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.

Let $$f(a,b,\lambda,\gamma)=\frac{1}{2}[(1+\cos(\gamma))\cos(\lambda(b-a))+(1-\cos(\gamma))\cos(\lambda(b+a))],$$ for $a\ge 0, b\ge 0$.

I am sure that $$\lim_{(a,b)\to (0,0)} \frac{\cos^{-1}(f(a,b,\lambda,\gamma))}{\cos^{-1}(f(a,b,1,\gamma))}=|\lambda|$$ uniformly for $0< \gamma\le \pi$, but do not have a convincing proof. Does anyone have a good proof? Note that $$f(a,b,\lambda,\gamma)=\cos(\lambda a)\cos(\lambda b)+\sin(\lambda a)\sin(\lambda b)\cos(\gamma).$$

share|improve this question
1  
As $f(a,-a,\lambda,\pi)=1$ for all $\lambda$ you don't get $\lim_{(a,b)\to(0,0)}\bigl({\it your\ expression}\bigr)=\lambda$ when $\gamma=\pi$ –  Christian Blatter Nov 12 '11 at 11:26
    
The OP chose to delete some comments above, making the point of some of mine difficult to grasp. I deleted these. Note also that the question is now substantially different from its previous version (although this, I think, is frowned upon on the site). –  Did Nov 14 '11 at 19:25
add comment

3 Answers 3

up vote 2 down vote accepted

Big ${\bf Edit}\ $:

TCL is right; the convergence is uniform with respect to $\gamma\in\ ]0,\pi[\ $.

Put

$$a+b:=r \quad(r\geq0),\quad b-a:=t\ r \quad(|t|\leq1), \quad q:=\sin{\gamma\over2}\ .$$

In terms of the new variables $r\to 0$ and $t$ we have

$$f(a,b,\lambda,\gamma)=(1-q^2)\cos(\lambda t r) + q^2 \cos(\lambda r) = 1 -{\lambda^2\over2}(q^2+t^2-q^2 t^2) r^2 + O(r^4)\ .$$

Now we are talking about an angle $\phi(a,b,\lambda,\gamma)=:\phi_\lambda$ that goes to $0$ when $(a,b)\to(0,0)$, and whose cosine is given by $f(a,b,\lambda,\gamma)$. It follows that

$$\sin^2\phi_\lambda=1-\cos^2\phi_\lambda= \lambda^2(q^2+t^2-q^2 t^2) r^2 +O(r^4)\ .\qquad(*)$$

As $q>0$ we have $q^2+t^2>0$; so there is an $u>0$ and a $v\in\bigl]{-{\pi\over2}},{\pi\over2}\bigr[$ with $t=u\sin v$, $\ q= u \cos v$. Introducing $u$ and $v$ into $(*)$ we get

$$\sin^2\phi_\lambda ={1\over8}\lambda^2 u^2\bigl(8-u^2+u^2\cos(4v)\bigr)r^2 + O(r^4)\ .$$

Note that $8-u^2+u^2\cos(4v)$ is bounded away from zero for small $u$. Therefore we can write

$${\sin^2\phi_\lambda \over \sin^2\phi_1}=\lambda^2 + O(r^2)\ ,$$

where the implicit constant in the $O$-term does not depend on $u$ and $v$, i.e., on $\gamma$. It follows that

$$\lim_{r\to 0+}{\phi_\lambda \over \phi_1}\ =\ \lambda\ ,$$

uniformly with respect to $\gamma$.

share|improve this answer
    
@Christian.Your proof is short. Unfortunately it does not prove uniform convergence for $0<\gamma<\pi.$ Notice that I have changed my original problem: $a,b\ge 0$. –  TCL Nov 12 '11 at 21:33
    
When $a\geq0$, $b\geq0$ the right end $\gamma=\pi$ is no problem anymore. But you will not get uniform convergence for $\gamma$ near zero, since $f(a,a,\lambda,0)=1$ for all $\lambda>0$. –  Christian Blatter Nov 13 '11 at 19:15
    
Let $F(a,b,\lambda,\gamma)=\frac{\cos^{_1}(\cdots)}{\cos^{-1}(\cdots)}$. $F$ is undefined at $(a,a,\lambda,0)$. That does not mean that it cannot converges to $\lambda$ uniformly for $0<\gamma\le\pi$. I don't find anything wrong with my given proof. If you find anything not right there, I would be happy to know. Of course, one way to definitively refute my proof is to show a sequence $(a_n,b_n,\gamma_n)$ such that $(a_n,b_n)$ converges to 0, $0<\gamma\le\pi$ and $F(a_n,b_n,\lambda,\gamma_n)$ bounded away from $\lambda$. –  TCL Nov 14 '11 at 2:14
    
@TCL: I have rewritten my post. –  Christian Blatter Nov 14 '11 at 12:05
add comment

Because: $$ f(a,b,\lambda, \gamma) = \cos(\lambda(a-b)) - 2 \sin(\lambda a) \sin(\lambda b) \sin^2 \left(\frac{\gamma}{2} \right) $$ Hence, for small $a$ and $b$: $$ f(a, b, \lambda, \gamma) \sim 1 - \frac{1}{2}\lambda^2 (a-b)^2 - 2 \sin^2 \left(\frac{\gamma}{2} \right) \lambda^2 a b + o(a^2,ab,b^2) $$ Taking the principal value of $\arccos(f)$, we get: $$ \arccos(f(a,b,\lambda, \gamma)) \sim \vert \lambda\vert \sqrt{ (a-b)^2 + 4 a b \cdot \sin^2 \left(\frac{\gamma}{2} \right) } $$ So it results, that the limit is $\vert \lambda \vert$.


