I. Product Calculus

E. Product Expansion Theorem

In this section, we will derive a taylor-like expansion theorem for discrete products, when the n^th anti-product of a function is known.The method used here will be used in the next unit to solve expansion theorems for additional calculuses.

Product Expansion Theorem:

f(x) =

( q(k) f(a) 1/dx )(1/k!)(x-a)^k  k=0

where q^(k) f(a) ^1/dx is the k^th AP of the function f(x) at x=a.

^ is the computer notation for powers (i.e. a^b = a^b). It is used here because HTML and ASCII are not very flexible in transliterating mathematical formulas.

The Derivation Method:

Assumption: If two functions have the same 0^th, 1^st, 2^nd, .., infinitith AP, THEN the functions are equivalent.

If we start with a constant a and perform successive indefinite iterations (indefinite CP's) on it, we get:

a = a
a ^dx = a^x b
a ^dxdx = a ^{(1/2 x^2)} b^x c
a ^{d^3 x} = a ^{(1/6 x^3)} b^{(1/2 x^2)} c^x c
a ^{d^4 x} = a ^{(1/24 x^4)} b^{(1/6 x^3)} c^{(1/2 x^2)} d^x e
...

The pattern becomes apparent. Note, a,b,c,d,e, ... are unknown constants. Not only that, but they are AP's of the expansion when x=0. Consider the 5^th equation:

a ^{d^4 x} = a ^{(1/24 x^4)} b^{(1/6 x^3)} c^{(1/2 x^2)} d^x e

If we set a, b, c, d and e to an arbitrary constants and set x=0, the result is:
a ^{(1/24 x^4)} b^{(1/6 x^3)} c^{(1/2 x^2)} d^x e ) = (1)(1)(1)(1)e = e.
Therefore, the 0^th successive AP is e.

Now do one AP on the expression and set x=0:
q( a ^{(1/24 x^4)} b^{(1/6 x^3)} c^{(1/2 x^2)} d^x e )
= q( a ^{(1/24 x^4)} )^1/dx q( b^{(1/6 x^3)} )^1/dx q( c^{(1/2 x^2)} )^1/dx q( d^x )^1/dx q(e)^1/dx
(product rule)
= a ^{(1/6 x^3)} b^{(1/2 x^2)} c^x d
= (1) (1) (1) d = d
The 1^st successive AP equals d.

Now do two AP's on the expression and set x=0:
q( a ^{(1/6 x^3)} b^{(1/2 x^2)} c^x d )
= q( a ^{(1/6 x^3)} )^1/dx q( b^{(1/2 x^2)} )^1/dx q( c^x )^1/dx q( d )^1/dx
= a ^{(1/2 x^2)} b^x c
= (1) (1) c.
The 2^st successive AP equals c.

Similarly, we find the 3^rd AP=d and the 4^th AP=e.
The fact that every term except the final constant cancels out when we set x=0 is crucial to the method.

We now have the ability to define a function with set successive AP's.

f(x) =

( kth AP )(1/k!)x^k  k=0

This is the corresponding formula to the Maclauren Series. We can make this a Taylor series by replacing x with (x-a) and placing the k^th AP at x=a--instead of at x=0--into the equation above. This results in the Product Expansion Therorem which we sought to derive.

We can replace the quotiential in the Product Expansion Theorem with its derivative equivalent (i.e. q^(k) f(a) ^1/dx with e  ^ln(f(x)) ):

f(x)=

e  (1/k!)(x-a)^k  ln(f(x))  k=0

It happens that this reverts into the basic Taylor's Theorem for Derivatives. since:

 
e  ^{(1/k!)(x-a)^k  ln(f(x))}
k=0

=  exp(

(1/k!)(x-a)k  ln(f(x)) ) k=0

which is simply the Taylor expansion of ln(f(x)).

This method could have been taken to prove the Product Expansion Theorem, but the longer method was used to illustrate the concepts we will use in the next unit to derive similar expansions for other calculuses.

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