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In most textbooks on the subject I have seen, Euler's formula (by which I mean $e^{ix}=\cos(x)+i\sin(x)$) is proved by applying either differential equations or the power series of sine and cosine. However, any of these two approaches would rely on the derivatives of sine and cosine. However, I have never seen a proof of these derivatives that does not somehow use Euler's formula.

Is it possible to give such a proof? And if it is, which kind of mathematics does it rely on? Does it rely on the definition of sine and cosine based on the unit circle? And if it does, a new problem arises: If we are to define sine and cosine based on the unit circle, we need the concept of radians and hence the concept of arc length. But this, in return, would require a ---- from a constructive POV --- relatively advanced level of integral calculus. And this suddenly makes the derivation of the formula rely on a large amount of mathematics that has to be constructed as well.

Gaussler
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  • I too feel that involving arc length integrals is more than is necessary. If you start with the Taylor series, you can directly compute its derivative. – user21820 Jul 10 '14 at 11:28
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    While I see lot of discussion over definition of sin and cos in answers and comments, I would like to know the particular definition of $e^{ix}$ you are using here. – Paramanand Singh Jul 29 '14 at 04:02

5 Answers5

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Consider the trigonometric circle, having for implicit equation: $$x^2+y^2=1$$ Compute the arc length, from the point $(1,0)$ to $(x,y)$, which is also the angle, as $r=1$: $$x\ dx+y\ dy=0$$ implies $$ds=\sqrt{dx^2+dy^2}=\frac{dy}{\sqrt{1-y^2}}$$ so that $$s(y)=\int_0^sds=\int_0^y\frac{dy}{\sqrt{1-y^2}}\ \big(=\arcsin y\to y=\sin s\big)$$ At this stage, we must abstain from using the $\arcsin$ and $\sin$ functions as such, as this would create a circular argument.

Let us compute the derivatives of the inverse function, $y(s)$: $$\frac{dy}{ds}=\sqrt{1-y^2}$$ $$\frac{d^2y}{ds^2}=\frac{d\sqrt{1-y^2}}{dy}\frac{dy}{ds}=-\frac y{\sqrt{1-y^2}}\sqrt{1-y^2}=-y$$ $$\frac{d^3y}{ds^3}=\frac{d(-y)}{dy}\frac{dy}{ds}=-\sqrt{1-y^2}$$ $$\frac{d^4y}{ds^4}=\frac{d(-\sqrt{1-y^2})}{dy}\frac{dy}{ds}=\frac y{\sqrt{1-y^2}}\sqrt{1-y^2}=y$$ and so on.

Clearly, the Taylor development from $s=0$ (i.e. $y=0$) is $$y(s)=s-\frac{s^3}{3!}+\frac{s^5}{5!}-\frac{s^7}{7!}...$$ Repeating with $x(s)$, we can show that the Taylor development of $x(s)+iy(s)$ formally coincides with the Taylor development of $e^{is}$. $$x(s)+iy(s)=1+is+\frac{(is)^2}2+\frac{(is)^3}{3!}+\frac{(is)^4}{4!}+\frac{(is)^5}{5!}...$$ The circular argument was avoided by using independent definitions of the trigonometric functions, based on the circle equation, and the fact that computation of the derivatives uses rational expressions only.

UPDATE

When handling the function $s(x)$, you see the strange constant $\int_0^1\frac{dx}{\sqrt{1-x^2}}\approx1.57079632679...$ appear. Denoting it as $\pi/2$ (for instance), we have the identity $e^{i\pi/2}=i$.

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Many modern texts define the sine, cosine, and exponential function directly in terms of their Taylor expansions. The definition are then very rigorous, requiring very little analysis to prove the basic properties etc. Euler's formula is then easily derived, as you say, through the series.

This would have been really great except that these definition being very hard to motivate (i.e., why the hell do we take just these coefficients, and then everything works, like magic). So, when I teach these functions properly I usually spend some time on the geometric definitions of the since and cosine function (in terms of the unit circle and radians) but everything is informal. Then one can give a pretty simple proof that the derivative of $\sin$ is $\cos$. The function $\exp$ can be motivated in various ways (i.e., through compound interest or as a solution to $x'=x$). Then you can informally run over the idea of series expansion and obtain (without justification (yet)) the series expansions of these functions, and note the Euler formula. At that point you say: well let's just define these functions and forget about the whole geometric motivation. Better yet, let's remember the geometric motivation and figure out we found a slick way of defining them rigorously without using somewhat vague geometric concepts.

