Calculating The Fine Structure Constant.

[Don Blazys]

[math]x[/math]_______[math]\varpi(x)[/math]____[math]B(x)*\left(\frac{\mu-e^{\frac{\pi}{2}}-\alpha}{\mu-e^{\frac{\pi}{2}}}\right)[/math]_____Diff_____%Error

10^1____3_______________5_____________________2_____.66666

10^2____57______________60____________________3_____.05263

10^3____622_____________628___________________6_____.00964

10^4____6,357___________6,364__________________7_____.00110

10^5____63,889__________63,910_________________21____.00032

10^6____639,946_________639,963________________17____.00002656

10^7____6,402,325________6,402,362______________37 ____.00000577915

10^8____64,032,121_______64,032,273_____________152___.00000237380

10^9____640,349,979______640,350,091____________112___.00000017490

10^10___6,403,587,409____6,403,587,428___________19____.00000000297

10^11___64,036,148,166___64,036,147,870__________-296__-.00000000462

10^12___640,362,343,980__640,362,343,856_________-124__-.00000000019

 

Here, we have the same function, exept with [math]k=e^{\frac{\pi}{2}}[/math] instead of [math]k=2*e[/math].

So far, the accuracy is mind blowing, but obviously, we need higher values of [math]\varpi(x)[/math]

in order to get a better idea of what the actual value of [math]k[/math] really is.

 

Don.

[IDMclean]

How about [math]e^{\frac{x}{4}}[/math]? Also, enlighten me on what the e and [math]\mu[/math] terms are again?

[Don Blazys]

To: IDMclean,

 

Quoting IDMclean:

How about [math]e^{\frac{x}{4}}[/math] ? Also, enlighten me on what the e and [math]\mu[/math] terms are again?

 

If [math]k[/math] is a predictable function, then it is some really crazy predictable function

that is either fluctuating somewhere between 4 and 6...who knows... :shrug:

or slooooowwwly descending towards 0...again...who knows... :shrug:

 

Thus, it can't possibly be [math]e^{\frac{x}{4}}[/math] .

 

Right now, it appears that [math]k[/math] is simply some constant whose value is

somewhere between 4 and 6.

 

Heck, maybe it's 5. :hyper:

 

The "term" (or multiplier) containing [math]\mu[/math] is a sort of "error term",

 

and [math]\mu[/math] itself is the "proton to electron mass ratio".

 

Don.

[IDMclean]

Is it possible that k is a complex number/has a complex solution?

[Pyrotex]

Is it possible that k is a complex number/has a complex solution?

I think k = 2 * e,

where e is the base of the natural logarithms.

In that case, e is a transcendental number (like pi) and is not the solution of any polynomial.

[Rade]

Please see this paper by T. Y. Thomas (1965):

RED SHIFT DISCRETIZATION IN THE EPOCHAL COSMOLOGY — PNAS

 

The first equation by A. G. Wilson related to red shift in clusters of galaxies has the number 137.5. If I am reading this paper correct, this equation of Wilson is related to the mass and radius of the universe. And, Thomas shows that the number 137 is related to two integers m and n that relate to the mass and radius of the universe, such that ~137 = [math]m^{2}n^{2}[/math].

 

Is it possible that the number 137 derives from the geometry of the universe itself, thus a reason why so many aspects of the physical universe are related to the number 137 ?

 

Perhaps if the Wilson equation was modified to the quantum value of the fine structure constant 137.035.... it may be more predictive of galaxy red shifts ?

 

Just looking for leads.

[Don Blazys]

To: Rade,

 

Quoting Rade:

Just looking for leads.

 

In Turtles thread: "Non-Figurate Numbers",

 

Post #180 by Modest relates polygonal numbers and Pascal numbers via "dimensionality".

 

Fascinating stuff!

 

Don.

[Don Blazys]

To: Pyrotex,

 

Quoting Pyrotex:

I think k = 2 * e.

 

I too am leaning towards that, but only because it "looks good". :dogwalk: :lightning :thumbs_up

 

Do you have a better reason?

 

Don.

[Don Blazys]

I updated the counting function on my website:

 

http://donblazys.com/on_polygonal_numbers.pdf

 

Don.

[IDMclean]

If the fine structure constant is irrational, maybe this little article will help with it's calculation: Bailey-Borwein-Plouffe formula

It's a spigot formula.

 

Since this discovery, many formulas for other irrational constants have been discovered of the general form

 

[math] \alpha = \sum^{\infty}_{k=0}[\frac{1}{b^k}\frac{p(k)}{q(k)}][/math]

 

where α is the constant and p and q are polynomials in integer coefficients and b ≥ 2 is an integer numerical base.

