# Inequality with trigonometric relations

performing a Von Neumann analysis, I arrived to the relation

$$$$E_m(t+\Delta t)=\underbrace{\left[1-\frac{r}{4}\left(3-4e^{-i\theta}+e^{-2i\theta}\right)\right]}_{G}E_m(t),$$$$ where $$\theta\in[-\pi,\pi]$$ and $$r>0$$. The question is to find the largest $$r$$ such that the factor $$G$$ has absolute value less or equal than 1, i.e., $$|G|\leq 1$$.

I know that I can estimate $$r$$ numerically, but the idea is to find some analytical relation. I have spent quite some time looking to that expression without success, so, any help will be appreciated.

Update: The inequality $$|G|\leq 1$$ has to be valid for all values of $$\theta$$.

• The expression $3-4e^{-i\theta}+e^{-2i\theta}$ is equal to zero if $\theta=0$, which is within your allowed range, so there is no upper bound on the value of $r$. Sep 22, 2022 at 0:18
• note that $\theta =\pi$ gives $G=1-2r$ so one needs $r \le 1$; are you sure that there is such $r$ as some computations suggest that for every $r \in (0, 1]$ there is $\theta_r$ (given by $\cos \theta_r=\frac{4-4r}{4-3r}$) st $|G(\theta_r)| >1$ Sep 22, 2022 at 2:30
• @Conrad, thank you a lot. I only ask you to write your comment as a solution for me to close the question :) Another thing, can you give me some hints on how you arrived to that expression? Sep 22, 2022 at 20:16
• Done as requested - computed the maximum value of $|G|^2$ for $0 \le r \le 1$; the same analysis works for $r >1$ where we now we still have a $\theta_r$ for $1 \le r \le 8/7$ and the maximum of $|G|^2$ is the bigger of $\frac{r^3}{4(4-3r)}+1, (2r-1)^2$ (which coincide at $r=8/7$ not surprisingly as $\theta_{8/7}=\pi$), so most likely the maximum is still $\frac{r^3}{4(4-3r)}+1$) while for $r >8/7$ there is no more $\theta_r$ as $|\frac{4-4r}{4-3r}| >1$ so the maximum of $|G|$ is just $2r-1$ Sep 23, 2022 at 0:54

Note that for $$z=e^{-i\theta}$$ one has $$G(z)=1-\frac{r}{4}(z^2-4z+3)$$ so by the polar C-R the maximum of $$|G(z)|$$ when $$|z|=1$$ is attained at points where $$\frac{zG'(z)}{G(z)} \ge 0$$ (which of course means $$\frac{zG'(z)}{G(z)}$$ real)

Now $$G'(z)=\frac{r}{2}(2-z)$$ is non zero on $$|z|=1$$ so the above is equivalent to $$\frac{G(z)}{zG'(z)} > 0$$ and simplifying $$r/2 >0$$ one gets $$\frac{G(z)}{2z-z^2}>0$$ hence one needs $$\Im (G(z) (2\bar z-\bar z^2)) =0$$ so using $$z\bar z=1$$ one needs $$\Im (2\bar z-\bar z^2-\frac{r}{4}(2z+6\bar z+4\bar z-3\bar z^2))=0$$

But $$z=e^{-i\theta}$$ so $$\Im (2z+10\bar z)=8\sin \theta$$ hence we get:

$$2\sin \theta -\sin 2\theta-2r\sin \theta +\frac{3r}{4} \sin 2\theta=0$$ which means $$\sin \theta =0$$ or $$2-2\cos \theta-2r+\frac{3r}{2}\cos \theta=0$$

In particular for $$\theta =\pi$$ one gets that $$z=-1$$ and $$G(-1)=1-2r$$ hence we need $$r \le 1$$.

For $$r=1$$ one gets $$\cos \theta =0$$ as the other roots of the maximum modulus equation so $$z =\pm i$$ and $$G(\pm i)=1-\frac{1}{4}(2 \pm 4i)$$ has modulus greater than $$1$$ so $$r<1$$ and we need to study the points where $$\cos \theta =\frac{4-4r}{4-3r}=f(r)$$ so $$\sin^2 \theta =1-f(r)^2, \cos 2\theta =2f(r)^2-1, \sin 2\theta=2f(r)\sin \theta$$

Now (separating in real and imaginary parts) $$|G(r,\theta)|^2=(1-\frac{r}{2}(1-\cos \theta)^2)^2+\frac{r^2(2-\cos \theta)^2\sin^2 \theta}{4}$$ so substituting as above, one gets

$$|G_{\max}(r)|^2 \ge (1-\frac{r^3}{2(4-3r)^2})^2+\frac{r^2}{4}(2-\frac{4-4r}{4-3r})^2(1-\frac{(4-4r)^2}{(4-3r)^2})$$ or by further simplifications

$$|G_{\max}(r)|^2 \ge \frac{(18r^2+32-r^3-48r)^2+4r^3(2-r)^2(8-7r)}{4(4-3r)^4}$$

But (with the help of Wolfram Alpha) we note that:

$$\frac{(18r^2+32-r^3-48r)^2+4r^3(2-r)^2(8-7r)}{4(4-3r)^4}=\frac{(2-r)^2(r+4)}{4(4-3r)}=\frac{r^3}{4(4-3r)}+1$$

which shows that $$|G(r, \theta)|^2 \ge \frac{r^3}{4(4-3r)}+1$$ for $$0 \le r \le 1$$ and actually the analysis above shows that is the maximum for $$0 \le r \le 1$$ and it is bigger than $$1$$