What's in an e: the origins of Euler's number

[fishsicles]

After reading this month's highlighted essay on the development of infinitesimal calculus, I felt like sharing one of my own favourite mathematical origin stories; and as a combinatorialist, it's hard to top Euler's number. Or, well, the important Euler's number. We name a lot of things after Euler.

Euler's number, e, is a number that most people first encounter in a high school mathematics course, alongside the introduction of exponential functions and the high school economics version of compound interest. It's a number around 2.718, and... that's usually the deepest that most people get into it, as the reasons behind e being what it is are well beyond the scope of a high school student's. This essay is written assuming that the reader has some familiarity with the factorial function and sigma/pi notation for sums and products, but doesn't delve into anything more esoteric than that.

Introduction: The Compound Interest Problem

Compound interest, in the apocryphal words of Einstein, is the most powerful force in the universe. It's how banks make money, and how everyone else loses money. The core concept is simple - given some amount of borrowed money, increase the amount owed by some percentage of the amount on a regular interval. We call the annual percentage increase the rate, and denote it r. After t years, assuming they didn't pay down the principal A₀, the borrower will be on the hook for

\[ A(t) = A_{0}\left(1+r\right)^{t} \]

However, lenders often don't want to only see their revenue once a year. Instead of a single loan that increases by 6% once a year, they may offer a loan that increases by 0.5% once a month. Intuitively - a dangerous word in mathematics, which a professor of mine once described as the science of unlearning intuition - this may seem to be "the same amount"; after all, it's a 0.5% increase each month. And in the case of non-compounding interest, it will be; but what if each month's interest was also applied to last month's interest?

\[ \begin{align*} A(t) ={} & A_{0}\left(1+0.06\right)^{1}\\ ={} & 1.06A_{0}\\ A(t) ={} & A_{0}\left(1+\frac{0.06}{12}\right)^{1\cdot 12}\\ \approx{} & 1.0617A_{0}& \end{align*} \]

This idea - of applying interest to the interest - is what we call compound interest. In the case of our 6% interest rate loan here, compounding monthly only saw about a 0.17% increase over compounding annually, but lenders aren't in the business of passing up even those small gains, so they'll try and compound as frequently as possible. But what's the limit of "as frequently as possible"? What if your interest compounded at every single moment?

Well, that would probably be n of around 5.85e50, depending on how we define "moments"; but if there's one thing real mathematicians and the pretenders in the world of economics can agree on, it's that physical constraints are just a set of overly narrow axioms, so we're going to operate as if time is continuous, and interest is continuous alongside it. In other words: what if n is infinite?

What Does "Continuity" Mean?

The human brain does not like the concept of infinity. We are very finite creatures, and mathematics is one of the few ways we ever really encounter infinity; and even then, there's a whole branch called finitism for people who hate breaking into heaven and playing with all of God's favourite toys. (I'm convinced the ultrafinitists are doing a bit, though.) Infinity and continuity are two sides of the same coin; one informal definition for continuity is that it can be divided up into infinitely many infinitely small pieces without being separated.

Key to this notion of continuity is the notion of limits. There are a lot of ways to define a limit, but the key concept of the limit is the behaviour of some function arbitrarily close to but not at a given value. For instance, the function y=x/x is obviously 1 at all points other than 0, but undefined at 0 - however, since it is 1 at all points arbitrarily close to 0, we define

\[\lim_{x\to 0} \frac{x}{x} = 1\]

There's a more formal definition involving a couple Greek letters and some very tiresome arithmetic that is inflicted on every analysis student until we accept that the limit laws are our friends and will save us trouble, but it's a bit out of scope here.
The limit we're looking at for continuously compounding interest, though, isn't about approaching a specific value. We can't get close to infinity; we can get bigger, but we'll always be infinitely far away. Rather than see what happens as we get arbitrarily close to some number of periods, we want to see what happens as we make our number of periods arbitrarily large.

To define how a limit works, we pick a very small number to call ε (the Greek letter epsilon); then the limit is at some value L if there exists some N such that

\[N<x \implies \left|f(x)-L\right| < \epsilon\]

If for any ε, such an N exists, then our limit at infinity exists, and is denoted

\[\lim_{x\to\infty} f(x) = L\]

Notably, these infinite limits work fine when the domain of f isn't continuous, just so long as it is unbounded; which is perfect for the natural numbers we need in our compound interest problem.

