How can we find F satisfying: exists a c such that F(x,x)+F(x,c−x)−F(c−x,x)−F(c−x,c−x)=0 for all x,y
Several quadratic polynomials in x,y satisfy the above property. I'm trying to generate more examples of functions that satisfy it. I will be very grateful for any other examples.
The property says there is one c that works for all x but given F we have freedom to find a c that works. For example:
- If F(x,y)=Ax2+By2+Cxy+Dx+Ey+F and A+C≠0 we set c=−2DA+C and one can easily verify F satisfy the functional equation.
- Also if F(x,y)=(A+By)x2+Cy2+Dxy+Ex+Fy+H there are two values of c.
- As Jeb pointed in the comments if F(x,y) is even in x and also even in y then c=0 works.
My first idea was to try to transform the equation into a PDE by differentiating wrt x but we do not get a PDE because we are evaluating F at different points: x and c−x. Honestly, I'm not even sure of what tags to use.
Answer
For any function, F(x,y), define the function g(x,c) satisfying
g(x,c)=F(x,x)+F(x,c−x)−F(c−x,x)−F(c−x,c−x)
We therefore require that ∃c s.t. ∀x, g(x,c)=0.
If we let x=z+c2 and ˆF(x,y)=F(x+c2,y+c2), then we have
ˆg(z,c)=g(z+c2,c)=ˆF(z,z)+ˆF(z,−z)−ˆF(−z,z)−ˆF(−z,−z)
Therefore, without loss of generality, we can consider the case of c=0, for which we require that ˆg(z,0)=0. In this case, we may split the even and odd components of our function in each of the two dimensions, thus giving us
ˆF(x,y)=A(x,y)+B(x,y)+C(x,y)+D(x,y)
satisfying
A(x,y)=A(−x,y)=A(x,−y)B(x,y)=−B(−x,y)=B(x,−y)C(x,y)=C(−x,y)=−C(x,−y)D(x,y)=−D(−x,y)=−D(x,−y)
From which (letting ˆg(x,c)=ˆg(x,0)=G(x)), we may determine that
G(x)=A(x,x)+A(x,−x)−A(−x,x)−A(−x,−x)+B(x,x)+B(x,−x)−B(−x,x)−B(−x,−x)+C(x,x)+C(x,−x)−C(−x,x)−C(−x,−x)+D(x,x)+D(x,−x)−D(−x,x)−D(−x,−x)=4B(x,x)
Therefore, the requirement is that B(x,x)=0. Recall that B(x,y) represents the component of ˆF(x,y) that is even in y and odd in x. This provides us with a "natural" method of creating a solution. Given a function ˆF(x,y), we can find B(x,y) as
B(x,y)=ˆF(x,y)−ˆF(−x,y)+ˆF(x,−y)−ˆF(−x,−y)4
and therefore, we may define a modified function
˜F(x,y)={3ˆF(x,x)+ˆF(−x,x)−ˆF(x,−x)+ˆF(−x,−x)4x=yˆF(x,y)otherwise
Note that this leads to a result identical to that given by Ewan Delanoy, it is merely expressed in a different manner. For a more "natural" solution, we can require that B(x,y)=0, and thus define
˜F(x,y)=3ˆF(x,y)+ˆF(−x,y)−ˆF(x,−y)+ˆF(−x,−y)4
From here, we may choose any c, and shift the function so that
F(x,y)=˜F(x−c2,y−c2)
and we have a solution for a given c. Notice that, for
F(x,y)=Ax2+By2+Cxy+Dx+Ey+F
with c=−2D2A+C we have
ˆF(x,y)=Ax2+By2+Cxy−2BD−CD+E(2A+C)2A+Cy+C2F+4ACF+4A2F−CDE−2ADE+BD2−AD2(2A+C)2
which, you can see, does not have the x term, which is the only term in a bivariate quadratic polynomial that is even in y and odd in x. Note the 2A+C on the denominator of c - the value given in the question is incorrect.
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