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I'm starting to study calculus and I've become very interested in Euler's number ($e$). I understand that the property of being its own derivative makes it the "natural" base to work on for studying rates of change.

However, I was wondering what would happen if we pretended not to know about the existence of $e$. Would trying to find the derivative of something like $a^x$ lead us into finding the definition of $e$ or is it possible to avoid $e$ altogether?

In this video it says that not using $e$ in calculus leads to some pretty crazy math. What does that math look like?

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closed as unclear what you're asking by Xander Henderson, Simply Beautiful Art, GAVD, J. M. is a poor mathematician, user223391 Oct 13 '17 at 2:04

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    $\begingroup$ It couldn't change anything. It's just a number. The number would still exist, somewhere between $2$ and $3$, and we would just rediscover $e$ anyway. It has too many nice properties to be ignored. $\endgroup$ – Squirtle Oct 11 '17 at 22:32
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    $\begingroup$ I don't understand. What makes that "interesting"? Note that there are other ways to arrive at $e$ (and the exponential function, which is really what you are talking about - not the number $e$). For example, the anti-derivative (indefinite integral) of $1/x$ is the natural logarithm (in base $e$) - up to an additive constant - and this has little to do with differential equations. $\endgroup$ – mathguy Oct 11 '17 at 22:33
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    $\begingroup$ Better to call it Euler's number. Euler's constant is something else $\endgroup$ – A.Γ. Oct 11 '17 at 22:59
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    $\begingroup$ The derivative of a function equal to Euler's number is 0, not itself. You mean that the exponential function with Euler's number as the base is its own derivative. (Yes, this is nitpicking and I knew what you meant. But this is a math site. Math is one of the few fields where we can have life without much ambiguity, so let's do so.) $\endgroup$ – jpmc26 Oct 12 '17 at 2:15
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    $\begingroup$ You would encounter e pretty fast. For example with interests rates. If you get 50% every 6 months instead of 100% every year, you get $2.25 from your $1 investement after a year. If you split the rates every hour, you get $2.718 after a year. $\endgroup$ – Eric Duminil Oct 12 '17 at 11:22
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What was meant in the video is that if you do calculus with $2^x$ for example, you get that

$$ \begin{aligned} \frac{d}{dx} 2^x = (\ln 2)2^x, \quad \int2^x dx = \frac{1}{\ln 2}2^x + C \end{aligned}$$

And of course, if you didn't know about $e$ and logarithms, $\ln 2$ would just be some crazy constant you have to figure out somehow. So the formulas for differentiation and integration are "not very nice". Compare this with $e^x$:

$$ \begin{aligned} \frac{d}{dx} e^x = e^x, \quad \int e^x dx = e^x + C \end{aligned}$$

Suddenly there's nothing to remember, no crazy constants, and differentiation and integration are as simple as they could possibly be.

Finally, if no-one had discovered $e$, I think that once calculus had been invented people would find it pretty fast. Once you know what differentiation is, it's natural to ask "is there a function that is its own derivative?" and this question will naturally lead to the function $e^x$.

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    $\begingroup$ I'd argue that you CAN'T do calculus without e; in your derivatives and integrals of $2^x$, the e is still present, embedded in the ln $\endgroup$ – iammax Oct 12 '17 at 4:03
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    $\begingroup$ @iammax You can use whatever approximation methods to figure out what that constant has to be. If you spend a long time trying to sketch tangent lines to the graph $2^x$, you'll end up with approximately $0.693$ times $2^x$. You don't need to know $e$ or the natural log to do this, and besides, people made tables of logarithms way before $e$ was known (or at least popular). $\endgroup$ – Joppy Oct 12 '17 at 6:11
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    $\begingroup$ @user121330: You can (and some sources do) define $\ln x$ without any reference to logarithms or $e$. $\endgroup$ – ruakh Oct 12 '17 at 6:36
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    $\begingroup$ @iammax: $ln(a)$ is the area under the hyperbola $1/x$ between $1$ and $a$. What's this $e$ you're talking about? $\endgroup$ – Eric Duminil Oct 12 '17 at 11:25
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    $\begingroup$ @iammax $e$ will always be hidden behind the scenes, yes, but I would argue that calculus is very much possible to do without being aware of the number or its significance. It gets more messy without it than with it, though, which would mean people would start to look for it quite quickly. For instance, the function $x\mapsto a^x$ has derivative $c_aa^x$ where $a\mapsto c_a$ is some increasing function of $a$ (of course, this function has interesting properties itself). We can relatively easily prove $c_2<1<c_3$, so asking which number $e$ makes $c_e = 1$ would be quite natural. $\endgroup$ – Arthur Oct 12 '17 at 11:35
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The usual method for not introducing $e$ early is to do everything else, including integral and the fundamental theorem of calculus. Next, define a new function for $x>0$ by $$ f(x) = \int_1^x \; \frac{1}{t} \; dt $$ Then the exponential function is the inverse function of $f(x),$ call that $\exp, $ then the constant becomes $\exp 1$

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    $\begingroup$ Just to add a few steps for those unfamiliar with this approach: with this we define ln as the f is this answer. One can show, without any reference to e, that this function has all the properties of what we know of the logarithm. And it's inverse has the property to be it's own derivative. $\endgroup$ – Pacopaco Oct 12 '17 at 5:00
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    $\begingroup$ This is equivalent of course to defining $e$ as the unique positive solution $x$ to the equation $$1 = \int_1^x \frac{1}{t} dt,$$ cf. the image Hyperbola_E. $\endgroup$ – Jeppe Stig Nielsen Oct 12 '17 at 7:21
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\begin{align} \frac d {dx} 2^x & = \lim_{\Delta x\to0} \frac{2^{x+\Delta x} - 2^x}{\Delta x} = \lim_{\Delta x\to 0} \left( 2^x \frac {2^{\Delta x} - 1}{\Delta x} \right) \\[10pt] & = 2^x \lim_{\Delta x\to0} \frac {2^{\Delta x} - 1}{\Delta x} \quad \text{This step is possible because $2^x$ is} \\ & \quad \text{ “constant” in the sense that it does not change as $\Delta x$ approaches $0.$} \\[10pt] & = \Big(2^x \times \text{a constant}\Big) \text{ where this time “constant” means} \\ & \qquad \text{ not changing as $x$ changes.} \end{align} Similarly $$ \frac d {dx} 3^x = \Big(3^x \times \text{a constant} \Big) $$ but it's a different constant.

