Functional equations

We all have a pretty good idea of how to define a function. For example: $$\begin{align*} f(x)&=3\sin(5x)\\[4pt] f(x)&=\frac{3x^2-1}{\sqrt{2+x^2}} \end{align*}$$ We can also define a function implicitly: $$\begin{align*} &2f(x)^2+x^2=9\\[4pt] &x^2f(x)-f(x)=2 \end{align*}$$ And we can even define a function by a differential equation: $$\begin{align*} &f'(x)=(x+1)f(x)\\[4pt] &f''(x)-2f'(x)=3\cos x \end{align*}$$ although we might need a bit of extra information to nail the precise function (such as the values of \(y\) and/or \(y'\) when \(x=0\). But we can also form an equation relating the different values of the same function. For example: $$\begin{align*} &f(x)=xf(x-1)\\[4pt] &f(f(x))=x \end{align*}$$ and sometimes, an equation like this, a functional equation, is enough to define a function completely, or at least to within the value of a constant or two.

The question to ask about a functional equation is: can we solve it? That is to say, can we find a more or less explicit definition of the function that is the subject of the functional equation.

functional equation 1

Suppose the only thing we know about a function is that $$f(x+y)=f(x)f(y)$$
How much can we find out about the function from this one property? As we will see, the answer is: nearly everything!

I'm quite sure you can!

The question itself doesn't really give us that much insight into the functional equation, so let's leave it aside for a moment and ask some rather more basic questions about a function that satisfies this functional equation.

One of the first things to think about is the effect of \(f\) on some easy values: \(0\) or \(1\), for example.

The trick here is to put \(y=0\) to get $$\begin{align*} &f(x+y)=f(x)f(y)\\[4pt] \Rightarrow &f(x)=f(x+0)=f(x)f(0)\\[4pt] \Rightarrow &f(0)=1 \end{align*}$$

Maybe we can't find \(f(1)\) quite so easily (give it a go!), but we can find other values of the function in terms of \(f(1)\), so we should do this.

$$f(2)=f(1+1)=f(1)f(1)=f(1)^2$$

We'll be using \(f(1)\) quite a bit, so let's give it a name: $$a=f(1)$$

$$\begin{align*} f(2)&=f(1+1)=f(1)f(1)=a^2\\[4pt] f(3)&=f(2+1)=f(2)f(1)=a^2a=a^3\\[4pt] f(4)&=f(3+1)=f(3)f(1)=a^3a=a^4\\[4pt] &\;\;\vdots\\ \end{align*}$$

$$\begin{align*} &1=f(0)=f(1+-1)=f(1)f(-1)\\[4pt] \Rightarrow &f(-1)=\frac{1}{f(1)}=\frac{1}{a}=a^{-1} \end{align*}$$

\(a\ne 0\) otherwise \(f(-1)\) would not exist.

$$\begin{align*} &f(-2)=f(-1+-1)=f(-1)f(-1)=f(-1)^2=\frac{1}{a^2}=a^{-2}\\[4pt] &f(-3)=f(-2+-1)=f(-1)^2f(-1)=f(-1)^3=\frac{1}{a^3}=a^{-3}\\[4pt] &f(-4)=f(-3+-1)=f(-1)^3f(-1)=f(-1)^4=\frac{1}{a^4}=a^{-4} \end{align*}$$

We're doing pretty well! We know that $$f(n)=a^n$$ whenever \(n\) is an integer. What about rational values of \(x\)?

$$\begin{align*} f(1)&=f\left(\frac{1}{2}+\frac{1}{2}\right)=f\left(\frac{1}{2}\right)^2\\[4pt] \Rightarrow f\left(\frac{1}{2}\right)&=\sqrt{f(1)}=\sqrt a=a^\frac{1}{2} \end{align*}$$

We must have \(a\ge 0\) otherwise \(f\left(\frac{1}{2}\right)\) would not exist (or at least, it would not be real).

