Mixing
Imaginary component of Dirichlet Eta Function's root with real component equal to 1/2
Clash Royale CLAN TAG#URR8PPP
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Let $$spacespacespace eta(z) = sum_a=1^infty frac1a^z cdot (-1)^a-1 $$
Now let $space$$z = sigma + it $ ,
$$eta(sigma + it) = sum_a=1^infty frac1a^sigma + it cdot (-1)^a-1 $$
May we split $spacespace$$cfrac1a^sigma + it$ $spacespace$ into its real and complex components:
$$frac1a^sigma + it = frac1a^sigma cdot left(frac1a^itright) = frac1a^sigma cdot left(frac1(e^lna)^itright) = frac1a^sigma cdot left(frac1e^(icdot tlna)right) = frac1a^sigma cdot e^-(icdot tlna)$$
$ \ $
$ \ $
$$ qquad qquad quad = frac1a^sigma cdot bigg( cos(-tlna) + icdot sin(-tlna) bigg) = frac1a^sigma cdot bigg( cos(tlna) - icdot sin(tlna) bigg) $$
Therefore,
$$spacespacespaceRe big(eta(sigma+it) big) = sum_a=1^infty frac1a^sigmacdot cos(tlna) cdot (-1)^a-1$$
$\$
$$Im big(eta(sigma+it) big) = sum_a=1^infty frac1a^sigmacdot sin(tlna) cdot (-1)^a $$
$ \ $
$ \ $
$ \ $
Where $Re$ and $Im$ respectively represent the real and complex components of its argument.
$ \ $
$ \ $
For $z$ to be a root of $eta$ we thus require that $Re big(eta(z)big) = 0 = Im big(eta(z)big) $.
Presumably there will be no such $z$ , with $Re(z) = sigmaneq frac12 $ satisfying the condition above (for sure if Riemann Hypothesis turns to be true).
$ \ $
So now I will focus on the numbers $z$, such that $Re(z) = frac12$.
$ \ $
Given that, the unknown of the problem will be $space$ $t = Im(z)$ $space$ for which $space$ $ eta(frac12 + it) = 0$.
Let me now reformulate what we're working on:
$ \ $
$$ eta(tfrac12 + it) = sum_a=1^infty frac1a^frac12 + it cdot (-1)^a-1 $$
Which leads us to:
$ \ $
$$ spacespacespace Re big(eta(tfrac12+it) big) = sum_a=1^infty frac1a^frac12cdot cos(tlna) cdot (-1)^a-1$$
$ \ $
and
$$ Im big(eta(tfrac12+it) big) = sum_a=1^infty frac1a^frac12cdot sin(tlna) cdot (-1)^a$$
$ \ $
$ \ $
Now letting $space$ $a^frac12 = sqrta $ $space$, and writing it by extend:
$ \ $
$$ Re big(eta(tfrac12+it) big) = frac1sqrt1 cdot cos(tln1) - frac1sqrt2 cdot cos(tln2) + frac1sqrt3 cdot cos(tln3) - dots $$
$ \ $
$$ Im big(eta(tfrac12+it) big) = -frac1sqrt1 cdot sin(tln1) + frac1sqrt2 cdot sin(tln2) + frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
Onward,
$ \ $
$$ space space Re big(eta(tfrac12+it) big) space = space 1 - frac1sqrt2 cdot cos(tln2) + frac1sqrt3 cdot cos(tln3) - dots $$
$ \ $
$$ space space space Im big(eta(tfrac12+it) big) space = space 0 space + frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
$ \ $
$ \ $
For $space$ $z = frac12 + it$ $space$ to be a root of $eta$, we ought to require that:
$ \ $
$$ space space 0 space = space 1 - frac1sqrt2 cdot cos(tln2) + frac1sqrt3 cdot cos(tln3) - dots $$
$ \ $
$$ space space 0 space = space 0 space + frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
$ \ $
$ \ $
$$therefore$$
$ \ $
$$ frac1sqrt2 cdot cos(tln2) - frac1sqrt3 cdot cos(tln3) + dots = 1 $$
$ \ $
$$ frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots = 0 $$
$ \ $
$ \ $
$ \ $
This system of equations doesn't look too alien, at least.
$ \ $
Let's dive into the first equation, taken it's left side to be a function $f$ of $t$
$ \ $
$$ f = frac1sqrt2 cdot cos(tln2) - frac1sqrt3 cdot cos(tln3) + dots $$
$ \ $
We will be then interested in finding the numbers $t$ for which $space$ $f = 1$ $space$ .
$ \ $
Using the expansion
$$cos(x) = 1 - fracx^22 + fracx^424 - space space cdots space space (-1)^j cdot fracx^2j(2j)! space space cdots $$
$ \ $
$ \ $
We can rewrite $f$ as
$ \ $
$$ f = frac1sqrt2 cdot bigg( 1 - frac(tln2)^22 + frac(tln2)^424 - cdots bigg) - frac1sqrt3 cdot bigg( 1 - frac(tln3)^22 + frac(tln3)^424 - cdots bigg) + cdots $$
$ \ $
which we can reassemble to be written as a infinite polynomial in $t$:
$ \ $
$$ f = bigg( frac1sqrt2 - frac1sqrt3 + cdots bigg) - fract^22 bigg( frac(ln2)^2sqrt2 - frac(ln3)^2sqrt3 + cdots bigg) + fract^424 bigg( frac(ln2)^4sqrt2 - frac(ln3)^4sqrt3 + cdots bigg)-cdots $$
$ \ $
Now we've reached a remarkable point.
$ \ $
The above polynomial's coefficients are no more than just constant numbers. And those numbers are pretty familiar.
$ \ $
The independent coefficient,
$$bigg( frac1sqrt2 - frac1sqrt3 + cdots bigg)$$
is equal to
$$ 1 - Re big(eta(tfrac12) big)$$
$ \ $
$ \ $
$ \ $
The quadratic coefficient,
$$ -frac12bigg( frac(ln2)^2sqrt2 - frac(ln3)^2sqrt3 + cdots bigg) $$
is equal to
$$ -frac12 bigg( fracd^2 Re big(eta(tfrac12+it) big)dt^2 bigg) _t=0 $$
$ \ $
$ \ $
$ \ $
And the quartic coefficient
$$ frac124bigg( frac(ln2)^4sqrt2 - frac(ln3)^4sqrt3 + cdots bigg)$$
is equal to
$$ frac124 bigg( - fracd^4 Re big(eta(tfrac12+it) big)dt^4 bigg) _t=0 $$
$ \ $
And so on with the subsequent coefficients.
