Proof of Euler's formula that doesn't use differentiation?

Question

So I saw a 'proof' of the sine and cosine angle addition formulae, i.e. $\sin(x+y)=\sin x\cos y+\cos x \sin y$, using Euler's formula, $e^{ix}=\cos x+i\sin x$. By multiplying by $e^{iy}$, you can get the desired result.

However, this 'proof' appears to be circular reasoning, as all proofs I have seen of Euler's formula involve finding the derivative of the sine and cosine functions. But to find the derivative of sine and cosine from first principles requires the use of the sine and cosine angle addition formulae.

So is there any proof of Euler's formula that doesn't involve finding the derivative of sine or cosine? I know you can prove the trigonometric formulas geometrically, but it is more laborious to do.

Answer 1

I will write down one of Eulers own proofs.

Euler starts with writing down De Moivres Formula (can be prooven by simple induction using some basic trig identities).

$$\cos(nx)+i\sin(nx)=\left( \cos(x)+i\sin(x)\right)^n$$

He says that $n$ is very large ($n \to \infty$) and $x$ is very small ($x\to 0$). The product of both will be a finite number called $\omega =nx$. Then he applies this as substitution for De Moivres Formula: $$\cos(\omega)+i\sin(\omega)=\left( \cos(\frac{\omega}{n})+i\sin(\frac{\omega}{n})\right)^n$$

Euler now applies the limit $n\to \infty$: $$\cos(\omega)+i\sin(\omega)=\lim_{n\to \infty}\left( \cos(\frac{\omega}{n})+i\sin(\frac{\omega}{n})\right)^n$$

using small angle aproximations $\cos(x)\approx 1$ and $\sin(x)\approx x$: $$\cos(\omega)+i\sin(\omega)=\lim_{n\to \infty}\left( 1+\frac{i\omega}{n}\right)^n=e^{i\omega}.$$

In the last line he applied the limit representation $e^x=\lim_{n\to \infty}(1+\frac{x}{n})^n$.

The proof might not be rigourus but it captures the main ideas in a very simple and beautiful way.

Answer 2

For all real $z$, we have: $$e^z=\lim_{N\to\infty}\left(1+\frac zN\right)^N$$ so it seems like a good definition for complex $z$ as well. Letting $z=ix$ and using De Moivre and L'Hôpital gets you $\cos x+i\sin x$.

Answer 3

There is a way to prove Euler's formula without using power series. Try integrating $\frac {1}{1+x^2}$ using partial fractions to get a formula for the complex logarithm. You then have to use polar conversion formulas.

Answer 4

As the commenters have pointed out, whether or not your proof is circular depends on how you define $\sin$ and $\cos$.

The standard proof of Euler's formula is this: define $$\sin(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}z^{2n+1}}{(2n+1)!}\qquad \cos(z)=\sum_{n=0}^{\infty}(-1)^{n}\frac{z^{2n}}{(2n)!}$$
and $$\exp(z)=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}$$ Show that these series converge absolutely for all $z$, then let $z=i\theta$ and deduce Euler's formula.

A common first approach is to begin with the geometry of the unit circle, and then reason geometrically about limits, but this runs into difficulty when you try to interpret $\sin$ or $\cos$ of a complex number, since you have lost your interpreteation. Since it is possible to begin with the series definition and deduce a geometric interpretation, but not vice versa, this is the way it is usually done.

Answer 5

First of all, how are you defining $e^z$ for complex $z$? For example I could just as easily define $e^z = e^{Re(z)}$ and I would get a continuous function on the complex numbers which agrees with the function $e^x$ on the real numbers, and it even still satisfies $e^{z+1} = e^z \cdot e^1$.

The "correct" extension of the exponential is characterized by the fact that it is the unique such extension which is differentiable as a complex function (i.e. holomorphic). Let's take this to be our definition of the complex exponential.

