Functional Equations
By Ryan Ng
Functional equations are a type of math problem where you need to solve for functions that satisfy a given relation. For example, f(x + y) = x + f(y), where you need to solve for f. They’re a type of question that looks difficult at first glance, but in reality can be simplified extensively.
Strategies
The first strategy is guessing solutions. This works more often than you think. For instance, in the above example, we could intuitively guess f(x) = x. Then, noticing that there are an equal number of functions on either side, we could also guess f(x) = c. Finally, combining both solutions also works as a solution, yielding f(x) = x + c.
The second strategy is plugging in 0, 1, or both at the same time. For example, take f(x + y) = f(x) + f(y). Plugging in y = 0, we have f(x) = f(x) + f(0). We’ve found here that f(0) = 0, which is a clue.
The third strategy is substituting the variables directly. We might, for example, replace x with 1 - x. Or we might substitute x = y, or vice versa We’ll work with this in examples below.
The fourth strategy is to make use of some symmetries in the equation between x and y. We’ll also look at this in some examples below.
Finally, we can sometimes use induction on the solutions.
Question 1
f(x - y) = f(x)f(y)
Plugging in y = 0, we get f(x) = f(x)f(0). Rearranging, we get f(x)(1-f(0))=0.
Since either term must be zero, our first solution is f(x) = 0. Our second family of solutions would need to have f(0) = 1.
For the second family of solutions, we’ll substitute y = x. This gives us f(x - x) = f(x)f(x) = f(0). So f(x)^2 = 1. This means f(x) = ±1.
Note that when we do a substitution like y = x and yield a restricted answer for the function like this, we can safely say they are the only solutions.
Question 2
f(x) - xf(1 - x) = x^2
Let’s try switch x with 1 - x. We will get
f(1 - x) - (1 - x)f(x) = (1 - x)^2
We’d like to eliminate the inconvenient extra function term f(1 - x). We need the term to be the same in both equations, which we can do by multiplying the second equation by x.
xf(1 - x) - x(1 - x)f(x) = x(1 - x)^2
Adding the first equation to this:
xf(1 - x) - x(1 - x)f(x) + f(x) - xf(1 - x) = x(1 - x)^2 + x^2
x(1 - x)f(x) + f(x) = x(1 - x)^2 + x^2
f(x)(1 - x(1-x)) = x((1-x)^2 + x)
f(x)(1 - x + x^2) = x(1 - 2x + x^2 + x) = x(1 - x + x^2)
f(x) = x
So it simplifies quite elegantly. Since we did this purely algebraically, we know this is the only solution.
An interesting extra question
This is not very related to the previous strategies, but I think it’s really cool. This might not be very pedagogical or pedantic — if anything doesn’t make sense, GPT-4o seems to be quite potent right now. Ask it to explain the intermediate steps to you.
f(x + h) = f(x)f(h)
We, of course, start by plugging in random numbers to find things out. Let’s plug in h = 0.
f(x) = f(x)f(0)
f(x)(1 - f(0)) = 0
Once again, either the function is identically zero — f(x) = 0 — or f(0) = 1. Now for our actual solution. Let’s first subtract f(x) from both sides of the original functional equation:
f(x + h) - f(x) = f(x)f(h) - f(x) = f(x)(f(h) - 1)
Dividing both sides by h:
(f(x + h) - f(x)) / h = f(x)(f(h) - 1)/h
If we limit h to zero on both sides, notice that the left is just the derivative of f(x).
f’(x) = f(x) * lim (h → 0) (f(h) - 1)/h
We found earlier that f(0) = 1, so we can actually get on the right side
f’(x) = f(x) * lim (h → 0) (f(h) - f(0))/h
Now, see that the limit on the right is just the derivative of f at 0, f’(0). So this simplifies to
f’(x)/f(x) = f’(0)
If we integrate both sides with respect to x, we get
ln(f(x)) = f’(0)x + c
f(x) = m * exp(f’(0) x)
We’ll write this as f(x) = m * exp(kx).
Substituting this into the original equation f(x + y) = f(x)f(y), we get
m * exp(kx + ky) = m^2 * exp(kx + ky)
So m must be 0 or 1. If m is zero, the entire function is identically zero, so it must be 1.
f(x) = exp(kx)
Now that you’ve learnt these strategies, try out these questions. These videos were created by a great YouTube mathematician called SyberMath.