$ \ $
We can thus write $f$ as:
$ \ $
$$ f = 1 - Re big(eta(tfrac12) big) + fract^22 bigg( fracd^2 Re big(eta(tfrac12+it) big)dt^2 bigg) _t=0 - fract^424 bigg(fracd^4 Re big(eta(tfrac12+it) big)dt^4 bigg) _t=0 + cdots $$
$ \ $
$ \ $
If we want $f = 1$, then $t$ must be the roots of $space$ $beth$ $space$, where
$ \ $
$$ beth = Re big(eta(tfrac12) big) - fract^22 bigg( fracd^2 Re big(eta(tfrac12+it) big)dt^2 bigg) _t=0 + fract^424 bigg(fracd^4 Re big(eta(tfrac12+it) big)dt^4 bigg) _t=0 - cdots $$
$ \ $
$ \ $
Compactly written as
$ \ $
$$ beth = sum_l=0^infty (-1)^l cdot fract^2l(2l)! cdot bigg( fracd^(2l) Re big(eta(tfrac12+it) big)dt^(2l) bigg) _t=0$$
$ \ $
$ \ $
Let's go through an analogous procedure now with the second of our two main equations.
$ \ $
First, taken it's left side to be a function $g$ of $t$
$ \ $
$$ g = frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
We will be then interested in finding the numbers $t$ for which $space$ $g = 0$ $space$ .
$ \ $
Using the expansion
$$sin(x) = x - fracx^36 + fracx^5120 - space space cdots space space (-1)^j cdot fracx^2j+1(2j+1)! space space cdots $$
$ \ $
$ \ $
We can rewrite $g$ as
$ \ $
$$ g = frac1sqrt2 cdot bigg( (tln2) - frac(tln2)^36 + frac(tln2)^5120 cdots bigg) - frac1sqrt3 cdot bigg( (tln3) - frac(tln3)^36 + frac(tln3)^5120 cdots bigg) + cdots $$
$ \ $
which we can reassemble to be written as an infinite polynomial in $t$:
$ \ $
$$ g = t bigg( frac(ln2)sqrt2 - frac(ln3)sqrt3 + cdots bigg) - fract^36 bigg( frac(ln2)^3sqrt2 - frac(ln3)^3sqrt3 + cdots bigg) + fract^5120 bigg( frac(ln2)^5sqrt2 - frac(ln3)^5sqrt3 + cdots bigg)$$
$ \ $
$$ - cdots $$
$ \ $
$ \ $
Those polynomial's coefficients are also familiar.
$ \ $
The linear coefficient,
$$ bigg( frac(ln2)sqrt2 - frac(ln3)sqrt3 + cdots bigg) $$
is equal to
$$ bigg( fracd Im big(eta(tfrac12+it) big)dt bigg) _t=0 $$
$ \ $
$ \ $
$ \ $
The cubic coefficient,
$$ - frac16 bigg( frac(ln2)^3sqrt2 - frac(ln3)^3sqrt3 + cdots bigg) $$
is equal to
$$ frac16 bigg( fracd^3 Im big(eta(tfrac12+it) big)dt^3 bigg) _t=0 $$
$ \ $
$ \ $
$ \ $
And the quintic coefficient,
$$ frac1120 bigg( frac(ln2)^5sqrt2 - frac(ln3)^5sqrt3 + cdots bigg) $$
is equal to
$$ frac1120 bigg( fracd^5 Im big(eta(tfrac12+it) big)dt^5 bigg) _t=0 $$
$ \ $
$ \ $
So on with the subsequent coefficients we can write $g$ as:
$ \ $
$$ g = t bigg( fracd Im big(eta(tfrac12+it) big)dt bigg) _t=0 + fract^36 bigg(fracd^3 Im big(eta(tfrac12+it) big)dt^3 bigg) _t=0 + fract^5120 bigg(fracd^5 Im big(eta(tfrac12+it) big)dt^5 bigg) _t=0 + cdots $$
$ \ $
$ \ $
As we want $g = 0$, then $t$ must be the roots of $space$ $gimel$ $space$, where
$ \ $
$$ gimel = t bigg( fracd Im big(eta(tfrac12+it) big)dt bigg) _t=0 + fract^36 bigg(fracd^3 Im big(eta(tfrac12+it) big)dt^3 bigg) _t=0 + fract^5120 bigg(fracd^5 Im big(eta(tfrac12+it) big)dt^5 bigg) _t=0 + cdots $$
$ \ $
Compactly written as
$ \ $
$$ gimel = sum_l=0^infty fract^2l+1(2l+1)! cdot bigg( fracd^(2l+1) Im big(eta(tfrac12+it) big)dt^(2l+1) bigg) _t=0$$
$ \ $
$ \ $
$ \ $
So far this is the most explicit representation of $t$, which is the imaginary component of a non-trivial root of $eta$(or $zeta$), I could reach.
$ \ $
To partially conclude this post, and to state what has been accomplished:
$ \ $
$ \ $
There are two infinite polynomials, $beth$ and $gimel$, defined as mentioned, such that their common roots happen to be the imaginary component of the (some, probably all) non-trivial roots of the Riemann Zeta function.
$ \ $
$ \ $
I'd like to thank you if you went along all through this point, and to say that any comments or suggestions will be greatly welcomed.
riemann-zeta dirichlet-series
asked Jul 27 at 22:00
Leonardo Bohac
648
add a comment |Â
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0
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Let $$spacespacespace eta(z) = sum_a=1^infty frac1a^z cdot (-1)^a-1 $$
Now let $space$$z = sigma + it $ ,
$$eta(sigma + it) = sum_a=1^infty frac1a^sigma + it cdot (-1)^a-1 $$
May we split $spacespace$$cfrac1a^sigma + it$ $spacespace$ into its real and complex components:
$$frac1a^sigma + it = frac1a^sigma cdot left(frac1a^itright) = frac1a^sigma cdot left(frac1(e^lna)^itright) = frac1a^sigma cdot left(frac1e^(icdot tlna)right) = frac1a^sigma cdot e^-(icdot tlna)$$
$ \ $
$ \ $
$$ qquad qquad quad = frac1a^sigma cdot bigg( cos(-tlna) + icdot sin(-tlna) bigg) = frac1a^sigma cdot bigg( cos(tlna) - icdot sin(tlna) bigg) $$
Therefore,
$$spacespacespaceRe big(eta(sigma+it) big) = sum_a=1^infty frac1a^sigmacdot cos(tlna) cdot (-1)^a-1$$
$\$
$$Im big(eta(sigma+it) big) = sum_a=1^infty frac1a^sigmacdot sin(tlna) cdot (-1)^a $$
$ \ $
$ \ $
$ \ $
Where $Re$ and $Im$ respectively represent the real and complex components of its argument.
$ \ $
$ \ $
For $z$ to be a root of $eta$ we thus require that $Re big(eta(z)big) = 0 = Im big(eta(z)big) $.
Presumably there will be no such $z$ , with $Re(z) = sigmaneq frac12 $ satisfying the condition above (for sure if Riemann Hypothesis turns to be true).
$ \ $
So now I will focus on the numbers $z$, such that $Re(z) = frac12$.
$ \ $
Given that, the unknown of the problem will be $space$ $t = Im(z)$ $space$ for which $space$ $ eta(frac12 + it) = 0$.
