Question about Euler's Identity

I’m trying to puzzle through some information which I think ends up explaining why Euler’s Identity holds. (Specifically what I’m reading is section 5.3 out of Penrose’s The Road to Reality.)

Quick question. Am I understanding correctly that anything raised to the power of pi-times-i equals 1?

-FrL-

No. e^ipi works because e^ix is equal to sin(x)+i*cos(x). When you insert pi for x the cosine term is zero and the sine term is one.

Okay, so something I didn’t understand in the text I read was where e came from. As far as I could tell, it was just pulled out of a hat. I couldn’t see why natural logarithms were being used instead of, say, base 10 logarithms.

However, here it may be that I am seeing more clearly where e is supposed to come in. One part of the derivation goes like this:

This appears to be where the natural log is introduced. Can someone explain why it can be introduced here? I recognize “dx/dy” from waaaaay back in High School as having something to do with differentials, but I do not remember how to interpret it here. To my recollection, a polynomial always came after it, and I don’t know what it means when there isn’t one.

Well anyway, is it possible to explain the derivation to me without requiring me to learn all of the Calculus?

-Kris

ETA: Maybe the key is here:

[quote=Wikipedia article on e]

More generally, the only functions equal to their own derivatives are of the form Cex, where C is a constant.

Well anyway, there’s a statement connecting e with derivatives. If I didn’t have to go pick up a crying baby now, I’d have time to see whether the connection is relevant to my query.

dy/dx refers to the rate of change of y with respect to x. That is, it’s the slope of a line in the x-y plane.

Are you interested in the derivation of e, or from Euler’s identity? Euler’s identity comes from Euler’s formula. There are some proofs of it here. I like the calculus proof. You should be able to understand it if you can follow why the derivative of a constant function is zero. Since the derivative is the slope of a line… what’s the slop of a constant?

Does that help?

No. Complex exponentiation gets to be a little tricky. As usual, Wikipedia has a good explanation, but as they mention, you have to be careful computing a[sup]b[/sup] if a is not a real number. Their writeup on the complex logarithm is probably as close as you’re going to get to a clear explanation of that particular horror.

Other way around: e^ix=cos(x)+isin(x). e^ipi=-1, not 1.

What they did in these steps
=> i dx/dy = 1/y
=> ix = ln y + c
is a little trick from differential equations called separation of variables. Basically you “mulitply” both sides by dy and then take the integral of both sides (I put multiply in scare quotes because the “multiplication” by dy is just a notational device–what really makes this work is the substitution rule )
So, filling in the steps, using “Int” in place of an integration symbol:
=> i dx/dy = 1/y
=> Int i dx = Int 1/y dy
=> i Int dx = Int 1/y dy
=> ix = ln |y| + c
Note that I was able to take the imaginary number i outside of the integral because it’s just a constant. The last step follows because the antiderivative of 1 is x (i.e Int dx = x), and the antiderivative of 1/y is ln |y|, where |y| represents the absolute value of y. You can get rid of the absolute value if you know your y variable only takes on positive values, and I suspect that is the case for this problem, but in general you should leave them there. Finally, we also have to have a constant c because it’s integration without any specified limits.

But yes, what’s important about e is that the derivative of e^x is e^x. If you have a base other than e, say b (with b>0), the derivative of b^x is ln(b) b^x. So for a fixed base the derivative of b^x is always a constant times itself. The precise value of b for when this constant is equal to 1 is the number e. More details on this here: e (mathematical constant) - Wikipedia

Thanks for the leads so far, everyone. It shall all come together for me in short order I think.

Special shout out to totoismomo. I feel certain that was (will turn out to have been) a very helpful post. Thanks!

-FrL-

I don’t feel like typing out the details, but I just showed this to my Calc II students on Thursday in terms of power series:

If you take the power series for e^x =1+x+(x^2)/2!+(x^3)/3!+…, and substitute ix instead of x, you get a power series which is exactly cos(x)+i*sin(x). Then plugging in pi for x gives the famous equation. Now the reason that e^x (and not 5^x, for example) has such a neat power series has to do with the fact that d/dx (e^x) = e^x…

Of course, if one were setting out the conventions of complex exponentiation anew, one could make e^(i*x) equal whatever one wanted it to equal. The quite natural condition which leads us to set it out as we do is our desire to preserve the property that the function e^x should be its own derivative and take the value 1 on the input 0, even when considered in the complex context.

It’s essentially the same as some of the proofs linked to before, but just to phrase it in (what I feel to be) a particularly simple manner, I like the following way of proving e^(ix) = cos(x) + isin(x), without using Taylor series and using from the theory of differential equations only the knowledge that certain kinds of systems of equations have unique solutions:

First, we decompose e^(ix), as a complex-valued function, into its real and imaginary components, setting it equal to a(x) + ib(x) for some real-valued functions a and b of x. Since e^0 = 1, it follows that a(0) = 1 and b(0) = 0.

Next, we obtain the derivative of e^(ix). Since e^x has as derivative e^x, it follows that e^(ix) has as derivative ie^(ix), which is therefore equal to i*(a(x) + ib(x)) = -b(x) + ia(x). But since this is the derivative of a(x) + i*b(x), it must be the case that the derivative of a(x) is -b(x) and the derivative of b(x) is a(x).

