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  #76  
Old 08-28-2017, 11:14 PM
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Default Re: Math trivia

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I'll now consider groups with continuous parameters, groups with elements M(a) where a is some set of real numbers. Such groups are called "Lie groups", after 19th cy. Swedish mathematician Sophus Lie ("Lee"). ...
That looks a lot like angular momentum in quantum mechanics. Interesting.
Yes indeed. Angular momentum = rotation generators.

I'll illustrate with the z-component in spherical coordinates: Lz = - i*d/(dφ) for azimuthal angle φ.

Consider exp(i*a*Lz) f(φ) for some function f.
It equals sum over n from 0 to infinity of 1/n! * an * (dnf(φ))(dφn)

That is a Taylor series, and it evaluates to f(φ + a). That is, rotate by azimuthal angle a.


When one works out the possible angular-momentum states, one works out the representations of the rotation group.
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  #77  
Old 08-29-2017, 02:06 PM
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Default Re: Math trivia

Just for fun:
Lobachevsky Lyrics - Tom Lehrer
Lobachevsky (song) - Wikipedia
Notice how he sings in a mock Russian accent and how he tries to imitate a Russian speaking English, like not using "the" and putting objects first in some sentences.

Non-Euclidean geometry - Wikipedia (1792 - 1856) and János Bolyai - Wikipedia (1802 - 1860) independently discovered that Euclid's fifth or parallel postulate was independent of his other axioms by constructing hyperbolic geometry, a system that obeys those axioms but not the parallel one. In hyperbolic geometry, one can have an infinite number of lines that do not intersect some other line, and not just one line (Euclid's fifth). A consequence is that the sum of the angles of a triangle is less than 180d, with comparable inequalities for more-angle polygons.

Their successors also worked out elliptic geometry, where there are no non-intersecting lines, and where the sum of the angles of a triangle is greater than 180d. One of them was Bernhard Riemann - Wikipedia, who, in 1854, generalized further to curvature as an arbitrary function of position.

This bit of math was used by Albert Einstein in General Relativity for describing gravity as space-time curvature.
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  #78  
Old 08-29-2017, 09:24 PM
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Default Re: Math trivia

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If you have ever worked through quantum-mechanical angular momentum, much of this should be familiar:

Angular-momentum operators = rotation generators
Angular-momentum magnitude = Casimir invariant of rotation
Etc.
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...
That looks a lot like angular momentum in quantum mechanics. Interesting.
:blush:
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  #79  
Old 08-31-2017, 04:36 AM
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Default Re: Math trivia

I'll now consider regular polygons (2D) and their counterparts in greater numbers of dimensions: polyhedra (3D), polychora (4D), and polytopes in general.

In two dimensions, there is an infinite family of regular polygons, starting with equilateral triangles and squares. For n >= 3, they have n vertices and n edges. In the notation that I will use below:
n, n

In three dimensions, there are only the 5 Platonic solids. They are:
Tetrahedron: 4, 6, 4 (triangle)
Cube: 8, 12, 6 (square)
Octahedron: 6, 12, 8 (triangle)
Dodecahedron: 20, 30, 12 (pentagon)
Icosahedron: 12, 30, 20 (triangle)

In order: vertices, edges, faces. They have the following dualities, from reflecting the list of contents:
Tetrahedron (self), like polygons
Cube - Octahedron
Dodecahedron - Icosahedron

In four dimensions, there are 6 of them. They are:
5-cell: 5, 10, 10, 5 (triangle, tetrahedron)
Tesseract: 16, 32, 24, 8 (square, cube)
16-cell: 8, 24, 32, 16 (triangle, tetrahedron)
24-cell: 24, 96, 96, 24 (triangle, octahedron)
120-cell: 600, 1200, 720, 120 (pentagon, dodecahedron)
600-cell: 120, 720, 1200, 600 (triangle, icosahedron)

Dualities:
5-cell (self)
Tesseract - 16-cell
24-cell (self)
120-cell - 600-cell

Four more than four dimensions, there are 3 of them. Instead of a list, I will give a count of k-dimensional objects for n dimensions (point = 0, line = 1, ...).
Simplex: (n+1)!/((k+1)!*(n-k)!)
Hypercube: 2n-k*n!/(k!*(n-k)!)
Cross-polytrope or orthoplex: 2k+1*n!/((k+1)!*(n-k-1)!)

