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Euler's identity, obviously, states that $e^{i \pi} = -1$, deriving from the fact that $e^{ix} = \cos(x) + i \sin(x)$. The trouble I'm having is that that second equation seems to be more of a definition than a result, at least from what I've read. It happens to be convenient. Similarly, the exact nature of using radians as the "pure-number" input to trig functions is a similar question of convenience -- would it be fundamentally wrong to define sine and cosine as behaving the same way as they do now, except over a period of $1$ rather than $2 \pi$? In such a system, $e^{i \pi} = \cos(\pi) + i\sin(\pi) = \cos(\pi - 3) + i\sin(\pi - 3)$, or transforming back into our $2\pi$-period system to get a result, $\cos(\pi\frac{(\pi - 3)}{1}) + i\sin(\pi\frac{(\pi - 3)}{1})$, which is approximately $0.903 + 0.430i$. (Hopefully I did that right.)

Since there are equally mathematically true systems where $e^{i \pi}$ gives you inelegant results, I'm asking whether the fact that $e^{i \pi} = -1$ really demonstrates some hidden connection between $e$ and $\pi$ and the reals and imaginaries, as it rests largely on what seem to me to be arbitrary definitions of convenience rather than fundamental mathematical truths.

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    $\begingroup$ You need to be a little careful here. It is not that $e$ and $\pi$ exist in some arbitrary fashion and a magical connection is found. Given a definition of $e$, one defines $\pi$ to be the smallest non zero number such that $e^{i 2 \pi} =1$. $\endgroup$ – copper.hat Dec 18 '15 at 3:26
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    $\begingroup$ If you can get your hands on a copy of Tristin Needham's Visual Complex Analysis, he makes a pretty strong case using the geometry of complex numbers that it's anything but arbitrary, in the first chapter. $\endgroup$ – pjs36 Dec 18 '15 at 4:11
  • $\begingroup$ @Why-Seven-Six this is an interesting topic! I did some weeks ago a question about the visualization of a more generic pattern of Euler's identity. It might give you another perspective, so just in case for your reference: math.stackexchange.com/questions/1436167/… $\endgroup$ – iadvd Dec 18 '15 at 12:34
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    $\begingroup$ By the way, $\pi$ is also defined independently and consistently as the ratio of the circumference and the diameter of a circle. This definition is independent of any possible hidden length scale. $\endgroup$ – lcv Dec 18 '15 at 13:34
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The real function $\exp: \mathbb R \to \mathbb R$ is the unique solution to the initial value problem $f'(x)=f(x)$ and $f(0)=1$. The complex function $\exp: \mathbb C \to \mathbb C$ is defined exactly the same way - as the unique complex-differentiable function satisfying the same initial value problem. The fact that such a solution exists over $\mathbb C$ is not obvious at all; it has to be proven. The easiest way to prove it is using the Taylor Series expansion of $\exp(z)$. In fact this is the unique way of extending $\exp$ from $\mathbb R$ to $\mathbb C$ while keeping the condition that it be differentiable.

It just so happens that, when $\exp$ is defined this way, we get the identity $\exp(ix) = \cos(x)+i\sin(x)$, where $\cos$ and $\sin$ use radians. There is no choice here; this is simply what comes out of defining $\exp$ in the only manner possible suited for analysis.

To see this using Taylor series, you need to know what the Taylor series for $\exp$, $\cos$, and $\sin$ are, where $\cos$ and $\sin$ use radians. I'll assume these are known, and they are:

$$\exp(z) = \sum_{k=0}^\infty \frac{z^k}{k!} = 1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!} + \cdots,$$

$$\sin(z) = \sum_{k=0}^\infty (-1)^k\frac{z^{2k+1}}{(2k+1)!} = z - \frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\cdots,$$

and

$$\cos(z) = \sum_{k=0}^\infty (-1)^k\frac{z^{2k}}{(2k)!} = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots.$$

The idea is simply to put $ix$ in for $\exp(z)$. I'll just show what you get with the first few terms:

$$\exp(ix) = 1+(ix)+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}+\frac{(ix)^5}{5!}+\cdots$$ $$=1+ix-\frac{x^2}{2!}-i\frac{x^3}{3!}+\frac{x^4}{4!}+i\frac{x^5}{5!}-\cdots$$ $$=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\right)+i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)$$

where I've collected real and imaginary parts in the last step. The observation is simply that the real part matches the taylor series for $\cos(x)$ and the imaginary part matches the taylor series for $\sin(x)$.

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  • $\begingroup$ Why would it fail to work in a system that doesn't use radians? $\endgroup$ – Why-Seven-Six Dec 18 '15 at 3:04
  • $\begingroup$ If $\cos$ and $\sin$ are defined using a different system of angles, the identity would just change to $\exp(ix) = \cos(ax)+i\sin(ax)$ for an appropriate scaling constant $a$. As for why radians basically fall out of the equations, I think you need to see some calculus, or at least Taylor series, to really understand what's going on here. $\endgroup$ – Dustan Levenstein Dec 18 '15 at 3:09
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    $\begingroup$ Taylor series is typically how Euler's identity is proven to a classroom. The taylor series of $cos(x) + isin(x)$ is the same as the series of $e^{ix}$. Because these functions are both analytic, this proves that they're the same. Of course, radian measure is necessary for the derivatives of trigonometric functions to make "sense". $\endgroup$ – Kaynex Dec 18 '15 at 3:19
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    $\begingroup$ @WillR Yes, that's it. I'm writing up the derivation in terms of taylor series right now. $\endgroup$ – Dustan Levenstein Dec 18 '15 at 3:20
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    $\begingroup$ Re: the use of radians...the measurement of radians are what makes true the formulae $d/dx(\sin(x)) = \cos(x)$ and $d/dx(\cos(x)) = -\sin(x)$ because those derivations rely on computing the limit $\lim_{\theta \rightarrow 0} \sin(\theta)/\theta = 1$, something that is only true when $\theta$ is the radian measurement. $\endgroup$ – tkr Dec 18 '15 at 15:31
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One of the more remarkable things about this identity is that it falls out of so many different definitions of the terms. Its not just a convenience or a happenstance, it arises from almost any valid definition of exponentiation.

