# B Generating the formula for the coefficients of an alternating series

#### Matt Benesi

Gold Member
ETA. Read the bottom post first. Well, and..
Obviously mathematicians know this identity.

At the x=b=c=n=2 point, pi exists. There are also connections to the Wallis product (pi/2).

Anyway, I simplified it to the n=2 case. And re-remembered my fascination with the Pidentity, where x=n=b=c=2. The leverage point in the whole function system. Where all primes are included in order, on the same side of the coefficient quotient (all on one side).
n=2 (simplified equation)

b= $\sqrt{2x}$
c= x^2-x

$$\\ Q_{x,k} = b^k \, \times \, \sqrt{x - \sqrt {c+\sqrt {c +...}}} \\ \\$$

k-->infinity, with x and n of 2:
b= $\sqrt{4}$ =2
c= 2^2 -2 =2

Re-iterated: I'd like the coefficients for series associated with constants generated by the following function at the limit k-->infinity.

At x=2,n=2, the point at which the following function generates pi, the series are sine and cosine series (I know these coefficients, a_m = m!/2).

First, here is the cosine series I refer to, for k radicals:

$$2 \times [\, 1- \frac {\pi^2}{2! \, 4^{1k}} + \frac {\pi^4}{4! 4^{2k}} -\frac {\pi^6}{6! \, 4^{3k}}+ \frac {\pi^8}{8!\, 4^{4k}} \,\, ...] = \sqrt{2 +\sqrt{2+...}}$$

Here is the function to generate constants (such as pi) with inputs x, n, and k as the number of radicals, x and n are reals>1:

$$Q_{x,n,k} =\sqrt[n] { \left( nx^{n-1} \right)^k \, \times \left[x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \sqrt[n]{x^n-x+...}}} \right]}$$

Above function explained:
$$Q_{x,n,2} =\sqrt[n] { \left( nx^{n-1} \right)^2 \, \times \left[x - \sqrt[n]{x^n-x} \right]}$$
$$Q_{x,n,3} =\sqrt[n] { \left( nx^{n-1} \right)^3 \, \times \left[x - \sqrt[n]{x^n-x+ \sqrt[n]{x^n-x}} \right]}$$

As k approaches infinity, you get some value determined by x and n.

$$\gamma= q_{x,n} = Q_{x,n,k \to \infty}$$

One of the values is pi at n=2, x=2:

$$\gamma= \pi \, = \, q_{2,2} \, = \, Q_{2,2,k \to \infty} =\sqrt { 4^k \, \times \left[2 - \sqrt[2]{2^2-2 + \sqrt{4-2 + \sqrt{2+\sqrt{2+...}}}} \right]}$$

All of the constants have associated functions that allow "fractional iteration" of radical addition.

$\sqrt[n] {x^n-x}$ continuous to
$\sqrt[n] {x^n-x + \sqrt[n]{x^n-x}}$ to
$\sqrt[n] {x^n-x + \sqrt[n]{x^n-x+ \sqrt[n]{x^n-x}}}$.

There are exact formulas for the coefficients of terms in the functions that allow smooth iteration in this case. In the x=2, n=2 case, the coefficients are directly related to sine/cosine coefficients.

$a_m$ coefficients of specific terms in the expansions

$b= nx^{n-1}$

$\gamma= q_{x,n}$

one form of "Cosine like" function (k radicals):

$$x- \frac {(Q_{x,n,k})^n}{b^k} \, = \,x- \frac {\gamma^2}{a_2 \, b^{1k}} + \frac {\gamma^4}{a_4 \, b^{2k}}-\frac {\gamma^6}{a_6 \, b^{3k}}+ \frac {\gamma^8}{a_8 \, b^{4k}} \,\, ... = \sqrt[n]{x^n-x +\sqrt[n]{x^n-x+...}}$$

The above reduces to 2*cos(pi/4^(k-1)) in the x=2, n=2 case.

a form if the "sine like" function with gamma/2 (used with coefficients in spoiler below):

$$\frac {(Q_{x,n,k})^n}{b^k} \, = \, \frac {(\frac{\gamma}{2})}{a_1 \, b^{1k}} - \frac {(\frac{\gamma}{2})^3}{a_3 \, b^{2k}} + \frac {(\frac{\gamma}{2})^5}{a_5 \, b^{3k}} -\frac {(\frac{\gamma}{2})^7}{a_7 \, b^{4k}}+ \frac {(\frac{\gamma}{2})^9}{a_9 \, b^{5k}} \,\, ... = \, x- \sqrt[n]{x^n-x +\sqrt[n]{x^n-x+...}}$$

Using the functions, one can increase k with non-integer values, and traverse smoothly between iterations of radicals- it allows one to smoothly connect addition of radicals in a naturally arising way: infinitely nested radicals smoothly connect all radical addition.

