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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 $f(a,b,\lambda,\gamma)=\cos(\lambda a)\cos(\lambda b)+\sin(\lambda a)\sin(\lambda b)\cos(\gamma).$

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    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).2011-11-14

3 Answers 3

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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 $0<\gamma\le\pi$.

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    Notice that $\gamma$ does not appear in both $o$ and $o_1$.2011-11-12
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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 $\lim_{a \to 0} \frac{\arccos(\cos(2 a \lambda))}{\arccos( \cos(2 a))} = \vert \lambda \vert$.

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    If $a=b$ and $\gamma=\pi$, the ratio of arccosines is identically $|\lambda|$ hence the limit is $|\lambda|$.2011-11-11
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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 $\gamma$.

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    @TCL: I have rewritten my post.2011-11-14