Calculating The Fine Structure Constant.

[Turtle]

some afterthoughts & asides:

 

I’m trying to calculate the count with which this thread concerns itself ,

“the number of polygonal numbers of rank greater than 2 under x”,

and appear to be misunderstanding it.

 

For example, when I count the third and greater triangular, square, through 34-gonal numbers less than 100, rather than the 57 that everyone else gets, I get 75.

...

i'd say the reason this didn't come up as problematic for calculating the cardinality of the figurate set in this thread earlier, is that Donk is using a sieve method rather then a generating expression.

...

more than you may want to know on it is in this thread: :hihi: >> Non-Figurate Numbers

 

the disparity craig mentions did come up, at least in my little pea brain as i struggled to see what donk & modest had done over in the non-figurate thread. :naughty: once i got my m-brane wrapped around it though, it's bloody brilliant! :doh: likewise i'm playing ketchup catsup catch-up here with don-the-insightful & the fellas. (not being misogynist here; just happen to know the respondents are of the male persuassion. would love to here from some gals on this. ;) )

 

anyway, i think the multiplicity of figurateness may give us yet more countably infinite sets to play with, whether here or over in the non-figurate thread if anyone has a mind to consider or pursue it. for example, is there a limit to the number of different figurate subsets that a number may belong to? what is the percentage/density of "multi-figurate" numbers compared to "singularly-figurate" numbers? does the fine structure constant show up here too? inquiring minds want to know. :shrug:

 

that is all. :cup: . . . . . .:eek:

[Don Blazys]

To: Turtle,

 

Quoting turtle:

I think the multiplicity of figurateness may give us yet more countably infinite sets to play with.

 

I think so too. Just the phrase "Polygonal numbers of rank >2"

(some use the phrase "order >2", but it's the same thing)

makes one wonder about rank >3, rank>4...and so on.

 

You weren't just "whistling Dixie" :Whistle: when you pointed out that

the properties separating figs from non-figs were relatively unexplored.

 

If it turns out that the function: [math]B(x)-B(x)*\alpha*\mu^{-1}=[/math]

 

[math]\left(x-(\alpha*\Pi*e+e)^{-1}*x-\frac{1}{2}*\sqrt{x-(\alpha*\Pi*e+e)^{-1}*x}\right)-\left(x-(\alpha*\Pi*e+e)^{-1}*x-\frac{1}{2}*\sqrt{x-(\alpha*\Pi*e+e)^{-1}*x}\right)*\alpha*\mu^{-1}[/math]

 

can be used to get more accurate determinations of the fine structure constant,

then the pebble that you so casually flicked in your " Non-Figurate Numbers" thread

could very well turn into an avalanche!

 

An entire mountain of knowledge and information landing right in our back yards!

 

Don.

[IDMclean]

I'm late to the game. I have a couple of straightforward questions: what exactly do we have to know about the numbers? Is it good enough to know if a given number in the set does or does not possess the property we're looking for? IE can we reduce this to a boolean sieve?

 

Can we algorithmically determine where the numbers are along the number line? Alternatively, can we algorithmically determine where the numbers are which do not possess the particular property we want? With prime numbers, you can't predict the position of prime numbers--given currently known formula--along the number line, but you can predict the position of the composites.

 

If we can both reduce it to a boolean sieve and we can predict the location of either the numbers matching the property or those that do not match, we can use a bit array set to an arbitrary length and jump through setting the bits to either 1 or 0 to produce the list.

[Don Blazys]

To: KickAssClown,

 

I'm not a "coder", so I don't understand your computer lingo at all,

but this much I do know...

 

Calculating how many polygonal numbers there are under a

given number [math]x[/math] is an exeedingly difficult problem.

 

To put it in context, computers have been able to count

all the prime numbers less than [math]x=10^{23}[/math],

but have not been able to count

all the polygonal numbers of rank >2 less than [math]10^{12}.[/math]

 

Donk is the hero who calculated all the polygonals of rank >2 less than [math]10^{11},[/math]

so I'm sure that he could fill you in on all the details and specifics

as to why such a seemingly simple generating formula

should result in calculations so arduous and daunting.

 

Craig D and Turtle are also quite knowledgable as to the difficulties involved.

