Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the integral $$ f(\alpha,\beta)= \int_0^{2\pi}\,dx \sqrt{1- \cos(\alpha x ) \cos(\beta x)}$$ as a function of the two parameters $\alpha,\beta$. I am interested in the asymptotic behavior for $\alpha, \beta \gg 1$.

For $\alpha = \beta$ the integral can be evaluated explicitly with the result $$ f(\alpha , \alpha) = \frac{2}{\alpha} \left[ \lfloor 2 \alpha\rfloor + \sin^2 \left(\frac\pi2 \{ 2 \alpha \} \right) \right]$$ with $\{ x \} = x - \lfloor x \rfloor$. For large $\alpha$ the function $f(\alpha, \alpha)$ thus approaches $4$.

If we see what happens if we keep $\beta$ large but fixed and vary $\alpha$, we see that $\beta \approx \alpha$ with $f(\alpha,\beta) \approx 4$ looks like a minimum of the function and it quickly approaches $2\pi$ (at least for $\alpha$ large) for $\beta$ sufficiently different from $\alpha$. However, there are oscillations on top of the mean value $2\pi$. In the figure you see a numerical evaluation of the integral for $\beta=20$ and $\alpha$ between 0 and 40.

numerical evaluation of f as a function of alpha with beta=20 fixed

  • Is the value of $f(\alpha,\beta)$ for $\alpha,\beta \gg 1$ and $\alpha \neq \beta$ indeed $2\pi$?
  • Why the value $f=4$ for $\alpha \approx \beta$ is lower than the generic value $2\pi$ (for $\alpha$, $\beta$ large)?
  • What is the period of the (fast) oscillations as a function of $\alpha$ with $\beta$ fixed which are visible in the plot?
  • What is the shape of the envelope? (it is a peaked function -> Lorentzian, or Gaussian, or ...?)
  • Does anybody know how to obtain a good asymptotic expression for $f(\alpha, \beta)$?
share|improve this question
    
Regarding your second question: when $\alpha=\beta$, the argument $1-\cos^2(\alpha x)$ of the square root is always at most 1, while it sometimes exceeds 1 when $\alpha\ne\beta$. To me that explains why the special value is less than the average value. –  Greg Martin Mar 26 '12 at 7:56
add comment

3 Answers

up vote 5 down vote accepted
+200

Set $\alpha-\beta = r$ and $\alpha+\beta =s$. So our integral is $$f(r,s) := \int_0^{2 \pi} \sqrt{1-(\cos (rx) + \cos (sx))/2} dx.$$ It looks like, for $r$ constant and $s$ large, a pretty good approximation is just to ignore the $s$ term. Set $$g(r) := \int_0^{2 \pi} \sqrt{1-\cos (rx)/2} dx.$$ The figure below shows $f$ (in red) and $g$ (in blue) for $s=40$. enter image description here

An even better approximation seems to be using $3$ terms of the Taylor series for the square root: $$\sqrt{1-\cos(rx)/2 - \cos(sx)/2} \approx$$ $$ \left(1-\cos(rx)\right)^{1/2} - \frac{\cos{sx}}{4} \left(1-\cos(rx)\right)^{-1/2} - \frac{\cos^2{sx}}{32} \left(1-\cos(rx)\right)^{-3/2}=$$ $$\left(1-\cos(rx)\right)^{1/2} - \frac{1}{64} \left(1-\cos(rx)\right)^{-3/2} + \cos(sx) (\mbox{something}) + \cos(2x)(\mbox{something}).$$ Here I have replaced $\cos^2(sx)$ by $(\cos(2sx)+1)/2$.

So $$f(r,s) \approx \int_0^{2 \pi} \left( \left(1-\cos(rx)\right)^{1/2} - \frac{1}{64} \left(1-\cos(rx)\right)^{-3/2} \right) dx +$$ $$\int_0^{2 \pi} \cos(sx) (\mbox{something}) dx + \int_0^{2 \pi} \cos(2sx) (\mbox{something}) dx.$$ The two terms below the line break will go to zero as $s \to \infty$, by the Riemann-Lebesgue lemma. (As a general rule of thumb, anytime that you have a highly oscillatory integral, try to use Riemann-Lebesgue.)

