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pi and e. (1 Viewer)

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no_arg

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I believe it's 4 colour conjecture NOT 4 colour theorem!
No proof has as yet been offered although certain cases have been verified by computer.
Proof by computer is not valid I'm afraid and significant work is still being undertaken to establish a proof.

Please don't get worked up Tywebb this is just mathematics.
 
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no_arg

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"However, because part of the proof consisted of an exhaustive analysis of many discrete cases by a computer, some mathematicians do not accept it."

"logic program Coq" ?? a program is not a proof


Susequent attempts mentioned in the article are all still computer based and unverifiiable

Still the 4 colour conjecture I'm afraid...
 
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And how's the status of the ... well, um, ... let's just call it the no_arg conjecture, &pi;<sup>4</sup>+&pi;<sup>5</sup>=e<sup>6</sup>? Have you proved it yet? I've disproved it, as have most other people on this thread!
 

no_arg

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Disproof?
What disproof?
Punching keys on a caculator will never prove anything!
What has happened to you Tywebb?
First of all a test for POI that can't even deal with sin(x) at x=0 and now serving as an apologist for computer scientists pretending to be mathematicians.
Oh well what more needs to be said?

"The fundamental things apply as time goes by"
 
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acmilan

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no_arg said:
...computer scientists pretending to be mathematicians.
Wow that sums me up completely :)

BSc(Computer Science)/BSc(Pure Mathematics) @ unsw :D
 
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no_arg said:
serving as an apologist for computer scientists pretending to be mathematicians.

Well, some of the best stuff done on the Riemann Hypothesis was done by Odlyzko. Whilst I accept that it might be possible to come up with a non-constructive disproof of the Riemann Hypothesis, Odlyzko's research may however one day come up with a counterexample - all done by computer. And I'd believe him, even if, and especially if, no_arg doesn't. And I'm still not tywebb.



http://www.dtc.umn.edu/~odlyzko
 
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withoutaface

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If a computer found a root that didn't lie on the strip you could just sub it in manually and prove it that way, couldn't you?
 
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withoutaface said:
If a computer found a root that didn't lie on the strip you could just sub it in manually and prove it that way, couldn't you?
It is known that they all lie in the critical strip. The Riemann Hypothesis says that all the nontrivial zeros of &zeta;(s) lie on the critical line, Re(s)=1/2. If that's what you meant to say, then theoretically, yes.

But you're a braver man than me if you try to do it by hand. Not even Odlyzko himself would do that. It would be a huge number. Incidently, Riemann himself did calculate the first few zeros by hand and I've seen the page from the Nachlass on which he did the first one. It's quite a mess, and very hard to understand. And that value was only pretty small (0.5&plusmn;i(14.134...)). Subsequently, further calculations by hand were done by Gram, Backlund and Hutchinson, but thereafter all the work was done by computers.

Gram, Backlund and Hutchinson used Euler-Maclaurin summation. But Riemann's method was different. He used the Riemann-Siegel formula. This formula is what most of the computer work is based on, until very recently when Gourdon used the Odlyzko-Schönhage algorithm (a more powerful generalisation of the Riemann-Siegel formula) to calculate the first 10 trillion zeros, and found they are all on the line.

Hardy proved there are infinitely many zeros on the line, so computers probably can't be used to actually prove the Riemann Hypothesis. Nevertheless, they might be able to be used to disprove it if they find one off the line.

But of course no_arg won't have a bar of it because it's all done by computers thesedays. The best anyone has done by hand, was by Hutchinson, and he proved by hand that the first 138 are on the line. Compare that with 10 trillion! Give me computers any day! I believe them! Especially because no_arg doesn't!
 
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Xayma

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no_arg said:
Surely you are not suggesting that putting numbers into a calculator could ever
constitute a proof? Pleeeeassse

It should be also pointed out that being approximately equal does not preclude things from also being equal! Many approximations in the literature are later refined to equality. The references given are unfortunately very dated and no longer reflect the state of play on these issues.
Depends on what you are trying to prove. If you are trying to prove that they aren't equal then yes. If you were trying to prove that they were then no. However, when the answer is outside the uncertainity, such that one side does not lie within the error (which has a known upper limit) of the other then you know they are unequal.
 
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Yeah. Anyway at least we know that &pi;<sup>4</sup>+&pi;<sup>5</sup> is not equal to e<sup>6</sup>! no_arg said they are equal, so I guess you could just deduce from this that they aren't!

no_arg said:
a test for POI that can't even deal with sin(x) at x=0
Really? Well how about this.

