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The Math Thread

Kharakov said:
There is a class of transcendental numbers that I'm revisiting. What I want to do is a bit twisted, but fun. I'll try and explain it, keep in mind that I don't know what I'm talking about, so don't expect strict terminology.

\( f(x,n,a) = \sqrt[n]{x^n-x + \sqrt[n]{x^n-x +\sqrt[n]{x^n-x + \cdots}}}\)

As the number of nestings (nested radicals???), a, approaches infinity, f(x,n,a) approaches x.

\( dx = n x ^{n-1} \approx \frac {x - f(x,n,a)} {x-f(x,n,a+1)} \) or \( \simeq\) ??

\( q= {dx}^{a+1} \, x - {dx}^{a+1} \, f(x,n,a)\)

Should the limit for q be 0 as a --> infinity? What about for x=2, n=2?


Challenge: create a generating function for x>1 and n>1, that creates integer q.
Click to expand...

I'm going to have to say that I don't know what you're talking about either. :p

George S said:
beero1000 said:
Change variables \(x = \frac{2^a}{\pi}\) and fiddle...

\(\displaystyle \lim_{a \to \infty} \left(\frac {1}{ \frac{\pi^2}{2^{2a+.5}}+\frac{\pi}{2^a} } - \frac{2^a}{\pi}\right) = \lim_{x \to \infty} \left(\frac {1}{\frac{1}{\sqrt{2}x^2}+\frac{1}{x}} - x\right) = \lim_{x \to \infty} \left(\frac{\sqrt{2}x^2}{1 + \sqrt{2}x} - x\right) = \lim_{x \to \infty} \left(\frac{\sqrt{2}x^2 - x(1+\sqrt{2}x)}{1 + \sqrt{2}x}\right) = \lim_{x \to \infty} \left(\frac{-x}{1 + \sqrt{2}x}\right) = - \frac{1}{\sqrt{2}}\)
Click to expand...

No, no! The correct answer is
\( - \frac{sqrt{2}}{2}\)
Pedantically yours,
George
Click to expand...

That is indeed a correct representation of the real number to which the limit converges.
 
beero1000 said:
Kharakov said:
There is a class of transcendental numbers that I'm revisiting. What I want to do is a bit twisted, but fun. I'll try and explain it, keep in mind that I don't know what I'm talking about, so don't expect strict terminology.



\( f(x,n,a) = \ \sqrt[n]{x^n-x + \sqrt[n]{x^n-x +\sqrt[n]{x^n-x + \cdots}}}\)

As the number of nestings (nested radicals???), a, approaches infinity, f(x,n,a) approaches x.

\( dx = n x ^{n-1} \approx \frac {x - f(x,n,a)} {x-f(x,n,a+1)} \) or \( \simeq\) ??

\( q= {dx}^{a+1} \, x - {dx}^{a+1} \, f(x,n,a)\)

Should the limit for q be 0 as a --> infinity? What about for x=2, n=2?



Challenge: create a generating function for x>1 and n>1, that generates integer rational q.
Click to expand...
I'm going to have to say that I don't know what you're talking about either. :p
Click to expand...

a is the number of nested radicals (# of nestings):


a=1:
\(\sqrt{y}\)

a=2:
\( \sqrt{y+\sqrt{y}}\)

a=3:
\( \sqrt{y+\sqrt{y+\sqrt{y}}}\)

a=6:
\( \sqrt{y+\sqrt{y+\sqrt{y+\sqrt{y+\sqrt{y+\sqrt{y}}}}}}\)



\( f(x,n,a) = \sqrt[n]{x^n-x + \sqrt[n]{x^n-x +\sqrt[n]{x^n-x + \cdots}}}\)

\(x = \ \lim_{a\to\infty} \ f(x,n,a) \)

\(z = \ \lim_{a\to\infty} \ \frac {x \ - \ f(x,n,a)} {x \ - \ f(x,n,a+1)} \ = \ n x ^{n-1} \)

\( q = \ \lim_{a\to\infty} \ z ^{a+1} \ x \ - \ z ^{a+1} \ f(x,n,a) \)


Challenge 1: find values of x and n that generate rational q (or prove that it is impossible to do so). Setting n and x to 2, you get q=pi^2, which isn't very rational.