Could we choose to make \(f\left(\frac{1}{2}\right)=-\sqrt{f(1)}\)? I suppose in theory at this point perhaps we could but it would get us into terrible difficulties later, so I'm going to put this possibility to one side.

$$\begin{align*} f(x+y+z)&=f((x+y)+z)=f(x+y)f(z)=f(x)f(y)f(z) \end{align*}$$

It's easy to extend this idea to get $$f\left(x_1+x_2+\dots+x_n\right)=f\left(x_1\right)f\left(x_2\right)\dots f\left(x_n\right)$$

$$\begin{align*} f(1)&=f\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)=f\left(\frac{1}{3}\right)^3\\[4pt] \Rightarrow f\left(\frac{1}{3}\right)&=\sqrt[3]{f(1)}=\sqrt[3]a=a^\frac{1}{3}\\[4pt] f\left(\frac{1}{4}\right)&=\sqrt[4]{f(1)}=\sqrt[4]a=a^\frac{1}{4}\\[4pt] &\phantom{a}\vdots\\[4pt] f\left(\frac{1}{n}\right)&=\sqrt[n]a=a^\frac{1}{n}\\[4pt] \end{align*}$$

$$\begin{align*} f\left(\frac{3}{2}\right)&=f\left(3\times\frac{1}{2}\right)=f\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{3}=\sqrt{f(1)}^{\;3}=a^\frac{3}{2} \end{align*}$$

$$\begin{align*} f\left(\frac{3}{n}\right)&=f\left(3\times\frac{1}{n}\right)=f\left(\frac{1}{n}\right)^{3}=\sqrt[n]{f(1)}^{\;3}= a^\frac{3}{n} \end{align*}$$

$$\begin{align*} f\left(\frac{m}{n}\right)&=f\left(m\times\frac{1}{n}\right)=f\left(\frac{1}{n}\right)^{m}=\sqrt[n]{f(1)}^{\;m}= a^\frac{m}{n} \end{align*}$$

Now we know that for any rational value of \(x\): $$f(x)=a^x$$ and because we can get as close as we like to any real number with some rational number, this must be true for all real values of \(x\).

Well, we have essentially solved the functional equation, at least we would have done if we knew that value of \(f(1)\).

We can also find the differential of the function in terms of the function just from the functional equation:

$$\begin{align*} f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\[4pt] &=\lim_{h\to0}\frac{f(x)f(h)-f(x)}{h}\\[4pt] &=\lim_{h\to0}f(x)\frac{f(h)-1}{h}\\[4pt] &=f(x)\lim_{h\to0}\frac{f(h)-1}{h}\\[4pt] &=f'(0)f(x)\\\\[4pt] \end{align*}$$

Now the question itself asks us to find \(f(0)\), which we have already seen is \(1\).

It then asks us to derive the differential equation \(f'(x)=kf(x)\), which we have just done, showing that \(k=f'(0)\).

It did not ask us to show that \(f(x)=a^x\) without using the differential equation. That's what we've done. It does, though, ask us to reach the same conclusion using the differential equation, which we have not done. So we'll do that now.

$$\begin{align*} &\frac{\mathrm{d}y}{\mathrm{d}x}=ky\\[4pt] \Rightarrow &\frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x}=k\\[4pt] \Rightarrow &\int \frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x}\;\mathrm{d}x=\int k\;\mathrm{d}x\\[4pt] \Rightarrow &\int \frac{1}{y}\;\mathrm{d}y=\int k\;\mathrm{d}x\\[4pt] \Rightarrow &\ln y=kx+c\\[4pt] \Rightarrow &y=e^{kx+c}\\[4pt] &\text{through }(0,\,1)\Rightarrow c=0\\[4pt] \Rightarrow &y=e^{kx}=\left(e^k\right)^x=a^x\text{ where }a=e^k=\text{value of }y\text{ when }x=1\\[4pt] \end{align*}$$

This is the same differential equation, so $$f(x)=a^x$$ where \(a=e^{f'(0)}=f(1)\).

You may say: "why go to all the trouble of figuring out the function without using the differential, when solving the differential equation is so easy?" to which I would reply: "because we can, and how cool is that?!"

functional equation 2

Now that we have explored a functional equation for exponentiation, could we do the same for its inverse: logarithms. That's where we are going next, though only in brief: I'll leave some of the detail to you.

