Given $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$, where $a,b$ are positive constants $(a>b)$, prove that $f(x)$ is a periodic function

Question

Given $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$, where $a,b$ are positive constants $(a>b)$, prove that $f(x)$ is a periodic function

I have done the following:

$f(a-x)=f(a+x)$....(1)

or, $f(-(x-a))=f(x+a). $

Putting $x=x+a$, we get $f(-x)=f(x+2a) ....(i)$

Similarly,

$f(b-x)=f(b+x).....(2) $

Putting $x=x+b,$ we get
$f(-x)=f(x+2b)......(ii)$

Equating $(i)$ & $(ii)$, we get:

$f(x+2a)=f(x+2b).$

Given, $a>b$

Therefore, $2a=2b+h.$

Putting $x=x-2a$, we get
$f(x+h)=f(x).$

Is this mathematically correct? Also, what other ways are there to solve it?

Please elaborate. Thank You. – Anubhab Giri Oct 05 '18 at 12:46 — Anubhab Giri, Oct 05 '18 at 12:46
Your approach looks fine to me – Vasili Oct 05 '18 at 12:51 — Vasili, Oct 05 '18 at 12:51

score 2 · Answer 1 · answered Oct 05 '18 at 13:07

I guess that $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$ hold for every $x$.

In particular, evaluating at $x+a$ instead of $x$, $$ f(-x)=f(x+2a) $$ Similarly $f(-x)=f(x+2b)$. Therefore $$ f(x+2a)=f(x+2b) $$ for every $x$. Evaluating at $x-2b$ instead of $x$, we obtain $$ f(x+2a-2b)=f(x) $$ and, setting $T=2a-2b$, $$ f(x)=f(x+T) $$ for every $x$.

Good job!

Given $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$, where $a,b$ are positive constants $(a>b)$, prove that $f(x)$ is a periodic function

1 Answers1