• Jul 8th 2011, 11:25 AM
pika
Well, there is a function f(x) that has two properties
(1) $f(x) \geq -1$ and $f(x) \leq 1$

(2) f(x + $\frac{13}{42}$) + f(x) = f(x + $\frac{1}{6}$) + f(x + $\frac{1}{7}$)

and I must prove that exists a T > 0 that f(x + T) = f(x) for all x $\in$ $\mathbb{R}$. Little help?
• Jul 8th 2011, 01:24 PM
Siron
Re: a problem about periodic functions
Is this one complete exercice?
This is what I think, given is that f(x)>=-1 and f(x)<=1 so that means that for every x, f(x) lies in the closed interval -1 to 1. Can you imagine a function with the same property/range? ...
• Jul 9th 2011, 09:23 AM
Opalg
Re: a problem about periodic functions
Quote:

Originally Posted by pika
Well, there is a function f(x) that has two properties
(1) $f(x) \geq -1$ and $f(x) \leq 1$

(2) $f\bigl(x + \tfrac{13}{42}\bigr) + f(x) = f\bigl(x + \tfrac{1}{6}\bigr) + f\bigl(x + \tfrac{1}{7}\bigr)$

and I must prove that exists a T > 0 that f(x + T) = f(x) for all $x\in\mathbb{R}$. Little help?

I think that this is quite tricky unless you know something about recurrence relations. I would start by proving this result:

Let $(a_n)_{n\geqslant0}$ be a bounded sequence of real numbers such that $a_{n+13} = a_{n+7} +a_{n+6} - a_n$ (for all n≥0). Then $(a_n)$ is periodic with period 42. (In other words, $a_{n+42} = a_n$ for all n.)

The result about the function f will then follow by taking $a_n = f\bigl(x+\tfrac n{42}\bigr)$ (and it will show that T=1).

To prove the result about the sequence, you need to go from the recurrence relation $a_{n+13} - a_{n+7} - a_{n+6} + a_n = 0$ to the auxiliary equation $\lambda^{13} - \lambda^7 - \lambda^6 + 1 = 0.$ Write this as $(\lambda^7-1)(\lambda^6-1) = 0$ to see that its roots are $\lambda=1$ (double root) together with all the other (complex) sixth and seventh roots of unity. The theory of recurrence relations tells you that the sequence $(a_n)$ is given by $\textstyle a_n = An+B + \sum_{\omega_k} C_k\omega_k^n$, where $A,\, B$ and the $C_k$s are constants, and the sum is taken over all the sixth and seventh roots of unity apart from 1 itself.

The fact that the sequence is bounded tells you that $A = 0$. Since each $\omega_k$ satisfies $\omega_k^{42} = 1$ it then follows that $(a_n)$ has period 42, as required.
• Jul 11th 2011, 11:11 AM
kalyanram
Re: a problem about periodic functions
Quote:

I think that this is quite tricky unless you know something about recurrence relations. I would start by proving this result:

Let be a bounded sequence of real numbers such that (for all n≥0). Then is periodic with period 42. (In other words, for all n.)

The result about the function f will then follow by taking (and it will show that T=1).

To prove the result about the sequence, you need to go from the recurrence relation to the auxiliary equation Write this as to see that its roots are (double root) together with all the other (complex) sixth and seventh roots of unity. The theory of recurrence relations tells you that the sequence is given by , where and the s are constants, and the sum is taken over all the sixth and seventh roots of unity apart from 1 itself.

The fact that the sequence is bounded tells you that . Since each satisfies it then follows that has period 42, as required.
That really is interesting theory Opalg, could you please tell me where I can read systematically about recurrence relations most of what you said was too quick for me.

Kalyan.