I'm asking this question from the point of view of an introductory non-rigorous calculus instructor. Calculus textbooks have different approaches about how to define $e$ and $\ln$. For example, my current textbook defines $e$ as the number such that

a) $\int_1^e \frac1x \, \mathrm{d}x= 1$.

I've also seen $e$ defined as the number such that

b) $\frac{\mathrm{d}}{\mathrm{d}x} e^x |_{x = 0} = 1$, or

c) $e := \lim_{n \to \infty} (1 + \frac1n)^n$, or

d) $e := \sum_{n=0}^\infty \frac{1}{n!}$

Of course no matter where you start, you can derive the others, but the distinction between which facts are definitions and which are derived feels very artificial, especially to the students! The questions:

  • Do you prefer one of these approaches, and if so, why?
  • Is it a good idea to explain the (arbitrarily) chosen definition and the subsequent derivations to students? I often wonder: why not just say $e$ is an important number that is about 2.718, and talk about its properties? Maybe the appropriate place for these worries is an analysis course.
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    $\begingroup$ I have always hated saying <foo> is important and go on about their properties instead of showing why it is important. As a student I don't care if it's above my level, as long as you attempt to show me its importance. $\endgroup$ Commented Jun 30, 2014 at 22:27
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    $\begingroup$ Please don't remove content from a question when editing it especially if you have enough rep to edit without peer review... the fact that I am teaching non-rigorous freshman calculus is important to the question. $\endgroup$ Commented Jul 1, 2014 at 10:29
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    $\begingroup$ Students are exposed to e and ln in high school, well before their first calculus class. Do you feel you need to 'undo' some harm or approach it as if for the first time? $\endgroup$ Commented Jul 1, 2014 at 13:08
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    $\begingroup$ @JoeTaxpayer That's part of the question. Our calculus books certainly seem to think they should start from scratch, always defining either e or ln and never referring to prerequisite knowledge. I think that's odd and worth discussing! $\endgroup$ Commented Jul 1, 2014 at 19:57

9 Answers 9


I think it is a good thing to talk about how there are some concepts where there are choices for where you start when definining them. It happens in linear algebra too, with the definition of linear dependence. You need to talk about how there is this web of connected properties, and it depends on what you're trying to achieve as to where you start. I do recommend prefacing any discussion of it with a general discussion on logs and exponentials using familar numbers, so they get a feel for how they ought to work first. This can even be done without graphs.

As to which definition of e I prefer...

Firstly, it's the functions ln(x) and $e^x$ that are the important thing, rather than the number $e$ itself per se. The number $e$ just crops up as a useful number to unify things.

I quite like the one where ln(x) is defined to be $\int_1^x \frac{1}{t} dt$, and e is the number for x that makes this area come out to 1. I have several reasons:

  1. There is a real need for this function to have an integral because no power of $x$ differentiates to give $\frac{1}{x}$.

  2. You can approximate e by using upper and lower sums of an easy function like 1/x.

  3. The area truly exists because you can draw it there on the page, whereas it's a bit dicey as to whether it's reasonable to differentiate a function like $a^x$ when you don't know what it even means when $x$ is irrational!

  4. Defining e using the area of a shape we can draw makes it feel a bit like $\pi$, which can be defined as the area of a circle and approximated by polygonal areas, and is one of the other famous irrationals we are familar with.

EDIT: For reference, here is the full sequence of definitions as I understand it (since it has been hotly debated in the comments):

As far as I understood it, it goes like this: for positive real $a$ you still define $a^1:=a$, $a^n:=a^{n−1}\times a$ for $n\in \mathbb{N}$, $a^0:=1$, $a^{−n}:=\frac{1}{a^n}$ for $n\in\mathbb{N}$, $a^\frac{1}{n}:=\sqrt[n]{a}$ for $n\in\mathbb{N}$ and $a^\frac{m}{n}:=(a^\frac{1}{n})^m$ for $n\in\mathbb{N}$ and $m\in\mathbb{Z}$.

Then you can still prove all the familiar properties of powers from these definitions, and that $\frac{d}{dx}x^r=rx^{r−1}$ for any rational $r$ using methods of first principles, without using logs at all. The issue is how you define $a^x$ for irrational $x$, because it needs a definition before you can differentiate $a^x$ or indeed define an inverse for the function.

