-
Features
Integer formula for transcendental function group
- Thread starter Kharakov
- Start date
4321lynx
Veteran Member
You may find the Phisics Forums helpful. They have a whole section of Mathematics.
https://www.physicsforums.com/
https://www.physicsforums.com/
Kharakov
Quantum Hot Dog
Yeah. Maybe. But I'll ask the people who know and hate me first, so they can ignore me. 
So, reworked the math, realized when I was looking at it, that I was drawing correlation between 2 different sets of related equations, and I confused the results. So here is the corrected rewrite of part of the question. Hopefully it's a bit more coherent. Hopefully it's coherent. Hopefully you're coherent.
I'm sure once someone says something along the lines of "that's Bernouli's so and so...." I'll drop the matter. Couldn't find the answer in Dixon's "A Chronology of Continued Square Roots and Other Continued Functions".
Excerpt:
QUESTION: Is there a general formula for the integer coefficients an of the following (presumably) transcendental functions?***
There is a set of numbers defined by (with k= to the number of radicals + 1):
\( Q_{x,n,k} = \left( nx^{n-1} \right)^k \, \times \left[x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \sqrt[n]{x^n-x+...}}} \right]\)
with constants associated with the numbers defined by:
\(q_{x,n} = Q_{x,n,k \to \infty}\)
with
\(q_{2,2} = \pi^2\)
When \(q_{x,n}\) is known, it appears as if there is a general formula for other \(Q_{x,n,k}\), with k being some natural number>1:
\(a_n=\) rational integer coefficients of terms (if x is rational and n is a natural number greater than 1) ***What is the formula for rational an, if there is a formula?
\(b= nx^{n-1}\)
\(\gamma= q_{x,n}\)
\(Q_{x,n,k} = \frac {\gamma}{a_1 \, b^{0k}} - \frac {\gamma^2}{a_2 \, b^{1k}} +\frac {\gamma^3}{a_3 \, b^{2k}}- \frac {\gamma^4}{a_4 \, b^{3k}} \,\, ...\)
The pi^2 case is easy to see, the coefficients are \(a_n= \frac{(2n)!}{2} = \, 1,12,360,20160... \)
\(q_{2,2}=\pi^2\)
\(Q_{2,2,k} = \,\,\ \frac{ pi^2}{ 4^{0k}} - \frac{pi^4}{12 \times 4^{1k}} +\frac{ pi^6}{360 \times 4^{2k}} - \frac{ pi^8}{20160 \times 4^{3k}} \dots \)
So
\(Q_{2,2,3} = \,\,\ \frac{ pi^2}{ 4^{0}} - \frac{pi^4}{12 \times 4^{3}} +\frac{ pi^6}{360 \times 4^{6}} - \frac{ pi^8}{20160 \times 4^{9}}\,\,\, \dots = \,\,\, 4^3 \, \times \left[2 - \sqrt{2 + \sqrt{2}} \right] \)
Wolfram Alpha Code:
Sample partial integer coefficient lists of a few other Q, with different associated q:
1,6,72,1404...
\(Q_{1.5,2,k} = \,\,\ \frac{q_{1.5,2}}{1 \time 3^{0k}} - \frac{q_{1.5,2}^2}{6 \times 3^{1k}} +\frac{q_{1.5,2}^3}{72\times 3^{2k}} - \frac{q_{1.5,2}^4}{1404\times 3^{3k}} + \frac{q _{1.5,2}^5}{\frac{673920\times 3^{4k}}{17}} - \dots \)
1,20,1200,148800....
\(Q_{2.5,2,k} = \,\,\ \frac{q_{2.5,2}}{ 5^{0k}} - \frac{q_{2.5,2}^2}{20 \times 5^{1k}} +\frac{q_{2.5,2}^3}{1200 \times 5^{2k}} - \frac{ q_{2.5,2}^4}{148800 \times 5^{3k}} +\frac{q_{2.5,2}^5}{32240000 \times 5^{4k}} \dots \)
1,30,3150, ?? 8127000/11 ??....
\(Q_{3,2,k} = \,\,\ \frac{q_{3,2}}{6^{0k}} - \frac{q_{3,2}^2}{30 \times 6^{1k}} +\frac{q_{3,2}^3}{3150\times 6^{2k}} - \frac{q_{3,2}^4}{\frac{8127000 \times 6^{3k}}{11}} + \frac{q_{3,2}^5}{\frac{31573395000\times 6^{4k}}{97}} - \dots \)
General formula for an?
So, reworked the math, realized when I was looking at it, that I was drawing correlation between 2 different sets of related equations, and I confused the results. So here is the corrected rewrite of part of the question. Hopefully it's a bit more coherent. Hopefully it's coherent. Hopefully you're coherent.
I'm sure once someone says something along the lines of "that's Bernouli's so and so...." I'll drop the matter. Couldn't find the answer in Dixon's "A Chronology of Continued Square Roots and Other Continued Functions".
Excerpt:
QUESTION: Is there a general formula for the integer coefficients an of the following (presumably) transcendental functions?***
There is a set of numbers defined by (with k= to the number of radicals + 1):
\( Q_{x,n,k} = \left( nx^{n-1} \right)^k \, \times \left[x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \sqrt[n]{x^n-x+...}}} \right]\)
with constants associated with the numbers defined by:
\(q_{x,n} = Q_{x,n,k \to \infty}\)
with
\(q_{2,2} = \pi^2\)
When \(q_{x,n}\) is known, it appears as if there is a general formula for other \(Q_{x,n,k}\), with k being some natural number>1:
\(a_n=\) rational integer coefficients of terms (if x is rational and n is a natural number greater than 1) ***What is the formula for rational an, if there is a formula?
\(b= nx^{n-1}\)
\(\gamma= q_{x,n}\)
\(Q_{x,n,k} = \frac {\gamma}{a_1 \, b^{0k}} - \frac {\gamma^2}{a_2 \, b^{1k}} +\frac {\gamma^3}{a_3 \, b^{2k}}- \frac {\gamma^4}{a_4 \, b^{3k}} \,\, ...\)
The pi^2 case is easy to see, the coefficients are \(a_n= \frac{(2n)!}{2} = \, 1,12,360,20160... \)
\(q_{2,2}=\pi^2\)
\(Q_{2,2,k} = \,\,\ \frac{ pi^2}{ 4^{0k}} - \frac{pi^4}{12 \times 4^{1k}} +\frac{ pi^6}{360 \times 4^{2k}} - \frac{ pi^8}{20160 \times 4^{3k}} \dots \)
So
\(Q_{2,2,3} = \,\,\ \frac{ pi^2}{ 4^{0}} - \frac{pi^4}{12 \times 4^{3}} +\frac{ pi^6}{360 \times 4^{6}} - \frac{ pi^8}{20160 \times 4^{9}}\,\,\, \dots = \,\,\, 4^3 \, \times \left[2 - \sqrt{2 + \sqrt{2}} \right] \)
Wolfram Alpha Code:
Sample partial integer coefficient lists of a few other Q, with different associated q:
1,6,72,1404...
