Find all real valued functions f(x) that satisfy the equation:
f(x)*f(y)= f(x+y) + x*y
Plug in various values and see if we get any results:
(x,x): f(x)^2 = f(2x) + x^2
(0,0): f(0)^2 = f(0) so f(0) could be 0 or 1
(0,y): f(0)*f(y) = f(y) so f(0) = 1 [one result]
(1,-1): f(1)*f(-1) = f(0) - 1, but f(0) = 1, so
f(1)*f(-1) = 0, either f(1) or f(-1) or both equal zero
(1,1): f(1)^2 = f(2) + 1
(1,2): f(1)*f(2) = f(3) + 2
(2,2): f(2)^2 = f(4) + 4
(1,3): f(1)*f(3) = f(4) + 3
In these last 4, replace f(1), ..., f(4) with variables such as a,b,c,d and you have 4 equations and 4 unknowns.
Wolfram Alpha finds two solutions for this:
(0,-1,-2,-3) or (2,3,4,5) suggesting f(x) = 1 + x or 1 - x
Alternate method: knowing f(0) = 1, if we try a first order polynomial,
f(x) = Ax + 1, plugging this into the equation gives
(Ax+1)(Ay+1) = A(x+y) + 1 + xy --> A^2xy = xy; so A = ±1, same result
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Posted by Larry
on 2024-01-01 11:09:52 |
Solution
