If f(a+b) = f(a) + f(b) + 2ab, what is f?

December 1, 2006 Leave a comment

OK, I think I have a proof of the following:
If f(a+b) = f(a) + f(b) + 2ab, then f(x) = x^2  + mx for some real m.

Proof:
Setting a=0 and b=0 , we have f(0+0) = f(0) + f(0) + 0;
in other words f(0) = 0.
Next, set a=x and b=-x , which yields f(x-x) = f(x) + f(-x) - 2x^2.
Since f(x-x) = f(0) = 0, we have f(x) + f(-x) = 2x^2, or
f(x) - x^2 = -[f(-x) - (-x)^2].
Let g(x) = f(x) - x^2 (the left side)
Then we see that f(x) = x^2  + g(x) and g is an odd function, i.e. g(x) = -g(-x).
Now,
f(a+b) = (a+b)^2 + g(a+b) = a^2 + b^2 + 2ab + g(a+b)
but also
f(a+b) = f(a) + f(b) + 2ab
Setting equal the right sides of the above two equations and simplifying,
we get
g(a+b) = g(a) + g(b),
which is the definition of linear function with g(0) = 0. (y-intercept = 0)

Categories: math
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