If we want, can implicitly assume it does have a consistent definition which works for all the other properties we observe about powers, and investigate the behaviour of this function and its inverse and their derivatives, calling the inverse $\log_a (x)$, but realising that there is a hole in our definition of it and a hole in our arguments about derivatives until we know what $a^x$ means for irrational $x$.

Now we define $\ln(x) := \int_1^x \frac{1}{t}dt$ which immediately gives us that $\frac{d}{dx} \ln(x) = \frac{1}{x}$. From this definition we are able to prove all the familiar log laws for $\ln$. For example, to prove that $\ln(x^r) = r\ln(x)$ we can do the following:

By the chain rule, $\frac{d}{dx} \ln(x^r) = \frac{1}{x^r}\cdot rx^{r-1} = \frac{r}{x}$. However $\frac{d}{dx} r\ln(x) = \frac{r}{x}$. Since these two functions have the same derivative, they differ by a constant and so $\ln(x^r) = r\ln(x) +C$ for some $C$. Substituting $x=1$ gives $\ln(1) = r\ln(1) + C$, so $C=0$ and $\ln(x^r) = r\ln(x)$. Note that this only works for $r$ rational, because we still don't know what $x^r$ means for $r$ irrational.

We define $\exp$ to be the inverse of $\ln$, and using the properties of $\ln$ we are able to prove that it behaves the same as the powers laws we know. For example, we can prove that $\exp(a+b) = \exp(a)\times\exp(b)$ and $\exp(rx) = \exp(x)^r$. However this last one is only currently proved for rational $r$.

We define the particular number $e$ as the number such that $ln(e)=1$ or equivalently $e := exp(1)$, because we notice that $\exp(x)$ is the same as $e^x$ for all rational $x$ (and we know what $e^x$ is for rational $x$ because we defined what rational powers meant already without reference to $\exp$ or $\ln$). So now we define $e^x := \exp(x)$ for irrational $x$ and we define $a^x := \exp(\ln(a)\times x)$ for irrational $x$. We don't need to define $a^x$ for rational $x$ because it already has a definition, but we notice that even for rational $x$ this new calculation gives the correct result. We also notice that now $\exp(x) = e^x$ for all real $x$.

Finally, we can prove how derivatives of power functions work, and we can define $\log_a$ as the inverse of the function sending $x$ to $a^x$. And finally we observe that $\log_e(x)$ is the same as $\ln(x)$ for all real $x$.

This seems like a roundabout way of doing things, but if we define $a^x$ for $x$ irrational as the limit of a sequence, then all our proofs have to involve how limits of sequences interact with these functions and their derivatives and they are therefore actually harder.

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    $\begingroup$ @Matt:@Matt: This definition of $a^x$ is perfectly valid and used in several textbooks. The usual approach is not to assume any previous definition of powers and provide the intuitive $a^n=\prod_{k=1}^na$ for natural $n$ as a theorem. Your second sentence makes no sense; you must use the definition of $a^x$ for any property involving $a^x$. $\endgroup$
    – Dennis
    Commented Jul 2, 2014 at 0:20
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    $\begingroup$ @Dennis I see your point. I generally prefer to think of $a^2=a\times a$ rather than its "true" definition being $a^2=\exp(2\ln(a))$ but I don't disagree with the self-consistency of it. Cheers. $\endgroup$
    – Matt
    Commented Jul 2, 2014 at 0:42
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    $\begingroup$ +1 @DavidButlerUofA Since you're on the logarithm path, it is worth including the identity: $\ln(xy)=\int_1^{xy}\frac{1}{t}dt=\int_1^x\frac{1}{t}dt+\int_x^{xy}\frac{1}{t}dt=\ln(x)+\int_1^{y}\frac{1}{xu}xdu=\ln(x)+\ln(y)$ where the 2nd to last equality uses the substitution $t=xu$. It is from this that one can prove $\ln(a^x)=x\ln(a)$ for $x\in\mathbb{N},\mathbb{Z}$,and $\mathbb{Q}$. Lastly, I would disagree with "then all our proofs have to involve how limits of sequences interact..." You need not resort back to the limit definition; just stand upon whatever theorem is most convenient. $\endgroup$
    – Matt
    Commented Jul 3, 2014 at 22:44
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    $\begingroup$ @DavidButlerUofA Nice one! Maybe this logarithm path is faster than the exponential path proposed in my answer. Regarding the issue of using limits, one reason I like it is because it is a simple introduction to what a real number "really" is at least from the perspective of many real analysis courses. That is, Cauchy sequences of rational numbers. But I am not going to split hairs on this point. :) $\endgroup$
    – Matt
    Commented Jul 3, 2014 at 23:23
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    $\begingroup$ @Matt I suppose, as we all agree, sometimes there are multiple pathways to definitions and you need to choose the one that works for your purposes, knowing that the others are there. This particular method is focussed on the behaviour of functions and their derivatives and integrals. If you wanted to do it without reference to this and build it fully on the numbers themselves and functions of them in the pure sense of correspondence between numbers, then sequences and series is probably the way to go, but it would be less easy to be rigorous for beginners. $\endgroup$ Commented Jul 3, 2014 at 23:34

