# Thread: Integer formula for transcendental function group

1. ## Integer formula for transcendental function group

whoops, please delete. I'll work on it.

2. You may find the Phisics Forums helpful. They have a whole section of Mathematics.

https://www.physicsforums.com/

3. 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:

sum ((-1)^(n+1)*pi^(2n)/((2n)!/2*4^(2*(n-1)))), n = 1 to infinity

Change the bolded part to change k from 2:
sum ((-1)^(n+1)*pi^(2n)/((2n)!/2*4^(2boooollllded2*(n-1)))), n = 1 to infinity

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?

4. 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?

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:

$c=x^n-x$

$r_1=\sqrt[n]{c}$
$r_2=\sqrt[n]{r_1 + c}$
...
$r_{k-2}=\sqrt[n]{r_{k-3} + c}$
$r_{k-1}=\sqrt[n]{r_{k-2} + c}$

I only go to k-1, because k= number of radicals +1

$Q_{x,n,k} = \left(nx^{n-1}\right)^k \times \, \left(x-r_{k-1}\right)$

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:

1,6,72,1404... for..$q_{1.5,2} =6.426755149863556627$

$Q_{1.5,2,k} = \,\,\ \frac{q_{1.5,2}}{1 \times 3^{0k}} - \frac{q_{1.5,2}^2}{6 \times 3^{1k}} +\frac{q_{1.5,2}^3}{72\times 3^{2k}} - \frac{q_{1.5,2}^4}{1404\times 3^{3k}} + \frac{q _{1.5,2}^5}{\frac{673920\times 3^{4k}}{17}} - \dots$

1,12,360,20160.... for .. $q_{2,2}=\pi^2=9.8696044010893586$

$Q_{2,2,k} = \,\,\ \frac{ pi^2}{ 4^{0k}} - \frac{pi^4}{12 \times 4^{1k}} +\frac{ pi^6}{360 \times 4^{2k}} - \frac{ pi^8}{20160 \times 4^{3k}} \dots$

1,20,1200,148800.... for .. $q_{2.5,2}=14.5044147020882738664771$

$Q_{2.5,2,k} = \,\,\ \frac{q_{2.5,2}}{ 5^{0k}} - \frac{q_{2.5,2}^2}{20 \times 5^{1k}} +\frac{q_{2.5,2}^3}{1200 \times 5^{2k}} - \frac{ q_{2.5,2}^4}{148800 \times 5^{3k}} +\frac{q_{2.5,2}^5}{32240000 \times 5^{4k}} \dots$

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

The combined starts of "sin like" and "cosine like" sequences for q's 1.5; 2; and 2.5 are... baring mistakes:

1.5 : ??? 6,6*2, 72,72*2,1872, 1404*2, 673920/11, 673920/17*2... ???? Doesn't look right to me.

2 (the pi one) : **3!, 4!/2, 5!, 6!/2... <--you get it...** all primes.. at least here. Most densist of primes on the critical line that intersects the exponential function???

2.5: 10, 20, 120, 1200, 62000/3, 148800 , 2321280/131, 32240000

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.

5. 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.

6. 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:

Try multiplying (2-sqrt(2+sqrt(2))*4^3 then do (2-sqrt(2+sqrt(2+sqrt(2)))*4^4... see what it approaches (ok, take the square root of it if you don't recognize the number). Better yet, just do:

sqrt(2-sqrt(2+sqrt(2)) *2*2*2
sqrt(2-sqrt(2+sqrt(2+sqrt(2)))) *2*2*2*2
....

More on #2 and pi later. Totally a coincidence that there is a #2 joke so close, yet so far from pi and e.

So far, the e case an coefficients is easiest to describe: an = n!.

7. 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.

8. 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.

9. 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:
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}

10. Can you run that with (not 1-, but 1+), just n=2 case:

$s_{k+1} = 1 + \sqrt{1- \frac{s_k}{x}}$

I might be able to do that with Maxima... but...

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