Riemann Hypothesis (1 Viewer)

[KeypadSDM]

Need a little Help with some *simple* algebra:

I need someone to help me prove:

O(x^1/2Ln[x]) = Int[x->oo]((t(t² - 1)(Ln[t]))^-1dt) - Ln[2] - Sum[All q](Li[x^q])

Where q are the non-trivial roots of the Riemann-Zeta function, (extended to all of compex space, z != 1)

Help?

 

[Grey Council]

Keypad, please save us from your bad humour.

lol, is this a serious post? Or are you making fun of a certain someone?

 

[KeypadSDM]

Well i'd like someone to prove it, it might be 'helpful'.

 

[George W. Bush]

Took me a little while, but I've got a solution.

I'll post it up tonight.

 

[KeypadSDM]

Originally posted by George W. Bush
Took me a little while, but I've got a solution.

I'll post it up tonight.

And I suppose the dog ate it?

 

[George W. Bush]

Whoops, I left it on my desk at work. Can't get it until Monday - hang in there, I hope your test isn't soon.

 

[KeypadSDM]

If by desk you mean fire, and by monday you mean thirty years, then I might actually believe you.

Then again, you are G W, you probably thought solving 5x = 4 was Riemanns.

 

[Grey Council]

ahh, i see
*has been enlightened*
This is one of Keypad's humourous posts.

Hrm, Buchanan, don't you think this is a tad bit too high for year 12 standard? I mean, hell, if most guys struggle with the 4u course (some parts of the 3u course as well) than what point to be leading into Riemanns? mmm

 

[McLake]

Originally posted by buchanan
Anyway, I've posted it here in extracurricular topics, so hopefully the mods won't get upset.

I, for one, am happy, and encourage your further ramblings on this and other subjects - it gives the many maths geniuses we have among us something to do ...

 

[Grey Council]

lol, fair enough. I take my comments back (albeit a bit reservedly )

 

[KeypadSDM]

Originally posted by buchanan
and this is the Riemann Hypothesis in explicit form.

Ahem, what?

 

[paper]

wow ...the greek alphabet is like so cool...

 

[Xayma]

Originally posted by buchanan
but if a not=1/2, we can see that it NEVER passes through the origin (and if you can prove that, then you've proved the Riemann Hypothesis):

And that makes it so much easier... (at least to understand what they are trying to prove)

 

[Xayma]

Well I dont plan on proving it anytime soon no matter how tempting the $1 million US is, and is that indexed for inflation or is its value dropping as time go by (ie making it better to solve it sooner rather then later)?

And did you get fed up with people calling you doc?

 

[Xayma]

And what exactly does the Reimann Hypothesis do? Isnt it something to do with prime numbers or something.

 

[CM_Tutor]

If it were ultimately proven to be false, that'll raise some problems for the number theory work that is based on the assumption that it is true.

 

[Xayma]

So if it is proved wrong, does it affect anything practical? Or just the number system and things like quantam mechanics (since it has something to do with nuclei)

 

[KeypadSDM]

Originally posted by buchanan
In particular, the equivalence keypad spoke earlier about, i.e., that pi(x)=Li(x)+O(x^1/2ln x) would be false and that would be disastrous for the theory of the distribution of prime numbers.

Like I would care. All number theories I would develop would simply rely on:

Pi(x) = Li(x) + O(xe^{-a(Ln[x]^1/2)})

 

[KeypadSDM]

Are you sure? Isn't that MUCH weaker than the R.H.?

 

[KeypadSDM]

Originally posted by buchanan
No keypad. Here I'm afraid you're mistaken. It is equivalent to RH.

O(x^1/2Ln[x])
==>
O(x^{1/2 + E})
==>
O(x^1/2(Ln[x])^N)

For any Value N.

Isn't that weaker?