Set $$h(r) := \int_0^{2 \pi} \left( \left(1-\cos(rx)\right)^{1/2} - \frac{1}{64} \left(1-\cos(rx)\right)^{-3/2} \right) dx.$$ Here is the above plot with $h$ added on (in green) enter image description here

Here is a plot of $f(r,40)-h(r)$ (note the small vertical range). I wouldn't trust this data too much -- the jaggedness is often a sign that we are getting close to Mathematica's numerical tolerance: enter image description here

I suspect that one should be able to show that there exists a function $F(y)$ such that $$\lim_{s \to \infty} f(r,s) = \int_{0}^{2 \pi} F(1-\cos(rx)/2) dx,$$ given by a convergent power series which starts $F(y) = y^{1/2} - y^{-3/2}/64+\cdots$.

share|improve this answer
    
Does this mean that the limit $\lim_{t\to\infty}[\lim_{s\to\infty} f(r,s)]$ does not exist? –  Fabian Mar 29 '12 at 14:47
    
I assume that $t$ is meant to be $r$. I don't know, but I would guess that that limit DOES exist. Looking at $g(r)=\int_{0}^{2 \pi} \sqrt{1-\cos(rx)/2} dx$, we have $g(r)=2 \pi \frac{1}{2 \pi r} \int_0^{2 \pi r} \sqrt{1 - \cos(u)/2}$ so $\lim_{r \to \infty} g(r)$ is $2 \pi$ times the average value of $\sqrt{1-\cos(u)/2}$, which exists. I would guess that $f(r,s)$ behaves the same way. –  David Speyer Mar 29 '12 at 15:17
    
According to Mathematica, the limit thus would be $2\sqrt{2} E(i \sqrt{2}) \approx 6.178$ which is different from @joriki's answer. –  Fabian Mar 29 '12 at 16:53
    
Sorry, I was probably unclear. I suspect that $\lim_{s \to \infty} f(r,s)$ exists; call that limit $\ell(r)$. It looks like $g$ is a pretty good approximation to that limit and that $h$ is a better one. I don't think that either of them are equal to $\ell$, or even that $\lim_{r \to \infty} g(r)$ is equal to $\lim_{r \to \infty} \ell(r)$. –  David Speyer Mar 29 '12 at 18:11
    
@Fabian: I don't think that "the" limit exists. It depends very much on how you go to $\infty$. My answer shows, that for $\alpha=r\beta$ there are solutions depending on $r$. Davids solution shows, that for $\alpha = \beta + c$ there are different solutions depending on $c$. You could think of other "paths" to $\infty$ like $\alpha=c \beta^2$ and they will probably result in other answers (most of which will at least be close to $2\pi \langle\sqrt{1-\cos(x)\cos(y)}\rangle_{x,y}$). –  example Mar 30 '12 at 17:04
show 6 more comments

Let's look at some special cases:

  1. b constant, $a\rightarrow\infty$: The limit is roughly $6.01987$. As $a$ becomes large, the number of different phases between the two cosines increases so that it approaches the average of $\sqrt{1-\cos(x)\cos(y)}$.

  2. $a=b\rightarrow\infty$: You already concluded correctly, that this converges to 4.

  3. (this one is almost visible in your plot - if it were of a higher resolution) $a=n\cdot b$ with $b\rightarrow\infty$ and $n \in \mathbb{N}$. These have unique values that differ from $6.01987...$.

    Eg. $n=2$ converges to $f(a,2a)=\frac{4\sqrt{10}}{3}\approx 4.22$

    $n=3$ converges to $f(a,3a)=2\sqrt{5}+\text{ArSinh}(2)\approx 5.91577$

It appears to me, that the right way to go to $\infty$ is by fixing a ratio $a/b=:r$. Once this ratio is set, any increase in a (and thus b) can be neglected after reaching the least common multiple (if this LCM is $c$ then clearly $f(c,rc)=f(2c,2rc)=f(3c,3rc)=\dots$ ). Such numbers have a finite limit value $\neq 6.0198...$ .

Any two numbers with $\text{LCM}(a,b)=\infty\,\Leftrightarrow\,r\notin\mathbb{Q}$ converge to the abovementioned $6.01987...$ by the same argument. To see this lets define a new function $$ f'(a,b,\phi) = \int_0^{2\pi}\!\!\!\!\!\sqrt{1-\cos(a x+\phi)\cos(b x)}\,\text{d}x $$ $$ \Rightarrow f'(a+1,r(a+1),\phi) = \frac{a}{a+1}f'(a,ra,\phi) + \frac{1}{a+1}f'(1,r,\phi'(\phi,r,a))$$ for fitting $\phi'$. The series of all $\phi,\phi',\phi'',\dots$ repeats for $r\in\mathbb{Q}$ after $c$ terms, so that the resulting limit is just equal to the average of these $c$ (different) integrals over one period of $\cos(a x)$. The larger the LCM is, the closer we come to the actual average of $6.01987...$, especially for $r\notin\mathbb{Q}$ the limit is identical to this number.

