Question #4cbca

2 Answers
Dec 16, 2016

Please see the explanation section below.

Explanation:

Note that f(0) = f(T)

Consider g(x) = f(x)-f(x+T/2)

Case 1:
If g(0)=0 , then we are done, because f(0) = f(0+T/2)

Case 2
If g(0) = f(0)-f(T/2) != 0 , then

g(T/2) = f(T/2)+f(T) = f(T/2)-f(0) = -(f(0)-f(T/2)) = -g(0).

So g(T/2) has sign opposite that of g(0).

g is continuous on [0,T/2].

Apply the Intermediate Value Theorem.

.

NOTE: If we do not have f is continuous, then the result fails for

f(x) = {(tanx,x != pi/2+pik, k" an integer"),(0,x = pi/2+pik, k" an integer"):}.

Dec 17, 2016

By definition, the period T > 0 of a periodic function is such that f(x+T) = f(x) and T is the least possible such value.

Explanation:

tan(x+2pi)=tan(x+pi)=tanx.

Here, the period is pi and not 2pi.

By definition, the period T > 0 of a periodic function is such that

f(x+T) = f(x) and T is the least possible such value.