$${Q_{2,2,3}}^2 = \,\,\ \frac{ pi^2}{ 4^{0}} - \frac{pi^4}{12 \times 4^{3}} +\frac{ pi^6}{360 \times 4^{6}} - \frac{ pi^8}{20160 \times 4^{9}}\,\,\, \dots = \,\,\, 4^3 \, \times \left[2 - \sqrt{2 + \sqrt{2}} \right]$$

so

$$\frac{{Q_{2,2,3}}^2}{4^3} = \,\,\ \frac{ pi^2}{ 4^{3}} - \frac{pi^4}{12 \times 4^{6}} +\frac{ pi^6}{360 \times 4^{9}} - \frac{ pi^8}{20160 \times 4^{12}}\,\,\, \dots = \,\,\,2 - \sqrt{2 + \sqrt{2}}$$

so

$$2 - \,\,\ \frac{ pi^2}{ 4^{3}} - \frac{pi^4}{12 \times 4^{6}} +\frac{ pi^6}{360 \times 4^{9}} - \frac{ pi^8}{20160 \times 4^{12}}\,\,\, \dots = \sqrt{2 + \sqrt{2}}$$

In the x=1.5, n=2 case we have $\gamma = 6.42675515$ b=3, for k roots of .75:

$$x- \frac{\gamma}{3^k} + \frac {\gamma^2}{6 \, \times \, 3^{2k} } - \frac {\gamma^3}{72 \, \times \, 3^{3k}} + \frac {\gamma^4}{1404\, \times \, 3^{4k} }- \frac {\gamma^4}{(673920/17 \, \times \, 3^{5k} } \\ +\frac {\gamma^6}{1510080 \, \times \, 3^{6k} } -\frac {\gamma^7}{\frac{178092794880}{2407} \, \times \, 3^{7k} } +\frac {\gamma^8}{\frac{18248946075360}{4051} \, \times \, 3^{8k} } \\ -\frac {\gamma^9}{ \frac{112670561875981824}{8887225} \, \times \, 3^{9k} } = \\ k roots \\ \sqrt{.75+\sqrt{.75+...}}$$

They are also periodic functions at n=2, x=2 for k<0.

I am interested in the semiperiodic properties of the other infinitely nested radical constants ($\gamma = q_{x,n}$) and functions at x and n other than 2.

For which, I require a generating function for a_m-- or a function f(x,n,k) that multiplied by cosine( g(x,n,k) ) gives the values I need.

I'd prefer both (because I want to check the a_m expansions against what I am given).

Thanks for any actual help. Anyone who suggests groots instead of kroots will be grooted.

for n=2:
a_1 is 1
a_2 is (2x)(2x-1)/2
a_3 is (2x)(2x-1)
a_4 is $\frac{\left( 2 x-1\right) \, \left( 2 x\right) \, \left( 2 x-1\right) \, \left( 2 x\right) \, \left( 2 x+1\right) }{2}$

General cases n=2, in "order":

x=2 : ... yup

x=2.5 : 1, 10, 20, 600, 1200, 62000/3, 148800, 58032000/131, 32240000...

x= 3: 1, 15, 30, 3150, 50400, 812700/11, 2889600....

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#### mfb

Mentor
One of the constants is Pi, at n=2, x=2.
Why? If $\displaystyle \pi = q_{2,2} = \lim_{k \to \infty} Q_{2,2,k}$ then $\displaystyle \pi^2 = q_{2,2}^2 = \lim_{k \to \infty} Q_{2,2,k} = \lim_{k \to \infty} \left( 2\right)^k \, \times \left[x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \sqrt[n]{x^n-x+...}}} \right]$ but that limit does not exist. In general you need $|nx^{n-1}| \leq 1$ for the limit to exist. If it is smaller than 1 then the limit will always be zero, if it is exactly one then things are interesting.

#### Matt Benesi

Gold Member
Why? If $\displaystyle \pi = q_{2,2} = \lim_{k \to \infty} Q_{2,2,k}$ then $\displaystyle \pi^2 = q_{2,2}^2 = \lim_{k \to \infty} Q_{2,2,k} = \lim_{k \to \infty} \left( 2\right)^k \, \times \left[x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \sqrt[n]{x^n-x+...}}} \right]$ but that limit does not exist. In general you need $|nx^{n-1}| \leq 1$ for the limit to exist. If it is smaller than 1 then the limit will always be zero, if it is exactly one then things are interesting.
No. I think you made an error somewhere, but I have to reply quickly.

As k -->infinity, $\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+...}} \to x$. You can look up various works involving nested roots/radicals in Dixon Jones' Chronology of Nested Roots. https://arxiv.org/pdf/1707.06139.pdf

2 is actually a pretty cool case here. It's the balancing point of all numbers (gives you the exponential function coefficients if you dissect its cosine and sine functions, because $q_{2,2} =\pi$). It's the only number 2^2-2=2... Technically... $2^\frac{n}{n-1}-2^\frac{1}{n-1} = 2^ \frac{1}{n-1}$, but it's the only integer case...

And it leads to exponential function coefficients a_m (well, maybe a_m*2... are exponential function coefficients....a_m = m!/2 I think.. for cosine at least)

#### mfb

Mentor
Ah, I misunderstood "and k being the number of radicals". Okay, then the limit can be interesting.

#### Matt Benesi

Gold Member
removing extra replies:
Ah, I misunderstood "and k being the number of radicals". Okay, then the limit can be interesting.
How would I phrase it so people aren't confused?

$$Q_{x,n,k} =\sqrt[n] { \left( nx^{n-1} \right)^k \, \times \left[x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \sqrt[n]{x^n-x+...}}} \right]}$$

k=1

$$Q_{x,n,1} =\sqrt[n] { \left( nx^{n-1} \right)^1 \, \times \left[x \right]}$$

k=2

$$Q_{x,n,2} =\sqrt[n] { \left( nx^{n-1} \right)^2 \, \times \left[x - \sqrt[n]{x^n-x} \right]}$$

etc.??

