

10292007, 03:46 PM


Sane (but only just)


Join Date: Oct 2007
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Re: Math trivia
On the bingo card problem, Knuth gives a, very complex, analysis of it in Concrete Mathematics.
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12262007, 10:58 PM


puzzler


Join Date: Aug 2004
Location: UK


Re: Math trivia
An interesting, though practically useless theorem, by Mills: It's often said that there isn't a formula for producing prime numbers. It may or may not be true that there isn't a practical formula that will produce prime numbers (though there are certainly simple algorithms which produce primes in a practical way).
Anyway, there is at least one formula that is proven to produce primes, though it is useless in practice as it relies on a 'seed' number which must be known to incredible accuracy. Our hero, who discovered this useless (as far as we know) formula in 1947, is one W. H. Mills. He proved there is a number, A, such that:
A3n is prime for all integer values of n.
Mills didn't even calculate a value for A  his proof showed that such a number must exist but didn't say what it was.
It has since been shown that there are an infinite number of possible values for A, the smallest of which is approximately 1.306377883863. This value generates an infinite series of primes beginning with 2, 11 and 1361.
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12292007, 08:17 AM


Member




Re: Math trivia
Quote:
Originally Posted by ceptimus
I shall post occasional mathematical 'factoids' here. Please feel free to add your own.
...
(Twin primes are the prime numbers that only differ by two, such as 5 and 7 or 71 and 73. There are conjectured to be an infinity of twin primes, though this has yet to be proven. A large pair are:
33218925 * 2^169690 +/ 1, which have 51,090 digits each.
...

It's interesting that they know the number of digits in the result.
In fact, you can compute the number of digits in any integer without having to carry out the complete computation or actually count the resulting digits at all.
Here's how to do it:
Based on the given data, the number of digits in the prime numbers above can be found by the formula:
Digits = floor((ln(33218925) + ln(2)*169689) / ln(10)) + 1
= 51090
in terms of natural logarithms.
The computation can be simplified by using base 10 logarithms instead, as:
Digits = floor((log(33218925) + log(2)*169689)) + 1
= 51090
Note for clarity:
ln = Natural logarithm
log = base 10 logarithm
For the general case, let
X = Integer (positive) in question
D = Number of digits in X
To find the unknown number of digits in X
D = floor(log(X)) + 1
EXAMPLE:
Given
X = 6^29000
log(X) = log(6)*29000
Then,
D = floor(log(X)) + 1 = floor(log(6^29000)) + 1
or
D = floor(22566.3862611) + 1 = 22567 digits in X
FOR INTEGERS IN OTHER BASES:
A base 10 integer (X) after conversion to base (B) has (D) digits.
where:
D = floor(log(X) / log(B)) + 1
EXAMPLE:
The base 10 value (13^967), when converted to base 2, would have (D) digits, where
D = floor(log(13^967) / log(2)) + 1
= floor(1077.18322168 / 0.301029995664) + 1
= 3578 + 1 = 3579 digits

12292007, 08:04 PM


Admin of THIEVES and SLUGABEDS




Re: Math trivia
Great to have another math person on the board. Welcome to FF, Brain_Droppings.

12292007, 09:37 PM


silky...


Join Date: Feb 2005
Location: XOXLIV&VMXOX


Re: Math trivia
You guys are so nerdy, I love it
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03062009, 05:19 PM