The thing is also that the definition of the trigonometric functions (when done geometrically) are a bit tricky since they rely on the notion of length, but length is a very subtle thing. It is possible though to define the trigonometric functions in terms of area (but I can't give you a reference I'm afraid).

Ittay Weiss
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  • It is actually very easy to motivate the definition of $\exp$ as its Taylor series, if we want a function whose derivative is itself, the main motivation being to be able to solve ODEs. Similarly $\cos$ and $\sin$ arise in the solution for simple harmonic oscillation, which motivates their Taylor series. Once we prove the convergence and differentiability of the Taylor series for $\exp$, we can later recover the geometric relationships rather easily as I showed at http://math.stackexchange.com/a/802678/21820. – user21820 Jul 10 '14 at 11:22
  • But even though length is subtle, we cannot really say that we have proved Euler's formula until we have a line of reasoning starting from the exponential function and ending with the arc length of the unit circle (which does require integration). But as I understand you, this should be possible? – Gaussler Jul 10 '14 at 12:23
  • I was thinking of another possible approach: To define the curve $\Bbb R\ni t\mapsto e^{it}$ and convince oneself that this is a unit-speed parametrization of the complex unit circle. We then define the radian angle at a point $p$ on the circle to be the number $\theta\in [0,2\pi)$ such that $e^{i\theta} = p$; it should be possible to come up with a proof that the curve is $2\pi$-periodic. We then see that $\cos(\theta)$ and $\sin(\theta)$, as defined by Euler's formulas, are exactly the real and imaginary parts of $e^{i\theta}$, respectively. – Gaussler Jul 10 '14 at 12:31
  • The fact that it is a smooth unit-speed parametrization of the complex unit circle follows if we prove that its absolute value is $1$. – Gaussler Jul 10 '14 at 12:39
  • @Gaussler: There's no need for any physical things like length or speed. You can see how I proved the period of $\exp$ in the post I linked in my first comment, which is roughly the same idea that you have, but one must be careful with the details. – user21820 Jul 11 '14 at 07:35
  • @user21820: I did not mean a physical thing, but to give a mathematical definition of arc length (and hence of angle) that did not use a lot of integral calculus. My definition of radians only relies on differntial calculus. The speed of the curve $\gamma\colon\Bbb R\to\Bbb R,\gamma(t) = e^{it}, t\in\Bbb R$ is defined as $\lVert d\gamma/dt\rVert$. Hence the radian angle of a point $p$ can be defined as above. – Gaussler Jul 11 '14 at 10:03
  • This definition is consistent with the definition of radians based on arc length, since the length of the curve $\gamma\big|{[0,\theta]}$ is $$\ell(\gamma\big|{[0,\theta]}) = \int_0^\theta\lVert d\gamma/dt\rVert d t = \int_0^\theta 1d t = \theta.$$ – Gaussler Jul 11 '14 at 10:04
  • I might make an answer myself later based on this definition in order to illustrate my point (I won't necessarily accept my own answer, though). – Gaussler Jul 11 '14 at 10:13
  • EDIT: I accidentally wrote $\gamma\colon\Bbb R\to\Bbb R$; it should, of course, be $\Bbb R\to\Bbb C$. – Gaussler Jul 11 '14 at 10:14
  • @Gaussler: Now that I read your comments again, it seems that you may not have grasped the issues fully. To be able to obtain all the properties of the exponential and trigonometric functions, you have to use integrals or follow the method I gave. Did you look at it? – user21820 Jul 11 '14 at 10:16
  • @user21820, see my answer below for a way around this obstacle. – Gaussler Jul 11 '14 at 13:58
  • @Gaussler: You yourself say that you assume that $e^{i2\pi}=1$. That is exactly what I prove in my linked post, and also the fact that it is the minimum period. So you did not in fact find a way around the obstacle. – user21820 Jul 11 '14 at 14:00
  • Well, I did read your answer, but the problem is that you define $pi$ based on $\cos$ and hence based on $\exp$, or am I wrong? This is as close to defining it as the period of $\Bbb R\ni t\mapsto\exp(it)$ as we get, since I assume that at that point, you use Euler's formula as your definition of $\cos$. Thus your proof that $e^{2\pi i} = 1$ is close to being a tautology (of course, it is not completely so). That being said, this may be one of the best ways of defining $\pi$. – Gaussler Jul 11 '14 at 14:27
  • But yes, I agree that our approaches are basically the same. What I wanted to demonstrate was that, if we can find a proof that $e^{2\pi i} = 1$, then integral calculus is not a part of the definition of angle. – Gaussler Jul 11 '14 at 14:29
  • And I agree that your post may the best way to prove the periodicity of the complex exponential function. – Gaussler Jul 11 '14 at 14:36
  • @Gaussler: Yes I had defined π based on $\cos$, but $\cos$ was defined as its Taylor series, which in turn can be motivated by the differential equation it solves. So I didn't actually define $\cos$ and $\sin$ by $\exp$. The reason I say that it is easier is because the properties of $\exp$ make the proof of periodicity of $\cos$ and $\sin$ come out naturally. Your method is fine if you define π using the first positive root of $\cos$ as well. I earlier thought that you were using the definition of π via arc length; sorry. – user21820 Jul 11 '14 at 15:23
  • I'm the one who's sorry; I should have read your post better. :-) – Gaussler Jul 11 '14 at 15:34
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I have never seen a proof of these derivatives that does not somehow use Euler's formula.