\(Q_{1.5,2,k} = \,\,\ \frac{q_{1.5,2}}{1 \time 3^{0k}} - \frac{q_{1.5,2}^2}{6 \times 3^{1k}} +\frac{q_{1.5,2}^3}{72\times 3^{2k}} - \frac{q_{1.5,2}^4}{1404\times 3^{3k}} + \frac{q _{1.5,2}^5}{\frac{673920\times 3^{4k}}{17}} - \dots \)
1,20,1200,148800....
\(Q_{2.5,2,k} = \,\,\ \frac{q_{2.5,2}}{ 5^{0k}} - \frac{q_{2.5,2}^2}{20 \times 5^{1k}} +\frac{q_{2.5,2}^3}{1200 \times 5^{2k}} - \frac{ q_{2.5,2}^4}{148800 \times 5^{3k}} +\frac{q_{2.5,2}^5}{32240000 \times 5^{4k}} \dots \)
1,30,3150, ?? 8127000/11 ??....
\(Q_{3,2,k} = \,\,\ \frac{q_{3,2}}{6^{0k}} - \frac{q_{3,2}^2}{30 \times 6^{1k}} +\frac{q_{3,2}^3}{3150\times 6^{2k}} - \frac{q_{3,2}^4}{\frac{8127000 \times 6^{3k}}{11}} + \frac{q_{3,2}^5}{\frac{31573395000\times 6^{4k}}{97}} - \dots \)
General formula for an?
Kharakov
Quantum Hot Dog
I found that each of the numbers generated by this function has a sine-like expansion associated with it as well. So you can take \(\frac{\sqrt{q_{x,n}}}{2}\) and use it in the sine-like function to extract sine-like behaviors. Weird. It's like the various sine-like functions and cosine-like functions have different rotation periods. I wonder if the behaviors are exactly the same for the different functions?
Please help map a few.
Here is a partial rewrite, with some addition pointers to describe what I'm doing a bit more thoroughly.
The following function generates values with 3 inputs, x, n, and integer k (with k= to the number of radicals +1):
\(Q_{x,n,k} = \left( nx^{n-1} \right)^k \, \times \left[x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \sqrt[n]{x^n-x+...}}} \right]\)
another way of looking at it is:
I use the following to identify the values that are generated by the above function as the number of radicals, k, approaches infinity:
\(q_{x,n} = Q_{x,n,k \to \infty}\)
For example: \(q_{2,2} = \lim_{k\to\infty} \,\, Q_{2,2,k} = \pi^2\)
In the following expansion for the cosine like function, so called because it is very similar to the Taylor for cosine for the pi^2 case, I use the following to denote parts of the expansion:
\(a_n\) is the coefficients of specific terms
\(b= nx^{n-1}\) is the nx^(n-1)
\(\gamma= q_{x,n}\)
\(Qcos_{x,n,k} = \frac {\gamma}{a_2 \, b^{0k}} - \frac {\gamma^2}{a_4 \, b^{1k}} +\frac {\gamma^3}{a_6 \, b^{2k}}- \frac {\gamma^4}{a_8 \, b^{3k}} \,\, ...\)
\(\Gamma= \frac{\sqrt{q_{x,n}}}{2}\)
\(Qsin_{x,n,k} =\frac {\Gamma}{a_1 \, b^{0k}} - \frac {\Gamma^3}{a_3 \, b^{1k}} + \frac {\Gamma^5}{a_5 \, b^{2k}} -\frac {\Gamma^7}{a_7 \, b^{3k}}+ \frac {\Gamma^9}{a_9 \, b^{4k}} \,\, ...\)
Sample partial integer coefficient lists and partial write ups of a few q_{x,2} cosine like functions:
Here is an example of the pi cosine like function (which looks off to me for some reason- should generate 2 instead of 4 at k=0, and I could massage it so it does, but I got those a_n!! maybe it's me... sigh):
\(Qcos_{2,2,3} = \,\,\ \frac{ pi^2}{ 4^{0}} - \frac{pi^4}{12 \times 4^{3}} +\frac{ pi^6}{360 \times 4^{6}} - \frac{ pi^8}{20160 \times 4^{9}}\,\,\, \dots = \,\,\, 4^3 \, \times \left[2 - \sqrt{2 + \sqrt{2}} \right] \)
A couple of combined lists for exponential like functions for
I'd like to figure out a generating scheme for the successive numbers. The first series of cosine numbers (other than one) appears to be 2x(2x+1):6,12,20,30... if you do x=1+.5*natural. The second is f(x):=(2*x-1)*(2*x) *(2*x-1)*(2*x)*(2*x+1)/2 for 72,360.... it also works for non integer x (experimented with 4x^2-2x=11), but it's harder to see factors.
Please help map a few.
Here is a partial rewrite, with some addition pointers to describe what I'm doing a bit more thoroughly.
The following function generates values with 3 inputs, x, n, and integer k (with k= to the number of radicals +1):
\(Q_{x,n,k} = \left( nx^{n-1} \right)^k \, \times \left[x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \sqrt[n]{x^n-x+...}}} \right]\)
another way of looking at it is:
I use the following to identify the values that are generated by the above function as the number of radicals, k, approaches infinity:
\(q_{x,n} = Q_{x,n,k \to \infty}\)
For example: \(q_{2,2} = \lim_{k\to\infty} \,\, Q_{2,2,k} = \pi^2\)
In the following expansion for the cosine like function, so called because it is very similar to the Taylor for cosine for the pi^2 case, I use the following to denote parts of the expansion:
\(a_n\) is the coefficients of specific terms
\(b= nx^{n-1}\) is the nx^(n-1)
\(\gamma= q_{x,n}\)
\(Qcos_{x,n,k} = \frac {\gamma}{a_2 \, b^{0k}} - \frac {\gamma^2}{a_4 \, b^{1k}} +\frac {\gamma^3}{a_6 \, b^{2k}}- \frac {\gamma^4}{a_8 \, b^{3k}} \,\, ...\)
\(\Gamma= \frac{\sqrt{q_{x,n}}}{2}\)
\(Qsin_{x,n,k} =\frac {\Gamma}{a_1 \, b^{0k}} - \frac {\Gamma^3}{a_3 \, b^{1k}} + \frac {\Gamma^5}{a_5 \, b^{2k}} -\frac {\Gamma^7}{a_7 \, b^{3k}}+ \frac {\Gamma^9}{a_9 \, b^{4k}} \,\, ...\)
Sample partial integer coefficient lists and partial write ups of a few q_{x,2} cosine like functions:
Here is an example of the pi cosine like function (which looks off to me for some reason- should generate 2 instead of 4 at k=0, and I could massage it so it does, but I got those a_n!! maybe it's me... sigh):
\(Qcos_{2,2,3} = \,\,\ \frac{ pi^2}{ 4^{0}} - \frac{pi^4}{12 \times 4^{3}} +\frac{ pi^6}{360 \times 4^{6}} - \frac{ pi^8}{20160 \times 4^{9}}\,\,\, \dots = \,\,\, 4^3 \, \times \left[2 - \sqrt{2 + \sqrt{2}} \right] \)
A couple of combined lists for exponential like functions for
I'd like to figure out a generating scheme for the successive numbers. The first series of cosine numbers (other than one) appears to be 2x(2x+1):6,12,20,30... if you do x=1+.5*natural. The second is f(x):=(2*x-1)*(2*x) *(2*x-1)*(2*x)*(2*x+1)/2 for 72,360.... it also works for non integer x (experimented with 4x^2-2x=11), but it's harder to see factors.
steve_bank
Diabetic retinopathy and poor eyesight. Typos ...
