*2017 May 31*

Find non-constant functions \(f\), \(g\), \(h\), \(j\), \(k\), and \(m\) that have the following properties:

\[f(x)+f(2x)=2x\]

\[g(2x)+2g(x)=0\]

\[h(x)-2h(\frac{2}{x})=\frac{1}{2x}\]

\[2j(x)+j(2x+1)=3x-5\]

\[k(3x)+7k(x)=x^2+4x-1\]

\[m(3x-1)-2m(2-x)=3x^2-x+10\]

An interesting topic to explore is the concept of a *functional equation* in which one solves for a function of a variable rather than for the variable itself.

Before I go into the examples I must first clarify a technique that I will frequently use. If we have \((f \circ g)(x)\) set equal to some function \(h(x)\), we can change \(x \to g^{-1}(x)\) to get

\[f(x)=(h \circ g^{-1})(x)\]

and the truth of the statement will still hold. I will refer to this as “making a substitution”.

An easy example is

\[\frac{1}{2}f(x)+4=x\]

where we can easily solve for \(f(x)\) to obtain \(f(x)=2x-8\). However, the complications arise when expressions containing \(x\) other than just \(x\) itself occur within the parentheses of \(f(x)\). For example, this is the case in the functional equation

\[f(\ln x)+3=\sin x\]

However, this still is not very difficult because we can make a substitution to get

\[f(x)=\sin(e^x)-3\]

Which is the solution. However, the trickiest ones occur when multiple instances of the function occur in the equation, each with different inputs, like in the equation

\[f(x)+f(2x)=2x\]

This puzzle stumped me for a while, and when I realized how simple it was, I was very frustrated with myself. In these types of puzzles, a very helpful strategy can be to guess that the function is of a certain type. For example, in this problem, we might guess that \(f\) is in the form

\[f(x)=ax\]

where \(a\) is a constant. We want to solve for \(a\), and if we can find a solution for \(a\) that is *independent of* \(x\), then we have solved the problem. Let’s set up the problem this way:

\[ax+2ax=2x\]

If we divide both sides of the equation by \(x\), then we have

\[a+2a=2\]

\[3a=2\]

\[a=\frac{2}{3}\]

So

\[f(x)=\frac{2}{3}x\]

The answer is so simple! In my original attempt, I got an answer of

\[f(x)=\frac{2}{3}x+\frac{1}{3}\cos(\frac{\pi}{\ln 2}\ln x)\]

Which is indeed a solution, but is far more complicated than it need be. Here is another functional equation:

\[g(2x)+2g(x)=0\]

we can rearrange this to get

\[g(2x)=-2g(x)\]

Now we can employ another useful strategy. It can often be useful to define the function to be some value at some point, and then solve for all other values and use a recurrence relation to solve for the function. If we let \(g(1)=1\), then we have

\[g(2)=-2\]

\[g(4)=4\]

\[g(8)=-8\]

\[g(16)=16\]

We can quickly observe the pattern

\[g(2^x)=(-2)^x\]

By substitution, we have

\[g(x)=(-2)^{\log_2 x}\]

\[g(x)=2^{\log_2 x}*(-1)^{\log_2 x}\]

\[g(x)=x(-1)^{\log_2 x}\]

This is a little bit messy, since we have a negative base raised to a power that can sometimes be irrational. However, the real purpose of the \((-1)^{\log_2 y}\) is to yield a number of opposite sign after every power of \(2\). Its fluctuation can be replicated by that of the sine function. This fluctuation occurs in the same way if we replace it with \(\sin(\pi \log_2 x)\) or \(\sin(\frac{\pi}{\ln 2}\ln x)\), since it will be \(1\) at every even power of \(2\) and \(-1\) at every odd power of \(2\). So we have

\[g(x)=x\sin(\frac{\pi}{\ln 2}\ln x)\]

Luckily enough, the property holds for all values of \(x\), and can be verified algebraically.

That problem reveals another handy technique. Often, in functional equation problems, a negative number to an irrational exponent may come up, and in these cases, the trigonometric functions can come in handy, since they demonstrate similar fluctuating behavior.

