Reply to Re: I gave up to the Computers National Olympiad
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Ok. Hope you'll understand my "English math".
1. The three points X(5,4,0), Y(3,0,2), Z(1,8,4), are the vertex of a cube. Find it's center.
2. m is a natural number, and the function F:M2(C)->M2(C), F(X)=X^m.
Prove that F is surjective if and only if m=1.
3. f:R->R is a function, and there exists L>0, so |f(x)-f(y)|>=L|x-y|, for every x,y, that belongs to R.
Prove that f is surjective if and only if f is continous.
4. Find all the continous functions g:R->R, with the following property: for every x that belongs to R, there exists a,b that belongs to (0,1), a+b=1, so g(x)=a*g(ax)+b*g(bx).(a and b aren't necessary the same for every x)
The first one was way too easy and the secod one was quite easy too.(I took 7/7 on both of them). The third one wasn't difficult, but it seems that I wasn't 100% rigurous(only 6.5/7). The last one was pretty difficult, but I managed to obtain 5 points(and I think the special prize I obtained was for this). I won't tell you yet how I solved them, I'll let you think a while on them.
1. The three points X(5,4,0), Y(3,0,2), Z(1,8,4), are the vertex of a cube. Find it's center.
2. m is a natural number, and the function F:M2(C)->M2(C), F(X)=X^m.
Prove that F is surjective if and only if m=1.
3. f:R->R is a function, and there exists L>0, so |f(x)-f(y)|>=L|x-y|, for every x,y, that belongs to R.
Prove that f is surjective if and only if f is continous.
4. Find all the continous functions g:R->R, with the following property: for every x that belongs to R, there exists a,b that belongs to (0,1), a+b=1, so g(x)=a*g(ax)+b*g(bx).(a and b aren't necessary the same for every x)
The first one was way too easy and the secod one was quite easy too.(I took 7/7 on both of them). The third one wasn't difficult, but it seems that I wasn't 100% rigurous(only 6.5/7). The last one was pretty difficult, but I managed to obtain 5 points(and I think the special prize I obtained was for this). I won't tell you yet how I solved them, I'll let you think a while on them.