Why is sin(x^2) not a periodic function?

2 Answers
Mar 19, 2018

While #sin(x)# repeats every time #x# changes by a fixed amount (#2pi#), #sin(x^2)# repeats only when #x^2# changes by #2pi#, and this does not happen at uniform intervals of #x#.

Explanation:

For a function #f(x)# to be periodic with a period #X#, we must have

#f(x+X)=f(x) qquad forall x#

This immediately implies that #f(x), f(x+X), f(x+2x), f(x+3X),... # etc. all have the same value. So, a periodic function of #x# repeats after equally spaced values of #x#.

Now, #sin(x^2)# attains the value 0, for example, when #x^2# attains the values 0, #pi#, #2pi#, #3pi#, etc. While these values are equally spaced, the corresponding #x# values are 0, #sqrt(pi)#, #sqrt(2pi)#, #sqrt(3pi)# etc. - and these are not equally spaced.

Mar 19, 2018

While #sin(x)# repeats every time #x# changes by a fixed amount (#2pi#), #sin(x^2)# repeats only when #x^2# changes by #2pi#, and this does not happen at uniform intervals of #x#.

Explanation:

For a function #f(x)# to be periodic with a period #X#, we must have

#f(x+X)=f(x) qquad forall x#

This immediately implies that #f(x), f(x+X), f(x+2x), f(x+3X),... # etc. all have the same value. So, a periodic function of #x# repeats after equally spaced values of #x#.

Now, #sin(x^2)# attains the value 0, for example, when #x^2# attains the values 0, #pi#, #2pi#, #3pi#, etc. While these values are equally spaced, the corresponding #x# values are 0, #sqrt(pi)#, #sqrt(2pi)#, #sqrt(3pi)# etc. - and these are not equally spaced.