I think I messed up the math in the OP anyway. The one part should not be:

/////////////////////////////////
$$Q_{2,2,3} = \,\,\ \frac{ pi^2}{ 4^{0}} - \frac{pi^4}{12 \times 4^{3}} +\frac{ pi^6}{360 \times 4^{6}} - \frac{ pi^8}{20160 \times 4^{9}}\,\,\, \dots = \,\,\, 4^3 \, \times \left[2 - \sqrt{2 + \sqrt{2}} \right]$$

so

$$Q_{2,2,3} = \,\,\ \frac{ pi^2}{ 4^{3}} - \frac{pi^4}{12 \times 4^{6}} +\frac{ pi^6}{360 \times 4^{9}} - \frac{ pi^8}{20160 \times 4^{12}}\,\,\, \dots = \,\,\,2 - \sqrt{2 + \sqrt{2}}$$

//////////////////////////

It should be:
$${Q_{2,2,3}}^2 = \,\,\ \frac{ pi^2}{ 4^{0}} - \frac{pi^4}{12 \times 4^{3}} +\frac{ pi^6}{360 \times 4^{6}} - \frac{ pi^8}{20160 \times 4^{9}}\,\,\, \dots = \,\,\, 4^3 \, \times \left[2 - \sqrt{2 + \sqrt{2}} \right]$$

so

$${Q_{2,2,3}}^2 = \,\,\ \frac{ pi^2}{ 4^{3}} - \frac{pi^4}{12 \times 4^{6}} +\frac{ pi^6}{360 \times 4^{9}} - \frac{ pi^8}{20160 \times 4^{12}}\,\,\, \dots = \,\,\,2 - \sqrt{2 + \sqrt{2}}$$

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#### Matt Benesi

Gold Member
ETA to fix question:
Anyone? Am I asking the question incorrectly?

Basically, I'm looking for coefficients of alternating series that allow one to smoothly iterate roots.

$\sqrt[n] {x^n-x}$ to whatever depth of roots you want $\sqrt[n] {x^n-x+\sqrt[n] {x^n-x+\sqrt[n] {x^n-x}}}$.

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#### Matt Benesi

Gold Member
I am still stuck on this. I would appreciate some input (wikipedia page, or whatever), as I'd like to use the results.

//// post 1 :
Re-explained (maybe better):

k is the number of radicals, x and n are reals>1:

$$Q_{x,n,k} =\sqrt[n] { \left( nx^{n-1} \right)^k \, \times \left[x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \sqrt[n]{x^n-x+...}}} \right]}$$

In other words:

$$Q_{x,n,2} =\sqrt[n] { \left( nx^{n-1} \right)^2 \, \times \left[x - \sqrt[n]{x^n-x} \right]}$$

$$Q_{x,n,3} =\sqrt[n] { \left( nx^{n-1} \right)^3 \, \times \left[x - \sqrt[n]{x^n-x+ \sqrt[n]{x^n-x}} \right]}$$

As k approaches infinity, you get some value determined by x and n:

$$q_{x,n} = Q_{x,n,k \to \infty}$$

One of the values is p at n=2, x=2: $q_{2,2} \,=\, \pi$

All of the constants have associated functions that allow us to smoothly iterate radicals. By smoothly iterate, I mean they provide continuity between iterations of radicals $\sqrt[n] {x^n-x}$ to $\sqrt[n] {x^n-x + \sqrt[n]{x^n-x}}$ to $\sqrt[n] {x^n-x + \sqrt[n]{x^n-x+ \sqrt[n]{x^n-x}}}$.

There are exact formulas for the coefficients of terms in the functions that allow smooth iterations between radicals. In the x=2, n=2 case, the coefficients are directly related to sine/cosine.

$a_m$ coefficients of specific terms in the expansions

$b= nx^{n-1}$

$\gamma= q_{x,n}$

one form of "Cosine like" function (k radicals):

$$\frac {(Q_{x,n,k})^n}{b^k} \, = \, \frac {\gamma^2}{a_2 \, b^{1k}} - \frac {\gamma^4}{a_4 \, b^{2k}} +\frac {\gamma^6}{a_6 \, b^{3k}}- \frac {\gamma^8}{a_8 \, b^{4k}} \,\, ... = x- \sqrt[n]{x^n-x +\sqrt[n]{x^n-x+...}}$$

a form if the "sine like" function with gamma/2 (used with coefficients in spoiler below):

$$\frac {(Q_{x,n,k})^n}{b^k} \, = \, \frac {(\frac{\gamma}{2})}{a_1 \, b^{1k}} - \frac {(\frac{\gamma}{2})^3}{a_3 \, b^{2k}} + \frac {(\frac{\gamma}{2})^5}{a_5 \, b^{3k}} -\frac {(\frac{\gamma}{2})^7}{a_7 \, b^{4k}}+ \frac {(\frac{\gamma}{2})^9}{a_9 \, b^{5k}} \,\, ... = \, x- \sqrt[n]{x^n-x +\sqrt[n]{x^n-x+...}}$$