(But if it weren't for the difficulties, then it wouldn't be any fun either!)

 

They are around here somewhere... :shrug:

 

Anyway, I hope your ideas work.

 

I can't help you, but you definitely have my "moral support",

and my thanks.

 

These truly fascinating numbers really do deserve a much higher count.

 

Don.

[modest]

I'm late to the game. I have a couple of straightforward questions: what exactly do we have to know about the numbers?

 

The conversation got started in 20236. A figurate number (aka polygonal number) is

[math]F = {(\frac{s}{2}-1)n^2-(\frac{s}{2}-2)n}[/math]

where s>2, n>2. A non-figurate number would then be any natural number [math]\neq[/math] F.

 

The idea was to find a figurate counting function analogous to the prime counting function that would approximate the cardinality of a set of figurate numbers smaller than n. Don believes the best candidate function has in it the fine structure constant.

 

Is it good enough to know if a given number in the set does or does not possess the property we're looking for? IE can we reduce this to a boolean sieve?

 

Can we algorithmically determine where the numbers are along the number line? Alternatively, can we algorithmically determine where the numbers are which do not possess the particular property we want? With prime numbers, you can't predict the position of prime numbers--given currently known formula--along the number line, but you can predict the position of the composites.

 

If we can both reduce it to a boolean sieve and we can predict the location of either the numbers matching the property or those that do not match, we can use a bit array set to an arbitrary length and jump through setting the bits to either 1 or 0 to produce the list.

 

Yes, if I understand you correctly then that is essentially the method that has been used. All F(s,n) : s>2, n>2 are punched out of an array, just like the sieve of Eratosthenes would do for the primes.

 

It is, on the other hand, not possible to look at a number, x, do a simple boolean operation,

x =? f(x)

and get a true response if the number is figurate.

 

~modest

 

EDIT: Sorry Don to repeat you. We posted about the same time and I didn't see your post

[IDMclean]

Modest, you caught my meaning dead on.

 

Just for clarification:

We're looking for the number of non-figurative numbers under an arbitrary limit.
All we need to know is whether a number is or is not figurative.

We can generate the figurative numbers directly, but there is no known function for generating the non-figurative numbers.
Figurative numbers are given by the function: 
[math] F(n, s) = {(\frac{s}{2}-1)n^2-(\frac{s}{2}-2)n}[/math] 
equivalently,
[math]F(n, s)=\frac{[(s-2)n^2-(s-4)n]}{2}[/math]
where [math]n, s \in \mathbb{N}[/math] satisfying [math]n, s>2[/math]

 

The main question which remains unanswered is can we jump to the position of the figurative numbers along the natural number line? IE do the figurative numbers have a one-to-one correspondence with the natural number line like the primes?

 

Once again with primes, my method is a variation of the sieve of Eratosthenes where rather than generating a integer list and eliminating each element which matches the form x*y. Instead, I generate a bitarray of n length with each element set to 1. I go to each element under n at index 2ij+i+j and set it to 0. The reasoning for this follows from

[math]f(n) = 2n+1[/math],

[math]f(i)*f(j) = x*y = f(2ij+i+j)\mbox{ where }n,i,j\in\mathbb{N}^+[/math]

 

Some additional questions to help us figure out if non-figuratives can be attack this way. What is the cardinality of figurative and non-figurative numbers? If the cardinality of the nonfigurative numbers is [math]\aleph_0[/math], we have a countably infinite list which would positively indicate that we can attack the problem by finding the index of the figurative numbers and falsifying them recursively.

 

Another question, can we reduce the formula for polygonal numbers to a one variable equation?

[Don Blazys]

To: IDMclean,

 

Quoting IDMclean:

The main question which remains unanswered is can we jump to the position of the figurative numbers along the natural number line?

 

The generating function repeats figurative numbers, and the ones that are repeated

(generated more than once) must then be eliminated.

 

Quoting IDMclean:

Another question, can we reduce the formula for polygonal numbers to a one variable equation?

 

That is algebraically (and conceptually) impossible.

 

Don.

[Don Blazys]

I have another "pretty wild" idea brewing in my mind.

 

How much would a "super-computer" powerfull enough to calculate... say... [math]\varpi(10^{23})[/math] cost ?

