## 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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