Hello.
I am trying to make the next problem:
If g and h are continuous and real functions and g(x-y)= g(x)g(y) + h(x)h(y) for all x and y real, proove that g and h are bounded.
I have realised that, for example g=cos and h=sin is a possible example.
Also, I have thought to give values to x and y for example: y=0 -->g(x)= g(x)g(0) + h(x)h(0); x=0 --> g(-y)=g(0)g(y) + h(0)h(y); y=x-->g(0)=g^2(x)+h^2(x)=g^2(y)+h^(y) and y=-x --> g(-2y)=g(x)g(-x) + h(x)h(-x) but I don't get any conclusion.
Could you help me?
Thanks for every aportation.
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Ok. I think that as g(0)=g^2(x)+h^2(x) we have that for all x the point (g(x), h(x)) is in the circle of center 0 and radius squared root of g(0) so they are bounded,
is it a good argument?
How could I find examples of g and h differents of "cos" and "sin"??
Thanks again.
I am trying to make the next problem:
If g and h are continuous and real functions and g(x-y)= g(x)g(y) + h(x)h(y) for all x and y real, proove that g and h are bounded.
I have realised that, for example g=cos and h=sin is a possible example.
Also, I have thought to give values to x and y for example: y=0 -->g(x)= g(x)g(0) + h(x)h(0); x=0 --> g(-y)=g(0)g(y) + h(0)h(y); y=x-->g(0)=g^2(x)+h^2(x)=g^2(y)+h^(y) and y=-x --> g(-2y)=g(x)g(-x) + h(x)h(-x) but I don't get any conclusion.
Could you help me?
Thanks for every aportation.
---
Ok. I think that as g(0)=g^2(x)+h^2(x) we have that for all x the point (g(x), h(x)) is in the circle of center 0 and radius squared root of g(0) so they are bounded,
is it a good argument?
How could I find examples of g and h differents of "cos" and "sin"??
Thanks again.
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