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Difference between revisions of "1975 Canadian MO Problems/Problem 7"

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To prove that <math>\sin(x^2)</math> is periodic, we need to check if there exists a positive number <math>p</math> such that <math>\sin((x + p)^2) = \sin(x^2)</math> for all <math>x</math>. Using the trigonometric property <math>\sin(a) = \sin(b) \iff a - b = 2n\pi</math> (where <math>n \in \mathbb{Z} </math>), this implies <math>(x + p)^2 - x^2 = 2n\pi</math> for some integer <math>n</math>. Expanding and simplifying, <math>(x + p)^2 - x^2 = 2px + p^2</math>, so the equation becomes <math>2px + p^2 = 2n\pi</math>. Rewriting, <math>p(2x) + p^2 = 2n\pi</math>. For this equation to hold for all <math>x</math>, the term <math>p(2x)</math> must vanish, which is only possible if <math>p = 0</math>. However, since <math>p > 0</math> is required for periodicity, no such <math> p</math> exists, meaning <math>\sin(x^2)</math> is not periodic. [Intuitively, the argument <math>x^2</math> of <math>\sin(x^2)</math> grows faster than linearly as <math>x</math> increases, causing the values of <math>\sin(x^2)</math> to fail to repeat in a regular pattern. Therefore, <math>\sin(x^2)</math> is not periodic.] <math>\blacksquare</math>
 
To prove that <math>\sin(x^2)</math> is periodic, we need to check if there exists a positive number <math>p</math> such that <math>\sin((x + p)^2) = \sin(x^2)</math> for all <math>x</math>. Using the trigonometric property <math>\sin(a) = \sin(b) \iff a - b = 2n\pi</math> (where <math>n \in \mathbb{Z} </math>), this implies <math>(x + p)^2 - x^2 = 2n\pi</math> for some integer <math>n</math>. Expanding and simplifying, <math>(x + p)^2 - x^2 = 2px + p^2</math>, so the equation becomes <math>2px + p^2 = 2n\pi</math>. Rewriting, <math>p(2x) + p^2 = 2n\pi</math>. For this equation to hold for all <math>x</math>, the term <math>p(2x)</math> must vanish, which is only possible if <math>p = 0</math>. However, since <math>p > 0</math> is required for periodicity, no such <math> p</math> exists, meaning <math>\sin(x^2)</math> is not periodic. [Intuitively, the argument <math>x^2</math> of <math>\sin(x^2)</math> grows faster than linearly as <math>x</math> increases, causing the values of <math>\sin(x^2)</math> to fail to repeat in a regular pattern. Therefore, <math>\sin(x^2)</math> is not periodic.] <math>\blacksquare</math>
  
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~sitar
 
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{{Old CanadaMO box|num-b=6|num-a=8|year=1975}}
 
{{Old CanadaMO box|num-b=6|num-a=8|year=1975}}

Latest revision as of 18:37, 17 November 2024

Problem 7

A function $f(x)$ is $\textit{periodic}$ if there is a positive integer such that $f(x+p) = f(x)$ for all $x$. For example, $\sin x$ is periodic with period $2\pi$. Is the function $\sin(x^2)$ periodic? Prove your assertion.

Solution

To prove that $\sin(x^2)$ is periodic, we need to check if there exists a positive number $p$ such that $\sin((x + p)^2) = \sin(x^2)$ for all $x$. Using the trigonometric property $\sin(a) = \sin(b) \iff a - b = 2n\pi$ (where $n \in \mathbb{Z}$), this implies $(x + p)^2 - x^2 = 2n\pi$ for some integer $n$. Expanding and simplifying, $(x + p)^2 - x^2 = 2px + p^2$, so the equation becomes $2px + p^2 = 2n\pi$. Rewriting, $p(2x) + p^2 = 2n\pi$. For this equation to hold for all $x$, the term $p(2x)$ must vanish, which is only possible if $p = 0$. However, since $p > 0$ is required for periodicity, no such $p$ exists, meaning $\sin(x^2)$ is not periodic. [Intuitively, the argument $x^2$ of $\sin(x^2)$ grows faster than linearly as $x$ increases, causing the values of $\sin(x^2)$ to fail to repeat in a regular pattern. Therefore, $\sin(x^2)$ is not periodic.] $\blacksquare$

~sitar .

1975 Canadian MO (Problems)
Preceded by
Problem 6 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed by
Problem 8


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