$$\begin{align*} &u(xy)=u(x)+u(y)\\[4pt] &u(x)=u(1x)=u(1)+u(x)\Rightarrow u(1)=0\\[4pt] &0=u(1)=u\left(x\times\frac{1}{x}\right)=u(x)+u\left(\frac{1}{x}\right)\Rightarrow u\left(\frac{1}{x}\right)=-u(x)\\[4pt] &u(x^2)=u(x)+u(x)=2u(x)\\[4pt] &u(x^{-2})=2u\left(\frac{1}{x}\right)=-2u(x)\\[4pt] &u(x^n)=nu(x)\\[4pt] &u(x)=u\left(\sqrt{x}\sqrt{x}\right)=2u\left(\sqrt{x}\right) \Rightarrow u\left(\sqrt{x}\right)=\frac{1}{2}u(x)\\[4pt] &u(x)=u\left(\sqrt[n]{x}^n\right)=nu\left(\sqrt[n]{x}\right) \Rightarrow u\left(x^\frac{1}{n}\right)=\frac{1}{n}u(x)\\[4pt] &u\left(x^\frac{m}{n}\right)=u\left(\left[x^\frac{1}{n}\right]^m\right)=mu\left(x^\frac{1}{n}\right)=\frac{m}{n}u(x) \end{align*}$$
And for the differential:
$$\begin{align*} u'(x)&=\lim_{h\to0}\frac{u(x+h)-u(x)}{h}\\[4pt] &=\lim_{h\to0}\frac{u(x+h)+u\left(\frac{1}{x}\right)}{h}\\[4pt] &=\lim_{h\to0}\frac{u\left(\frac{x+h}{x}\right)}{h}\\[4pt] &=\lim_{h\to0}u\left(\left[\frac{x+h}{x}\right]^\frac{1}{h}\right)\\[4pt] &=u\left(\lim_{h\to0}\left[1+\frac{h}{x}\right]^\frac{1}{h}\right)\\[4pt] &=u\left(e^\frac{1}{x}\right)\\[4pt] &=\frac{1}{x}u(e) \end{align*}$$

This uses the property $$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$ which you can find out plenty more about on my own website.

Here are some questions about \(f\) and \(u\) for you to ponder at leisure.

functional equation 3

Now for something that looks (and I guess is) quite a bit harder.

$$\begin{align*} &g(x)=g(x+0)=\frac{g(x)+f(0)}{1+g(x)g(0)}\\[4pt] \Rightarrow &g(x)\left(1+g(x)g(0)\right)=g(x)+g(0)\\[4pt] \Rightarrow &g(x)^2g(0)=g(0)\\[4pt] \Rightarrow &g(0)=0 \end{align*}$$

$$\begin{align*} g(2)&=\frac{g(1)+g(1)}{1+g(1)^2}\\[4pt] &=\frac{2g(1)}{1+g(1)^2} \end{align*}$$

Write \(g(1)=a\). $$\begin{align*} &(a-1)^2\ge 0\\[4pt] \Rightarrow &a^2-2a+1\ge 0\\[4pt] \Rightarrow &a^2+1\ge 2a\\[4pt] \Rightarrow &1\ge \frac{2a}{1+a^2}\\[4pt] \Rightarrow &g(2)\le 1\\[2em] &(a+1)^2\ge 0\\[4pt] \Rightarrow &a^2+2a+1\ge 0\\[4pt] \Rightarrow &a^2+1\ge -2a\\[4pt] \Rightarrow &1\ge -\frac{2a}{1+a^2}\\[4pt] \Rightarrow &-1\le \frac{2a}{1+a^2}\\[4pt] \Rightarrow &g(2)\ge -1\\[2em] \Rightarrow &-1\le g(2)\le 1 \end{align*}$$

$$\begin{align*} &g(b)=1\\[4pt] \Rightarrow &g(x+b)=\frac{g(x)+1}{1+g(x)}=1\\[4pt] \Rightarrow &g(x)=1\text{ for any }x \end{align*}$$ So it cannot be the case that \(1\) is in the range of \(g\) (unless \(g\) is just a constant function). It all works out similarly for \(-1\), which also cannot be in the range of \(g\).

Could it be that the range of \(g\) goes outside \(\{y|-1 < y <1\}\) at all?

The proof is identical to the proof that \(-1< g(2) < 1\). This means that \(g\) can never take us outside \(\{y|-1 < y < 1\}\), because if \(|g(d)|>1\) then there is no possible value of \(g\left(\frac{d}{2}\right)\).