Your math is beyond me, and actually gives me a head ache.
But when you said sine like series the Fourier Transform comes to mind. Given a function the FT translates a function into a sin-cos series. It appears everywhere in applied math and technology.
But when you said sine like series the Fourier Transform comes to mind. Given a function the FT translates a function into a sin-cos series. It appears everywhere in applied math and technology.
Kharakov
Quantum Hot Dog
It's not beyond anyone, least of all an engineer.
If I said "you take the carburetor, hook it up to intakes on the engine block, hook up each of the 6 spark wires from the distributor cap, hook up the battery, pull the choke, turn the key, and press the throttle pedal to see if it revs, if it turns over push in the choke", your mind would understand what I'm talking about.
What I described has less steps.
I'm saying put 2 numbers (x and n) into an equation to get one number c:
c=x^n-x
Use c in a repetitive function for k-1 steps (one step, pistons moving don't count as multiple steps):
\( r_0= 0 \\ r_1 = \sqrt[n]{r_0 +c} \\ r_2 = \sqrt[n]{r_1 + c} \\ ... \\ r_{k-1} = \sqrt[n]{r_{k-2} +c}\)
When you take the limit as k-->infty, which is basically saying make the number of steps really big, you get some specific value associated with specific (x and n), which are very close to x. In other words, when you repeat the operation rnext = square root (rlast +c) a bunch of times, you get a number that is arbitrarily close to x, depending on how many times you do the operation.
Next, you take the difference between x and rk-1 and multiply it by the derivative of x^n taken to the kth power to get your constant (such as pi^2):
Qx,n,k = (nxn-1)k * (x-rk-1)
You have your constant in 3 steps:
1) create c
2) make rk-1
3) multiply (x- rk-1) by (nxn-1)k
This constant is then used in functions (I'll tell you how I generate the functions at another time) that are either exactly like sine and cosine expansions (in the case of n and x being equal to 2, which makes c= to 2 as well... who does #2 work for?), or they have very similar behaviors and structure to sine and cosine.
example x=2,n=2:
So far, the e case an coefficients is easiest to describe: an = n!.
If I said "you take the carburetor, hook it up to intakes on the engine block, hook up each of the 6 spark wires from the distributor cap, hook up the battery, pull the choke, turn the key, and press the throttle pedal to see if it revs, if it turns over push in the choke", your mind would understand what I'm talking about.
What I described has less steps.
I'm saying put 2 numbers (x and n) into an equation to get one number c:
c=x^n-x
Use c in a repetitive function for k-1 steps (one step, pistons moving don't count as multiple steps):
\( r_0= 0 \\ r_1 = \sqrt[n]{r_0 +c} \\ r_2 = \sqrt[n]{r_1 + c} \\ ... \\ r_{k-1} = \sqrt[n]{r_{k-2} +c}\)
When you take the limit as k-->infty, which is basically saying make the number of steps really big, you get some specific value associated with specific (x and n), which are very close to x. In other words, when you repeat the operation rnext = square root (rlast +c) a bunch of times, you get a number that is arbitrarily close to x, depending on how many times you do the operation.
Next, you take the difference between x and rk-1 and multiply it by the derivative of x^n taken to the kth power to get your constant (such as pi^2):
Qx,n,k = (nxn-1)k * (x-rk-1)
You have your constant in 3 steps:
1) create c
2) make rk-1
3) multiply (x- rk-1) by (nxn-1)k
This constant is then used in functions (I'll tell you how I generate the functions at another time) that are either exactly like sine and cosine expansions (in the case of n and x being equal to 2, which makes c= to 2 as well... who does #2 work for?), or they have very similar behaviors and structure to sine and cosine.
example x=2,n=2:
So far, the e case an coefficients is easiest to describe: an = n!.
Last edited:
steve_bank
Diabetic retinopathy and poor eyesight. Typos ...
Nice use of metaphor. In the 70s I tried a math program. It was not for me. Chains of pure abstract logical chains give me a headache.
I enjoyed learning applied math and went through proofs necessary to have an understanding.
I enjoyed learning applied math and went through proofs necessary to have an understanding.
Kharakov
Quantum Hot Dog
The only easy exponential function case, for these types of constants, appears to be at the critical line n=2 (or 1/2, depending on which way you write it).
If you write it as:
\(c=x^{\frac{1}{n}} -x\)
and
\(r_1= \left(r_0 + x\right)^n\)
...
You end up with the part that has every natural (thus every prime) encoded in it, at n=1/2. This is the only place I am aware of in which every natural is encoded into the equations (it's the one that generates the exponential function a_n=1,2!,3!,4!,5!,6!... and pi stuff). I wouldn't be surprised if #2 is directly related to all primes.. with this equation. Who does #2 work for?
The other ones appear to be missing specific primes in their a_n. This doesn't mean you don't find most primes, it just means all of them don't appear to be in the other q_x,n at first glace, since I don't know the actual formula for their terms, other than the a_n case.
Cosine pattern I've got up to pi^8 case. Beginning to find that pattern. Maybe. I might be wrong. Again.
If you write it as:
\(c=x^{\frac{1}{n}} -x\)
and
\(r_1= \left(r_0 + x\right)^n\)
...
You end up with the part that has every natural (thus every prime) encoded in it, at n=1/2. This is the only place I am aware of in which every natural is encoded into the equations (it's the one that generates the exponential function a_n=1,2!,3!,4!,5!,6!... and pi stuff). I wouldn't be surprised if #2 is directly related to all primes.. with this equation. Who does #2 work for?
The other ones appear to be missing specific primes in their a_n. This doesn't mean you don't find most primes, it just means all of them don't appear to be in the other q_x,n at first glace, since I don't know the actual formula for their terms, other than the a_n case.
Cosine pattern I've got up to pi^8 case. Beginning to find that pattern. Maybe. I might be wrong. Again.
lpetrich
Contributor
I will try to express Kharakov's series in a form that makes it easier to do small-factor expansions.
\(r_0 = 0\)
\(r_{k+1} = (x^n - x + r_k)^{1/n}\)
Set
\(r = x (1 - s)\)
Then,
\(s_0 = 1\)
\(s_{k+1} = 1 - (1 - x^{1-n}s_k)^{1/n}\)
This has the nice result that
\(Q_{x,n,k} = x (nx^{n-1})^k s_k\)
For ease of calculation, I did x^(1-n) -> z and power 1/n -> power p.
