How fundamental is Euler's identity, really?

Question

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.

Answer 1

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 $\sin(x)$.

Answer 2

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.

Answer 3

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:

• Why are $e$ and $\pi$ so common as results of seemingly unrelated fields?

• Am I wrong in thinking that $e^{i \pi} = -1$ is hardly remarkable?

• Is it possible to intuitively explain, how the three irrational numbers $e$, $i$ and $\pi$ are related?

• Factorial in power series; intuitive/combinatorial interpretation?

Answer 4

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.