## Functional Equation (Canada 1969)

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mutasimmim
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### Functional Equation (Canada 1969)

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 $f(n+1)>f(n)$.

*Mahi*
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### Re: Functional Equation (Canada 1969)

Hint:
The set $S_n = \{n^q | q \in \mathbb Q\}$ is dense $\forall n \in \mathbb N, n > 1$.
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Nur Muhammad Shafiullah | Mahi

mutasimmim
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### Re: Functional Equation (Canada 1969)

Hints for two different solutions:

Nirjhor
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### 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.

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*Mahi*
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### Re: Functional Equation (Canada 1969)

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 $$f(n)=n$$.
Wrong. For example the following set of functions defines a solution to $f(mn) = f(m)f(n), f: \mathbb N \mapsto \mathbb N$.
$f(\prod p_i ^{e_i}) = \prod f(p_i)^{e_i}$
$f(p_i) = F(p_i)$
Where the function $F: \mathbb P \mapsto \mathbb P$ is injective.
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Nur Muhammad Shafiullah | Mahi

Nirjhor
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### Re: Functional Equation (Canada 1969)

*Mahi* wrote:
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 $$f(n)=n$$.
Wrong. For example the following set of functions defines a solution to $f(mn) = f(m)f(n), f: \mathbb N \mapsto \mathbb N$.
$f(\prod p_i ^{e_i}) = \prod f(p_i)^{e_i}$
$f(p_i) = F(p_i)$
Where the function $F: \mathbb P \mapsto \mathbb P$ is injective.
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?

(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.

*Mahi*
Posts: 1175
Joined: Wed Dec 29, 2010 12:46 pm
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### 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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Nur Muhammad Shafiullah | Mahi

Nirjhor
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### 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.

*Mahi*
Posts: 1175
Joined: Wed Dec 29, 2010 12:46 pm
Location: 23.786228,90.354974
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### 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).
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Nur Muhammad Shafiullah | Mahi

Nirjhor
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Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

### 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.