Sidenote: The limit need not depend continuously on $a/b$ even though $f(a,b)$ is continuous in both arguments.

share|improve this answer
    
You posted this answer after my answer, but you didn't comment on why you disagree with it. Why do you think that my argument against $2\pi$ is incorrect; and does it seem to you in the image that the asymptotic value is closer to $2\pi$ than to $6.01987$? I don't find your "on average so to say" argument for $2\pi$ convincing; the average of a function is generally not the function value at the average argument. –  joriki Mar 28 '12 at 12:35
    
Ok, sorry, I actually missed that (and misunderstood that point in your answer). When I considered this I just approximated $\sqrt{1-x}\approx 1-0.5 x+\dots$ which would have made the average of the function roughly the function value at the average argument. My point about the difference of $a/b\in\mathbb{Q}$ vs. $a/b\notin\mathbb{Q}$ remains valid though. One should just add, that the limit in the latter case is not $2\pi$ but rather 6.01987... –  example Mar 28 '12 at 12:52
    
I only got notified of your comment because my comment was the only one so far under this answer; usually you have to use the @username idiom to ping someone if you want them to get notified of a comment that isn't under one of their posts. It's not clear to me what you mean by "The limit value need not depend continuously on $a$ and $b$" -- if $a$ and $b$ are still free to vary, then what limit is this? –  joriki Mar 28 '12 at 13:07
    
@joriki I just edited my answer to give the correct average in the case of $a/b\notin\mathbb{Q}$. It now looks much more like your answer, sry again that I didn't see that earlier (or I might just have commented you answer instead of writing a new one) When talking about the difference between $a/b\in\mathbb{Q}$ or $\notin\mathbb{Q}$ you argued with continuity (comments under your answer). This is correct for finite $a$ and $b$ but not as you take the limit $\rightarrow\infty$. In fact I calculated some examples of function values for ratios $a/b$ that do not converge to anything near 6.01987 –  example Mar 28 '12 at 13:35
    
I didn't (or at least didn't intend to) claim that the limit for fixed ratio $\alpha/\beta$ (not sure why you renamed them $a$ and $b$) is always $6.01987$. This is just the limit as $\alpha\to\infty$ for fixed $\beta$, and it seems we agree on that. Regarding the continuity argument, I still don't understand what you mean because you didn't answer the question in my last comment: What do you mean by "The limit value need not depend continuously on a and b" -- if a and b are still free to vary, then what limit is this? –  joriki Mar 28 '12 at 13:54
show 4 more comments

Greg has already indicated in a comment why the value for $\alpha=\beta$ is lower. In this case the phases of the two cosines are maximally correlated and their product is non-negative; in fact the integrand simplifies to $|\sin\alpha x|$ in this case. For $\alpha,\beta,|\alpha-\beta|\gg1$, on the other hand, the phases of the cosines are approximately uncorrelated (in fact, for incommensurable $\alpha,\beta$ they would come arbitrarily close to every pair of phases if we extend the integral to infinity). I don't think the asymptotic value is $2\pi$; it should be $2\pi$ times the average of $\sqrt{1-\cos x\cos y}$ over full periods of $x$ and $y$. This is approximately $6.01987$, which seems to agree with your image.

Regarding the frequency of the oscillations, it looks like it's simply $1$, which would make sense, since we add one full period of $\cos\alpha x$ when we increase $\alpha$ by $1$. The integrand can be written as $\sqrt{1-(\cos(\alpha+\beta)x+\cos(\alpha-\beta)x)/2}$, and both of these cosines are at their maximum at $2\pi$ when $\alpha$ is an integer (because you've chosen $\beta$ as an integer). On that basis, my guess for the shape of the envelope would be $I\pm c/|\alpha-\beta|$, where $I$ is the average value and $c$ some constant.

share|improve this answer
    
Do you expect a difference between $\alpha/\beta \in \mathbb{Q}$ and $\alpha/\beta \notin \mathbb{Q}$? –  Fabian Mar 26 '12 at 17:33
    
@Fabian: No, that was just a side comment; the integral should depend continously on $\alpha$ and $\beta$, so there can't be such a difference. –  joriki Mar 27 '12 at 2:13
    
@Fabian: User example makes a valid point about the limit for fixed $\alpha/\beta$; my comment on the integral depending continuously on $\alpha$ and $\beta$ doesn't preclude that limit from behaving differently for rational or irrational $\alpha/\beta$. –  joriki Mar 28 '12 at 14:23
    
I wish a bounty could be shared. Your answer helped a lot but David's answer was more useful for me... –  Fabian Apr 2 '12 at 14:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.