Let me now reformulate what we're working on:
$ \ $
$$ eta(tfrac12 + it) = sum_a=1^infty frac1a^frac12 + it cdot (-1)^a-1 $$
Which leads us to:
$ \ $
$$ spacespacespace Re big(eta(tfrac12+it) big) = sum_a=1^infty frac1a^frac12cdot cos(tlna) cdot (-1)^a-1$$
$ \ $
and
$$ Im big(eta(tfrac12+it) big) = sum_a=1^infty frac1a^frac12cdot sin(tlna) cdot (-1)^a$$
$ \ $
$ \ $
Now letting $space$ $a^frac12 = sqrta $ $space$, and writing it by extend:
$ \ $
$$ Re big(eta(tfrac12+it) big) = frac1sqrt1 cdot cos(tln1) - frac1sqrt2 cdot cos(tln2) + frac1sqrt3 cdot cos(tln3) - dots $$
$ \ $
$$ Im big(eta(tfrac12+it) big) = -frac1sqrt1 cdot sin(tln1) + frac1sqrt2 cdot sin(tln2) + frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
Onward,
$ \ $
$$ space space Re big(eta(tfrac12+it) big) space = space 1 - frac1sqrt2 cdot cos(tln2) + frac1sqrt3 cdot cos(tln3) - dots $$
$ \ $
$$ space space space Im big(eta(tfrac12+it) big) space = space 0 space + frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
$ \ $
$ \ $
For $space$ $z = frac12 + it$ $space$ to be a root of $eta$, we ought to require that:
$ \ $
$$ space space 0 space = space 1 - frac1sqrt2 cdot cos(tln2) + frac1sqrt3 cdot cos(tln3) - dots $$
$ \ $
$$ space space 0 space = space 0 space + frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
$ \ $
$ \ $
$$therefore$$
$ \ $
$$ frac1sqrt2 cdot cos(tln2) - frac1sqrt3 cdot cos(tln3) + dots = 1 $$
$ \ $
$$ frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots = 0 $$
$ \ $
$ \ $
$ \ $
This system of equations doesn't look too alien, at least.
$ \ $
Let's dive into the first equation, taken it's left side to be a function $f$ of $t$
$ \ $
$$ f = frac1sqrt2 cdot cos(tln2) - frac1sqrt3 cdot cos(tln3) + dots $$
$ \ $
We will be then interested in finding the numbers $t$ for which $space$ $f = 1$ $space$ .
$ \ $
Using the expansion
$$cos(x) = 1 - fracx^22 + fracx^424 - space space cdots space space (-1)^j cdot fracx^2j(2j)! space space cdots $$
$ \ $
$ \ $
We can rewrite $f$ as
$ \ $
$$ f = frac1sqrt2 cdot bigg( 1 - frac(tln2)^22 + frac(tln2)^424 - cdots bigg) - frac1sqrt3 cdot bigg( 1 - frac(tln3)^22 + frac(tln3)^424 - cdots bigg) + cdots $$
$ \ $
which we can reassemble to be written as a infinite polynomial in $t$:
$ \ $
$$ f = bigg( frac1sqrt2 - frac1sqrt3 + cdots bigg) - fract^22 bigg( frac(ln2)^2sqrt2 - frac(ln3)^2sqrt3 + cdots bigg) + fract^424 bigg( frac(ln2)^4sqrt2 - frac(ln3)^4sqrt3 + cdots bigg)-cdots $$
$ \ $
Now we've reached a remarkable point.
$ \ $
The above polynomial's coefficients are no more than just constant numbers. And those numbers are pretty familiar.
$ \ $
The independent coefficient,
$$bigg( frac1sqrt2 - frac1sqrt3 + cdots bigg)$$
is equal to
$$ 1 - Re big(eta(tfrac12) big)$$
$ \ $
$ \ $
$ \ $
The quadratic coefficient,
$$ -frac12bigg( frac(ln2)^2sqrt2 - frac(ln3)^2sqrt3 + cdots bigg) $$
is equal to
$$ -frac12 bigg( fracd^2 Re big(eta(tfrac12+it) big)dt^2 bigg) _t=0 $$
$ \ $
$ \ $
$ \ $
And the quartic coefficient
$$ frac124bigg( frac(ln2)^4sqrt2 - frac(ln3)^4sqrt3 + cdots bigg)$$
is equal to
$$ frac124 bigg( - fracd^4 Re big(eta(tfrac12+it) big)dt^4 bigg) _t=0 $$
$ \ $
And so on with the subsequent coefficients.
$ \ $
We can thus write $f$ as:
$ \ $
$$ f = 1 - Re big(eta(tfrac12) big) + fract^22 bigg( fracd^2 Re big(eta(tfrac12+it) big)dt^2 bigg) _t=0 - fract^424 bigg(fracd^4 Re big(eta(tfrac12+it) big)dt^4 bigg) _t=0 + cdots $$
$ \ $
$ \ $
If we want $f = 1$, then $t$ must be the roots of $space$ $beth$ $space$, where
$ \ $
$$ beth = Re big(eta(tfrac12) big) - fract^22 bigg( fracd^2 Re big(eta(tfrac12+it) big)dt^2 bigg) _t=0 + fract^424 bigg(fracd^4 Re big(eta(tfrac12+it) big)dt^4 bigg) _t=0 - cdots $$
$ \ $
$ \ $
Compactly written as
$ \ $
$$ beth = sum_l=0^infty (-1)^l cdot fract^2l(2l)! cdot bigg( fracd^(2l) Re big(eta(tfrac12+it) big)dt^(2l) bigg) _t=0$$
$ \ $
$ \ $
Let's go through an analogous procedure now with the second of our two main equations.
$ \ $
First, taken it's left side to be a function $g$ of $t$
$ \ $
$$ g = frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
We will be then interested in finding the numbers $t$ for which $space$ $g = 0$ $space$ .
$ \ $
Using the expansion
$$sin(x) = x - fracx^36 + fracx^5120 - space space cdots space space (-1)^j cdot fracx^2j+1(2j+1)! space space cdots $$
$ \ $
$ \ $
We can rewrite $g$ as
$ \ $
$$ g = frac1sqrt2 cdot bigg( (tln2) - frac(tln2)^36 + frac(tln2)^5120 cdots bigg) - frac1sqrt3 cdot bigg( (tln3) - frac(tln3)^36 + frac(tln3)^5120 cdots bigg) + cdots $$
$ \ $
which we can reassemble to be written as an infinite polynomial in $t$:
$ \ $
$$ g = t bigg( frac(ln2)sqrt2 - frac(ln3)sqrt3 + cdots bigg) - fract^36 bigg( frac(ln2)^3sqrt2 - frac(ln3)^3sqrt3 + cdots bigg) + fract^5120 bigg( frac(ln2)^5sqrt2 - frac(ln3)^5sqrt3 + cdots bigg)$$
$ \ $
$$ - cdots $$
$ \ $
$ \ $
Those polynomial's coefficients are also familiar.