The value of the limit is $\vert \lambda \vert$ assuming that the $(0,0)$ point is not approached along the path, where $(a-b)^2 + 4 a b \cdot \sin^2 \left(\frac{\gamma}{2}\right) = 0$, which could happen only for $\gamma = \pi$ and $a=b$. In that case $ f(a,a, \lambda, \pi) = \cos( 2 \lambda a)$, and the limit equals $\lim_{a \to 0} \frac{\arccos(\cos(2 a \lambda))}{\arccos( \cos(2 a))} = \vert \lambda \vert$.

share|improve this answer
1  
Uniformly? $ $ $ $ –  Did Nov 11 '11 at 16:19
2  
If $a=b$ and $\gamma=\pi$, the ratio of arccosines is identically $|\lambda|$ hence the limit is $|\lambda|$. –  Did Nov 11 '11 at 17:48
add comment

This is to prove the uniform convergence using Sasha's approach. Write $g(a,b,\gamma)$ for $(a-b)^2+4ab\sin^2(\gamma/2)$. Then $$\lim_{(a,b)\to (0,0)}\frac{\cos^{-1}(f(a,b,\lambda,\gamma)}{\cos^{-1}(f(a,b,1,\gamma)}=|\lambda|\lim_{(a,b)\to (0,0)}\sqrt{\frac{g(a,b,\gamma)+o((a-b)^2)+o(ab)\sin^2(\gamma/2)} {g(a,b,\gamma)+o_1((a-b)^2)+o_1(ab)\sin^2(\gamma/2)}},$$ where $o,o_1$ are independent of $\gamma$. Here we are using the fact that $$\lim_{x\to 0}\frac{\cos^{-1}(1-x)}{\sqrt{2x}}=1.$$ Now divide the numerator and the denominator of the fraction inside the square root sign by $g(a,b,\gamma)$ (which is $>0$ since $(a,b)\neq (0,0)$ and $\gamma>0$). Note that $o((a-b)^2)=0$ if $a=b,$ and $o((a-b)^2)/g(a,b,\gamma)\le o((a-b)^2)/(a-b)^2\to 0.$

Also $$o(ab)\sin^2(\gamma/2)/g(a,b,\gamma)=\frac{o(ab)\sin^2(\gamma/2)}{(a-b)^2+4ab\sin^2(\gamma/2)}\le o(ab)/ab\to 0.$$ Same for $o_1$. So the limit converges uniformly for $0<\gamma\le\pi$ to $|\lambda| $ as $(a,b)\to (0,0)$.

Actually I also need to show that the two limits $$\lim_{(a,b)\to (0,0)}\frac{\cos^{-1}(f(a,b,\lambda,\gamma))}{\sqrt{2(1-f(a,b,\lambda,\gamma))}}$$ and $$\lim_{(a,b)\to (0,0)}\frac{\cos^{-1}(f(a,b,1,\gamma))}{\sqrt{2(1-f(a,b,1,\gamma))}}$$ also converge to 1 as $(a,b)\to (0,0)$ uniformly for $0<\gamma\le\pi$. But this is easier since it is equivalent to both $f(a,b,\lambda,\gamma)$ and $f(a,b,1,\gamma)$ converging to 1 uniformly for $0<\gamma\le\pi$.

share|improve this answer
    
Why are $o$ and $o_1$ independent o[n] $\gamma$ (I guess you mean uniform with respect to $\gamma$)? –  Did Nov 12 '11 at 10:21
    
$\cos(a-b)=1-(a-b)^2/2+o((a-b)^2), \sin(a)\sin(b)=(a-a^3/6+\cdots)(b-b^3/6+\cdots)=ab+o(ab)$, so $o$ and $o_1$ are independent of $\gamma$. –  TCL Nov 12 '11 at 12:48
    
*so $o$ and $o_1$ are independent o[n] $\gamma$*... Is this an act of faith or a mathematical statement? @Christian's remark that things go awry at $\gamma=\pi$ should have sobering effects on your enthousiasm, don't you think? –  Did Nov 12 '11 at 13:12
    
I don't understand your $o[n]\gamma$. –  TCL Nov 12 '11 at 13:34
    
Just read independent on $\gamma$. –  Did Nov 12 '11 at 13:45
show 3 more comments

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.