I. Generalized Calculus

B. Iterations from Anti-Iterations

Deriving the Continued Product from the AP

In Unit I, Section III, the anti-product iteration was computed from the continuous product iteration. In essence, half of the Fundamental Therom of Product Calculus was derived:

q [  f(x) ^dx ] ^1/dx = f(x).

Now, we will do the reverse, compute the iteration from the anti-product limit, and show that:

[qf(x)^1/dx] ^dx = f(x) c. (constant c).

recall qf(x)1/d = [f(x+d)/f(x)]^1/d = [f(x)/f(x-d)]^1/d (as d-->0)

when we multiply qf(x)^1/d qf(x-d)^1/d, the f(x-d)^1/d terms cancel:

f(x)1/d  f(x-d)1/d

f(x-d)1/d  f(x-2d)1/d

= f(x)1/d  f(x-2d))1/d

with qf(x)^1/d qf(x-d)^1/d qf(x-2d)^1/d ... qf(f ^-1(1))^1/d, many more intermdediates cancel:

f(x)1/dx  f(x-d)1/d

f(x-d)1/d  f(x-2d)1/d

(x-d)1/d  f(x-2d)1/d

...

(f -1(1)+d)1/d  f(f-1(1))1/d

= f(x)1/d  f(f-1(1))1/d

The expansion on the left is equivalent to the continuous product

   x
= qf(x)^1/d
   f ^-1(1)

The specific f ^-1(1) limit was chosen because at this point, f(f^-1(1)) = 1, and the right, expansion result above simplifies to:

[f(x)/f(f ^-1(1))]^1/d = f(x)^1/d. Therefore, appending the right side to the left yields:

x
qf(x)^1/d = f(x)^1/d.
f ^-1(1)

x
[qf(x)^1/d] ^d = f(x)^1/d. (raise both sides to the d^th power to get the origional equation.)
f ^-1(1)

Since (a..x) g(x)^dx =  (a..s) g(x)^dx (s..x) g(x)^dx = c  (s..x) g(x)^dx = c g(x)^dx  for some constant c,

x  [qf(x)1/d] d =  f -1(1)

c [qf(x)1/dx]dx

The 2nd Derivative

Recall from the last section,
d²f(x)/dx² = [ f(x+2dx) - 2f(x+dx) + f(x) ] / d².

A double integral could be performed to get the inverse, but, instead, a method similar to the above will be taken:

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