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If i know, that $f(x)$ is a periodic function with period $T$, how should i prove, that $(f(x))^2$ has period $T_1 : T_1 \le T$.

I tried to use periodic function determination: $f(x)=f(x+T)$, but it wont help.

Git Gud
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Vladislav Brylinskiy
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1 Answers1

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Then it is obvious, because $f^2(x+T)=\{f(x+T)\}^2=\{f(x)\}^2=f^2(x)$. Hence $f^2$ repeats itself after every interval of length $T$. Thus the period of $f^2$ divides $T$. Hence $T_1\leq T$.

Samrat Mukhopadhyay
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QED
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  • I think you wanted to say "Hence $f^2\cdots$". – Samrat Mukhopadhyay Sep 13 '13 at 18:16
  • What do you mean with $a$ divides $b$ in $\Bbb R$? – Git Gud Sep 13 '13 at 18:17
  • @GitGud Sorry for using the word "divides", I only meant $T=mT_1$ for some integer $m\geq1$. – QED Sep 13 '13 at 18:21
  • Thanks @SamratMukhopadhyay for making the required edit. – QED Sep 13 '13 at 18:22
  • If $f^2$ repeats itself after every interval of length $T$, then $T$ is it's period by the definition. How can i prove, that there is period $T_1$ which "divides" $T$? – Vladislav Brylinskiy Sep 13 '13 at 18:29
  • @VladislavBrylinskiy Just set $T_1=T$ to solve your problem. – Git Gud Sep 13 '13 at 18:31
  • Sorry, but i do not understand, how can i prove that $T_1$ can be smaller than $T$? Thank you for answers. – Vladislav Brylinskiy Sep 13 '13 at 18:39
  • The period of a function is the smallest such interval after which the function repeats itself. Suppose $T_1$ is the period of the function and we know that the function repeats itself after every interval of length $T$, then we claim that $T=mT_1$ for some integer $m\geq 1$. This is because there exists $m_0$ such that $m_0T_1\leq T< (m_0+1)T_1$. Then either $T=m_0T_1$, or $f$ repeats after every interval of length $T-m_0T_1<T_1$, which the minimality of $T_1$. – QED Sep 13 '13 at 18:41
  • Vladislav: You don't know that $T_1$ necessarily is smaller than $T$, but you know that it's no greater (since $T$ is 'a' period of $f^2$, the least period is $T$ or less). To show that $T_1$ can be less than $T$, all you need to do is find an example. For a simple one, try $f(x)=\sin(x)$. – Steven Stadnicki Sep 13 '13 at 19:19
  • One thing to be a little careful about is talking about "the" period of a periodic function as the smallest interval of repetition. What if $f$ only takes the two values $\pm1$? What is "the" period of $f^2$? – Barry Cipra Sep 13 '13 at 19:32