User:J D Bowen/sandbox2

=Homework Ch. 7 & 8=

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.

pg. 130
We are asked to demonstrate Eulers criterion.

Since the square-roots of 1 are $$1, -1 \ (\text{mod} \ p) \ $$, and since $$a^{p-1}= 1 \ (\text{mod} \ p), a^{(p-1)/2} \ $$ is either $$1, -1 \ (\text{mod} \ p) \ $$, and so Eulers criterion is equivalent to stating that a is a quadratic residue modulo p if and only if $$a^{(p-1)/2} =1 \ (\text{mod} \ p) \ $$.

We prove each direction separately.

($$\Rightarrow \ $$) Assume a is a quadratic residue modulo p. We pick k such that $$k^2=a \ (\text{mod} \ p) \ $$. Then

$$a^{(p-1)/2}=k^{p-1}=1 \ (\text{mod} \ p) \ $$.

($$\Leftarrow \ $$) Assume $$a^{(p-1)/2} =1 \ (\text{mod} \ p) \ $$. Then let $$y \ $$ be a primitive root modulo p, so that a can be written as $$y^j \ $$. In particular, $$y^{j(p-1)/2} =1 \ (\text{mod} \ p) \ $$. By Fermat's little theorem, $$p-1 | j(p-1)/2 \ $$, so j must be even. Let $$k=y^{j/2} \ (\text{mod} \ p) \ $$. We have $$k^2=y^j= a \ (\text{mod} \ p) \ $$.

pg. 134
We are asked to prove lemma 3, namely:

Suppose p and q are distinct odd primes. Set

$$p^{*}=(-1)^{(p-1)/2}p \ $$.

Show the Gauss sum g satisfies

$$g^{q-1}=\left({ \frac{p^{*}}{q} }\right) \ (\text{mod} \ q) \ $$.

The Gauss sum $$g=\hat{h}_p(-1) = \sum_{a=1}^{p-1} \left({\frac{a}{p}}\right)\text{exp}\left({\frac{2\pi ia}{p}}\right) \ $$, and we have use of lemma 2, which tells us

$$g^2=(-1)^{(p-1)/2}p \ $$.

Since q is odd, we have

$$g^{q-1} = (g^2)^{(q-1)/2} = (-1)^{(p-1)(q-1)/2}p^{(q-1)/2} \ $$

and of course, by Eulers criterion, we have

$$\left({ \frac{p^{*}}{q} }\right) = ((-1)^{(p-1)/2}p)^{(q-1)/2} =(-1)^{(p-1)(q-1)/2}p^{(q-1)/2}$$

since $$p^*=(-1)^{(p-1)/2}p \in \left\{{p,q-p}\right\} \ $$, neither of which can divide q.

pg. 136
The use of p and q being prime in the proof of the QRL is in the application of lemma 3 (see proof, above); when we divided p out of the congruence (trivially), and when we applied lemma 4, which relies on q being prime for the division of the binomial coefficients $$qCn$$ by q.

Attempt 1
The Gauss sum is defined

$$G_p(x) = \sum_{k\in\mathbb{Z}_p} \text{exp}\left({ \frac{2\pi i k^2x}{p} }\right) \ $$

Show that if p does not divide x, then

$$\hat{h}_p(-x)=G_p(x) \ $$

We compute the transform:

$$\hat{h}_p(-x) = \sum_{a=0}^{p-1} h_p(a)\text{exp}\left({\frac{2\pi i x a}{p} }\right) = \sum_{a=1}^{p-1} \left({\frac{a}{p}}\right) \text{exp}\left({\frac{2\pi i x a}{p} }\right) \ $$

which is simply

$$= \sum_{a=1}^{p-1} \begin{cases} \text{exp}\left({\frac{2\pi i x a}{p} }\right) \text{ if } \exists k:k^2=a \pmod{p}\\ -\text{exp}\left({\frac{2\pi i x a}{p} }\right) \text{ else } \end{cases} \ $$

(From here on we simply use $$m\in\mathbb{Z}_p \ $$, with the understanding the sum omits $$m=0 \ $$). Since exactly half of $$\mathbb{Z}_p$$ are quadratic residues and each has two possible solutions,

$$G_p(x)=2\sum_{a:a=k^2,k\in\mathbb{Z}_p} \text{exp}\left({ \frac{2\pi i ax}{p} }\right) \ $$

and so

$$G_p(x)-\hat{h}_p(-x)=\sum_{a:a=k^2,k\in\mathbb{Z}_p} \text{exp}\left({ \frac{2\pi i ax}{p} }\right)-\sum_{b:b\neq k^2,k\in\mathbb{Z}_p} -\text{exp}\left({ \frac{2\pi i bx}{p} }\right) \ $$.

So we need only show that

$$\sum_{b:b\neq k^2,k\in\mathbb{Z}_p} -\text{exp}\left({ \frac{2\pi i bx}{p} }\right)=\sum_{a:a=k^2,k\in\mathbb{Z}_p} \text{exp}\left({ \frac{2\pi i ax}{p} }\right) \ $$

and the proof will be complete.


 * hmmm... once we consider that n  is a quadratic residue if and only if p-n is not, this seems to lead to


 * $$\overline{\sum_{b:b\neq k^2,k\in\mathbb{Z}_p} \text{exp}\left({ \frac{2\pi i bx}{p} }\right)}=\sum_{a:a=k^2,k\in\mathbb{Z}_p} \text{exp}\left({ \frac{2\pi i ax}{p} }\right) \ $$,


 * which is not what we want...

Attempt 2
The Gauss sum is defined

$$G_p(x) = \sum_{k\in\mathbb{Z}_p} \text{exp}\left({ \frac{2\pi i k^2x}{p} }\right) \ $$

Show that if p does not divide x, then

$$\hat{h}_p(-x)=G_p(x) \ $$

We take the transform from lemma 1 and the calculation of $$\hat{h}_p(-1)$$ (equation 6):

$$\hat{h}_p(-x) = h_p(x)\hat{h}(-1)= \left({ \frac{x}{p} }\right) \sum_{a=1}^{p-1}\left({ \frac{a}{p} }\right) \text{exp}\left({\frac{2\pi i xa}{p} }\right) = \sum_{a=1}^{p-1}\left({ \frac{xa}{p} }\right) \text{exp}\left({\frac{2\pi i xa}{p} }\right) $$