Solving Functional EquationsDate: 08/11/2005 at 20:31:32 From: Dmitry Subject: a*f(f(x))+b*f(x)+c=g(x) I am currently a 9th grade student and have been teaching myself calculus. During my studies, I came across the chain rule. I already knew how to use functions to the second power as substitution. However, if one would have an equation like... f(f(x)) + 5f(x) + 7 = x^2 + 6x; How would one go about solving something like this? I have no idea how to approach this. If a function, for instance, is in the form of a*f^2(x) + b*f(x) + c = 0, it is simple to solve by using f as a variable. However, in this case, I really am clueless. My idea was to differentiate both sides, obtaining... f'(f(x))*f'(x) + f'(x) = 2x + 6... I presume this would be a differential equation which I do not yet have sufficient knowledge to solve. My question is: is there an easier way? And, if not, can you show me how one would solve something like this? Date: 08/12/2005 at 20:11:26 From: Doctor Vogler Subject: Re: a*f(f(x))+b*f(x)+c=g(x) Hi Dmitry, Thanks for writing to Dr. Math. That's not an easy question to answer. That kind of an equation is called a "functional equation," and finding all functions that satisfy a particular functional equation is often very difficult. Some equations are easy, such as the one you mentioned, which is simply a quadratic polynomial in f(x). Some are harder. Let's look at a couple which are in between, and see what we can do. Suppose we had f(f(x)) = r*f(x) for some real number r. Then if we let y = f(x), this becomes f(y) = ry. You might then jump to the conclusion that f(x) = rx is the only solution. But actually, we only need f(x) = rx for every x in the range of f. For example, you might have rx if x = r^n for some integer n > 0 f(x) = { 0 otherwise Then this function also works. Now suppose that we had f(f(x)) = r for some real number r. We can do the same as above, and we conclude that f(y) = r for every y in the range of f. For example, we can choose any subset S of the real numbers, and then set f(y) = r for every y in the set S, and then, for every x not in S, we can set f(x) to be anything in S. Now suppose that we had f(f(x)) = x. It turns out that all of the solutions to this equation can be found as follows: Pair up all of the real numbers into pairs {a, b} and singletons {c}. Then define f(a) = b and f(b) = a for each pair, and define f(c) = c for each singleton. For example, you can pair up x with x+1 when floor(x) is even, and with x-1 when floor(x) is odd. But what happens if we also make a few extra conditions? It takes some challenging complex analysis to prove that every analytic function of a complex variable that satisfies f(f(x)) = x is, in fact, a rational function. And it doesn't take as much work to show that every rational function that satisfies that equation is either f(x) = x, or f(x) = k/x, for some constant a, or f(x) = (x + a)/(bx - 1) for some constants a and b. Along similar lines, it is challenging to show, and requires the Axiom of Choice, that there are real functions of a real variable which satisfy f(x+y) = f(x) + f(y) for which f(x)/x is not a constant. On the other hand, if the function f is differentiable, then it's a simple matter to show that f(x)=rx for some constant r. If the function is continuous, then you can prove the same thing. But there are other functions (discontinuous ones) that satisfy the functional equation. Next we should try to solve f(f(x)) = x+1 or perhaps f(f(x)) = 2x. And then things start to get difficult. Of course, one solution to the first is f(x) = x + 1/2, and one solution to the second is f(x) = x*sqrt(2), but are there other solutions? What are they? That's not an easy question to answer. The best ideas that I can suggest are to guess a form for the function, such as a first-degree rational function ax + b f(x) = ------ cx + d and then try to see what coefficients (if any) will solve the equation. Another form is a power series f(x) = a + bx + cx^2 + dx^3 + ... and any analytic function can be expressed in this form. These methods don't work, however, for discontinuous solutions. But here is one way to analyze your equation. Let's find all rational functions that solve it. (A "rational function" means a polynomial in x divided by another polynomial in x.) It doesn't take too much work to show that if the degree of the function f(x) is d (the degree of a rational function means the larger of the degrees of the numerator and the denominator), then the degree of f(f(x)) is d^2. Now our equation is f(f(x)) = x^2 + 6x - 7 - 5f(x). So the degree of the left side is d^2. What is the degree on the right side? Well, if the degree of the numerator of f(x) is less than or equal to the degree of the denominator, then the degree of the right side will be d+2. If the degree of the numerator of f(x) is one more than that of the denominator, then the degree of the right side will be d+1. If the degree of the numerator is at least two more than that of the denominator, then the degree of the right side will be d, with one exception: If the degree of the numerator is exactly two more, and the leading coefficients are opposites, then the x^2 will cancel with the largest terms, and you will get a degree smaller than d. Now consider that the degree of the right side has to equal the degree of the left side. So that means that d^2 = d+2 or d^2 = d+1 or d^2 <= d. But since d is a positive integer, the last equation only holds when d is 1, and the first equation only holds when d is 2, and the middle equation never holds. But then if we have d = 1, then the degree of the numerator certainly can't be two less than that. So we have only one possibility, which is that f(x) is a degree-two rational function ax^2 + bx + c f(x) = -------------. x^2 + dx + e Now substitute this expression into the equation f(f(x)) + 5f(x) = x^2 + 6x - 7, multiply by denominators to get a single big polynomial in x, and then since they have to be the same functions on both sides of the equation, that means that the coefficients have to be the same. That will give you several equations in the coefficients a, b, etc. Then see if there are any solutions to these equations. If not, then there are no rational functions that solve your functional equation. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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