# A limit involving inverse cosine function

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

-
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

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 $\gamma$.

-
@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

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 $\lim_{a \to 0} \frac{\arccos(\cos(2 a \lambda))}{\arccos( \cos(2 a))} = \vert \lambda \vert$.

-
Uniformly?   –  Did Nov 11 '11 at 16:19
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

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 $0<\gamma\le\pi$.

-
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