User:J D Bowen/sandbox2

=Homework Ch. 7=

pg 126
We are asked to complete the proof of the isoperimetric inequality by demonstrating that all of the terms in the sum

$$4\sum_{j=0}^{k-1} \sin \left({ \frac{\pi j}{k} }\right) \left({ \sin \left({\frac{\pi j}{k} }\right) - \tan \left({\frac{\pi}{k}}\right)\cos\left({\frac{\pi j}{k}}\right) }\right) |\hat{z}(j)|^2 \ $$

are non-negative. Of course, since $$|\hat{z}(j)|^2 \ $$ is always non-negative and since $$\sin\left({ \frac{\pi j}{k} }\right) \ $$ is non-negative for $$j\in\left\{{0,1,\dots,k-1}\right\} \ $$, we need only show

$$\sin \left({\frac{\pi j}{k} }\right) - \tan \left({\frac{\pi}{k}}\right)\cos\left({\frac{\pi j}{k}}\right)\geq 0 \ $$,

at least for $$j\in\left\{{1,\dots,k-1}\right\} \ $$ (we need not consider the term $$j=0 \ $$, since for this the coefficient of this whole expression, $$\sin (\pi j/k) \ $$, vanishes).

For $$j=1 \ $$, this expression reduces 0 fairly easily.

If we make the fairly reasonable assumption that $$k> 2 \ $$ (reasonable since we are dealing with a polygon with k vertices), we can reduce this expression with trigonometric identities:

$$\sin \left({\frac{\pi j}{k} }\right) - \tan \left({\frac{\pi}{k}}\right)\cos\left({\frac{\pi j}{k}}\right)

=\left({ \frac{ \cos (\tfrac{\pi}{k}) }{ \cos (\tfrac{\pi}{k}) } }\right)\sin \left({\frac{\pi j}{k} }\right) - \left({ \frac{ \sin (\tfrac{\pi}{k}) }{ \cos (\tfrac{\pi}{k}) } }\right)\cos\left({\frac{\pi j}{k}}\right)

=\frac{\sin \left({ (j-1)\frac{\pi }{k} }\right)}{\cos(\tfrac{\pi }{k})} \ $$

As we stated before, we do not have to consider the term $$j=0 \ $$ any longer, which is fortunate, because then $$\sin \left({ (j-1)\frac{\pi }{k} }\right) \geq 0 $$, since the sine's argument will always be less than $$\pi \ $$. Additionally, since $$k>2, \cos(\tfrac{\pi}{k})>0 \ $$ and so the quotient as a whole is >0. This completes the proof.