Duality:
Simplex (self)
Hypercube - Orthoplex

Simplex: triangle, tetrahedron, 5-cell, ...
Hypercube: square, cube, tesseract, ...
Orthoplex: square, octahedron, 16-cell, ...

So in summary, the number of regular polytopes:
infinite, 5, 6, 3, 3, 3, 3, 3, 3, 3, 3, ...
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Old 08-31-2017, 06:47 AM
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Default Re: Math trivia

I'll now consider plane tilings and their generalizations to n-space tilings, all for flat spaces.

2D: triangle, square, hexagon
3D: cube
4D: tesseract (8-cell), 16-cell, 24-cell
5D and more: hypercube

Polytopes can be interpreted as space tilings for convex spaces, and there is a third possibility: hyperbolic spaces. One can distinguish them by how the circumference of an expanding circle behaves.

If it increases as (2pi)*(radius), then the space is flat. If it increases more slowly, then it is convex, while if it increases faster, then it is hyperbolic.

For radius r,
  • Flat: 2pi * r
  • Convex / Spherical: 2pi * R * sin(r/R)
  • Fully-symmetric hyperbolic: 2pi * R * sinh(r/R)
Let's see what happens to a growing disk.

In the flat case, it stays flat, as one would expect.

In the convex case, it starts curling toward one side, and its edge eventually shrinks and becomes a point.

In the hyperbolic case, it gets more and more crinkly as one goes.
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  #81  
Old 08-31-2017, 07:05 AM
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Default Re: Math trivia

I'll show what regular polyhedra and plane tilings are possible. They are a sort of island in a sea of hyperbolic tilings.

Let's start with a regular triangle, an equilateral one. Its vertex angles are 180d/3 = 60d.

3 triangles: 180d -- tetrahedron
4 triangles: 240d -- octahedron
5 triangles: 300d -- icosahedron
6 triangles: 360d -- triangular tiling
7 triangles: 420d -- hyperbolic

Now a regular quadrangle, a square. Its vertex angles are 360d/4 = 90d

3 squares: 270d -- cube
4 squares: 360d -- square tiling
5 squares: 450d -- hyperbolic

Now a regular pentagon. Its vertex angles are 540d/5 = 108d

3 pentagons: 324d -- dodecahedron
4 pentagons: 432d -- hyperbolic

Now a regular hexagon. Its vertex angle = 720d/6 = 120d

3 hexagons: 360d -- hexagonal tiling
4 hexagons: 480d -- hyperbolic

Now a regular heptagon. Its vertex angle = 900d/7 = (900/7)d

3 heptagons: (2700/7)d -- hyperbolic
(Numerical value: 385.714d)


This construction can be generalized using "Schläfli symbols".
(none) -- point
{} -- line
{p} -- polygon
{p,q} -- polyhedron
{p,q,r} -- polychoron
...

For {p,...,q}, one finds a shape where r of the objects meet at each meeting point, and one gets {p,...,q,r}


Thus, a n-gon is {n} and the regular polyhedra and tilings are
{3,3} - tetrahedron
{3,4} - octahedron
{3,5} - icosahedron
{3,6} - triangular tiling
{4,3} - cube
{4,4} - square tiling
{5,3} - dodecahedron
{6,3} - hexagonal tiling


These symbols also express stellated polytopes, like star polygons. One uses fractional values p/q, where p is the total number and q is the skip value as one goes around. Thus a pentagram is {5/2}, with total number 5 and skip value 2.

Instead of 1, 2, 3, 4, 5,
a pentagram does 1, 3, 5, 2, 4

Duality is easy to find for Schläfli symbols -- reverse the order.

Last edited by lpetrich; 08-31-2017 at 07:17 AM.
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  #82  
Old 08-31-2017, 08:58 AM
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Default Re: Math trivia

So here's a list of all non-hyperbolic regular polytopes and tilings, using their Schläfli symbols.