I think a lot depends on context -- what course you are teaching and what the characteristics of that course are. If by "Freshman college calculus" you mean what I think you mean -- namely, a non-rigorous course that treats a lot of things using intuitive and heuristic arguments -- then I wonder if "define" is even the right verb to use. Students come into such a class having already encountered $e$; they may not feel like they really understand it well, but they should know it is a number about equal to 2.718, and they know it is used to describe exponential growth and decay, and that it has something to do with "continuous compounding". They have some familiarity with $f(x)=\ln x$, although the motivation for using $e$ as the base of a logarithm may seem far from "natural".

If that is the background of the students coming in, then I would not define $e$. Rather, I would do (have done) something roughly like this:

We want to find the derivative of functions of the form $y=a^x$, where $a$ is some positive real number. So we compute $$ \lim_{h \to 0} \frac{a^{x+h}-a^x}{h} = a^x \lim_{h \to 0} \frac{a^h-1}{h}$$ and we notice that the limit does not depend on $x$ at all; it's just some constant that depends only on the choice of $a$. Let's denote that constant (temporarily) by the symbol $K_a$. Then we have $$y=a^x \implies y'=K_a a^x$$

This is a really interesting (and surprising) formula. When we take the derivative of a power function, the exponent decreases; but when we take the derivative of an exponential function, we end up with the same exponential function, times some constant that depends only on the base of the exponent.

So what is that constant? Well, students can do some numerical approximation with a graphing calculator or a spreadsheet: Choose a single specific value for $a$ and make a table of values of $\frac{a^h-1}{h}$ for smaller and smaller values of $h$ to estimate the corresponding value of $K_a$. Compare results with other students to compile a table of values of $K_a$ for different values of $a$. Graph them. Any ideas what the relationship is?

At this point I reveal the "amazing fact":

For any choice of $a$, the constant $K_a$ turns out to be exactly $\ln a$.

Students can, at the very least, convince themselves that this claim is consistent with what they found above.

Now as a consequence of this amazing fact, we have that if a happens to be equal to e, then the exponential function is its own derivative.

Now here is where my teaching sequence may differ from others: I make no attempt to prove this amazing fact -- to do so you would have to already have a rigorous definition of $\ln a$ available, and in this type of course we are dealing with students who have no such definition to work with. Nor do I take the amazing fact as a definition of $\ln a$; one could certainly do that, but then you would have to prove that this definition syncs up in some way with what they already know about $\ln$.

Instead, I take the position that $\ln$ and $e$ are basically already "out there" in their background knowledge, not really defined but available to them to use. They know some properties about them, but none of them are "primitive", nor are any of them "derived". And I am adding one more property to that constellation, making it seem plausible through numerical examples but not trying to do anything more rigorous than that.

All of that seems to me to make sense in a 1st-year, non-rigorous college Calculus course.

Is this were a Precalc course? Then we'd be having a completely different conversation. We'd be talking about compound interest, and what it means to compound interest twice per year, or three times per year, or $N$ times per year. We'd see the effective growth rate $(1 + 1/N)^N$ appear naturally in that context, and we'd explore what happens when $N$ gets very large. Eventually we'd be convinced (but would not prove) that as $N$ gets very large this value seems to converge to some number about equal to $2.718$, and we'd call that number $e$, and then talk about how whenever something is growing (or decaying) continuously (rather than at discrete intervals) it makes sense to express that growth (or decay) in terms of $e$.

