06/07/2014, 11:38 AM
About method 2 from post 1.
There is a difference between f(x) f(x-id(1)) f(x - id(2)) ... f(x - id(q))
Where q satisfies q is a positive integer such that x - i(q) >= 0
AND the continuum product.
example :
2^17.5 - 1 equals the continuum product of 2^n from 0 to 16.5.
2^17.5 - 1 =/= 2^16.5 + 2^15.5 + 2^14.5 + ... 2^0.5.
If f(z) is real-analytic for Re(z) > -1 then the ratio between the continuum product (CP) and the product f(x) f(x-id(1)) f(x - id(2)) ... f(x - id(q)) remains bounded on the real line ... " usually".
Therefore method 2 is conjectured to give a result that is
O ( sexp(x)^(1+o(1)) ) * O(C^x) if the parameter k goes to +oo.
Method 1 however is exact , but it might have a natural boundary , in which case the functional equation sexp(z+1) = exp(sexp(z)) starts too loose meaning.
Hope that was insightfull.
regards
tommy1729
There is a difference between f(x) f(x-id(1)) f(x - id(2)) ... f(x - id(q))
Where q satisfies q is a positive integer such that x - i(q) >= 0
AND the continuum product.
example :
2^17.5 - 1 equals the continuum product of 2^n from 0 to 16.5.
2^17.5 - 1 =/= 2^16.5 + 2^15.5 + 2^14.5 + ... 2^0.5.
If f(z) is real-analytic for Re(z) > -1 then the ratio between the continuum product (CP) and the product f(x) f(x-id(1)) f(x - id(2)) ... f(x - id(q)) remains bounded on the real line ... " usually".
Therefore method 2 is conjectured to give a result that is
O ( sexp(x)^(1+o(1)) ) * O(C^x) if the parameter k goes to +oo.
Method 1 however is exact , but it might have a natural boundary , in which case the functional equation sexp(z+1) = exp(sexp(z)) starts too loose meaning.
Hope that was insightfull.
regards
tommy1729