An Alternate Derivation of Euler's Identity

Miscellaneous and Personal Stuff I Must Share
1

We all know and love it: e[sup]ix[/sup] = cos(x) + i·sin(x). But the usual derivation using the Taylor’s series always felt too convenient and not intuitive enough. Something is missing from the connection between an exponential function and the trig functions. So here is another derivation.

We’ll start simple, by noticing a pattern:
-1[sup]0[/sup] = 1
-1[sup]0.5[/sup] = i
-1[sup]1[/sup] = -1
-1[sup]1.5[/sup] = -i
-1[sup]2[/sup] = 1

Before you ask, I’ve cheated already. You ask why I chose i instead of -i for the square root. Well, because it makes such a nice repeating pattern that it would be inelegant if I chose anything else.

Look how it circles around the complex plane, repeating itself as the exponent goes up by 2. It almost looks like we can fit some trig functions to it–in fact, we most certainly can:
-1[sup]t[/sup] = cos(πt) + i·sin(πt)

Well, that looks familiar. But there are some questions, like how we can justify that our fit works for any number t, not just the ones at half-integers. Well, we can try one thing:
-1[sup]a+b[/sup] =
-1[sup]a[/sup]·-1[sup]b[/sup] =
(cos(πa) + i·sin(πa))(cos(πb) + i·sin(πb)) =
(cos(πa)cos(πb) - sin(πa)sin(πb)) + i(cos(πa)sin(πb) + sin(πa)cos(πb))

Wait–that looks very familiar. Those are just trig identities! In fact, we just get:
(cos(πa)cos(πb) - sin(πa)sin(πb)) + i(cos(πa)sin(πb) + sin(πa)cos(πb)) =
cos(π(a+b)) + i·sin(π(a+b))

That’s exactly what we’d have expected if we’d just substituted the a+b into the trig functions to begin with. What this means is that the relationship follows the normal rules of arithmetic, which basically means it’s true for all real numbers. We can generate solutions for -1[sup]0.25[/sup], -1[sup]0.125[/sup], -1[sup]0.0625[/sup], etc. and then add them together to get whatever number we want.

I’m going to do a little substitution here (x=πt) to finish off the proof:
(-1[sup]1/iπ[/sup])[sup]ix[/sup] = cos(x) + i·sin(x)

Same thing, just massaged a bit. Let’s do the derivative of both sides to see what happens:
d/dx((-1[sup]1/iπ[/sup])[sup]ix[/sup]) = d/dx(cos(x) + i·sin(x)) =>
i·(-1[sup]1/iπ[/sup])[sup]ix[/sup]·log(-1[sup]1/iπ[/sup]) = -sin(x) + i·cos(x)

Multiply by -i:
-i·i·(-1[sup]1/iπ[/sup])[sup]ix[/sup]·log(-1[sup]1/iπ[/sup])= -i(-sin(x) + i·cos(x)) =
(-1[sup]1/iπ[/sup])[sup]ix[/sup]·log(-1[sup]1/iπ[/sup]) = cos(x) + i·sin(x)

Now substitute the original equation back in the left:
(cos(x) + i·sin(x))·log(-1[sup]1/iπ[/sup]) = cos(x) + i·sin(x)

But… we can just divide that out instead, so we have:
log(-1[sup]1/iπ[/sup]) = 1

And so:
-1[sup]1/iπ[/sup] = e[sup]1[/sup]

A little substitution back into this formula from before:
(-1[sup]1/iπ[/sup])[sup]ix[/sup] = cos(x) + i·sin(x)

And we have:
e[sup]ix[/sup] = cos(x) + i·sin(x)

Et voila. With a little playing around, you can find this weird identity:
log(-1) = iπ

Anyway, I hadn’t seen this derivation before, though I certainly doubt I’m the first to come up with it.

2

“Who was Euler, really? Our team of experts from ancestry.com decided to find out.”

3

It was my understanding that there would be no math.

4

Historical question:

Given the Taylor series for exp(x), sin(x) and cos(x), this Euler’s Identity appears immediately if you have the imagination to write the series for exp(i·x).

Sir Isaac Newton had written down all three of those Taylor series, IIUC; and had worked with imaginary numbers. Why didn’t he notice Euler’s Identity? Was it due to some aversion to imaginary numbers?

5

My constructive criticism is that this seems a bit confusing and not rigorous. Right from the start, complex exponentiation is used, but this is never defined. We can take that as read, as well as the derivative of the complex exponential function, but, later on, natural logarithms are applied to complex numbers, yet these are multiple-valued.

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I’m curious about the answer myself. It does seem that Newton, along with Descartes, was of the era that viewed imaginary numbers as only semi-legitimate; you could do valid math with them, but if they appeared at all it was somehow a sign that things went very wrong, and if they appeared in the final answer, things had gone very wrong. Newton was a physical scientist, so that’s not a completely wrong-headed thing from his perspective. Planets orbit along the real-valued solutions to Ax[sup]2[/sup] + By[sup]2[/sup] = 0, not the complex solutions. An equation that contains imaginary values from the very start but have been seen as highly misguided.

