Functional Equation (Thailand Math POSN 2nd round)

Write a full solution.

1.) Find all functions f:NNf: \mathbb{N} \rightarrow \mathbb{N} that satisfy the inequality for all nNn \in \mathbb{N}

f(n+1)>f(f(n))f(n+1) > f(f(n))

2.) Let f:NNf: \mathbb{N} \rightarrow \mathbb{N} be a function that satisfy the equation for all m,nNm,n \in \mathbb{N}

f(m+f(n))=n+f(m+58)f(m+f(n)) = n+f(m+58)

Prove that there exists only 1 solution of functional equation above, and find the value of f(1)+f(2)++f(15)f(1)+f(2)+\dots + f(15).

3.) Find all functions f:R0+R0+f: \mathbb{R_{0}^{+}} \rightarrow \mathbb{R_{0}^{+}} such that for all x,yR0+x,y \in \mathbb{R_{0}^{+}}

f(x+f(y2))=f(x)+y2f(x+f(y^{2})) = f(x) + y^{2}

Where R0+\mathbb{R_{0}^{+}} is set of non-negative real numbers.

4.) Find all functions f:RRf: \mathbb{R} \rightarrow \mathbb{R} such that for all x,y,zRx,y,z \in \mathbb{R}

f(x2(z2+1)+f(y)(z+1))=1f(z)(x2+f(y))z((1+z)x2+2f(y))f(x^{2}(z^{2}+1)+f(y)(z+1)) = 1 - f(z)(x^{2}+f(y)) - z((1+z)x^{2}+2f(y))

5.) Let f:RRf: \mathbb{R} \rightarrow \mathbb{R} be a function that satisfy the equation for all x,yRx,y \in \mathbb{R}

f3(x+y)+f3(xy)=(f(x)+f(y))3+(f(x)f(y))3f^{3}(x+y)+f^{3}(x-y) = (f(x)+f(y))^{3}+(f(x)-f(y))^{3}

Prove that f(x+y)=f(x)+f(y)f(x+y) = f(x)+f(y) for all x,yRx,y \in \mathbb{R}.

This note is a part of Thailand Math POSN 2nd round 2015

Note by Samuraiwarm Tsunayoshi
4 years, 2 months ago

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Samuraiwarm Tsunayoshi, I wanted to ask that for Q-5, can we use a particular f(x)f(x) that satisfies the given condition to prove the required statement?

If so, then one can note that the class of functions f:RRf:\Bbb{R}\mapsto\Bbb{R} defined by f(x)=λx , λf(x)=\lambda x~,~\lambda is some constant satisfies the given condition. Then, the proof becomes trivial since,

f(x+y)=λ(x+y)=λx+λy=f(x)+f(y)f(x+y)=\lambda (x+y)=\lambda x + \lambda y = f(x)+f(y)

Prasun Biswas - 4 years, 2 months ago

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Nope, you have to prove that no other functions exist.

Samuraiwarm Tsunayoshi - 4 years, 2 months ago

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