Here is another example which can be solved with another very useful method:

\[h(x)-2h(\frac{2}{x})=\frac{1}{2x}\]

If we start with this equation and make the substitution \(x \to \frac{2}{x}\), we get

\[h(\frac{2}{x})-2h(x)=\frac{1}{4}x\]

This can be solved like a system of equations. Let us multiply the first equation by \(2\) to get

\[2h(x)-4h(\frac{2}{x})=\frac{1}{x}\]

and then add our two simultaneous equations to get

\[h(\frac{2}{x})-2h(x)+2h(x)-4h(\frac{2}{x})=\frac{1}{4}x+\frac{1}{x}\]

\[-3h(\frac{2}{x})=\frac{1}{4}x+\frac{1}{x}\]

we substitute \(x \to \frac{2}{x}\) again to get

\[-3h(x)=\frac{1}{2x}+\frac{1}{2}x\]

and now we can solve for \(h(x)\):

\[h(x)=-\frac{1}{6}(x+\frac{1}{x})\]

In fact, if we have a functional equation of the form

\[af(x)+b(f\circ g)(x)=h(x)\]

where \(g\) is *n-involutory*, it can be solved in a similar way.

Here is a very interesting method that can be applied to solving polynomial functional equations. First we start with the very simplest case… a polynomial of degree one. Suppose we have

\[2j(x)+j(2x+1)=3x-5\]

This can be easily solved if we assume that \(j(x)\) takes the form \(ax+b\), where \(a, b\) are constants. If we assume this, we can substitute and solve for \(a\) and \(b\):

\[2(ax+b)+a(2x+1)+b=3x-5\]

\[2ax+2b+2ax+a+b=3x-5\]

\[4ax+a+3b=3x-5\]

So we have that

\[4a=3\]

\[a+3b=-5\]

By solving the system, we get \(a=\frac{3}{4}\) and \(b=-\frac{23}{12}\), so

\[j(x)=\frac{3}{4}x-\frac{23}{12}\]

Okay, that one was easy. How about this:

\[k(3x)+7k(x)=x^2+4x-1\]

That’s a second degree polynomial. We could assume that \(k(x)\) takes the form \(ax^2+bx+c\), but… that would take a *really* long time to work out. There is an easier way to do this, and it involves another trick. Watch what happens when we differentiate both sides:

\[3k'(3x)+7k'(x)=2x+4\]

Look at that! Now all we have to do is solve for \(k'(x)\), which will be linear. We can say that \(k'(x)=ax+b\) for some constants \(a\) and \(b\), so we have

\[3(3ax+b)+7(ax+b)=2x+4\]

\[9ax+3b+7ax+7b=2x+4\]

\[16ax+10b=2x+4\]

and we have \(a=\frac{1}{8}\) and \(b=\frac{2}{5}\). Then

\[k'(x)=\frac{1}{8}x+\frac{2}{5}\]

and, by antidifferentiation, we get

\[k(x)=\frac{1}{16}x^2+\frac{2}{5}x+c\]

where \(c\) is some constant. We can now solve for \(c\) using our original functional equation. We have

\[k(3x)+7k(x)=x^2+4x-1\]

which means that

\[c+7c=-1\]

\[8c=-1\]

\[c=-\frac{1}{8}\]

and so our final answer will be

\[k(x)=\frac{1}{16}x^2+\frac{2}{5}x-\frac{1}{8}\]

Here is a more difficult example.

\[m(3x-1)-2m(2-x)=3x^2-x+10\]

By differentiating, we get

\[3m'(3x-1)+2m'(2-x)=6x-1\]

Again, let \(m(x)=ax+b\), and so

\[3a(3x-1)+3b+2a(2-x)+2b=6x-1\]

\[9ax-3a+3b+4a-2ax+2b=6x-1\]

\[7ax+a+5b=6x-1\]

Which gives us \(a=\frac{6}{7}\) and \(b=-\frac{5}{7}\). Then we have

\[m'(x)=\frac{6}{7}x-\frac{5}{7}\]

and

\[m(x)=\frac{3}{7}x^2-\frac{5}{7}x+c\]

From our original functional equation, we can see that

\(c-2c=10\)

\(c=-10\)

and so we have, as a final answer,

\[m(x)=\frac{3}{7}x^2-\frac{5}{7}x-10\]

This can be applied to any functional equation in which the answer is a polynomial, regardless of the degree of the polynomial. Perhaps it could even be applied when some of the terms have fractional or negative exponents.

Okay, that’s about all I’ve got on functional equations for now. There are also some functional equation problems asking for all functions displaying a certain property regarding \(f(x+y)\) or \(f(xy)\), but I’ll save those for a later post. Another topic that I may explore later are functional *systems* of equations in which one must solve for two or more functions, given two equations.