## Generating Approximate Pythagorean Angles (ADDENDUM) – Simplified Method

Tuesday, 20 October 2015

In a recent post I described a method of generating the simplest primitive Pythagorean triple (a,b,c) where one of the angles of the triangle with sides a, b and c is θ° to within some (small) error bound Δθ°.

One of the steps was, given the cosine C of the angle [from step (2)]:

(3) Calculate the Farey ratio approximant …

$R = \sqrt{\dfrac{1-C}{1+C}}$

Now, that function

$f(C) = \sqrt{\dfrac{1-C}{1+C}}$

seemed semi-familiar, resembling functions that occur in trigonometric or hyperbolic identities.
An example is:

$\cos( \tan^{-1}(t)) = \sqrt{\dfrac{1}{t^2 + 1}}$

A little further investigation, and reading around, including the Wikipedia articles on trigonometric identities, and in particular on those of the tangent half-angle, revealed that the Farey ratio approximant does in fact correspond directly to a simple trigonometric function of the angle:

$\tan \left( \dfrac{\theta}{2} \right) = \dfrac{1 - \cos \theta}{\sin \theta} = \dfrac{1 - \cos \theta}{\sqrt{1 - \cos^2 \theta}} = \dfrac{\sqrt{(1 - \cos \theta)^2}}{\sqrt{(1 - \cos \theta)(1 + \cos \theta)}} = \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta}}$

The slightly simplified method follows.

## Generating Approximate Pythagorean Angles (IV) – Derivation and Proof of The Method

Saturday, 10 October 2015

In the previous post is a table of values.

Suppose you wish to find the simplest primitive Pythagorean triangle (a,b,c) where one of the angles is θ° to within some (small) error bound Δθ°.

Here’s the derivation of the method which was given in an earlier post.

## Generating Approximate Pythagorean Angles (III) – A Table for (1/100)°

Saturday, 10 October 2015

The previous post provides a worked example of the method.
The next post provides a derivation and proof of the method.

### Short Table

 θ° ± 0.01° a b c θ° ±%ε 5° 33425 2928 33553 5.006° 62.8% 10° 6351 1120 6449 10.001° 12.9% 15° 1419 380 1469 14.992° −82.8% 20° 66005 24012 70237 19.991° −90.9% 25° 16272 7585 17953 24.992° −79.8% 30° 2911 1680 3361 29.990° −98.4% 35° 7623 5336 9305 34.992° −84.8% 40° 20424 17143 26665 40.009° 86.1% 45° 4059 4060 5741 45.007° 70.6%

A longer table follows

## Generating Approximate Pythagorean Angles (II) – A Worked Example

Saturday, 10 October 2015

The previous post describes the method.
In the next post is a table of values.

As an example, let

θ = 24°
Δθ = 0.001°

## Generating Approximate Pythagorean Angles (I) – The Method

Saturday, 10 October 2015

The next post provides a worked example of the method.

ADDENDUM [20-10-2015]: A slight simplification of the method below is described in an addendum post.

Suppose you wish to find the simplest primitive Pythagorean triple (a,b,c) where one of the angles of the triangle with sides a, b and c is θ° to within some (small) error bound Δθ°.

A Pythagorean triple (a,b,c) is such that:

a,b,c ∈ ℕ₀ (i.e. are natural numbers ≥ 0), and
a² + b² = c²

A primitive Pythagorean triple (a,b,c) is one such that also

a ⊥ b  (i.e. a and b are coprime, i.e. have no common factors),
a ⊥ c, and
b ⊥ c

that is, a, b and c are pairwise coprime.

The method follows.

## Reversal of Indices in Infinite Triangular Sums

Friday, 31 January 2014

We have:

$\sum_{i=0}^{\infty} \sum_{j=i}^{\infty} \varphi(i,j) = \sum_{j=0}^{\infty} \sum_{i=0}^{j} \varphi(i,j)$

because

$\sum_{i=0}^{\infty} \sum_{j=i}^{\infty} \varphi(i,j) = \sum_{\begin{array}{c}i,j \\ 0 \le i \le j < \infty\end{array}} \varphi(i,j) = \sum_{j=0}^{\infty} \sum_{i=0}^{j} \varphi(i,j)$

In an earlier post, I showed the finite version of this result:

$\sum_{i=0}^{n} \sum_{j=i}^{n} \varphi(i,j) = \sum_{j=0}^{n} \sum_{i=0}^{j} \varphi(i,j)$

This might be seen to hold as both sums are

$\sum_{\begin{array}{c}i,j \\ 0 \le i \le j \le n\end{array}} \varphi(i,j)$

## Fibonacci Formulae

Thursday, 21 November 2013

Whilst doodling with the Fibonacci sequence

 n Fn 0 1 2 3 4 5 6 7 8 9 10 … 0 1 1 2 3 5 8 13 21 34 55 …

I found some interesting formulae:

• $F_n = \displaystyle\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor} \binom{n-1-k}{k}$
• $F_{2m} = - \dfrac{1}{2} \displaystyle\sum_{k=0}^{2m-1} (-1)^k \binom{2m}{k} F_k$
• $\displaystyle\sum_{k=0}^{2m} (-1)^k \binom{2m+1}{k} F_k = 0$

The first of these is not new, but I did not find the other two on the web.

## An Intuitive Representation of the Eisenstein Integers

Wednesday, 6 November 2013

${\mathbb Z} [ \omega ]$

where ω is a primitive cube root of 1 given by

$\omega = \dfrac{-1 + {\mathrm {i}} \sqrt{3}}{2} = \sqrt[3]{1}$

are often represented in the form

$\{ a + b \cdot \omega \;|\; a, b \in {\mathbb Z}\}$
(noting that $\omega^2 = - (1 + \omega)$)

and perhaps abbreviated by pair notation such as

$\{ [ a ; b ]^{*} \;|\; a, b \in {\mathbb Z}\}$

(where the asterisk is to distinguish this notation from that introduced below)

so that

$[ a ; b ]^{*} + [ c ; d ]^{*} = [a+b \;;\; c+d]^{*}$

$[ a ; b ]^{*} \times [ c ; d ]^{*} = [(a c - b d) \;;\; (b c + a d - b d) ]^{*}$

Here is a more intuitive representation that is simpler to manipulate and reason about.

## Sublime Text Tip: Indicating Current and Edited Tabs

Friday, 25 October 2013

To colour the current tab green, and any tab of an edited file red, add the following to your theme file.

## Hyperbolic Mnemonic

Friday, 11 October 2013

If you have problems remembering which is which with sinh and cosh, or their graphs, or which way around their definitions are:

$\sinh x = \dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}$

$\cosh x = \dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}$

then it might help if you notice that the graph of sinh is somewhat S-shaped (though backwards), and cosh is C-shaped (though open upwards).

[Well, unless you’re Russian, where the Cyrillic letter Es (‘С’) corresponds to our ‘S’; that might throw a graphemic spanner in the works.]