If f(x)=sin(x), then f''(x)=-sin(x), so f''(0)=0, and f'''(x)=-cos(x) and so f'''(0)=-1&ne;0. Hence (0,0) is an inflection.

I accept that there are some functions for which the third derivative method either does not work, or for which it is inappropriate. But this isn't one of them.

no_arg said:
"The fundamental things apply as time goes by"
Yes. Yes. Well, you said it, no_arg.
 
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KFunk

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buchanan said:
... hmmmm ... Yeah. That's right no_arg. I'll put money on it. I bet you $50 you're wrong! Or couple of cartons of VB. Do you accept?
I also have some reputable sources at hand. It seems that if you "...supply liquor to a minor ... [the maximum penalty is] ... $5,500."<sup>1</sup> Shall we crack open the grange?

<sup>1.</sup>Department of Gaming and Racing "Young People and the NSW Liquor Laws." Sydney, DGR. pp2-3, 2003.
 

brett86

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no_arg said:
Pi^4+Pi^5=e^6 is an exact formula
just to clarify, what no_arg said is NOT TRUE

π<sup>4</sup> + π<sup>5</sup> is clearly not equal to e<sup>6</sup> which is why no_arg could not produce any proof for his argument

in maths many things may look like they are correct but we must remember that they are only correct when there is a proper mathematical proof backing them up

there is no mathematical proof that justifies saying π<sup>4</sup> + π<sup>5</sup> = e<sup>6</sup>

EDIT: no_arg probably knows that π<sup>4</sup> + π<sup>5</sup> ≠ e<sup>6</sup>. i dont think he is really stupid enough to believe π<sup>4</sup> + π<sup>5</sup> = e<sup>6</sup>. he probably was saying it to irritate others
 
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no_arg

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in maths many things may look like they are correct but we must remember that they are only correct when there is a proper mathematical proof backing them up



I couldn't agree more!!

So where is the proof that π4 + π5 is not equal to e6 ??

Please don't say that you checked it on a calculator and got different answers somewhere in the tail of the decimal expansion. The previous example:

"For example 1/(sqrt(10001)-100) is exactly equal to sqrt(10001)+100
THis is easily verified by rationalising the denominator.

But when you evalaute on a calculator

1/(sqrt(10001)-100)=200.0050001 and

sqrt(10001)+100=200.0049999"

clearly indicates that identical irrationals can easily display differently on a calculator!

Let's see the implementation of some mathematical skill and the production of a proof!!!
 
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acmilan

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Essentially you have to prove that
,

correct? Or rather that the above is false.
 

Xayma

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no_arg said:
in maths many things may look like they are correct but we must remember that they are only correct when there is a proper mathematical proof backing them up



I couldn't agree more!!

So where is the proof that π4 + π5 is not equal to e6 ??

Please don't say that you checked it on a calculator and got different answers somewhere in the tail of the decimal expansion. The previous example:

"For example 1/(sqrt(10001)-100) is exactly equal to sqrt(10001)+100
THis is easily verified by rationalising the denominator.

But when you evalaute on a calculator

1/(sqrt(10001)-100)=200.0050001 and

sqrt(10001)+100=200.0049999"

clearly indicates that identical irrationals can easily display differently on a calculator!

Let's see the implementation of some mathematical skill and the production of a proof!!!
The error in those overlap meaning they can be equal. You can never prove them equal using a calculator, however, you could prove them unequal (in this case you couldn't because they aren't) using a calculator if you show that they do not lie in the upper bound of the error.

The answers on the calculator only differ by 2.5*10<sup>-7</sup>, you can calculate the error margin and I strongly suspect it would lie in there.

For example using the windows calculator we get answers of

1/(&radic;(10001)-100)=200.00499987500624960940234489433
&radic;(10001)+100=200.00499987500624960940234169938

More accurate but not completly, had these differd by 2 in the 5th decimal place you could say there were not equal because it is outside the error margin.
 
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no_arg

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Pretty scary isn't it.
The two numbers above look totally different on a machine yet they are the same!

Just to make matters worse there is no garantee that answers from a computer are even correct! You can stop fussing about error bounds...the calculations themselves are flawed!!

http://support.intel.com/support/processors/pentium/sb/CS-013005.htm

This is of course why calculation must never a substitute for proof
 
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withoutaface

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no_arg said:
Pretty scary isn't it.
The two numbers above look totally different on a machine yet they are the same!

Just to make matters worse there is no garantee that answers from a computer are even correct!

http://support.intel.com/support/processors/pentium/sb/CS-013005.htm
Work them out by hand then and prove that there is no discrepency up until the 10th decimal place and you may have an argument.
EDIT: And who says Xayma is even using an Intel processor?
 
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