Wikipedia:Reference desk/Archives/Mathematics/2011 December 5

= December 5 =

Maths Question.
please let me know is integration difficult of differentiation? — Preceding unsigned comment added by ACCA Student (talk • contribs) 07:04, 5 December 2011 (UTC)
 * What? Shadowjams (talk) 07:10, 5 December 2011 (UTC)
 * We are not sure what you were asking, but integration is often considered to be more difficult than differentiation because, on average, for most of the functions you are likely to meet, integration produces a more complicated function than differentiation.   D b f i r s   08:23, 5 December 2011 (UTC)
 * Differentiability is a stronger condition on a function than integrability. But the derivative of every elementary function is elementary and there's a simple algorithm to find its elementary form, but the integral of an elementary function can be nonelementary and it can be hard to find it even if it isn't. -- Meni Rosenfeld (talk) 09:10, 5 December 2011 (UTC)
 * From another more abstract point of view, integration is much easier than differentiation, in that the class of integrable functions is much larger than the class of differentiable functions, and much more stable --e.g. compare the usual convergence theorems for sequences or series, in both cases. --pm a 14:26, 9 December 2011 (UTC)

Logic in quantum theory
The law of no contradiction is disabled in this theory, because it enable P&~P. But it's still need the law ~(P&~P) to derive theorems and make the whole theory consistent. So how could I understand this? TUOYUTSENG (talk) 08:29, 5 December 2011 (UTC)


 * There are a number of different interpretations of quantum mechanics, but I doubt any of them actually allow logical contradictions like that. If you are talking about quantum superposition, then the usual Copenhagen interpretation describes this in a probabilistic way.  That is, if a particle is said to be in a superposition of two states, this simply means that if its state is measured, it will be found to be in one of those states with some probability p, and in the other state with probability 1-p. 81.98.43.107 (talk) 12:50, 5 December 2011 (UTC)


 * Scientific laws aren't proved using logical deduction. They are constructed in an ad hoc way and then tested in experiment. This is different from most mathematical proofs which depend on elementary logical rules and frequently use e.g. proof by contradiction which depends upon the law of the excluded middle. A mathematical law isn't true because it's been shown to be true by a rigorous logical argument, but because it accords with experiments.  (See e.g. Grounds of validity of scientific reasoning, Validity, Falsifiability.)
 * Additionally, quantum physicists understand which parts of quantum theory are deterministic and which are non-deterministic, and as mentioned above they understand the probability distributions for the non-deterministic parts, so they can reason about what will happen, formulate hypotheses, and test them. --Colapeninsula (talk) 10:38, 6 December 2011 (UTC)


 * Hello. I don't really know enough about this subject to answer your questions, but Wikipedia has an article on quantum logic. The Stanford Encyclopedia of Philosophy has an article on "Quantum Logic and Probability Theory." Also, there is a discussion of the relation of logic to the foundations of quantum mechanics in the book Understanding Quantum Mechanics by Roland Omnès. 96.46.201.210 (talk) 04:33, 7 December 2011 (UTC)

Equivalence of two congruences
Can someone show me that 2p ≡ 2 (mod p2) is equivalent to saying 2p &minus; 1 &equiv; 1 (mod p2) or explain how that can be shown? I don't seem to see it right now. Toshio Yamaguchi (talk) 10:10, 5 December 2011 (UTC)
 * If p is not 2, then you can divide both sides by 2. If p is 2, then you can check that the statement is trivially true.--121.74.125.249 (talk) 10:13, 5 December 2011 (UTC)
 * Ah, yes. $${2^p - 2} \equiv {0} \pmod{p^2}$$ and after division by 2 one gets $${2^{p-1} - 1} \equiv {0} \pmod{p^2}$$. Thanks. Toshio Yamaguchi (talk) 10:38, 5 December 2011 (UTC)


 * Is it implied here that p is prime? -- 203.82.66.201 (talk) 00:57, 6 December 2011 (UTC) Lurking WP:RD/MA, trying to learn something.
 * p must be odd, as if it's even and greater than zero then 2p is divisible by 4, as is p2 so neither 2p ≡ 2 (mod p2) nor 2p &minus; 1 &equiv; 1 (mod p2) can be true.-- JohnBlackburne wordsdeeds 02:42, 6 December 2011 (UTC)
 * But if the statements are both false for a given value of p then they are indeed equivalent for that value of p, no ? In other words, there is no value of p, either even or odd, for which one statement is true and the other is false. Gandalf61 (talk) 09:29, 6 December 2011 (UTC)
 * Since these two statements are in fact the same, yes, there is no value of p for which only one of the two statements is true. Toshio Yamaguchi (talk) 10:17, 6 December 2011 (UTC)
 * Well, the statements are not trivially the same because the "division by 2" argument can only be used if p is odd. If p is even then you have to use John Blackburne's argument to show that they are both false and so still equivalent (in terms of truth values). Gandalf61 (talk)