 

Could one be "rented"? Is it possible to "buy" computer time on a "super-computer" ?

 

Don.

[Don Blazys]

I updated the my website to show exactly how the proton to electron mass ratio

occurs in the Polygonal Number of Rank Greater Than 2 Counting Function.

 

You can find it here:

 

http://donblazys.com/on_polygonal_numbers.pdf

 

Don.

[Turtle]

I have another "pretty wild" idea brewing in my mind.

 

How much would a "super-computer" powerfull enough to calculate... say... [math]\varpi(10^{23})[/math] cost ?

 

Could one be "rented"? Is it possible to "buy" computer time on a "super-computer" ?

 

Don.

 

nice thinkin' lincoln! :smart: apparently, the answer is in the affirmative. :)

 

Supercomputing For Rent - Forbes.com

Supercomputing For Rent

...

Today Exa owns a network of 3,500 Intel and amd chips and rents thousands more, all housed in IBM's Poughkeepsie, N.Y. data center. Layering on its modeling software, Exa sells that processing as a service over the Internet, charging about a dollar per processor core per hour. That makes Exa one of a small number of companies offering supercomputers as a "cloud computing" option...

[modest]

Just for clarification:

We're looking for the number of non-figurative numbers under an arbitrary limit.
All we need to know is whether a number is or is not figurative.

We can generate the figurative numbers directly, but there is no known function for generating the non-figurative numbers.
Figurative numbers are given by the function: 
[math] F(n, s) = {(\frac{s}{2}-1)n^2-(\frac{s}{2}-2)n}[/math] 
equivalently,
[math]F(n, s)=\frac{[(s-2)n^2-(s-4)n]}{2}[/math]
where [math]n, s \in \mathbb{N}[/math] satisfying [math]n, s>2[/math]

 

That sounds right

 

The main question which remains unanswered is can we jump to the position of the figurative numbers along the natural number line? IE do the figurative numbers have a one-to-one correspondence with the natural number line like the primes?

 

Hummm.... I should think both the figurates and the non-figurates are a countably infinite set. According to wiki "Every subset of a countable set is countable. In particular, every infinite subset of a countably infinite set is countably infinite." The non-figs contain the primes so must be infinite. The figs likewise should be infinite and both are a subset of the natural numbers so both should be countable.

 

I'm not sure where that gets us.

 

Once again with primes, my method is a variation of the sieve of Eratosthenes where rather than generating a integer list and eliminating each element which matches the form x*y. Instead, I generate a bitarray of n length with each element set to 1. I go to each element under n at index 2ij+i+j and set it to 0. The reasoning for this follows from

[math]f(n) = 2n+1[/math],

[math]f(i)*f(j) = x*y = f(2ij+i+j)\mbox{ where }n,i,j\in\mathbb{N}_0[/math]

 

I'm afraid I don't follow. I don't know how x*y is used in the sieve of Eratosthenes, and I don't see how 2ij+i+j would give the primes or composites. I'm notoriously slow with this type of thing though.

 

Some additional questions to help us figure out if non-figuratives can be attack this way. What is the cardinality of figurative and non-figurative numbers?

 

Countably infinite... I'm pretty sure.

 

If the cardinality of the nonfigurative numbers is [math]\aleph_0[/math], we have a countably infinite list which would positively indicate that we can attack the problem by finding the index of the figurative numbers and falsifying them recursively.

 

I'm not sure what you mean by finding the index of the figurates. The main problem, as Don says, is that the only function we have 1) creates multiple entries of the same number and 2) has 2 variables. So the numbers generated from:

[math]F(s,n) = \left(\frac{s}2 -1\right)n^2 -\left(\frac{s}2 -2\right)n[/math]

are non-sequential and have multiple entries:

Another question, can we reduce the formula for polygonal numbers to a one variable equation?

That has been the holy grail of what we've been looking for. So far, no joy

 

~modest

[IDMclean]

[...] I should think both the figurates and the non-figurates are a countably infinite set. According to wiki "Every subset of a countable set is countable. In particular, every infinite subset of a countably infinite set is countably infinite." The non-figs contain the primes so must be infinite. The figs likewise should be infinite and both are a subset of the natural numbers so both should be countable.

 

I'm not sure where that gets us.