7

Thanks–I agree with the critique. I would say that this isn’t exactly meant as a true proof, but a kind of more direct connection between the trig functions and the exponential. Yes, the Taylor series proof makes the equivalence very obvious, but to me it’s hard to not see it as a giant coincidence. Did we just get lucky that the series for sin/cos interleaves in just the right way to reproduce e[sup]ix[/sup]? Also, it’s not inherently obvious that this is even legal. Infinite series are tricky like that. Maybe I’m an ultrafinitist and don’t believe in infinite series at all :).

Arguably, the first part of the derivation isn’t necessary at all; it just explains the motivation. Taking the derivative is really the crucial step, because we’ve defined e as “that number such that the slope of e[sup]x[/sup] at x is e[sup]x[/sup]”. In that respect, -1[sup]1/iπ[/sup] must be equivalent to e, if not equal.

At any rate, this was by no means intended to be rigorous. I had been puttering around with -1[sup]t[/sup], saw that it behaved very much like e[sup]ix[/sup], and wondered if I could more or less fill in the gaps between the two.

8

Sir, this is the Wendy’s drive-through. Are you gonna order or what?

9

Sorry, I was confused. I thought we were in Mathematical Proofs and Solutions I Must Share.

10

I wonder what you think about BetterExplained’s explanation.

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Paging Indistinguishable

(is he still around? <search> Not since August, apparently: A pity)

12

That’s not bad at all. The essence of it is not so far off from mine, but with this difference:
For theirs, e comes from the idea of continuous compounding–what you get if, instead of multiplying by one lump sum of something, you do many multiplies of a fraction of that something. That’s just another definition of e (i.e., (1+1/n)[sup]n[/sup] for large n). And so they get a continuous rotation out of many multiplies by a tiny rotation.

For my proof, I describe the motion of the point around the complex plane directly, using the trig functions. At first I’m just trying to pass a curve through these points, but then show that it follows the normal trig identity rules and so holds true for intermediate numbers as well. And I start with powers of -1, not e, so it’s more obvious that -1 just represents a 180 degree rotation, and a 360 degree rotation is -1*-1=1, etc.

So they’re just two ways of showing how the exponential really does represent a rotation. I’m a little skeptical over whether anyone would have made the “continuous compounding” connection directly, not knowing Euler’s identity to start with–but maybe the same is true for mine as well. Hard to say.

13

One derivation that appears in some calculus textbooks and is reproduced in Wikipedia proceeds as follows: first set e[sup]ix[/sup] = r(cos θ + i sin θ) where r and θ are some functions of x to be determined. Differentiating both sides yields ir(cos θ + i sin θ) = (cos θ + i sin θ) dr/dx + r(-sin θ + i cos θ) dθ/dx. Solving this for dr/dx and dθ/dx shows that dθ/dx = 1 and dr/dx = 0, which, together with e[sup]0[/sup] = 1, imply that r = 1 and θ = x.

14

That’s a nice one. I like it because the trig functions don’t come out of nowhere; they just come from changing the coordinate system to polar. That said, it’s still a bit indirect. And does seem to make the assumption that e[sup]ix[/sup] results in a complex number and not something else, like a quaternion.

15

Sorry, I got stuck at this first equation. It’s not an equality as written. Do you intend to mean (-1)[sup]0[/sup] = 1? Missing parenthesis makes your mathematics very difficult to read. Where else have you dropped symbols that would change the meaning if included?

16

I think that the unary negative is considered to have a stronger binding than the exponent, particularly when in front of a numeric literal. I would expect most to interpret -1[sup]2[/sup] as (-1)[sup]2[/sup], not -(1[sup]2[/sup]). Some texts use a raised minus symbol to eliminate this possible confusion: [sup]-[/sup]1[sup]2[/sup]. Not everyone is familiar with that usage, though, so it may actually decrease clarity.

In any case, I’m happy to clarify any parts where you get stuck. The lack of LaTeX support on this board does mean that expressing math is more difficult that it should be.

17

Well, Wikipedia says I’m wrong:
Unary minus sign
There are differing conventions concerning the unary operator − (usually read “minus”). In written or printed mathematics, the expression −3[sup]2[/sup] is interpreted to mean 0 − (3[sup]2[/sup]) = − 9,

Apologies. For the above, assume that a unary minus in front of the base of an exponential binds to the base, not to the result.

18

I think that the key realization is that i is just a perfectly mundane rotation by 90º, not some weird mysterious “fake number” that’s worth panicking over. Which would have been a much better starting point for the study of what we now call “complex numbers”. But it’s not the starting point we historically had.

19

Agreed, at least with respect to understanding the complex numbers at all. But with regard to Euler’s Identity specifically, the e makes it weird. The π makes sense if we’re talking rotations. But the e is… magical.

That’s what motivated me to use this starting point instead:
(-1)[sup]t[/sup] = cos(πt) + i·sin(πt)

While less beautiful than Euler’s Identity, this form (IMHO) makes up for it in directness. First, it’s easy to spot-check against a few known values. Furthermore, it’s easy to see the t=1 case as a 180 degree rotation with no imaginary component at all. And then show that it easily divides into 90 degree rotations, or any other smaller value, and that the rotations all combine in exactly the way we expect. All you need are the trig identities we learned in high school.

From there, it takes a little work to massage it into the e[sup]ix[/sup] form. The connection between e and π is magical, but perhaps a little too magical, because the connection to rotations is slightly lost.

That’s just my view, at any rate. I still fully endorse understanding things in as many ways as possible in order to truly grok that thing.