This implies that it agrees with any taylor series expansion centered at a real value, so long as it is in the radius of convergence (which in this particular case is infinite). Hence our definition gives us the usual power series representation (i.e. the taylor series at 0). It also tells us that the functional equation $e^{z+w}=e^ze^w$, and the differential equation $\frac{d}{dz}e^z = e^z$ are satisfied for complex values as well.

Now that we have a definition and some properties lets give a geometric proof. Consider the curve $t \rightarrow e^{it}$ in the complex plane. Its image lies in the unit circle $|z|=1$, since $|e^{it}|^2 = e^{it}e^{-it} = e^0 = 1$. Moreover, its derivative with respect to $t$ is $ie^{it}$ which is also of magnitude $1$.

So $e^{it} = x(t)+iy(t)$ is a unit speed parameterization of the unit circle $\{x+iy | \ x^2+y^2=1\}$ which starts at the point corresponding to $(x,y)= (1,0)$ and starting with $y'(0) >0$. The $x$ and $y$ coordinates of such a unit speed parameterization are precisely the geometric definition of cosine and sine.

Answer 6

Another possible definition of sine and cosine is as solutions to the equation $$ u''+u=0, $$ with $\sin$ the one with $u(0)=0$, $u'(0)=1$, and $\sin$ defined as $-\cos'$ (uniqueness theorems implies this gives a workable definition). It is also apparent that if $u$ satisfies the equation, so does $au'$: $$ 0=a(u''+u)' = (au')''+(au'). $$ So we have the differential equation $$ \frac{d}{dx}(\cos{x}+i\sin{x}) = i(\cos{x}+i\sin{x}), $$ ($(\cos+i\cos')' = \cos'+i\cos''=\cos'+i\cos=i(\cos-i\cos')$) so $\cos{x}+i\sin{x}$ is a solution of $u'=iu$ with $u(0)=1$. But we know (from properties of the exponential, wherever you want to get them from) that $e^{ix}$ is also such a solution. Hence the uniqueness theorem gives $$ e^{ix}=\cos{x}+i\sin{x}. $$

Now, the addition formulae can also be proved by differentiation: we have that $$ u(x) = \sin{(x+y)} $$ is a solution to $u''+u=0$ with $u(0)=\sin{y}$, $u'(0)=\cos{y}$. Then uniqueness for differential equations implies that $$ u(x) = u(0)\cos{x}+u'(0)\sin{x} = \sin{y}\cos{x}+\cos{y}\sin{x}, $$ because $\cos$ and $\sin$ are linearly independent. You can do $\cos{(x+y)}$ in the same way.

Answer 7

First note that the trigonometric addition formulas can be proved with pure geometric arguments (without use of the Euler formula) as you can see here.

Using this result we can find the derivatives of the functions $\sin x$ and $\cos x$ for $x \in \mathbb{R}$ and their Taylor expansions.

$$ \sum_{k=0}^\infty\dfrac{(-1)^k\,x^{2k}}{(2k)!}= 1-\dfrac{(x)^2}{2!}+\dfrac{(x)^4}{4!}-\dfrac{(x)^6}{6!}+ \cdots = \cos(x) $$ $$ \sum_{k=0}^\infty\dfrac{(-1)^k\,x^{2k+1}}{(2k+1)!}= x-\dfrac{(x)^3}{3!}+\dfrac{(x)^5}{5!}+\cdots =\sin(x) $$

Now, using the series definition of the exponential function for a complex variable, we find: $$ \begin{split} e^{ib}= \sum_{k=0}^\infty\dfrac{(ib)^k}{k!}&=1+ib+\dfrac{(ib)^2}{2!}+\dfrac{(ib)^3}{3!}+\cdots+\dfrac{(ib)^n}{n!}+\cdots\\ &=1-\dfrac{(b)^2}{2!}+\dfrac{(b)^4}{4!}-\dfrac{(b)^6}{6!}+ \cdots +i\left[ b-\dfrac{(b)^3}{3!}+\dfrac{(b)^5}{5!}+\cdots \right] \end{split} $$ and this is the Euler's formula.