- 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 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. - --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...>

wrote:>

ponder-

> 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

> ed one myself. I call it the ideal zero asymmetrical range of a

the

> 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

> 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

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

> 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.

it.

>

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

>

de-

> 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

> finition could be as fundamental for complex numbers as are the

real

> number definitions for monoids, groups, rings, etc. in mathematics.

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

>

> 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

>

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

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

I haven't heard from Dr. Rob and think that he will either find 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.

>

corrections to my original letter or skip the question entirely.

Bill finally I clarified it for those in the group and myself