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 Forum index » Diversions » Perplex City Puzzle Cards » PXC: Silver Puzzle Cards
[Puzzle] Silver #238 Riemann
Moderators: AnthraX101, bagsbee, BrianEnigma, cassandra, Giskard, lhall, Mikeyj, myf, poozle, RobMagus, xnbomb
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AngusA
Boot


Joined: 02 Jan 2006
Posts: 44
Location: London, UK

This site has some of the lower values, either as numbers or in relation to pi. http://en.wikipedia.org/wiki/Riemann_zeta_function

PostPosted: Thu Feb 23, 2006 11:10 pm
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Guin
Unfettered


Joined: 11 Jan 2006
Posts: 400
Location: Antartica

So I have had a few spare hours while I cover the office and did a little digging around.

Something about this card that has struck are Errors - and was wondering if this isnt relevant to the answer. I cant test out one of my ideas as my card is at home so I will try that out later (but why go to the trouble to point out the error in Riemann's initials when the card is riddled with grammar and spelling mistakes?)

anyway, Found here [url] http://www.krellinst.org/UCES/archive/classes/CNA/dir2.2/uces2.2.html [/url] is the following:

Quote:
Riemann Sum Error Theorem. If f'(x) is continuous on [a,b], then

Proof. The proof follows from applying (3) with a = xi and i between 1 and n.

Note that in (3) the M1 is over each subinterval, and the M1 in the theorem is over the larger interval [a,b].

Example. Consider the integral

Then f'(x) = -3x2 /(1 + x3 )2. On the interval [0,1] |f'(x)| is bounded by 3, and so we may estimate M1 by 3. Therefore,

|Error| (3(1 - 0)/2)h.

If the error was required to be less than 0.001, then this would force h (2/3).001, or n should be at least (3/2)1000.


Okay means pretty much bugger all to me but if we are looking for errors it could be a start? (okay laugh at the foolish penguin!)

I also wondered if anyone had tried the following (If I get the time before my jolly off down to London I will - but card and code at home GuiN in office Evil or Very Mad )

Quote:
Proof of the Riemann hypothesis is number 8 of Hilbert's problems and number 1 of Smale's problems.
found here http://mathworld.wolfram.com/RiemannHypothesis.html

It feels too easy though but somehow seems relevant. Anyway I am going gaga waiting to set off Sad[/url]
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PostPosted: Fri Feb 24, 2006 12:17 pm
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jpwelton
Greenhorn

Joined: 13 Dec 2005
Posts: 6
Location: Berkshire

I tried the line about Smale and Hilbert, that doesnt work, but the other boxed part, just doesnt seem to make sense - its as though parts are missing from the explanation. eg "Consider the integral ... Then"
I expected there to be something in the ... part!

PostPosted: Fri Feb 24, 2006 2:23 pm
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Guin
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Joined: 11 Jan 2006
Posts: 400
Location: Antartica

the math thing justs makes me want to drink heavily lol means nothing to me - all numbers and gibberish Shocked
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PostPosted: Fri Feb 24, 2006 3:31 pm
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d4nny
Kilroy


Joined: 14 Feb 2006
Posts: 2
Location: Texas , US

possible

ok so heres what i did: I counted and recorded every character in the text that layed on a prime (no punctuations marks, just letters and numbers)... needless to say ALOT. with THAT list i recorded the characters that stood on a prime. i did this about 6 times, until i could not break it down anymore, and this is what i got

on the 5th cycle : romaeere
on the 6th cycle: omer
and finaly : me

* now technically this could be broken down to the letter "e" , but i would say that the proof we need is "me"

maybe a drunk man's theory, but it is backed with some work.... oh and if this crap is right, will someone lend me a card or code or something, cuz i dont even have this card....

PostPosted: Sat Feb 25, 2006 11:11 am
Last edited by d4nny on Mon Jun 09, 2008 1:08 am; edited 2 times in total
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donstobbart
Veteran


Joined: 20 Nov 2005
Posts: 96
Location: Cumbria

I tried that "ME" for you, and ......It didn't work, sorry mate. Sad

Been looking in to Riemann himself, my theory being that there is no question on the card, and after looking into the Riemann Zeta Function, I really don't think it's asking us to prove the function. There are new primes being discovered constantly, and each of these would need to be checked to try to prove or disprove the theory. There is also a school of thought amongst mathematicians that the theory is unprovable.

Also to get the correct answer for the theory accepted by the website, someone must have already done it, and put the answer into Mind Candys site.

So, where from here. To me It seems that the Title of the card is always a reference to the answer (think Blue 155, Sadako Sasaki) so its something to do with Riemann, but I really don't think its the Zeta Function at all. (But PLEASE someone prove me wrong!!!) So I was looking at his other work. I know diddly squat about maths but his most famous work was Riemannian Geometry. Maybe we should be looking in another direction?

Don
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PostPosted: Sat Feb 25, 2006 12:33 pm
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fretty
Decorated

Joined: 19 Nov 2004
Posts: 281
Location: South Yorkshire, England

I think that it is something to do with famous statements in history with proof, e.g. the Fermat statement.

PostPosted: Sat Feb 25, 2006 12:46 pm
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d4nny
Kilroy


Joined: 14 Feb 2006
Posts: 2
Location: Texas , US

thanx for the check don !

guess its back to the drawing board..

PostPosted: Sat Feb 25, 2006 6:39 pm
Last edited by d4nny on Mon Jun 09, 2008 1:07 am; edited 1 time in total
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unrealt87
Boot


Joined: 29 Nov 2005
Posts: 66
Location: Guildford!

Hey peoples Very Happy

Just got back from the london event, damn what a tiring day! Drool

Anyway, i'm not sure if this another red herring, and feel free to trout if you wish, but i got from the big boss himself from MC, that this card we HAVE to solve the function, one way or the other.