Mathematica code:
Result (k = 0 to 6, coefficients of z after dividing by (p*z)^k):
{1}
{1, (1 - p)/2, ((-2 + p)*(-1 + p))/6, -((-3 + p)*(-2 + p)*(-1 + p))/24, ((-4 + p)*(-3 + p)*(-2 + p)*(-1 + p))/120, -((-5 + p)*(-4 + p)*(-3 + p)*(-2 + p)*(-1 + p))/720, ((-6 + p)*(-5 + p)*(-4 + p)*(-3 + p)*(-2 + p)*(-1 + p))/5040}
{1, (1 - p)/2, -((-1 + p)*(1 + p))/3, ((-1 + p)*(-6 - 7*p + 11*p^2))/24, -((-1 + p)*(24 + 29*p - 41*p^2 + 14*p^3))/120, -((-2 + p)*(-1 + p)*(1 + p)*(-60 - 43*p + 91*p^2))/720, ((-2 + p)*(-1 + p)*(360 + 617*p - 294*p^2 - 767*p^3 + 624*p^4))/5040}
{1, (1 - p)/2, -((-1 + p)*(1 + p))/3, -((-1 + p)*(1 + p)*(6 + p))/24, ((-1 + p)*(-24 - 29*p - 19*p^2 + 46*p^3))/120, ((-1 + p)*(-120 - 146*p - 131*p^2 + 314*p^3 + 59*p^4))/720, -((-1 + p)*(1 + p)*(720 + 154*p + 741*p^2 - 2611*p^3 + 1686*p^4))/5040}
{1, (1 - p)/2, -((-1 + p)*(1 + p))/3, -((-1 + p)*(1 + p)*(6 + p))/24, -((-1 + p)*(1 + p)*(24 + 5*p + 14*p^2))/120, ((-1 + p)*(-120 - 146*p - 131*p^2 - 46*p^3 + 419*p^4))/720, -((-1 + p)*(720 + 874*p + 895*p^2 + 440*p^3 - 2185*p^4 + 636*p^5))/5040}
{1, (1 - p)/2, -((-1 + p)*(1 + p))/3, -((-1 + p)*(1 + p)*(6 + p))/24, -((-1 + p)*(1 + p)*(24 + 5*p + 14*p^2))/120, ((-1 + p)*(1 + p)*(-120 - 26*p - 105*p^2 + 59*p^3))/720, ((-1 + p)*(-720 - 874*p - 895*p^2 - 440*p^3 - 335*p^4 + 1884*p^5))/5040}
{1, (1 - p)/2, -((-1 + p)*(1 + p))/3, -((-1 + p)*(1 + p)*(6 + p))/24, -((-1 + p)*(1 + p)*(24 + 5*p + 14*p^2))/120, ((-1 + p)*(1 + p)*(-120 - 26*p - 105*p^2 + 59*p^3))/720, -((-1 + p)*(1 + p)*(720 + 154*p + 741*p^2 - 301*p^3 + 636*p^4))/5040}
\(r_0 = 0\)
\(r_{k+1} = (x^n - x + r_k)^{1/n}\)
Set
\(r = x (1 - s)\)
Then,
\(s_0 = 1\)
\(s_{k+1} = 1 - (1 - x^{1-n}s_k)^{1/n}\)
This has the nice result that
\(Q_{x,n,k} = x (nx^{n-1})^k s_k\)
For ease of calculation, I did x^(1-n) -> z and power 1/n -> power p.
Mathematica code:
Code:
pwrx[x_, p_, n_] := Module[{t = 1, k},
t + Sum[t *= ((p - k + 1)/k)*x, {k, 1, n}]
]
pwrxm1[x_, p_, n_] := Module[{t = 1, k},
Sum[t *= ((p - k + 1)/k)*x, {k, 1, n}]
]
khrknext[x_, p_, n_, arg_] := Module[{k, s},
{k, s} = arg;
k += 1;
s = Normal[Series[- pwrxm1[-x*s, p, Floor[n/k]+1], {x, 0, n + k}]];
{k, s}
]
khrk[x_, p_, n_] :=
CoefficientList[#,
x] & /@ ((Last /@
NestList[khrknext[x, p, n, #] &, {0, 1}, n])/(p*x)^
Range[0, n]) // Factor
khrk[x, p, 6]
TableForm[InputForm /@ %]Result (k = 0 to 6, coefficients of z after dividing by (p*z)^k):
{1}
{1, (1 - p)/2, ((-2 + p)*(-1 + p))/6, -((-3 + p)*(-2 + p)*(-1 + p))/24, ((-4 + p)*(-3 + p)*(-2 + p)*(-1 + p))/120, -((-5 + p)*(-4 + p)*(-3 + p)*(-2 + p)*(-1 + p))/720, ((-6 + p)*(-5 + p)*(-4 + p)*(-3 + p)*(-2 + p)*(-1 + p))/5040}
{1, (1 - p)/2, -((-1 + p)*(1 + p))/3, ((-1 + p)*(-6 - 7*p + 11*p^2))/24, -((-1 + p)*(24 + 29*p - 41*p^2 + 14*p^3))/120, -((-2 + p)*(-1 + p)*(1 + p)*(-60 - 43*p + 91*p^2))/720, ((-2 + p)*(-1 + p)*(360 + 617*p - 294*p^2 - 767*p^3 + 624*p^4))/5040}
{1, (1 - p)/2, -((-1 + p)*(1 + p))/3, -((-1 + p)*(1 + p)*(6 + p))/24, ((-1 + p)*(-24 - 29*p - 19*p^2 + 46*p^3))/120, ((-1 + p)*(-120 - 146*p - 131*p^2 + 314*p^3 + 59*p^4))/720, -((-1 + p)*(1 + p)*(720 + 154*p + 741*p^2 - 2611*p^3 + 1686*p^4))/5040}
{1, (1 - p)/2, -((-1 + p)*(1 + p))/3, -((-1 + p)*(1 + p)*(6 + p))/24, -((-1 + p)*(1 + p)*(24 + 5*p + 14*p^2))/120, ((-1 + p)*(-120 - 146*p - 131*p^2 - 46*p^3 + 419*p^4))/720, -((-1 + p)*(720 + 874*p + 895*p^2 + 440*p^3 - 2185*p^4 + 636*p^5))/5040}
{1, (1 - p)/2, -((-1 + p)*(1 + p))/3, -((-1 + p)*(1 + p)*(6 + p))/24, -((-1 + p)*(1 + p)*(24 + 5*p + 14*p^2))/120, ((-1 + p)*(1 + p)*(-120 - 26*p - 105*p^2 + 59*p^3))/720, ((-1 + p)*(-720 - 874*p - 895*p^2 - 440*p^3 - 335*p^4 + 1884*p^5))/5040}
{1, (1 - p)/2, -((-1 + p)*(1 + p))/3, -((-1 + p)*(1 + p)*(6 + p))/24, -((-1 + p)*(1 + p)*(24 + 5*p + 14*p^2))/120, ((-1 + p)*(1 + p)*(-120 - 26*p - 105*p^2 + 59*p^3))/720, -((-1 + p)*(1 + p)*(720 + 154*p + 741*p^2 - 301*p^3 + 636*p^4))/5040}
Kharakov
Quantum Hot Dog
Love the \(x \, \times \, \sqrt[n]{1-x^{1-n}+\sqrt[n]{1-x^{1-n}+\dots}}\) trick.
Looks like you ran it with the + instead of -, but I'm not sure you have the expansions that I'm looking for, which are at limit k->infinity.
Where in those expansions do you take into account x?
Mine end up, at n=2 (or p=1/2):
a_3(x):=4*x^2-2*x; 3! at x=2
a_4(x):=a_3(x)*2; 4!/2 at x=2
a_5(x):= ??? 5! at x=2
a_6(x):=16*x^5-8*x^4-4*x^3+2*x^2; 6!/2 at x=2
a_7(x):= has some divisor term for the x=2.5 case (62000/3). I'm thinking (x+.5) for whatever reason.
