Until it will be withdrawn in a while and then replaced by yet another version...
To be fair, the last few versions are not exactly major revisions.
To be truly fair, you’re phuking insane.
Lol, says someone who doesn't understand that if A(x)/x < 1/k for arbitrarily large k, then A(x)/x tends to 0 as x tends to infinity.
Go and study Calculus 1, loser.
To be fair, you're mentaIly iIl.
The latest version seems correct to me.
Until it will be withdrawn in a while and then replaced by yet another version...
To be fair, the last few versions are not exactly major revisions.
To be truly fair, you’re phuking insane.
It's amazing how these nutters still have the nerve to insult TK, yet their ignorance in simple calculus was brutally exposed.
To be fair, you're mentaIly iIl.
Nah, he isn't but you are.
You don't understand simple calculus and it was brutally exposed. Shameful.To be fair, any person who talks about himself in 3rd-person needs psychiatric help.
You can insinuate whatever you wish about my identity, dude, I couldn't care less.
It doesn't change the fact that you're pathetically and shamefully ignorant about simple calculus, and therefore not qualified to say anything on this topic.
You been cyber abusing TK for too long now and its now irritating, not to mention the level of your incompetence in simple calculus.
To be fair, you're mentaIly iIl.
Nah, he isn't but you are.
You don't understand simple calculus and it was brutally exposed. Shameful.To be fair, any person who talks about himself in 3rd-person needs psychiatric help.
The one who needs pychiatric help, is the person who tries to criticise an analytic number theory paper, yet they are thoroughly unaware of stuff that is taught in the first week of a Calculus 1 class. Absolutely pathetic.
To be fair, you're mentaIly iIl.
Nah, he isn't but you are.
You don't understand simple calculus and it was brutally exposed. Shameful.To be fair, any person who talks about himself in 3rd-person needs psychiatric help.
You can insinuate whatever you wish about my identity, dude, I couldn't care less.
It doesn't change the fact that you're pathetically and shamefully ignorant about simple calculus, and therefore not qualified to say anything on this topic.
You been cyber abusing TK for too long now and its now irritating, not to mention the level of your incompetence in simple calculus.
To be fair, you are a few slices short of a loaf.
To be fair, you're mentaIly iIl.
Nah, he isn't but you are.
You don't understand simple calculus and it was brutally exposed. Shameful.To be fair, any person who talks about himself in 3rd-person needs psychiatric help.
The one who needs pychiatric help, is the person who tries to criticise an analytic number theory paper, yet they are thoroughly unaware of stuff that is taught in the first week of a Calculus 1 class. Absolutely pathetic.
To be fair, you are a few bricks shy of a load.
Until it will be withdrawn in a while and then replaced by yet another version...
To be fair, the last few versions are not exactly major revisions.
To be truly fair, you’re phuking insane.
It's amazing how these nutters still have the nerve to insult TK, yet their ignorance in simple calculus was brutally exposed.
To be fair, they are insulting YOU, who claims to not be TK.
You confusion of this matter indicates you lack the lQ required to comprehend written English, in addition to being off your rocker.
Until it will be withdrawn in a while and then replaced by yet another version...
To be fair, the last few versions are not exactly major revisions.
To be truly fair, you’re phuking insane.
It's amazing how these nutters still have the nerve to insult TK, yet their ignorance in simple calculus was brutally exposed.
To be fair, they are insulting YOU, who claims to not be TK.
You confusion of this matter indicates you lack the lQ required to comprehend written English, in addition to being off your rocker.
You disqualified yourself from saying anything in tbis thread the very moment you pathetically and shockingly failed to understand why A(x)/x tends to 0 as x tends to infinity, given that A(x)/x < 1/k for arbitrarily large k.
Trying to cover this pathetic level of incompetence by insulting TK, makes you look so foolish.
To be fair, the last few versions are not exactly major revisions.
To be truly fair, you’re phuking insane.
It's amazing how these nutters still have the nerve to insult TK, yet their ignorance in simple calculus was brutally exposed.