Formulas in this form are known as BBP-type formulas.[3] Certain combinations of specific p, q and b result in well-known constants, but there is no systematic algorithm for finding the appropriate combinations and known formulas are discovered through experimental mathematics.

[IDMclean]

If the primes are a subset of the nonfiguratives, does it stand to reason that a counting function for the nonfiguratives would include the counting function for primes?

Given that, would this (if it's not already) be of use to us: Logarithmic integral function.

[IDMclean]

[math]

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

[/math]

 

[math]

\left(\frac{(-\mu+k+\alpha)\sqrt{x-(\frac{x}{\alpha*\Pi*e+e})}}{2^2 * \mu * \alpha*\Pi*e - 2^2 * k * \alpha*\Pi*e + 2^2 * e * \mu - 2^2*e*k}\right)

[/math]

 

[math]

\frac{k\sqrt{x-(\frac{x}{\alpha*\Pi*e+e})}}{(2^2 * \mu * \alpha*\Pi*e)+(2^2 * e * \mu)}

[/math][math]

+ \frac{\mu\sqrt{x-(\frac{x}{\alpha*\Pi*e+e})}}{(2^2 * k * \alpha*\Pi*e) + (2^2*e*k)}

[/math][math]

+ \frac{\alpha\sqrt{x-(\frac{x}{\alpha*\Pi*e+e})}}{(2^2 * e * \mu)-(2^2*e*k)}

[/math][math]

+ \frac{\sqrt{x-(\frac{x}{\alpha*\Pi*e+e})}}{(2^2 * \mu *\Pi*e) - (2^2 * k *\Pi*e) - (2 * \alpha*\Pi*e) - (2 * e)}

[/math]

 

That was hairy. I hope I didn't make a mistake.

[Don Blazys]

To: IDMclean,

 

If the primes are a subset of the nonfiguratives,

does it stand to reason that a counting function for the nonfiguratives

would include the counting function for primes?

Given that, would this (if it's not already) be of use to us:

Logarithmic integral function.

 

x minus the counting function for figuratives (non-trivial polygonals)

gives us the counting function for non-figuratives.

 

Then, the counting function for non-figuratives

minus the counting function for primes (the "logarithmic integral" Li(x))

gives us the counting function for non-figuratives-non-primes.

 

However, for this purpose,

I already have a counting function for primes

that appears to be more accurate than Li(x),

and is perhaps somewhat more "suitable" in that

unlike Li(x), which is mostly an overestimate,

it appears to be mostly an underestimate.

 

It also has an advantage over Li(x) in that

it has the general form x-C

where C is the counting function for composites.

 

The reason that I haven't posted it yet is because it uses four constants,

and I'm only "tentatively certain" as to what two of them are.

The other two remain a complete mystery,

and I can only estimate what their values really are.

 

Quoting IDMclean:

That was hairy. I hope I didn't make a mistake.

 

That is indeed a most impressive display of algebraic fireworks!

 

I espesially like the way that you broke it up into four terms,

each with the radical in the numerator preceded by the coefficients

k, mu, alpha and (implied) unity.

 

Simply beautiful!

 

I have so little time for posting these days, but I will check it the first chance I get.

 

Don.

[Don Blazys]

The problem of closely estimating the number of

figurates and non-figurates is almost solved.

 

All we really need is to determine the value of k.

Pyrotex and I believe that k=2*e,

but the only way to be sure is to determine [math]\varpi(x)[/math]

to perhaps [math]\varpi(10^{14})[/math] or so.

 

Donks code in some spare, unused computer would give us that

in a matter of months... or, at worst... years.

 

For a short while, the conversation was on

buying, renting, or somehow aquiring the use of

a "supercomputer" or an "array" of computers,

but that conversation quickly "fizzled".

 

That's really a shame, because by the end of this year

(or the beginning of the next), a new estimate of

the fine structure constant will be released,

and I think that it would be great fun to get our own

"Hypography estimate of the fine structure constant"

using the function in this thread, before that estimate is released.

 

I am not a coder, and have very little time for math these days,

but what we have is indeed extraordinarily accurate, and in my opinion,

Turtles numbers deserve a counting function that is complete.

 

Don.

[IDMclean]

Wolfram Alpha's analysis of our Math

 

I tried for about two and half hours to get it to count the number of figuratives under x, but I couldn't figure it out.

[Don Blazys]

Nvidia Tesla Personal Supercomputer - Wikipedia, the free encyclopedia

[Don Blazys]

These "personal supercomputers" cost from $1500.00 to $60,000.00.

 

So, how much would I have to spend

in order to calculate w(10^24)

in a reasonable amount of time?

 

Don.