$ c^2 = a^2 + b^2 $ factors as $ (a+ib)(a-ib) $ so let $ a+ib = (p+iq)^2 = p^2-q^2 + i(2pq) $ and $ a-ib = (p-iq)^2 = p^2-q^2 - i(2pq) $ now $ c = (p+iq)(p-iq) = p^2 + q^2 $.

In conclusion, if we pick $ p $, $ q$ then ($ p^2 - q^2 $, $ 2pq $, $ p^2 + q^2 $) forms a pythagorean triple.

Number theory notes

# Natural Deductions

Expository preaching, non-rigorous proofs

## Saturday, 5 June 2010

## Saturday, 22 May 2010

## Thursday, 22 April 2010

### Cauchy Product

We have two convergent series,

$ A_n = \sum^n_{k=0} a_k $ & $ B_n = \sum^n_{k=0} b_k $.

Theorem: Cauchy Product $ C_\infty = A_\infty B_\infty = \sum_{k=0} \sum_{i=0..k} a_i b_{k-i} $.

Demonstration: The product is the sum of the sums anti-diagonals of this table:

$

\begin{tabular}{ c | c | c | c | c }

(a_0 b_0) & (a_1 b_0) & (a_2 b_0) & (a_3 b_0) & a_4 b_0 & \\ \hline

(a_0 b_1) & (a_1 b_1) & (a_2 b_1) & a_3 b_1 & a_4 b_1 & \\ \hline

(a_0 b_2) & (a_1 b_2) & a_2 b_2 & a_3 b_2 & a_4 b_2 & \cdots \\ \hline

(a_0 b_3) & a_1 b_3 & a_2 b_3 & a_3 b_3 & a_4 b_3 & \\ \hline

a_0 b_4 & a_1 b_4 & a_2 b_4 & a_3 b_4 & a_4 b_4 & \\ \hline

& & \vdots & & \ddots & \\

\end{tabular}

$

Bracketed is $ C_3 $, now $A_n B_n \leq C_{2n} \leq A_{2n} B_{2n} $ so $ C_\infty $ converges. $ \qed $

Cauchy Product - Planet Math

The theorem of Mertens about the Cauchy product of infinite series - Harold P. Boas

$ A_n = \sum^n_{k=0} a_k $ & $ B_n = \sum^n_{k=0} b_k $.

Theorem: Cauchy Product $ C_\infty = A_\infty B_\infty = \sum_{k=0} \sum_{i=0..k} a_i b_{k-i} $.

Demonstration: The product is the sum of the sums anti-diagonals of this table:

$

\begin{tabular}{ c | c | c | c | c }

(a_0 b_0) & (a_1 b_0) & (a_2 b_0) & (a_3 b_0) & a_4 b_0 & \\ \hline

(a_0 b_1) & (a_1 b_1) & (a_2 b_1) & a_3 b_1 & a_4 b_1 & \\ \hline

(a_0 b_2) & (a_1 b_2) & a_2 b_2 & a_3 b_2 & a_4 b_2 & \cdots \\ \hline

(a_0 b_3) & a_1 b_3 & a_2 b_3 & a_3 b_3 & a_4 b_3 & \\ \hline

a_0 b_4 & a_1 b_4 & a_2 b_4 & a_3 b_4 & a_4 b_4 & \\ \hline

& & \vdots & & \ddots & \\

\end{tabular}

$

Bracketed is $ C_3 $, now $A_n B_n \leq C_{2n} \leq A_{2n} B_{2n} $ so $ C_\infty $ converges. $ \qed $

Cauchy Product - Planet Math

The theorem of Mertens about the Cauchy product of infinite series - Harold P. Boas

### Infinity factorial and Wallis' formula

In his second letter to Hardy, Ramanujan stated:

$ 1 + 2 + 3 + 4 + \cdots = -\frac{1}{12}$

Hardy could tell Ramanujan was a genius from this equation because he must have developed the theory of the zeta function, and its functional equation.

