Cauchy and the (other) mean value theorem.

I just realised that there's a neat relationship between Cauchy's integral formula and the mean value theorem for harmonic functions of two variables, so for the benefit of anybody else who hasn't already noticed it, here it is.

First, let's remember what the Cauchy integral formula tells us.

If $f$ is a holomorphic function on a simply connected domain, $D$, $a \in D$, and $C$ is a simple closed contour in $D$ surrounding $a$, then $$f(a) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z-a} dz.$$

I confess, this always seems like magic to me.

One way of seeing that it is true is to note that for a holomorphic function continuous deformation of the countour doesn't change the integral: therefore we can take $C$ to be an extremely small circle centred on $a$, so that $f(z)$ is very close to $f(a)$ all around the contour. Then if we parameterise $C$ as $a+\varepsilon e^{i \theta}$, where $0 \leq \theta < 2\pi$, the integral is approximately $$\frac{f(a)}{2 \pi i} \oint_C \frac{1}{z-a} dz = \frac{f(a)}{2 \pi i} \int_0^{2 \pi} i d\theta = f(a)$$ and the approximation can be made arbitrarily good by choosing $\varepsilon$ small enough.

And even with that, it still seems like magic.

Now, let's take a look at this in terms of real and imaginary parts. We can think of the complex number $z$ as $x+iy$, and then thinking about $f$ as a function of $x$ and $y$. As is customary, I write: $$f(z) = u(x,y)+iv(x,y)$$ where $f$ being holomorphic tells us that $u$ and $v$ satisfy the Cauchy-Riemann equations, $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}$$ and as a consequence $$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} = 0 = \frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}$$ so that both $u$ and $v$ are harmonic functions.

It's less obvious, but equally true that if $u$ is a harmonic function of $x$ and $y$, then there is another harmonic function $v$ such that $$f(z)=f(x+iy) = u(x,y) + iv(x,y)$$ is holomorphic.

So now let's see what the Cuachy integral formula says, in terms of $u$ and $v$.

To this end, let $C$ be a circular contour of radius $R$, centred on $a$, and parameterised by $\theta$ as $$a+Re^{i\theta}, \qquad 0 \leq \theta < 2\pi$$

Then we have $$\begin{split} \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z-a}dz &= \frac{1}{2 \pi i} \int_0^{2 \pi} \frac{f(z(\theta))}{Re^{i\theta}} iRe^{i\theta} d\theta \\ &= \frac{1}{2\pi} \int_0^{2\pi} u(x(\theta),y(\theta))d\theta + i \frac{1}{2\pi} \int_0^{2\pi} v(x(\theta),y(\theta))d\theta \end{split}$$

So the real part of $f(a)$ is the mean value of $u(x,y)$ on a(ny) circle centred on $a$, and the imaginary part is the mean value of $v(x,y)$ on a(ny) circle centred on $a$.

But the real part of $f(a)$ is just $u(Re(a),Im(a))$, and similarly for $v$!

This result, that the value of $u(x,y)$ for harmonic $u$ is the mean value of $u$ on any circle centred on $(x,y)$, is the mean value theorem for harmonic functions.

We can go backwards: assuming the mean value theorem for harmonic functions, we can deduce the Cauchy integral formula (at least, if we are happy that contour integrals of holomorphic functions are unchanged by a deformation of the contour).

So it turns out that Cauchy's integral formula, a fundamental result in complex analysis, is equivalent to a real analysis result about the properties of harmonic functions.

In retrospect, I suppose it should be obvious in some sense that Cauchy's integral formula relies on the properties of harmonic functions, since holomorphic functions are built out of harmonic ones. But I hadn't put the two ideas together like this before.

Trig functions with (almost) no triangles

In this post I'm going to look at what we can say about the trigonometric functions, $\sin$ and $\cos$ without actually doing any classical trigonometry, in other words, without using triangles or ratios of sides - if we know enough other maths. Note that this is not a "we should teach it this way" article it's an "isn't it great how mathematics all hangs together" article.

Definitions

We start off with a little Euclidean geometry, as described in Cartesian coordinates.

Consider the unit circle in the Euclidean plane, centred on the origin, so given by $$x^2+y^2=1$$

and define $\cos(t)$ and $\sin(t)$ as the $x$ and $y$ coordinates of the point that you get to on this circle if you start at the point $(1,0)$ and travel $t$ units of length (inches if the circle is one inch in radius, meters if it is one meter in radius, and so on) anti-clockwise around the circle.

This is how I define the functions $\cos$ and $\sin$; at some point I should make contact with the usual definitions in terms of ratios of side lengths in right-angled triangles. But not yet.

But we can immediately notice that since I've defined these functions as the $x$ and $y$ coordinates of points on this circle, we know that $$\sin^2(t)+\cos^2(t)=1$$ and since the circumference of the circle is $2\pi$, both functions are periodic with period $2\pi$.

I can also extend my definition to negative values of $t$, by thinking of this as distance in the clockwise direction from the same starting point.

Then we also see immediately from the symmetry of the circle that $$\sin(-t)=-\sin(t), \qquad \cos(-t)=\cos(t)$$

Derivatives

And now a tiny nod towards differential geometry.

