Euler’s Formula without Calculus pt. 3 – The Natural Logarithm

Let’s recapitulate the previous two posts (one and two).

Our goal was to find a reasonable way to expand the exponential function to the domain of complex numbers. We began by demanding the following properties hold when x_1, x_2 are positive real numbers and z_1, z_2 are arbitrary complex numbers.

x^0 = 1 .

x^{z_1}*x^{z_2} = x^{z_1 + z_2} .

\left(x^{z_1}\right)^{z_2} = x^{z_1*z_2} .

(x_1*x_2)^z = x_1^z*x_2^z .

I was curious whether Euler’s famous formula, which students typically see in a college calculus or differential equations course, was useful or meaningful outside the context of calculus.

From those four rules, we deduced that

e^{\imath \theta} = \cos(n\theta) + \imath\sin(n\theta)

for some real number n.

Then I claimed that

n=1

was the simplest choice, which was perhaps unfounded from the standpoint of someone who is not interested in calculus. My justification at the time was that only by choosing n to be one can you obtain the simple formula

\cos(\imath\theta) = \textrm{cosh}(\theta) .

Outside of calculus, though, there’s no particular reason to believe the hyperbolic and circular trig functions should bear any such relation. At least, no reason that’s completely obvious to me.

An easy way to fix n with calculus is to demand

\frac{d}{d\theta}e^{\imath\theta} = \imath e^{\imath\theta}

which translates in component form to

\begin{array}{rr} {} & \frac{d}{d\theta}\left[\cos(n\theta) + \imath\sin(n\theta)\right] \\ = & n\left[-\sin(n\theta) + \imath \cos(n\theta)\right] \\ = & \imath\left[\cos(n\theta) + \imath \sin(n\theta)\right] \end{array} ,

which implies

n = 1.

The first equality follows from differentiating the trig functions, while the second comes from translating \imath e^{\imath\theta} directly into component form.

So fixing n comes from the way that e is special as an exponent. To see what sort of insight we can get into Euler’s formula without calculus, we’ll need to know in what ways e has some special character outside of calculus.

We’ll start with some definitions of e. The first one I learned in school was

e = \lim_{n\to \infty}\left(1 + \frac{1}{n}\right)^n

As my friend Ian recently pointed out to me (while on a long distance run), we can use the binomial theorem, along with the approximation

n \approx n-1 \approx n-2 \approx \ldots \approx n-k , \,\,\,\,\, n>>k

to find the more practical definition

e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + ... = \sum_{k=0}^\infty \frac{1}{k!}

which is easier to evaluate to a given accuracy. You may also recognize it as a Taylor series expansion of e^x evaluated at x=1.

Or, defining

\ln(x) \equiv \lim_{h\to 0} \frac{x^h - 1}{h}

we can define e by

\ln(e) = 1.

Wikipedia catalogs a remarkable array of alternative definitions to these.

What is still unclear to me is whether this is of any particular interest outside a setting of calculus. e may arise when asking a question about compounded interest or certain discrete gambling problems. In those cases e arises in the limit as the number of compounds in a unit time or the number of plays of the game increase without bound. That is, e arises when we can approximate things as continuous, so that we’ve migrated to the domain of calculus.

The only ways I have on hand to prove the equivalence of all my definitions of e are essentially calculus – like. My most direct plan is to substitute

m = \frac{1}{n}

and redefine e as

e = \lim_{m\to 0} (1+m)^{1/m}

then substitute this into the definition of the natural logarithm to obtain

\ln(e) = \lim_{h\to 0} \frac{\left(\lim_{m\to 0} (1+m)^{1/m}\right)^h - 1}{h} .

By allowing m to be some arbitrary analytic function of h and expanding in a Taylor series

m = m(h) = m(0) + m'(0)h + O(h^2) = m'(0)h

I can make a substitution in the previous formula, reduce it to just one limit, and show that the answer does indeed come to one for arbitrary m'(0), but this is far from a calculus-free demonstration, and probably not even right by mathematical standards. I sure wouldn’t want to do a delta-epsilon proof that the method is valid.