Challenge 2: find a non-trivial function z = z (x,n) that causes q to converge to pi^2. In other words, not the following:



\(\ \lim_{a\to\infty} \ z = \left ( \frac {\pi^2}{x-f(x,n,a)} \right )^{\frac{1}{a+1}}\)

 
Scratch off challenge 2. Anything other than z= nx^(n-1) doesn't converge as you ramp up a. Doy.
 
I'm leery of convergence here, but if we suppose x > 1 and n > 1, we will probably (hopefully?) be ok.

Since q as a function of x is continuous (I think) and increases without bound as x goes to infinity, it has to cross rational values.
 
beero1000 said:
I'm leery of convergence here, but if we suppose x > 1 and n > 1, we will probably (hopefully?) be ok.

Since q as a function of x is continuous (I think) and increases without bound as x goes to infinity, it has to cross rational values.
Click to expand...
Q does cross rational values:


q= 9 @ x ~ 1.890037864286 n=2
q=pi^2 @ x =2 n=2
q=10 @ x ~ 2.015898407895 n=2
q=12 @ x ~ 2.244454948781 n=2



There are inflection points as well, for each n (because square root of numbers less than 1 increases size of numbers, x^n-x is less than 1 for specific values of x). The inflection points for specific n are close to x=(1+2^(1/n))/2, and at first glance get closer to that value as n-->infinity.



n=1.000...1 x=1.00....1 q=1
n=2 x~ 1.1900419016802 q~ 5.3538327682514
n=3 x~ 1.1216979914246 q~ 7.7852669557221
n=4 x~ 1.0893966984437 q~ 10.230368441016
n=19 x~ 1.016977237659 q~ 49.4960400660355




I'm still looking for an inverse function, along the lines of log/exp, that allows one to generate rationals with Q.
 
Kharakov said:
beero1000 said:
I'm leery of convergence here, but if we suppose x > 1 and n > 1, we will probably (hopefully?) be ok.

Since q as a function of x is continuous (I think) and increases without bound as x goes to infinity, it has to cross rational values.
Click to expand...
Q does cross rational values:


q= 9 @ x ~ 1.890037864286 n=2
q=pi^2 @ x =2 n=2
q=10 @ x ~ 2.015898407895 n=2
q=12 @ x ~ 2.244454948781 n=2



There are inflection points as well, for each n (because square root of numbers less than 1 increases size of numbers, x^n-x is less than 1 for specific values of x). The inflection points for specific n are close to x=(1+2^(1/n))/2, and at first glance get closer to that value as n-->infinity.



n=1.000...1 x=1.00....1 q=1
n=2 x~ 1.1900419016802 q~ 5.3538327682514
n=3 x~ 1.1216979914246 q~ 7.7852669557221
n=4 x~ 1.0893966984437 q~ 10.230368441016
n=19 x~ 1.016977237659 q~ 49.4960400660355




I'm still looking for an inverse function, along the lines of log/exp, that allows one to generate rationals with Q.
Click to expand...

Sure. If it helps, you can look at a series expansion for q(x). For n = 2, I get \(q(x) = 2x^2 + \frac{x}{2} + \frac{1}{2} + \frac{13}{32x} + \frac{12}{32x^2} + \frac{81}{256x^3} + \frac{75}{256x^4} + \dots\).
 
Did that in the past, should re gander it.

I noticed that it tends to the following values (x>1.9,n>1.9) so if this triggers a memory:

\(r(x,n) = x^{n+1}\ +\ x^n (n-x)\ +\ \frac{xn-x}{2}\)

x,n: blah blah blah
whoops??? wrong numbers or formula... be back later.
2,2: 9 + .86960440...
3,3: 84 + .32510815402
4,4: 1,030 + .079091298
4,4: 15,635 + .012815892
5,5: 279,951 + .001500491

4,2: 34 +.631414536
5,3: 380. +.183607775
6,4: 5193 +.034848296


In other words, the lower x and n are, the more it drifts away from r (x,n), which doesn't really help create an inverse function ini the slightest. I need to focus on inverting the function, instead of playing around...
 