To be fair, they are insulting YOU, who claims to not be TK.
You confusion of this matter indicates you lack the lQ required to comprehend written English, in addition to being off your rocker.You disqualified yourself from saying anything in tbis thread the very moment you pathetically and shockingly failed to understand why A(x)/x tends to 0 as x tends to infinity, given that A(x)/x < 1/k for arbitrarily large k.
Trying to cover this pathetic level of incompetence by insulting TK, makes you look so foolish.
To be fair, you disqualified yourself by being phvking insane.
Since I know some math, I'll assume you're talking to me.
Your latest revision is much better, but there's still a problem with the conclusion drawn from (16). Let S denote the LHS of (14), and B denote the RHS of (16).
You want to show that, for all eps > 0, there exists x_0 s.t. x > x_0 implies S/x < eps. In this statement, x_0 must be a fixed constant (i.e., it can depend on NOTHING but eps).
To accomplish this, you try to show that, for all eps > 0, x > x_0 = c_2(exp(c_2)) implies S/x < B/x = 5c_3/ln(2c_2) < eps. This would be fine if x_0 and 5c_3/ln(2c_2) were not functionally related. However, they clearly ARE related -- through c_2. To try to avoid this problem, you argue that the c_2 in 5c_3/ln(2c_2) doesn't matter, because c_1 can be chosen arbitrarily large. However, choosing c_1 arbitrarily large is equivalent to choosing c_2 arbitrarily large, which changes the value of x_0 you started with.
In this case, epsilon = (5c_3)/(log (2c_2)), so x_0 = c_{2}exp(c_2) is indeed dependent only on epsilon.
The derivation of (16) can actually be done by replacing "for x > (c_2)exp(c_2)" by "for all large enough x".
By the way, please educate your colleagues why the left-hand side of (12) is o(x).
For the moment, let's focus on the proof.
What is the argument for going from (15) to (16)?
Thanks for being only interested in facts.
Okay, the update is, the sum on the extreme right-hand side of (15) is actually >>1 for any given c_2. This, unfortunately, invalidates the proof. That being said, the bound in (12) remains true, and it could be interesting in its own right.
However, I have temporarily withdrawn this paper.
May we now shift attention to the first paper:
https://figshare.com/articles/preprint/Untitled_Item/14776146
By arguing only for sigma in C, one is able to invoke the Mean-Value Theorem for integrals in the evaluation of f(sigma) and g(sigma), thus circumventing the issues raised by some people on MJR and here. Kindly see the latest version.
May we now shift attention to the first paper:
https://figshare.com/articles/preprint/Untitled_Item/14776146
By arguing only for sigma in C, one is able to invoke the Mean-Value Theorem for integrals in the evaluation of f(sigma) and g(sigma), thus circumventing the issues raised by some people on MJR and here. Kindly see the latest version.
I sent an email to your yahoo account some time ago. Please check your spam filter.
May we now shift attention to the first paper:
https://figshare.com/articles/preprint/Untitled_Item/14776146
By arguing only for sigma in C, one is able to invoke the Mean-Value Theorem for integrals in the evaluation of f(sigma) and g(sigma), thus circumventing the issues raised by some people on MJR and here. Kindly see the latest version.I sent an email to your yahoo account some time ago. Please check your spam filter.
Thanks for your email and your genuine interest in my paper, hadn't checked my spam folder in weeks. I replied you just now.
Why should the defining series for q converge ?Because for Re(s) = sigma > 1, the log derivative of zeta(s) is << 2^{-sigma}, whilst the Möbius function is always bounded by 1.
You have a typo, on the integral definition of q. It must not have a negative sign.
Negative *coefficient*, i mean.
Why should the defining series for q converge ?Because for Re(s) = sigma > 1, the log derivative of zeta(s) is << 2^{-sigma}, whilst the Möbius function is always bounded by 1.
You have a typo, on the integral definition of q. It must not have a negative sign.
Corrected, thanks.