Answer:

$cos~x + cos~y$

Solution:

Approximate the x and y intercepts of the loop as $\frac{\pi}{3}$ (since $2^{-\frac{8}{5}} \simeq \frac{1}{3.03}$). Then we need $h(\pm \frac{\pi}{3}, \pm \frac{\pi}{3}) = 1$, so to turn those $\pm\frac{\pi}{3}$s into something reasonable we probably need $cos~x$ (there are other ways, of course, but it was hinted that we need trig and also it takes care of both positive and negative very compactly: $cos~\frac{\pi}{3} = cos~\frac{-\pi}{3} = \frac{1}{2}$). It's likely to be symmetric in $x$ and $y$ so try out $cos~x + cos~y$ for now. We've got the loop; now to check what that looks like when $h = 0$ so we can modify it for the square and woah we're already done.

Answer:

$\cos~x + \cos~y$

Solution:

Approximate the x and y intercepts of the loop as $\frac{\pi}{3}$ (since $2^{-\frac{8}{5}} \simeq \frac{1}{3.03}$). Then we need $h(\pm \frac{\pi}{3}, \pm \frac{\pi}{3}) = 1$, so to turn those $\pm\frac{\pi}{3}$s into something reasonable we probably need $\cos~x$ (there are other ways, of course, but it was hinted that we need trig and also it takes care of both positive and negative very compactly: $\cos~\frac{\pi}{3} = \cos~\frac{-\pi}{3} = \frac{1}{2}$). It's likely to be symmetric in $x$ and $y$ so try out $\cos~x + \cos~y$ for now. We've got the loop; now to check what that looks like when $h = 0$ so we can modify it for the square and woah we're already done.

Answer:

$$cos~x + cos~y$$

Solution:

Approximate the x and y intercepts of the loop as $$\frac{\pi}{3}$$ (since $$2^{-\frac{8}{5}} \simeq \frac{1}{3.03}$$). Then we need $$h(\pm \frac{\pi}{3}, \pm \frac{\pi}{3}) = 1$$, so to turn those $$\pm\frac{\pi}{3}$$s into something reasonable we probably need $$cos~x$$ (there are other ways, of course, but it was hinted that we need trig and also it takes care of both positive and negative very compactly: $$cos~\frac{\pi}{3} = cos~\frac{-\pi}{3} = \frac{1}{2}$$). It's likely to be symmetric in $$x$$ and $$y$$ so try out $$cos~x + cos~y$$ for now. We've got the loop; now to check what that looks like when $$h = 0$$ so we can modify it for the square and woah we're already done.

Answer:

$cos~x + cos~y$

Solution:

Approximate the x and y intercepts of the loop as $\frac{\pi}{3}$ (since $2^{-\frac{8}{5}} \simeq \frac{1}{3.03}$). Then we need $h(\pm \frac{\pi}{3}, \pm \frac{\pi}{3}) = 1$, so to turn those $\pm\frac{\pi}{3}$s into something reasonable we probably need $cos~x$ (there are other ways, of course, but it was hinted that we need trig and also it takes care of both positive and negative very compactly: $cos~\frac{\pi}{3} = cos~\frac{-\pi}{3} = \frac{1}{2}$). It's likely to be symmetric in $x$ and $y$ so try out $cos~x + cos~y$ for now. We've got the loop; now to check what that looks like when $h = 0$ so we can modify it for the square and woah we're already done.