In this excellent note by Dror Bar-Natan, we start with the following equations:

$ 1 \times 2 \times 3 \times 4 \times \cdots = \sqrt{2 \pi}$ $ 1 + 1 + 1 + 1 + \cdots = -\frac{1}{2}$

and are asked to derive the Wallis' formula for $\pi$. This will use a couple of double factorial identities.

$ \frac{\pi}{2} $

$ = \frac{1}{4} \times 2 \pi $

$ = {2^{-\frac{1}{2}}}^4 \times {\infty !}^2 $

$ = \frac{{2^{-\infty}}^4 {\infty !}^4}{{\infty !}^2} $

$ = \frac{{\infty !!}^4}{{\infty !}^2} $

$ = \frac{2 \times 2 \times 2 \times 2 \times 4 \times 4 \times 4 \times 4 \times 6 \times 6 \times 6 \times 6 \cdots}{1 \times 2 \times 2 \times 3 \times 3 \times 4 \times 4 \times 5 \times 5 \times 6 \times 6 \times 7 \times 7 \cdots} $

$ = \frac{2}{1} \frac{2}{3} \frac{4}{3} \frac{4}{5} \frac{6}{5} \frac{6}{7} \cdots \qed $

What is infinity factorial (and why might we care)?

Srinivasa Ramanujan - G. H. Hardy

The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation - Terence Tao

On the Number of Primes Less Than a Given Magnitude - Bernhard Riemann

Double Factorial - Mathworld

$ 1 + 2 + 3 + 4 + \cdots = -\frac{1}{12}$

Hardy could tell Ramanujan was a genius from this equation because he must have developed the theory of the zeta function, and its functional equation.

In this excellent note by Dror Bar-Natan, we start with the following equations:

$ 1 \times 2 \times 3 \times 4 \times \cdots = \sqrt{2 \pi}$ $ 1 + 1 + 1 + 1 + \cdots = -\frac{1}{2}$

and are asked to derive the Wallis' formula for $\pi$. This will use a couple of double factorial identities.

$ \frac{\pi}{2} $

$ = \frac{1}{4} \times 2 \pi $

$ = {2^{-\frac{1}{2}}}^4 \times {\infty !}^2 $

$ = \frac{{2^{-\infty}}^4 {\infty !}^4}{{\infty !}^2} $

$ = \frac{{\infty !!}^4}{{\infty !}^2} $

$ = \frac{2 \times 2 \times 2 \times 2 \times 4 \times 4 \times 4 \times 4 \times 6 \times 6 \times 6 \times 6 \cdots}{1 \times 2 \times 2 \times 3 \times 3 \times 4 \times 4 \times 5 \times 5 \times 6 \times 6 \times 7 \times 7 \cdots} $

$ = \frac{2}{1} \frac{2}{3} \frac{4}{3} \frac{4}{5} \frac{6}{5} \frac{6}{7} \cdots \qed $

What is infinity factorial (and why might we care)?

Srinivasa Ramanujan - G. H. Hardy

The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation - Terence Tao

On the Number of Primes Less Than a Given Magnitude - Bernhard Riemann

Double Factorial - Mathworld

### Euler Zeta Function

Definition: $ \zeta(s) = 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \cdots $ (1).

Theorem: $ \zeta(s) = \frac{1}{1 - \frac{1}{2^s}} \times \frac{1}{1 - \frac{1}{3^s}} \times \frac{1}{1 - \frac{1}{5^s}} \times \frac{1}{1 - \frac{1}{7^s}} \times \cdots$ (2).

Demonstration: Expand each $\frac{1}{1 - p^s} = 1 + p^s + p^{2s} + p^{3s} + \cdots$ in (2) and multiply out the infinite product. We are left with the sum of reciprocals of every product of powers of primes (to the s), The Fundamental Theorem of Arithmetic shows that this gives each natural number in one way and concludes the proof. $\square$

Equation (1) shows us that $ \zeta(1) $ is the harmonic series which diverges, but (2) could not diverge unless there were an Infinitude of Primes. We pointed the zeta laser cannon at the number 1 and fired it, this instantly killed Euclids theorem about the number of primes.