I'm most interested in the n=2, x=2 case, because I think that's the only one in which all primes are in the series expansion.... however, it might be relatively simple to find an exception with Mathematica.... which I don't have.
Looks like you ran it with the + instead of -, but I'm not sure you have the expansions that I'm looking for, which are at limit k->infinity.
Where in those expansions do you take into account x?
Mine end up, at n=2 (or p=1/2):
a_3(x):=4*x^2-2*x; 3! at x=2
a_4(x):=a_3(x)*2; 4!/2 at x=2
a_5(x):= ??? 5! at x=2
a_6(x):=16*x^5-8*x^4-4*x^3+2*x^2; 6!/2 at x=2
a_7(x):= has some divisor term for the x=2.5 case (62000/3). I'm thinking (x+.5) for whatever reason.
I'm most interested in the n=2, x=2 case, because I think that's the only one in which all primes are in the series expansion.... however, it might be relatively simple to find an exception with Mathematica.... which I don't have.
Last edited:
Kharakov
Quantum Hot Dog
Rewrite and "simplification".
x and n are arbitrarily chosen reals >1 k is the (integer) number of radicals, ranging from 1 to infinity
\( Q_{x,n,k} =\sqrt[n]{\left( nx^{n-1} \right)^k }\, \times \sqrt[n]{x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \sqrt[n]{x^n-x+...}}} }\)
If k is 3, the number of radicals is 3:
\( Q_{x,n,3} =\sqrt[n]{ \left( nx^{n-1} \right)^3}\, \times \sqrt[n]{x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x }} }\)
the limit as k approaches infinity results in some constant, written as \(q_{x,n}\)
\(q_{x,n} = Q_{x,n,k \to \infty}\)
There is one constant generated that we have all heard of:
\(q_{2,2} = \pi \)
All q constants at k-->infinity have functions that generate lower k constants. For example:
\(a_n\) is the coefficients of specific terms
\(b= nx^{n-1}\) is the nx^(n-1)
\(\gamma= q_{x,n}\)
The cosine like extraction function is below:
\(Q_{x,n,k} = \frac {\gamma^2}{a_2 \, b^{0k}} - \frac {\gamma^4}{a_4 \, b^{1k}} +\frac {\gamma^6}{a_6 \, b^{2k}}- \frac {\gamma^8}{a_8 \, b^{3k}} \,\, ...\)
The sine like extraction function is below:
\(Qsin_{x,n,k} =\frac {\frac{\gamma}{2}}{a_1 \, b^{0k}} - \frac {{\frac{\gamma}{2}}^3}{a_3 \, b^{1k}} + \frac {\frac{\gamma}{2}^5}{a_5 \, b^{2k}} -\frac {\frac{\gamma}{2}^7}{a_7 \, b^{3k}}+ \frac {\frac{\gamma}{2}^9}{a_9 \, b^{4k}} \,\, ...\)
So I'm thinking, because of the behavior of cosine and sine functions, that smoothly increasing and decreasing k (with non integer values) will result in fluctuations, instead of smooth transitions from a certain amount of nested radicals to the next:
\(Q_{2,2,3} =\sqrt{ \,\,\, 4^3 \, \times \left[2 - \sqrt{2 + \sqrt{2}} \right]} \)
to
\(Q_{2,2,4} = \sqrt{\,\,\, 4^4 \, \times \left[2 - \sqrt{2 + \sqrt{2+sqrt{2}}} } \right] \)
Using the cosine like function, one can increase k with non-integer values, and traverse smoothly between iterations of radicals:
\(Q_{2,2,k} = \,\,\ \frac{ pi^2}{ 4^{0}} - \frac{pi^4}{12 \times 4^{k}} +\frac{ pi^6}{360 \times 4^{2k}} - \frac{ pi^8}{20160 \times 4^{3k}}\,\,\, \dots\)
However, does this result in fluctuations? Does the cosine like function return oscillating values, like the regular cosine function, or does it steadily decrease? The sine like function makes me question my brain saying "yeah, it's basically cosine of pi/4^k so it will vary like cosine".
Is this a mathematically sound way to smoothly move from one radical to the next?
x and n are arbitrarily chosen reals >1 k is the (integer) number of radicals, ranging from 1 to infinity
\( Q_{x,n,k} =\sqrt[n]{\left( nx^{n-1} \right)^k }\, \times \sqrt[n]{x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \sqrt[n]{x^n-x+...}}} }\)
If k is 3, the number of radicals is 3:
\( Q_{x,n,3} =\sqrt[n]{ \left( nx^{n-1} \right)^3}\, \times \sqrt[n]{x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x }} }\)
the limit as k approaches infinity results in some constant, written as \(q_{x,n}\)
\(q_{x,n} = Q_{x,n,k \to \infty}\)
There is one constant generated that we have all heard of:
\(q_{2,2} = \pi \)
All q constants at k-->infinity have functions that generate lower k constants. For example:
\(a_n\) is the coefficients of specific terms
\(b= nx^{n-1}\) is the nx^(n-1)
\(\gamma= q_{x,n}\)
The cosine like extraction function is below:
\(Q_{x,n,k} = \frac {\gamma^2}{a_2 \, b^{0k}} - \frac {\gamma^4}{a_4 \, b^{1k}} +\frac {\gamma^6}{a_6 \, b^{2k}}- \frac {\gamma^8}{a_8 \, b^{3k}} \,\, ...\)
The sine like extraction function is below:
\(Qsin_{x,n,k} =\frac {\frac{\gamma}{2}}{a_1 \, b^{0k}} - \frac {{\frac{\gamma}{2}}^3}{a_3 \, b^{1k}} + \frac {\frac{\gamma}{2}^5}{a_5 \, b^{2k}} -\frac {\frac{\gamma}{2}^7}{a_7 \, b^{3k}}+ \frac {\frac{\gamma}{2}^9}{a_9 \, b^{4k}} \,\, ...\)
So I'm thinking, because of the behavior of cosine and sine functions, that smoothly increasing and decreasing k (with non integer values) will result in fluctuations, instead of smooth transitions from a certain amount of nested radicals to the next:
\(Q_{2,2,3} =\sqrt{ \,\,\, 4^3 \, \times \left[2 - \sqrt{2 + \sqrt{2}} \right]} \)
to
\(Q_{2,2,4} = \sqrt{\,\,\, 4^4 \, \times \left[2 - \sqrt{2 + \sqrt{2+sqrt{2}}} } \right] \)
Using the cosine like function, one can increase k with non-integer values, and traverse smoothly between iterations of radicals:
\(Q_{2,2,k} = \,\,\ \frac{ pi^2}{ 4^{0}} - \frac{pi^4}{12 \times 4^{k}} +\frac{ pi^6}{360 \times 4^{2k}} - \frac{ pi^8}{20160 \times 4^{3k}}\,\,\, \dots\)
However, does this result in fluctuations? Does the cosine like function return oscillating values, like the regular cosine function, or does it steadily decrease? The sine like function makes me question my brain saying "yeah, it's basically cosine of pi/4^k so it will vary like cosine".
Is this a mathematically sound way to smoothly move from one radical to the next?
