Question

(c) (x-y) from (x+y)​

Answer 1

Answer:

This depends on a couple of things: first, the conditions you want on your functions (do they need to be continuous? smooth? analytic?) and second, the zero locus of f.

So, first, let's suppose that f and g are analytic. Setting x=y, we see that

2f(x,x)g(x,x)=02f(x,x)g(x,x)=0

for all x, or in other words

2f|Δg|Δ=02f|Δg|Δ=0

where ΔΔ is the diagonal. Now remember that an analytic function of one variable vanishes either on a discrete set of points of everywhere. Consequently, we have either f|Δ=0f|Δ=0 or g|Δ=0g|Δ=0.

Consider first the case where g|Δ=0g|Δ=0. It follows from analytic function theory in several variables that g(x,y)=(x−y)h(x,y)g(x,y)=(x−y)h(x,y) for some analytic function h. Then we have

f(x,y)(x−y)h(x,y)+f(y,x)(y−x)h(y,x)=0;f(x,y)(x−y)h(x,y)+f(y,x)(y−x)h(y,x)=0;

factoring,

(x−y)[f(x,y)h(x,y)−f(y,x)h(y,x)]=0(x−y)[f(x,y)h(x,y)−f(y,x)h(y,x)]=0

So for all x≠yx≠y we have

f(x,y)h(x,y)=f(y,x)h(y,x).f(x,y)h(x,y)=f(y,x)h(y,x).

Write s(x,y):=f(x,y)h(x,y)s(x,y):=f(x,y)h(x,y), and notice that this function is analytic. Then the above equation says that s(x,y)−s(y,x)=0s(x,y)−s(y,x)=0 on ΔcΔc; since s(x,y)−s(y,x)s(x,y)−s(y,x) is an analytic function vanishing on a dense open set, it vanishes identically, i.e. s(x,y)=s(y,x)s(x,y)=s(y,x). So s is a symmetric analytic function.

This shows that when f is analytic and f|Δ≠0f|Δ≠0, every analytic g solving this functional equation is of the form

g(x,y)=(x−y)f(x,y)s(x,y)g(x,y)=(x−y)f(x,y)s(x,y)

where s is a symmetric analytic function. Conversely, suppose that s is a symmetric analytic function such that s/f extends to an analytic function everywhere. Then if we take g to be of the form above, g is analytic and

f(x,y)g(x,y)+f(y,x)g(y,x)f(x,y)g(x,y)+f(y,x)g(y,x)

=f(x,y)x−yf(x,y)s(x,y)+f(y,x)y−xf(y,x)s(y,x)=f(x,y)x−yf(x,y)s(x,y)+f(y,x)y−xf(y,x)s(y,x)

=(x−y)[s(x,y)−s(y,x)]=0=(x−y)[s(x,y)−s(y,x)]=0

And then we could consider the case where f|Δ=0f|Δ=0, and so forth.

If you let f be much worse than analytic, say C∞C∞, then you're going to have to be substantially more careful, because (IIRC) every closed set is the vanishing locus of some smooth function

Answer 2

Answer:

(x-y) (x+y)

x(x+y)-y(x+y)

x^2+xy-xy-y^2

x^2-y^2

Step-by-step explanation:

hope it help you