1D: {}

2D convex: {p}
2D stellated: {p/q}

3D convex: {3,3}, {3,4}, {4,3}, {3,5}, {5,3}
3D stellated: {3,5/2}, {5/2,3}, {5,5/2}, {5/2, 5}
2D plane: {3,6}, {6,3}, {4,4}

4D convex: {3,3,3}, {3,3,4}, {3,4,3}, {4,3,3}, {3,3,5}, {5,3,3}
4D stellated: {3,3,5/2}, {5/2,3,3}, {3,5,5/2}, {3,5/2,5}, {5/2,3,5}, {5,3,5/2}, {5,5/2,3}, {5/2,5,3}, {5,5/2,5}, {5/2,5,5/2}
3D plane: {4,3,4}

5D convex: {3,3,3,3}, {3,3,3,4}, {4,3,3,3}
(no stellated ones for 5D and higher)
4D plane: {3,3,4,3}, {3,4,3,3}, {4,3,3,4}

n-D convex: {3,...,3}, {3,...,3,4}, {4,3,...,3}
n-D plane: {4,3,...,3,4}


Counting them:
1D: 0
2D: (infinite), (infinite)
3D: 5, 4, 3
4D: 6, 10, 1
5D: 3, 0, 3
nD: 3, 0, 1
(n>5)

The space-tiling vertices' coordinates can be expressed in forms with some very interesting properties.

The hypercubic {4,3...3,4} tiling exists in all numbers of dimensions >= 2, and its vertices are {integers}


The triangular tiling {3,6} has vertices {1,0}*n1 + (1/2,sqrt(3)/2}*n2 where n1 and n2 are integers.

The hexagonal tiling {6,3} is like the triangular tiling, having that tiling's vertices an offset of those vertices: n1 and n2 = integers + 1/3


The 16-cell tiling {3,3,4,3} has vertices {4 integers} with an even sum. This is {4 even}, {4 odd}, and {2 even, 2 odd}. Offsetting, one gets {4 integers} with an odd sum, {3 even, 1 odd}, {1 even, 3 odd}.

The 24-cell tiling {3,4,3,3} has vertices offset from the 16-cell vertices by the 24-cell vertices: {1 (+-1), 3 0} and {4 (+- 1/2)}

This is {4 integers} with either an even or an odd sum, and {4 integers} + 1/2.
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  #83  
Old 09-16-2017, 09:39 PM
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Now some more group theory. As I'd mentioned, the rotation groups SO(n) have spinor representations (spinor = spin vector). Those reps are sometimes called groups Spin(n).

Spinors were first developed by physicist Wolfgang Pauli around 1930 to describe the spins of electrons. This including "Pauli matrices" for electrons' spin operators. Someone or other later discovered that he had reinvented quaternions, because Pauli matrices can be used to create a matrix representation of quaternions.

Generalizing to an arbitrary number of dimensions, we find a difference between even and odd numbers. Spin(2n) describes spinors with size 2[sup]n[/usp], but each one splits into two irreducible ones with sizes both 2n-1. Spin(2n+1) describes irreducible spinors with size 2n.

Can any spinors be described with real matrices? For some numbers of dimensions, they can. For this, I'll introduce whether or not a group representation is real. Consider an irreducible representation (irrep) D(a) for group elements a. Now try to find a matrix Z that can map D(a) onto its complex conjugate:

Z.D(a).Z-1 = D*(a)

If all of the D(a)'s elements are real, then one can set Z = I and one is done. But if of some of the D(a)'s elements are non-real complex, then it is more challenging. If it is not possible, then the irrep is called "complex", while if it is possible, there are two special cases for Z:
Z.Z* = Z*.Z = +1 or -1

The +1 case means that the irrep is real, while the -1 case means that the irrep is "pseudo-real" or "quaternionic".

The three cases can be abbreviated real: R, pseudo-real: H, complex: C. The H is after William Rowan Hamilton, the discoverer of quaternions.