puzzler


Join Date: Aug 2004
Location: UK


Re: Math trivia
Pick's Theorem and area measurement: In 1899, Georg Pick discovered a remarkable method of easily calculating the area of irregular polygons. Consider the two examples shown below.
What we do is count the number of corners of a polygon and the number of dots on the unit lattice that it encloses.
The figure on the left has six corners and encloses seventeen dots. By Pick's Theorem its area is:
A = 17 + 6 / 2  1 = 19
The figure on the right has seven corners and encloses thirteen dots, so its area is:
A = 13 + 7 / 2  1 = 15.5
In general the area of any figure is :
A = enclosed dots + corners / 2  1
Note that if one of the edges passes straight through a dot then you count that dot as an extra 'corner'  think of it as a corner with an angle of 180 degrees. There are no such 'corners' on the two examples above.
If you've not come across Pick's Theorem before, then it's an interesting exercise to attempt to discover a proof. The theorem turns out to have important applications in number theory and other mathematics  it's not just useful for measuring areas.
For the method to work, the corners of the polygon have to be located on dots  this is equivalent to saying that the coordinates of the corners must be expressible as integers. If the corners don't have integer coordinates then it may be possible to choose a different scale where they do. If such a scale is not possible we can always choose a fine enough scale so that the area of the polygon can be approximated to any desired accuracy.
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03072009, 05:06 AM


select custom_user_title from user_info where username='Goliath';


Join Date: Jul 2004
Location: Kansas City, MO
Gender: Male


Re: Math trivia
Suppose that the vertices of our polygonwhen traversed counterclockwise around the polygonhave coordinates (x 1, y 1), (x 2, y 2), ...., (x n, y n) for some positive integer n which is at least 3 (so if n=3 you have a triangle, if n=4 a quadrilateral, etc).
As a beautiful consequence of Green's Theorem, areas that are enclosed by a curve (on a plane) can be expressed in terms of the line integral around the curve. So, it turns out that the area of our polygon above can be written as:
(1/2)*[(x 2y 1x 1y 2)+(x 3y 2x 2y 3)+(x 4y 3x 3y 4)+...+(x 1y nx ny 1)].
(Sorry for the crappy subscripts.)
What's also kinda neat is that, as a corollary, the quantity above is never negative (it's an area!), which isn't immediately obvious, since it contains a lot of sums and differences.
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03072009, 07:08 AM

God Made Me A Skeptic


Join Date: Jul 2004
Location: Minnesota


Re: Math trivia
That is genuinely fucking awesome.
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03072009, 11:01 AM


puzzler


Join Date: Aug 2004
Location: UK


Re: Math trivia
Quote:
Originally Posted by Goliath
...,the quantity above is never negative (it's an area!), which isn't immediately obvious, since it contains a lot of sums and differences.

It would be negative if we traversed the polygon clockwise though  even though it's still an area!
Question: Where does this counterclockwise being the positive way to turn come from? It can seem mysterious at first.
Answer: It arises from our conventions of having:  x increase when moving to the right
 y increase when moving upwards
 writing the x coordinate before the y coordinate
If we change an odd number of these conventions then the positive rotation direction becomes clockwise. An example is in computer graphics where it is common to have the origin at the top left corner of the display, with y increasing downwards.
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03072009, 04:58 PM


select custom_user_title from user_info where username='Goliath';


Join Date: Jul 2004
Location: Kansas City, MO
Gender: Male


Re: Math trivia
Quote:
Originally Posted by ceptimus
It would be negative if we traversed the polygon clockwise though  even though it's still an area!

Yep, which is why I set things up so that we traversed the polygon counterclockwise. Good catch, though, as line integrals switch sign when the orientation of the curve is reversed.
Quote:
Question: Where does this counterclockwise being the positive way to turn come from? It can seem mysterious at first.
Answer: It arises from our conventions of having: x increase when moving to the right
 y increase when moving upwards
 writing the x coordinate before the y coordinate
If we change an odd number of these conventions then the positive rotation direction becomes clockwise. An example is in computer graphics where it is common to have the origin at the top left corner of the display, with y increasing downwards.