If this were a rigorous (i.e. proof-based) Analysis course? The sequence of topics in such a course likely has convergence of sequences ("limit as $N \to \infty$") before either limit at a point (which necessarily precedes derivatives) and also before convergent series (which necessarily precedes integrals), so the first opportunity to define $e$ would be as $\lim_{N \to \infty} (1 + \frac{1}{N})^N$. I would start there (making sure to motivate it by connecting it to what students remember about compound interest and other contexts of continuous growth). Then later, when we are covering other types of limits, use that definition to prove both that $\lim_{h \to 0} \frac{e^h-1}{h} = 1$. Then go on from there, revisiting the topic when we cover derivatives, series and integrals, each time proving the "new" property in terms of the previously established ones.


When I teach calculus I, my initial working definition for $e$ is that it is the real number implicitly defined by: $$ \lim_{h \rightarrow 0} \frac{e^h-1}{h}=1 $$ Naturally, this allows me to derive that $\frac{d}{dx} e^x = e^x$ as $$ \lim_{h \rightarrow 0} \frac{e^{x}e^h-e^x}{h}=e^x\lim_{h \rightarrow 0} \frac{e^{h}-1}{h}=e^x. $$ Of course, we know this. At this point, I usually say something like, you know, if we wanted, we could have defined $e$ to be the number for which $\frac{d}{dx}e^x = e^x$. That is an interesting equation, it says the slope of the tangent line to $y=e^x$ is the same as the value of the function at the point of tangency. In the world of all possible exponential functions the base $e$ is very special.

In what follows below, this is the sequence I make the comments, but, you could rearrange them depending on what sequence of topics you follow.

  • Later, when I discuss the second derivative test and give an introduction to Taylor polynomials I show $e^x = 1+x+ \frac{1}{2}x^2 + \cdots$. This allows me to explicitly calculate $e$ with just a little arithmetic. Often at this point, I admit that this is our third definition and I know of at least 5 more or less distinct definitions. That said, the power series definition is near and dear to my heart as it is the method I use to define exponentials in supermath and other formal systems... (this I usually abstain from saying)
  • Later, when I treat indeterminant powers I show $e = \lim_{n \rightarrow \infty} \left( 1+\frac{1}{n}\right)^n$. At this point, if I have my wits about me, I will foolishly attempt to explain how this ties into continuous compounding of interest. Indeed, continuous multiplication is how the exponential is thought of in physics and a few other places. So, it's worthwhile to mention this.
  • Later, after we've covered FTC part I, I point out how $\frac{d}{dx} \int_1^x \frac{dt}{t} = \frac{1}{x}$ suggests that $\int_1^x \frac{dt}{t} = \ln (x)$ as we have already shown that $\frac{d}{dx} \ln (x) = \frac{1}{x}$ via implicit differentiation. Then, I'll show how to derive the various laws of logs as consequences of differential and integral calculus. I then confess that there are other instructors who take the integral as the definition of the log and so the exponential is introduced as the inverse function of the log. However, I tell them that was not our approach since I thought they all could use a whole semester to practice logs and exponential functions. Of course, my approach also has holes ( I don't prove laws of exponents, I just assume we know them and $a^x$ is somehow reasonable for $a>0$, admittedly, there is more substructure here, but I haven't found a class with sufficient curiosity to push the issue)

Naturally, there are other ideas to plan. Intuition for exponential functions is worth cultivating as it is one of those constructions which appears in a multitude of abstract fields of mathematics. I also like this topic because it gives us one topic which spans the entire course of calculus both integral and differential.


A textbook I have used first defines $$\ln x =\int_1^x \frac 1 t \, \mathrm{d}t,$$ and then defines the number $e$ as in your Definition a).

Next, the exponential function $e^x$ is defined as the inverse of the function $\ln x$. (They show that this definition of $e^x$ coincides with the usual one from repeated multiplication when $x$ is an integer.) Finally, for $a>0$, they define exponentiation with real exponent as $$a^x=e^{(\ln a)x}$$

This seems a bit backwards, but I can't find anything wrong with it. One advantage could be that formal properties like $$a^x a^y=a^{x+y}$$ can be proven more directly than the way it's done in the Real Analysis textbooks I have seen.