Let's see how the spinor reps stack up:
  • 8n + 0: RR
  • 8n + 1: R
  • 8n + 2: CC
  • 8n + 3: H
  • 8n + 4: HH
  • 8n + 5: H
  • 8n + 6: CC
  • 8n + 7: R
This period-8 is sometimes called a "Bott periodicity". The CC ones are conjugates of each other, while the RR and HH are two separate self-conjugate ones each.

For Spin(n1,n2), the counterpart of SO(n1,n2), the reality is like for Spin(|n2-n1|). Thus, a massless spin-1/2 particle in our SO(3,1) space has CC and thus divides up into two parts that are mirror images of each other: a left-handed part and a right-handed part.
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  #84  
Old 09-16-2017, 10:09 PM
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Default Re: Math trivia

I'll now get into the Cayley-Dickson construction, because it gives a hierarchy of mathematical entities that lose properties as one goes for the first few.

From an algebra, one constructs the next algebra by taking ordered pairs of elements of the original one and applying various rules to it. One usually starts with the real numbers, and I will do so here, though one may use other algebraic fields.

Addition: (a1,b1) + (a2,b2) = (a1+a2,b1+b2)
Scalar multiplication: c*(a,b) = (c*a,c*b)

Conjugation: (a,b)* = (a*, -b)
So the first element's sign is unchanged and all the others have reversed signs.
It undoes itself: (a*)* = a

Multiplication: (a1,b1).(a2,b2) = (a1.a2 - (b2*).b1, b2.a1 + b1.(a2*))
This can be generalized by multiplying the (b2*).b1 term by something like -1, but I won't do that here.

It is distributive over addition. Also, (a.b)* = (b*).(a*)

Norming: N(a) = a.(a*) = (a*).a = sum of squares of all components of a.

Reciprocal: a-1 = (1/N(a))*(a*)

For real numbers, N((1,1)) = 2
For GF(2), addition and multiplication modulo 2, N((1,1)) = 0
So for finite fields, one has to be careful.
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Old 09-16-2017, 10:59 PM
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Default Re: Math trivia

First, a bit of clarification. The multiplicative identity (1) = (1,0,0,...,0)
Norming is a.(a*) = (a*).a = N(a)*(1)


Now for what properties drop out as one advances in the Cayley-Dickson construction. I'll start with a big list:

Multiplication is
Commutative: a.b = b.a
Associative: (a.b).c = a.(b.c)
Alternative: like associative but with two of the variables equal: (a.a).b = a.(a.b), (a.b).a = a.(b.a), (b.a).a = b.(a.a)
Power-associative: it does not matter what order one evaluates the powers in it: ap+q = ap.aq for all p and q
Well-normed: N(a.b) = N(a)*N(b)
Lacking zero divisors: a.b = 0 implies at least one of a = 0 or b = 0

Also,
Self-conjugate: a = a*

The real numbers (size 1) satisfy all these properties: comm, assoc, alt, pwra, norm, nozd, scjg.

The next C-D step gives the complex numbers (size 2). They are not self-conjugate, leaving comm, assoc, alt, pwra, norm, nozd.

The next C-D step gives the quaternions (size 4). Their multiplication is not commutative, leaving assoc, alt, pwra, norm, nozd.

The next C-D step gives the octonions (size 8). Their multiplication is not associative, though it is alternative, leaving alt, pwra, norm, nozd.

The non-associativity means that it has no matrix representation.

The next C-D step gives the sedenions (size 16). Their multiplication is not alternative or well-normed, and they have some zero divisors. However, it is still power-associative, leaving pwra.

Zero divisors are counterintuitive, but they can exist in some cases. Taking the square of matrix {{0,1},{0,0}} gives the zero matrix {{0,0},{0,0}}. Likewise, component-by-component multiplication of {1,0} and {0,1} gives {0,0}.

All subsequent C-D steps have no changes in properties.