Continuously Compounding Interest

Returning to compound interest, we can denote the idea of continuously compounding interest using the limit as n grows without bound:

\[A(t) = \lim_{n\to\infty} A_{0} \left(1+\frac{r}{n}\right)^{nt}\]

Limits have a bunch of handy properties that I'm not going to prove in detail, and that are common features of introductory calculus courses. The relevant ones for the moment are the product and power laws, which allow us to rewrite the above limit as

\[A(t) = A_{0}\left(\lim_{n\to\infty} \left(1+\frac{r}{n}\right)^{n}\right)^{t}\]

which allows us to set aside the principal and the amount of time in favour of focusing just on what happens to any amount of money in a single year. If you remember the continuously compounding interest formula from high school, you might already see its skeleton in the above; but we still need to clean up

\[\lim_{n\to\infty}\left(1+\frac{r}{n}\right)^{n}\]

First, we can rewrite this by distributing the power of n; since n is an ordinary natural number inside the limit, we can use the binomial theorem to expand the power:

\[\begin{align*}\lim_{n\to\infty}\left(1+\frac{r}{n}\right)^{n} = {} & \lim_{n\to\infty}\sum_{k=0}^{n} \binom{n}{k} 1^{n-k} \left(\frac{r}{n}\right)^{k}\\ = {} & \lim_{n\to\infty} \sum_{k=0}^{n} \binom{n}{k} \frac{r^{k}}{n^{k}}\\ = {} & \lim_{n\to\infty} \sum_{k=0}^{n} \frac{n!}{k!\left(n-k\right)!} \frac{r^{k}}{n^{k}}\\ = {} & \lim_{n\to\infty} \sum_{k=0}^{n} \frac{n!}{n^{k}\left(n-k\right)!} \frac{r^{k}}{k!} \end{align*}\]

We can't just sum this as we might a conventional infinite series, because n occurs inside our terms alongside k. But since we've grouped the n terms all together, let's take a closer look at it:

\[\frac{n!}{n^{k}\left(n-k\right)!}\]

Dividing a larger factorial by a smaller gives the products of all the numbers greater than the lower bound, and less than or equal to the upper. Using product notation, we can then rewrite this inner term as

\[\frac{n!}{n^{k}\left(n-k\right)!} = \frac{1}{n^{k}} \prod_{j=1}^{k} \left(n-k+j\right) = \prod_{j=1}^{k} \frac{n-k+j}{n} = \prod_{j=1}^{k} \left(1-\frac{k-j}{n}\right)\]

As n grows without bound, every single factor of every single term of the sum that contains n approaches 1, meaning the entire product essentially eliminates itself as n increases. This is such an important property of the factorial function that one way of deriving the gamma function (the most significant generalised factorial) starts with this property and works backwards. We can apply it here to rewrite our limit as

\[\begin{align*}\lim_{n\to\infty}\left(1+\frac{r}{n}\right)^{n} = {} & \lim_{n\to\infty} \sum_{k=0}^{n} \frac{n!}{n^{k}\left(n-k\right)!} \frac{r^{k}}{k!}\\ = {} & \lim_{n\to\infty} \sum_{k=0}^{n} \frac{r^{k}}{k!}\\ = {} & \sum_{k=0}^{\infty} \frac{r^{k}}{k!} \end{align*}\]

Our limit has now been brought down to a summation in one variable - an infinite series expansion of some function of r. In fact, it's a very special kind of function.

Exponential Functions, Formally

In high school algebra, you were probably told that an exponential function is a function of the form Ab^x. This is close to true, and works outside formal settings, but there's numerous equivalent ways to describe "exponential functions" as a family. Most of them involve calculus, but one that doesn't is defined by the property

\[f(a)\cdot f(b) = f(a+b) \iff f(x) = \left(f(1)\right)^{x}\]

The proof that these two criteria are equivalent is relatively straightforward; I wouldn't be a proper mathematics writer if I didn't leave at least something to the reader.

Let's use our infinite series from the last section, but with a particular argument, and revisit our old friend the binomial theorem:

\[\begin{align*}f(a+b) = \sum_{k=0}^{\infty} \frac{(a+b)^{k}}{k!} = {} & \sum_{k=0}^{\infty} \frac{1}{k!} \left(\sum_{j=0}^{k} \binom{k}{j} a^{k-j}b^{j}\right)\\ = {} & \sum_{k=0}^{\infty} \frac{1}{k!} \left(\sum_{j=0}^{k} \frac{k!}{j! (k-j)!} a^{k-j}b^{j}\right)\\ = {} & \sum_{k=0}^{\infty} \sum_{j=0}^{k} \frac{a^{k-j}}{(k-j)!} \frac{b^{j}}{j!}\\ = {} & \sum_{k=0}^{\infty} \sum_{i+j=k} \frac{a^{i}}{i!} \frac{b^{j}}{j!}\\ = {} & \left(\sum_{k=0}^{\infty} \frac{a^{k}}{k!}\right)\left(\sum_{k=0}^{\infty} \frac{b^{k}}{k!}\right) = f(a)\cdot f(b) \end{align*}\]

So we know that our function of r is an exponential function, defined by some base f(1). What's that base? Well, we didn't have a name for it for a long time. But Leonhard Euler had already used a and wanted to use a vowel, so he called it e, and it's been that ever since.

\[e = \sum_{k=0}^{\infty} \frac{1}{k!} \approx 2.718\]

 

[Nyvis]

This is really interesting because it does introduce e in a completely different way than the way I was introduced to it through the progression of math class. In my own learning, it first came up when learning about calculus and looking for the solution to f'(x) = f(x).

