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In course of a particular research I ran into these two inequalities which I would like to have some help with.

$r,R>0$ for both the questions.

  • Is there a function $M(r)$ for which this inequality is satisfied?

$$\cos^{-1}\left (\frac{4M(r)}{r}-1 \right ) + \sqrt{1- \left ( \frac{4M(r)}{r}-1 \right )^2} > \pi$$

  • Let $t_0$ be the minimum positive solution of $a \cosh (\omega t) + b \sinh (\omega t) = 0$.

Then for $-B<0$ and $t>0$, is it possible to choose the values of variables $a,b$ and $g>0$ such that the expression multiplying $-B$ is negative for $t \leq t_0$ and $r>gR$?

$$-B\frac {(a\sqrt{r^2-g^2R^2} \mp bgr\omega)\cosh (\omega t)+ (b\sqrt{r^2-g^2R^2} \mp agr\omega)\sinh(\omega t)}{r\sqrt{r^2-g^2R^2}(a \cosh(\omega t) + b \sinh (\omega t))^{2g+1}}$$

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Please put a backslash before function names like "cos" and "cosh"; that will make it easier to read ($\cos x$ instead of $cos x$). –  Hans Lundmark Jan 19 '11 at 13:42
    
@Hans @Anirbit Fixed. Also made displayed equation actually displayed to improve legibility. –  Willie Wong Jan 19 '11 at 15:30
    
Aren't the tags real-analysis and algebra-precalculus kind of contradictory? –  Aryabhata Jan 19 '11 at 22:12

2 Answers 2

For the first, let $(\frac{4M(r)}{r}-1)=z$ You want $\cos^{-1}z+\sqrt{1-z^2}>\pi$ Wolfram Alpha says the inverse cosine near -1 is less than $\pi-\sqrt{2(z+1)}$ so you want $\sqrt{1-z^2}-\sqrt{2(z+1)}>0$ or $\sqrt{z+1}(\sqrt{1-z}-\sqrt{2})>0$, which is never true because $z>-1$. If you pick a different branch of inverse cosine, it can be true easily.

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Can you explain what is that "Alpha" in the first line? –  Anirbit Jan 19 '11 at 17:44
    
@Anirbit: I assume Wolfram Alpha –  Willie Wong Jan 19 '11 at 17:48
    
@Willie: Correct. Added –  Ross Millikan Jan 19 '11 at 18:21

First inequality

In order for the inequality to make sense, I assume that $M(r)\le r/2$. Let $\theta$ be such that $$\cos\theta=\frac{4M(r)}{r}-1.$$ Then what you want is $\theta+|\sin\theta|>\pi$. This cannot be achieved with the usual branch of $\cos^{-1}(x)$, which takes values in $[0,\pi/2]$ for $x\in[0,1]$ and in $[\pi/2,\pi]$ for $x\in[-1,0]$.

Edit You con take for instance $\arccos_{1}x=2\pi+\cos^{-1}x$, where $\cos^{-1}$ is the usual branch. Then $$\arccos_{1}x+\sqrt{1-x^2}>\pi\quad\forall x \in[-1,1].$$

Then any $M(r)$ such that $M(r)\le r/2$ will do.

I have also included absolte values around the $\sin\theta$ in the previous paragraph.

Second inequality

If $a,b>0$, then $a\cosh t+b\sinh t\ge a>0$ for all $t\ge0$, so that $t_0$ is not defined.

Edit The O.P. has edited the question in the sense that $a$ and $b$ are such that $a\cosh t+b\sinh t=0$ has a positive solution. The question remains open.

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looks like the OP edited the question so that $a,b$ are not restricted to be positive. –  Willie Wong Jan 19 '11 at 17:46
    
What would be the branch of $\cos^{-1}$ that you would suggest? I guess the following works, $[\pi, \frac{3\pi}{2}]$ for $x \in [-1,0]$ and in $[\frac{3\pi}{2},2\pi]$ for $x \in [0,1]$. This I guess doesn't put any further restriction on $M(r)$. –  Anirbit Jan 19 '11 at 17:50
    
Sorry for my weird notation. I only meant $g>0$. Of course if $a$ and $b$ are both positive then it makes no sense. Also one would need to adjust the values of $a$ and $b$ such that $t_0$ exists. (In general it may not) Hence assume that $a$ and $b$ are of such chosen values for which there is a well-defined $t_0$. –  Anirbit Jan 19 '11 at 17:53

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