Now for power-associativity. I'll start with
a1 = P1*(1) + Q1*b
a2 = P2*(1) + Q2*b
a12 = P12*(1) = Q12*b
where
b is anti-self-conjugate or imaginary-like: b* = - b

Then,
P12 = P1*P2 - Q1*Q2*N(b)
Q12 = P1*Q2 + Q1*P2
or
(P12 + i*nb*Q12) = (P1 + i*nb*Q1) * (P2 + i*nb*Q2)
where nb = sqrt(N(b))

To get the nth power of a*(1) + b, one starts with P = a and Q = 1 and finds
(P(n) + i*nb*Q(n)) = (a + i*nb)n
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  #86  
Old 11-18-2021, 06:41 PM
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Lincolnshire sheep counting.

Some shepherds in rural parts of Britain still use Brythonic Celtic words to count their sheep. Today, it's just a fun thing to do, but it was standard usage for sheep counting until the eighteenth century, even though the words weren't used to count things other than sheep for over a thousand years before then. Modern Welsh (if there is such a thing) still uses a very similar system, and the words sound much like the Lincolnshire ones.

There are lots of variations, but the Lincolnshire version is the most fun, because if you add a dik to a pimp, you get a bumfit.
  1. Yan
  2. Tan
  3. Tethera
  4. Pethera
  5. Pimp
  6. Sethera
  7. Lethera
  8. Hovera
  9. Covera
  10. Dik
  11. Yan-a-dik
  12. Tan-a-dik
  13. Tethera-dik
  14. Pethera-Dik
  15. Bumfit
  16. Yan-a-bumfit
  17. Tan-a-bumfit
  18. Tethera-bumfit
  19. Pethera-bumfit
  20. Figgot
You can see that the system repeats after the word for ten, by adding one, two, three, or four, to ten - and then the same again after the new word for fifteen.

The system stops at twenty. When shepherds wanted to count flocks much larger than twenty, they would put a pebble in their pocket, or similar, for each set of twenty, and then start over. Once they'd counted all the sheep, they would use the same counting system to count the pebbles, so it's a kind-of base-twenty counting system, though containing vestiges of base-five and base-ten.

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  #87  
Old 11-20-2021, 06:00 AM
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Default Re: Math trivia

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Originally Posted by ceptimus View Post
Lincolnshire sheep counting.

Some shepherds in rural parts of Britain still use Brythonic Celtic words to count their sheep. Today, it's just a fun thing to do, but it was standard usage for sheep counting until the eighteenth century, even though the words weren't used to count things other than sheep for over a thousand years before then. Modern Welsh (if there is such a thing) still uses a very similar system, and the words sound much like the Lincolnshire ones.

There are lots of variations, but the Lincolnshire version is the most fun, because if you add a dik to a pimp, you get a bumfit.
  1. Yan
  2. Tan
  3. Tethera
  4. Pethera
  5. Pimp
  6. Sethera
  7. Lethera
  8. Hovera
  9. Covera
  10. Dik
  11. Yan-a-dik
  12. Tan-a-dik
  13. Tethera-dik
  14. Pethera-Dik
  15. Bumfit
  16. Yan-a-bumfit
  17. Tan-a-bumfit
  18. Tethera-bumfit
  19. Pethera-bumfit
  20. Figgot
You can see that the system repeats after the word for ten, by adding one, two, three, or four, to ten - and then the same again after the new word for fifteen.

The system stops at twenty. When shepherds wanted to count flocks much larger than twenty, they would put a pebble in their pocket, or similar, for each set of twenty, and then start over. Once they'd counted all the sheep, they would use the same counting system to count the pebbles, so it's a kind-of base-twenty counting system, though containing vestiges of base-five and base-ten.

Learning to count sheep (the Lincolnshire way) - YouTube
I learned it like that except that twenty was jiggit. Which leads to the question, "Which Doors song mentions sheep, and how many?"
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  #88  
Old 11-20-2021, 11:42 AM
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Default Re: Math trivia

I tried to learn this method of counting but I kept falling asleep.
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  #89  
Old 11-20-2021, 02:13 PM
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Default Re: Math trivia

It looks to me like the original words for two and three might have been tether and pether.