 

[fishsicles]

This is really interesting because it does introduce e in a completely different way than the way I was introduced to it through the progression of math class. In my own learning, it first came up when learning about calculus and looking for the solution to f'(x) = f(x).

Click to shrink...

For me, the infinite sum definition of e will always be the most important one, but it's one of those numbers that has so many different appearances in so many seemingly distinct problems that all turn out to be the same thing. I'm fond of this definition because it's one of the more "natural" (ignoring that continuously compounding interest occurring in reality is the sort of thing that would wake the ghost of Robespierre) cases, as well as one of the older ones (though the oldest is the natural logarithm).

Of course, the differential equation and series definitions are equivalent (to the point that I honestly don't remember how to solve the differential equation without using series, though I'm sure there's some method or another; I'm afraid the combinatorics is terminal.)

\[\begin{align*}f^{\prime}(x) = {} & f(x)\\ \frac{d}{dx} \sum_{n=0}^{\infty} a_{n}x^{n} = {} & \sum_{n=0}^{\infty} a_{n}x^{n}\\ \sum_{n=1}^{\infty} na_{n}x^{n-1} = {} & \sum_{n=0}^{\infty} a_{n}x^{n}\\ na_{n} = {} & a_{n-1}\\ a_{n} = {} & \frac{a_{n-1}}{n} = \frac{a_{0}}{n!}\\ f(x) = {} & a_{0} \sum_{n=0}^{\infty} \frac{x^{n}}{n!} = a_{0}e^{x}\end{align*}\]

 

[Nyvis]

For me, the infinite sum definition of e will always be the most important one, but it's one of those numbers that has so many different appearances in so many seemingly distinct problems that all turn out to be the same thing. I'm fond of this definition because it's one of the more "natural" (ignoring that continuously compounding interest occurring in reality is the sort of thing that would wake the ghost of Robespierre) cases, as well as one of the older ones (though the oldest is the natural logarithm).

Of course, the differential equation and series definitions are equivalent (to the point that I honestly don't remember how to solve the differential equation without using series, though I'm sure there's some method or another; I'm afraid the combinatorics is terminal.)

\[\begin{align*}f^{\prime}(x) = {} & f(x)\\ \frac{d}{dx} \sum_{n=0}^{\infty} a_{n}x^{n} = {} & \sum_{n=0}^{\infty} a_{n}x^{n}\\ \sum_{n=1}^{\infty} na_{n}x^{n-1} = {} & \sum_{n=0}^{\infty} a_{n}x^{n}\\ na_{n} = {} & a_{n-1}\\ a_{n} = {} & \frac{a_{n-1}}{n} = \frac{a_{0}}{n!}\\ f(x) = {} & a_{0} \sum_{n=0}^{\infty} \frac{x^{n}}{n!} = a_{0}e^{x}\end{align*}\]

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Oh I agree, this is neater math than just filling a slot in the calculus sheet.

But I'm an engineer so

 

[fishsicles]

But I'm an engineer so

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My condolences. Hopefully it's treatable?

(God, the best part of being a mathematician is all the jokes we have about every other discipline. I have a print of this on my wall.)

 

[Nyvis]

My condolences. Hopefully it's treatable?

(God, the best part of being a mathematician is all the jokes we have about every other discipline. I have a print of this on my wall.)

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Well, the only way to treat it would be to go back to school and get a PhD and there's zero way I'm putting myself back into the grind of what's basically a piece of glorified multiyear homework. So no I'm afraid I'm stuck with it.

 

[fishsicles]

Well, the only way to treat it would be to go back to school and get a PhD and there's zero way I'm putting myself back into the grind of what's basically a piece of glorified multiyear homework. So no I'm afraid I'm stuck with it.

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As a mathematics PhD dropout, it's not about degrees, it's about appreciation for the art.

(When I went to get my policy masters, an engineer friend said "Did you think you were too employable as a mathematician?")

 

[Aranfan]

First, this is great.

Second:

(to the point that I honestly don't remember how to solve the differential equation without using series, though I'm sure there's some method or another; I'm afraid the combinatorics is terminal.)