  • $\begingroup$ This would only work if transcendentals are done after both differentiation and integration have been introduced. $\endgroup$
    – user507
    Commented Jul 1, 2014 at 13:07

I have few remarks which may be enlightening. I have not seen most of these spelled out anywhere for some reason.

I do not prefer one definition over the others: in fact I think this is a very nice place to try to get students to compare and contrast various definitions, and to derive each definition from the other.

Here are my observations:

  1. $e^x = [e^{\frac{x}{n}}]^n$. For very large values of $n$, $e^{\frac{x}{n}} \approx 1+\frac{x}{n}$ by the tangent line approximation to $y=e^t$ at $t=0$. So we obtain $e^x \approx (1+\frac{x}{n})^n$. This approximation should get better and better as $n$ increases without bound, motivating the possibility that $e^x = \displaystyle\lim_{n \to \infty} (1+\frac{x}{n})^n$. Proving this is a nice challenge.

  2. Another way to $(1+\frac{x}{n})^n$ is Euler's Method. Apply Euler's method to the diff EQ $f'(x)=f(x)$ with $f(0)=1$ to the interval $[0,x]$, with $n$ subintervals. We obtain:

$\begin{align*} f\left(\frac{x}{n}\right) &\approx 1+\frac{x}{n}\\ f\left(\frac{2}{n}\right) &\approx f\left(\frac{x}{n}\right)+\frac{x}{n}f\left(\frac{x}{n}\right) = \left(1+\frac{x}{n}\right)f\left(\frac{x}{n}\right) \approx \left(1+\frac{x}{n}\right)^2\\ &\vdots\\ f(x) &\approx \left(1+\frac{x}{n}\right)^n \end{align*}$

  1. Assuming existence and uniqueness of ODE's is all that is needed to derive $e^{a+b} = e^ae^b$ and $e^{ax} = (e^x)^a$. For the first, observe that $f_1(x) = e^{a+x}$ and $f_2(x) = e^ae^x$ both satisfy the DE $y'=y$ with initial condition $y'(0)=e^a$. For the second note that $f_1(x) = e^{ax}$ and $f_2(x) = (e^x)^a$ both satisfy the DE $y'=ay$ with initial condition $y'(0)=1$.

  2. You may also enjoy this post where I spell out how solving equations like $2^x=3$ is connected to finding areas under $\frac{1}{t}$, and how this is all linked with the number $e$. This story can also be turned into a very nice activity, where the students discover all of these connections themselves.


My personal oppinion is that the number e itself has virtually no significance and one should rather focus on the exponential function. Depending on which course you are teaching, it can show up in various ways.

If you are giving an introduction to analysis, defining limits, series, etc. than c) and d) are very nice, but of course the exponential series $$ \exp(x)= \sum_{n=0}^\infty \frac{x^n}{n!} $$ is much more important than only its value at $x=1$. It can be highlighted that this is a very important if not the stereotype of a power series, one can discuss radius of convergence, etc.

If the focus is on functions and derivatives or differential equations, one may of course ask for a function that solves $$ f'=f, f(0)=1 $$ to again find $\exp$. (Notice that your version of b) is not a definition, e.g. $f(x)=x$ also fulfils $f'(0)=1$.)

My favourite version for lower grades is to introduce exponentiation by integers in the obvious way: multiple addition -> multiplication with integers and similarly multiple multiplication -> exponentiation. Then one can introduce $\log_b$ as the inverse of $b^x$, explain the calculus of exponentials and logarithms, in particular $$ b^x =c^{xlog_c(b)} $$ which highlights the insignificance of the basis and mention the preferred choices of $c=10$ for convenience and $c=e$. This leads then naturally to the question why $c=e$ is actually preferred for e.g. implementation in calculators, which could bring you back to the exponential series. (If pupils are actually interested: https://math.stackexchange.com/questions/18445/fastest-way-to-calculate-ex-upto-arbitrary-number-of-decimals )

  • $\begingroup$ Among all numbers a, consider the derivatives of y = a^x at x = 0. There is a unique base such that the derivative is 1. $\endgroup$ Commented Jul 1, 2014 at 10:30
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    $\begingroup$ @Chris: Sure. But what's the point? ;) In my opinion f'=f is more interesting. $\endgroup$
    – Echsecutor
    Commented Jul 1, 2014 at 10:35
  • $\begingroup$ I'm clarifying that in fact (b) from the OP is a perfectly valid definition of e if you get the quantifiers right. Obviously f'=f is more interesting. $\endgroup$ Commented Jul 1, 2014 at 19:53

Consider the question: What is the form of the function $f:\mathbb{R}\rightarrow\mathbb{R}$ that has the following two properties?

