SOLUTION: Stanford University Polar Coordinates & ?Orientations Diffeomorphism Proof
The solutions are fitted. Please ask if bigwig is unclear.
𝑥 3 − 3𝑥𝑦 2
𝑓(𝑥, 𝑦) = 𝑟 cos(3𝜃) = 2
, 𝑓(0,0) = 0.
𝑥 + 𝑦2
Proof. It is simpler to use polar coordinates short. For (𝑥1 , 𝑦1 ) = (0,0) we have
|𝑓(𝑥1 , 𝑦1 ) − 𝑓(𝑥2 , 𝑦2 )| = |𝑓(𝑥2 , 𝑦2 )| = |𝑟 cos(3𝜃)| ≤ 𝑟.
For (𝑟1 , 𝜃1 ) not at the derivation and any (𝑟2 , 𝜃2 ) deliberate a track from (𝑟2 , 𝜃2 ) to (𝑟1 , 𝜃1 ) consisting of two
parts: 𝑟 = 𝑟2 until we get the similar 𝜃 and then 𝜃 = 𝜃1 until we get the similar 𝑟:
We see |𝑓(𝑟1 , 𝜃1 ) − 𝑓(𝑟2 , 𝜃2 )| ≤ |𝑓(𝑟2 , 𝜃1 ) − 𝑓(𝑟2 , 𝜃2 )| + |𝑓(𝑟2 , 𝜃1 ) − 𝑓(𝑟1 , 𝜃1 )|.
Estimate the pristine summand:
|𝑓(𝑟2 , 𝜃1 ) − 𝑓(𝑟2 , 𝜃2 )| = 𝑟2 |cos(3𝜃1 ) − cos(3𝜃1 )| ≤ 3𝑟2 |𝜃2 − 𝜃1 |.
Note that it is 3 times the extension of the pristine multiply of the track.
Estimate the promote summand:
|𝑓(𝑟2 , 𝜃1 ) − 𝑓(𝑟1 , 𝜃1 )| = |(𝑟2 − 𝑟1 ) cos(3𝜃1 )| ≤ |𝑟2 − 𝑟1 |.
It is simply the extension of the promote multiply of the track.
Now we deficiency to appreciate the sum by 𝐾dist((𝑟1 , 𝜃1 ), (𝑟2 , 𝜃2 )) for some trustworthy 𝐾. Deliberate the bold
blue triangle. The angle betwixt pristine two sides is frequently oblite (we can frequently choose
𝜃1 , 𝜃2 : |𝜃2 − 𝜃1 | ≤ 𝜋). Therefore
dist((𝑟1 , 𝜃1 ), (𝑟2 , 𝜃2 )) = 𝑎2 + 𝑏 2 − 2𝑎𝑏 cos ∠𝑎𝑏 ≥ 𝑎2 + 𝑏 2 ,
dist((𝑟1 , 𝜃1 ), (𝑟2 , 𝜃2 )) ≥ √𝑎2 + 𝑏 2 .
One further being to do is to appreciate the pristine incurvation extension 𝐿1 by 𝑎. It is further or near plain that 𝐿1 ≤ 2 𝑎
(the whack contingency when ...