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This is actually a separable differential equation:

\[ a\frac{dy}{dx}=y \\ ady=ydx \\ \frac{dy}{y}=\frac{dx}{a} \\ \int\frac{dy}{y}=\int\frac{dx}{a} \\ \ln(y)=\frac{x}{a}+c \\ \exp(\ln(y))=\exp(\frac{x}{a}+c) \\ y=\exp(\frac{x}{a})\exp(c) \\ y=C\exp(\frac{x}{a}) \]

Where "a" is a constant equal to unity that's just there for dimensional consistency.

 

[fishsicles]

Ah, yeah, that triggers the memories now that I see it. God it's been a long time since I took diff. eq., and almost as long a time since I taught it.

 

[fishsicles]

We've covered the origins of Euler's number, but you may be familiar with a more... esoteric formula involving it. Does this look familiar?

\[e^{\pi i} + 1 = 0\]

It's been called the archetypal example of mathematical beauty, and it's comprised of what are often called the five most important numbers in mathematics - the additive identity, multiplicative identity, complex unit, and the transcendental constants e and π. It's often given a dedicated day in introductory treatments of complex analysis, and occasionally shown to students earlier - and on first appearance, the response is usually a situation-appropriate translation of "what the fuck?"

Complex Exponents

"Real" and "imaginary" are one of the greatest misnomers in mathematics. Almost every real number is undefinable - meaning there exists no finite formula that can specify that number. "Almost every" in mathematics means that if you pick a real number completely at random, you have a 0% chance of getting a definable number; i.e., for every definable number, there are infinitely many undefinable numbers. Compared to these nominally real numbers that we cannot even describe, "the square root of negative one" is pretty straightforward. (And for those less convinced by abstract reasoning, "imaginary" numbers occur in physical problems all the time, most notably those involving electricity and magnetism.)

Putting i into an exponent may seem unintuitive for people used to the colloquial definition of exponentiation, which is "multiply something by itself n times", but this breaks down just as much for the typical real exponents. Even given that roots allow fractional exponents, how does one raise something to the √2th power?

Fortunately, we just solved this on the last post - we can define exponential functions using the infinite series representation, and put i - or any other number where integer exponentiation is defined - inside the series.

\[\begin{align*}e^{x} = {} & \sum_{n=0}^{\infty} \frac{x^{n}}{n!}\\ e^{ix} = {} & \sum_{n=0}^{\infty} \frac{\left(ix\right)^{n}}{n!}\\ = {} & \sum_{n=0}^{\infty} i^{n}\frac{x^{n}}{n!} \end{align*}\]

Now we can do a bit of trickery. Since n necessarily alternates between even numbers (n = 2k) and odd numbers (n = 2k+1), we can rewrite this sum as

\[\begin{align*}e^{ix} = {} & \sum_{n=0}^{\infty} i^{n}\frac{x^{n}}{n!}\\ = {} & \sum_{k=0}^{\infty} i^{2k} \frac{x^{2k}}{\left(2k\right)!} + i^{2k+1} \frac{x^{2k+1}}{\left(2k+1\right)!}\\ = {} & \sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k}}{\left(2k\right)!} + i(-1)^{k} \frac{x^{2k+1}}{\left(2k+1\right)!}\\ = {} & \sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k}}{\left(2k\right)!} + i \sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k+1}}{\left(2k+1\right)!} \end{align*}\]

We call sums of this form - those containing a (-1)^k factor - as alternating sums, because the sign of each term is opposite the sign of the previous term. Note that neither sum contains i; since the real numbers are closed under addition and multiplication, the results of these sums must also be real numbers, meaning we are in the standard a + bi form for general complex numbers. The real part of our result is the alternating sum of the even-power terms of the exponential function, while the imaginary part is the alternating sum of the odd terms.

For now, we'll call the real part a and the imaginary part b.

On the Alternating Sums

If you're familiar with Taylor expansions, you probably already see where we're going, but I want to do this with as little calculus as possible. It's not exactly obvious what I'm doing at first, but bear with me; we're going to start with the squares of both sums.

