- #1

Matt Benesi

- 134

- 7

The properties arise from

Simply raise both sides to the n

Divide through by x

What we will do is explore the properties of finitely iterated nestings. The variable

For this example, a=4: [itex]f(x,n,a)=f(x,n,4)=\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}}}}} [/itex]

The interesting properties arise when we subtract a nested function from the number it equals at infinite nestings (x).

With this particular variety of nesting: [itex]f(x,n,a) = x-\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1...}}}}} [/itex]

We end up approaching [itex]n^{-a}x\ln{x}[/itex] as

There are many multiplicative nested type equations we can try, such as:

[itex]x=\log_B [ B^x\log_B [B^x \log_B [B^x...]]][/itex] so [itex]f(x,B,a)=x-\log_B [ B^x\log_B [B^x \log_B [B^x...]]][/itex]

which has the interesting property:

[tex]\dfrac{f(x,B,a)}{f(x,B,a+1)}\to x \ln B[/tex]

Of course, there are whole other types of nested functions using addition/subtraction, in addition to multiplication (if you mix them). You can do cosine and cosine

This next formulas approach the derivative of the inner functions when taking [itex]\dfrac{f(...,a)}{f(...,a+1)}[/itex] for higher

For this one, the inner function is x^n:

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

so [tex]\dfrac{f(x,n,a)}{f(x,n,a+1)}\to nx^{n-1}[/tex]

For this one, the inner function is B

[tex]f(x,B,a)=x-\log_B [B^x-x+\log_B [B^x-x+\log_B [B^x-x+\log_B [...]]]][/tex]

so [tex]\dfrac{f(x,B,a)}{f(x,B,a+1)}\to B^{x}\ln B[/tex]

as

In fact, all of the basic formulas that use

Combining the functions results in approaching the derivative of the combined inner function:

[tex]f(x,B,a)=x-\sqrt[n]{\log_B [ B^{x^n}-x+ \sqrt[n] {\log_B [B^{x^n}-x+ \sqrt[n]{ \log_B [B^{x^n}-x+...}}}]]][/tex]

Note that it is set up to take x^n first, then take B^(x^n) next (as if it were infinitely iterated so that it is algebraically sound). The "[" symbol doesn't show up to clearly under the radical. Anyways...

As with the other -x + ... functions, this one approaches the derivative of the inner function [itex]B^{x^n}[/itex]

[tex]\dfrac{f(x,B,a)}{f(x,B,a+1)}\to n\,{x}^{n-1}\,{y}^{{x}^{n}}\,log\left( y\right)[/tex]

For all of these functions, the exact approached value (so for x^n as the inner function, nx^(n-1) ) for f(...,a)/f(...,a+1) can be taken to the a

Note that a value more or less than the value that f(...,a)/f(...,a+1) approaches causes the calculated constant to diverge towards 0 or infinity as a increases [UNLESS you use the exact constant (such as the derivatives, or n for the first example, x ln (B) for the second, etc.)].

**nested functions such as:***infinitely*[itex]x=\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}...}}}}[/itex]

You can solve it algebraically to verify that x is equal to the nested function (all of the following functions can be solved in a similar manner).Simply raise both sides to the n

^{th}power: [itex]x^n=x^{n-1}\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}...}}}[/itex]Divide through by x

^{n-1}: [itex]x=\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}...}}}[/itex]What we will do is explore the properties of finitely iterated nestings. The variable

*a*is used to denote the number of nestings.For this example, a=4: [itex]f(x,n,a)=f(x,n,4)=\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}}}}} [/itex]

The interesting properties arise when we subtract a nested function from the number it equals at infinite nestings (x).

With this particular variety of nesting: [itex]f(x,n,a) = x-\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1}\sqrt[n]{x^{n-1...}}}}} [/itex]

We end up approaching [itex]n^{-a}x\ln{x}[/itex] as

*a*gets larger. It approaches the value quicker for higher*n*and*x*.*f(x,n,a)/f(x,n,a+1)*approaches*n*as*a*increases.There are many multiplicative nested type equations we can try, such as:

[itex]x=\log_B [ B^x\log_B [B^x \log_B [B^x...]]][/itex] so [itex]f(x,B,a)=x-\log_B [ B^x\log_B [B^x \log_B [B^x...]]][/itex]

which has the interesting property:

[tex]\dfrac{f(x,B,a)}{f(x,B,a+1)}\to x \ln B[/tex]

Of course, there are whole other types of nested functions using addition/subtraction, in addition to multiplication (if you mix them). You can do cosine and cosine

^{-1}together, x^{n}and x^{1/n}and other combinations.This next formulas approach the derivative of the inner functions when taking [itex]\dfrac{f(...,a)}{f(...,a+1)}[/itex] for higher

*a*.For this one, the inner function is x^n:

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

so [tex]\dfrac{f(x,n,a)}{f(x,n,a+1)}\to nx^{n-1}[/tex]

For this one, the inner function is B

^{x}:[tex]f(x,B,a)=x-\log_B [B^x-x+\log_B [B^x-x+\log_B [B^x-x+\log_B [...]]]][/tex]

so [tex]\dfrac{f(x,B,a)}{f(x,B,a+1)}\to B^{x}\ln B[/tex]

as

*a*increases (or for larger B and x).In fact, all of the basic formulas that use

*- x + (the repeated formula...)*appear to approach the derivative of the inner formula for f(...,a)/f(...,a+1) except in conditions when the functions and inverse functions used have limited well defined domains (such as cosine and cosine^{-1}).Combining the functions results in approaching the derivative of the combined inner function:

[tex]f(x,B,a)=x-\sqrt[n]{\log_B [ B^{x^n}-x+ \sqrt[n] {\log_B [B^{x^n}-x+ \sqrt[n]{ \log_B [B^{x^n}-x+...}}}]]][/tex]

Note that it is set up to take x^n first, then take B^(x^n) next (as if it were infinitely iterated so that it is algebraically sound). The "[" symbol doesn't show up to clearly under the radical. Anyways...

As with the other -x + ... functions, this one approaches the derivative of the inner function [itex]B^{x^n}[/itex]

[tex]\dfrac{f(x,B,a)}{f(x,B,a+1)}\to n\,{x}^{n-1}\,{y}^{{x}^{n}}\,log\left( y\right)[/tex]

For all of these functions, the exact approached value (so for x^n as the inner function, nx^(n-1) ) for f(...,a)/f(...,a+1) can be taken to the a

^{th}power (number of iterations) and multiplied by f(...,a) to create a constant. Haven't found a rational one yet. Not a lot of them are listed over at the http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html", although for the x^n-x+... one, with x and n equal to 2, you end up with pi^2/4.Note that a value more or less than the value that f(...,a)/f(...,a+1) approaches causes the calculated constant to diverge towards 0 or infinity as a increases [UNLESS you use the exact constant (such as the derivatives, or n for the first example, x ln (B) for the second, etc.)].

Last edited by a moderator: