Answer:
f(x) = cos 2πx
Thus the fundamental period T = 1
Explanation:
From the information given:
The periodic function f(x + T) = f(x) T ≠ 0 provided that T is the smallest positive value for which f(x + T) = f(x) holds, then T is called the fundamental period of f.
Now, for the given function f(x) = cos(2πx)
suppose T = 1/4 since T ≠ 0
Then:
f(x+ 1/4) = cos(2π (x + 1/4) )
f(x+ 1/4) = cos (2πx + π/2)
f(x+ 1/4) = -sin 2πx (Not satisfied)
suppose T = 1/2 since T ≠ 0
Then:
f(x+ 1/2) = cos(2π (x + 1/2) )
f(x+ 1/2) = cos (2πx + π)
f(x+ 1/2) = -cos 2πx (Not satisfied)
suppose T = 1 since T ≠ 0
Then:
f(x+ 1) = cos(2π (x + 1) )
f(x+ 1) = cos (2πx + 2π)
f(x+ 1) = cos 2πx
f(x+ 1) = f(x)
f(x) = cos 2πx
Therefore, the fundamental period T = 1 since f(x) = cos 2πx