Can $lim_ntoinftymathcal Ileft(expleft(frac2picdot ilog_n(p_n#)right)right)$ dominate any real-valued function? [closed]
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I'm pretty sure the imaginary part $mathcal I$ of $f(n)=g(x)cdotexpleft(frac2picdot ilog_n(p_n#)right)$ converges as $ntoinfty$, and probably to $0$, for any real-valued function $g:Bbb NtoBbb R$.
I stumbled upon this result studying the Collatz conjecture and it seemed potentially interesting in respect of prime number theory. Is this a difficult result to prove and does it have any significance?
I'm more certain for monotonic $g$ but then I'm inexperienced in such matters.
$f:Bbb Nto Bbb C$
$p_n#$ is the $n^th$ primorial
Show that for all such $displaystyle g:lim_ntoinfty mathcal I(g(f(n)))=0$
Algebraically I have that $mathcal I(f(x)/g(x))to expleft(dfrac2picdot ilog(n)vartheta(nlog(n))right)to0$ so the question is really about this part's ability to dominate any real-valued function on integers by multiplication.
The remainder of the proof is just an argument at the moment; I haven't translated it into algebra. I assume for now I'm reinventing the wheel but if this is new or interesting I can work on translating it into algebra.
complex-analysis analysis convergence prime-numbers chebyshev-function
asked Sep 3 at 8:21
Robert Frost
3,9491036
closed as unclear what you're asking by Did, Jyrki Lahtonen, Brian Borchers, José Carlos Santos, Alon Amit Sep 9 at 0:03
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-1
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I'm pretty sure the imaginary part $mathcal I$ of $f(n)=g(x)cdotexpleft(frac2picdot ilog_n(p_n#)right)$ converges as $ntoinfty$, and probably to $0$, for any real-valued function $g:Bbb NtoBbb R$.
I stumbled upon this result studying the Collatz conjecture and it seemed potentially interesting in respect of prime number theory. Is this a difficult result to prove and does it have any significance?
I'm more certain for monotonic $g$ but then I'm inexperienced in such matters.
$f:Bbb Nto Bbb C$
$p_n#$ is the $n^th$ primorial
Show that for all such $displaystyle g:lim_ntoinfty mathcal I(g(f(n)))=0$
Algebraically I have that $mathcal I(f(x)/g(x))to expleft(dfrac2picdot ilog(n)vartheta(nlog(n))right)to0$ so the question is really about this part's ability to dominate any real-valued function on integers by multiplication.
The remainder of the proof is just an argument at the moment; I haven't translated it into algebra. I assume for now I'm reinventing the wheel but if this is new or interesting I can work on translating it into algebra.
complex-analysis analysis convergence prime-numbers chebyshev-function
asked Sep 3 at 8:21
Robert Frost
3,9491036
closed as unclear what you're asking by Did, Jyrki Lahtonen, Brian Borchers, José Carlos Santos, Alon Amit Sep 9 at 0:03
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$Im(g(n)cdotexpleft(frac2picdot ilog_n(p_n#)right))=g(n)sindfrac2pilog_n(p_n#)$ doesn't converge for general $g.$
– mfl
Sep 3 at 8:48
@mfl you mean general $g:Bbb NtoBbb R$? Do you have an example or a reason? The reason I think this is I think $lim_ntoinftytextImageleft(expleft(dfrac2picdot ilog_x(p_n#)right)right)=Bbb R>0$ but I may be mistaken hence the question.
– Robert Frost
Sep 3 at 9:51
3
Consider $g(n)=dfrac(-1)^nsindfrac2pilog_n(p_n#).$ Then $Im(g(n)cdotexpleft(frac2picdot ilog_n(p_n#)right))=(-1)^n$ that doesn't converge.
– mfl
Sep 3 at 12:19
2
No. Consider $g(n)=dfrac2sindfrac2pilog_n(p_n#)$ if $n$ is odd and $g(n)=dfrac3sindfrac2pilog_n(p_n#)$ if $n$ is even. Of course, if you assume $g$ is bounded then you have convergence to $0.$
– mfl
Sep 3 at 12:29
2
$g(n)=dfracnsindfrac2pilog_n(p_n#)$ satisfies monotonicity. However the imaginary part goes to $+infty.$
– mfl
Sep 3 at 12:37
 |Â
show 4 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I'm pretty sure the imaginary part $mathcal I$ of $f(n)=g(x)cdotexpleft(frac2picdot ilog_n(p_n#)right)$ converges as $ntoinfty$, and probably to $0$, for any real-valued function $g:Bbb NtoBbb R$.
I stumbled upon this result studying the Collatz conjecture and it seemed potentially interesting in respect of prime number theory. Is this a difficult result to prove and does it have any significance?