So basically this means that if this is true, you will get three thing from solving this card:

1. The generous 60 points on offer
2. A million dollars
3. A lot of angry mathemations after you Razz

Or the dude coulda been telling fibs to put off off the scent.

Ok, i'm done here, let the trouting commence...

PostPosted: Sat Feb 25, 2006 7:40 pm
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donstobbart
Veteran


Joined: 20 Nov 2005
Posts: 96
Location: Cumbria

Ok, so I've already said I know nothing about maths, so.....If there is anyone out there with a brain, could we maybe just need to prove this using mathematical induction. What I have read it seems that you can "prove" something without actually doing it. Like this

Quote:
Theore. For any positive integer n, 1 + 2 + ... + n = n(n+1)/2.

Proof. (Proof by Mathematical Induction) Let's let P(n) be the statement "1 + 2 + ... + n = (n (n+1)/2." (The idea is that P(n) should be an assertion that for any n is verifiably either true or false.) The proof will now proceed in two steps: the initial step and the inductive step.

Initial Step. We must verify that P(1) is True. P(1) asserts "1 = 1(2)/2", which is clearly true. So we are done with the initial step.

Inductive Step. Here we must prove the following assertion: "If there is a k such that P(k) is true, then (for this same k) P(k+1) is true." Thus, we assume there is a k such that 1 + 2 + ... + k = k (k+1)/2. (We call this the inductive assumption.) We must prove, for this same k, the formula 1 + 2 + ... + k + (k+1) = (k+1)(k+2)/2.

This is not too hard: 1 + 2 + ... + k + (k+1) = k(k+1)/2 + (k+1) = (k(k+1) + 2 (k+1))/2 = (k+1)(k+2)/2. The first equality is a consequence of the inductive assumption.



I know I am grasping at straws, but maybe.....
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Hindu Prince Gautama Siddharta


PostPosted: Sat Feb 25, 2006 9:14 pm
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ensign
Kilroy

Joined: 25 Feb 2006
Posts: 2

OK, just got this card and have read everything on this forum and over at perplexcity.

I think this will be a red herring but try everything I say! (btw serious kudos to the guy who microwaved his card. Proof that the line between insanity and genius is very fine indeed.)

I suggest there might be more to the little black dot and the holographic nature of the card. Does someone have a lazer pointer to shine on the card a la:

http://www.holomex.com.mx/eng/seguridad/imagen_oculta.html

I note the holographic paper can be seen through the white of the card in the right light.

A holographic image in the silver margin would fit with allusion to fermat's theorum.

Let me know how you get on.

PostPosted: Sat Feb 25, 2006 9:48 pm
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ensign
Kilroy

Joined: 25 Feb 2006
Posts: 2

Reply to donstobbart:

Only did arts at uni but had inspired maths teacher at school. From dim and rusty memory:

This proof by induction is a basic proof technique that we learned fairly much by rote for easy marks in the exam. I am sure that using it on Reimann is equivalent to trying to assemble a 747 using two sticks and a pilchard (maths majors back me up!).

Either way I think your wasting yout time learning any more than the wikipedia basics on this. Heaps more lateral thinking from everyone.

I suspect that when we find how simple the the solution is we will have a moral imperitive to find the puzzle's author and ritually disembowl him.

Cruel but fair.

PostPosted: Sat Feb 25, 2006 10:14 pm
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locqust
Unfettered


Joined: 14 Oct 2005
Posts: 545
Location: Gloucestershire UK

Going back to the funny gradient that Don found a while back and that me and a few people played with to highlight. My team looked at it last night in the pub and we all agree that something is there but the confusion still reigns on whether this is just printing errors or actually something to look at.

I did use my camera on it again last night with the night vision on, (just in case anybody was wondering what the hell I was doing to it in the pub)
and again nothing, I know its been tried before but, ahh what the hell Smile

My dad did suggest that idea Don, about trying to prove that a prime is a prime by using that equation and comparing to a non-prime also with the equation. But my maths isnt brilliant and to be honest I'm not entirely sure if we could put the math symbols on the answer box!

Riemann came up with many theories over his life, which I'm not going to discuss in full here Smile plenty of other sites that do that for you! But this equation was only one of maybe 30 that he came up with and countless ideas for things like the "ether" and newtonian physics as well as theoretical maths.

I still somehow cant see this card being what it says, which would you take? 60 odd points or a million dollars? Very Happy
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PostPosted: Sun Feb 26, 2006 8:29 am
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CherokeeParkes
Decorated


Joined: 03 Nov 2005
Posts: 236
Location: Belfast, Northern Ireland

As wild tangents go here's a thought that others have menditoned. Like the origami crane could this be something else. I decided to look at the article on Riemann in Wikipedia since answers to other cards can be found on this site.

http://en.wikipedia.org/wiki/Riemann

So here's a link to their article. If there is a hint of what this card is looking for, might it be here?

PostPosted: Sun Feb 26, 2006 5:21 pm
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locqust
Unfettered


Joined: 14 Oct 2005
Posts: 545
Location: Gloucestershire UK

As there is no guest designer mentioned on the card I assume it designed by the academy, which would explain why they chose Riemann, Cherokee's link mentions the "hypercube" (basically a precursor to curved or flat time-space theory that cosmologists work on now)

Ok going back to the gradient thing again (again and again Smile )I believe someone thought it was 333 cubed or tretretretre in the image. Well that does kinda fit with a Riemann theory regarding the hypercube (sorta, my brain is beginning to fry while working this out in my head and remembering my cosmology coursework)

So maybe another path to go down?
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PostPosted: Sun Feb 26, 2006 5:35 pm
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