$ \ $
The linear coefficient,
$$ bigg( frac(ln2)sqrt2 - frac(ln3)sqrt3 + cdots bigg) $$
is equal to
$$ bigg( fracd Im big(eta(tfrac12+it) big)dt bigg) _t=0 $$
$ \ $
$ \ $
$ \ $
The cubic coefficient,
$$ - frac16 bigg( frac(ln2)^3sqrt2 - frac(ln3)^3sqrt3 + cdots bigg) $$
is equal to
$$ frac16 bigg( fracd^3 Im big(eta(tfrac12+it) big)dt^3 bigg) _t=0 $$
$ \ $
$ \ $
$ \ $
And the quintic coefficient,
$$ frac1120 bigg( frac(ln2)^5sqrt2 - frac(ln3)^5sqrt3 + cdots bigg) $$
is equal to
$$ frac1120 bigg( fracd^5 Im big(eta(tfrac12+it) big)dt^5 bigg) _t=0 $$
$ \ $
$ \ $
So on with the subsequent coefficients we can write $g$ as:
$ \ $
$$ g = t bigg( fracd Im big(eta(tfrac12+it) big)dt bigg) _t=0 + fract^36 bigg(fracd^3 Im big(eta(tfrac12+it) big)dt^3 bigg) _t=0 + fract^5120 bigg(fracd^5 Im big(eta(tfrac12+it) big)dt^5 bigg) _t=0 + cdots $$
$ \ $
$ \ $
As we want $g = 0$, then $t$ must be the roots of $space$ $gimel$ $space$, where
$ \ $
$$ gimel = t bigg( fracd Im big(eta(tfrac12+it) big)dt bigg) _t=0 + fract^36 bigg(fracd^3 Im big(eta(tfrac12+it) big)dt^3 bigg) _t=0 + fract^5120 bigg(fracd^5 Im big(eta(tfrac12+it) big)dt^5 bigg) _t=0 + cdots $$
$ \ $
Compactly written as
$ \ $
$$ gimel = sum_l=0^infty fract^2l+1(2l+1)! cdot bigg( fracd^(2l+1) Im big(eta(tfrac12+it) big)dt^(2l+1) bigg) _t=0$$
$ \ $
$ \ $
$ \ $
So far this is the most explicit representation of $t$, which is the imaginary component of a non-trivial root of $eta$(or $zeta$), I could reach.
$ \ $
To partially conclude this post, and to state what has been accomplished:
$ \ $
$ \ $
There are two infinite polynomials, $beth$ and $gimel$, defined as mentioned, such that their common roots happen to be the imaginary component of the (some, probably all) non-trivial roots of the Riemann Zeta function.
$ \ $
$ \ $
I'd like to thank you if you went along all through this point, and to say that any comments or suggestions will be greatly welcomed.
riemann-zeta dirichlet-series
asked Jul 27 at 22:00
Leonardo Bohac
648
What’s your question?
– Szeto
Jul 27 at 23:52
It’s not a specific question, indeed. If I’d to put in more directly it would be : “How can we find a closed formula for the non-trivial roots of the Riemann Zeta function, given that it has real part equal to 0.5â€Â. By doing so, I expect some ideias on how to prove (or disprove) the Riemann Hypothesis will be obtained.
– Leonardo Bohac
Jul 28 at 0:03
1
I recall when I was a student, I also attempted to break zeta function into real and imaginary parts, and then tried to prove/disprove RH. I didn’t know derivatives so what you have done is surely more advanced. However, I somehow think that the series you established are simply some taylor series. Moreover, the rearrangements of sums are not always permitted; I didn’t check validity, but the validity surely depends on the type of convergence of zeta/eta function.
– Szeto
Jul 28 at 0:26
My first thoughts were to split Zeta into real and imaginary, too. Though I found out that it wouldn't be very useful, since the Zeta function (the simple one) does not converge for Re(z) < 1. As I wanted to investigate Re(z) = 1/2, the Dirichlet Eta function turned out to be better to work on. Let me know if you find incorrectness among my work done so far.
– Leonardo Bohac
Jul 28 at 5:33
At the moment I don’t have the time. I may be will check later. By the way, I suggest you to rewrite your question more compactly so that the question won’t be too long; I am sure that many users lost interest due to its length.
– Szeto
Jul 28 at 5:41
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Let $$spacespacespace eta(z) = sum_a=1^infty frac1a^z cdot (-1)^a-1 $$
Now let $space$$z = sigma + it $ ,
$$eta(sigma + it) = sum_a=1^infty frac1a^sigma + it cdot (-1)^a-1 $$
May we split $spacespace$$cfrac1a^sigma + it$ $spacespace$ into its real and complex components:
$$frac1a^sigma + it = frac1a^sigma cdot left(frac1a^itright) = frac1a^sigma cdot left(frac1(e^lna)^itright) = frac1a^sigma cdot left(frac1e^(icdot tlna)right) = frac1a^sigma cdot e^-(icdot tlna)$$
$ \ $
$ \ $
$$ qquad qquad quad = frac1a^sigma cdot bigg( cos(-tlna) + icdot sin(-tlna) bigg) = frac1a^sigma cdot bigg( cos(tlna) - icdot sin(tlna) bigg) $$
Therefore,
$$spacespacespaceRe big(eta(sigma+it) big) = sum_a=1^infty frac1a^sigmacdot cos(tlna) cdot (-1)^a-1$$
$\$
$$Im big(eta(sigma+it) big) = sum_a=1^infty frac1a^sigmacdot sin(tlna) cdot (-1)^a $$
$ \ $
$ \ $
$ \ $
Where $Re$ and $Im$ respectively represent the real and complex components of its argument.
$ \ $
$ \ $
For $z$ to be a root of $eta$ we thus require that $Re big(eta(z)big) = 0 = Im big(eta(z)big) $.
Presumably there will be no such $z$ , with $Re(z) = sigmaneq frac12 $ satisfying the condition above (for sure if Riemann Hypothesis turns to be true).
$ \ $
So now I will focus on the numbers $z$, such that $Re(z) = frac12$.
$ \ $
Given that, the unknown of the problem will be $space$ $t = Im(z)$ $space$ for which $space$ $ eta(frac12 + it) = 0$.
