac{1}{e}$. (And also for one hit.) This is the most compelling occurrence of the constant $e$ "in nature" that I know of.

Some details filled in: The formula for $D_{n,0}$, the number of derangements of $n$ items, can be derived by defining $C_S$ to be the set of permutations of $1,2,\ldots,n$ in which each element of the set $S$ is a fixed point, so that $\lvert C_S\rvert=(n-\lvert S\rvert)!$. By inclusion-exclusion we have \begin{align} D_{n,0} &=n!-\sum_i\lvert C_{\{i\}}\rvert+\sum_{i<j}\lvert C_{\{i,j\}}\rvert-\sum_{i<j<k}\lvert C_{\{i,j,k\}}\rvert+\ldots+(-1)^n\lvert C_{\{1,2,\ldots,n\}}\rvert\\ &=n!-n(n-1)!+\binom{n}{2}(n-2)!-\binom{n}{3}(n-3)!+\ldots+(-1)^n\\ &=\frac{n!}{0!}-\frac{n!}{1!}+\frac{n!}{2!}-\frac{n!}{3!}+\ldots+(-1)^n\frac{n!}{n!}. \end{align} The probability that a random permutation of $n$ items is a derangement is therefore $$ \frac{D_{n,0}}{n!}=\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\ldots+(-1)^n\frac{1}{n!}. $$

That the probabilities for all possible numbers of fixed points sum to $1$ must, of course, work out for finite $n$. Let's see how this goes: \begin{align} \sum_{m=0}^n\frac{D_{n,m}}{n!} &= \sum_{m=0}^n\frac{1}{m!}\sum_{k=0}^{n-m}(-1)^k\frac{1}{k!}\\ &= \sum_{t=0}^n\sum_{m=0}^t\frac{1}{m!}(-1)^{t-m}\frac{1}{(t-m)!}\\ &= \sum_{t=0}^n\frac{1}{t!}\sum_{m=0}^t\binom{t}{m}(-1)^{t-m}\\ &= \sum_{t=0}^n\frac{1}{t!}\delta_{t,0}=1. \end{align} In the second line we have introduced a new summation index $t=m+k$; in the fourth line $\delta_{a,b}$ is the Kronecker delta; it arises from the evaluation of the alternating sum of binomial coefficients.

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This is a similar answer to Raciquel's: To look at Poisson processes. As a concrete example of the sort of thing we might want to model with a Poisson process, consider traffic accidents per day in a large city. Suppose that our records show that there are an average of $\lambda$ accidents per day. Then the Poisson model says that the probability of having $k$ accidents in a given day is $\tfrac{\lambda^k e^{-\lambda}}{k!}$. The fact that the probabilities add up to $1$ is thus equivalent to $e^{-\lambda} \sum_{k=0}^{\infty} \tfrac{\lambda^k}{k!} = 1$, so $\sum_{k=0}^{\infty} \tfrac{\lambda^k}{k!} = e^{\lambda}$.

Let's see how to motivate the Poisson model without using the Taylor series of $e^x$, so that this is not circular: Suppose that there are $N$ cars. Since there are $\lambda$ accidents per day, the probability of a particular car having an accident on a particular day is $\lambda/N$. For a given set of $k$ cars, if accidents are independent, the chance of these $k$ cars having accidents and no other cars doing so is $$(\lambda/N)^k (1-\lambda/N)^{N-k}.$$ There are $\binom{N}{k} = \tfrac{N(N-1)(N-2) \cdots (N-k+1)}{k!}$ sets of $k$ cars, so the chance that $k$ cars have accidents on a given day is $$\frac{N(N-1)(N-2) \cdots (N-k+1)}{k!} (\lambda/N)^k (1-\lambda/N)^{N-k} \approx \frac{N^k}{k!} (\lambda/N)^k(1-\lambda/N)^N \approx \frac{\lambda^k}{k!} e^{-\lambda}.$$

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  • $\begingroup$ @user24096 I think that this is short of a rigorous proof, but not circular. It is very common in intro calc classes to compute $\lim_{N \to \infty} (1+k/N)^N = e^k$ my making the substitution $N=Mk$, so the limit turns into $\lim_{N \to \infty} (1+1/M)^{kM} = \left( \lim_{N \to \infty} (1+1/M)^M \right)^k = e^k$. This is already a little fishy if $k$ is not a positive integer, since the limit $\lim_{M \to \infty} (1+1/M)^M = e$ is usually introduced for integer $M$. It is more fishy when $k$ is negative, as in this post. (continued) $\endgroup$ Commented Jun 8, 2023 at 20:57
  • $\begingroup$ One could justify it by showing that $1-\lambda/N$ is close enough to $(1+1/N)^{-\lambda}$ to justify the interchange, but, for most students, I think that would be too confusing. I was imagining a student for whom $\lim_{N \to \infty} (1+k/N)^N = e^k$ was already familiar, but who wasn't familiar with the Taylor expansion yet. $\endgroup$ Commented Jun 8, 2023 at 20:59

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