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Topic: Combinations of Pi and Sqrt(2) (Read 21693 times) 

Icarus
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Re: Combinations of Pi and Sqrt(2)
« Reply #50 on: Feb 3^{rd}, 2008, 12:13pm » 
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Yes, please do. SWF has given two examples of numbers that can be expressed as A2 + B, even though it is not at all clear that this is so, much less what the values of A and B would be. He has also given what is a very good algorithm for finding A and B, though it does fall in the "Brute Force" category, and for reasons also discussed in this thread, is not guaranteed to give you a definitive result in any finite amount of time.

« Last Edit: Feb 3^{rd}, 2008, 12:13pm by Icarus » 
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Hippo
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Re: Combinations of Pi and Sqrt(2)
« Reply #51 on: Feb 3^{rd}, 2008, 3:46pm » 
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BTW: It seems to me X=A+B pi decomposition would be of the same complexity as X=A sqrt(2) + B pi decomposition. Am I wrong?

« Last Edit: Feb 3^{rd}, 2008, 3:47pm by Hippo » 
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Icarus
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Re: Combinations of Pi and Sqrt(2)
« Reply #52 on: Feb 3^{rd}, 2008, 5:08pm » 
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Well, certainly X = A + B(/2) is of the same complexity. I don't see why itself would be any simpler.

« Last Edit: Feb 3^{rd}, 2008, 5:08pm by Icarus » 
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Mickey1
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Re: Combinations of Pi and Sqrt(2)
« Reply #53 on: Nov 14^{th}, 2010, 6:25am » 
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If you allow a newbie (as well as a mathematical amateur): If, as I assume, a transcendental number t added to an integer will also be transcendeltal, and the same is true for algebraic numbers, there will be no welldefined algebraic and transcendental part of the sum. That would take away any approach based on these fundamental principles leading me to assume that other approaches might have some component of what you might call brute force.


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SWF
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The equations I posted on the first page have disappeared, so I recreated them here and added a new one. I like the new one best because the right hand side contains only rational numbers, while the others have either pi or sqrt(2) on the right hand side. Remember, the intent here is to fully define expressions for which the correct values of A and B are not immediately obvious.


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Mickey1
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Re: Combinations of Pi and Sqrt(2)
« Reply #55 on: Feb 18^{th}, 2012, 3:21am » 
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1 How can X be given at all, being transcendental? The only way seems to be if X is known in an indirect way such as for the natural logarithm base e (or simply given by a certain combination of A and B which would make the problem very easy to solve). 2 How about the “other than by brute force” statement, indicating that such a method would be available? The right hand side part is countable, =RHS(N), and the solution is unique since (A1A2)pi+ (B1B2)sqrt(2) =0 implies the contradiction that sqrt(2) would be a transcendental number. But how are we to identify that the X=RHS(N) in a finite number of steps, assuming we were given X in a similar way as e above (and that X was indeed =RHS)? It would lead to a Goedel type statement that X must be =RHS(N) for some N but we cannot prove it (i.e. verify the solution).

« Last Edit: Feb 22^{nd}, 2012, 3:11am by Mickey1 » 
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Mickey1
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Re: Combinations of Pi and Sqrt(2)
« Reply #56 on: Feb 29^{th}, 2012, 12:30am » 
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Let me have another go at a ”solution”, in three steps: 1) X = 5*sqrt(2) + 3*Pi. It seems that A=5, B=3 is a (unique) solution. 2) We assume that the number is given in by an infinite decimal representation. I now attack the problem in a way similar to, or rather inspired by, Cantor’s diagonal argument: First, the digital part of X is considered (assuming it exists). It is clear to the calculation experts of the forum that infinitely many combinations exist, e.g. combining a large positive A with a large negative B, which would have the same digital part. I now multiply both side of the equation with 10 and repeat the argument, to show that the digital plus the first decimal have the same property (allowing infinitely many combinations which all have an identical digital and the first decimal part). The process can continue to infinity since, if it stopped, (I suggest without proof that) it would imply that there was a “tail” of decimals from A*pi (TApi) and B*sqrt(2) (TBs) and rational numbers r1 and r2, so that TApi=r1*TBs, implying in turn the same relation between the two original right hand side terms: A*pi = r2* B*sqrt(2), a contradiction since pi would have to be algebraic and sqrt(2) simultaneously must be transcendental. 3) The contradiction between 1 and 2 underlines the problem of the formulation of the question. According to 2), there can be no solution at all, not even for the obvious case, 1).


