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