I. Product Calculus

D. The Anti-Product and Quotientials

Deriving the Anti-Product

The anti-product is the third product calculus building block to be examined. It is a continuous limit. As defined in the preface, the continuous limit is the inverse of the continuous iteration. Our goal is to find the opperation opp( ) which satifies the condition:

opp(	x f(t) dt ) = f(x)

The CP is indefinite, so we may rewrite it:
x
f(t) ^dt
s
for an unknown lower-bound s. Expand the CP by it's iterative definition:
x
f(t) ^dt = f(x)^dt f(x-dt)^dt f(x-2dt)^dt ... f(s) = g(x)
^s
and define this as a function g(x)

A special result occurs when we try dividing g(x)/g(x-dt):

g(x)/g(x-dt) =

f(x)dt f(x-dt)dt f(x-2dt)dt ... f(s)           f(x-dt)dt f(x-2dt)dt ... f(s)

All the terms except f(x)^dt cancel out, leaving:
g(x)/g(x-dt) = f(x)^dt

rearranged as:

f(x) = [ g(x)/g(x-dt) ]^1/dt

In simplied terms, this is denoted:

qg(x) 1/dt = [ g(x)/g(x-dt) ]1/dt.

This is our inverse function of the continuous product, named the anti-product. The quantity g(x)/g(x-d) may be represented as the quotiential qg which means the quotient of infinitely close terms. qg is an infinitesimal, different than the differential dt, in that is is infinitely close to 1, as opposed to infinitely close to 0.
 

Geometric Interpretation of the Anti-Product

We define a new infinitesimal, qy, which instead of being infinitely close to zero as in dx, is infinitly close to one. Note, qy is not the length of the vertical increment of the curve over a dx--that is dy--it is instead the quotient of both endpoints of the increment. These concepts are clarified by the definition:

dx = x₂ - x₁, and qy = y₂ / y₁.

qy is termed the quotiential.

Now we can define the product calculus' counterpart of the derivative:

Definition of the Anti-Product:
The anti-product of f(x) at a point x=c is the limit of the quotient in the y-values of two infinitismaly close points on the curve, centered around the point x=c, raised to the reciprical of the difference in x-values of the two points.

The above can be expressed:

qf(x) ^1/dx = lim [ f(x2)/f(x1) ] ^1/(x2-x1).   (or more simply...)
 

qf(x) 1/dx = lim (d-->0) [ f(x+d)/f(x) ] 1/d.

An Example

Consider the function f(x) = x². We will compute its anti-product at x=2 by inputing successively smaller values of d in the above defintion.
 
 

dx = x2 - x1	qy = y2 / y1	qy 1/dx
1	= 32/22  = 2.25	2.25
0.1	= 2.12/22  = 1.1025	2.65330
0.01	1.010025	2.71152
0.001	1.00100025	2.71760
1 /	---	2.71828...  = e

 
The limit converges to e. We can write:

q(x²) ^1/dx | _(x=2) = e.

We can derive algebraically with our elementary theory of limits that
 

q(xn) 1/dx = en/x.

 
Proof:
q(xⁿ) ^1/dx = lim _(d-->0) [ (x+d)ⁿ / xⁿ ] ^1/d = lim _(d-->0) [ ((x+d)/x)ⁿ) ] ^1/d = lim _(d-->0) (1 + d/x)^n/d.
(set u=d/x)
q(xⁿ) ^1/dx = lim _(ux-->0) (1 + u)^n/(ux) = lim _(u-->0) (1+u)^n/(ux) = [ lim _(u-->0) (1+u)^1/u ]^n/x
(by the definition of e, the limit is substituted...)
q(xⁿ) ^1/dx = e^n/x.    Q.E.D.

Table of Anti-Products

We can derive the following AP's (anti-products) and use them for our convienence:
 
 

f(x)	qf(x) 1/dx
xn	en/x
ax	a
ln(x)	e 1/(x ln x)
sin(x)	e cot(x)
cos(x)	e -tan(x)
tan(x)	e 1/(cos x sin x)

Before we get too carried away tabulating AP's for every possible function, a short-cut is available which permits the derivation of AP's from their derivatives:
 

qf(x) 1/dx = e  ln f(x) = e[1/f(x)]  f(x).

