$ 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
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
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.
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