\[\begin{align*} a^2 = \left(\sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k}}{\left(2k\right)!}\right)^{2} = {} & \sum_{n=0}^{\infty} \sum_{j+k=n} (-1)^{k} \frac{x^{2k}}{\left(2k\right)!} (-1)^{j} \frac{x^{2j}}{\left(2j\right)!}\\ = {} & \sum_{n=0}^{\infty} (-1)^{n} x^{2n} \sum_{j+k=n} \frac{1}{\left(2k\right)!\left(2j\right)!}\\ = {} & \sum_{n=0}^{\infty} (-1)^{n} x^{2n} \sum_{k=0}^{n} \frac{1}{\left(2k\right)!\left(2(n-k)\right)!}\\ = {} & \sum_{n=0}^{\infty} (-1)^{n} x^{2n} \frac{2^{2n-1}}{(2n)!}\\ \\ b^2 \left(\sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k+1}}{\left(2k+1\right)!}\right)^{2} = {} & \sum_{n=0}^{\infty} \sum_{j+k=n} (-1)^{k} \frac{x^{2k+1}}{\left(2k+1\right)!} (-1)^{j} \frac{x^{2j+1}}{\left(2j+1\right)!}\\ = {} & \sum_{n=0}^{\infty} (-1)^{n}x^{2(n+1)} \sum_{j+k=n} \frac{1}{\left(2k+1\right)!\left(2j+1\right)!}\\ = {} & -\sum_{n=0}^{\infty} (-1)^{n+1}x^{2(n+1)} \sum_{k=0}^{n} \frac{1}{\left(2k+1\right)!\left(2(n-k)+1\right)!}\\ = {} & -\sum_{n=1}^{\infty} (-1)^{n}x^{2n} \sum_{k=0}^{n-1} \frac{1}{\left(2k+1\right)!\left(2(n-1-k)+1\right)!}\\ = {} & -\sum_{n=1}^{\infty} (-1)^{n}x^{2n} \frac{2^{2n-1}}{(2n)!} \end{align*}\]

Working out the factorial sum that serves as the last step for each square is this post's exercise for the reader; again, it's reasonably straightforward as long as you're comfortable manipulating finite sums and factorials. The two squares are very similar, and this gives us a very crucial property:

\[\begin{align*} a^2+b^2 = {} & \sum_{n=0}^{\infty} (-1)^{n} x^{2n} \frac{2^{2n-1}}{(2n)!} -\sum_{n=1}^{\infty} (-1)^{n}x^{2n} \frac{2^{2n-1}}{(2n)!}\\ = {} & 1 + \sum_{n=1}^{\infty} (-1)^{n} x^{2n} \frac{2^{2n-1}}{(2n)!} -\sum_{n=1}^{\infty} (-1)^{n}x^{2n} \frac{2^{2n-1}}{(2n)!}\\ = {} & 1 \end{align*}\]

If we were to plot this set of possible values as ordered pairs (x,y), we see a very familiar shape.


This is the unit circle. For any point (a,b) on this circle - necessarily including every possible value of our imaginary exponent - there exists some angle θ such that

\[\begin{align*}(a,b) = {} & (\cos\theta, \sin\theta)\\ e^{ix} = {} & a+bi = \cos\theta + i\sin\theta \end{align*}\]

And as it turns out, θ = x. But that's for the next post, I think.

 

[Aranfan]

Even given that roots allow fractional exponents, how does one raise something to the √2th power?

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Oh that's easy. Since:

\[ (n^x)^y=n^{xy} \]

All you do is say that raising something to √2 is what you need to do twice to square it.

 

[Nyvis]

Oh that's easy. Since:

\[ (n^x)^y=n^{xy} \]

All you do is say that raising something to √2 is what you need to do twice to square it.

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Many such cases. Math really like to take intuitive but discrete concepts and extend them to be continuous but unintuitive.

 

[fishsicles]

Oh that's easy. Since:

\[ (n^x)^y=n^{xy} \]

All you do is say that raising something to √2 is what you need to do twice to square it.

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Don't make me put e^e in there.

 

[Aranfan]

A little bit more seriously, my favorite way of showing that "cis (θ)" is exponential to just differentiate it and show it fits the chain rule for the exponential function.

 

[fishsicles]

A little bit more seriously, my favorite way of showing that "cis (θ)" is exponential to just differentiate it and show it fits the chain rule for the exponential function.

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Honestly, the main reason I'm doing it this way is that I'm taking "no calculus" as a challenge run of sorts.

 

[Aranfan]

Honestly, the main reason I'm doing it this way is that I'm taking "no calculus" as a challenge run of sorts.

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Ye, that's half of what makes this thread so fun.

 

[puskas78]

When I was a teenager I noticed that the value of "x√x" (xth root of x) is the highest if "x" is e. 2 and 4 are the same, 3 is a bit higher, and e has the highest result. I don't know what exactly this means, but it has to mean something

 

[fishsicles]

When I was a teenager I noticed that the value of "x√x" (xth root of x) is the highest if "x" is e. 2 and 4 are the same, 3 is a bit higher, and e has the highest result. I don't know what exactly this means, but it has to mean something

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This is actually pretty straightforward to prove, though I need to break the no-calculus rules - but since that's necessary for maxima anyway, let's go for it. (I'm still trying to figure out a good way to prove that e^(ix) = cis(x), but it's tricky - it's relatively straightforward to show that the angle corresponding to e^(ix) is additive and zero when x=0, but proving linearity from there relies on continuity...)

