# Thread: Prove that function is periodic?

1. ## Prove that function is periodic?

Prove that if the graph of the function $\displaystyle y = f(x)$, defined throughout the number scale, is symmetrical about the two lines $\displaystyle x = a\ \mbox{and}\ x = b,\ (a<b)$, then the function is a periodic one.

2. ## Proof by Induction

(i) Suppose $\displaystyle f(a-x)=f(a+x)$ for all $\displaystyle x$
(ii) Suppose $\displaystyle f(b-x)=f(b+x)$ for all $\displaystyle x$
(iii) Define $\displaystyle \Delta=b-a$

Base Case $\displaystyle (n=1)$
$\displaystyle f(a)=f(b-\Delta)=f(b+\Delta)=f(a+2\Delta)$

General Case
Suppose $\displaystyle f(a)=f(a+2n\Delta)$ for some $\displaystyle n$. Then $\displaystyle f(a)=f(b+(2n-1)\Delta)=f(b+(1-2n)\Delta)=f(a+(2-2n)\Delta)=f(a+2(n+1)\Delta)$ . So if $\displaystyle f(a)=f(a+2n\Delta)$ then $\displaystyle f(a)=f(a+2(n+1)\Delta)$.

Induction
Therefore, $\displaystyle f(a)=f(a+2n\Delta)$ for all $\displaystyle n$. So the function has a period of $\displaystyle 2\Delta$.

Similarly, the argument can be made in the reverse direction.

3. Originally Posted by Media_Man
Therefore, $\displaystyle f(a)=f(a+2n\Delta)$ for all $\displaystyle n$. So the function has a period of $\displaystyle 2\Delta$.

Similarly, the argument can be made in the reverse direction.
Well generally the period must be a constant... here $\displaystyle \Delta$ depends on 'a'

Something is wrong

4. Originally Posted by Isomorphism
Well generally the period must be a constant... here $\displaystyle \Delta$ depends on 'a'

Something is wrong
But you fix a and b before proving that it is periodic.
So it's not really a problem

It is a constant with respect to the variable. But a and b act like constants.

5. Originally Posted by Moo
But you fix a and b before proving that it is periodic.
So it's not really a problem

It is a constant with respect to the variable. But a and b act like constants.
a and b are already fixed . It has been shown that $\displaystyle f(\text{\LARGE{a}}) = f(\text{\LARGE{a}}+2n\Delta)$ for a specific value of $\displaystyle \text{\LARGE{a}}$.

One has to show $\displaystyle f(x) = f(x+\Delta)$ for any value of x, which does not depend on $\displaystyle \Delta$. Here $\displaystyle x$ is $\displaystyle a$ and $\displaystyle \Delta = b - a$

6. Originally Posted by Isomorphism
a and b are already fixed . It has been shown that $\displaystyle f(\text{\LARGE{a}}) = f(\text{\LARGE{a}}+2n\Delta)$ for a specific value of $\displaystyle \text{\LARGE{a}}$.

One has to show $\displaystyle f(x) = f(x+\Delta)$ for any value of x, which does not depend on $\displaystyle \Delta$. Here $\displaystyle x$ is $\displaystyle a$ and $\displaystyle \Delta = b - a$
Oh sorry, thought you were pointing out something else
The problem is not the period, it's the fact that he set x=a, isn't it ?

7. ## Correction

Isomorphism is absolutely right. The definition of periodic is not simply "there exists a $\displaystyle \Delta$ for which $\displaystyle f(x)=f(x+n\Delta)$ for some x." This has to be true for all $\displaystyle x$. The proof is the same though.

(i) Suppose $\displaystyle f(a-x)=f(a+x)$ for all $\displaystyle x$
(ii) Suppose $\displaystyle f(b-x)=f(b+x)$ for all $\displaystyle x$
(iii) Define $\displaystyle \Delta=b-a$

Base Case $\displaystyle (n=1)$
$\displaystyle f(a+x)=f(a-x)=f(b-\Delta-x)=f(b+\Delta+x)=f(a+x+2\Delta)$

General Case
Suppose $\displaystyle f(a+x)=f(a+x+2n\Delta)$ for some $\displaystyle n$. Then $\displaystyle f(a+x)=f(b+x+(2n-1)\Delta)=$$\displaystyle f(b-x+(1-2n)\Delta)=f(a-x+(2-2n)\Delta)=f(a+x+2(n+1)\Delta)$ . So if $\displaystyle f(a+x)=f(a+x+2n\Delta)$ then $\displaystyle f(a+x)=f(a+x+2(n+1)\Delta)$.

Induction
Therefore, $\displaystyle f(a+x)=f(a+x+2n\Delta)$ for all $\displaystyle n\geq 1$. Letting $\displaystyle x'=a+x$, $\displaystyle f(x')=f(x'+2n\Delta)$, so the function has a period of $\displaystyle 2\Delta$.

Again, a separate, albeit identical proof is necessary to prove the function is periodic in the negative direction.

8. It is not difficult;
$\displaystyle f(a-x)=f(a+x)$
Replace $\displaystyle x$ by $\displaystyle a-x$
$\displaystyle f(x)=f(2a-x)$

Similarly
$\displaystyle f(x)=f(2b-x)$

Therefore,
$\displaystyle f(2a-x)=f(2b-x)$

Replace $\displaystyle x$ by $\displaystyle 2a-x$
$\displaystyle f(x)=f(2b-2a+x)$

Therefore,period is $\displaystyle T=2(b-a)$ though it still remains to be proved that $\displaystyle 2(b-a)$ is the least value of $\displaystyle T$ for which $\displaystyle f(x+T)=f(x)$.

Originally Posted by Media_Man

Again, a separate, albeit identical proof is necessary to prove the function is periodic in the negative direction.
Consider $\displaystyle f:[0,\infty)\rightarrow R$
$\displaystyle f(x)=\sin x$
Is $\displaystyle f(x)$ not periodic

9. Originally Posted by pankaj
Therefore,period is $\displaystyle T=2(b-a)$ though it still remains to be proved that $\displaystyle 2(b-a)$ is the least value of $\displaystyle T$ for which $\displaystyle f(x+T)=f(x)$.
Thats not required. The question merely demands the function to be periodic. It does not ask for a certain period period

10. Originally Posted by pankaj
Therefore,period is $\displaystyle T=2(b-a)$ though it still remains to be proved that $\displaystyle 2(b-a)$ is the least value of $\displaystyle T$ for which $\displaystyle f(x+T)=f(x)$.
This is also not always the case. Consider $\displaystyle f(x)=sin(x)$. This has two lines of symmetry at $\displaystyle x=\frac{\pi}{2}$ and $\displaystyle x=\frac{3\pi}{2}$ giving it a period of $\displaystyle 2\pi$ . But applying your algorithm with $\displaystyle a=\frac{\pi}{2}$ and $\displaystyle b=\frac{17\pi}{2}$ would certainly not reveal the "lowest" possible period.

You could say that the lowest value of $\displaystyle \Delta$ satisfying $\displaystyle f(x)=f(x+\Delta)$ for all $\displaystyle x$ also satisfies $\displaystyle \Delta |2(b-a)$ (or $\displaystyle \frac{2(b-a)}{\Delta}$ is a whole number).

11. Originally Posted by Isomorphism
Thats not required. The question merely demands the function to be periodic. It does not ask for a certain period period
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.