Using the functions, one can increase k with non-integer values, and traverse smoothly between iterations of radicals- it allows one to smoothly connect addition of radicals in a naturally arising way: infinitely nested radicals smoothly connect all radical addition.

$${Q_{2,2,3}}^2 = \,\,\ \frac{ pi^2}{ 4^{0}} - \frac{pi^4}{12 \times 4^{3}} +\frac{ pi^6}{360 \times 4^{6}} - \frac{ pi^8}{20160 \times 4^{9}}\,\,\, \dots = \,\,\, 4^3 \, \times \left[2 - \sqrt{2 + \sqrt{2}} \right]$$

so

$$\frac{{Q_{2,2,3}}^2}{4^3} = \,\,\ \frac{ pi^2}{ 4^{3}} - \frac{pi^4}{12 \times 4^{6}} +\frac{ pi^6}{360 \times 4^{9}} - \frac{ pi^8}{20160 \times 4^{12}}\,\,\, \dots = \,\,\,2 - \sqrt{2 + \sqrt{2}}$$

so

$$2 - \,\,\ \frac{ pi^2}{ 4^{3}} - \frac{pi^4}{12 \times 4^{6}} +\frac{ pi^6}{360 \times 4^{9}} - \frac{ pi^8}{20160 \times 4^{12}}\,\,\, \dots = \sqrt{2 + \sqrt{2}}$$

They are also periodic functions at n=2, x=2 for k<0.

I am interested in the semiperiodic properties of the other infinitely nested radical constants and functions at x and n other than 2. For which, I require a generating function for a_m-- or a function f(x,n,k) that multiplied by cosine( g(x,n,k) ) gives the values I need. I'd prefer both (because I want to check the a_m expansions against what I am given).

for n=2:
a_1 is 1
a_2 is (2x)(2x-1)/2
a_3 is (2x)(2x-1)
a_4 is $\frac{\left( 2 x-1\right) \, \left( 2 x\right) \, \left( 2 x-1\right) \, \left( 2 x\right) \, \left( 2 x+1\right) }{2}$

General cases n=2, in "order":

x= 1.5: 1, 3, 6, 18, 72, 234, 1404, 42120/11, 637820/17, (2^3*3^5*5*11^2*13)/223....

x=2 : 1,6,12,120,360,5040,20160..... == 1, 3!, 4!/2, 5!, 6!/2, 7!, 8!/2, 9!, 10!/2, 11!.... (the only decomposition functions with all primes distributed in monotonic order appear to be along the x=2, n=2 divide- intuitively, it makes sense that the pi function has all primes encoded within it....)

x=2.5 : 1, 10, 20, 600, 1200, 62000/3, 148800, 58032000/131, 32240000

x= 3: 1, 15, 30, 3150, 50400, 812700/11, 2889600....

Thanks for any help.

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#### suremarc

So I’ve read this thread again, which was tiresome to decipher. It sounds like you are essentially looking for a formula for fractional iterates of $f_x(c)=\sqrt{x^n-x+c}$ about the fixed point $c=x$. The approach to this would be to solve Schroder’s equation for that fixed point, then use the solution to calculate the Taylor series with respect to the iterate parameter. See this Wikipedia article for more information.

#### Matt Benesi

Gold Member
That doesn't help- except that I learned I could edit posts and clean up the thread.... which is sort of useful. If I get good answers to my questions.

Can you explain how to use Schröder's equation? Can you write out an example of its use, since you are familiar with its use, and this is the first time I've heard of it?

To others: Isn't there some general solution out there? There is for every other piece of simplistic math.

Why not this specific one?

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#### suremarc

Can you explain how to use Schröder's equation? Can you write out an example of its use, since you are familiar with its use, and this is the first time I've heard of it?
Sure. Schröder's equation says, given $h(x)$, find a function $\Psi(x)$ and a number $s$ such that $\Psi(h(x)) = s\Psi(x)$. Then we have $h^t(x)=\Psi^{-1}(s^t\Psi(x))$. This gives you a power series in $s^t$, which you could analyze with some sort of integral transform to extract information about its growth rate. From there you could work out some equations for $q_{x,n}$ and $Q_{x,n,k}$.
Solutions to Schröder's equation can be calculated to however many terms you like, using a thing called a Carleman matrix, but the procedure is complicated and ugly. I refer you to this paper I found, which shows how to calculate the terms.

Another thing I'd like to mention is that the expression in OP doesn't make sense when $n\neq 2$. You seem to have extrapolated the term $b=nx^{n-1}$, which coincides with the correct value for $n=2$, but the correct expression is $nx$. This amounts to the inverse of the derivative of $f(c)=\sqrt[n]{x^n-x+c}$, which represents the ratio by which $f$ tends to contract space around $x$. If $b$ were any other value, the expression would vanish or diverge.