I'm more certain for monotonic $g$ but then I'm inexperienced in such matters.
$f:Bbb Nto Bbb C$
$p_n#$ is the $n^th$ primorial
Show that for all such $displaystyle g:lim_ntoinfty mathcal I(g(f(n)))=0$
Algebraically I have that $mathcal I(f(x)/g(x))to expleft(dfrac2picdot ilog(n)vartheta(nlog(n))right)to0$ so the question is really about this part's ability to dominate any real-valued function on integers by multiplication.
The remainder of the proof is just an argument at the moment; I haven't translated it into algebra. I assume for now I'm reinventing the wheel but if this is new or interesting I can work on translating it into algebra.
complex-analysis analysis convergence prime-numbers chebyshev-function
asked Sep 3 at 8:21
Robert Frost
3,9491036
I'm pretty sure the imaginary part $mathcal I$ of $f(n)=g(x)cdotexpleft(frac2picdot ilog_n(p_n#)right)$ converges as $ntoinfty$, and probably to $0$, for any real-valued function $g:Bbb NtoBbb R$.
I stumbled upon this result studying the Collatz conjecture and it seemed potentially interesting in respect of prime number theory. Is this a difficult result to prove and does it have any significance?
I'm more certain for monotonic $g$ but then I'm inexperienced in such matters.
$f:Bbb Nto Bbb C$
$p_n#$ is the $n^th$ primorial
Show that for all such $displaystyle g:lim_ntoinfty mathcal I(g(f(n)))=0$
Algebraically I have that $mathcal I(f(x)/g(x))to expleft(dfrac2picdot ilog(n)vartheta(nlog(n))right)to0$ so the question is really about this part's ability to dominate any real-valued function on integers by multiplication.
The remainder of the proof is just an argument at the moment; I haven't translated it into algebra. I assume for now I'm reinventing the wheel but if this is new or interesting I can work on translating it into algebra.
complex-analysis analysis convergence prime-numbers chebyshev-function
complex-analysis analysis convergence prime-numbers chebyshev-function
asked Sep 3 at 8:21
Robert Frost
3,9491036
asked Sep 3 at 8:21
Robert Frost
3,9491036
edited Sep 3 at 12:30
asked Sep 3 at 8:21
Robert Frost
3,9491036
asked Sep 3 at 8:21
Robert Frost
3,9491036
asked Sep 3 at 8:21
Robert Frost
3,9491036
3,9491036
closed as unclear what you're asking by Did, Jyrki Lahtonen, Brian Borchers, José Carlos Santos, Alon Amit Sep 9 at 0:03
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Did, Jyrki Lahtonen, Brian Borchers, José Carlos Santos, Alon Amit Sep 9 at 0:03
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$Im(g(n)cdotexpleft(frac2picdot ilog_n(p_n#)right))=g(n)sindfrac2pilog_n(p_n#)$ doesn't converge for general $g.$
– mfl
Sep 3 at 8:48
@mfl you mean general $g:Bbb NtoBbb R$? Do you have an example or a reason? The reason I think this is I think $lim_ntoinftytextImageleft(expleft(dfrac2picdot ilog_x(p_n#)right)right)=Bbb R>0$ but I may be mistaken hence the question.
– Robert Frost
Sep 3 at 9:51
3
Consider $g(n)=dfrac(-1)^nsindfrac2pilog_n(p_n#).$ Then $Im(g(n)cdotexpleft(frac2picdot ilog_n(p_n#)right))=(-1)^n$ that doesn't converge.
– mfl
Sep 3 at 12:19
2
No. Consider $g(n)=dfrac2sindfrac2pilog_n(p_n#)$ if $n$ is odd and $g(n)=dfrac3sindfrac2pilog_n(p_n#)$ if $n$ is even. Of course, if you assume $g$ is bounded then you have convergence to $0.$
– mfl
Sep 3 at 12:29
2
$g(n)=dfracnsindfrac2pilog_n(p_n#)$ satisfies monotonicity. However the imaginary part goes to $+infty.$
– mfl
Sep 3 at 12:37
 |Â
show 4 more comments
1
$Im(g(n)cdotexpleft(frac2picdot ilog_n(p_n#)right))=g(n)sindfrac2pilog_n(p_n#)$ doesn't converge for general $g.$
– mfl
Sep 3 at 8:48
@mfl you mean general $g:Bbb NtoBbb R$? Do you have an example or a reason? The reason I think this is I think $lim_ntoinftytextImageleft(expleft(dfrac2picdot ilog_x(p_n#)right)right)=Bbb R>0$ but I may be mistaken hence the question.
– Robert Frost
Sep 3 at 9:51
3
Consider $g(n)=dfrac(-1)^nsindfrac2pilog_n(p_n#).$ Then $Im(g(n)cdotexpleft(frac2picdot ilog_n(p_n#)right))=(-1)^n$ that doesn't converge.
