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Prove f(x) = sin(x^2) isn't periodic.

f is periodic if f(x + t) = f(x) for all x in f's domain.

I know about 1 way of proving it by derivative and boundedness but is this another way?

If f is periodic then

sin(x^2) = sin((x + t)^2)

Finding t:

(I don't know the proper way of doing this, maybe I ought to add + 2kpi)

x^2=x^2 +2xt + t^2

(I could square root it but it's the same)

t(t + 2x) = 0

t1 = 0 or t2 = -2x

t > 0 so t1 isn't.

t(x) = -2x, but t can't be dependant on x, as it needs to be constant, so t2 isn't a solutio . So I have proved the hipothesis.

f is periodic if f(x + t) = f(x) for all x in f's domain.

I know about 1 way of proving it by derivative and boundedness but is this another way?

If f is periodic then

sin(x^2) = sin((x + t)^2)

Finding t:

(I don't know the proper way of doing this, maybe I ought to add + 2kpi)

x^2=x^2 +2xt + t^2

(I could square root it but it's the same)

t(t + 2x) = 0

t1 = 0 or t2 = -2x

t > 0 so t1 isn't.

t(x) = -2x, but t can't be dependant on x, as it needs to be constant, so t2 isn't a solutio . So I have proved the hipothesis.