Yep. Another reason is that if we view the curve in 3D space and lying on the xyplane, the usual counterclockwise orientation (when viewed from above) allows for the binormal vector to point upward (i.e. in the direction of the positive zaxis). This allows for a natural discussion of Stokes' Theorem when using compact surfaces given by a function of two variables (surfaces that need not be stuck on the xyplane anymore), as the counterclockwise (when viewed from above) orientation allows for all of our normal vectors to point "upwards."
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03122009, 05:37 PM


scatterbrain


Join Date: Aug 2007
Location: Mi'kmaq Land
Gender: Male


Re: Math trivia
Here’s one of my favourite bits of math trivia.
The five wellknown Platonic solids are the only possible regular polyhedra (3D polytopes), if we adopt the strongest possible sense of the word “regular”. Generalizing this result to other numbers of dimensions, there are exactly six regular polychorons (4D polytopes), and if n is any integer greater than 4, there are exactly three regular ndimensional polytopes.
A regular ndimensional polytope can be described in terms of the (n1)dimensional polytope species to which its “faces” belong, along with the (n1)dimensional polytope species to which its “vertex figures” belong. Roughly, a “vertex figure” is the shape you expose when you snip off (or file down) a vertex.
To see what I mean, let’s stick with three dimensions for a moment.
Tetrahedron: 4 faces, 4 vertices. Faces are triangles. Vertex figures are triangles.
Cube: 6 faces, 8 vertices. Faces are squares. Vertex figures are triangles.
Octahedron: 8 faces, 6 vertices. Faces are triangles. Vertex figures are squares.
Dodecahedron: 12 faces, 20 vertices. Faces are pentagons. Vertex figures are triangles.
Icosahedron: 20 faces, 12 vertices. Faces are triangles. Vertex figures are pentagons.
The first three polyhedra in the list above are 3D examples of families of polytopes that exist in any number of dimensions: simplices (singular simplex), hypercubes, and crosspolytopes (I hate that term). (In two dimensions, an equilateral triangle is a simplex, and a square is both a hypercube and a crosspolytope.)
nD simplex: n+1 faces, n+1 vertices. Faces are (n1)D simplices. Vertex figures are (n1)D simplices.
nD hypercube: 2n faces, 2n vertices. Faces are (n1)D hypercubes. Vertex figures are (n1)D simplices.
nD crosspolytope: 2nfaces, 2n vertices. Faces are (n1)D simplices. Vertex figures are (n1)D crosspolytopes.
If n > 4, there are no regular ndimensional polytopes other than those belonging to these three families.
What about n = 4?
There’s a polychoron analogous to a dodecahedron: 120 faces, 600 vertices. Its faces are dodecahedra. Its vertex figures are tetrahedra.
There’s a polychoron analogous to an icosahedron: 600 faces, 120 vertices. Its faces are tetrahedra. Its vertex figures are icosahedra.
There’s also a polychoron that isn’t analogous to anything in 3D: 24 faces, 24 vertices. Its faces are octahedra, and its vertex figures are cubes.
This last beastie is really cool, because a 4D Euclidean space can be infinitely tiled by it. 4D is thus something like 2D, because you have more than one way to tile the space with infinite congruent copies of a single regular polytope – you aren’t stuck with using hypercubes. (In 2D, of course, you could use triangles or hexagons; you don’t have to use squares.)

