Suppose that $\tau>0$ and $r>0$ are real parameters, and define $\alpha:\mathbb{R}^{+}\mapsto\mathbb{R}$ to be $$ \alpha(\omega):=\frac {[(\omega^2+r)\cos^2(\omega\tau)+\frac{\omega}{2}(1-r)\sin(2\omega\tau)-r](\omega^2+r^2)} {[(r^2-\omega^2)\cos^2(\omega\tau)+\omega r\sin(2\omega\tau)-r^2]r}\;. $$ If it exists at all, what is a lower bound for this quantity? So far, I have found (I could be wrong) the following lower bounds for the numerator and the denominator, respectively. $$ [(\omega^2+r)\cos^2(\omega\tau)+\frac{\omega}{2}(1-r)\sin(2\omega\tau)-r](\omega^2+r^2)\geq -\left(r+\frac{\omega}{2}|1-r|\right)\left(\omega^2+r^2\right)\;,\\ [(r^2-\omega^2)\cos^2(\omega\tau)+\omega r\sin(2\omega\tau)-r^2]r\geq -(\omega+r)r^2\;. $$ Is there a way to use these results to find a lower bound for $\alpha(\omega)$?
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- $\begingroup$ What's the backstory for this expression? There may be a geometric interpretation that helps show a bound (or lack thereof). $\endgroup$Blue– Blue2025-11-02 06:58:01 +00:00Commented Nov 2 at 6:58
- $\begingroup$ It is part of a special pair of `parametric equations' that are parameterized by $\omega>0$. The other one is $$\beta(\omega):=\frac{\omega(1+r)}{r\sin(\omega\tau)-\omega\cos(\omega\tau)}\;.$$ For fixed $r$ and $\tau$, one can plot these parametric equations in the $\alpha\beta$-plane to see what they look like. The limit as $\omega\rightarrow +\infty$ of $\alpha(\omega)$ is $-\infty$. $\endgroup$user775349– user7753492025-11-02 07:47:09 +00:00Commented Nov 2 at 7:47
- $\begingroup$ Please add such context into the body of the question. (Comments are easily overlooked and may be hidden.) FYI: The expression reduces to $$\alpha(w) = \frac{w^2+r^2}{r}\frac{ \phantom{r}\sin wt + w\cos wt}{r \sin wt - w\cos wt} $$ $\endgroup$Blue– Blue2025-11-02 07:49:02 +00:00Commented Nov 2 at 7:49
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