Functional Equation (Canada 1969)
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Find all functions $f:N \rightarrow N$ such that $f(2)=2$ and for any two $m,n ; f(mn)=f(m)f(n)$ and $f(n+1)>f(n)$.
Re: Functional Equation (Canada 1969)
Hint:
The set $S_n = \{n^q | q \in \mathbb Q\}$ is dense $\forall n \in \mathbb N, n > 1$.
The set $S_n = \{n^q | q \in \mathbb Q\}$ is dense $\forall n \in \mathbb N, n > 1$.
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- Posts:107
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Re: Functional Equation (Canada 1969)
Hints for two different solutions:
Re: Functional Equation (Canada 1969)
See that \((m,n)=(2,1)\) gives \(f(1)=1\) and \((m,n)=(2,2)\) gives \(f(4)=4\). Now \[2=f(2)<f(3)<f(4)=4\] so \(f(3)=3\). Suppose \(f(n)=n\) for \(n\le 2k\). Then \[f(2k+2)=f\left(2(k+1)\right)=f(2)f(k+1)=2k+2.\] And so \[2k=f(2k)<f(2k+1)<f(2k+2)=2k+2\] hence \(f(2k+1)=2k+1\). Therefore \(\boxed{f(n)=n}\) for all \(n\in\mathbb{N}\) by strong induction.
Last edited by Nirjhor on Sun Sep 21, 2014 10:34 pm, edited 1 time in total.
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
Re: Functional Equation (Canada 1969)
Wrong. For example the following set of functions defines a solution to $f(mn) = f(m)f(n), f: \mathbb N \mapsto \mathbb N$.Nirjhor wrote:The given is Cauchy's Functional Equation. Hence \(f(x)=x^n\) for any fixed \(n\in\mathbb{N}\). Now \(f(2)=2^n=2\) gives \(n=1\) so \(f(n)=n\).
\[f(\prod p_i ^{e_i}) = \prod f(p_i)^{e_i}\]
\[f(p_i) = F(p_i)\]
Where the function $F: \mathbb P \mapsto \mathbb P$ is injective.
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
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Re: Functional Equation (Canada 1969)
Didn't understand what you wanted to say. The equality \[f\left(\prod p_i^{e_1}\right)=\prod f\left(p_i\right)^{e_i}\] is obvious from the given multiplicativity condition. If you're saying \(F:\mathbb{P}\mapsto\mathbb{P}\) is an arbitrary injective function, how does that satisfy the conditions given?*Mahi* wrote:Wrong. For example the following set of functions defines a solution to $f(mn) = f(m)f(n), f: \mathbb N \mapsto \mathbb N$.Nirjhor wrote:The given is Cauchy's Functional Equation. Hence \(f(x)=x^n\) for any fixed \(n\in\mathbb{N}\). Now \(f(2)=2^n=2\) gives \(n=1\) so \(f(n)=n\).
\[f(\prod p_i ^{e_i}) = \prod f(p_i)^{e_i}\]
\[f(p_i) = F(p_i)\]
Where the function $F: \mathbb P \mapsto \mathbb P$ is injective.
(I've edited my comment and added an inductive solution)
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
Re: Functional Equation (Canada 1969)
For example, $F(2)=3, F(3)=2$, and $F(p)=p$ for all other prime $p$ yields a solution.
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Re: Functional Equation (Canada 1969)
I was considering the constraint that \(f\) is strictly increasing.
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
Re: Functional Equation (Canada 1969)
And I meant the part "$f:\mathbb N \mapsto \mathbb N, f(mn)=f(m)f(n) \Rightarrow$ Cauchy" is wrong, and provided a counter example. Even if you add the strictly increasing part, you'd have to prove it differently (like you did in the edit).
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Re: Functional Equation (Canada 1969)
Got it. So is there any other solution to the given problem?
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.