Wagstaff Conjecture - Page 15

CRGreathouse · 2011-01-10, 19:47

Quote:

Originally Posted by davar55

Our key
difference is my claims based on my conjecture (1), that the ratio is in
fact bounded above, which denies the fundamental assumption of the
current conjecture, namely that the distribution of primes can be "modeled"
as a random (poisson or otherwise) process. They can not, except as an
approximation. The primes, just as the integers, are immutable, not random.

So I suppose you believe that prime gaps are bounded above as well?

science_man_88 · 2011-01-10, 20:33

${q_n}=2^{e^{-\gamma}}^n \gt 1$ ? that's what i got out of it.

CRGreathouse · 2011-01-10, 20:49

Quote:

Originally Posted by science_man_88

${q_n}=2^{e^{-\gamma}}^n \gt 1$ ? that's what i got out of it.

Go back and do it again, then.

science_man_88 · 2011-01-10, 22:03

Quote:

Originally Posted by CRGreathouse

Go back and do it again, then.

http://upload.wikimedia.org/math/2/a...b3c9172fe1.png is what I can see it relating to on that page, that raised to the power of nis my best guess though I'm unsure of a symbol in the image so I'm unclear on what it could mean.

CRGreathouse · 2011-01-10, 22:17

Quote:

Originally Posted by science_man_88

http://upload.wikimedia.org/math/2/a...b3c9172fe1.png is what I can see it relating to on that page, that raised to the power of nis my best guess though I'm unsure of a symbol in the image so I'm unclear on what it could mean.

The image you linked to is a property of the function. The definition is given in the first (short!) paragraph.

science_man_88 · 2011-01-10, 22:28

Quote:

Originally Posted by CRGreathouse

The image you linked to is a property of the function. The definition is given in the first (short!) paragraph.

then I'm clueless because all i get with reading that first paragraph is that $2^{e^-{\gamma}} + o(1)$ grows faster than $\sqrt[n] q_n$ in which case unless it's an upper bound they should intersect in my mind.

CRGreathouse · 2011-01-10, 23:41

Quote:

Originally Posted by science_man_88

then I'm clueless because all i get with reading that first paragraph is that $2^{e^-{\gamma}} + o(1)$ grows faster than $\sqrt[n] q_n$ in which case unless it's an upper bound they should intersect in my mind.

Rewrite the equation I have until it's in the form in the definition ("foo = o(bar)"), then substitute the appropriate functions into the definition. What do you get?

science_man_88 · 2011-01-11, 00:58

${\sqrt[n]{q_n}} - {2^{e^{-\gamma}}}\lt 1$ ?

CRGreathouse · 2011-01-11, 01:39

Quote:

Originally Posted by science_man_88

${\sqrt[n]{q_n}} - {2^{e^{-\gamma}}}\lt 1$ ?

Where do you even see a "<"? Clearly you're not using the definition (or, rather, you're not using either of the definitions).

science_man_88 · 2011-01-11, 02:01

Quote:

Originally Posted by CRGreathouse

Where do you even see a "<"? Clearly you're not using the definition (or, rather, you're not using either of the definitions).

I don't see how to do grows faster than and that's my understanding from what I read.

davar55 · 2011-01-11, 21:29

Quote:

Originally Posted by CRGreathouse

So I suppose you believe that prime gaps are bounded above as well?

No, two different issues.

Prime gaps (differences) are known to increase without an absolute upper bound.

In the MPE case, it's the ratio of consecutive terms I claim is bounded,
not the differences.

2011-01-10, 20:33	#156
science_man_88 "Forget I exist" Jul 2009 Dumbassville 8384₁₀ Posts	${q_n}=2^{e^{-\gamma}}^n \gt 1$ ? that's what i got out of it. Last fiddled with by science_man_88 on 2011-01-10 at 20:35

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2011-01-11, 00:58	#162
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶·131 Posts	${\sqrt[n]{q_n}} - {2^{e^{-\gamma}}}\lt 1$ ?