 

That gets us home. If we have a countably infinite set, we have a set which corresponds one for one to the natural numbers. This is good news, it means we can compute the sequence without fail. Also, it means we can array them into a sequence and find the figuratives by index. There exists an algorithm which decides "is the nth number of the integers a figurative number?"

 

A caveat, it may require two indexes, in which case it's a 2D array.

 

I'm afraid I don't follow. I don't know how x*y is used in the sieve of Eratosthenes, and I don't see how 2ij+i+j would give the primes or composites. I'm notoriously slow with this type of thing though.

The sieve of Eratosthenes works by taking each integer > 1 and eliminating all x*y. So it gets rid of (2*2, 2*3, ... x*y) etc. My refinement is that rather than generating the numbers, I generate the smaller indexes of the numbers. I ask "where is the nth composite on the odd integer number line?" Rather than storing numbers, I store bits. Simple 1 or 0. This allows for ~32x (standard integers are 32-bits) the compactness compared to storing a list of integers. 2ij + i + j is the n of any composite following the form of 2n+1. So I can ask where is the 0th composite along the odd integer number line (3, 5, 7, 9, ... 2n+1)? It's the 4th element of the sequence, n = 2(1)(1) + 1 + 1 = 4. It's equivalent to asking what is the number that is the product of 3*3 except, I ask where rather than what.

 

[math]2ij+i+j = \begin{array}{c|cccccc}

i,j & i=1 & i=2 & i=3 & i=4 & ... & i = n\\ \hline

j=1 & 4 & 7 & 10 & 13 & & \\

j=2 & 7 & 12 & 17 & 22 & & \\

j=3 & 10 & 17 & 24 & 31 & & \\

j=4 & 13 & 22 & 31 & 40 & & \\

... & & & & & & \\

j=n &3n+1 & 5n+2 & 7n+3 & 9n+4 & ... & 2(n^2+n) \end{array}[/math]

[math]f(2ij+i+j) = x*y = \begin{array}{c|cccc}

x,y & x=3 & x=5 & x=7 & x=9\\ \hline

y=3 & 9_{4} & 15_{7} & 21_{10} & 27_{13}\\

y=5 & 15_{7} & 25_{12} & 35_{17} & 45_{22}\\

y=7 & 21_{10} & 35_{17} & 49_{24} & 63_{31}\\

y=9 & 27_{13} & 45_{22} & 63_{31} & 81_{40}\end{array}[/math]

 

If we can produce a similar table for the figurative numbers, I can write a program for very quickly sieving the figuratives out that will be extremely compact in memory.

[Don Blazys]

To: Modest, Turtle, Donk and IDMclean,

 

Quoting Turtle:

nice thinkin' lincoln! apparently, the answer is in the affirmative.

 

Thanks Turtle.

 

Please don't tell anybody, but I really don't know anything about computers.

 

(For pete's sake, I still type with one finger!)

 

Does this mean that Donk can "hook into" their "super computer" from home or work?

 

Can one of you "coders" please do the necessary research to realistically estimate

what a determination of [math]\varpi (x)[/math] would actually cost for values of [math]x[/math] ranging from: [math]10^{20}[/math] to...say... [math]10^{100}[/math] ?

 

Maybe it's affordable !

 

Don.

[JMJones0424]

Maybe it's affordable !

 

I have no idea what process one must go through to submit a project, but has anyone considered Berkeley Open Infrastructure for Network Computing - Wikipedia, the free encyclopedia as a platform for allowing donation of free cycles of hypography users for computing your problem?

 

Homepage - BOINC

 

According to the project list

BOINC can be used by anybody for their own distributed computing projects, either public or private.

[Don Blazys]

To: JMJones0424,

 

That is exiting.

 

I'm sure that someone here can get in contact with them

and figure it out...for here is where the :smart: reside!

 

Since my counting function for "non-trivial polygonals" also involves

the proton to electron mass ratio [math]\mu[/math],

its usefullness in calculating the fine structure constant [math]\alpha[/math] is limited.

 

Also, since the "random fluctuations" in [math]\varpi(x)[/math]

seem to be increasing (at an unknown rate),

it's hard to say how high we need to calculate it

before those "random fluctuations" are sufficiently negligible.