Dirichlet's theorem states on the arithmetic progression (for $(a,k) = 1$) $ a, a + k, a + 2k, a + 3k, \cdots $ there are infinitely many primes. Let us recalibrate our weapon so it can operate on these arithmetic progressions rather than $ 1, 2, 3, 4, \cdots $.

Definition: $ \zeta(s,\chi) = \frac{\chi(1)}{1} + \frac{\chi(2)}{2^s} + \frac{\chi(3)}{3^s} + \frac{\chi(4)}{4^s} + \cdots$ (3).

Lemma: $ \zeta(s,\chi) = \frac{1}{1 - \frac{\chi(2)}{2^s}} \times \frac{1}{1 - \frac{\chi(3)}{3^s}} \times \frac{1}{1 - \frac{\chi(5)}{5^s}} \times \frac{1}{1 - \frac{\chi(7)}{7^s}} \times \cdots$ (4).

Theorem: Dirichlet's theorem on arithmetic progressions.

Demonstration: Pick $ \chi $ so that it corresponds to the arithmetic progression, only finitely many terms are removed from (4) so divergence again lets us conclude the proof. $\square$

How Euler discovered the zeta function - Keith Devlin

Introduction to Analytic Number Theory, Lecture Notes

Theorem: $ \zeta(s) = \frac{1}{1 - \frac{1}{2^s}} \times \frac{1}{1 - \frac{1}{3^s}} \times \frac{1}{1 - \frac{1}{5^s}} \times \frac{1}{1 - \frac{1}{7^s}} \times \cdots$ (2).

Demonstration: Expand each $\frac{1}{1 - p^s} = 1 + p^s + p^{2s} + p^{3s} + \cdots$ in (2) and multiply out the infinite product. We are left with the sum of reciprocals of every product of powers of primes (to the s), The Fundamental Theorem of Arithmetic shows that this gives each natural number in one way and concludes the proof. $\square$

Equation (1) shows us that $ \zeta(1) $ is the harmonic series which diverges, but (2) could not diverge unless there were an Infinitude of Primes. We pointed the zeta laser cannon at the number 1 and fired it, this instantly killed Euclids theorem about the number of primes.

Dirichlet's theorem states on the arithmetic progression (for $(a,k) = 1$) $ a, a + k, a + 2k, a + 3k, \cdots $ there are infinitely many primes. Let us recalibrate our weapon so it can operate on these arithmetic progressions rather than $ 1, 2, 3, 4, \cdots $.

Definition: $ \zeta(s,\chi) = \frac{\chi(1)}{1} + \frac{\chi(2)}{2^s} + \frac{\chi(3)}{3^s} + \frac{\chi(4)}{4^s} + \cdots$ (3).

Lemma: $ \zeta(s,\chi) = \frac{1}{1 - \frac{\chi(2)}{2^s}} \times \frac{1}{1 - \frac{\chi(3)}{3^s}} \times \frac{1}{1 - \frac{\chi(5)}{5^s}} \times \frac{1}{1 - \frac{\chi(7)}{7^s}} \times \cdots$ (4).

Theorem: Dirichlet's theorem on arithmetic progressions.

Demonstration: Pick $ \chi $ so that it corresponds to the arithmetic progression, only finitely many terms are removed from (4) so divergence again lets us conclude the proof. $\square$

How Euler discovered the zeta function - Keith Devlin

Introduction to Analytic Number Theory, Lecture Notes

Labels:
arithmetic progression,
dirichlet,
euclid,
euler,
prime,
zeta function

### This is not a test!

` \sqrt[5]{1} = -\frac{1}{4} \quad+\frac{1}{4}\cdot\sqrt{5}\quad+\frac{1}{2}\cdot\sqrt{-\frac{5}{2}-\frac{1}{2}\cdot\sqrt{5}} `

Ripped out of Constructing Regular Polygons -- Primitive Roots of Unity.

Subscribe to:
Posts (Atom)