Inspired by Sandeep Bhardwaj

Algebra Level 3

If f ( x ) f(x) and g ( x ) g(x) have the same fundamental period, then so does f ( x ) + g ( x ) f(x) + g(x) .

False, it could have a smaller fundamental period False, the fundamental period is larger True

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2 solutions

Pranshu Gaba
Jun 26, 2015

Let h ( x ) = f ( x ) + g ( x ) h(x) = f(x) + g(x) .

Counterexample 1: Let f ( x ) = sin x f(x) = |\sin x| , g ( x ) = cos x g(x) = |\cos x| .

Fundamental period of f f is π \pi and fundamental period of g g is also π \pi . However fundamental period of h = π 2 < π h = \frac{\pi }{2}< \pi , therefore the statement is false.

Counterexample 2: Let f ( x ) = f(x) = any periodic function with fundamental period say t t . Then let g ( x ) = f ( x ) + C x g(x) = -f(x) +C ~ \forall ~ x and some constant C C . f f and g g will have the same fundamental period t t , however h h will be a constant function. In this case, h h has no fundamental period, so the statement is false.


Suppose fundamental period of f f and g g is P P . Then P P is the minimum possible value such that f ( x ) = f ( x + P ) f(x) = f(x+P) and g ( x ) = g ( x + P ) g(x) = g(x+P) . Adding these two equations, we can also say that h ( x ) = h ( x + P ) h(x) = h(x+P) . Therefore h h is periodic with period P P , so the fundamental period of h h is definitely not greater than P P .

We can make three cases:

  • Case 1: h h does not have a fundamental period. This case occurs when f ( x ) = g ( x ) + C x f(x) = -g(x) + C ~ \forall ~ x .

  • Case 2: Fundamental period of h h is half of fundamental period of f f and g g .

    • If fundamental period of f f and g g is P P and there exists 0 < p < P 0 < p < P such that f ( x + p ) = g ( x ) f(x + p) = g(x) and g ( x + p ) = f ( x ) x g(x + p) = f(x) ~ \forall ~ x , then the fundamental period of h h becomes p p . This is because h ( x + p ) = f ( x + p ) + g ( x + p ) = g ( x ) + f ( x ) = h ( x ) 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 p is P 2 \frac{P}{2} . From the above equations, we obtain f ( x ) = f ( x + 2 p ) f(x) = f(x + 2p) where p p takes the minimum value possible. We also know the f ( x ) = f ( x + P ) f(x) = f(x + P) where P P is minimum. Hence P = 2 p P = 2p .

  • Case 3: Fundamental period of h h is equal to fundamental period of f f and g 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". _\square

Moderator note:

Great counterexamples!

All that we know is that the fundamental period will be a period of h h , and we don't know for certain that it will be the fundamental period.

Prashant Kr
Jun 27, 2015

If f(x) and g(x) are periodic functions with period A and B respectively ,then h(x) =f(x)+g(x) has period as : a. L.C.M. of {A,B} ; if f(x) and g(x) cannot be interchanged by adding a least positive number less than the L.C.M. Of {A,B} . b. K; if f(x) and g(x) can be interchanged by adding a least possitive number k(k<L .C.M. of {A,B} ) These 2 points can be used to perdict the answer .

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