# Question about derivative approximation through nested functions

1. Oct 26, 2011

### Matt Benesi

Is there a name for these types of derivative approximation?

I've encountered several different derivative approximation formulas, all of which work.

For this series of formulas, the variable "a" denotes the nesting depth (a refers only to the nested part of the equation, not the initial $x -$):

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

This next formulas approach the derivative of the inner functions* when taking $\dfrac{f(...,a)}{f(...,a+1)}$ or sometimes ***$\dfrac{1- \dfrac {f\left(...,a\right)}{f\left(...,a+1\right)}}{1- \dfrac{f\left(...,a+1\right)}{f\left(...,a+2\right)}}$ for higher a. Note that whatever pattern the equation follows, it will continue to follow if you delve deeper into it, but please don't let this aside distract you for now (you can find the derivative in it multiple times by subtracting the derivative from the approximate derivative, dividing this value by a deeper a derivative).

*For the following equation, the inner function is x^n:
$$f(x,n,a)=x-\sqrt[n]{x^{n}-x+\sqrt[n]{x^{n}-x+\sqrt[n]{x^{n}-x+\sqrt[n]{x^{n}-x+...}}}}$$
so $\dfrac{f(x,n,a)}{f(x,n,a+1)}\to nx^{n-1}$ OR $\dfrac{1-\dfrac{f(x,n,a)}{f(x,n,a+1)}}{1-\dfrac{f(x,n,a+1)}{f(x,n,a+2)}}\to nx^{n-1}$

For this one, the inner function is Bx:
$$f(x,B,a)=x-\log_B [B^x-x+\log_B [B^x-x+\log_B [B^x-x+\log_B [...]]]]$$
so *** $$\dfrac{f(x,B,a)}{f(x,B,a+1)}\to B^{x}\ln B$$
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:
$$f(x,n,B,a)=x-\sqrt[n]{\log_B [ B^{x^n}-x+ \sqrt[n] {\log_B [ B^{x^n}-x+ \sqrt[n]{ \log_B [ B^{x^n}-x+... }}} ]]]$$

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 $B^{x^n}$
***$$\dfrac{f(x,n,B,a)}{f(x,n,B,a+1)}\to n\,{x}^{n-1}\,{B}^{{x}^{n}}\,log\left( B\right)$$

***Note that for all of these functions, you can end up with such close values of f(...,a) and f(...,a+1) that you sometimes need to subtract out 1, then calculate the derivative with the new values. You can always tell when this happens, because if you get 1.000....00XXXXX... (XXX's are beginning of different digits) when you divide out f(...,a)/f(...,a+1), it's time to subtract 1 from it, and subtract 1 from the next set f(...,a+1)/f(...,a+2), and then divide the new values out to get your approximate derivative.

2. Oct 28, 2011

### Stephen Tashi

Perhaps there will be more comments if we phrase this differently. The word "nested" suggests iteration of a function, but what you are doing is recursively defining a sequence of functions.

The general pattern appears to be:

Let $f(x)$ be a given function with inverse function $f^{-1}(x)$

Define $f_n(x)$ recursively by:

$f_1(x) = f^{-1}( f(x) - x)$
$f_n(x) = f^{-1}( f(x) - x - f_{n-1}(x))$

A conjecture is that at each point x, the derivative of $f(x)$ is

$f'(x) = \lim_{n \rightarrow \infty} \frac { A + B(x - f_{n+1}(x)) }{A + B(x - f_{n}(x)) }$

where A and B are some constants that work for all x. In your examples A is 0 or 1 and B is 1 or -1.