$$\sin'x~=~\lim_{h\to 0}\frac{\sin(x+h)-\sin x}h~=~\lim_{h\to 0}\frac{\big(\sin x\cos h+\cos x\sin h\big)-\sin x}h~=$$

$$=~\sin x~\underbrace{\lim_{h\to 0}\frac{\cos h-1}h}_{\cos'0}~+~\cos x~\underbrace{\lim_{h\to 0}\frac{\sin h}h}_{\sin'0}~=~-\sin x~\lim_{h\to 0}\frac{2\sin^2\frac h2}h~+~\cos x~\lim_{h\to 0}\frac{\sin h}h$$

Now all we have to do is prove that $\dfrac{\sin h}h\to1$ as $h\to0$, which can be done using geometry and squeezing:

$\qquad\qquad\qquad$

It follows from the figure that $\sin\theta<\theta<\tan\theta~=>~1<\dfrac{\theta}{\sin\theta}<\dfrac1{\cos\theta}$ . By letting $\theta\to0$, the value of the desired limit is thus established.

Lucian
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    And how can one prove that the $\sin$ in Euler's formula is the same thing as the $\sin$ in geometry ? –  Jul 10 '14 at 17:55
  • @YvesDaoust: Did you not read the link ? – Lucian Jul 10 '14 at 17:59
  • @YvesDaoust: As far as I am able to understand, the OP seems to be complaining about the fact that all proofs for the derivatives of the sine and cosine function which he personally is familiar with rely on Euler's formula. This one doesn't. Now, with an independent proof in place, he can use Taylor's series for all three functions, and notice the truthfulness of Euler's observation, without fearing circularity. – Lucian Jul 10 '14 at 18:11
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    It is a true question. Do you know how one makes the link between trigonometry in the geometric sense and in the analytic sense ? –  Jul 10 '14 at 20:01
  • @YvesDaoust: I'm sorry, but I don't really understand what your questions have to do with anything. – Lucian Jul 10 '14 at 20:13
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    The problems with this argument are, apart from understanding which definitions of sine and cosine we're using, that we use a geometrically intuitive, yet highly complex argument for the inequality at the end. Furthermore, we use some trigonometric formulas which I would also personally derive form Euler's formula. However, that being said, it may be possible to derive these differently. But I must say that the argument does become a bit informal at times; however, it is indeed an interesting way of formulating it, particularly if it can be formalized without too much extra mathematics. – Gaussler Jul 11 '14 at 10:12
  • @Gaussler: Would you consider the section at the end of this article to satisfactorily complete the informalities in question, or would you say that the argument, even with these added mentions, is still not rigorous enough ? – Lucian Jul 11 '14 at 16:41
  • I think that Lucian has done the right thing here. The definition of sine which he uses is fully rigorous if one establishes a formal definition of area / length of a circle which is done via integrals. If one wishes to avoid circle, then one may use the integrals directly in definition of arcsin or arctan. I am unable to understand the fuss (as the comments indicate) over his simple and clear answer. +1 from my end. – Paramanand Singh Jul 29 '14 at 03:50
  • @Gaussler: why do u think that trig formulas are derived using Euler's formula. They are most easily derived using the fact that the coordinates of a point on circle $x^2+y^2=1$ can be expressed as$(\cos\theta,\sin\theta)$. – Paramanand Singh Jul 29 '14 at 03:58
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Since I asked the question, I came up with the following solution, which, I think, may be the most beautiful formalisation of a proof of the Euler's formula. It only works, however, if we assume that we already know that $e^{2\pi i} = 1$. If anyone can find a simple proof of this that does not use Euler, say so.

