II. Generalized Calculus

A. More Anti-Iterations

As we have seen, an anti-iteration is the inverse of a continuous iteration, expressable as an infinitesimal limit and is the continuous limit building block of a calculus. We have come across two of these, the derivative and the anti-product (AP):

f(x) = [ f(x+dx) - f(x) ] / dx

qf(x)1/dx = [ f(x+dx)/f(x) ]1/dx

(as dx-->0)

We have also shown that each has an identity function, defined by the restrictions:

indentity(x) 1/dx = c (constant)
q identity(x) ^1/dx = c (constant)

where it is found that the identity(x) for the derivative is = mx + b, and the identity(x) for the AP is = b m^x. In both cases, m is the constant which it's anti-iteration equals and b is an arbitrary constant having no bearing on the outcome of the anti-iteration.

The restrictions for being an identity function can be generalized.

Ati_x identity(x) = c (constant).

where Ati_x means some anti-iteration with respect to x.

There are obviously many other anti-iterations, including ones based on successive derivatives, such as the following ATI:

Ati( ) =

d2f(x)  ----- =  dx2

( f(x+d) - f(x) )/d =

f(x+2d) - 2f(x+d) + f(x)  d2

(as d-->0)

Unlike the other two, this is a relationship between three infinitely close samplings of f(x). Additional multi-sampled API's can be made, indepent of derivatives, but we will not deal with them in this section. Instead, we will limit ourselves to an Ati expressing the relationship between only two infinitely close samplings of f(x). In derivates and AP's, we have limited ourselves to the relationship between f(x) and f(x+dx):

f(x) = [ f(x+dx) - f(x) ] / dx

qf(x)1/dx = [ f(x+dx)/f(x) ]1/dx

but we will no longer. Ati's with relationships between other infintely close samplings such as f(x^q) and f(x) will also be acceptable, such as the following.

Ati f(x) = [ f(x^q) / f(x)^q ]^1/(1-q).

Under those restrictions, we can create a simple formula that will derive an Ati formula, given it's identity function. Outside of those restrictions, I do not have a formula.

Let identity(x) be the known identity function, having an arbitrary constant b and an ati of m.
Let identity-1(x) be the inverse function of identity(x)
Let c(x) be the relation between both sampled points x₁ and x₂, such that x₂ = c(x₁)
Let f(x) be an arbitrary function.
Let Ati_x[f(x)] be the anti-iteration of f(x) with respect to x and which corresponds to the given identity function.
Let [ ... ]_{res m} be the solution of the equation when solved for m.
THEN

Atix[f(x)] = lim (d-->0) [ identity-1(f(c(x))) = identity-1(f(x)) + d ]res m.

To put meaning to the above, let us take take the derivative Ati from its identity function. We know that indentity(x) = b + mx. Therefore identity^-1(x) = (x-b)/m. Now we choose to set c(x) = x+d. This is because we want the Ati to relate points at x-values of x₁ and x₂=c(x₁)=x₁+d. That is the simplest relation possible. We now substitute in the above formula:

Ati_x[f(x)] = lim_(d-->0) [(f(x+d)-b)/m = (f(x)-b)/m +d]_{res m}.
now we must solve the expression [(f(x+d)-b)/m = (f(x)-b)/m +d] for m.
[(f(x+d)-b)/m = (f(x)-b)/m +d]
[f(x+d)/m-b/m = f(x)/m-b/m +d]
[f(x+d)/m = f(x)/m +d]
[ m = (f(x+d)-f(x))/d ] now substitute...

Ati_x[f(x)] = lim_(d-->0) (f(x+d)-f(x))/d.

Similarly, if we had the identity function identity(x)=b m^x, an choose the same c(x)=x+d, then:

identity(x)=b m^x.
identity-1(x)=ln(x/b)/ln(m).
c(x)=x+d.
(substitute...)

Ati_x[f(x)] = lim_(d-->0) [ ln(f(x+d)/b)/ln(m) = ln(f(x)/b)/ln(m) +d ]_{res m}.
= lim [ ln(f(x+d)/f(x) = d ln(m) ]_{res m}.
= lim [ f(x+d)/f(x) = m^d ]_{res m}.
= lim [ m = (f(x+d)/f(x))^1/d ]_{res m}.
= lim_(d-->0) (f(x+d)/f(x))^1/d = qf(x)^1/dx.
This is the AP anti-iteration. Recal that,
q(b m^x)^1/dx = q(m^x)^1/dx = m.

The following is a sampling of derivable ATI's:

identity(x)	x+d	Ati	d/dx form
v+x	x+d	<undefined>
v+x	x*q	f(xq)-qf(x)  1-q	f(x)-x f '(x)
v+x	xq	[f(xq) = v+(f(x)-v)q]res v	x f '(x)/f(x)
vx	x+d	f(x+d)-f(x)  d	f '(x)
vx	x*q	<undefined>
vx	xq	(f(xq)/f(x)q)1/(1-q)	f x -x f '(x)/f(x)
vx	x+d	(f(x+d)/f(x))1/d	e  ln f(x)
vx	x*q	<undefined>
vx	xq	f(x)[ (ln(f(x))/ln(x^q) ]^(1/(1-q)]
xv	x+d	[f(x+d) = (f(x)1/v + d)v]res v
xv	x*q	ln f(xq) - ln f(x)  ln q

(as q-->1 and d-->0)
Some of the equivalent derivative forms have been computed when they could be.
 
This speaks nothing about more complicated Ati's not restricted to the above-given limitations. Such an Ati would not ordinarily be expressable in terms of a 1^st derivative expression, or neccessarily as a finite n^th derivative expression, I suspect.

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