Furthermore, $q_{x,n}$ is not uniquely defined because it depends on the initial value for the sequence of iterates. If I take $\lim_{k\rightarrow\infty}\sqrt[n]{(nx)^k(x-f^k(c))}$ at $c=x$, the term is zero for all values of k, and therefore the limit is zero. If I take the limit at $c=0$, I get a different number; when $x=2$ and $n=2$, this happens to be $\pi$. In between $c=0$ and $c=x$ the limit varies smoothly between the two values. When $x=2$ and $n=2$, it is closely approximated by an arccosine curve, within a few thousandths. You should plot it yourself and see the results, it's pretty interesting.

To others: Isn't there some general solution out there? There is for every other piece of simplistic math.

Why not this specific one?
If only!! There are many, many extremely simple functions which give rise to extremely complicated behavior under iteration, for example the logistic map. Even functions which aren't chaotic often lack a closed form solution for the nth iterate. Because what you're dealing with is fundamentally related to iterated functions, you shouldn't expect a general solution for much of anything.

Edit: upon looking at OP again, it looks as if you stumbled upon the solution to Schröder’s equation for the special case $x=n=2$. If you replace $\gamma$ with $2\cos^{-1}(c)$, you get something very close to Schröder’s equation. It doesn’t match exactly, which makes me think I did something wrong, but I’m too tired to look at it more. I will look at it again later.

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#### Matt Benesi

Gold Member
Sure. Schröder's equation says, given $h(x)$, find a function $\Psi(x)$ and a number $s$ such that $\Psi(h(x)) = s\Psi(x)$. Then we have $h^t(x)=\Psi^{-1}(s^t\Psi(x))$. This gives you a power series in $s^t$, which you could analyze with some sort of integral transform to extract information about its growth rate. From there you could work out some equations for $q_{x,n}$ and $Q_{x,n,k}$.
I was hoping to see you post a working example of Schröder’s equation- I know the wikipedia rehash is... the easiest thing to throw out there, but.

Maybe something a bit more like the example I posted for x=1.5: explanations leading up to it, then it in use, results being in line with goals (or serendipitous), so I can question parts of it. Get a good understanding of what you're doing.

I'll look at the paper more. I was just hoping for a live walkthrough, so I could ask questions about various points.

Another thing I'd like to mention is that the expression in OP doesn't make sense when $n\neq 2$. You seem to have extrapolated the term $b=nx^{n-1}$, which coincides with the correct value for $n=2$, but the correct expression is $nx$. This amounts to the inverse of the derivative of $f(c)=\sqrt[n]{x^n-x+c}$, which represents the ratio by which $f$ tends to contract space around $x$. If $b$ were any other value, the expression would vanish or diverge.
Remember we're doing (x-....)/( x-..........) to get the inner derivative, not ..../........

For certain continuous nested inverse functions that approach a fixed point, you get something that approaches the "inner function derivative". Stephen Tashi showed it is true with a Taylor approximation years ago, after I found it in my mathematical ramblings.

For example: If b and n are within domain in which n=output of function at infinite iterates and number of iterates approaches infinity/(infinity+1): $\frac{n-log_b [b^n-n +log_b [b^n-n+....}{n-log_b [b^n-n +log_b [b^n-n+log_b [b^n-n+.... }\, \to \, b^n \, Ln(b)$. In this case, the "inner function" is b^n.

Same thing with the $x= \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + ...}}$ functions: the ratio of successive iterations of (x-function) approaches the derivative of x^n ($nx^{n-1}$) as the number of iterations approaches infinity.

Furthermore, $q_{x,n}$ is not uniquely defined because it depends on the initial value for the sequence of iterates.
The initial value is x^n-x. Double check where I said "Above function explained". Is it clear that radical k=2 has x^n-x as a seed value?

When $x=2$ and $n=2$, it is closely approximated by an arccosine curve, within a few thousandths. You should plot it yourself and see the results, it's pretty interesting.
I'm not getting how you did this. Sounds interesting though. Maybe I could plot all of them for various x and n and compare the results to other functions. Visual math search! Wow.. I am easily excited.

Because what you're dealing with is fundamentally related to iterated functions, you shouldn't expect a general solution for much of anything.
But there is. In fact, I think I figured out a way of extracting specific enough results to find the exact pattern for various cases (especially n=2). I just wanted the already existing solution, because I was going to use it for fractal artwork. You know "the Bernoulli Gouldfarb equation for termination coefficients of nested functions" or something like that.

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#### Matt Benesi

Gold Member
I'm still plodding through this. Making charts. Since I'm pretty much certain the formulas exist out there somewhere, I'd really like the general coefficient formulas for something I'd like to make in GLSL. Might be a pipe dream, but I think smooth variable strength "cosine" and "sine" functions would look cool for certain applications.