03222011, 07:20 PM


puzzler


Join Date: Aug 2004
Location: UK


Re: Math trivia
The harmonic series (and modifications thereof):
It's interesting that sometimes, we can add an infinite number of numbers together and get a finite result. Perhaps the most obvious example is where each term in the series is half as big as the one before:
1/2 + 1/4 + 1/8 + 1/16 + ...
This is the series made famous by Zeno, and although it can seem confusing at first, it's pretty obvious that, in this case, the sum of all the terms is one. Each extra term added takes us 'half the remaining distance' towards one, but it's clear that even after an infinite number of terms we can never get past one.
Now lets look at the harmonic series:
1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ...
Each term is smaller than the one before, so it looks like it might approach some total as we add more and more terms, but never pass that total.
So we begin adding terms and find that after the first 10 terms we reach 2.929, after 100 terms we've got to 5.187 and after 200 terms we have 5.878  perhaps we're heading towards 2 x PI which would be 6.283?
But no, after 300 terms that value is passed and the total keeps creeping up  albeit very slowly  after the first thousand terms the total is still less than 7.5
The amazing thing is that although the rate of increase of the total quickly slows to a snaillike crawl, it's a snail that never gives up, so the total will eventually pass 10, 100, a billion, any number you can name  the total is, in fact, infinite.
Now let's modify the series slightly  pick any number you like, say 314159 in honour of PI. Cross out all the numbers in the harmonic series that contain that number. So the first term we cross out is at the 314160th place and the next is one million places after that: 1/1314159 then another million places: 1/2314159 and so on...
The total was infinite before, so you might think that after crossing out these numbers the total would still be infinite. But no! Schmelzer and Baillie discovered an efficient way of working out such results, and found that the total with all the '314159 numbers' missing is 2302582.33386378260789202376 (rounded to the last decimal place!)
In general, you can choose any rule you like for crossing out numbers, and it will be enough to topple the harmonic series from having an infinite sum to a finite one (mathematicians would say it changes from being a 'divergent series' to a 'convergent' one).
One reason why this is so, is that once we get to really huge numbers  ones with a few billion digits  then there is a very good chance that they will contain, say, '314159' somewhere  so we end up crossing out most of the numbers as we get further into the list.
But even if we make our chosen string of digits much longer  say we only cross out those numbers where the digit '4' occurs a billion times in a row  then that is still enough to change the divergent harmonic series into a convergent one!
Last edited by ceptimus; 03222011 at 09:27 PM.

03232011, 06:05 AM


Warlord of Mars


Join Date: Jan 2005
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Gender: Male


Re: Math trivia
Great thread. Cept!
Here's an interesting, and to me very disturbing, consequence of Cantor's transfinites. In short, there are uncountably many real numbers that will never be named. There are, in fact, so many of these unnamed numbers relative to the numbers we can name that if we had a dart with an infinitely sharp point (such that if we threw it at the number line, it would hit exactly one number) and then threw it at the real number line, the probability that it would hit a named number is exactly 0!
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04202011, 06:04 PM


select custom_user_title from user_info where username='Goliath';


Join Date: Jul 2004
Location: Kansas City, MO
Gender: Male


Re: Math trivia
Quote:
Originally Posted by John Carter
Great thread. Cept!
Here's an interesting, and to me very disturbing, consequence of Cantor's transfinites. In short, there are uncountably many real numbers that will never be named. There are, in fact, so many of these unnamed numbers relative to the numbers we can name that if we had a dart with an infinitely sharp point (such that if we threw it at the number line, it would hit exactly one number) and then threw it at the real number line, the probability that it would hit a named number is exactly 0!

It's even worse than that! To make things simpler, and just look at the closed interval [0,1]. So, the probability of randomly picking a number in a subinterval of [0,1] is merely the length of the interval (for example, the probability of picking a number in (1/2,3/4) would be 1/4). In general, the probability of picking a number in a given subset of [0,1] would be the Lebesgue Measure of the subset.* It turns out that any countable setin particular, the subset of rational numbers in [0,1]has a Lebesgue measure of zero. Therefore, the probability of choosing a rational number is zero.
In fact, there are only countably many algebraic numbers (ie numbers that are the root of a polynomial equation with integer coefficients), so the probability of picking an algebraic number in [0,1] is also zero. Numbers that are not algebraic are called transcendental. There are very few transcendental numbers that we've named (pi, e, a few others (along with their obvious sums and multiples)), but almost all numbers are transcendental.
*  Not every set is Lebesgue measurable if one accepts the axiom of choice.
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04222011, 01:43 AM


Now in six dimensions!


