Re: integral zeta

hagman wrote :

On 31 Aug., 12:35, David C. Ullrich
<ullr...@xxxxxxxxxxxxxxxx> wrote:
On Thu, 30 Aug 2007 17:07:54 EDT, tommy1729
<tommy1...@xxxxxxxxx>
wrote:



david wrote :

On Wed, 22 Aug 2007 09:10:37 EDT, tommy1729
<tommy1...@xxxxxxxxx>
wrote:

quasi wrote:

To tommy1729:

If you insist that primitives are uniquely
determined
simply by the
concept of "just set C=0", you are going to
run
into
some problems.

For example, let f(x )= e^x - x.

Let's integrate f(x) two different ways ....

method (1) [standard Calc 1 approach]:

int f(x) dx = e^x - x^2/2 + C

Setting C=0 yields e^x - x^2/2

method (2) [using power series]:

e^x = 1 + x + x^2/2 + x^3/6 + x^4/24 + ...

f(x) = e^x - x = 1 + x^2/2 + x^3/6 +
x^4/24 +
...

int f(x) dx = (x + x^3/6 + x^4/24 + x^5/120 +
...
... ) + C

Setting C=0 yields x + x^3/6 + x^4/24 +
x^5/120 +
0 + ...

which equals e^x - x^2/2 - 1.

So which one is the unique primitive?

quasi

intresting argument. :-)

first note : i have already defined "integral
zeta"
by a series.

although perhaps there might be discussion
about the
analytic continuation or riemann surface ....

but anyways "integral zeta" is already
defined...

First, it _had not_ been defined at the point
where
you started insulting people for not knowing
what you
meant. Second, that series does not converge
except
for Re(z) > 1, and it has many _different_
continuations
to larger regions - hence that series does _not_
specify where the zeroes in the plane are -
where the
zeroes in the plane are depends on what
particular
continuations we're talking about. Saying that
there
"might be discussion about" analytic
continuation
is progress, but it shows you still have no idea
what you're talking about - the question of
non-uniqueness of analytic continuation is
crucial
here.

in your argument, perhaps a good way to clarify
is
this

f(0) = INTEGRAL f(0)dx if possible by changing
C.

so f(x) = exp(x) - x

f(0) = 1 = integral f(0)

therefore the integral is method(1).

another reason for avoiding series is the
taylor
expansion of exp(x^2) at 0 ...

you know what happens there dont you ?

a bit off topic perhaps but can you give nice
examples of an integral which can take 3 or more
values ...

just for fun :-)

regards
tommy1729

************************

David C. Ullrich

ok , so you FINALLY accepted the series.

Huh? What does it mean to "accept" a series?

the continuation is arguable yes.

i admit that.

but there are zero's that are in common with any
possible continuation ; "critical" ones.

correct me if im wrong about that

( would surprise me ; the wrong , not the
correction )

( "critical" is not the same here as usual ; not
referring to a line but the zero' s that are in
common to all continuations )

i never claimed to be an expert on continuation
though ...

In fact you know nothing whatever about it, but
that
doesn't stop you from making statements about it.
I'll give you a hint, so you can realize exactly
how absurd you're being here:

Hint: What you're claiming is this: If f is a
function
defined in a certain region and g is defined in the
same region by g = f + 2 pi i then f and g have
certain zeroes in common.


Indeed.
And to allow tommy1729 to learn something:
zeta(s) has one pole, of residuum 1.

Loosely speaking, this allows us to look at "integral
zeta mod 2pi".
In other words, the research question is:

DEFINITION1 :
f:U->C is called a tommy-function if
- U is a region in C
- 0 in U, 1 not in U
- f holomorphic on U
- f(0)=0
- f'(s)=zeta(s) fr all s in U

DEFINITION 2:
If f:U->C is a tommy-function, a point z in C is
called
a tommy-point for f if z in U and f(z) is an integer
multiple of 2pi.

tommy's objective is to study tommy-points.

Trivial LEMMA: tommy-pointnesss does not depend on
the choice of f (and/or U).

Which allows

DEFINITION 3:
z in C\{1} is called a tommy-point if there exists
a tommy-function of which z is a tommy-point

So, now there is something defined.
Next question: What is interesting about
tommy-points?

hagman


hahaha hagman

tommy-points lol

your making stuff up i never said , just like david.

i assume its a parody :-)

if not , well then you made that up ( tommy points) ...

and yes zeta(1) has a pole since it is the harmonic series 1+1/2+1/3+...


you have outlined a clearer similar version of the topic , yet ullrich ignores it ...

just as he ignored the series i gave him ...

instead he invents stuff ala tommy-points :-)

regards
tommy1729
.