Kharakov
Quantum Hot Dog
So, I've made a tiny bit of progress. I still write it the way I write it instead of Loren's method, because I like the x=b=c=n=2 case which generates pi, and the easiest to find coefficients (a_m =(2m!)/2). Yup.. more #2. It's also a "circular" function in the sense that its points connect at k=infinity. It wraps around to x-x... or the zero point.
I am still curious about the general formula for coefficients for the series. I am focusing on the n=2 case, although I've discovered the 1st 2 coefficients for all cases, first 3 for n=2 and 3.
The n=2 case is specific.
There are 2 types of functions. An iterative forwards function (directly calculate as many radicals as one wishes):
k= number of radicals
b= \(\sqrt{2x}\)
c= x^2-x
\(\\ Q_{x,k} = b^k \, \times \, \sqrt{x - \sqrt {c+\sqrt {c +...}}} \\ \\\)
k=3 radicals:
\( Q_{x,3} = b^3 \, \times \, \sqrt{x- \sqrt{c+\sqrt{c}}}\\\)
And a smooth "backwards" function, starting at k= infinity (which oscillates perfectly at the x=2 line for k<0, maybe because of distribution of primes? Or maybe because this is the only function like this where x=b=c=n. Loren??? It's like a zero point for this network of functions, creates double amplitude cosine oscillations (or 2 * cosine).
\(k \, \to \, \infty \)
\( \\ q_x =b^k \,\times\, \sqrt {x- \sqrt{c + \sqrt {c + \sqrt{c +...}}}}\)
at x=2, you get pi. Also, this is the only place that the function is balanced perfectly. x=b=c (=n too, but... we're doing the simple case in which we don't regard n)
Backwards function, which allows smooth iterative addition of radicals, which isn't the reason I want the coefficients- I want the oscillating functions for some artwork. So I can say I did some stuff. With the help of whoever helped me. I'm not bothering to define the coefficients for sine too, because there are a million sine varieties. The simplest case is the 2*cosine case.
\( {(\frac {Q_{x,k}}{b^k})}^2 = \frac {{q_x}^2}{a_1 \, b^{2k}} + \frac {{q_x}^4}{a_2 \, b^{4k}} -\frac {{q_x}^6}{a_3 \, b^{6k}}+ \frac {{q_x}^8}{a_4 \, b^{8k}} -\,\, ... = x- \sqrt{c+\sqrt{c+...} \)
Not to be circular, but you should be able to see how at k--> infinity, how you get the constants associated with specific x, if you put them into the function. Circular reasoning corner.
At k=0, you get \( \frac {{q_x}^2}{a_1 } - \frac {{q_x}^4}{a_2 } +\frac {{q_x}^6}{a_3 }- \frac {{q_x}^8}{a_4 } +\,\, ... = x^2 \)
At k=1, you get \( \frac {{q_x}^2}{a_1 \, b^2}-\frac {{q_x}^4}{a_2 \, b^4} +\frac {{q_x}^6}{a_3 \, b^6}- \frac {{q_x}^8}{a_4 \, b^8} +\,\, ... = x \)
At k=2, you get \( \frac {{q_x}^2}{a_1 \, b^4} - \frac {{q_x}^4}{a_2 \, b^8} +\frac {{q_x}^6}{a_3 \, b^{12}}- \frac {{q_x}^8}{a_4 \, b^{16}} +\,\, ... = x - sqrt{c}\)
Anyway. So I've found a general formula for the smooth "backwards" function, well, for some of its coefficients. I haven't found the general formula.
Ok, so this connects infinite iterations of a function smoothly to single iterations. It allows one to extract, from the infinite iteration point, single iterations.
Not only that, pi exists at the x=b=c=n=2. Point. This mathematical coincidence is amazing.. it's a bit like Euler's thingy, but it has number 2 in it. And who does number 2 work 4? Fourier transforms, of coursier.
Pi, a cornerstone of all mathematics, exists at the leverage point of this function system. As does 2. There is a nice leverage point of #2s through all n, in which b= 2^(n/(n-1)) -2^(n-1). So n=3 has b= 2^(3/2) -2^(1/2)=2^(1/2). They actually make some nice simple cosine like series for all n (surprisingly simple expansions, easy to extract terms for). But the main leverage point is where all the 2s meet, and make pi. At n=2.
Which is crazy. e^i pi=-1? I see 2 2 2 2 2. Terrible twos.
I have yet to connect it to the Wallis product. Which is Pi/2. However, it definitely will have a connection to it as well, as pi/2 is the point at which the "sine" version of the leverage point of the function has coefficients 1/(2m-1)!. Ok, there you go. The connection to the Wallis product. It and the cosine point, are the only ones in which all coefficients are easily represented with factorials. Every other coefficient, for other n and x, ARE way more complicated.
So, I am just asking for someone who already has the answer, because me learning the answer the hard way is probably not going to happen, because I am on the fricken street, to tell me the general formula for coefficients.
Here they are:
a2: 2x (2x-1)
a4 : x (4x^2-1)
a6: 4x (8x^3-1) / (2x +5)
a8: x(16x^4-1) (2x+5) /(8x^2+6x+7)
a10: 2x (32x^5-1) (8x^2+6x+7) / (48x^4+64x^3+56x^2 +28x +21)
Do you know how I am extracting coefficients? The dumb hard way.
I am still curious about the general formula for coefficients for the series. I am focusing on the n=2 case, although I've discovered the 1st 2 coefficients for all cases, first 3 for n=2 and 3.
The n=2 case is specific.
There are 2 types of functions. An iterative forwards function (directly calculate as many radicals as one wishes):
k= number of radicals
b= \(\sqrt{2x}\)
c= x^2-x
\(\\ Q_{x,k} = b^k \, \times \, \sqrt{x - \sqrt {c+\sqrt {c +...}}} \\ \\\)
k=3 radicals:
\( Q_{x,3} = b^3 \, \times \, \sqrt{x- \sqrt{c+\sqrt{c}}}\\\)
And a smooth "backwards" function, starting at k= infinity (which oscillates perfectly at the x=2 line for k<0, maybe because of distribution of primes? Or maybe because this is the only function like this where x=b=c=n. Loren??? It's like a zero point for this network of functions, creates double amplitude cosine oscillations (or 2 * cosine).
\(k \, \to \, \infty \)
\( \\ q_x =b^k \,\times\, \sqrt {x- \sqrt{c + \sqrt {c + \sqrt{c +...}}}}\)
at x=2, you get pi. Also, this is the only place that the function is balanced perfectly. x=b=c (=n too, but... we're doing the simple case in which we don't regard n)
Backwards function, which allows smooth iterative addition of radicals, which isn't the reason I want the coefficients- I want the oscillating functions for some artwork. So I can say I did some stuff. With the help of whoever helped me. I'm not bothering to define the coefficients for sine too, because there are a million sine varieties. The simplest case is the 2*cosine case.
\( {(\frac {Q_{x,k}}{b^k})}^2 = \frac {{q_x}^2}{a_1 \, b^{2k}} + \frac {{q_x}^4}{a_2 \, b^{4k}} -\frac {{q_x}^6}{a_3 \, b^{6k}}+ \frac {{q_x}^8}{a_4 \, b^{8k}} -\,\, ... = x- \sqrt{c+\sqrt{c+...} \)
Not to be circular, but you should be able to see how at k--> infinity, how you get the constants associated with specific x, if you put them into the function. Circular reasoning corner.