Let me now reformulate what we're working on:
$ \ $
$$ eta(tfrac12 + it) = sum_a=1^infty frac1a^frac12 + it cdot (-1)^a-1 $$
Which leads us to:
$ \ $
$$ spacespacespace Re big(eta(tfrac12+it) big) = sum_a=1^infty frac1a^frac12cdot cos(tlna) cdot (-1)^a-1$$
$ \ $
and
$$ Im big(eta(tfrac12+it) big) = sum_a=1^infty frac1a^frac12cdot sin(tlna) cdot (-1)^a$$
$ \ $
$ \ $
Now letting $space$ $a^frac12 = sqrta $ $space$, and writing it by extend:
$ \ $
$$ Re big(eta(tfrac12+it) big) = frac1sqrt1 cdot cos(tln1) - frac1sqrt2 cdot cos(tln2) + frac1sqrt3 cdot cos(tln3) - dots $$
$ \ $
$$ Im big(eta(tfrac12+it) big) = -frac1sqrt1 cdot sin(tln1) + frac1sqrt2 cdot sin(tln2) + frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
Onward,
$ \ $
$$ space space Re big(eta(tfrac12+it) big) space = space 1 - frac1sqrt2 cdot cos(tln2) + frac1sqrt3 cdot cos(tln3) - dots $$
$ \ $
$$ space space space Im big(eta(tfrac12+it) big) space = space 0 space + frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
$ \ $
$ \ $
For $space$ $z = frac12 + it$ $space$ to be a root of $eta$, we ought to require that:
$ \ $
$$ space space 0 space = space 1 - frac1sqrt2 cdot cos(tln2) + frac1sqrt3 cdot cos(tln3) - dots $$
$ \ $
$$ space space 0 space = space 0 space + frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
$ \ $
$ \ $
$$therefore$$
$ \ $
$$ frac1sqrt2 cdot cos(tln2) - frac1sqrt3 cdot cos(tln3) + dots = 1 $$
$ \ $
$$ frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots = 0 $$
$ \ $
$ \ $
$ \ $
This system of equations doesn't look too alien, at least.
$ \ $
Let's dive into the first equation, taken it's left side to be a function $f$ of $t$
$ \ $
$$ f = frac1sqrt2 cdot cos(tln2) - frac1sqrt3 cdot cos(tln3) + dots $$
$ \ $
We will be then interested in finding the numbers $t$ for which $space$ $f = 1$ $space$ .
$ \ $
Using the expansion
$$cos(x) = 1 - fracx^22 + fracx^424 - space space cdots space space (-1)^j cdot fracx^2j(2j)! space space cdots $$
$ \ $
$ \ $
We can rewrite $f$ as
$ \ $
$$ f = frac1sqrt2 cdot bigg( 1 - frac(tln2)^22 + frac(tln2)^424 - cdots bigg) - frac1sqrt3 cdot bigg( 1 - frac(tln3)^22 + frac(tln3)^424 - cdots bigg) + cdots $$
$ \ $
which we can reassemble to be written as a infinite polynomial in $t$:
$ \ $
$$ f = bigg( frac1sqrt2 - frac1sqrt3 + cdots bigg) - fract^22 bigg( frac(ln2)^2sqrt2 - frac(ln3)^2sqrt3 + cdots bigg) + fract^424 bigg( frac(ln2)^4sqrt2 - frac(ln3)^4sqrt3 + cdots bigg)-cdots $$
$ \ $
Now we've reached a remarkable point.
$ \ $
The above polynomial's coefficients are no more than just constant numbers. And those numbers are pretty familiar.
$ \ $
The independent coefficient,
$$bigg( frac1sqrt2 - frac1sqrt3 + cdots bigg)$$
is equal to
$$ 1 - Re big(eta(tfrac12) big)$$
$ \ $
$ \ $
$ \ $
The quadratic coefficient,
$$ -frac12bigg( frac(ln2)^2sqrt2 - frac(ln3)^2sqrt3 + cdots bigg) $$
is equal to
$$ -frac12 bigg( fracd^2 Re big(eta(tfrac12+it) big)dt^2 bigg) _t=0 $$
$ \ $
$ \ $
$ \ $
And the quartic coefficient
$$ frac124bigg( frac(ln2)^4sqrt2 - frac(ln3)^4sqrt3 + cdots bigg)$$
is equal to
$$ frac124 bigg( - fracd^4 Re big(eta(tfrac12+it) big)dt^4 bigg) _t=0 $$
$ \ $
And so on with the subsequent coefficients.
$ \ $
We can thus write $f$ as:
$ \ $
$$ f = 1 - Re big(eta(tfrac12) big) + fract^22 bigg( fracd^2 Re big(eta(tfrac12+it) big)dt^2 bigg) _t=0 - fract^424 bigg(fracd^4 Re big(eta(tfrac12+it) big)dt^4 bigg) _t=0 + cdots $$
$ \ $
$ \ $
If we want $f = 1$, then $t$ must be the roots of $space$ $beth$ $space$, where
$ \ $
$$ beth = Re big(eta(tfrac12) big) - fract^22 bigg( fracd^2 Re big(eta(tfrac12+it) big)dt^2 bigg) _t=0 + fract^424 bigg(fracd^4 Re big(eta(tfrac12+it) big)dt^4 bigg) _t=0 - cdots $$
$ \ $
$ \ $
Compactly written as
$ \ $
$$ beth = sum_l=0^infty (-1)^l cdot fract^2l(2l)! cdot bigg( fracd^(2l) Re big(eta(tfrac12+it) big)dt^(2l) bigg) _t=0$$
$ \ $
$ \ $
Let's go through an analogous procedure now with the second of our two main equations.
$ \ $
First, taken it's left side to be a function $g$ of $t$
$ \ $
$$ g = frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
We will be then interested in finding the numbers $t$ for which $space$ $g = 0$ $space$ .
$ \ $
Using the expansion
$$sin(x) = x - fracx^36 + fracx^5120 - space space cdots space space (-1)^j cdot fracx^2j+1(2j+1)! space space cdots $$
$ \ $
$ \ $
We can rewrite $g$ as
$ \ $
$$ g = frac1sqrt2 cdot bigg( (tln2) - frac(tln2)^36 + frac(tln2)^5120 cdots bigg) - frac1sqrt3 cdot bigg( (tln3) - frac(tln3)^36 + frac(tln3)^5120 cdots bigg) + cdots $$
$ \ $
which we can reassemble to be written as an infinite polynomial in $t$:
$ \ $
$$ g = t bigg( frac(ln2)sqrt2 - frac(ln3)sqrt3 + cdots bigg) - fract^36 bigg( frac(ln2)^3sqrt2 - frac(ln3)^3sqrt3 + cdots bigg) + fract^5120 bigg( frac(ln2)^5sqrt2 - frac(ln3)^5sqrt3 + cdots bigg)$$
$ \ $
$$ - cdots $$
$ \ $
$ \ $
Those polynomial's coefficients are also familiar.