I worked from here:

to here:

For cosine like series (q_x,n)^2:

coefficient 1= $\frac{n\, {{x}^{n}}-x}{n-1}$
coefficient 2= $\frac{\left( n\, {{x}^{n}}-x\right) 3 \left( n\, {{x}^{n-1}}+1\right) }{\left( n-2\right) \, \left( n\, {{x}^{n-1}}+1\right) +n+1}$
coefficient 3 top part = $4 x\, \left( {{n}^{3}}\, {{x}^{3 \left( n-1\right) }}-1\right) \, \left( \left( n-2\right) \, \left( n\, {{x}^{n-1}}+1\right) +n+1\right)$

$$\begin{array}{|c|c|c|c|} \hline &a_2=\frac{x(x^{n-1}-1}{n-1} & a_4 = \frac{3x (x^{n-1}-1)(x^{n-1}+1)}{(n-2)n x^{n-1} + 2n-1}& a_6 = \frac{ 4x (n^4x^{n+1} -1) ((n-2)n x^{n-1} + 2n-1)}{???} & a_8&\dots\\ \hline n=2 & \frac{x (2x-1)}{1} & \frac{3x(2x-1)(2x+1)}{0x^1 +3} & \frac{ 4x(8x^3-1) (0x^1+3)}{3(2x+5)} & \dots\\ \hline n=3 & \frac{x (3x^2-1)}{2} & \frac{3x(3x^2-1)(3x^2+1)}{3x^2 +5} & \frac{ 4x(27x^6-1)(3x^2+5)}{?} & \dots\\ \hline n=4 & \frac{x (4x^3-1)}{3} & \frac{3x(4x^3-1)(4x^3+1)}{8x^3 +7} & \frac{ 4x(64x^9-1)(8x^3+7)}{?} & \dots\\ \hline \dots &\dots & \dots&\dots &\dots&\dots&\dots\\ \hline \end{array}$$

To where I'm at now:

TLDR
Seeing some basic patterns in the coefficients emerging. I only really need the n=2 case, but... maybe it would be better to develop and all encompassing formula and leave it out there.

Anyways, these coefficients, in my mind, aren't only for smooth iteration of radical addition with specific constants associated. They are for smooth oscillating functions that I hope will make some interesting artwork. The point? The same thing as a nicely mowed lawn. Make something nice. Maybe make something really cool to look at, in between meals, talking, farming, working, and creation of artistic narratives to tie us together. That's why I'd like the general formula. It's a good enough reason to motivate me, when I expect nothing from life.. but finishing this one project.

$$\begin{array}{|c|c|c|c|} \hline &a_2=\frac{2x(x^{n-1}-1)}{2(n-1)} & a_4 = \frac{3x (x^{2(n-1)}-1)}{(n-2)n x^{n-1} + 2n-1}& a_6 = \frac{ 4x (x^{3(n-1)}-1)\,\, ((n-2)n x^{n-1} + 2n-1)}{???} & a_8&\dots\\ \hline n=2 & \frac{2x (2x-1)}{2} & \frac{3x(4x^2-1)}{0x^1 +3} & \frac{ 4x(8x^3-1) (0x^1+3)}{6x+15} & \frac{ x(16x^4-1) (2x^1+5)}{8x^2+6x+7} & \dots\\ \hline n=3 & \frac{2x (3x^2-1)}{4} & \frac{3x(9x^4-1)}{3x^2 +5} & \frac{ 4x(27x^6-1)(3x^2+5)}{4(27x^4+15x^2+10) } & \dots\\ \hline n=4 & \frac{2x (4x^3-1)}{6} & \frac{3x(16x^6-1)}{8x^3 +7} & \frac{ 4x(64x^9-1)(8x^3+7)}{?} & \dots\\ \hline \dots &\dots & \dots&\dots &\dots&\dots&\dots\\ \hline \end{array}$$

Last edited:

#### Matt Benesi

Gold Member
Coefficients for n=2 case, up to coefficient 5. Once again, any general form would be appreciated. If you recognize the pattern, let me know. At x=2, this reduces to cosine.

And yeah, it's something simple you learned in high school. If you ever decide to stop messing with me and leading me down wrong paths, it would be appreciated, if unexpected.

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)

I think about 8 more terms might be enough for what I want to try, but maybe not.

#### suremarc

I find it incredible that you're attacking me, but since you are so skeptical, I wrote a program in Mathematica to generate your next 8 terms using the method I described:
 k = 15; x =.; n = 2; f[y_] := (x^n - x + y)^(1/n); g[y_] := f[y + x] - x; M = Table[ D[g[y]^(i - 1), {y, j - 1}]/((j - 1)!), {i, 1, k}, {j, 1, k}] /. y -> 0; M = Simplify[M, Element[x, Reals] && x > 0 && Element[n, Integers]]; L = M[[2, 2]] V = IdentityMatrix[k]; For[i = 1, i <= k, i++, For[j = i + 1, j <= k, j++, V[[i, j]] = Sum[V[[i, l]]*M[[l, j]], {l, i, j - 1}]/(L^(i - 1) - L^(j - 1)); ] ] V = Simplify[V]; Psi = V[[2]] PsiInv = Simplify[Inverse[V]][[2]] 