  1. $f'(x)=f(x)$
  2. $f(0)=1$

The unique answer is of course $f(x)=e^x$ however the proof is longer than people generally expect if they haven't considered this before, (for instance, we don't assume a priori that the solution is in the form of an exponential) but it consists of a series of steps that are mostly within the grasp of freshman calculus students. The only step that may perhaps have to be put off until later in the course is proof of absolute convergence. Here it is in outline form:

  • Derive the Maclaurin series $f(x)=\sum_{n=0}^\infty\frac{x^n}{n!}$.
  • Prove absolute convergence (can be put off until later.)
  • Derive $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb{R}$ by multiplying out $\sum_{n=0}^\infty\frac{(x+y)^n}{n!}$ and invoking the binomial theorem.
  • Prove $f(xy)=f(y)^x$ for all $y\in\mathbb{R}$, as $x$ varies over successively larger sets $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}$.
  • Define $e\equiv f(1)$. Thus $f(x)=e^x$.

This may seem like a lot of overhead for a seemingly elementary result, but I have asked this to a number of math grads and asked it here. There doesn't seem to be a substantially shorter proof that doesn't take for granted a known result about $e$ or the natural logarithm.

Do you prefer one of these approaches, and if so, why?

Not speaking as a professional educator, but as an occasional tutor, I prefer the above $f'(x)=f(x)$ approach as I feel it is both the most "natural" motivation, as well as one which is most directly applicable to real-life applications (e.g. Newton's law of cooling, etc.). The other results as listed in the question follow fairly readily from this approach. It is also a good introduction to more formalized/advanced proofs.

Is it a good idea to explain the (arbitrarily) chosen definition and the subsequent derivations to students? I often wonder: why not just say e is an important number that is about 2.718, and talk about its properties? Maybe the appropriate place for these worries is an analysis course.

One can begin to compile a list of all the places where $e$ shows up. E.g. question to college freshmen: Did you know that in order to create mp3 files, one must use $e$ in the fourier transform to encode the sound as frequencies? (get as in-depth on this point as desired.) Another: It is one of the most important numbers in mathematics. The one equation which relates the 5 most important constants is: $$ e^{i\pi} + 1 = 0 $$

This list can go on and on...

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    $\begingroup$ If you assume existence and uniqueness of solutions to ODE, then one can prove $f(ax) = [f(x)]^a$ for the solution of $f'(x)=f(x)$, $f(0)=1$,by verifying that both $g_1(x) = f(ax)$ and $g_2(x)=[f(x)]^a$ both satisfy $g'(x) = ag(x)$ with $g(0) = 1$, so they must coincide. No need to follow the long and cumbersome sequence you outline above. $\endgroup$ Commented Oct 28, 2014 at 23:44
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    $\begingroup$ Yes certain assumptions allow one to skip over steps that might be considered "cumbersome". Also your method says nothing about the actual value of $e\equiv f(1)=2.718...$. In any case I appreciate your point, as in a first introduction this may be an appropriate and expedient way to present the material without sweeping too much under the rug like is all too often done. +1 to your answer. $\endgroup$
    – Matt
    Commented Oct 29, 2014 at 17:10
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    $\begingroup$ if you look at my answer, you can see that starting with the differential equation, Euler's method gives $e = \displaystyle\lim_{n \to \infty} (1+\frac{1}{n})^n$. I hope you did not interpret "cumbersome" as a "diss": I very much like your answer! $\endgroup$ Commented Oct 29, 2014 at 17:14
  • $\begingroup$ @StevenGubkin Yes I was referring to your comment in which you are suggesting that by assuming existence/uniqueness of ODE solutions, then there is "No need to follow the long and cumbersome sequence [I] outline above." This tells you nothing about the value of $e$ - that requires a separate step, as you are describing, which involves about as much work as the first two steps I am proposing, in my opinion. $\endgroup$
    – Matt
    Commented Oct 29, 2014 at 17:26

I prefer d), which fits with the all-important definition $$\exp(x):=\sum_{n=0}^\infty \frac{x^n}{n!}$$

It is a straightforward definition. It makes obvious sense. The proof that the sum converges is about as easy as it can be. By contrast:

Definition a), via the integral of $1/x$ is indirect.

