Some random thoughts, posted to work on.

[Kharakov]

hmm.

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for email:
\(f1(b,x,k,n) = x_n = log_b \sqrt[k]{b^{xk}-x +log_b \sqrt[k]{b^{xk}-x + log_b \sqrt[k]{b^{xk}-x + \cdots }}}\)

\(f2(b,x,k,n)=\ \lim_{n\to\infty} \ \frac{x-x_n}{x-x_{n+1}}\ =\ \frac{d}{dx} b^{xk} \ \ = \ k b ^{kx} log (b)\)

\(f3(x,k,n) = \ x_n = \sqrt[k]{x^k-x + \sqrt[k]{x^k-x + \cdots }}\)

\(f4(x,k,n) = \ \lim_{n\to\infty} \ \frac{x-x_n}{x-x_{n+1}}\ = \ \frac{d}{dx} x^{k} \ \ = \ k x ^{k-1}\)

With a= value of inner function, in the case of f3, the inner function is x^k-x, for f1 it is b^(kx)-x:

\( \frac {(x-x_n)^{n+1}}{(x-x_{n+1})^{n}} \ \to \ C\)

\(f1(b,x,n,a) = x_a = log_b \sqrt[n]{b^{xn}-x +log_b \sqrt[n]{b^{xn}-x + log_b \sqrt[n]{b^{xn}-x + \cdots }}}\)

a= number of recursions, iterations, nestings, or whatever

as a --> infinity

\(f2(b,x,n,a)=\ \frac{x-x_a}{x-x_{a+1}}\ =\ \frac{d}{dx} b^{xn} \ \ = \ n b ^{nx} log (b)\)

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\(f2(b,x,n,a)=\ \frac{x-f1(b,x,n,a)}{x-f1(b,x,n,a+1)}\ =\ \frac{d}{dx} b^{xn} \ \ = \ n b ^{nx} log (b)\)

For manipulating derivative of above:

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Set \(b = \sqrt[x]{x}\) then \( f2(...) = \ n x^{n-1} \ log(x) \) if x=e=b=n or whatever

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\(f1(b,x,n,a) = \ x_a = log_{\sqrt[x]{x}} \sqrt[n]{x^n-x + log_{\sqrt[x]{x}} \sqrt[n]{x^n-x + \cdots }}\)

set \(n= \frac{1}{log(b)}\) ... \(f2(...) = \ n b ^{nx} log (b) \ = \ \sqrt[x]{x}^{\frac{x}{log(\sqrt[x]{x})} } = \ x^{\frac{1} {log(\sqrt[x]{x})} } = \ x^{\frac {x} {log(x)}} = \ e^x \)
also....
\(f1(...) = \ x = log_{\sqrt[x]{x}} \sqrt[n]{e^x-x + log_{\sqrt[x]{x}} \sqrt[n]{e^x-x + \cdots }}\)

so the interior function is the same as the derivative of the interior function (well.... haha)..
set x=e..... you get

\(f1(...) = \ e = log_{\sqrt[e]{e}} \sqrt[\frac{1}{log_{\sqrt[e]{e}}}] {e^e-e + log_{\sqrt[e]{e}} \sqrt[\frac{1}{log_{\sqrt[e]{e}}}] {e^e-e + \cdots }}\)

with an interior function = e^e, derivative of interior function = e^e, and the function itself=e.

Of course, n= x/log(x), with x=e, n=e..... note that the above is:

\( e = \ log(e^e -e + log (e^e-e + \cdots\)

and the derivative of the inner function if e^e, the function =e exactly the same as

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

So the log function and the nth root function are the balancing point between 2 different types of infinitely nested clamping functions... which is sort of obvious.

Set \( x = \ \frac {log_b (x^{n-1})}{n} \) then \( f2(...) = \ n x^{n-1} \ log(b)\) if b=e=x=n

Set \( x = \ \frac{log_x(x^{n-1})}{n}=\ \frac{n-1}{n}\) and \( n = \ \frac{1}{log(x)}\) and b=x then \( f2(...) = \ \frac{\sqrt[log(x)]{x}}{x}= \ \frac {e}{x}\) is it even possible?? no.... but it's funny.

Set \( x = b^2+b \) and \(n = 1/b\) then \( f2(...) = \ b^{b} \ log(b)\) if b=e....

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\(x = log_{\sqrt[x]{x}} \sqrt[n]{x^n-x + log_{\sqrt[x]{x}} \sqrt[n]{x^n-x + \cdots }}\)

\(\frac{x-f1(b,x,n,a)}{x-f1(b,x,n,a+1)}\ = \ n x ^{n-1} log (x)\)

At x=e it looks a lot like the derivative of x^n... at x=n=e you have the balance point....

The above part might be the same as (check):

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

corrected:

\(1 = log_x \sqrt[n]{x^n-x + log_{\sqrt[x]{x}} \sqrt[n]{x^n-x + \cdots }}\)
wrong (hidden):

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\(1 = log_x \sqrt[n]{x^n-1 + log_x \sqrt[n]{x^n-1 + \cdots }}\)

\(f3(x,n,a) = \ x_a = \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \cdots }}\)


\(f4(x,n,a) = \ \frac{x-x_a}{x-x_{a+1}}\ = \ n x ^{n-1}\)

Set so "derivatives" = e^e:

f1 (b=e, x=e^2+e, n=1/e, a...)

f1 (b=e^(1/e), x=e, n=e, a...)

f3 (x=e, n=e, a...)

Each ended up with same values for derivatives, of course, but multiplying:

[x- f.(...)] * derivative^(a) = \(\frac{[x - f.(..., a)] ^{ a+1}} { [x- f.(...,a+1)]^{a}}\) gives you a convergent value (other than the derivative) as well.

corrected formula for above:

\( \frac {(x-x_a)^{a+1}}{(x-x_{a+1})^{a}} \ \to \ C_{(a,k)}\)

Had a fantasy about there being some other convergent value besides pi^2/4 for x=2,n=2, for the 4th formula (which is basically (with x=n=2) a geometric formula for pi, if you throw in the last square root).


Other properties? As you increase the inputs to the various functions (x>2, n>2, b>2) [x- f.(...)] * derivative^(a) = \(\frac{[x - f.(..., a)] ^{ a}} { [x- f(...,a+1)]^{a-1}}\) approaches x.

Reading Tito Piezas III article: https://sites.google.com/site/tpiezas/0014 showed me that things converge to different, more interesting numbers if you take the derivative to (a) instead of (a+1). There are ways to extract integers from the constants generated by these functions, other than the pi function (although the one is... quite compute intensive)! Ohh, and the other thing generated is 2 ln(2), which is maybe a bit transcendental itself (like the pi^2/4)... maybe . I still want to generate an integer with an input to one of these functions.....