Let's think about the circle, where I now parameterise it by the $t$ I defined above: so we can associate with any value of $t$, the point $$\pmb{r}(t) = (\cos(t),\sin(t))$$ (and I commit the usual sin of identifying a point in the plane with its position vector).

So, what can I say about $\dot{\pmb{r}}$ (where I use the overdot to denote differentiation with respect to $t$, in the finest Newtonian tradition).

Again, since each point lies on the unit circle, we have $$\pmb{r}(t).\pmb{r}(t)=1$$ so, differentiating, we have $$2\pmb{r}(t).\dot{\pmb{r}}(t)=0$$ and the tangent to the circle at any point is (as we already know from Euclidean geometry) orthogonal to the radius to that point.

From this, we know that $$\dot{\pmb{r}}(t) = k(-\sin(t),\cos(t))$$ for some $k$.

We can say more, though. Since $t$ is distance along the circle, so that for all $t$ $$t = \int_{0}^t \|\dot{\pmb{r}}(u) \| du,$$ then we must have $$\| \dot{\pmb{r}}(t) \| = 1.$$

Finally, we note that when $t=0$, the tangent vector points vertically upwards. This finally leaves us with $$\dot{\pmb{r}}(t)=(-\sin(t),\cos(t))$$ or, in other words, $$\frac{d}{dt}\sin(t) = \cos(t), \qquad \frac{d}{dt}\cos(t)=-\sin(t).$$

Calculation

Now we take a little detour through some real analysis.

For no apparent reason, I introduce two functions defined by power series, $$\text{SIN}(t) = \sum_{n=0}^\infty (-1)^{n}\frac{t^{2n+1}}{(2n+1)!}, \qquad \text{COS}(t) = \sum_{n=0}^\infty (-1)^{n} \frac{t^{2n}}{(2n)!}.$$

These are new functions, and as far as we know so far, distinct from $\sin$ and $\cos$ as I defined them up above. But we'll see in a moment that the choice of names is not a coincidence, and that there is a very close relationship between these power series and the geometrically defined $\cos$ and $\sin$.

We can immediately see that $\text{SIN}(0)=0$ and $\text{COS}(0)=1$.

But since the two functions are defined by power series, we can differentiate both term by term and find that $$\frac{d}{dt}\text{SIN}(t) = \text{COS}(t), \qquad \frac{d}{dt}\text{COS}(t)=-\text{SIN}(t).$$

Differentiating again, we find that each of $\sin$, $\cos$, $\text{SIN}$ and $\text{COS}$ satisfies the ordinary differential equation $$\ddot{X} = -X.$$ Furthermore, both $\sin$ and $\text{SIN}$ satisfy the same initial conditions at $t=0$, namely that the value of the function is $0$, and its derivative is $1$; similarly for $\cos$ and $\text{COS}$. We therefore deduce that $$\sin(t)=\text{SIN}(t), \qquad \cos(t)=\text{COS}(t).$$

This gives us two things:

• An effective way of calculating the functions (at least for small values of $t$).
• Another handle on the functions, which we are just about to use.

Addition Formulae

Finally we make use of just a little complex analysis.

Looking at the power series we now have for $\sin$ and $\cos$, and remembering the power series for the exponential function, we have $$\exp(it)=\cos(t)+i\sin(t).$$

But now we have $$\begin{split} \cos(s+t)+i\sin(s+t) &= \exp(i(s+t))\\ & = \exp(is)\exp(it)\\ & = \cos(s)\cos(t)-\sin(s)\sin(t)\\ & +i(\sin(s)\cos(t)+\cos(s)\sin(t)) \end{split}$$ from which we extract $$\cos(s+t)=\cos(s)\cos(t)-\sin(s)\sin(t), \qquad \sin(s+t)=\sin(s)\cos(t)+\cos(s)\sin(t).$$

And finally, some triangles

It's time to make contact with the triangle notion of these trigonometric functions, so consider a right angled triangle, and let $\theta$ be the radian measure of the angle at one of its vertices, $V$, (not the right angle). Place this triangle in the first quadrant, with the vertex $V$ at the origin, and the adjacent side along the $x$-axis. (This may require a reflection.) Call the vertex on the $x$ axis $U$ and the remaining vertex $W$.

But we can scale the triangle by dividing each side length by the length of the hypotenuse, which doesn't affect any of the ratios of side lengths.

The traditional definition of the sine of angle $\theta$ is that it is the length of the adjacent side to $V$ divided by the length of the hypotenuse; similarly $\cos(\theta)$ is the length of the side opposite to $V$ divided by the length of the hypotenuse. Then in our rescaled triangle, the vertex $U'$ corresponding to $U$ lies on the unit circle, and (by the definition of radian measure of angles) the angle $\theta$ is the distance around the circle from the point $(1,0)$.

But now, the length of the hypotenuse is $1$, and so the ratio for the sine of $\theta$ is just our $\sin(\theta)$, and similarly for $\cos(\theta)$.

So, finally, we see that $\sin$ and $\cos$ have all the old familiar properties.

I love it when a bunch of different bits of maths come together.

Acknowledgement

Thanks for @ColinTheMathmo for some useful comments and corrections.