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Antipodal points

Posted on: September 6th, 2015 by
2

Let f(t) be a continuous periodic function with period T, i.e.

  1. f(t) is continuous over reals.
  2. f(t+T) = f(t) for all real t.

Show that there always exists some time instant t_0 for which f(t_0) = f(t_0+T/2).


2 Responses to Antipodal points

  1. Shaleen had this to say about that:

    Consider a function L from R->R where L(x) = f(x+T) - f(x).
    Since f is continuous over R, so is L.

    If L(0)=0, we are done and have our antipodal point.
    If not, there exists some x for which L(0) < 0 0 > L(T/2)). Therefore for some value of x between 0 and T/2, L(x)=0 which becomes our antipodal point.

  2. Jeetesh M had this to say about that:

    Consider a continuous function, P(x) = f(x + T/2) - f(x)
    Then,
    P(x+T/2) = f(x+T) - f(x+T/2)
    So, P(x) + P(x+T/2) = f(x+T) - f(x) = 0
    which means P(x+T/2) = -P(x)
    If P(x) = 0 for all x, then we are done.
    Otherwise, by Intermediate Value Theorem, there exists y such that y > x, y < x+T/2 & P(y) = 0 f(y+T/2) = f(y)


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