$ \ $
The linear coefficient,
$$ bigg( frac(ln2)sqrt2 - frac(ln3)sqrt3 + cdots bigg) $$
is equal to
$$ bigg( fracd Im big(eta(tfrac12+it) big)dt bigg) _t=0 $$
$ \ $
$ \ $
$ \ $
The cubic coefficient,
$$ - frac16 bigg( frac(ln2)^3sqrt2 - frac(ln3)^3sqrt3 + cdots bigg) $$
is equal to
$$ frac16 bigg( fracd^3 Im big(eta(tfrac12+it) big)dt^3 bigg) _t=0 $$
$ \ $
$ \ $
$ \ $
And the quintic coefficient,
$$ frac1120 bigg( frac(ln2)^5sqrt2 - frac(ln3)^5sqrt3 + cdots bigg) $$
is equal to
$$ frac1120 bigg( fracd^5 Im big(eta(tfrac12+it) big)dt^5 bigg) _t=0 $$
$ \ $
$ \ $
So on with the subsequent coefficients we can write $g$ as:
$ \ $
$$ g = t bigg( fracd Im big(eta(tfrac12+it) big)dt bigg) _t=0 + fract^36 bigg(fracd^3 Im big(eta(tfrac12+it) big)dt^3 bigg) _t=0 + fract^5120 bigg(fracd^5 Im big(eta(tfrac12+it) big)dt^5 bigg) _t=0 + cdots $$
$ \ $
$ \ $
As we want $g = 0$, then $t$ must be the roots of $space$ $gimel$ $space$, where
$ \ $
$$ gimel = t bigg( fracd Im big(eta(tfrac12+it) big)dt bigg) _t=0 + fract^36 bigg(fracd^3 Im big(eta(tfrac12+it) big)dt^3 bigg) _t=0 + fract^5120 bigg(fracd^5 Im big(eta(tfrac12+it) big)dt^5 bigg) _t=0 + cdots $$
$ \ $
Compactly written as
$ \ $
$$ gimel = sum_l=0^infty fract^2l+1(2l+1)! cdot bigg( fracd^(2l+1) Im big(eta(tfrac12+it) big)dt^(2l+1) bigg) _t=0$$
$ \ $
$ \ $
$ \ $
So far this is the most explicit representation of $t$, which is the imaginary component of a non-trivial root of $eta$(or $zeta$), I could reach.
$ \ $
To partially conclude this post, and to state what has been accomplished:
$ \ $
$ \ $
There are two infinite polynomials, $beth$ and $gimel$, defined as mentioned, such that their common roots happen to be the imaginary component of the (some, probably all) non-trivial roots of the Riemann Zeta function.
$ \ $
$ \ $
I'd like to thank you if you went along all through this point, and to say that any comments or suggestions will be greatly welcomed.
riemann-zeta dirichlet-series
asked Jul 27 at 22:00
Leonardo Bohac
648
Let $$spacespacespace eta(z) = sum_a=1^infty frac1a^z cdot (-1)^a-1 $$
Now let $space$$z = sigma + it $ ,
$$eta(sigma + it) = sum_a=1^infty frac1a^sigma + it cdot (-1)^a-1 $$
May we split $spacespace$$cfrac1a^sigma + it$ $spacespace$ into its real and complex components:
$$frac1a^sigma + it = frac1a^sigma cdot left(frac1a^itright) = frac1a^sigma cdot left(frac1(e^lna)^itright) = frac1a^sigma cdot left(frac1e^(icdot tlna)right) = frac1a^sigma cdot e^-(icdot tlna)$$
$ \ $
$ \ $
$$ qquad qquad quad = frac1a^sigma cdot bigg( cos(-tlna) + icdot sin(-tlna) bigg) = frac1a^sigma cdot bigg( cos(tlna) - icdot sin(tlna) bigg) $$
Therefore,
$$spacespacespaceRe big(eta(sigma+it) big) = sum_a=1^infty frac1a^sigmacdot cos(tlna) cdot (-1)^a-1$$
$\$
$$Im big(eta(sigma+it) big) = sum_a=1^infty frac1a^sigmacdot sin(tlna) cdot (-1)^a $$
$ \ $
$ \ $
$ \ $
Where $Re$ and $Im$ respectively represent the real and complex components of its argument.
$ \ $
$ \ $
For $z$ to be a root of $eta$ we thus require that $Re big(eta(z)big) = 0 = Im big(eta(z)big) $.
Presumably there will be no such $z$ , with $Re(z) = sigmaneq frac12 $ satisfying the condition above (for sure if Riemann Hypothesis turns to be true).
$ \ $
So now I will focus on the numbers $z$, such that $Re(z) = frac12$.
$ \ $
Given that, the unknown of the problem will be $space$ $t = Im(z)$ $space$ for which $space$ $ eta(frac12 + it) = 0$.
Let me now reformulate what we're working on:
$ \ $
$$ eta(tfrac12 + it) = sum_a=1^infty frac1a^frac12 + it cdot (-1)^a-1 $$
Which leads us to:
$ \ $
$$ spacespacespace Re big(eta(tfrac12+it) big) = sum_a=1^infty frac1a^frac12cdot cos(tlna) cdot (-1)^a-1$$
$ \ $
and
$$ Im big(eta(tfrac12+it) big) = sum_a=1^infty frac1a^frac12cdot sin(tlna) cdot (-1)^a$$
$ \ $
$ \ $
Now letting $space$ $a^frac12 = sqrta $ $space$, and writing it by extend:
$ \ $
$$ Re big(eta(tfrac12+it) big) = frac1sqrt1 cdot cos(tln1) - frac1sqrt2 cdot cos(tln2) + frac1sqrt3 cdot cos(tln3) - dots $$
$ \ $
$$ Im big(eta(tfrac12+it) big) = -frac1sqrt1 cdot sin(tln1) + frac1sqrt2 cdot sin(tln2) + frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
Onward,
$ \ $
$$ space space Re big(eta(tfrac12+it) big) space = space 1 - frac1sqrt2 cdot cos(tln2) + frac1sqrt3 cdot cos(tln3) - dots $$
$ \ $
$$ space space space Im big(eta(tfrac12+it) big) space = space 0 space + frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
$ \ $
$ \ $
For $space$ $z = frac12 + it$ $space$ to be a root of $eta$, we ought to require that:
$ \ $
$$ space space 0 space = space 1 - frac1sqrt2 cdot cos(tln2) + frac1sqrt3 cdot cos(tln3) - dots $$
$ \ $
$$ space space 0 space = space 0 space + frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
$ \ $
$ \ $
$$therefore$$
$ \ $
$$ frac1sqrt2 cdot cos(tln2) - frac1sqrt3 cdot cos(tln3) + dots = 1 $$
$ \ $
$$ frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots = 0 $$
$ \ $
$ \ $
$ \ $
This system of equations doesn't look too alien, at least.
$ \ $
Let's dive into the first equation, taken it's left side to be a function $f$ of $t$
$ \ $
$$ f = frac1sqrt2 cdot cos(tln2) - frac1sqrt3 cdot cos(tln3) + dots $$
$ \ $
We will be then interested in finding the numbers $t$ for which $space$ $f = 1$ $space$ .
$ \ $
Using the expansion
$$cos(x) = 1 - fracx^22 + fracx^424 - space space cdots space space (-1)^j cdot fracx^2j(2j)! space space cdots $$
$ \ $
$ \ $
We can rewrite $f$ as
$ \ $
$$ f = frac1sqrt2 cdot bigg( 1 - frac(tln2)^22 + frac(tln2)^424 - cdots bigg) - frac1sqrt3 cdot bigg( 1 - frac(tln3)^22 + frac(tln3)^424 - cdots bigg) + cdots $$
$ \ $
which we can reassemble to be written as a infinite polynomial in $t$:
$ \ $
$$ f = bigg( frac1sqrt2 - frac1sqrt3 + cdots bigg) - fract^22 bigg( frac(ln2)^2sqrt2 - frac(ln3)^2sqrt3 + cdots bigg) + fract^424 bigg( frac(ln2)^4sqrt2 - frac(ln3)^4sqrt3 + cdots bigg)-cdots $$
$ \ $
Now we've reached a remarkable point.