 

(If they never become "sufficiently negligible",

then we will have to calculate the "maximums" of [math]\varpi(x)[/math],

and use them along with their associated values of [math]x[/math] to calculate [math]\alpha[/math].)

 

However, since [math]\mu[/math] has been

calculated to more decimal places than [math]\alpha[/math],

we should be able to at least confirm the current value of [math]\alpha[/math]

and perhaps even calculate it to a few extra decimal places

given sufficiently high determinations of [math]\varpi(x)[/math].

 

Don.

[CraigD]

I have another "pretty wild" idea brewing in my mind.

 

How much would a "super-computer" powerfull enough to calculate... say... [math]\varpi(10^{23})[/math] cost ?

 

Could one be "rented"? Is it possible to "buy" computer time on a "super-computer" ?

Can one of you "coders" please do the necessary research to realistically estimate

what a determination of [math]\varpi (x)[/math] would actually cost for values of [math]x[/math] ranging from: [math]10^{20}[/math] to...say... [math]10^{100}[/math] ?

A key feature of counting the polygonal numbers is that, because programs to do it (eg: see Another non-filter approach to generating polygonal numbers) can use “spigot” algorithms, the calculation can be divided into many small ranges – that it, parallelized This means it can be done on many small, non-super computers, and the resulting counts summed. So an expensive supercomputer isn’t necessary, and is actually a waste of supercomputing resources, which are best used for problems that can’t or can’t easily be parallelized.

 

A variety of parallelized approaches can give us a “virtual supercomputer”, such as:

• Using a commercial “cloud computing” service. From my limited reading about them, I’ll hazard a guess that providers like Amazon, which lets you compile and execute any code you want, are better choices than ones like Google, that require you only run prorams in interpreted languages like Python.
  
• A “beowolf cluster” of inexpensive machine, such as pallet loads of old company commodity PCs destine for the recycling dumpster. Requires a big enough room, a spike in you electric bill, maybe adding electric circuits to your home or office, and someone tech-y enough to install Linux on hundreds of PCs and get them all talking to your PC – or finding and borrowing a cluster that’s already been built but isn’t being used much.
  
• ”Charitable” distributed computing, like SETI@home, which invites hundreds or thousands of people to install a program you write that
  • contacts your server to get an “assignment”,
  • processes it when the machine is idle,
  • sends the result back to your server, and repeats.

Regardless of the approach chose, a first step – which should let us get a count for another power of ten or two – is to code an efficient spigot algorithm. I suspect the algorithm of mine I prototyped in the very numerically inefficient MUMPS interpreted language is nearly as efficient as possible – time permitting, I’ll write a faster, compiled language (eg: C) version in the next day or so, or if someone wants to write one before me, explain anything about the algorithm that isn’t clear from the posted prototype code.

 

I updated the my website to show exactly how the proton to electron mass ratio

occurs in the Polygonal Number of Rank Greater Than 2 Counting Function.

...

http://donblazys.com/on_polygonal_numbers.pdf

Nice paper, Don :thumbs_up

 

Reading it, I had some thoughts and suggestions, which I'll post later.

Hummm.... I should think both the figurates and the non-figurates are a countably infinite set.

It’s trivially true that the 2+ sided figurate numbers is an infinite set, as {6, 9, 12, ...} are figurate numbers.

 

While my intuition says the non-figurate numbers are also infinite, I’ve not proven it of seen a proof, so can’t rule out the possibility that there are a finite number of non-figurate numbers (and consequently, a greatest figurate number). Good mathematical rigor requires such a proof. :shrug: :confused:

There exists an algorithm which decides "is the nth number of the integers a figurative number?"

Yes, at least one – see Another non-filter approach to generating polygonal numbers, also linked to above.

 

My most algorithm does about [imath]\frac{\ln(x)}3[/imath] iterations (each iteration having 12 multiplication, 2 division, 9 additions, and 1 costly square root operation) to determine if x is non-figurate. The largest non-figurate number I’ve tried using my program, [imath]10^{17}+3[/imath], took 1105207 iterations.

Can one of you "coders" please do the necessary research to realistically estimate

what a determination of [math]\varpi (x)[/math] would actually cost for values of [math]x[/math] ranging from: [math]10^{20}[/math] to...say... [math]10^{100}[/math] ?