Here's 15 terms (the first two being trivial of course):
$$0 \\ 1 \\ -\frac{1}{2 x-4 x^2} \\ \frac{1}{2 (1-2 x)^2 x^2 (1+2 x)} \\ \frac{5+2 x}{8 x^3 (-1+2 x)^3 \left(1+4 x+8 x^2+8 x^3\right)} \\ \frac{7+6 x+8 x^2}{8 (1-2 x)^4 x^4 (1+2 x)^2 \left(1+4 x^2\right) \left(1+2 x+4 x^2\right)} \\ \frac{21+28 x+56 x^2+64 x^3+48 x^4}{16 x^5 (-1+2 x)^5 (1+2 x)^2 \left(1+4 x^2\right) \left(1+2 x+4 x^2\right) \left(1+2 x+4 x^2+8 x^3+16 x^4\right)} \\ \frac{33+60 x+148 x^2+256 x^3+432 x^4+384 x^5+512 x^6+128 x^7}{16 (1-2 x)^6 x^6 (1+2 x)^3 \left(1+2 x+4 x^2\right)^2 \left(1+8 x^2+8 x^3+32 x^4+32 x^5+128 x^6+256 x^8\right)} \\ \frac{429+990 x+2820 x^2+6048 x^3+12976 x^4+21120 x^5+34304 x^6+41856 x^7+52224 x^8+45568 x^9+27648 x^{10}+2048 x^{11}}{128 x^7 (-1+2 x)^7 (1+2 x)^3 \left(1+4 x^2\right) \left(1-2 x+4 x^2\right) \left(1+2 x+4 x^2\right)^2 \left(1+2 x+4 x^2+8 x^3+16 x^4\right) \left(1+2 x+4 x^2+8 x^3+16 x^4+32 x^5+64 x^6\right)} \\ \frac{715+2002 x+6380 x^2+15840 x^3+38960 x^4+79808 x^5+164480 x^6+276352 x^7+474624 x^8+695808 x^9+986112 x^{10}+1099776 x^{11}+1335296 x^{12}+1032192 x^{13}+688128 x^{14}+131072 x^{15}}{128 (1-2 x)^8 x^8 (1+2 x)^4 \left(1+4 x^2\right)^2 \left(1-2 x+4 x^2\right) \left(1+2 x+4 x^2\right)^2 \left(1+16 x^4\right) \left(1+2 x+4 x^2+8 x^3+16 x^4\right) \left(1+2 x+4 x^2+8 x^3+16 x^4+32 x^5+64 x^6\right)} \\ \frac{2431+8008 x+28028 x^2+77880 x^3+211200 x^4+496320 x^5+1160192 x^6+2393984 x^7+4811264 x^8+8904704 x^9+15927296 x^{10}+25892864 x^{11}+41525248 x^{12}+59195392 x^{13}+80183296 x^{14}+100237312 x^{15}+114229248 x^{16}+107479040 x^{17}+92536832 x^{18}+48234496 x^{19}+14680064 x^{20}}{256 x^9 (-1+2 x)^9 (1+2 x)^4 \left(1+4 x^2\right)^2 \left(1-2 x+4 x^2\right) \left(1+2 x+4 x^2\right)^3 \left(1+16 x^4\right) \left(1+2 x+4 x^2+8 x^3+16 x^4\right) \left(1+8 x^3+64 x^6\right) \left(1+2 x+4 x^2+8 x^3+16 x^4+32 x^5+64 x^6\right)} \\ \frac{4199+15912 x+60424 x^2+184184 x^3+539968 x^4+1400352 x^5+3569024 x^6+8298240 x^7+18809088 x^8+39772160 x^9+81849344 x^{10}+158185472 x^{11}+299859968 x^{12}+532553728 x^{13}+920403968 x^{14}+1500348416 x^{15}+2370502656 x^{16}+3493462016 x^{17}+4992532480 x^{18}+6505889792 x^{19}+8226078720 x^{20}+9206497280 x^{21}+9709813760 x^{22}+8321499136 x^{23}+6408896512 x^{24}+3053453312 x^{25}+939524096 x^{26}}{256 (1-2 x)^{10} x^{10} (1+2 x)^5 \left(1+4 x^2\right)^2 \left(1-2 x+4 x^2\right) \left(1+2 x+4 x^2\right)^3 \left(1+16 x^4\right) \left(1-2 x+4 x^2-8 x^3+16 x^4\right) \left(1+2 x+4 x^2+8 x^3+16 x^4\right)^2 \left(1+8 x^3+64 x^6\right) \left(1+2 x+4 x^2+8 x^3+16 x^4+32 x^5+64 x^6\right)} \\ \frac{29393+125970 x+514488 x^2+1697696 x^3+5317312 x^4+14906528 x^5+40650688 x^6+102775296 x^7+252708608 x^8+588984320 x^9+1331216384 x^{10}+2874906624 x^{11}+6064906240 x^{12}+12258697216 x^{13}+24152276992 x^{14}+45857964032 x^{15}+84758429696 x^{16}+150767009792 x^{17}+261238030336 x^{18}+434696617984 x^{19}+702951718912 x^{20}+1090808446976 x^{21}+1632393756672 x^{22}+2326160998400 x^{23}+3195891875840 x^{24}+4132261855232 x^{25}+5062558482432 x^{26}+5744921411584 x^{27}+6039529324544 x^{28}+5567351357440 x^{29}+4554812817408 x^{30}+2847563317248 x^{31}+1352914698240 x^{32}+300647710720 x^{33}}{1024 x^{11} (-1+2 x)^{11} (1+2 x)^5 \left(1+4 x^2\right)^2 \left(1-2 x+4 x^2\right) \left(1+2 x+4 x^2\right)^3 \left(1+16 x^4\right) \left(1-2 x+4 x^2-8 x^3+16 x^4\right) \left(1+2 x+4 x^2+8 x^3+16 x^4\right)^2 \left(1+8 x^3+64 x^6\right) \left(1+2 x+4 x^2+8 x^3+16 x^4+32 x^5+64 x^6\right) \left(1+2 x+4 x^2+8 x^3+16 x^4+32 x^5+64 x^6+128 x^7+256 x^8+512 x^9+1024 x^{10}\right)} \\ \frac{52003+248710 