We want to prove that \(g(x+h) > g(x)\) when \(h > 0\), that is $$\frac{g(x)+g(h)}{1+g(x)g(h)} > g(x)$$. $$\begin{align*} &\frac{g(x)+g(h)}{1+g(x)g(h)} > g(x)\\[4pt] \Leftrightarrow &g(x)+g(h) > g(x)+g(x)^2g(h)\\[4pt] \Leftrightarrow &g(h) > g(x)^2g(h)\\[4pt] \end{align*}$$ which is true because \(|g(x)| < 1\)

Let's summarize what we know so far about our function \(g\).

$$\begin{align*} &g(0)=0\\[4pt] &-1 < g(x) < 1\\[4pt] &a < b\Rightarrow g(a)< g(b) \end{align*}$$ So we have a pretty good idea of the shape of the graph \(y=g(x)\).

To refine the picture, we'll find the differential.

$$\begin{align*} g'(x)&=\lim_{h\to0}\frac{g(x+h)-g(x)}{h}\\[4pt] &=\lim_{h\to0}\frac{\frac{g(x)+g(h)}{1-g(x)g(h)}-g(x)}{h}\\[4pt] &=\lim_{h\to0}\frac{g(h)\left(1+f(x)^2\right)}{h(1-g(x)g(h))}\\[4pt] &=\lim_{h\to0}\frac{g(h)}{h}\times \lim_{h\to0}\frac{1+g(x)^2}{1-g(x)g(h)}\\[4pt] &=g'(0)\left(1-g(x)^2\right) \end{align*}$$

So long as \(1-g(x)^2>0\), \(|g'(x)|<|g'(0)|\), so while \(-1 < g(x) < 1\), the graph is always getting less steep as \(x\) increases in a positive or negative direction. The closer \(|g(x)|\) is to \(1\), the closer \(g'(x)\) is to \(0\), but we already know that \(g(x)\) cannot get to \(\pm 1\).

$$\begin{align*} &\frac{\mathrm{d}y}{\mathrm{d}x}=k\left(1-y^2\right)\\[4pt] \Rightarrow &\frac{1}{\left(1-y^2\right)}\frac{\mathrm{d}y}{\mathrm{d}x}=k\\[4pt] \Rightarrow &\frac{1}{(1+y)(1-y)}\frac{\mathrm{d}y}{\mathrm{d}x}=k\\[4pt] \Rightarrow &\frac{1}{2}\left(\frac{1}{1+y}+\frac{1}{1-y}\right)\frac{\mathrm{d}y}{\mathrm{d}x}=k\\[4pt] \Rightarrow &\frac{1}{2}\int \left(\frac{1}{1+y}+\frac{1}{1-y}\right)\frac{\mathrm{d}y}{\mathrm{d}x}\;\mathrm{d}x=k\int\mathrm{d}x\\[4pt] \Rightarrow &\frac{1}{2}\int \left(\frac{1}{1+y}+\frac{1}{1-y}\right)\;\mathrm{d}y=k\int\mathrm{d}x\\[4pt] \Rightarrow &\frac{1}{2}\left(\ln(1+y)-\ln(1-y)\right)=kx+c\\[4pt] \Rightarrow &\ln\frac{1+y}{1-y}=2kx+2c\\[4pt] \Rightarrow &\frac{1+y}{1-y}=e^{2kx+2c}=Ae^{2kx} \end{align*}$$
But the curve goes through \((0,\,0)\), so:

$$\begin{align*} &A=1\\[4pt] \Rightarrow &\frac{1+y}{1-y}=e^{2kx}\\[4pt] \Rightarrow &1+y=e^{2kx}(1-y)\\[4pt] \Rightarrow &y\left(e^{2kx}+1\right)=e^{2kx}-1\\[4pt] \Rightarrow &y=\frac{e^{2kx-1}}{e^{2kx+1}}\\[4pt] &\phantom{y}=\frac{e^{kx}-e^{-kx}}{e^{kx}+e^{-kx}} \end{align*}$$

You may recognize this as \(\tanh(kx)\), and if so, you could have used the substitution \(u=\tanh^{-1}y\) instead of partial fractions in solving the differential equation.

This time, we couldn't reasonably get an explicit formula for the function without the differential, and it turns out (of course) that everything we did without it we could have done more easily with it. But again, look what we would have missed!

functional equation 4

Suppose I change the functional equation ever so slightly, so that
$$f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$$ Investigating this one turns out to be quite a bit more interesting, if it's possible to imagine such a thing!

If you are curious to know more about this and about functional approaches to trigonometric functions, head over to mathsinsight.co.uk and look for the article "functional trigonometry".