Functional EquationDate: 02/18/2008 at 00:46:51 From: Jack Subject: solve for f(x) If 2f(x) + f(1-x) = x^2 for all x, then f(x) = ? First I tried to solve for x, and get x alone. Once I got x alone, I went further but just ended up confusing myself. Any help would be appreciated. Date: 03/16/2008 at 21:43:31 From: Doctor Ali Subject: Re: solve for f(x) Hi, Jack! Thanks for writing to Dr. Math. Let's start with what you said: 2f(x) + f(1-x) = x^2 This equality is TRUE for all x's. So, we can change x to anything and the equality remains TRUE. For example, by what you said, we can deduce that whatever f(x) may be, the following equalities are TRUE: Substitute zero for x. 1) 2f(0) + f(1) = 0 Substitute one for x. 2) 2f(1) + f(0) = 1 Note that 1 and 2 are both expressions with respect to f(0) and f(1) and both of them are TRUE. We can therefore see them as a system of simultaneous equations in terms of f(0) and f(1). Let’s continue: Assume that f(0) = A and f(1) = B and let’s re-write the expressions: 1) 2A + B = 0 2) 2B + A = 1 Solve the equations for A and B and you’ll get: A = -1/3 B = 2/3 Now we know that whatever f(x) may be, f(0) has to be -1/3 and f(1) has to be 2/3. So, what's the use? We know only two special values of x. But let’s generalize the method. As you may have noticed, when we substituted 1 for x, we got an expression with respect to f(0) and f(1), the two values we got in the first substitution and not f(2) or something else that was not in the first one. Having both expressions involve f(0) and f(1) allowed us to go on to the step of solving the system of equations. So that's the key idea. We said that we can change x into anything. Here “anything” can be some kind of expression itself. For example, it can be “2x”, “1/x” or whatever. Among all of these expressions, we are looking for the one that, when simplified, gives us a system of equations and doesn’t make new variables. By solving the system, we can find the answer. Let’s try your equation again: 2f(x) + f(1-x) = x^2 What should I change x into in order to get the system? Remember that we want to wind up with f(x) and f(1-x) again after our substitution so that we will have a system of equations with the new equation and the original one. Is it “2x”? Let’s try: 2f(2x) + f(1-2x) = 4x^2 Wow! Now I have two other unknowns with f(2x) and f(1-2x). So 2x didn’t work. What about “x+1”? 2f(x+1) + f(1-(x+1)) = (x+1)^2 Or 2f(x+1) + f(-x) = x^2 + 2x + 1 We get f(-x) and f(x+1) but those aren't in the original equation. Let’s try “1-x”: 2f(1-x) + f(1-(1-x))=(1-x)^2 Or 2f(1-x) + f(x) = x^2 - 2x + 1 It worked. Now we have f(x) and f(1-x), as in the original expression. What’s good about this? Remember the system of simultaneous equations. 1) 2f(x) + f(1-x) = x^2 2) 2f(1-x) + f(x) = x^2 - 2x + 1 In order not to confuse ourselves, let’s assume that A = f(x) and B = f(1-x). 1) 2A + B = x^2 2) 2B + A = x^2 - 2x + 1 Solve the system to get A, which is f(x). Note: This kind of equation is called a "FUNCTIONAL EQUATION", which means that you are not trying to find x's that satisfy the equality (as we do in ordinary equations), but instead are trying to find FUNCTIONS like f(x) that satisfy the equality. Please write back if you still have any difficulties. Bye! - Doctor Ali, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/