We consider the curve $\gamma\colon\Bbb R\to\Bbb C$ given by $\gamma(t) = e^{it}$ for all $t\in\Bbb R$. It is obviously smooth because $\exp$ is. We define $\cos$ and $\sin$ by Euler's formulas.

Lemma 1: $\gamma$ is a parametrization of a part of the complex unit circle $S^1$.

Proof: We have $\lvert \gamma(t)\rvert =\gamma(t)\overline{\gamma(t)}= e^{it}e^{-it} = e^0 = 1$.

Lemma 2: $\gamma$ is $2\pi$-periodic.

Proof: Since $e^{it}e^{is} = e^{i(t+s)}$ for all $s,t\in\Bbb R$, it follows from the fact that $e^{2\pi i} = 1$, which we assumed to know. (We haven't proved, however, that $\gamma$ could not have a shorter period).

Lemma 3: $\gamma$ paremetrizes all of the unit circle.

Proof: Let $p\in S^1$ be a point on the unit circle. We must show that $\gamma$ hits $p$. Since $\overline{\gamma(t)} = \gamma(-t)$ for all $t\in\Bbb R$, we may assume without loss of generality that the imaginary part of $\gamma$ is $\ge 0$, hence $p\in S^1_+ = \{s\in S^1\mid\operatorname{Im}(s)\ge 0\}$, the upper complex unit circle. It is not hard to show that all points on $S^1_+$ are uniquely determined by their real values, hence we only have to show that $\operatorname{Re}(\gamma)$ takes the value $r:=\operatorname{Re}(p)$. Now the real part $\operatorname{Re}(\gamma)$ is a smooth function $\Bbb R\to [-1,1]$ with $\gamma(0) = 1$, so if it does not take the value $r$ on $[0,2\pi)$, $\operatorname{Re}(\gamma)$ must have a lower bound, hence $\operatorname{Re}(\gamma)(t)\ge b$ for alle $t$ for some $b\in(-1,1)$ (if $b = -1$, $\operatorname{Re}(\gamma)$ would be surjective onto $[-1,1]$ by continuity, which is a contradiction). This is also true for the restriction of $\gamma$ to the compact set $[0,2\pi]$, hence $\operatorname{Re}(\gamma(s)) = b$ for some $s\in[0,2\pi)$. Then $\operatorname{Re}(\gamma'(s)) = 0$. But $\gamma$ has the constant absolute value 1, hence $\gamma$ and $\gamma'$ are perpendicular (differentiate $1 = \langle\gamma,\gamma\rangle$ and get $0 = 2\langle\gamma',\gamma\rangle$); but the only points on $S^1_+$ with a purely imaginary tangent vector are $\pm i$. Hence $p$ is perpendicular to $\pm i$. But this implies that $p = \pm 1$. $p$ obviously cannot be $1$, and if $p = -1$, the same continuity argument as before shows that $\operatorname{Re}(\gamma)$ is surjective onto $[-1,1]$. In either case, we get a contradiction.

Definition 4: Let $p\in S^1$. The angle of $p$ is the (minimal) number $\theta$ in $[0,2\pi)$ such that $\gamma(\theta) = p$.

Now $\cos(\theta)$ and $\sin(\theta)$ are the real and imaginary parts of the number $p$, by definition of $\cos$ and $\sin$. Notice that our definition of angle does not rely on the arc length as defined by integration methods. It is, however, consistent with the angle as defined through arc lengths:

Proposition 5: The angle of a point $p\in S^1$ is the arc length from $1$ to $p$.

Proof: The arc length is, by definition, the length of any surjective smooth curve on $S^1$ from $1$ to $p$; $\gamma\big|_{[0,\theta]}$ is such a curve, where $\theta$ is the angle of $p$. We have that the length of $\gamma\big|_{[0,\theta]}$ is $$ \ell(\gamma\big|_{[0,\theta]}) = \int_0^\theta\lVert\gamma'\rVert d t = \int_0^\theta 1 d t = \theta. $$

Hence we have gone from the complex exponential function to the arc length on the unit circle, which was what I requested. Thus if someone has a good argument that $e^{2\pi i} = 1$, we are done.

Gaussler
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defining $\cos$ and $\sin$ as the projections of a radius vector on two axes separated by a right angle, the derivatives can be computed from the compound angle formulae, which may be proved geometrically.

David Holden
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