Join Date: Jan 2005
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Gender: Male


Re: Math trivia
What sorts of numbers (presumably nonalgebraic, if any) in [0,1] have a nonzero probability of being chosen?
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12292011, 12:54 PM


puzzler


Join Date: Aug 2004
Location: UK


Re: Math trivia
Simpson's paradox or the YuleSimpson effect:
Imagine there are two hospitals carrying out a certain operation:  The first hospital claims that it has a lower death rate for the operation than the second.
 The second hospital breaks down its claims for the operation by sex. It claims it has a lower death rate than the first hospital for men, and it also has a lower death rate for women!
Both claims are true!
Does that seem preposterous? Here are the numbers (with highest death rates bolded):
Hospital  Men  Women  Both  A  8/200 (4%)  8/600 (1.33%)  16/800 (2%)  B  57/1500 (3.8%)  6/600 (1%)  63/2100 (3%) 
(figures for this example taken from the book: Professor Stewart's Hoard of Mathematical Treasures by Ian Stewart)
A reallife example was the passage of the Civil Rights Act of 1964. Overall, a larger fraction of Republican legislators voted in favour of the Act than Democrats. However, when the congressional delegations from the northern and southern States are considered separately, a larger fraction of Democrats voted in favour of the act in both regions.
Tables below shows "winning votes" bolded:
House  Democrat  Republican  Northern  94% (145/154)  85% (138/162)  Southern  7% (7/94)  0% (0/10)  Both  61% (152/248)  80% (138/172) 
Senate  Democrat  Republican  Northern  98% (45/46)  84% (27/32)  Southern  5% (1/21)  0% (0/1)  Both  69% (46/67)  82% (27/33) 
(Information copied from the Wikipedia page.)
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12292011, 03:14 PM


Re: Math trivia
J.D.

12292011, 10:39 PM


Clutchenheimer


Join Date: Jul 2004
Location: Canada
Gender: Male


Re: Math trivia
Simpson's Paradox moreover gets promoted from freaky quirk of some situations to general badass problem in statistical reasoning, once we take on board the general empirical availability of a partition that will create this just this sort of reversal of inequalities. I.e., P(X)>P(Y) within the partitions, and P(Y)>P(X) in aggregate. This is one of the phenomena I harp on when I'm selling my students on the idea that good statistical reasoning requires much more than formal virtuosity in calculating probabilities. Which axiom or theorem of probability tells you how to partition the data perspicuously? I ask them.
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06272017, 12:34 AM


forever in search of dill pickle doritos




Re: Math trivia
RISE ZoMBIE THREAD!
This is pretty neat:

07172017, 06:21 PM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
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Re: Math trivia
Quote:
Originally Posted by John Carter
Here's an interesting, and to me very disturbing, consequence of Cantor's transfinites. In short, there are uncountably many real numbers that will never be named. There are, in fact, so many of these unnamed numbers relative to the numbers we can name that if we had a dart with an infinitely sharp point (such that if we threw it at the number line, it would hit exactly one number) and then threw it at the real number line, the probability that it would hit a named number is exactly 0!

One ought to be more precise about what is a name.
A reasonable definition might be: a string of symbols drawn from a finitesized alphabet. If the strings have a finite maximum length, then the total number is finite, even if it is very large by ordinary standards. If the strings have no maximum length, then the total number can be shown to be aleph0, the cardinality or number of members of the countable sets.
Here are some countable sets:  The positive integers
 The nonnegative integers
 The integers
 The rational numbers
 The algebraic numbers
 The computable numbers (those that can be approximated to arbitrary precision by running a finitesized Turing machine for an arbitrary number of steps)
 All ordered finitelength ntuples of countableset members
 All finitelength subsets of a countable set

07172017, 06:35 PM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
Now consider power sets. A set S's power set P(S) is the set of all subsets of it. Georg Cantor showed that P(S) has a higher cardinality than S. He did so by showing that if it did not, then one gets a contradiction.
So element a of S would be mapped onto subset A of S. Let's consider the set of all a's that are not members of their corresponding A's, a set that I'll call N.
Let's suppose some n corresponds to N, then since n is not in n, then n must belong in N. But if that is the case, then n violates the definition of N, and n cannot be in N.
Thus, card(P(S)) > card(S), or P(S) > S.
One can construct a sequence of power sets starting with the smallest infinite cardinality, aleph0. This sequence is sometimes called the beth numbers. Thus,  beth0 = aleph0
 beth1 = C, the continuum cardinality, that of the real numbers
 beth2 = cardinality of all functions from real numbers to real numbers
We don't have identifications for any higher beth numbers.