Now we have the problem of ascertaining what these "constants" are.

And instead of $2$ or $3$ as the base, for which number as base would the "constant" be equal to $1\text{?}$

Answering that last question is how the number $e$ would be discovered if we didn't already know about it.

And once we've done that, the laws of exponents plus the chain rule would would tell us that the two "consants" mentioned above are $\log_e 2$ and $\log_e 3.$

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Just to be contrarian, imagine that $e$ has not been discovered but it is known that $\arcsin$ can be analytically continued to some subset of $\mathbb C$ containing the positive imaginary axis. Then using the substitution $u=t-1/t$, we can evaluate the integral $$ \int\frac{dt}t=\int\frac{du}{\sqrt{4+u^2}}=-i\arcsin(iu/2)+C=-i\arcsin(i(t-1/t)/2)+C. $$ In other words, you can (awkwardly) get by without the exponential function using trignometric functions, which have been known longer.

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Calculus with exponentials and logarithms leads inexorably to Euler’s constant.

Straight from the definition of derivative you learn that

$$(a^x)’ = L(a)a^x$$

Where $L(a)$ is some constant that depends on $a$. A little fooling (which I will insert in a bit) will convince you that $L(a)$ is a logarithm with a base between 2 and 4. By definition, $e$ is this base.

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Nothing would change really, because $e$ isn't that useful anyway.

That may seem a bit of a crazy statement, but bear with me. $x\mapsto e^x$ is of course an extremely useful function, just, it doesn't actually hinge much on Euler's number.

Starting from algebra, it doesn't make that much sense to calculate the derivative $\frac{\partial}{\partial x} b^x$ in the first place, for any basis $b\neq1$, because $b^x$ simply isn't defined for $x\in\mathbb{R}$, and you need real arguments in order for derivatives to make sense.

You can, however, leave “iterated-multiplication exponentiation” completely aside and just define the function $\exp$ with the specific goal of fulfulling $\frac{\partial}{\partial x} (\exp x) = \exp x$. After all, that is a Lipschitz-continuous ordinary differential equation; hence if we choose $\exp 0 = 1$ as the starting condition, the Picard-Lindelöf theorem tells us that this uniquely determines the function $\exp$ for all real arguments.

You can then (using Taylor expansion or integral formulas) go on to observe that $\exp$ behaves in every regard as if it was a function of the form $x\mapsto b^x$. Namely, you have $$ \exp(p+q) = \exp p \cdot \exp q $$ and can (using the starting condition) therefore, at least for rational $n$, always write $$ \exp n = (\exp 1)^n $$ (which we know as $\exp n = e^n$, but that's just a trivial shorthand definition of $e = \exp 1$.)

By introducing also logarithms as the inverse to $\exp$, you can then write any power function in terms of $\exp$ and that readily allows you to differentiate any such function. But that doesn't at all require that you've ever introduced the convention of what $e^x$ means. So, you'd just get slightly longer formulas because you'd occasionally need to write out $\exp 1$, but the maths as such wouldn't change at all.


This isn't in fact that much of a problem because $b^x$ is readily defined on all rationals, these are dense in $\mathbb{R}$, and you can show that $b^x$ maps Cauchy sequences to Cauchy sequences, which can also be used to define arbitrary-basis realexponential functions as the unique continuous extension of the rational exponentials. Problem is, the limits of the resulting Cauchy sequence are not in $\mathbb{Q}$, and computing them isn't very practical without using $\exp$ as a proxy.

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  • $\begingroup$ This is the correct answer. $\endgroup$ – Walter Oct 12 '17 at 20:14
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If you are concerned only with algebraic and trigonometric functions and their derivatives it is quite possible that you never encounter the number $e$ and that should be the way to go. But once you start integrals, it is just not possible not to know $e$.

Also note that introducing $e$ has a cost and corresponding reward in terms of presenting the related theory of exponential and logarithmic functions. It is upto individual textbooks to decide when to pay that cost (at the time of presenting derivatives or at the time of doing integrals). From the point of view of simplicity it makes sense to wait until one reaches the anti-derivative of $f(x) =1/x$ and define the logarithm as an integral and exponential function as inverse of logarithm.

The approach which you mention in your question is difficult. It involves defining $a^{x} $ without $e$ and evaluating its derivative. This will also lead to the logarithm but with some effort. I have outlined this approach in this answer.

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$$\frac{d}{dx} a^x = k a^x \text{, where } k = \lim_{h \rightarrow 0} \frac{a^h - 1}{h}$$

If we want some natural basis, we simply find a such that $k = 1$:

$$a = \lim_{h \rightarrow 0} \sqrt[h]{h + 1} = e$$

which would change

$$\frac{d}{dx} a^x = \log_e(a) a^x \text{, and } \frac{d}{dx} e^x = e^x$$

Calculating $e$ from this formula isn't great, but the goal was only to show that $e$ dribbles out whether you want it to or not.

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