12/01/2010, 06:16 PM

very general calculus

*********************

from

a [r] b = b [r] a = exp^[r]( exp^[-r](a) + exp^[-r](b) )

http://math.eretrandre.org/tetrationforu...hp?tid=520

" the distributive property "

it follows that

a generalization of infinite sums and infinite products in calculus is

f(z) = [ Tommy r ] ( exp^[r](1) + a_i z^n )

where [ Tommy r ] means

[ Tommy r ] b_i = b_0 [r] b_1 [r] b_2 [r] b_3 [r] ...

for instance

f(z) = [ Tommy -1 ] ( 0 + a_i z^n )

is a power series.

f(z) = [ Tommy 0 ] ( 1 + a_i z^n )

is the infinite product of ( 1 + a_i z^n)

defined for abs z at most 1.

and in general

for r > 0

f(z) = [ Tommy r ] ( exp^[r](1) + a_i z^n )

defined for abs z at most exp^[r](1).

tommy1729

*********************

from

a [r] b = b [r] a = exp^[r]( exp^[-r](a) + exp^[-r](b) )

http://math.eretrandre.org/tetrationforu...hp?tid=520

" the distributive property "

it follows that

a generalization of infinite sums and infinite products in calculus is

f(z) = [ Tommy r ] ( exp^[r](1) + a_i z^n )

where [ Tommy r ] means

[ Tommy r ] b_i = b_0 [r] b_1 [r] b_2 [r] b_3 [r] ...

for instance

f(z) = [ Tommy -1 ] ( 0 + a_i z^n )

is a power series.

f(z) = [ Tommy 0 ] ( 1 + a_i z^n )

is the infinite product of ( 1 + a_i z^n)

defined for abs z at most 1.

and in general

for r > 0

f(z) = [ Tommy r ] ( exp^[r](1) + a_i z^n )

defined for abs z at most exp^[r](1).

tommy1729