## knowing the zeta function

Expand Messages
• I wrote to Dr. Math on 9/6/06... today. He asked... What is your question? I wrote: I see the many definitions that math books are filled with and ponder- ed
Message 1 of 2 , Sep 6, 2006
• 0 Attachment
I wrote to Dr. Math on 9/6/06... today.

I wrote:

I see the many definitions that math books are filled with and ponder-
ed one myself. I call it the ideal zero asymmetrical range of a
complex function. It happens only when the following condition is
met. The function f(x) has an IZAR when f(a+bi) = a*g(bi) where the
function g(x) strictly cannot be written as g(a+bi). I believe that
if this condition has been met that the range of the function f(x)
must have an ideal zero asymmetrical range about the value 'a' on its
complex graph.

He said... Tell me what you find most difficult or confusing about it.

I wrote:

It is my belief that Riemann saw this condition but wasn't able to
define it, because the zeta function had gained so much popularity
with its connection to prime numbers. I want to believe that my de-
finition could be as fundamental for complex numbers as are the real
number definitions for monoids, groups, rings, etc. in mathematics.

He said... show me your work...

I replied...

After evaluating the zeta function for z(1/2+it), we get...

2 4
------- + ------------- + ...
(1+2it) (1-4t^2 +4it)

and 1/2 * g(it) is the only way to factor out 1/2 from f(1/2+it) so
by definition the ideal zero asymmetrical range must exist. It's my
belief that several manipulations of the zeta function have been made
in an effort to solve the problem and that definitions similar to the
zero domain for real functions, etc. haven't been explored in the
sense of the range of complex functions. It's been overlooked.

I'm waiting for Dr. Rob's stimulating conversation... I won't be
writing for a while... math is not my idea of fun... just under-
standing. Bill Semper paratus.
• ... ponder- ... the ... ...omit the restriction. The function should have been f(1/a + bi) = a * g(1+abi) ... it. ... de- ... real ... my ... made ... the ...
Message 2 of 2 , Sep 6, 2006
• 0 Attachment
wrote:
>
> I wrote to Dr. Math on 9/6/06... today.
>
>
> I wrote:
>
> I see the many definitions that math books are filled with and
ponder-
> ed one myself. I call it the ideal zero asymmetrical range of a
> complex function. It happens only when the following condition is
> met. The function f(x) has an IZAR when f(a+bi) = a*g(bi) where
the
> function g(x) strictly cannot be written as g(a+bi).

...omit the restriction.

The function should have been f(1/a + bi) = a * g(1+abi)

> I believe that
> if this condition has been met that the range of the function f(x)
> must have an ideal zero asymmetrical range about the value 'a' on

...and the value of asymmetrical interest should have been '1/a'

> its complex graph.
>
> He said... Tell me what you find most difficult or confusing about
it.
>
> I wrote:
>
> It is my belief that Riemann saw this condition but wasn't able to
> define it, because the zeta function had gained so much popularity
> with its connection to prime numbers. I want to believe that my
de-
> finition could be as fundamental for complex numbers as are the
real
> number definitions for monoids, groups, rings, etc. in mathematics.
>
> He said... show me your work...
>
> I replied...
>
> After evaluating the zeta function for z(1/2+it), we get...
>
> 2))))))))))))))4
> ------- + ------------- + ...
> (1+2it) (1 + 2it)^2

... I changed this series to agree with the above changes ...

>
> and 2 * g(1+2it) is the only way to factor out 2 from f(1/2+it) so

... this was also corrected to yield more clarity ...

> by definition the ideal zero asymmetrical range must exist. It's
my
> belief that several manipulations of the zeta function have been
> in an effort to solve the problem and that definitions similar to
the
> zero domain for real functions, etc. haven't been explored in the
> sense of the range of complex functions. It's been overlooked.
>
> I'm waiting for Dr. Rob's stimulating conversation... I won't be
> writing for a while... math is not my idea of fun... just under-
> standing. Bill Semper paratus.
>

I haven't heard from Dr. Rob and think that he will either find the
corrections to my original letter or skip the question entirely.

Bill finally I clarified it for those in the group and myself
Your message has been successfully submitted and would be delivered to recipients shortly.