First, we should note that formally,

\[ y = \sqrt[x]{x} = x^{\frac{1}{x}} \]

We can take the natural logarithm of both sides to get

\[ \ln y = \frac{1}{x}\ln x \]

and use implicit differentiation to get

\[ \begin{align*} \frac{1}{y} \frac{dy}{dx} = {} & \frac{1}{x^{2}} - \frac{1}{x^{2}} \ln x\\ = {} & \frac{1}{x^{2}} \left(1-\ln x\right)\\ \frac{dy}{dx} = {} & \frac{y}{x^{2}} \left(1-\ln x\right)\\ = {} & x^{\frac{1}{x}-2} \left(1-\ln x\right) \end{align*} \]

We know that local maxima and minima can only occur where the derivative is undefined or zero. There are two points of interest:

\[ \begin{align*} x^{\frac{1}{x}-2} \mbox{ undefined at } x = {} & 0\\ 1-\ln x = 0 \mbox{ at } x = {} & e \end{align*} \]

The function itself is undefined at 0, and the latter gives a maximum value of the eth root of e.

This is, however, only the maximum for real-valued x. The complex plane isn't normally ordered the way that the real line is, ideas like "minimum" and "maximum" don't really make sense - and most ways to generalise them give unintuitive results like -1 and 1 having the same magnitude. That said, there's at least one handy real-valued complex counterexample via Euler's identity:

\[ \sqrt[i]{i} = \sqrt[i]{e^{\frac{\pi}{2}i}} = e^{\frac{\pi}{2}} > e^{\frac{1}{e}} \]

 

[fishsicles]

I've been thinking a bit more about the whole x√x problem, which I'm calling "the self root of x". After a bit of algebra, I've determined the necessary and sufficient criteria for a complex number z to have a real self root x.

The Other Specific Cases

It's trivial to show that all positive real numbers have real self roots. For the negative reals, we have

\[ \begin{align} \sqrt[-a]{-a} = {} & (-a)^{\frac{1}{-a}}\\ ={} & \left(-\frac{1}{a}\right)^{\frac{1}{a}}\\ ={} & (-1)^{\frac{1}{a}} \frac{1}{\sqrt[a]{a}} \end{align} \]

which yields

\[\sqrt[-a]{-a}\in\mathbb{R} \implies {} a = \frac{1}{n} : n\in\mathbb{N}\]

This alone is sufficient to establish that self roots are unbounded; self roots from odd n decrease without bound, and those from even n increase without bound.

I briefly mentioned the self root of i in the footnote of the prior post, so before moving on to the most general case, I'd also like to address what happens when the real term is zero. This gives us

\[ \begin{align} 0+bi = {} & x^{0+bi} = e^{(b\ln x)i}\\ = {} & \mathop{\mathrm{cis}} b\ln x\\ \cos b\ln x = {} & 0\\ \sin b\ln x = {} & b = \pm 1\\ z = {} & \pm i \end{align} \]

Thus, the only imaginary numbers with real self roots are i and -i.

Complex Numbers with Positive Real Self Roots

The more interesting case is found when both the real and imaginary terms are nonzero. First, let's recall that much like points in the real plane, complex numbers can be represented both in Cartesian and polar forms:

\[ \begin{align} z = {} & a+bi\\ z = {} & \left|z\right|\mathop{\mathrm{cis}} \theta = \left|z\right|e^{\theta i} = e^{\ln\left|z\right| + \theta i} \end{align} \]

Requiring a real self root gives us an additional form,

\[ z = x^{z} = x^{a+bi} = e^{(a+bi)\ln x} \]

Combining the polar form with the above gives

\[ \begin{align} e^{\ln\left|z\right| + \theta i} = {} & e^{(a+bi)\ln x}\\ \ln\left|z\right| + \theta i = {} & (a+bi)\ln x \end{align} \]

In the case where x is positive, the natural logarithm is also real, giving us

\[\ln\left|z\right| = a\ln x\]

The expression for the modulus of z is particularly significant, as we can solve it for the natural logarithm of x, and replace it in our real self root form of z:

\[ \begin{align} z = {} & e^{(a+bi)\ln x}\\ = {} & e^{(a+bi) \cdot \frac{1}{a} \ln \left|z\right|}\\ = {} & e^{(1+\frac{b}{a}) \ln\left|z\right|}\\ = {} & \left|z\right|^{1+\frac{b}{a}i} \end{align} \]

We can rearrange this to give

\[\frac{z}{\left|z\right|} = \left|z\right|^{\frac{b}{a}i}\]

The left side of our equation is the unit complex number collinear with the origin and z,

\[\begin{align} z = {} & \left|z\right| e^{i\theta}\\ \frac{z}{\left|z\right|} = {} & e^{i\theta} = \mathop{\mathrm{cis}}\theta \end{align}\]

and it should be reasonably straightforward to see that

\[\frac{b}{a} = \tan\theta\]

