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Timothy Little a écrit :
> kentonyee@xxxxxxxxxxx wrote:
> > Find all functions V(x, y) that satisfy
> > V(x, y) + V(a, c) = V( (x+V(a,c)), y+c ).
> > for all real numbers x, y, a, c.
>
> V(x,y) + V(a,c) = V(x + V(a,c), y + c)
>
> Let f(x) = V(x,0). Then f(x) + f(y) = f(x + f(y)). Given that you've
> crossposted to sci.physics, I assume that V must be continuous over
> some region, however small.
>
> With that condition, there are only two solutions for f: f(x) = 0, or
> f(x) = x.
>
> 1) Assume f(x) = x. Then V(x,y) + V(a,0) = V(x,y) + a = V(x + a, y),
> and so V(x,y) = x + g(y) for some function g. However, by setting y=0
> in the original equation we find that x + V(a,c) = V(x + V(a,c), c)
> and so g(y) = 0 for all y. Hence V(x,y) = x.
>
> 2) Assume f(x) = 0 for all x. Then V(a,c) = V(x + V(a,c), c). Hence
> V(a,c) = g(c) for some continuous function g having g(0) = 0.
>
>
> - Tim
Dear Tim,
I also assume that V must be continuous over some region.
I might propose:
v(x,y)=x*sin(2*Pi*x/a)+a*(1-sin(2*Pi*x/a))*y/c , (1)
In fact a general solution is :
v(x,y)=K*x+(1-K)*a*y/c ; (2)
K is a constant or invariant for {x->x+a} ,{y->y+c},{x->x+a;y->y+c},
MY METHOD:
when {x->x+v(a,c),y->y+c} , function v(x,y)/v(a,c)-> ( )+1
i.e (+1) incremented ,
Alain.
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