$ \ $
The above polynomial's coefficients are no more than just constant numbers. And those numbers are pretty familiar.
$ \ $
The independent coefficient,
$$bigg( frac1sqrt2 - frac1sqrt3 + cdots bigg)$$
is equal to
$$ 1 - Re big(eta(tfrac12) big)$$
$ \ $
$ \ $
$ \ $
The quadratic coefficient,
$$ -frac12bigg( frac(ln2)^2sqrt2 - frac(ln3)^2sqrt3 + cdots bigg) $$
is equal to
$$ -frac12 bigg( fracd^2 Re big(eta(tfrac12+it) big)dt^2 bigg) _t=0 $$
$ \ $
$ \ $
$ \ $
And the quartic coefficient
$$ frac124bigg( frac(ln2)^4sqrt2 - frac(ln3)^4sqrt3 + cdots bigg)$$
is equal to
$$ frac124 bigg( - fracd^4 Re big(eta(tfrac12+it) big)dt^4 bigg) _t=0 $$
$ \ $
And so on with the subsequent coefficients.
$ \ $
We can thus write $f$ as:
$ \ $
$$ f = 1 - Re big(eta(tfrac12) big) + fract^22 bigg( fracd^2 Re big(eta(tfrac12+it) big)dt^2 bigg) _t=0 - fract^424 bigg(fracd^4 Re big(eta(tfrac12+it) big)dt^4 bigg) _t=0 + cdots $$
$ \ $
$ \ $
If we want $f = 1$, then $t$ must be the roots of $space$ $beth$ $space$, where
$ \ $
$$ beth = Re big(eta(tfrac12) big) - fract^22 bigg( fracd^2 Re big(eta(tfrac12+it) big)dt^2 bigg) _t=0 + fract^424 bigg(fracd^4 Re big(eta(tfrac12+it) big)dt^4 bigg) _t=0 - cdots $$
$ \ $
$ \ $
Compactly written as
$ \ $
$$ beth = sum_l=0^infty (-1)^l cdot fract^2l(2l)! cdot bigg( fracd^(2l) Re big(eta(tfrac12+it) big)dt^(2l) bigg) _t=0$$
$ \ $
$ \ $
Let's go through an analogous procedure now with the second of our two main equations.
$ \ $
First, taken it's left side to be a function $g$ of $t$
$ \ $
$$ g = frac1sqrt2 cdot sin(tln2) - frac1sqrt3 cdot sin(tln3) + dots $$
$ \ $
We will be then interested in finding the numbers $t$ for which $space$ $g = 0$ $space$ .
$ \ $
Using the expansion
$$sin(x) = x - fracx^36 + fracx^5120 - space space cdots space space (-1)^j cdot fracx^2j+1(2j+1)! space space cdots $$
$ \ $
$ \ $
We can rewrite $g$ as
$ \ $
$$ g = frac1sqrt2 cdot bigg( (tln2) - frac(tln2)^36 + frac(tln2)^5120 cdots bigg) - frac1sqrt3 cdot bigg( (tln3) - frac(tln3)^36 + frac(tln3)^5120 cdots bigg) + cdots $$
$ \ $
which we can reassemble to be written as an infinite polynomial in $t$:
$ \ $
$$ g = t bigg( frac(ln2)sqrt2 - frac(ln3)sqrt3 + cdots bigg) - fract^36 bigg( frac(ln2)^3sqrt2 - frac(ln3)^3sqrt3 + cdots bigg) + fract^5120 bigg( frac(ln2)^5sqrt2 - frac(ln3)^5sqrt3 + cdots bigg)$$
$ \ $
$$ - cdots $$
$ \ $
$ \ $
Those polynomial's coefficients are also familiar.
$ \ $
The linear coefficient,
$$ bigg( frac(ln2)sqrt2 - frac(ln3)sqrt3 + cdots bigg) $$
is equal to
$$ bigg( fracd Im big(eta(tfrac12+it) big)dt bigg) _t=0 $$
$ \ $
$ \ $
$ \ $
The cubic coefficient,
$$ - frac16 bigg( frac(ln2)^3sqrt2 - frac(ln3)^3sqrt3 + cdots bigg) $$
is equal to
$$ frac16 bigg( fracd^3 Im big(eta(tfrac12+it) big)dt^3 bigg) _t=0 $$
$ \ $
$ \ $
$ \ $
And the quintic coefficient,
$$ frac1120 bigg( frac(ln2)^5sqrt2 - frac(ln3)^5sqrt3 + cdots bigg) $$
is equal to
$$ frac1120 bigg( fracd^5 Im big(eta(tfrac12+it) big)dt^5 bigg) _t=0 $$
$ \ $
$ \ $
So on with the subsequent coefficients we can write $g$ as:
$ \ $
$$ g = t bigg( fracd Im big(eta(tfrac12+it) big)dt bigg) _t=0 + fract^36 bigg(fracd^3 Im big(eta(tfrac12+it) big)dt^3 bigg) _t=0 + fract^5120 bigg(fracd^5 Im big(eta(tfrac12+it) big)dt^5 bigg) _t=0 + cdots $$
$ \ $
$ \ $
As we want $g = 0$, then $t$ must be the roots of $space$ $gimel$ $space$, where
$ \ $
$$ gimel = t bigg( fracd Im big(eta(tfrac12+it) big)dt bigg) _t=0 + fract^36 bigg(fracd^3 Im big(eta(tfrac12+it) big)dt^3 bigg) _t=0 + fract^5120 bigg(fracd^5 Im big(eta(tfrac12+it) big)dt^5 bigg) _t=0 + cdots $$
$ \ $
Compactly written as
$ \ $
$$ gimel = sum_l=0^infty fract^2l+1(2l+1)! cdot bigg( fracd^(2l+1) Im big(eta(tfrac12+it) big)dt^(2l+1) bigg) _t=0$$
$ \ $
$ \ $
$ \ $
So far this is the most explicit representation of $t$, which is the imaginary component of a non-trivial root of $eta$(or $zeta$), I could reach.
$ \ $
To partially conclude this post, and to state what has been accomplished:
$ \ $
$ \ $
There are two infinite polynomials, $beth$ and $gimel$, defined as mentioned, such that their common roots happen to be the imaginary component of the (some, probably all) non-trivial roots of the Riemann Zeta function.
$ \ $
$ \ $
I'd like to thank you if you went along all through this point, and to say that any comments or suggestions will be greatly welcomed.
riemann-zeta dirichlet-series
asked Jul 27 at 22:00
Leonardo Bohac
648
edited Jul 30 at 11:58
asked Jul 27 at 22:00
Leonardo Bohac
648
asked Jul 27 at 22:00
Leonardo Bohac
648
asked Jul 27 at 22:00
Leonardo Bohac
648
648
What’s your question?