So this brings us to a pair of differential equations of a certain form relating a and b, along with their “initial values”; by the theory of differential equation, these will determine a unique solution [essentially because, starting at the initial values, one could straightforwardly deterministically “plot out” the rest of a solution, using the information from the differential equations]. And, as it happens, taking a(x) as cos(x) and b(x) as sin(x), we see that they do in fact clearly satisfy the differential equations and the initial values. Thus, we can conclude, e^(ix) = cos(x) + isin(x). Q.E.D.

As for why e^x should be its own derivative and take value 1 on the input 0, well… that’s, let us say, essentially the definition of e^x. “But why should there exist any or a unique e having that property for exponentiation with it as a base?”, you may ask. Well, let us restrict ourselves to the real context for a moment. By essentially the same reasons as invoked above in square brackets, there is a unique function exp from reals to reals satisfying exp(0) = 1 and being its own derivative. We can essentially define e as exp(1). Now, what remains to be shown is that exp(x) = e^x = exp(1)^x. As it happens, a function f(x) is everywhere equal to f(1)^x if and only if f is continuous, f(0) = 1, and f(c+x) = f©f(x) [the latter two conditions inductively show that f produces the value f(1)^x on all naturals x, which straightforwardly inductively extends to this holding for all integers x, and then for all rationals x; the continuity criterion takes it from there to all reals]. exp is continuous because it is, a fortiori, differentiable; furthermore, exp(0) = 1 because that was one of the constraints which defined it. Finally, exp satisfies the last condition because exp(c+x)/exp© is also, by straightforward differentiation rules, its own derivative, and clearly takes value 1 on the input 0; thus, exp(c+x)/exp© = exp(x), giving us the third condition [we can word this slightly differently to get around the worry that exp© might be 0, throwing the division into undefined territory, but, at any rate, this cleanly expresses the idea]. We can conclude, at least in the real context, that exp(x) does indeed equal exp(1)^x, which means, having defined e as exp(1) and combining this with what was shown above, that there is a unique value e such that e^x is its own derivative. Extending this to the complex context is then just a matter of obtaining an appropriate definition of exponentiation with a complex power that maintains the necessary properties, and what was done above in “proving” e^(ix) = cos(x) + isin(x) is essentially to demonstrate that the appropriate general definition of exponentiation is the one essentially given by that equation. [That is, where b^(c + id) = cos(ln(b)c) + isin(ln(b)d). This works quite well if one is just thinking about b as a positive real; more generally, one must confront the question of what the natural logarithm is to mean in more general contexts, which will generally have to involve some notion of a multi-valued function or branch cuts or so on (because, for example, e^0 and e^(2pi*i) take on the same value)].

Whoops, sorry. I meant to say b^c * (cos(ln(b)d) + isin(ln(b)*d)).

Also, I apparently use the word “essentially” a lot.

The answers above seem quite thorough. I’m going to go ahead and be less thorough (leaving out exactly how to get the general formula e[sup]ix[/sup] = cos(x) + isin(x)) in hopes of helping you focus on the main points.

The meaning of "derivative"
For a function f(x), we can generate a new function by plotting f(x) vs. x and finding the slope at every point. This slope function is called the “derivative of f with respect to x” (sometimes written df/dx).

Likewise, the “second derivative” is the slope of this slope function.

Why e appears in the identity
e appears in the identity because f(x) = e[sup]x[/sup] has the important property that df/dx = f(x)
Taking further derivatives likewise leaves the original function unchanged.

In fact, if we rescale the x-axis so that f(x) = e[sup]ax[/sup], the derivative is likewise rescaled:
df/dx = a
f(x)
Taking further derivatives produces further factors of a. In particular, the second derivative is a[sup]2[/sup]*f(x)

Why i appears in the identity
Basically, if we want to rewrite f(x) = e[sup]ax[/sup] in terms of sines and cosines we have to use i. This can be seen by the fact that the second derivative of sin(x) is -sin(x), and the second derivative of cos(x) is -cos(x). In order for the second derivative of e[sup]ax[/sup] to match, we have to have a[sup]2[/sup] = -1. Thus, we use a = i.

Why pi appears in the identity
Once we’ve rewritten e[sup]i*x[/sup] in terms of sines and cosines, it shouldn’t be too surprising that pi shows up, given that it’s half the period of sin(x) and of cos(x). In particular, the identity follows from the fact that sin(pi) = 0 and cos(pi) + 1 = 0.

Summary[ul]
[li]e[sup]x[/sup], sin(x), and cos(x) all have the property that the second derivative is the original function times a numerical factor. (In the first case, the factor is 1, for the other two cases it’s -1).[/li][li]If we include an i in the exponential, we can make this the same numerical factor (since i[sup]2[/sup] = -1), in which case we can write e[sup]x[/sup] in terms of sin(x) and cos(x).[/li]sin(x) and cos(x) have special values (0 and -1, respectively) if x = pi[/ul]

Among other things, I’ve glossed over the fact that the first derivatives of sin(x) and cos(x) give each other (times 1 and -1, respectively), which is why we can also match the fact that the first derivative of e[sup]a*x[/sup] gives itself (times a). I focused on the second derivative because I felt that’s where it’s easiest to see how i comes in.