# $e$, the Story of a Number Let: $M(t) = 2^t$ We then have the derivative of $M$ with respect to $t$: $\frac{d M(t)}{dt} = \lim_{dt \to 0} \frac{2^{t + dt} - 2^t}{dt} = \lim_{dt \to 0}\frac{2^t 2^{dt} - 2^t}{dt} = \lim_{dt \to 0} \frac{2^t(1 - 2^{dt})}{dt}$ $\frac{d M(t)}{dt} = \lim_{dt \to 0} \frac{2^t(1 - 2^{dt})}{dt}$ The *incredibly interesting* fact about the above derivative is that we have two terms: 1. The term that depends on $t$, the time that we started at 2. The term that depends on $dt$, the small nudge $\frac{d M(t)}{dt} = \lim_{dt \to 0} 2^t \Big( \frac{1 - 2^{dt}}{dt} \Big)$ So, what we can see is that the derivative of $2^t$ is just *itself*, multiplied by *some constant*. In other words, the derivative is *proportional* to itself. When the base is $2$ the proportionality constant is: $\Big( \frac{1 - 2^{dt}}{dt} \Big) = 0.6931472...$ Note: to see how we evaluate the limit of this fraction as $dt \rightarrow 0$, see [here](https://math.stackexchange.com/questions/359023/using-the-limit-definition-to-find-the-derivative-of-ex) An interesting question to ask would be: > What base would yield a *proportionality constant* of 1? This would mean that the derivative of $a^t$ was not just proportional to itself, but *equal* to itself. In other words, find $a$ such that: $a^t = \frac{da(t)}{dt}$ There is indeed a constant where that is the case! It is: $2.71828...$ This number, which is the base $a$ for which the derivative of $a^t$ is equal to itself, is often give a name: $e$: $e = 2.71828...$ This fact is *what defines* $e$. Again let us make that clear: > The number $e$ does not just happen to show up here. It is, in a sense, what *defines* the number $e$. If we ask "why does $e$ of all numbers have this property?", it is similar to asking "why does $\pi$ of all numbers happen to be the ratio of a circumference of a circle to it's diameter?"; this is at it's heart *what defines* $\pi$. All exponential functions are proportional to their own derivative. But $e$ alone is that base such that the proportionality constant is $1$. Note that our "mystery" proportionality constant when our base was equal to $2$ was just the natural log of $2$ (see [here](https://youtu.be/m2MIpDrF7Es?t=652)). It is this fact that causes $e$ to show up so frequently! We can write *any base* in terms of $e$: $2^t = e^{ln(2) t} = e^{(0.6931472...) t}$ $3^t = e^{ln(3) t} = e^{(1.0986...) t}$ ### A choice we make When you see something written as $e$ to some constant times t: $e^{ct}$ That is a *choice* that we make to write it that way. The number $e$ is *not* fundamental to that function itself. What is nice about writing it like this is that it gives the constant in the exponent a nice readable meaning. ### Proof We may wish to *prove* that the following limit does indeed converge: $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$ For references that show how this limit does indeed converge, see: * [Source 1](https://socratic.org/questions/how-do-you-solve-the-following-limit-e-x-1-x-as-x-approaches-zero) * Appendix 2, e the story of a number (needed in above reference * [here](https://proofwiki.org/wiki/Derivative_of_Exponential_at_Zero) * Note: The most straight forward way to prove this is to make use of the logarithm ([here](https://math.stackexchange.com/questions/3544121/prove-the-statement-lim-limits-h-to0-fracbh-1h-1-iff-b-e?rq=1)) or to utilize the taylor series for $e^x$. ### Misunderstandings of $e^x$ We often think about $e^x$ as a mysterious constant of nature multipled $x$ times: $e^x = \overbrace{e \cdot e \dots e}^{\text{x times}}$ This comes about because exponentiation was originally thought of as an extension of *repeated* multiplication. We run into trouble with this when we hit: $e^{\pi i} = -1$ It is challenging to reason about the extension of repeated multiplication into the complex numbers. However, notationally this is not a great way of representing the exponential function. This could be represented in terms of: * The problem it solves (a very simple