is sum of two periodic function is also periodic give reason


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Question Number 25800 by rita1608 last updated on 15/Dec/17

$i s s u m o f t w o p e r i o d i c f u n c t i o n i s$ $a l s o p e r i o d i c g i v e r e a s o n$
	Answered by kaivan.ahmadi last updated on 15/Dec/17

	$yes . if T_{f} = T_{1} and T_{g} = T_{2} then we have$ $T_{(f + g)} = T = [T_{1}, T_{2}] .$ $T = {mT}_{1} = {kT}_{2}; m, k \in Z .$ $f + g (x + T) = f (x + T) + g (x + T) = f + g (x)$
	Commented by rita1608 last updated on 15/Dec/17

	$b u t t h e s u m o f t w o p e r i o d i c f u n c t i o n s$ $l i k e x - [x] a n d ∣ s i n x ∣ i s n o t p e r i o d i c$ $t h e n h o w i t i s p o s s i b l e t h a t s u m o f$ $t w o p e r i o d i c f u n c t i o n s w i l l b e p e r i o d i c$
	Commented by kaivan.ahmadi last updated on 15/Dec/17

	$\frac{T_{1}}{T_{2}} must be rational otherwise we cant$ $calculate [T_{1}, T_{2}]$
	Commented by kaivan.ahmadi last updated on 15/Dec/17

	$\frac{1}{π} \notin Q$
	Answered by sushmitak last updated on 15/Dec/17

	$Not always$ $\cos (x) is periodic with period 2 π$ $\cos (π x) is periodic with period 2 .$ $\cos (x) + \cos (π x) is not periodic$
	Commented by kaivan.ahmadi last updated on 15/Dec/17

	$since \frac{2}{2 π} = \frac{1}{π} \notin Q$ $this proposition is true when \frac{T_{1}}{T_{2}} is rational .$
	Commented by rita1608 last updated on 15/Dec/17

	$t h a n k s$
	Commented by rita1608 last updated on 15/Dec/17

	$t h a n k s$
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