Combining this with the above results gives

\[\begin{align}\left|z\right|^{(\tan \theta)i} = {} & e^{\theta i}\\ (\tan\theta)\ln\left|z\right| = {} & \theta\\ \ln\left|z\right| = {} & \theta\cot\theta \end{align}\]

This gives us the modulus as a function of the angle measure - a polar function of one variable! We can substitute this back into our polar form for z; thus any z with positive real self root must be able to be expressed in the form

\[z = e^{\theta\cot\theta + \theta i} = e^{\theta\cot\theta} \mathop{\mathrm{cis}}\theta\]

for some real angle measure θ. While the algebra we took to get here relies on nonzero a, it's easy to verify that the function's range also includes the ±i cases we found earlier; it also has a removable discontinuity at e, though this does not properly describe the degenerate case for positive real values.

Plotting this as a polar function of theta between negative and positive pi gives a closed curve, if you're willing to close the discontinuity at (0,e) and add a limit point at (0,0). Shifting your angle slice by a full period will share the ±i cases and the limit point at (0,0), but diverges to infinity in the "middle" because the limit of θ cot θ no longer converges at even multiples of pi.

Sufficiency

Now it is a matter of working backwards. Evaluating the self root of z directly is troublesome; however, we can note that

\[\begin{align} \ln x = \ln\sqrt[z]{z} = {} & \frac{1}{z} \ln z\\ = {} & e^{-\theta\cot\theta - \theta i} \left(\theta\cot\theta + \theta i\right)\\ = {} & e^{-\theta\cot\theta} \left(\cos(-\theta) + i\sin(-\theta)\right)\left(\theta\cot\theta + \theta i\right)\\ = {} & \theta e^{-\theta\cot\theta} \left(\cos\theta- i\sin\theta\right)\left(\frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\sin\theta}i\right)\\ = {} & \frac{\theta}{\sin\theta} e^{-\theta\cot\theta}(\cos\theta - i\sin\theta)(\cos\theta + i\sin\theta)\\ = {} & \frac{\theta}{\sin\theta} e^{-\theta\cot\theta}(\cos^2\theta + \sin^2\theta)\\ = {} & \frac{\theta}{\sin\theta} e^{-\theta\cot\theta} \end{align}\]

As this is a product of real numbers, it must be a real number; and as ln x is a real number, x must be a positive real number.

I plan on poking around the negative case as I next find time. I suspect it will proceed similarly to the positive case, but conclude using tangent instead of cotangent.

 

[fishsicles]

I suspect it will proceed similarly to the positive case,

Click to shrink...

The first rule of mathematics is that intuition is never correct.

So, in the negative case, we can do all the same stuff up until

\[\begin{align*} \ln\left|z\right|+ \theta i = {} & (a+bi)\ln (-x)\\ = {} & (a+bi)(\ln x + \pi i)\\ = {} & (a\ln x -b\pi) + (b\ln x + a\pi)i\\ \ln x = {} & \frac{1}{a}\left(\ln\left|z\right| + b\pi\right)\\ \ln(-x) = {} & \frac{1}{a}\left(\ln\left|z\right| + b\pi\right) + \pi i \end{align*} \]

makes the elegant substitution used in the positive case turn into a clusterfuck:

\[\begin{align*} z = {} & e^{(a+bi)(\ln(-x))}\\ \ln z = {} & (a+bi)\left(\frac{1}{a}\left(\ln\left|z\right| + b\pi\right) + \pi i\right)\\ = {} & \ln\left|z\right| + b\pi + a \pi i + \frac{b}{a} i \ln\left|z\right| + \frac{b^2}{a} \pi i - b\pi\\ = {} & \left(1+\frac{b}{a}i\right) \ln\left|z\right| + \left(a+\frac{b^2}{a}\right) \pi i\\ = {} & \left(1+i\tan\theta\right) \ln\left|z\right| + \frac{1}{a} \left(a^2+b^2\right) \pi i\\ = {} & \left(1+i\tan\theta\right) \ln\left|z\right| + \frac{1}{\left|z\right| \cos\theta} \left|z\right|^{2} \pi i\\ = {} & \left(1+i\tan\theta\right) \ln\left|z\right| + \pi \left|z\right| i \sec\theta\\ \ln\left|z\right| + i\theta = {} & \ln\left|z\right| + (\tan\theta \ln\left|z\right| + \pi\left|z\right| \sec\theta) i\\ \theta = {} & (\tan\theta) \ln\left|z\right| + (\pi \sec\theta) \left|z\right| \end{align*}\]

Solving for the modulus in terms of the angle means solving an equation of the form

\[ax + b\ln x = {} c\]