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All equations and explanations:: editing...

a is the number of nested radicals (# of nestings):


a=1:
\(\sqrt{y}\)

a=2:
\( \sqrt{y+\sqrt{y}}\)

a=3:
\( \sqrt{y+\sqrt{y+\sqrt{y}}}\)

a=6:
\( \sqrt{y+\sqrt{y+\sqrt{y+\sqrt{y+\sqrt{y+\sqrt{y}}}}}}\)



\( f(x,n,a) = \sqrt[n]{x^n-x + \sqrt[n]{x^n-x +\sqrt[n]{x^n-x + \cdots}}}\)

\(x = \ \lim_{a\to\infty} \ f(x,n,a) \)

\(z = \ \lim_{a\to\infty} \ \frac {x \ - \ f(x,n,a)} {x \ - \ f(x,n,a+1)} \ = \ n x ^{n-1} \)

\( q = \ \lim_{a\to\infty} \ z ^{a+1} \ x \ - \ z ^{a+1} \ f(x,n,a) \)

r(x,n) generates an approximation of q, which is closer for higher x,n:

\(r(x,n) =\ n x^n \ + \ \frac{xn-x}{2} \ = \ x \ \left ( z + \frac{n-1}{2} \right) \)



x,n: r + (q-r)

2,2: 9 + .86960440...
3,3: 84 + .32510815402
4,4: 1,030 + .079091298
5,5: 15,635 + .012815892
6,6: 279,951 + .001500491

4,2: 34 +.631414536
5,3: 380. +.183607775
6,4: 5193 +.034848296

2,3: 26 + .55643749401
3,4: 328 + .64303611196 (really 328.5 + .14303611196 )
4,5: 5128 + .02507592815

 
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Yes, r(x,n) is the positive-exponent part of the series expansion so for large x you get a good approximation (the constant term vanishes after n = 2). In particular, you get something like \(q(x,n) = r(x,n) + O(\frac{1}{x^{n-2}})\) for n > 2.

You can use the  Lagrange inversion formula to compute the inverse from the series expansion.
 
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beero1000 said:
Sure. If it helps, you can look at a series expansion for q(x). For n = 2, I get \(q(x) = 2x^2 + \frac{x}{2} + \frac{1}{2} + \frac{13}{32x} + \frac{12}{32x^2} + \frac{81}{256x^3} + \frac{75}{256x^4} + \dots\).
Click to expand...

How did you arrive at this series for q? I've done the Taylor for \( \sqrt[n]{x^n-x}\), but couldn't imagine creating a series for q?
 
Kharakov said:
beero1000 said:
Sure. If it helps, you can look at a series expansion for q(x). For n = 2, I get \(q(x) = 2x^2 + \frac{x}{2} + \frac{1}{2} + \frac{13}{32x} + \frac{12}{32x^2} + \frac{81}{256x^3} + \frac{75}{256x^4} + \dots\).
Click to expand...

How did you arrive at this series for q? I've done the Taylor for \( \sqrt[n]{x^n-x}\), but couldn't imagine creating a series for q?
Click to expand...

It should just be the the standard Laurent series centered 'at infinity'. q converges so quickly that coefficients get nailed down almost immediately, so the approximation series for each new a will give more coefficients of q.
 
How did you generate the series?
 
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The math involves computing a Laurent series for q(1/y,n,a) centered at 0, and then evaluating that at y = 1/x to get q(x,n,a) centered at infinity. You can see how to compute  Laurent series in a bunch of places online, but it's a hassle to do by hand in this instance.

I used Mathematica for the actual computations, e.g. Series[(2*x)^(7 + 1)*(x - Nest[Power[(# + x^2 - x), 1/2] &, 0, 7]), {x, Infinity, 5}]. This returns the correct coefficients up to x-5:

\(2x^2 + \frac{x}{2} + \frac{1}{2} + \frac{13}{32x} + \frac{12}{32x^2} + \frac{81}{256x^3} + \frac{75}{256x^4} + \frac{2109}{8192x^5} + O(\frac{1}{x^6})\dots\)

You should be able to paste the command into wolfram-alpha if you don't have Mathematica. It should be relatively obvious how to modify it for different a or n, but I can parse it if something doesn't make sense.
 
Ok, alpha works. Nice, discovered a couple problems with some other math I was doing. Thanks.
 
You might be able to do similar stuff with other inverse functions (extract a value from infinite nestings)...

\(n = log_x \left ( x^n-n + log_x \left (x^n -n \cdots \)

\(\frac {\sqrt[n]{x^n-x+ \sqrt[n]{x^n-x + \cdots}}}{x} = \frac{log_x \left ( x^n-n + log_x \left (x^n -n \cdots}{n} \)

at x= n = e we have~~

\(e = \sqrt[e]{e^e-e+ \sqrt[e]{e^e-e + \cdots}} = log \left ( e^e-e + log \left (e^e -e \cdots \)
 
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