So, for instance, consider the definition $$e^x = \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^n$$ which is historically where $e$ first arose, in the work of Jacob Bernoulli.

So now we can ask: does this definition lead to Euler's identity? Or, more explicitly, is $$\lim_{n\to\infty} \left(1 + \frac{ix}{n}\right)^n = \cos(x) +i\sin(x)\ ?$$ Of course here, and later, we use the radian version of the trig functions, and $x\in \mathbb{R}$.

To answer this, lets assume that $|zw| = |z||w|$ and $\arg (zw) = \arg(z) +\arg(w) (\mod 2\pi)$. We can derive these identities using algebra, and results from geometry that are more than 2000 years old. Furthermore, these functions are continuous, which is obvious for $|\cdot|$ and is true for $\arg$ in the correct topology.

We can now calculate the modulus of the relevant limit. $$\begin{align*} |e^{ix}| &= \left| \lim_{n\to\infty} \left(1 + \frac{ix}{n}\right)^n \right|\\ &= \lim_{n\to\infty} \left| \left(1 + \frac{ix}{n}\right)^n \right| \\ &= \lim_{n\to\infty} \left| \left(1 + \frac{ix}{n}\right)\right|^n \\ &= \lim_{n\to\infty} \left( 1 + \frac{x^2}{n^2} \right)^{n/2} \\ &= \lim_{n\to\infty} \left(\left( 1 + \frac{x^2}{n^2} \right)^{n^2}\right)^{1/2n} \\ &= \lim_{n\to\infty} \left(e^{x^2}\right)^{1/2n} \\ &= 1 \end{align*}$$ The first line is our definition, the second is justified by continuity, the third by our modulus identity, the fourth by the definition of modulus, and from there we play with exponents and use our definition of the exponential (there's another way to do it with logs, but this ought to be fine).

We can also calculate the argument. $$\begin{align*} \arg(e^{ix}) &= \arg\left( \lim_{n\to\infty} \left(1 + \frac{ix}{n}\right)^n \right)\\ &= \lim_{n\to\infty} \arg\left(\left(1 + \frac{ix}{n}\right)^n \right)\\ &= \lim_{n\to\infty} n \arg\left(1 + \frac{ix}{n}\right)\\ &= \lim_{n\to\infty} n \arctan\left(\frac{x}{n}\right) \\ &= \lim_{h\to\infty} \frac{ \arctan(xh) - \arctan(0) }{ h } \\ &= \left. \frac{\text{d}\arctan'(xt)}{\text{d}t}\right\vert_{t=0} \\ &= x. \end{align*}$$ the justifications here are much the same as before, with a little calculus thrown in at the end.

Taking our two results together, and using a little more geometry, we have that $$e^{ix} = \lim_{n\to\infty} \left(1 + \frac{ix}{n}\right)^n = \cos(x) + i \sin(x)$$ and by implication $$e^{i\pi} = \lim_{n\to\infty} \left(1 + \frac{i\pi}{n}\right)^n = -1$$. So, this isn't just some arbitrary thing, it appears with all the definitions of exponentiation that can be easily extended to the complex numbers.

Anyway, I hope this adds something to your understanding @Why-Seven-Six.

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  • $\begingroup$ I was about to add a link to this answer, but it is pretty close to what you have above. $\endgroup$ – robjohn Dec 18 '15 at 14:11
  • $\begingroup$ +1. I think Wikipedia has a good illustration of this somewhere. $\endgroup$ – Dustan Levenstein Dec 18 '15 at 15:28
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    $\begingroup$ Yeah, its right here. en.wikipedia.org/wiki/Euler's_identity#/media/… $\endgroup$ – user24142 Dec 18 '15 at 15:31
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    $\begingroup$ @robjohn I find this argument so much more visceral and believable for a high-school/first-year-undergrad. It doesn't require the understanding of the primacy of power series that the other argument leans on so heavily. $\endgroup$ – user24142 Dec 18 '15 at 15:35
  • $\begingroup$ @user24142: yes, this is a pre-calculus approach. No derivatives or power series required. It also works from what I think of as the definition of $e^x=\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n$. $\endgroup$ – robjohn Dec 18 '15 at 15:53
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The use of radians in trig functions is a consequence of the calculus, because measuring angles in radians gives you $\cos'(x) = -\sin(x)$ and $\sin'(x) = \cos(x)$, and, when you go to second derivatives, $\cos''(x)= - \cos(x)$ and $\sin''(x) = - \sin(x)$. This is really convenient, certainly, but also suggests something fundamental about using radians. I've always viewed Euler's identity as being a curiosity, but not the relationship between the exponential function and trigonometric functions; these become more obvious when you study linear differential equations. For example, the solution of $f''(x)-f(x)=0$ is $A\exp(x) +B\exp(-x)+C\sin(x) +D\cos(x)$, where $A, B, C$ and $D$ are arbitrary constants.

If you know the Taylor expansions of the exponential and trigonometric functions, try plugging in ix as the argument of the exponential and see what you get.

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I'm asking whether the fact that $e^{i \pi} = -1$ really demonstrates some hidden connection
between $e$ and $\pi$ and the reals and imaginaries, as it rests largely on what seem to me to
be arbitrary definitions of convenience rather than fundamental mathematical truths.


See my answers on the following four posts, and tell me what you think:

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