Whenever we have a linear-shaped equation involving a value and its logarithm, the first goal should be to wrangle it into a form compatible with the product logarithm - that is, the inverse function (family) of

\[ f(x) = xe^{x}\]

This takes a bit of algebra:

\[\begin{align*} \ln x = {} & \frac{-ax + c}{b}\\ \ln \frac{1}{x} = {} & \frac{ax-c}{b}\\ \frac{1}{x} = {} & e^{\frac{a}{b} x - \frac{c}{b}}\\ e^{\frac{c}{b}} = {} & x e^{\frac{a}{b} x}\\ \frac{a}{b} e^{\frac{c}{b}} = {} & \left(\frac{a}{b} x\right) e^{\frac{a}{b} x}\\ \frac{a}{b} x = {} & W\left(\frac{a}{b} e^{\frac{c}{b}}\right)\\ x = {} & \frac{b}{a} W\left(\frac{a}{b} e^{\frac{c}{b}}\right) \end{align*}\]

With our expression solved in the general case, it now falls to plug in the values in our specific one:

\[\begin{align*} a = {} & \pi\sec\theta\\ b = {} & \tan\theta\\ c = {} & \theta\\ \left|z\right| = {} & \frac{\tan\theta}{\pi\sec\theta} W\left(\frac{\pi\sec\theta}{\tan\theta} e^{\frac{\theta}{\tan\theta}}\right)\\ = {} & \frac{\sin\theta}{\pi} W\left(\pi \csc \theta \, e^{\theta\cot\theta}\right) \end{align*}\]

Is this correct? Fuck if I know. Wrangling it into any kind of solution at all took a lot of dredging the brain as is, and I don't envy working backwards to prove it. I tested it with pi/4 and got a working result, so at the very least it's not completely off base.

No fancy viz this time because Wolfram|Alpha doesn't handle polar plots with singularities well and Desmos doesn't know what a product log is.

 

[fishsicles]

The proof of sufficiency for the negative self root case ended up being relatively straightforward, given some key observations. Since the modulus of z is by definition real, then we know that

\[N(\theta) = W\left(\pi\csc\theta\, e^{\theta\cot\theta}\right)\]

must be real, which means we both have a restriction on theta, and more importantly, we don't have to worry about it contributing a potential imaginary term when we're working backwards to prove that

\[z = \frac{\sin \theta}{\pi} N(\theta) e^{i\theta}\]

has a negative real self root.

We're going to use the same general approach we used to prove sufficiency for the positive case, which is to say we're going to show that

\[\ln \sqrt[z]{z} = \frac{1}{z} \ln z\]

has an imaginary component equivalent to πi. To start, though, it's better to look at ln z directly:

\[\begin{align*}\ln z = {} & \ln \left(\frac{\sin\theta}{\pi} W\left(\pi\csc\theta\, e^{\theta\cot\theta}\right)e^{i\theta}\right)\\ = {} & \ln\sin\theta - \ln\pi + \ln W\left(\pi\csc\theta\, e^{\theta\cot\theta}\right) + i\theta\\ = {} & \ln\sin\theta - \ln\pi + \ln \left(\pi\csc\theta\, e^{\theta\cot\theta}\right) - W\left(\pi\csc\theta\, e^{\theta\cot\theta}\right) + i\theta\\ = {} & \ln\sin\theta - \ln\pi + \ln\pi - \ln\sin\theta + \theta\cot\theta - N(\theta) + i\theta\\ = {} & \theta\cot\theta - N(\theta) + i\theta\end{align*}\]

Now, we can find the natural log of the self root:

\[\begin{align*} \ln (-x) = \frac{1}{z} \ln z = {} & \frac{\pi}{\sin \theta} \frac{1}{N(\theta)} e^{-i\theta} \left(\theta\cot\theta - N(\theta) + i\theta\right)\\ = {} & \frac{\pi}{N(\theta)} \frac{\cos(-\theta) + i\sin(-\theta)}{\sin\theta} \left(\theta\cot\theta + i\theta - N(\theta)\right)\\ = {} & \frac{\pi}{N(\theta)} (\cot\theta - i)(\theta(\cot\theta + i) - N(\theta))\\ = {} & \frac{\pi\theta}{N(\theta)} (\cot\theta - i)(\cot\theta + i) - \frac{\pi}{N(\theta)} (\cot\theta - i)N(\theta)\\ = {} & \frac{\pi\theta}{N(\theta)} \left(\cot^{2}\theta + 1\right) - \pi\cot\theta + \pi i \end{align*}\]

Since we know that N and theta are real-valued, every term other than πi is real valued (reals are closed under multiplication and cotangent), and thus the logarithm has imaginary component πi - in turn implying that the self root is a negative real number.