x+1085280 x^2+3840096 x^3+12748112 x^4+38142720 x^5+110096448 x^6+297174528 x^7+777740544 x^8+1944133120 x^9+4721329152 x^{10}+11024812032 x^{11}+25149571072 x^{12}+55451041792 x^{13}+119471800320 x^{14}+249982550016 x^{15}+511600623616 x^{16}+1017703170048 x^{17}+1982365171712 x^{18}+3755020910592 x^{19}+6962182881280 x^{20}+12564811481088 x^{21}+22168922488832 x^{22}+38002432671744 x^{23}+63695237414912 x^{24}+103517201104896 x^{25}+163980442075136 x^{26}+251246594228224 x^{27}+374255397109760 x^{28}+535713418313728 x^{29}+743357907206144 x^{30}+982007765008384 x^{31}+1247374466875392 x^{32}+1491977149349888 x^{33}+1689554234900480 x^{34}+1743997240344576 x^{35}+1680603523055616 x^{36}+1391706842857472 x^{37}+1001655092903936 x^{38}+550305569701888 x^{39}+223200860438528 x^{40}+30786325577728 x^{41}}{1024 (1-2 x)^{12} x^{12} (1+2 x)^6 \left(1+4 x^2\right)^3 \left(1+2 x+4 x^2\right)^4 \left(1+16 x^4\right) \left(1-4 x^2+16 x^4\right) \left(1-2 x+4 x^2-8 x^3+16 x^4\right) \left(1+8 x^3+64 x^6\right) \left(1+4 x^2+8 x^3+16 x^4+64 x^6\right)^2 \left(1+2 x+4 x^2+8 x^3+16 x^4+32 x^5+64 x^6\right) \left(1+2 x+4 x^2+8 x^3+16 x^4+32 x^5+64 x^6+128 x^7+256 x^8+512 x^9+1024 x^{10}\right)} \\ \frac{185725+980628 x+4547840 x^2+17131920 x^3+59935200 x^4+189751552 x^5+575619200 x^6+1640694528 x^7+4521121536 x^8+11944266752 x^9+30661671936 x^{10}+76064624640 x^{11}+184211517440 x^{12}+433407426560 x^{13}+997415403520 x^{14}+2239325929472 x^{15}+4926191370240 x^{16}+10592819347456 x^{17}+22349943668736 x^{18}+46156451151872 x^{19}+93583849816064 x^{20}+185928505098240 x^{21}+362715675623424 x^{22}+693542057934848 x^{23}+1302440980774912 x^{24}+2397203851116544 x^{25}+4331031465295872 x^{26}+7666432603062272 x^{27}+13311279397601280 x^{28}+22620266676879360 x^{29}+37669500295839744 x^{30}+61302111165480960 x^{31}+97592729390809088 x^{32}+151511757414727680 x^{33}+229489585913069568 x^{34}+337937372698116096 x^{35}+483941659189444608 x^{36}+670474219158503424 x^{37}+898528598799941632 x^{38}+1156986399094734848 x^{39}+1429159507434405888 x^{40}+1675607342219001856 x^{41}+1860545198010925056 x^{42}+1919043614655119360 x^{43}+1830484550107529216 x^{44}+1561763908279074816 x^{45}+1172413646744059904 x^{46}+716072340751908864 x^{47}+348466021167792128 x^{48}+99079191802150912 x^{49}+7881299347898368 x^{50}}{2048 x^{13} (-1+2 x)^{13} (1+2 x)^6 \left(1+4 x^2\right)^3 \left(1+2 x+4 x^2\right)^4 \left(1+16 x^4\right) \left(1-4 x^2+16 x^4\right) \left(1-2 x+4 x^2-8 x^3+16 x^4\right) \left(1+8 x^3+64 x^6\right) \left(1+4 x^2+8 x^3+16 x^4+64 x^6\right)^2 \left(1+2 x+4 x^2+8 x^3+16 x^4+32 x^5+64 x^6\right) \left(1+2 x+4 x^2+8 x^3+16 x^4+32 x^5+64 x^6+128 x^7+256 x^8+512 x^9+1024 x^{10}\right) \left(1+2 x+4 x^2+8 x^3+16 x^4+32 x^5+64 x^6+128 x^7+256 x^8+512 x^9+1024 x^{10}+2048 x^{11}+4096 x^{12}\right)} \\$$
I'll be damned if there's a general formula for that.

Last edited:

#### Matt Benesi

Gold Member
I can't use that code, don't have mathematica. I can use the coefficients though. That will help (assuming they are correct). I do like the way I organized them better (in the post a couple up). It makes their growth more apparent (only keeping in the last part of the coefficient, so one can see what is added to the coefficient each time).

Wish I could test your code.

<< Mentor Note -- Post has been edited >>

Last edited by a moderator:

Mentor

#### berkeman

Mentor
Thread re-opened after edits. Thanks for everybody's patience.

"Generating the formula for the coefficients of an alternating series"

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