Results 1 to 4 of 4

Math Help - a problem about periodic functions

  1. #1
    Newbie
    Joined
    Jul 2011
    Posts
    1

    a problem about periodic functions

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    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? ...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7

    Re: a problem about periodic functions

    Quote Originally Posted by pika View Post
    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_ks 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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member kalyanram's Avatar
    Joined
    Jun 2008
    From
    Bangalore, India
    Posts
    142
    Thanks
    14

    Re: a problem about periodic functions

    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. periodic functions
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: May 26th 2010, 08:31 PM
  2. Sum of periodic functions
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 24th 2009, 11:22 AM
  3. Periodic Functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 4th 2009, 07:43 AM
  4. Periodic Functions
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: February 8th 2009, 07:07 AM
  5. Periodic Functions
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: March 1st 2007, 05:58 PM

Search Tags


/mathhelpforum @mathhelpforum