There are other ways that are more lucid but less direct and more explicitly rooted in calculus. I could define e^x by its Taylor series, from which it would follow that it is its own derivative, and the other definitions of e would fall out naturally enough. Or I might define the logarithm as an integral (a common definition in calculus), show that it’s the inverse function of the exponential of a certain base, and derive the properties of e that way. But my point is that none of these make any sense outside of calculus.

That’s why, when I attempted to show Euler’s formula without calculus, I got hung up. Euler’s formula depends on e being a special number. Without calculus, it isn’t.

That, at least, is the way things seem to me. The internet is a lot smarter than me, and has a lot more minds. (For example, the internet knows I should technically say, “smarter than I”.) So if e does matter when its friends delta and epsilon are on vacation, let me know.

Finally, I’ll clarify a point I made at the end of the previous post, since Nikita rightly asked for more explanation on it.

I examined the behavior of

x^\imath

as x \to 0. This can be rewritten as

\left(e^{\ln(x)}\right)^\imath = e^{\imath \ln(x)} .

Because it is of the form

e^{\imath \theta} = \cos(\theta) + \imath \sin(\theta) ,

it is a point on the unit circle in the complex plane. That is, if you make an “x-axis” with the real part of the number, and a “y-axis” with the imaginary part, your “x-coordinate” is \cos(\theta) and your “y-coordinate” is \sin(\theta). Those points are the points on the unit circle.

But we have

\theta = \ln(x)

and as x gets very small, \theta blasts away to minus infinity. If we imagine

x^\imath = e^{\imath \theta}

as a dot on the unit circle and slowly start to change the value of x, we’ll also change the value of \theta and the dot will move. As x becomes close to zero, a small change in x produces an enormous change in \theta = \ln(x). In fact, the change in \theta is about 1/x * dx, with dx the change in x.

Allow x to count down towards zero with time, starting at x=10 with ten seconds on a timer. Then x falls smoothly as the time runs out. Instead of watching x during this countdown, you watch the point on the unit circle corresponding to x^\imath. At first it moves just slowly, but in the last second it moves faster, then faster, then faster, until it moves with such rapidity that it flies around the unit circle infinitely many times in the last flitting moment of its existence, and then explodes when x=0 and the exponential is undefined.

4 Responses to “Euler’s Formula without Calculus pt. 3 – The Natural Logarithm”

  1. Nik Says:

    Oh cool.
    Lightbulb moment.

  2. Ben Zinberg Says:

    Pretty cool. I’d agree that e is really only important in analysis. Also, I think that without an analytically defined exponential function exp(x), it is pretty difficult to mathematically articulate our “intuitive understanding” of exponents. Integral exponents can be defined by induction, but then what is (-4)^pi?

  3. meichenl Says:

    Hi Ben,

    (-4)^pi is definitely tough, but 4^3.141592 makes good sense. It is just a number such that, if you take it to the 100,000th power, yields 4^3141592.

  4. Colin Robinson Says:

    “So if e does matter when its friends delta and epsilon are on vacation, let me know.”

    I think the best way to explain why e matters without direct mention of calculus concepts is (as Wikipedia says in its page on natural logarithm)

    “for any numeric value close to the number one, the natural logarithm can be mentally computed by subtracting the number one from the numeric value. For example, the natural logarithm of 1.01 is 0.01 to an accuracy better than 5 parts per thousand. With similar accuracy one can assert that the natural logarithm of 0.99 is minus 0.01. The accuracy of this concept increases as one approaches the number one ever more closely, and reaches completeness of accuracy precisely there.”

    In other words, for small values of theta, e ^ theta approximates to 1 + theta. The relevance of this to Euler’s formula is apparent when we consider that sine theta approximates to theta for small values of theta expressed in radians. So cos theta + sine theta approximates to 1 + theta.

    If this characteristic of e is generalised to complex numbers, then it follows that in your equation

    e ^ i theta = cos (n theta) + i sine (n theta),

    the factor “n” must equal 1.

    To express these points with full rigour, one would admittedly need to call epsilon and delta back from their vacation.

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