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March 31

A maximal theory for solving physical problems

Hi everyone

This is a problem that I've wondered about and tried to solve for a very long time, but it's a bit abstract and difficult to explain. I'll start with the motivation for solving it - In the past I tried to find some general algorithm for solving all problems in mathematics. Ultimately I concluded that no such algorithm exists; in fact, I discovered Entscheidungsproblem (which I think can be re-framed into exactly this) and the proofs that it was not possible to solve Entscheidungsproblem. Giving up on that, I then considered if there was a "best possible approximation" for such an algorithm. I formally framed it like this: Does there exist an algorithm q such that for all algorithms p and problems x we have "p can solve x" implies "q can solve x". I'm still a bit unsure about this, but again I believe the answer is "no". Given any algorithm which can solve a subset of all problems, you can create a new improved algorithm from it just by adding the answer to a problem that it can't solve to the algorithm. You can limit the problems to ones for which there exist correctness proofs, which Marcus Hutter has done. However, for any particular problem, this algorithm takes over 5 times as long to solve as the most efficient algorithm for that problem, so it can again be improved by replacing it with the most efficient algorithm for this problem.

Can anything like this be done when limited to physical or synthetic problems? Widener (talk) 05:36, 31 March 2012 (UTC)[reply]

Can the "algorithm" be a meta-algorithm which chooses the correct algorithm for each problem ? StuRat (talk) 05:41, 31 March 2012 (UTC)[reply]

I don't see why not. Widener (talk) 05:45, 31 March 2012 (UTC)[reply]

In that case, I'd say sure, we can make a meta-algorithm which will search through all the possible solution algorithms and find the appropriate one for each math problem. This would be quite complex, as there are many ways to solve a math problem, and it isn't always obvious which one is correct. In some cases it may be necessary to try multiple methods to find the one that works.

Are you sure this is possible? Is the set of all the possible solution algorithms recursively enumerable (or even countable)? Widener (talk) 14:20, 31 March 2012 (UTC)[reply]

In fact, what you described looks very similar to Marcus Hutter's algorithm. Widener (talk) 14:22, 31 March 2012 (UTC)[reply]

Recognizing what type of problem you have would fall into the artificial intelligence area (especially if it starts as a story problem). The computer that could do this would be something like Watson. It might also need to ask follow up questions, like whether we want imaginary number solutions. StuRat (talk) 14:04, 31 March 2012 (UTC)[reply]

That hutter1.net link seems to be broken. 81.98.43.107 (talk) 11:29, 31 March 2012 (UTC)[reply]

Fixed it. Widener (talk) 13:17, 31 March 2012 (UTC)[reply]

Apart from infinite memory, one can make a physical model of a turing machine or equivalent, so couldn't Gödel or Entscheidungsproblem be translated into a physical problem, leading to the same conclusion? Ssscienccce (84.197.178.75) (talk) 13:43, 1 April 2012 (UTC)[reply]

That is a good point; it reminds me of Stephen Hawking's criticism of a theory of everything. Memory may be significant though. Widener (talk) 05:43, 2 April 2012 (UTC)[reply]

Differential Equations

Given the location of the (in my case, two, but I'd be interested in the general case as well) regular singular points of a second order linear ordinary differential equation, how do you go about constructing the differential equations with said singularities? In my particular example, I attempted to use the form of the second order linear ODE with three regular singular points and take the exponents (i.e. the roots of the associated indicial equation) as zero and 'work down' to two singular points, which happen to be zero and infinity, but then I realised that the equation I was given purposefully excludes consideration of the point at infinity. Can anyone help me? Thanks. 131.111.216.115 (talk) 15:56, 31 March 2012 (UTC)[reply]

According to our article Regular singular point and Wolfram, the regular singularities are give by the functional coefficients having simple poles at the at the required points. If you want singularities as p₁, …, p_n then why not try

{\frac {\operatorname {d} ^{2}y}{\operatorname {d} \!x^{2}}}+\left(\prod _{i=1}^{n}(x-p_{i})^{-1}\right){\frac {\operatorname {d} \!y}{\operatorname {d} \!x}}+y=0\,?

— Fly by Night (talk) 21:45, 31 March 2012 (UTC)[reply]

Thank you for your reply but, perhaps before addressing it, I should make clear precisely what I am attempting to do. I have to find all differential equations of the form

w''(z)+p(z)w'(z)+q(z)w(z)=0

such that zero and the point at infinity are regular singular points of the equation, meaning that p(z) and q(z) have, at most, a simple pole and a double pole respectively at these points, while every other point is an ordinary point. Now, to your answer: I can see how this would be helpful, after setting n=2, but it will only work for one case, e.g. we are assuming here that q(z) is analytic everywhere, which will not be true in general. Is this the best method or could it be done more efficiently? Thanks. 131.111.216.115 (talk) 09:35, 1 April 2012 (UTC)[reply]

Highly leveraged small financial bets as an alternative to lottery tickets

I'm sure this issue has been addressed before. I have a Contract for difference account which I have yet to ever use. Can someone point me in the direction of any articles / information on using such accounts to make small, highly leveraged bets, instead of traditional lottery tickets (and the relative mathematics and odds involved)? — Preceding unsigned comment added by 58.111.224.202 (talk) 23:40, 31 March 2012 (UTC)[reply]

Leveraged products such as CFDs appear attractive because they offer the potential for large gains on a small initial deposit. However, they also carry the risk of large losses. With a lottery ticket your maximum loss is limited to the price of the tickets that you buy. But with a CFD your maximum potential loss is very large indeed, and can certainly be many times your initial deposit. Another difference is that a lottery ticket is a "fire and forget" purchase, whereas a CFD account needs constant attention, as you must be continually deciding whether to close out your positions or roll them over. So two questions to ask yourself before you start trading CFDs are (i) can I afford large losses and (ii) can I afford the time required to manage my positions. Gandalf61 (talk) 01:05, 1 April 2012 (UTC)[reply]

(From the OP, on a different computer) Aren't those two problems solved simply by setting a Stop price on your position? That being a given, how else do the two compare, favourably or otherwise? 114.78.181.219 (talk) 01:59, 1 April 2012 (UTC)[reply]

Average return on investment in lottery tickets is almost always negative (I'm not sure of any case where it's actually in your interest to buy a lottery ticket, but theoretically, it could happen). Average return on investment in any stock fund is almost always positive (with the occasional exception, like if Bernie Madoff with your money). You might want to consider penny stocks, as they offer the possibility of high returns but with a minimal investment, much like the lottery. StuRat (talk) 01:09, 2 April 2012 (UTC)[reply]

The return can be positive if a lottery produces no winner and the prize money rolls over -- especially if that happens repeatedly, as with the MegaMillions lottery that was recently covered in the US news media. In that case, a $1 investment produced an expected return of around $2 -- although the chances of winning were estimated at around 1 in 176 million. Looie496 (talk) 22:49, 2 April 2012 (UTC)[reply]

Can you provide a source for those calcs ? I just did them myself, and figured it had a (negative) return of around 50 cents on a dollar. Did they count taxes, inflation for the installment plan, and having to split the prize with the other winners ? StuRat (talk) 23:42, 2 April 2012 (UTC)[reply]

It's well-known that draws can be positive in certain situations, but it would take many lifetimes to win the prize. Anyways, this isn't my question. 23:56, 2 April 2012 (UTC) — Preceding unsigned comment added by 58.111.224.202 (talk)