How can one tell the period of sin⁡(x32) ? Is it (2π)23?

The solutions ofsin⁡((x+T)32)=sin⁡(x32)are(x+T)32=x32+2mπand from there,T=(x32+2mπ)23−xwhich proves that there is no constant T that works.(The same computation with sin⁡(x) gives T=(x+2mπ)−x

Leaving aside the fact that the function is only defined for x≥0, you can easily see thatf′(x)=32xcos⁡(x32)If T&gt;0 is a period, we should have f(T)=f(0)=0 and also f′(T)=f′(0)=0. Then we havesin⁡(T32)=cos⁡(T32)=0which is a contradiction.

The solutions ofsin⁡((x+T)32)=sin⁡(x32)are(x+T)32=x32+2mπand from there,T=(x32+2mπ)23−xwhich proves that there is no constant T that works.(The same computation with sin⁡(x) gives T=(x+2mπ)−x

Leaving aside the fact that the function is only defined for x≥0, you can easily see thatf′(x)=32xcos⁡(x32)If T&gt;0 is a period, we should have f(T)=f(0)=0 and also f′(T)=f′(0)=0. Then we havesin⁡(T32)=cos⁡(T32)=0which is a contradiction.

Explained: "How can one tell the period of sin⁡(x32)..."

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Calculus and Analysis
Precalculus
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monkeyman130yb

Answered question

2022-03-24

How can one tell the period of

\sin (x^{\frac{3}{2}})

? Is it

{(2 π)}^{\frac{2}{3}}

Answer & Explanation

pastuh7vka

Beginner2022-03-25Added 13 answers

The solutions of

\sin ({(x + T)}^{\frac{3}{2}}) = \sin (x^{\frac{3}{2}})

are

{(x + T)}^{\frac{3}{2}} = x^{\frac{3}{2}} + 2 m π

and from there,

T = {(x^{\frac{3}{2}} + 2 m π)}^{\frac{2}{3}} - x

which proves that there is no constant T that works.
(The same computation with

\sin (x)

gives

T = (x + 2 m π) - x

Regan Gallegos

Beginner2022-03-26Added 9 answers

Leaving aside the fact that the function is only defined for

x \geq 0

, you can easily see that

f^{'} (x) = \frac{3}{2} \sqrt{x} \cos (x^{\frac{3}{2}})

T > 0

is a period, we should have

f (T) = f (0) = 0

and also

f^{'} (T) = f^{'} (0) = 0

. Then we have

\sin (T^{\frac{3}{2}}) = \cos (T^{\frac{3}{2}}) = 0

which is a contradiction.

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