0

The original question is Baby Rudin's problem 6.13(c), where I am asked to compute the upper and lower limits for $$\frac12(\cos(x^2)-\cos((x+1)^2))+\mathcal O(\frac1x)$$ as $x\to+\infty$.

The question is equivalent to the one asked in the title.

From a graphical point of view (as is shown below, with $x$ ranging from $20\pi$ to $30\pi$), the answers are clearly $1$ and $-1$.

enter image description here

But how on earth am I supposed to construct a fitting sequence $x_n$? When I was trying to obtain the upper limit I tried the form $$x_n=\frac12 \pi-\frac12+2k_n\pi$$ with two conditions $$k_n\to 0 \mod 1\quad\text{and}\quad x_n^2\to0\mod 2\pi $$ but it was so hard to find a proper $k_n$ that I had to give up.

I really need some help here. I'd be grateful if you would enlighten me. Thanks in advance.

Vim
  • 13,905
  • Try $x_n=(\pi-1)/2+2 A_n\pi+d_n$ where $x_n^2=2 B_n \pi +e_n$ where $A_n,B_n$ are positive integers and $d_n,e_n \to 0$. – DanielWainfleet Oct 31 '15 at 16:27
  • What do you mean by $A_n,B_n$ are positive integers?.. Are the two equations consistent? @user254665 – Vim Oct 31 '15 at 16:41
  • Yes.This is a number-theory problem.It helps to know that for any real $r$ there are infinitely many integers $a,b$ for which $|r-a/b|<1/b^2$. – DanielWainfleet Oct 31 '15 at 18:22

1 Answers1

3

Lemma: Let $a_1< a_2 < \cdots\to \infty,$ with $a_{n+1}-a_n \to 0.$ Then $\cos a_n$ is dense in $[-1,1].$

Proof: It suffices to show $e^{ia_n}$ is dense on the unit circle. But think about about it: As $n\to \infty, e^{ia_n}$ makes infinitely orbits around the circle (because $a_n \to \infty$), in steps of arc length $a_{n+1}-a_n.$ Those arc lengths $\to 0.$ Thus if $A$ is any open arc on the circle, $e^{ia_n}$ has to land in $A$ infinitely many times; you can't jump over an arc if the steps are less than the length of that arc.

Corollary: If $b> 0,$ then $\{\cos (b\sqrt n + 1) :n\in \mathbb {N}\}, \{\cos (b\sqrt {2n+1} + 1): n\in \mathbb {N}\} )$ are both dense in $[-1,1].$

Let $f(x) = \cos x^2 -\cos (x+1)^2.$ With $x_n = \sqrt {2\pi n}, y_n = \sqrt {\pi (2n+1)},$ use the corollary to see the upper limit of $f(x_n)$ is $2,$ and the lower limit of $f(y_n)$ is $-2.$

zhw.
  • 107,943