At k=0, you get \( \frac {{q_x}^2}{a_1 } - \frac {{q_x}^4}{a_2 } +\frac {{q_x}^6}{a_3 }- \frac {{q_x}^8}{a_4 } +\,\, ... = x^2 \)
At k=1, you get \( \frac {{q_x}^2}{a_1 \, b^2}-\frac {{q_x}^4}{a_2 \, b^4} +\frac {{q_x}^6}{a_3 \, b^6}- \frac {{q_x}^8}{a_4 \, b^8} +\,\, ... = x \)
At k=2, you get \( \frac {{q_x}^2}{a_1 \, b^4} - \frac {{q_x}^4}{a_2 \, b^8} +\frac {{q_x}^6}{a_3 \, b^{12}}- \frac {{q_x}^8}{a_4 \, b^{16}} +\,\, ... = x - sqrt{c}\)
Anyway. So I've found a general formula for the smooth "backwards" function, well, for some of its coefficients. I haven't found the general formula.
Ok, so this connects infinite iterations of a function smoothly to single iterations. It allows one to extract, from the infinite iteration point, single iterations.
Not only that, pi exists at the x=b=c=n=2. Point. This mathematical coincidence is amazing.. it's a bit like Euler's thingy, but it has number 2 in it. And who does number 2 work 4? Fourier transforms, of coursier.
Pi, a cornerstone of all mathematics, exists at the leverage point of this function system. As does 2. There is a nice leverage point of #2s through all n, in which b= 2^(n/(n-1)) -2^(n-1). So n=3 has b= 2^(3/2) -2^(1/2)=2^(1/2). They actually make some nice simple cosine like series for all n (surprisingly simple expansions, easy to extract terms for). But the main leverage point is where all the 2s meet, and make pi. At n=2.
Which is crazy. e^i pi=-1? I see 2 2 2 2 2. Terrible twos.
I have yet to connect it to the Wallis product. Which is Pi/2. However, it definitely will have a connection to it as well, as pi/2 is the point at which the "sine" version of the leverage point of the function has coefficients 1/(2m-1)!. Ok, there you go. The connection to the Wallis product. It and the cosine point, are the only ones in which all coefficients are easily represented with factorials. Every other coefficient, for other n and x, ARE way more complicated.
So, I am just asking for someone who already has the answer, because me learning the answer the hard way is probably not going to happen, because I am on the fricken street, to tell me the general formula for coefficients.
Here they are:
a2: 2x (2x-1)
a4 : x (4x^2-1)
a6: 4x (8x^3-1) / (2x +5)
a8: x(16x^4-1) (2x+5) /(8x^2+6x+7)
a10: 2x (32x^5-1) (8x^2+6x+7) / (48x^4+64x^3+56x^2 +28x +21)
Do you know how I am extracting coefficients? The dumb hard way.
Code:
f5(x,n,k):=[b:abs(bfloat(x^n-x)),r:(b)^(1b0/n),
for i:2 thru k-1 do[
r:(r+b)^(1b0/n)],
r:(x-r),
pi2:r*(n*x^(n-1b0))^(k),
disp (pi2)
]$
kk:60;
nn:2b0;
xx:3b0;
f5(xx,nn,kk*6)$
z252:(pi2);
f5(xx,nn,kk)$
pi2:(pi2);
zzz:nn*xx^(nn-1b0);
a:z252-pi2;
a2:a*zzz^kk;
z252^2b0/a2;
factor(%);
b:z252^2b0/((3*5)/2);
b1:b-a2;
a3:b1*zzz^kk;
z252^3b0/a3;
factor(%);
b2:z252^3b0/((5^2*7)/2)-a3;
a4:b2*zzz^kk;
z252^4b0/a4;
factor(%);
b3:z252^4b0/((3^3*5^3*7*13)/(2^2*43))-a4;
/// some coefficient code:
cof1(x,n):=2*factor((n*x^n- x)/((n-1)));cof1(2,2);
cof2(x,n):=factor( (n*x^(n)-x)*3*(n*x^(n-1)+1)/ ((n-2)*(n*x^(n-1)+1)+n+1));
cof21(x):= factor(4*x^2-2*x);
cof22(x):= factor(2*x^2-x)*(2*x+1);
cof23(x):=factor((4*x*(8*x^3-1))/(2*x+5));
cof24(x):=factor(x*(16*x^4-1)*(2*x+5)/(8*x^2+6*x+7));
cof25(x):=factor((2*x*(32*x^5-1)*(8*x^2+6*x+7))/ (48*x^4+64*x^3+56*x^2 +28*x +21));
Last edited:
steve_bank
Diabetic retinopathy and poor eyesight. Typos ...
Is that all there is to it?
Kharakov
Quantum Hot Dog
Ohh, it's silly. I thought that I was onto something new, because of my lack of education.
So to me, all these things seem brand new. Like when I discovered Taylor Series (didn't know what they were). I was like "holy fuck, these infinite series compress down into a single equation, WTF!!@$"
Fucking retarded, literally, since my education was basically stopped the year I got the shit kicked out of me by a bully every day until I got that hernia operation to repair the internal damage from the beatings ("my dad says nobody will catch me if I hurt you this way, it doesn't leave marks, what are you going to do?"). Funny how one douchebag can crush someone's potential and ruin their life. I found the guy, and his kids, on facebook though. And my life fucking sucks. Let me think about my options. Maybe he's going to be a grandfather.
So to me, all these things seem brand new. Like when I discovered Taylor Series (didn't know what they were). I was like "holy fuck, these infinite series compress down into a single equation, WTF!!@$"
Fucking retarded, literally, since my education was basically stopped the year I got the shit kicked out of me by a bully every day until I got that hernia operation to repair the internal damage from the beatings ("my dad says nobody will catch me if I hurt you this way, it doesn't leave marks, what are you going to do?"). Funny how one douchebag can crush someone's potential and ruin their life. I found the guy, and his kids, on facebook though. And my life fucking sucks. Let me think about my options.
Maybe he's going to be a grandfather. steve_bank
Diabetic retinopathy and poor eyesight. Typos ...
Ohh, it's silly. I thought that I was onto something new, because of my lack of education.
So to me, all these things seem brand new. Like when I discovered Taylor Series (didn't know what they were). I was like "holy fuck, these infinite series compress down into a single equation, WTF!!@$"
Fucking retarded, literally, since my education was basically stopped the year I got the shit kicked out of me by a bully every day until I got that hernia operation to repair the internal damage from the beatings ("my dad says nobody will catch me if I hurt you this way, it doesn't leave marks, what are you going to do?"). Funny how one douchebag can crush someone's potential and ruin their life. I found the guy, and his kids, on facebook though. And my life fucking sucks. Let me think about my options.
Did the catharsis get it all out? One believes in oneself or one does not. AE faced rejection. But he did not give up. Ali was down and get got back up. As I recall the guy who cracked the Fermat problem worked for years alone in his attic..
IMO a life spent pursuing math is well lived.
https://en.wikipedia.org/wiki/Andrew_Wiles#Proof_of_Fermat's_Last_Theorem
Kharakov
Quantum Hot Dog
Dude. Bank of Steve. I'm in no position to do anything amazing with what I know. I have outdated freeware, that... while it allows me to do amazing things that people could (allegedly) only imagine doing 20 years ago... is seriously hampered compared to payware like Mathematica.
