If and have the same fundamental period, then so does .
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Let h ( x ) = f ( x ) + g ( x ) .
Counterexample 1: Let f ( x ) = ∣ sin x ∣ , g ( x ) = ∣ cos x ∣ .
Fundamental period of f is π and fundamental period of g is also π . However fundamental period of h = 2 π < π , therefore the statement is false.
Counterexample 2: Let f ( x ) = any periodic function with fundamental period say t . Then let g ( x ) = − f ( x ) + C ∀ x and some constant C . f and g will have the same fundamental period t , however h will be a constant function. In this case, h has no fundamental period, so the statement is false.
Suppose fundamental period of f and g is P . Then P is the minimum possible value such that f ( x ) = f ( x + P ) and g ( x ) = g ( x + P ) . Adding these two equations, we can also say that h ( x ) = h ( x + P ) . Therefore h is periodic with period P , so the fundamental period of h is definitely not greater than P .
We can make three cases:
Case 1: h does not have a fundamental period. This case occurs when f ( x ) = − g ( x ) + C ∀ x .
Case 2: Fundamental period of h is half of fundamental period of f and g .
If fundamental period of f and g is P and there exists 0 < p < P such that f ( x + p ) = g ( x ) and g ( x + p ) = f ( x ) ∀ x , then the fundamental period of h becomes p . This is because h ( x + p ) = f ( x + p ) + g ( x + p ) = g ( x ) + f ( x ) = h ( x ) . This case was illustrated in counterexample 1.
We can further show that the only possible value of p is 2 P . From the above equations, we obtain f ( x ) = f ( x + 2 p ) where p takes the minimum value possible. We also know the f ( x ) = f ( x + P ) where P is minimum. Hence P = 2 p .
Case 3: Fundamental period of h is equal to fundamental period of f and g
As we saw in the examples 1 and 2, case 3 is not always true, sometimes case 2 can be true as well, so the answer is "False, it could have a smaller fundamental period". □