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SWF
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Re: Combinations of Pi and Sqrt(2)
« Reply #57 on: Mar 6^{th}, 2012, 7:23pm » 
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on Feb 18^{th}, 2012, 3:21am, Mickey1 wrote:1 How can X be given at all, being transcendental? The only way seems to be if X is known in an indirect way such as for the natural logarithm base e (or simply given by a certain combination of A and B which would make the problem very easy to solve). 
 Have you read this thread? There are 3 examples given immediately before your post. Obviously, listing all the digits is not the way to do it, but it can be defined mathematically, for example as an infinite series, an integral, or as the solution to transcendental equations. Quote: 2 How about the “other than by brute force” statement, indicating that such a method would be available? 
 A method is given on page 2 of this thread. Knowing that X is expressible in the desired form, another way of dealing with X is to have the digits revealed a few at a time. For each step of the algorithm, a best estimate based on the given digits is obtained. If that isn't the correct answer, the most significant digit of the error between between X and the estimate could be provided so another estimate can be made accounting for the additional information. Eventually that error would be zero.


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Mickey1
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Re: Combinations of Pi and Sqrt(2)
« Reply #58 on: Mar 14^{th}, 2012, 12:15pm » 
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I now have read the threads more carefully which doesn’t mean that I understood them all. I was first confused by the “Given X” passage since X cannot be given in a finite form (except indirectly). I then, later, interpreted the question to refer to a situation like: if X= 5*sqrt(2) + 3*Pi then 2X= 10*sqrt(2) + 6*Pi. Also, we accept squaring both the left and right hand side etc. for the same reason ( even we if still do not accept X given in infinite terms). The reason we would accept this, is not based on accepting infinitely long numbers, but based on the Peano axioms (assuming they can transported without too many contradictions through the Dedekind cut and apply to real numbers). Now, take these following statements as questions from an interested amateur: 1 As I interpret your stepwise process, it can identify one (perhaps the lowest?) possible A and B within a bounded environment of X. Since the right hand side gives a countable set, you could disregard the first answer coming up if you detected a difference at all between the given X and the right hand side of the equation. This process might take you through all possible values and sooner or later to the right one. (A more cumbersome approach would be to test all A and B values). 2. The verification process would still require infinite steps. Even if X was given in a finite form, the suggested solution would be infinite. 3. If one doesn’t accept the infinite verification process, the alternative is that X is given “as an infinite series, an integral, or as the solution to transcendental equations”, i.e. in a finite form. 4. In this case the solution would depend on (mathematical) pattern recognition, and we might not need a algorithm or any other approach managing the equation’s right hand side. 5 (This I believe has not been mentioned by the other thread authors, perhaps to their credit?): The possible forms of finite forms of infinite series etc are not only infinite, but I believe them to be similar to “all subsets of the natural numbers” i.e. uncountable. To any series can be added 11 or 1 – (1/2 + 1/4…) etc. I therefore launch the conjecture that no systematic approach of formula recognition can be sufficient to supply the solution.


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SWF
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Re: Combinations of Pi and Sqrt(2)
« Reply #59 on: Mar 14^{th}, 2012, 7:56pm » 
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Mickey1, I think many of the arguments you are making involve common characteristics of transcendental numbers or even irrationals, not just A*pi+B*sqrt(2). You can't specify such a number by writing out all of its decimal digits. on Mar 14^{th}, 2012, 12:15pm, Mickey1 wrote: 2. The verification process would still require infinite steps. Even if X was given in a finite form, the suggested solution would be infinite. 
 The way I think of this problem is that the person who provided X knows what A and B are, and challenges you to find A and B given X (or an approximation to X). The verification process involves checking the values of A and B, not the decimal expansion of X. Quote:3. If one doesn’t accept the infinite verification process, the alternative is that X is given “as an infinite series, an integral, or as the solution to transcendental equations”, i.e. in a finite form. 
 Do you mean "infinite form"? Quote:5.... I therefore launch the conjecture that no systematic approach of formula recognition can be sufficient to supply the solution. 
 This is true of expressions for integers too. Part of the discussion here has been how to fully express an X whose A and B are not immediately obvious. So trying to disguise expressions is part of the game. How about this one (does it count as a finite form?): X = 2*Tan^{1}( cot( exp(sinh^{1}( sqrt(17) ))  sqrt(17) ) )


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Mickey1
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Re: Combinations of Pi and Sqrt(2)
« Reply #60 on: Mar 15^{th}, 2012, 2:26pm » 
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3. "finite form"? No, I mean finite form i.e. using a finite number of words in a fixed language, hence the need  and possibility  to use recognition techniques. And you are right in that I am mainly looking at the general problem rather than the given one. I did note that the equation would be (or at least look) simpler if both sides were taken as exp(iX)=exp(i(A*pi +B*sqrt(2)), but I think that point was made early on. Perhaps new ideas can emerge about the actual problem but I actually doubt it. In any case it would probably not come from me.


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