differential equation-this yields more insight than trying to force the idea of exponentiation as repeated multiplication to apply to complex numbers) What $e^x$, exponentiation, really addresses is things where the rate at which something changes depends on it's own value; specifically, it depends on it's own value linearly. $e$ shows up when you have exponential growth and decay, when the relation that somethings rate of change has to itself is a simple scaling. A similar law also describes circular motion. Because we have *bad notation* we use the residue of how it shows up in the context of self reinforcing growth (like in the context of a population growing, or compound interest), the constant associated with that ($e$) is awkwardly placed into the context of where rotation comes about because they both come from pretty similar equations. So we end up seeing the $e$ and the $\pi$ juxtaposed a little bit more closely than they would be in a natural representation. Here is how 3blue1brown would describe the two: * They have a very important function in common. We can call this function $exp$. This is the exponential function. * When you plug in $1$ you get this nice constant called $e$ which shows up in probability, calculus, etc. * If you try and move in the imaginary direction it is periodic and the period is $\tau$. * These are the two constants associated with the same central function, $exp$, but for kind of unrelated/orthogonal reasons (one is what happens when you move in the real direction, one is what happens when you move in the imaginary direction) For more see conversation [here](https://youtu.be/U_lKUK2MCsg?t=250) ### $e^{\pi i} = -1$ ### Misunderstandings of $e^x$ When thinking about $e^x$, a lot of people think that it means taking a number, $e$, where we multiply it by itself $x$ times: $e^x = \overbrace{e \cdot e \dots e}^{\text{x times}}$ There is some notion of extending the idea of repeated multiplication to fractions and irrational numbers. However, this is **not what** $e^x$ **means**! Instead, what has emerged in math is that we use $e^x$ to be the *shorthand* for another function that we will call $exp$. It is defined to be a certain polynomial: $exp(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!}$ In other words this is a *series*: $exp(x) = \sum_{i=0}^{\infty} \frac{x^i}{i!}$ Right away we can see that this is going to work for all kinds of $x$ that we can plug in. Anything where we know how to raise $x$ to a whole number power and divide by a number, we can come up with a nice meaning for this $exp$ function. Now, we can evaluate $exp(x)$ at $x=1$: $exp(x) = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \dots$ And we can say that this series *converges* to a specific value, $2.71828...$ (the Andrew Jackson number-2 terms, 7 years, elected in 1828). So, $exp(1) = e$. Now, one property that we will see this $exp$ function possesses is that if we *add* two numbers in our input, it is the same as *multiplying* them in our output: $exp(3 + 4) = exp(3) \cdot exp(4)$ $exp(a + b) = exp(a) \cdot exp(b)$ So we can state that $exp$ has the property: > **Adding** in the input corresponds to **multiplying** in the output. To be clear, this is *not at all obvious* if you simply look at the polynomial. However, we can show this several ways: $exp(5) = exp(1 + 1 + 1 + 1 + 1) = \overbrace{exp(1) \dots exp(1)}^{5 \; times} = exp(1)^5 = e^5$ $exp(\frac{1}{2} \cdot exp(\frac{1}{2}) = exp(\frac{1}{2} + \frac{1}{2}) = exp(1) = e$ $exp(-1) \cdot exp(1) = exp(0) = 1$ What we want to discuss next is *why* it is reasonable to use the shorthand $e^x$ when talking about our function $exp(x)$. It is because the special property of $exp(x)$ that we have been discussing allows us to use the number $1$ to access nearly all of the real numbers! In other words, this property allows us to express *everything* that the polynomial function $exp$ can output *in terms of* what it can output at the number $x=1$, i.e. $exp(1)$. Specifically, we can write it as: $exp(x) = exp(1)^x = e^x$ This is making explicit use of the fact that we *can define* the number $e$ to be "where is the polynomial $exp$ when evaluated at $x=1$?" The challenge that we run into here is that this only makes sense for *real numbers*. As soon as we start introducing things like complex numbers this breaks down. So, the *key takeaway here is*: > The convention of writing $exp(x)$ as $e^x$ makes sense for real numbers; it is easy to understand. *However*, it breaks down and is difficult to understand/misleading and we extend it to the complex numbers. #### Misunderstandings of Euler's Equation There is a often a large misunderstanding about Euler's Equation: $e^{\pi i} = -1$ It *overstates* the relationship between $e$ and $\pi$. What it really stems from is this; We have this (polynomial) function $exp$ that we have defined, and when you plug in real values it *makes sense* to write $exp(x)$ in terms of $exp(1)$ raised to the $x$, i.e. $exp(1)^x$. We then can that value $exp(1)$, $e$. $exp(x)$ then has this other property that when we plug in imaginary numbers, the claim that $exp(xi)$ is making is that it is **periodic**. It is periodic with a period of $2\pi i$. In other words, when you increase the input by $2 \pi i$ you get back to where you started (via walking along a unit circle). We can think about the relation between these two branches of $exp$ as **orthogonal** to each other in a sense. #### Euler's Equation for different bases To ensure that we fully understand Euler's Equation, it is helpful to consider how it handles different bases. Let us quickly recall $exp(x) = e^x$: $exp(x) = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!}$ And now if we consider a base of $2$, we can make use of the following property: $2 = e^{ln(2)}$ $2^x = \Big(e^{ln(2)}\Big)^x = e^{ln(2)x}$ Then we can use our taylor series of $exp(x)$, only now we will plug in $ln(2)x$ for the value of $x$: $exp\big(ln(2)x\big) = e^{ln(2)x} = 1 + ln(2)x + \frac{(ln(2)x)^2}{2!} + \frac{(ln(2)x)^3}{3!} + \dots + \frac{(ln(2)x)^n}{n!}$ $exp\big(ln(2)x\big) = 2^x = 1 + xln(2) + \frac{x^2ln(2)^2}{2!} + \frac{x^3 ln(2)^3}{3!} + \dots + \frac{x^n ln(2)^n}{n!}$ Now, with this new series, we can see how plugging in $\pi i$ would change, compared to when our base is $e$: $e^{\pi i} = -1$ $2^{\pi i} = -0.5202... + 0.82148... i$ Now, note that this still leaves us on the unit circle (radius 1), but now instead of having rotated 180 degrees (sitting at -1 in the real dimension, 0 in the imaginary dimension), we have rotated ~125 degrees. Note that if we use 3 as our base we will go slightly *past* 180 degrees of rotation (landing in the 3rd quadrant): $3^{\pi i} = -0.9523... - 0.30486... i$ Notice that the reason for this-that bases less than $e$ rotate less than 180 degrees and bases greater than $e$ rotate greater than 180 degrees-is because of the $ln$ terms in the taylor series expansion: $ln(x) < 1, \;\;\; \forall \;\;\; x < e$ $ln(x) > 1, \;\;\; \forall \;\;\; x > e$ So these terms will either increase or decrease our amount of rotation. ### Proof of following equivalence We often see that $exp(x)$ is defined in multiple ways, such as: $exp(x) = \lim_{n \to \infty} (1 + \frac{x}{n})^n$ And $exp(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!}$ The equivalence is proven via my notes [here](https://photos.google.com/photo/AF1QipPBwTCP0OooI9GzHF5s3ZHPRfpNBB1CgEg59xbh), making use of the [binomial theorem](binomial%20theorem). For more see [here](https://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function). --- References: * [What is so special about Euler's number $e$](https://youtu.be/m2MIpDrF7Es?t=484) * [Derivative of Exponential](https://www.nathanieldake.com/Mathematics/06-Functions-02-Inverse-functions-exponentials-and-logarithms.html#Appendix-A:-Derivative-of-Exponential) * [Lockdown math 3b1b - Eulers formula](https://www.youtube.com/watch?v=ZxYOEwM6Wbk&t=67s) * [Imaginary interest rates](https://www.youtube.com/watch?v=IAEASE5GjdI) * Euler’s formula - 03172021, Notability