Definition b), via the derivative of $e^x$, only makes sense if you've already defined both differentiation and exponentiation over arbitrary real bases.

Definition c), the limit of $(1+1/n)^n$, is both a more difficult limit than the sum, and not as central a property as the others.

So I prefer d).

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    $\begingroup$ Totally straightforward, from the sky. When I took calculus I it was presented with (a) and I liked it very much. Indirect? Yes. Proving it's a series depends on commenting about series and how to use them, which is not typically calculus I, at least it wasn't until calculus III that I dealt with power series. $\endgroup$ Commented Jul 1, 2014 at 5:55
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    $\begingroup$ I learned d) in sixth grade, and it made sense then. An occasional gift from the sky is ok by me. $\endgroup$
    – user173
    Commented Jul 1, 2014 at 11:59
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    $\begingroup$ I've reached my threshold for gifts from the sky, it happened way too often. $\endgroup$ Commented Jul 1, 2014 at 14:44

When I taught high school Calculus, I talked about how e is a transcendental number and explained that it went on forever, never repeating, and that any finite length number sequence you search for could be found inside it. I had a student give me their date of birth and then looked it up in the pi and e search engine and found it in both pi and e (with 2 billion digits the odds of a 6 digit number being found it in is rather high): http://www.subidiom.com/pi/

which rather impressed them. I then had them consider whether e is found inside pi or pi inside of e - and that rather boggled their minds. I also presented the difference between an algebraic and trascendental irrational number.

I also had them consider e ^ pi and pi ^ e and whether those algebraic and transcendental. Also that e ^ pi is transcendental, but we don't know about pi ^ e. (Which is strange, because it feels like it should be, but no one has been able to prove it according to most sources I've looked at...) Here's some other fun transcendental numbers: http://sprott.physics.wisc.edu/pickover/trans.html

That seemed to be "good enough" for high school calculus - they were convinced that it was a number similar to pi and that it was important enough to have its own button on their calculator. And they could see that it worked similar to other logarithm bases to solve exponential functions - like, you could use ln instead of log and still get the same answers, and in fact you could use any base to solve the types of things we were doing. I didn't really rigorously prove it's derivative, but using the graphing calculator they could see rather readily that e^x and ln x are inverse functions and that the derivative of ln x was 1/x. (Most graphing calculators let you graph, say, the 1st derivative of a function, using a numerical approach).

I think if I had gone on to teach second semester calculus I might have used the area under the curve definition, or perhaps the limit of the compound interest formula. But the big thing, I think, is what you hope to get from the definitions - like, are there particular results you're hoping to prove that need a certain definition?

If not, maybe it's okay to have a bit of mystery around e and to merely pique students interest in learning more about it. Like, I'd rather have students going "oh wow, e is really fascinating - let's go read a book on it" (say, Eli Maor's excellent Story of E) then "I know a formal definition of e, but it put me to sleep."

  • 5
    $\begingroup$ The claim that "any finite length number sequence you search for could be found inside it" (where by "it" I presume you mean "its representation as a decimal in base 10") is commonly believed to be true, but is not known to be so. It is not even known whether every digit appears infinitely many times. $\endgroup$
    – mweiss
    Commented Jul 3, 2014 at 17:53
  • $\begingroup$ That's part of the fascination of pi and e for me - the fact that so many things that seem obvious are not necessarily proveable. Like, why can't we prove pi^e is transcendental? $\endgroup$
    – James S.
    Commented Jul 4, 2014 at 18:50
  • $\begingroup$ But yes, I agree, qualifying that with "it is thought this is true" makes sense from a math perspective. $\endgroup$
    – James S.
    Commented Jul 4, 2014 at 18:51
  • $\begingroup$ Do we know if pi and e are evenly distributed - like, if we go far enough will the digits start to even out in number? $\endgroup$
    – James S.
    Commented Jul 4, 2014 at 18:51

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