As the words were often combined with 'a-dik', ("and ten"), when counting thirteen and fourteen, I suspect the 'a' migrated to the end of the tether and pether words in the same way that the 'n' moved when "a nuncle" became "an uncle", and "a nadder" (the snake) became "an adder".

It sounds more rhythmical and poetic to count tethera, pethera, ... so I guess that's why it happened, and then the 'a' was also added to the ends of the words for six, seven, eight, and nine, even though they're never used with 'a-dik' - that kept the counting more song-like. :shrug:

I suspect the same didn't happen with Yan (one) becoming Yanna, and Tan (two) becoming Tanna, because the words one and two are so often used when not actually counting - so the shorter form survived for those.
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  #90  
Old 11-25-2021, 03:23 PM
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The history of pi

The first recorded use of the Greek letter :pi: for the ratio of a circle's circumference to its diameter, probably chosen to stand for 'periphery' (περιφέρεια) or 'perimeter' (περίμετρος) was in a book published in 1706 by the Welshman, William Jones. His book was titled, Synopsis Palmariorum Matheseos with the snappy sub-title, helpful for non-Latin speakers: A New Introduction to the Mathematics, for the Use of some Friends who have neither Leisure, Convenience, nor, perhaps, Patience, to search into the many different Authors, and turn over so many tedious Volumes, as is unavoidably required to make but tolerable progress in the Mathematics.

The symbol did not gain immediate acceptance, and only became the standard notation some thirty years later, when it was adopted by the famous Swiss mathematician, Leonhard Euler.

:pi: is both irrational, and transcendental. In mathematics, an irrational number is one that cannot be written exactly using ordinary numbers - it has an infinite non-repeating decimal representation. An irrational number, such as the square root of two, may have an infinite decimal representation (1.414213562...) but it can be represented concisely using an algebraic formula: x2 = 2

In contrast, a transcendental number, such as :pi: cannot be exactly represented by a simple formula, and can only be represented by an infinite series of terms, such as the Leibniz formula:

:pi:/4 = 1 - 1/3 + 1/5 -1/7 + 1/9 - 1/11 ...

There are not many known transcendental numbers. The first number proven to be transcendental (rather than being specially constructed to be so) was e in 1873, and :pi: itself was only proved transcendental in 1882.

It has been a common pastime for mathematicians, and latterly supercomputer programmers, to discover more and more digits in the decimal representation of :pi: the current record stands at 62.8 trillion digits.
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  #91  
Old 11-25-2021, 03:54 PM
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Default Re: Math trivia

Quote:
Originally Posted by ceptimus View Post
His book was titled, Synopsis Palmariorum Matheseos with the snappy sub-title, helpful for non-Latin speakers: A New Introduction to the Mathematics, for the Use of some Friends who have neither Leisure, Convenience, nor, perhaps, Patience, to search into the many different Authors, and turn over so many tedious Volumes, as is unavoidably required to make but tolerable progress in the Mathematics.
We need more subtitles in today's books.
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  #92  
Old 11-26-2021, 02:20 AM
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Default Re: Math trivia

We owe Leonhard Euler so much. :wriggle:
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  #93  
Old 12-01-2021, 10:34 PM
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Default Re: Math trivia

From surprising spirals appearing in polar plots of prime numbers, through rational approximations for pi, to Euler's totient function and Dirichlet's theorem.

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  #94  
Old 12-01-2021, 11:02 PM
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Default Re: Math trivia

3b1b is definitely in my top 3 favorite youtube channels. It may be my #1.

Spoiler alert:

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  #95  
Old 01-22-2023, 08:17 PM
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Default Re: Math trivia

Abraham Fraenkel, originally from Germany, a professor of logic, was sitting on a stationary bus in Tel Aviv one day. The bus was timetabled to leave at nine in the morning, but at five-past nine, it still hadn't moved.

Fraenkel waved his timetable at the driver.

The driver, obviously not impressed by this show of impatience, asked, "What are you - a German or a professor?"

Fraenkel, always keen to give accurate and succinct answers, responded, "Do you mean the inclusive or, or the exclusive or?"
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  #96  
Old 01-23-2023, 01:16 AM
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Default Re: Math trivia

Yes
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