Functional Equation (Thailand Math POSN 2nd round)

Write a full solution.

1.) Find all functions f:N→Nf: \mathbb{N} \rightarrow \mathbb{N} that satisfy the inequality for all n∈Nn \in \mathbb{N}

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

2.) Let f:N→Nf: \mathbb{N} \rightarrow \mathbb{N} be a function that satisfy the equation for all m,n∈Nm,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,y∈R0+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:R→Rf: \mathbb{R} \rightarrow \mathbb{R} such that for all x,y,z∈Rx,y,z \in \mathbb{R}

f(x2(z2+1)+f(y)(z+1))=1−f(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:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be a function that satisfy the equation for all x,y∈Rx,y \in \mathbb{R}

f3(x+y)+f3(x−y)=(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,y∈Rx,y \in \mathbb{R}.

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

#Algebra #FunctionalEquations

Note by Samuraiwarm Tsunayoshi
6 years, 2 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

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:R↦Rf:\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 - 6 years, 2 months ago

Log in to reply

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

Samuraiwarm Tsunayoshi - 6 years, 2 months ago
×

Problem Loading...

Note Loading...

Set Loading...