– Szeto
Jul 27 at 23:52
It’s not a specific question, indeed. If I’d to put in more directly it would be : “How can we find a closed formula for the non-trivial roots of the Riemann Zeta function, given that it has real part equal to 0.5â€Â. By doing so, I expect some ideias on how to prove (or disprove) the Riemann Hypothesis will be obtained.
– Leonardo Bohac
Jul 28 at 0:03
1
I recall when I was a student, I also attempted to break zeta function into real and imaginary parts, and then tried to prove/disprove RH. I didn’t know derivatives so what you have done is surely more advanced. However, I somehow think that the series you established are simply some taylor series. Moreover, the rearrangements of sums are not always permitted; I didn’t check validity, but the validity surely depends on the type of convergence of zeta/eta function.
– Szeto
Jul 28 at 0:26
My first thoughts were to split Zeta into real and imaginary, too. Though I found out that it wouldn't be very useful, since the Zeta function (the simple one) does not converge for Re(z) < 1. As I wanted to investigate Re(z) = 1/2, the Dirichlet Eta function turned out to be better to work on. Let me know if you find incorrectness among my work done so far.
– Leonardo Bohac
Jul 28 at 5:33
At the moment I don’t have the time. I may be will check later. By the way, I suggest you to rewrite your question more compactly so that the question won’t be too long; I am sure that many users lost interest due to its length.
– Szeto
Jul 28 at 5:41
add a comment |Â
What’s your question?
– Szeto
Jul 27 at 23:52
It’s not a specific question, indeed. If I’d to put in more directly it would be : “How can we find a closed formula for the non-trivial roots of the Riemann Zeta function, given that it has real part equal to 0.5â€Â. By doing so, I expect some ideias on how to prove (or disprove) the Riemann Hypothesis will be obtained.
– Leonardo Bohac
Jul 28 at 0:03
1
I recall when I was a student, I also attempted to break zeta function into real and imaginary parts, and then tried to prove/disprove RH. I didn’t know derivatives so what you have done is surely more advanced. However, I somehow think that the series you established are simply some taylor series. Moreover, the rearrangements of sums are not always permitted; I didn’t check validity, but the validity surely depends on the type of convergence of zeta/eta function.
– Szeto
Jul 28 at 0:26
My first thoughts were to split Zeta into real and imaginary, too. Though I found out that it wouldn't be very useful, since the Zeta function (the simple one) does not converge for Re(z) < 1. As I wanted to investigate Re(z) = 1/2, the Dirichlet Eta function turned out to be better to work on. Let me know if you find incorrectness among my work done so far.
– Leonardo Bohac
Jul 28 at 5:33
At the moment I don’t have the time. I may be will check later. By the way, I suggest you to rewrite your question more compactly so that the question won’t be too long; I am sure that many users lost interest due to its length.
– Szeto
Jul 28 at 5:41
What’s your question?
– Szeto
Jul 27 at 23:52
What’s your question?
– Szeto
Jul 27 at 23:52
It’s not a specific question, indeed. If I’d to put in more directly it would be : “How can we find a closed formula for the non-trivial roots of the Riemann Zeta function, given that it has real part equal to 0.5â€Â. By doing so, I expect some ideias on how to prove (or disprove) the Riemann Hypothesis will be obtained.
– Leonardo Bohac
Jul 28 at 0:03
It’s not a specific question, indeed. If I’d to put in more directly it would be : “How can we find a closed formula for the non-trivial roots of the Riemann Zeta function, given that it has real part equal to 0.5â€Â. By doing so, I expect some ideias on how to prove (or disprove) the Riemann Hypothesis will be obtained.
– Leonardo Bohac
Jul 28 at 0:03
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I recall when I was a student, I also attempted to break zeta function into real and imaginary parts, and then tried to prove/disprove RH. I didn’t know derivatives so what you have done is surely more advanced. However, I somehow think that the series you established are simply some taylor series. Moreover, the rearrangements of sums are not always permitted; I didn’t check validity, but the validity surely depends on the type of convergence of zeta/eta function.
– Szeto
Jul 28 at 0:26
I recall when I was a student, I also attempted to break zeta function into real and imaginary parts, and then tried to prove/disprove RH. I didn’t know derivatives so what you have done is surely more advanced. However, I somehow think that the series you established are simply some taylor series. Moreover, the rearrangements of sums are not always permitted; I didn’t check validity, but the validity surely depends on the type of convergence of zeta/eta function.
– Szeto
Jul 28 at 0:26
My first thoughts were to split Zeta into real and imaginary, too. Though I found out that it wouldn't be very useful, since the Zeta function (the simple one) does not converge for Re(z) < 1. As I wanted to investigate Re(z) = 1/2, the Dirichlet Eta function turned out to be better to work on. Let me know if you find incorrectness among my work done so far.
– Leonardo Bohac
Jul 28 at 5:33
My first thoughts were to split Zeta into real and imaginary, too. Though I found out that it wouldn't be very useful, since the Zeta function (the simple one) does not converge for Re(z) < 1. As I wanted to investigate Re(z) = 1/2, the Dirichlet Eta function turned out to be better to work on. Let me know if you find incorrectness among my work done so far.
– Leonardo Bohac
Jul 28 at 5:33
At the moment I don’t have the time. I may be will check later. By the way, I suggest you to rewrite your question more compactly so that the question won’t be too long; I am sure that many users lost interest due to its length.
– Szeto
Jul 28 at 5:41
At the moment I don’t have the time. I may be will check later. By the way, I suggest you to rewrite your question more compactly so that the question won’t be too long; I am sure that many users lost interest due to its length.
– Szeto
Jul 28 at 5:41
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What’s your question?
– Szeto
Jul 27 at 23:52
It’s not a specific question, indeed. If I’d to put in more directly it would be : “How can we find a closed formula for the non-trivial roots of the Riemann Zeta function, given that it has real part equal to 0.5â€Â. By doing so, I expect some ideias on how to prove (or disprove) the Riemann Hypothesis will be obtained.
– Leonardo Bohac
Jul 28 at 0:03
1
I recall when I was a student, I also attempted to break zeta function into real and imaginary parts, and then tried to prove/disprove RH. I didn’t know derivatives so what you have done is surely more advanced. However, I somehow think that the series you established are simply some taylor series. Moreover, the rearrangements of sums are not always permitted; I didn’t check validity, but the validity surely depends on the type of convergence of zeta/eta function.
– Szeto
Jul 28 at 0:26
My first thoughts were to split Zeta into real and imaginary, too. Though I found out that it wouldn't be very useful, since the Zeta function (the simple one) does not converge for Re(z) < 1. As I wanted to investigate Re(z) = 1/2, the Dirichlet Eta function turned out to be better to work on. Let me know if you find incorrectness among my work done so far.
– Leonardo Bohac
Jul 28 at 5:33
At the moment I don’t have the time. I may be will check later. By the way, I suggest you to rewrite your question more compactly so that the question won’t be too long; I am sure that many users lost interest due to its length.
– Szeto
Jul 28 at 5:41