Question

if the domain of sin(x) is restricted to [-2pi,2pi], is it still a periodic function

Answer 1

nope,

Answer 2

why not? there are two cycles right?

Answer 3

oh, yeah, you are right. My stupid comment. :)

Answer 4

but I thought it's not periodic just like you did though :/

Answer 5

despite there are two cycles

Answer 6

2 is enough!!

Answer 7

well, i'm still confused about the definition.
def: A function is said to be periodic if there exists a non-zero real number P such that f(x+P) = f(x) for all x in the domain.

Answer 8

so if the period is 2pi (which it is), the value of sin(2pi + 2pi) is undefined , so we can't say sin(2pi + 2pi) = sin(2pi)

Answer 9

somehow the definition implies there are infinitely many cycles

Answer 10

Hey!! why not? sin 4pi = sin 2pi =0

Answer 11

who says sin (4pi) is undefined?

Answer 12

well, because 4pi isn't in the domain [-2pi,2pi]

Answer 13

out of domain?

Answer 14

it doesn't mean sin(4pi) is undefined, to me!!

Answer 15

If you didn't have the domain restriction, then yes it would be periodic. The domain restriction makes the function not be periodic.

Answer 16

@Loser66, well sin(4pi) is undefined. If you draw a vertical line at x = 4pi, it doesn't intersect the function.

@jim_thompson5910 I though so too.

Answer 17

@jim_thompson5910 I don't think so. It is still http://mathworld.wolfram.com/PeriodicFunction.html

Answer 18

it is doubly periodic function

Answer 19

@jim_thompson5910 so somehow being period implies there are infinitely many cycles, which I need someone to confirm that.

if it is not periodic, then the negation of the definition is: For all non-zero value P, there exists a element x in the domain such that f(x + P) ≠ f(x)

Answer 20

"being periodic"

Answer 21

Then again I can't find any issues if I plug in values from the domain. So it appears sin(x) is periodic

Answer 22

x = 4pi isn't in the domain of [-2pi,2pi] but we don't plug in x = 4pi

x = 2pi is the value being plugged in which is in the domain

Answer 23

remember that the domain is [-2pi,2pi] instead of (-inf,inf)

Answer 24

right. We plug in x = 2pi but we still need to value of sin(4pi) right?

Answer 25

well, let's just take a look at what it means to be not periodic.

Any non-zero value P besides P = 2pi will satisfy the statement.

for example if P = pi, then choose x = pi/2
sin(pi/2 + pi) ≠ sin(pi/2).

But if P = 2pi, that what value of x will work? clearly, we are back the same issue of being out of the domain

Answer 26

doex x = 2pi work?

can I say sin(2pi + 2pi) ≠ sin(pi) ? If this is true, then the function is not periodic

Answer 27

Even the domain is [-2pi,2pi], It doesn't mean the range of the function is undefined. And the range of sin (x) is [-1,1] for all x. And that range make the function periodic, not the domain, to me

Answer 28

sorry, sin(2pi + 2pi) ≠ sin(2pi)