Prove that if the graph of the function , defined throughout the number scale, is symmetrical about the two lines , then the function is a periodic one.
(i) Suppose for all
(ii) Suppose for all
(iii) Define
Base Case
General Case
Suppose for some . Then . So if then .
Induction
Therefore, for all . So the function has a period of .
Similarly, the argument can be made in the reverse direction.
Isomorphism is absolutely right. The definition of periodic is not simply "there exists a for which for some x." This has to be true for all . The proof is the same though.
(i) Suppose for all
(ii) Suppose for all
(iii) Define
Base Case
General Case
Suppose for some . Then . So if then .
Induction
Therefore, for all . Letting , , so the function has a period of .
Again, a separate, albeit identical proof is necessary to prove the function is periodic in the negative direction.
This is also not always the case. Consider . This has two lines of symmetry at and giving it a period of . But applying your algorithm with and would certainly not reveal the "lowest" possible period.
You could say that the lowest value of satisfying for all also satisfies (or is a whole number).
Well,I wanted to know that does a function periodic function need to satisfy the following criterion:
f(x+T)=f(x), and
f(x-T)=f(x).
Actually I had read that for f to be periodic the condition f(x+T)=f(x) is to be satisfied where T is a positive constant independent of x.
But I had not read about f(x-T)=f(x) that is why I stated the example in particular.