Australia TST 2013 Exam2 Q3

Find all values of n \in \mathbb {N} ^+ such that there exists a function f: \mathbb {R} \to \mathbb {R} that satisfies the following properties:

1), f(x)=0 if and only if x=0.

2), \forall x \in \mathbb R,\, f(x) + f(2x) + ... + f(nx) =0.

Answer: \forall n \geq 2, n\in \mathbb {N}^+.

Proof: n=1 does not fit into the criteria because it requires f(x)=0,\forall x \in \mathbb R. Fix n \geq 2,n\in \mathbb N ^+.

Bertrand’s Postulate (a.k.a. Tchebyshev’s theorem): \forall m \in \mathbb {N}^+, \exists p, m<p \leq 2m, with p prime.

Now we aim to find a p such that p\leq n <2p.

For n=2, take p=2. For n odd, n\geq 3, apply Tchebyshev’s theorem to \left\lceil\frac{n}{2}\right\rceil , there exists a prime p, \frac {n+1}{2} < p \leq n+1. But now n+1 is even and n+1 \geq 4 thus n+1 is not prime, and thus we have p \leq n, which shows that p is the desired prime number as 2p > n+1 >n. For n even, apply Tchebyshev’s theorem to \frac {n}{2} to yield a prime p, \frac {n}{2} <p \leq n which gives p\leq n <2p. The p has been chosen for all n.

Let \mathbb Q ^*= \mathbb Q \backslash \{0\}, \mathbb R ^*= \mathbb R \backslash \{0\}. Let \mathbb Q_n = \{q\in \mathbb Q^* | q \textnormal { can be represented as }\frac {\alpha} {\beta},\, \alpha, \beta \in \mathbb Z, \textnormal{ where all prime factors of }\alpha \textnormal{ and }\beta \textnormal { do not exceed } n\}.

(E.g. let n=3, then Q_3 = \{q|q=\pm \frac {2^\eta}{3^\xi}, \eta,\xi \in \mathbb N ^+ \cup \{0\}\} \cup \{q|q=\pm \frac {3^\eta}{2^\xi}, \eta,\xi \in \mathbb N ^+ \cup \{0\}\}.

Define an equivalence relation on R^*: for x,y \in \mathbb R^*, x \sim y if and only if \frac {x} {y} \in \mathbb Q_n. It is easy to verify that this is indeed an equivalence relation. Partition R^* into equivalence classes K_{\gamma}, by this relation. Let \gamma be indexed by the set \Gamma.

Axiom of Choice: If K_{\gamma} is a family of sets indexed by the set \Gamma, then the Cartesian product \prod _ {\gamma \in \Gamma} K_{\gamma} is non-empty.

Apply Axiom of Choice to the equivalence classes of R^*, and we get a family of elements \kappa _\gamma, with each \gamma \in \Gamma corresponding to exactly one \kappa _ {\gamma}.

Let v_p(q) be the p-adic evaluation of q\in Q^*. (That is, reduce q to simplest form \frac {\alpha}{\beta}, v_p(q)= \textnormal{highest index of }p \textnormal{ dividing } \alpha if p does not divide \beta, and v_p(q)= - \textnormal{highest index of }p \textnormal { dividing } \beta if p|\beta.

Let f(\kappa _ \gamma) =1,\forall \kappa _\gamma. For x\in \mathbb R ^*, suppose x\in K_g, g \in \Gamma. Define f(x)= (-n+1)^{v_p(\frac{x}{\kappa _g})}. (E.g. if n=3, p=3, x= \frac {2\pi}{3}, x \in K_g, \kappa _g =\pi, f(x)= (-2)^{(v_3(\frac 23))}=-\frac 12.Similarly, f(2x)=f(\frac {4\pi}{3})= -\frac 12. This gives f(x)+f(2x)+f(3x)=(-\frac 12) + (-\frac 12) + 1=0.) We set f(0)=0.

Indeed, we show first show that the function satisfies the conditions.

First, we see that f(x) does not vanish for non-zero x.

We claim that if x,y \in K_g, f(x)/f(y)= (-n+1)^{v_p(\frac{x}{y})}. This is obvious by substituting the definitions of of f(x) and f(y), and then LHS=(-n+1)^{v_p(\frac{x}{\kappa_g})-v_p(\frac{y}{\kappa_g})}=(-n+1)^{v_p(\frac{x}{y})}=RHS.

Now, we verify the conditions. Condition 1 is already checked. If x=0, f(x)+f(2x)+...+f(nx)=0+0+...+0=0. If x \ne 0, let x \in K_g. For jx with j\ne p, we have v_p(\frac {jx}{x})=v_p(j)=0, and thus f(jx)=f(x). Since p\leq n < 2p, there are (n-1) such j's. f(px)=(-n+1)^{v_p(\frac{px}{x})}f(x)=(-n+1)f(x).

Thus f(x)+f(2x)+...+f(nx)= (n-1)f(x)+(-n+1)f(x)=0. We have found the desired function, for each of n\in \mathbb N^+,n\ge 2.


Comment: to be written later.





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