It's easy to prove that (countable set)  (finite subset) = countable set = aleph0 (A0).
From hw8ans.pdf is a rather simple proof of something a bit more difficult:
(real numbers)  (countable subset of them) = real numbers = C
For the real numbers R, consider a countable subset of them, S, and find a countable subset T of (R  S), all elements of R not in S. Now define a bijection f between R and (R  S):
f(x in (R  (S union T))) = x
f(s(n)) = t(2n)
f(t(n)) = t(2n1)
where n = 1, 2, 3, ..., all s(n) are in S, and all t(n) are in T. This bijection thus pushes the elements of S into (R  S), moving elements of T out of the way for them.
Thus, R  S = R = C

07172017, 06:50 PM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
Now for some cardinality arithmetic.
For sets A and B such that A intersect B = {}, A union B = A + B
For sets A and B with a and b members of them, then all (a,b) = A*B
Now consider functions f such that f(a) = b for all a in A and b in B. Then,
all functions f = BA
Here is a simple example. For ordered pairs drawn from a shared set A, define an indexing function: index((a1,a2),1) = a1 and index((a1,a2),2) = a2. Fold the (a1,a2) into the definition of the function, and we get f(1) = a1 and f(2) = a2. Thus, f goes from {1,2} to A, and its cardinality is A2, as one would expect.
The set of all subsets of A is equivalent to the set of all functions which accept a member of A and return whether or not it is in the subset. Such a function thus goes from A to {true,false}. Thus a power set's cardinality
P(A) = 2A
By composition of functions, the exponentiationcomposition identity (AB)C = AB*C is true for infinite as well as for finite sets.

07172017, 07:05 PM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
One can find 2aleph0 by considering binary representations of numbers between 0 and 1. Their cardinality turns out to be C. That is likewise true for any other number base b:
baleph0 = C
The openended real line segment (0,1) has the same cardinality as the closedended one [0,1] by a pushingintheends argument. The real numbers are easily mapped onto (0,1) and vice versa:
x <> (1 + x/sqrt(1+x2))/2
So both (0,1) and [0,1] in the reals have cardinality C.
One can find C*C by considering an ordered pair of numbers from [0,1] and interleaving the digits. One gets another real number, making C*C = C. That is also true of any finitelength ordered ntuples, by the same argument.
If one finds all (real number, integer), one also gets C, because 1 < aleph0 < C.
The number of infinite series of rational numbers, (aleph0)aleph0 is also C.
The number of permutations of positive integers is the number of their selfbijections: (aleph0)! = C.
Interestingly, Caleph0 = C. Thus, the total number of continuous functions from real numbers to real numbers is C, and that is also true of all the infinite sequences of real numbers.

07182017, 03:17 PM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
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Re: Math trivia
Another exponentiation identity also true of infinite sets is
(A*B)C = AC * BC
This can be seen from a function f(c) = (a,b) with a in A, b in B, and c in C. It can be expressed as f(c) = (f1(c),f2(c)).
Now for some addition results.
aleph0 + aleph0 = aleph0 (interleaving)
aleph0 + C = C (pushing in an infinite series)
C + C = C (dividing a line segment in two)

07192017, 02:01 AM


This is the title that appears beneath your name on your posts.


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Re: Math trivia
Very.. interesting, I mean...
I think I'm going to thank these posts as soon as I figure out what to do with this information.

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