Proof:
ln qf(x) ^1/dx = ln lim _(d-->0) [f(x+d)/f(x)] ^1/d = lim ln [f(x+d)/f(x)] ^1/d = lim [ ln f(x+d) - ln f(x) ] / d  =  ln f(x). (from the definition of the derivative)
(take the exp of both sides)
qf(x) ^1/dx = e  ^{ln f(x)} = e^{[1/f(x)]  f(x)}.    Q.E.D.
(We made some assumptions with respect to f(x), such as ln f(x) exists.)

The following identities for AP's can be derived by a number of methods:
 

Formula	Description
q(c f(x))1/dx = qf(x) 1/dx	Constant Rule
q(f(x)+g(x))1/dx = (q(ef)1/dx q(eg)1/dx)1/(f+g)  = (e df/dx e dg/dx)1/(f+g)	Addition Rule
q(f(x)g(x)) 1/dx = qf(x) 1/dx qg(x) 1/dx	Product Rule
q(f(x)g(x))1/dx = f(x) g(x) (qf 1/dx)g(x)	Power Rule
qg(f(x)) 1/dx = (qg 1/df) df/dx	Chain Rule

Tangent Identity Curve

(product curve tangents on a sine wave)
 
The identity function of a calculus is the function whose continuous limit is a constant. The identity function of the product calculus is the one which satisfies the constraint:
qf(x) ^1/dx = b (a constant)

The function y = a b^x satisfies this condition and is the identity function. The identity function of elementary calculus is a line, y = a + bx, since  (a + bx) = b. Note the similarity: the constant a has no bearing on value of the anti-iteration, since  (a + bx) =  (bx) and q(ab^x) ^1/dx = q(b^x) ^1/dx. The constant b equals the value of the anti-iteration:  (a + bx) = b and q(ab^x) ^1/dx = b. We can often identify a calculus by its identity curve; elementary calculus is the a + bx calculus or linear calculus, and the product calculus is the y = a b^x calculus.

Since we have the form of the identity curve, we can fit a tangent identity curve on a function at given points, as shown in the above graph.

Graph of Tangent Product Curve on f(x) at x=c:
g(x) = b^(x-c) f(c). where p=the qf(x)^1/dx at the point x=c.

The tangent curve may seem trivial--fitting a curve with the same AP on a function--but later on we will fit curves with similar successively higher AP's--2nd, 2rd, 4th, etc. AP--until the fitted curve equals the origional curve. That will essentially lead us to the product calculus equivalent of Taylor's Theorem, and from there will lead us to a generalized Taylor's Theorem for all calculus's.

An Example

Consider the function y = x! It can be represented in product notation as:
x
k = x!
k=1

Euler's Gamma Function is the continuous equivalent of the factorial.

(x+1) = x!

Now we will derive a continuous approximation function for these two functions.
Looking at the discrete product, we can see that with each increase in x, the height of the curve is multiplied by its new x-value. Therefore, we can estimate q(x!)^1/dx at x to be approximately x. To be more exact, we will say q(x!)^1/dx at x is approximately c+1/2. Knowing the AP of our approximation function, we can get the origional function by taking the AP's CP (Fundamental Theorem for Product Calculus). Note the similar appearance of the CP and the Discrete Product.

x
(t + 1/2) ^dt = exp(  ln(t + 1/2) dt )
1
(CP --> integral conversion)

= exp( [ x ln(x) - x ]

x+1/2  x+1/2)

)

= exp( [ (x+1/2) ln(x+1/2) - (x+1/2) ] - [ (3/2) ln(3/2) - (3/2) ] )
= (2/9)(3)(2x+1) e^1-x (x + 1/2)^x

Our approximation formula is:

x! = (x+1) = (2/9)(3)(2x+1) e1-x (x + 1/2)x

On testing the approximation formula against the true formula:
 

x	x! = (x+1)	Approx. Formula
1/2	0.886	0.897
1	1	1
2	2	1.979
3	6	5.909
4	24	23.574
5	120	117.674
150	5.713x10 262	5.562x10 262

The approximation formula is similar to Stirling's Formula.
 
 

Next, we advance to connect the Discrete Product, Continuous Product and Anti-Product into one cohesive theorem.

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