I won’t try making a money estimate (I you knew my professional history at this, you’d be glad ;)), but assuming there’s not an algorithm qualitatively more efficient than the best I’ve seen so far, we can estimate an upper limit on how high we can practically count.

 

As I note above, the best algorithm I know needs about [imath]\frac{\ln(x)}3[/imath] iterations (compute cycles) to check a candidate non-figurate number [imath]x[/imath], so the number of cycles needed to count all non-figurate numbers [imath]<n[/imath] is about [imath]\frac{n\ln(n)}3[/imath].

 

The world has about [imath]10^9[/imath] processors of average about [imath]10^{12}[/imath] cycles/s (compared to all the processors in the world, the world’s few supercomputers, capable of about [imath]10^{15}[/imath] cycles/s, are negligible).

 

There are about [imath]10^7[/imath] seconds in 4 months, so the world can perform a maximum of about [imath]10^{28}[/imath] iterations in 4 months.

 

Solving [math]10^{28} = \frac{n\ln(n)}3[/math] for [imath]n[/imath] gives [math]n = e^{\omega(3 \cdot 10^{28})} \dot= 4.9 \times 10^{26}[/math], where [imath]\omega(z)[/imath] is the product log (“Lambert omega”), function

 

So, assuming the best case of every CPU in the world being well-suited to our number-checking algorithm, all the computing power in the world could calculate the non-figurate numbers less than on the order of [imath]n = 10^{26}[/imath] ([imath]\varpi(10^{26})[/imath]) in a few days, weeks, or months.

 

[imath]\varpi(n)[/imath] values much higher than this seem beyond what will be possible within our lifetimes. Calculating [imath]\varpi(10^{100})[/imath] seems to me beyond physically possible given all the matter and time in the universe.

[modest]

The sieve of Eratosthenes works by taking each integer > 1 and eliminating all x*y. So it gets rid of (2*2, 2*3, ... x*y) etc.

 

Of course, I don't know what I was thinking :doh:

 

2ij + i + j is the n of any composite following the form of 2n+1. So I can ask where is the 0th composite along the odd integer number line (3, 5, 7, 9, ... 2n+1)? It's the 4th element of the sequence, n = 2(1)(1) + 1 + 1 = 4. It's equivalent to asking what is the number that is the product of 3*3 except, I ask where rather than what.

 

I see. I had suspected that, but after plugging i=0, j=3 into 2(2ij + i + j )+1 and getting seven I was convinced otherwise. I believe where you said [math]\mathbb{N}_0[/math]:

[math]f(n) = 2n+1[/math],

[math]f(i)*f(j) = x*y = f(2ij+i+j)\mbox{ where }n,i,j\in\mathbb{N}_0[/math]

It should be [math]\mathbb{N}_1[/math]. I'm cleared up now though

 

If we can produce a similar table for the figurative numbers, I can write a program for very quickly sieving the figuratives out that will be extremely compact in memory.

 

The best we have is [math]F(s,n) = \left(\frac{s}2 -1\right)n^2 -\left(\frac{s}2 -2\right)n

[/math]. It makes sense to me how you've made an alternate, smaller, index for the primes, but I don't see an easy way to do that with the figurates.

 

 

 

My most algorithm does about [imath]\frac{\ln(x)}3[/imath] iterations (each iteration having 12 multiplication, 2 division, 9 additions, and 1 costly square root operation) to determine if x is non-figurate.

 

I believe these are the equations you're working with:

[math]x_n = \left(\frac{s}2 -1\right)n^2 -\left(\frac{s}2 -2\right)n[/math]

[math]n = \frac{\sqrt{(8s -16)x +(s-4)^2 } +s-4}{2s -4}[/math]

[math]s = \frac{2x +2n^2 -4n}{n^2 -n}[/math]

While I can't get rid of the square root, you might do marginally better with these:

[math]F = \frac{1}{2} \left( n^2s + 2n - sn \right)[/math]

[math]n = \frac{\sqrt{8Fs+s^2-4s+4}-s+2}{2s}[/math]

[math]s = \frac{2(F-n)}{n(n-1)}[/math]

[math]s > 0, n > 2[/math]

 

~modest