In the February edition of SUMS magazine on the Integral website, I have an article on functional equations. These turn out to be very interesting ways to define a variety of functions, including circular (that is, trigonometric) functions.
For example, imagine that there is only one thing that you know about a function \(f\):
$$f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$$The question is: how much can we find out about the function from this single property, and what other information might we need to define the function precisely. Here goes!
Use the formula to find \(f(0)\).
$$\begin{align*} &f(x)=f(x+0)=\frac{f(x)+f(0)}{1-f(x)f(0)}\\[4pt] \Rightarrow &f(x)\left(1-f(x)f(0)\right)=f(x)+f(0)\\[4pt] \Rightarrow &-f(x)^2f(0)=f(0)\\[4pt] \Rightarrow &f(0)=0 \end{align*}$$
Find \(f(2x),\,f(3x),\text{ and }f(4x)\)
$$\begin{align*} f(2x)&=\frac{2f(x)}{1-f(x)^2}\\[4pt] f(3x)&=f(2x+x)=\frac{f(2x)+f(x)}{1-f(2x)f(x)}\\[4pt] &=\frac{\frac{2f(x)}{1-f(x)^2}+f(x)}{1-\frac{2f(x)}{1-f(x)^2}f(x)}\\[4pt] &=\frac{3f(x)-f(x)^3}{1-3f(x)^2}\\[4pt] f(4x)&=f(3x+x)=\frac{f(3x)+f(x)}{1-f(3x)f(x)}\\[4pt] &=\frac{\frac{3f(x)-f(x)^3}{1-3f(x)^2}+f(x)}{1-\frac{3f(x)-f(x)^3}{1-3f(x)^2}f(x)}\\[4pt] &=\frac{4f(x)-4f(x)^3}{1-6f(x)^2+f(x)^4} \end{align*}$$
Can we tell that the function is periodic?
On the face of it, this is surely impossible. As it turns out, though, it's actually quite possible, and it's why I asked you to find the multiple angle formulas.
First of all, let's imagine that \(1\) is in the range of \(f\). This is obviously not an essential property of a function, but we'll go with it for now. Later, we'll see why it must be true for our function. So we can write
$$f(a)=1$$
for some value of \(a\).
Find \(f(2a),\,f(3a),\text{ and }f(4a)\)
$$\begin{align*} f(2a)&=\frac{2f(a)}{1-f(a)^2}=\frac{2}{0}=\text{undefined}\\[4pt] f(3a)&=\frac{3f(a)-f(a)^3}{1-3f(a)^2}=\frac{2}{-2}=-1\\[4pt] f(4a)&=\frac{4f(a)-4f(a)^3}{1-6f(a)^2+f(a)^4}=\frac{0}{-4}=0 \end{align*}$$
Continue this sequence.
$$\begin{align*} f(0)=0\\[4pt] f(a)=1\\[4pt] f(2a)&=\text{undefined}\\[4pt] f(3a)&=-1\\[4pt] f(4a)&=0\\[4pt] f(5a)&=f(4a+a)=\frac{f(4a)+f(a)}{1-f(4a)f(a)}=1\\[4pt] f(6a)&=f(5a+a)=\text{undefined}\\[4pt] f(7a)&=f(4a+3a)=-1\\[4pt] f(8a)&=f(4a+4a)=0\\[4pt] &\vdots \end{align*}$$
What have a I done here that's a little bit sneaky?
All the results for multiples of \(x\) over \(2x\) depend on the formula for \(f(2x)\), which is undefined for \(x=a\). This means that all the higher formulas are slightly questionable when \(x=a\). However, they are true for values of \(x\) as close to \(a\) as we like, so (assuming the function is reasonably well behaved) they are also true when \(x=a\).
So far, we have seen that the functions of multiples of \(a\) form a periodic sequence, but that's not quite the same as saying that the function is itself periodic.
Find \(f(x+4a)\), and interpret this result.
$$\begin{align*} f(x+4a)&=\frac{f(x)+f(4a)}{1-f(x)f(4a)}=f(x) \end{align*}$$ which means the function is periodic with period \(4a\).
Remarkably, then, we have shown just from the functional equation for \(f(x+y)\) and the fact that \(1\) is in the range of \(f\) that the function is periodic. That seems pretty incredible to me!
We have yet to show that \(1\) must be in the range of \(f\), but this will become clear as we figure out the shape of the graph of \(f(x)\).
Find \(f'(x)\).
$$\begin{align*} f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\[4pt] &=\lim_{h\to0}\frac{\frac{f(x)+f(h)}{1-f(x)f(h)}-f(x)}{h}\\[4pt] &=\lim_{h\to0}\frac{f(h)\left(1+f(x)^2\right)}{h(1-f(x)f(h))}\\[4pt] &=\lim_{h\to0}\frac{f(h)}{h}\times \lim_{h\to0}\frac{1+f(x)^2}{1-f(x)f(h)}\\[4pt] &=f'(0)\left(1+f(x)^2\right) \end{align*}$$
How does this tell us that \(1\) is in the range of \(f\)?
$$\begin{align*} &1+f(x)^2\ge 1\\[4pt] \end{align*}$$ This means that the graph is always at least as steep as it is at the origin. That is, it's steeper than \(y=f'(0)x\), which will certainly cross \(y=1\) at some point so long as \(f'(0)\ne 0\) [in which case the function would always be \(0\)]. So the function itself must intersect the line \(y=1\), even if \(f'(0)<0\).
We can also see now that once \(|f(x)|>1\), the graph will get steeper very quickly indeed. We also know that, by choosing a value of \(x\) as close as we like to \(2a\), we can make \(|f(x)|\) as big as we like.
What extra information do we need to ensure that \(f(x)=\tan x\)?
We need to know that \(f'(0)=1\) and not only that \(f\left(\frac{\pi}{4}\right)=1\) but that \(\frac{\pi}{4}\) is the smallest positive solution of the equation \(f(x)=1\).
If you are slightly curious about functional equations, you will find more examples in UKMT and BMO problem sets.
You could also try finding functional equations of the same kind for \(\cot\) and \(\coth\), and see how much you can figure out about the functions from these functional equations alone.
You might even want to try linked functional equations for \(\sin\) and \(\cos\). Perhaps I'll write that up for another article one day. In the meantime, enjoy your investigations!