02/22/2023, 11:44 PM
Ok this idea does have issues and I guess I should adress them.
If you take the product formula for the derivative of tet(x) a bit too far , although that has logical justifications (!) , we run into this
tet(-1) = 0
tet(0) = 1
tet ' (0) = tet(0) * tet ' (-1).
But tet(0) = 1 , so
tet ' (0) = 1 * tet ' (-1) = tet ' (-1)
Since tet ' (0) = tet ' (-1) this means the function cannot be convex in [-1,0].
SO sufficiently-differentiable solutions must have an inflection point in [-1,0] !!
This means our unique zero of the second derivative is not so unique afterall.
Or maybe we should set
sexp ' ' ( - 1/2) = 0
to make the Riemann fans happy
I have to think about these issues.
Although these issues are very old ofcourse.
My apologies if I made things sound " completed " , they are not.
However the remark about the theta function theta(x) has analogues here !
So theta(x) is not so free to choose as you might have believed !
regards
tommy1729
If you take the product formula for the derivative of tet(x) a bit too far , although that has logical justifications (!) , we run into this
tet(-1) = 0
tet(0) = 1
tet ' (0) = tet(0) * tet ' (-1).
But tet(0) = 1 , so
tet ' (0) = 1 * tet ' (-1) = tet ' (-1)
Since tet ' (0) = tet ' (-1) this means the function cannot be convex in [-1,0].
SO sufficiently-differentiable solutions must have an inflection point in [-1,0] !!
This means our unique zero of the second derivative is not so unique afterall.
Or maybe we should set
sexp ' ' ( - 1/2) = 0
to make the Riemann fans happy
I have to think about these issues.
Although these issues are very old ofcourse.
My apologies if I made things sound " completed " , they are not.
However the remark about the theta function theta(x) has analogues here !
So theta(x) is not so free to choose as you might have believed !
regards
tommy1729
