is-sum-of-two-periodic-function-is-also-periodic-give-reason-

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