07/21/2021, 09:52 PM
For the external proof with error1(s) I have the idea to mainly use the absolute value.
To manage the imaginary parts , the idea is that we take the correct log branches in a consistant way and then we always take the same branch for the same neighbourhood ... thus a branch jump of at most 1 down or 1 up from the infinitesimal neigbourhood.
This leads to the partial error term o(L) ( little -o notation for absolute value bound )
where L satisfies L - 2pi = ln(L)
< follows from L = 2 pi + ln(2 pi + ln(2 pi + ... ) >
L = 8.4129585032844...
Notice that ln(1) is never 0 , it must be another branch.
ln(0) never occurs.
ln(z) for abs(z) < 1 is just - ln(1/z) and ln(ln(z)) = ln(- ln(1/z) ) = ln(ln(1/z)) + pi i
Just to show that the imaginary parts or error parts can also come from the (positive ) reals or the absolute values.
***
This gives the first estimate of tetration for suitable s ( Re(s) > 4 , Re(s) >> Im(s) )
lim n to +oo
abs(tet(s + s_0)) = abs ( ln^[n] ( Tom(s+n) ) ) < (o(A) + o(1/A)) * ( abs(Tom(s)) + abs(1/Tom(s)) + o(B) + o(L) + 1 ) + o(B) + o(L).
where o(L) is as above and
A = abs( t(s)*t(s+1)*t(s+2)*...*t(s+n) )
B = abs ( ln(t(s)) + ln(t(s+1)) +ln(t(s+2)) + ... + ln(t(s+n)) )
s_0 is the usual suitable constant.
As James noted A and B converge (fast) for n going to +oo so that is good.
this is a sketch of the proof.
notice v1 = exp(u) and v2 = exp( u * t(s)) have the same branch as inverse :
v1 = ln(u) , v2 = ln(u)^(1/t(s))
where the ln's have the same branches.
since t(s) is close to 1 and 1/t(s) valid , it follows v1 is relatively close to v2.
ln(v1) = ln ( ln(u) ) , ln(v2) = ln( ln(u) )/t(s)
ln(ln(v1)) = ln^[3](u) , ln(ln(v2)) = ln^[3](u) - ln(t(s)).
etc
ofcourse abs( a - b ) =< abs(a) + abs(b).
This should help you understand the proof , the branches and the absolute value error terms.
regards
tommy1729
To manage the imaginary parts , the idea is that we take the correct log branches in a consistant way and then we always take the same branch for the same neighbourhood ... thus a branch jump of at most 1 down or 1 up from the infinitesimal neigbourhood.
This leads to the partial error term o(L) ( little -o notation for absolute value bound )
where L satisfies L - 2pi = ln(L)
< follows from L = 2 pi + ln(2 pi + ln(2 pi + ... ) >
L = 8.4129585032844...
Notice that ln(1) is never 0 , it must be another branch.
ln(0) never occurs.
ln(z) for abs(z) < 1 is just - ln(1/z) and ln(ln(z)) = ln(- ln(1/z) ) = ln(ln(1/z)) + pi i
Just to show that the imaginary parts or error parts can also come from the (positive ) reals or the absolute values.
***
This gives the first estimate of tetration for suitable s ( Re(s) > 4 , Re(s) >> Im(s) )
lim n to +oo
abs(tet(s + s_0)) = abs ( ln^[n] ( Tom(s+n) ) ) < (o(A) + o(1/A)) * ( abs(Tom(s)) + abs(1/Tom(s)) + o(B) + o(L) + 1 ) + o(B) + o(L).
where o(L) is as above and
A = abs( t(s)*t(s+1)*t(s+2)*...*t(s+n) )
B = abs ( ln(t(s)) + ln(t(s+1)) +ln(t(s+2)) + ... + ln(t(s+n)) )
s_0 is the usual suitable constant.
As James noted A and B converge (fast) for n going to +oo so that is good.
this is a sketch of the proof.
notice v1 = exp(u) and v2 = exp( u * t(s)) have the same branch as inverse :
v1 = ln(u) , v2 = ln(u)^(1/t(s))
where the ln's have the same branches.
since t(s) is close to 1 and 1/t(s) valid , it follows v1 is relatively close to v2.
ln(v1) = ln ( ln(u) ) , ln(v2) = ln( ln(u) )/t(s)
ln(ln(v1)) = ln^[3](u) , ln(ln(v2)) = ln^[3](u) - ln(t(s)).
etc
ofcourse abs( a - b ) =< abs(a) + abs(b).
This should help you understand the proof , the branches and the absolute value error terms.
regards
tommy1729