– mfl
Sep 3 at 12:19
2
No. Consider $g(n)=dfrac2sindfrac2pilog_n(p_n#)$ if $n$ is odd and $g(n)=dfrac3sindfrac2pilog_n(p_n#)$ if $n$ is even. Of course, if you assume $g$ is bounded then you have convergence to $0.$
– mfl
Sep 3 at 12:29
2
$g(n)=dfracnsindfrac2pilog_n(p_n#)$ satisfies monotonicity. However the imaginary part goes to $+infty.$
– mfl
Sep 3 at 12:37
1
1
$Im(g(n)cdotexpleft(frac2picdot ilog_n(p_n#)right))=g(n)sindfrac2pilog_n(p_n#)$ doesn't converge for general $g.$
– mfl
Sep 3 at 8:48
$Im(g(n)cdotexpleft(frac2picdot ilog_n(p_n#)right))=g(n)sindfrac2pilog_n(p_n#)$ doesn't converge for general $g.$
– mfl
Sep 3 at 8:48
@mfl you mean general $g:Bbb NtoBbb R$? Do you have an example or a reason? The reason I think this is I think $lim_ntoinftytextImageleft(expleft(dfrac2picdot ilog_x(p_n#)right)right)=Bbb R>0$ but I may be mistaken hence the question.
– Robert Frost
Sep 3 at 9:51
@mfl you mean general $g:Bbb NtoBbb R$? Do you have an example or a reason? The reason I think this is I think $lim_ntoinftytextImageleft(expleft(dfrac2picdot ilog_x(p_n#)right)right)=Bbb R>0$ but I may be mistaken hence the question.
– Robert Frost
Sep 3 at 9:51
3
3
Consider $g(n)=dfrac(-1)^nsindfrac2pilog_n(p_n#).$ Then $Im(g(n)cdotexpleft(frac2picdot ilog_n(p_n#)right))=(-1)^n$ that doesn't converge.
– mfl
Sep 3 at 12:19
Consider $g(n)=dfrac(-1)^nsindfrac2pilog_n(p_n#).$ Then $Im(g(n)cdotexpleft(frac2picdot ilog_n(p_n#)right))=(-1)^n$ that doesn't converge.
– mfl
Sep 3 at 12:19
2
2
No. Consider $g(n)=dfrac2sindfrac2pilog_n(p_n#)$ if $n$ is odd and $g(n)=dfrac3sindfrac2pilog_n(p_n#)$ if $n$ is even. Of course, if you assume $g$ is bounded then you have convergence to $0.$
– mfl
Sep 3 at 12:29
No. Consider $g(n)=dfrac2sindfrac2pilog_n(p_n#)$ if $n$ is odd and $g(n)=dfrac3sindfrac2pilog_n(p_n#)$ if $n$ is even. Of course, if you assume $g$ is bounded then you have convergence to $0.$
– mfl
Sep 3 at 12:29
2
2
$g(n)=dfracnsindfrac2pilog_n(p_n#)$ satisfies monotonicity. However the imaginary part goes to $+infty.$
– mfl
Sep 3 at 12:37
$g(n)=dfracnsindfrac2pilog_n(p_n#)$ satisfies monotonicity. However the imaginary part goes to $+infty.$
– mfl
Sep 3 at 12:37
 |Â
show 4 more comments
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1
$Im(g(n)cdotexpleft(frac2picdot ilog_n(p_n#)right))=g(n)sindfrac2pilog_n(p_n#)$ doesn't converge for general $g.$
– mfl
Sep 3 at 8:48
@mfl you mean general $g:Bbb NtoBbb R$? Do you have an example or a reason? The reason I think this is I think $lim_ntoinftytextImageleft(expleft(dfrac2picdot ilog_x(p_n#)right)right)=Bbb R>0$ but I may be mistaken hence the question.
– Robert Frost
Sep 3 at 9:51
3
Consider $g(n)=dfrac(-1)^nsindfrac2pilog_n(p_n#).$ Then $Im(g(n)cdotexpleft(frac2picdot ilog_n(p_n#)right))=(-1)^n$ that doesn't converge.
– mfl
Sep 3 at 12:19
2
No. Consider $g(n)=dfrac2sindfrac2pilog_n(p_n#)$ if $n$ is odd and $g(n)=dfrac3sindfrac2pilog_n(p_n#)$ if $n$ is even. Of course, if you assume $g$ is bounded then you have convergence to $0.$
– mfl
Sep 3 at 12:29
2
$g(n)=dfracnsindfrac2pilog_n(p_n#)$ satisfies monotonicity. However the imaginary part goes to $+infty.$
– mfl
Sep 3 at 12:37