Not only that, I only have a narrow focus, on a narrow piece of mathematics, that is umbrella covered by some higher order stuff. While others in my knowledge group haven't done the legwork yet (because it's not available online for me... which means nothing), it doesn't mean someone else hasn't taken it to its end. In fact, I'd be very surprised if someone hasn't found all the implications of this already. If there are any. Which.. for some reason, I think there are.
There is only one solution to the general form equations at n=2, x=2 (which means b=2 and c=2), in which all primes are ordered perfectly in the decomposition function (2* cos(pi/2^k) for x=2,n=2).
What could this mean? If someone smart finds f(x,n,k) and g(x,n,k) such that:
\( g(x,n,k) \, \times \, cos ( \,f(x,n,k) \, ) \,\, = \, \sqrt[n]{ c + \sqrt[n]{c +...}} \)
We might have a function suite that can create maps of primes (a function suite in which primes play a big role in development of the oscillating functions).
Not that big of a deal.
g(x,n,k)=2 at n=2, x=2, b=2, c=2. f(x,n,k) = \(\frac{\pi}{2^k}\) at the 2s point.
So what the fuck is up with all the 2s? Is #2 really poo in the end? I'll ask a dog. Am I going to learn anything useful, or should I just do as many drugs as possible? I don't know.
Someone (I don't know their real name) way more informed (and technically... gifted... they have Mathematica, although maybe I could replicate this with Maxima?) than me solved the general case for n=2 (for a few terms).
their code:
It's interesting how it collapses down to 2m!/2 at the n=2, x=2 case. But in ALL other cases. It does not have monotonic distribution of primes. Just at the 2222222 point. 2.
I think Loren thinks this is worthless number two because he knows the answers already. That just because there is only one case in which primes have a monotonic distribution, it's not really that big of a deal. Or it is, but everyone learns this in elementary school, and I'm a moron. So I was picked on and beat up, because I'm supposed to do the mopping, garbage removal, etc. Which I do. Because I'm on the bottom (not because I'm a loser.. because of my lot in life. That's all.).
Like, we can't write code that will generate prime maps, where we can visualize were primes are using these functions. Like we can't write out the g(x,k) and f(x,k) functions.. and use them to show where there are prime "spikes".
Like we can't break encryption, world wide. Knock the shit down. Because those hiding stuff really are doing it for our best interests. Not because they want to hide stuff and live the good life. They really want to be nice. But they need to hide stuff. Because good people can't be trusted with the truth.
Not only that, I only have a narrow focus, on a narrow piece of mathematics, that is umbrella covered by some higher order stuff. While others in my knowledge group haven't done the legwork yet (because it's not available online for me... which means nothing), it doesn't mean someone else hasn't taken it to its end. In fact, I'd be very surprised if someone hasn't found all the implications of this already. If there are any. Which.. for some reason, I think there are.
There is only one solution to the general form equations at n=2, x=2 (which means b=2 and c=2), in which all primes are ordered perfectly in the decomposition function (2* cos(pi/2^k) for x=2,n=2).
What could this mean? If someone smart finds f(x,n,k) and g(x,n,k) such that:
\( g(x,n,k) \, \times \, cos ( \,f(x,n,k) \, ) \,\, = \, \sqrt[n]{ c + \sqrt[n]{c +...}} \)
We might have a function suite that can create maps of primes (a function suite in which primes play a big role in development of the oscillating functions).
Not that big of a deal.
g(x,n,k)=2 at n=2, x=2, b=2, c=2. f(x,n,k) = \(\frac{\pi}{2^k}\) at the 2s point.
So what the fuck is up with all the 2s? Is #2 really poo in the end? I'll ask a dog. Am I going to learn anything useful, or should I just do as many drugs as possible? I don't know.
Someone (I don't know their real name) way more informed (and technically... gifted... they have Mathematica, although maybe I could replicate this with Maxima?) than me solved the general case for n=2 (for a few terms).
their code:
It's interesting how it collapses down to 2m!/2 at the n=2, x=2 case. But in ALL other cases. It does not have monotonic distribution of primes. Just at the 2222222 point. 2.
I think Loren thinks this is worthless number two because he knows the answers already. That just because there is only one case in which primes have a monotonic distribution, it's not really that big of a deal. Or it is, but everyone learns this in elementary school, and I'm a moron. So I was picked on and beat up, because I'm supposed to do the mopping, garbage removal, etc. Which I do. Because I'm on the bottom (not because I'm a loser.. because of my lot in life. That's all.).
Like, we can't write code that will generate prime maps, where we can visualize were primes are using these functions. Like we can't write out the g(x,k) and f(x,k) functions.. and use them to show where there are prime "spikes".
Like we can't break encryption, world wide. Knock the shit down. Because those hiding stuff really are doing it for our best interests. Not because they want to hide stuff and live the good life. They really want to be nice. But they need to hide stuff. Because good people can't be trusted with the truth.
Last edited:
steve_bank
Diabetic retinopathy and poor eyesight. Typos ...
Did you say freeware? Find Scilab. It originated in European labs and is now public domain. It is maintained with periodic updates and there are user forums. I It is better than Matlab an expensive tool. It is used in Europe and has grown in the USA.
I used it for years as my primary math and simulation tool. A scripted language, good graphics, and a rich math library. Plenty of examples and good searchable documentation.
Binaries for most platforms.
I used it for years as my primary math and simulation tool. A scripted language, good graphics, and a rich math library. Plenty of examples and good searchable documentation.
Binaries for most platforms.
Kharakov
Quantum Hot Dog
I'll give it a go. I've done maxima for a while, have lots of code written.
I don't do the higher level stuff though- I still put together little ideas in my head, while others have learned various methods that allow them to dance around the stuff I do. Not that this is a problem- it's just apparent to me that I don't have a clear path to learning what I need to learn to do the same things, or whether or not it is even worth while. Ask a mathematician, and they enjoy solving problems, but are they getting to the point that they can solve everything- physical problems as well?
I don't do the higher level stuff though- I still put together little ideas in my head, while others have learned various methods that allow them to dance around the stuff I do. Not that this is a problem- it's just apparent to me that I don't have a clear path to learning what I need to learn to do the same things, or whether or not it is even worth while. Ask a mathematician, and they enjoy solving problems, but are they getting to the point that they can solve everything- physical problems as well?
steve_bank
Diabetic retinopathy and poor eyesight. Typos ...
Scilab is a high end tool even though it is free. For a project you can create custom pull down menus and create a custom loadable shell. You can call functions from a pull down menu.
The graphics takes a bit to learn. You can create data input windows and detect mouse position in a window. Graphics animation.
Then there is the simulator. You can select math boxes from a pallet, connect with lines, enter in equations, set the time step and it will solve the set of equations.
It is not as good but there is Euler also from Europe.
The graphics takes a bit to learn. You can create data input windows and detect mouse position in a window. Graphics animation.
Then there is the simulator. You can select math boxes from a pallet, connect with lines, enter in equations, set the time step and it will solve the set of equations.
It is not as good but there is Euler also from Europe.