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

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June 5

a simple ODE

I have a question about solving a simple linear ODE

${\dot {Y}}=\alpha (v(Y)-s(Y))$

where $\alpha >0$ and v(Y) and s(Y) are linear functions of Y that intersect at one point.

Taking the simplest example

${\dot {Y}}=\alpha (v-s)Y$

gives a solution of $Y(t)=e^{\alpha (v-s)t}$ which will go to infinity if $v>s$ and zero (which is the point of intersection) if $v<s$.

What if zero is not the point of intersection, so that $v(Y)=v_{0}+vY$ and $s(Y)=s_{0}+sY$?

Then the point of intersection is at $Y={\frac {s_{0}-v_{0}}{v-s}}$ and the ODE is

${\dot {Y}}=\alpha (v_{0}-s_{0}+(v-s)Y)$

How do I show the solution to the ODE in this case? I want to be able to show that depending on the parameters of the linear functions, the solution will tend to a stable equilibrium at that point of intersection. — Preceding unsigned comment added by 130.102.78.164 (talk) 00:13, 5 June 2011 (UTC)

This is a case of Linear differential equation#Nonhomogeneous equation with constant coefficients but in this case you can find a particular solution by assuming Y is constant. The general solution is then found by adding solutions to the homogeneous case and you've already done that.--RDBury (talk) 06:34, 5 June 2011 (UTC)

what if any statement implied its opposite?

what if there were a proof that didn't require any axioms (was true for all systems), that from any statement, its opposite followed? --86.8.139.65 (talk) 18:56, 5 June 2011 (UTC)

That would mean that the rules of inference you are using are not a very good choice. --Tango (talk) 19:27, 5 June 2011 (UTC)

What do you mean by "opposite"? What's the "opposite" of 2 + 3 = 5? Michael Hardy (talk) 20:17, 5 June 2011 (UTC)

In this case I would say the opposite is "two plus three very much does not equal five". I don't know how to mark that up. 86.8.139.65 (talk) 20:36, 5 June 2011 (UTC)

Is this like the Liar paradox? Grandiose (me, talk, contribs) 20:40, 5 June 2011 (UTC)

Then that would also apply to the statement that says that: "from any statement, its opposite follows". Count Iblis (talk) 22:34, 5 June 2011 (UTC)

Ex falso quodlibet 83.134.160.239 (talk) 09:42, 7 June 2011 (UTC)

I think that this may be different. The OP was asking about the possibility of there existing a $\phi$ such that $\phi \to \neg \phi$. The link you mention is about something different. Ex falso quodlibet is about being able to imply anything from an assumed contradiction, e.g. $(\phi \land \neg \phi )\to \psi$. Although if there exists a φ such that $\phi \to \neg \phi$ then we would have $\phi \land \neg \phi$ which would then imply any statement you could imagine by ex falso quodlibet. — Fly by Night (talk) 15:40, 7 June 2011 (UTC)

For many statements there's no problem with them implying their opposite. If we then showed the opposite implied the first thing again you start having problems. For instance if we showed that ants are three meters tall implied ants are not three meters tall then there's no problem. Dmcq (talk) 20:50, 7 June 2011 (UTC)

Could you give some examples? If (all) ants are three meters tall then then "(all) ants are not three meters tall" is clearly false, so you have $\phi \not \to \neg \phi$. They all cause problems: "if A is true then A is false" is what the OP was asking about. — Fly by Night (talk) 21:13, 7 June 2011 (UTC)

For instance, let $\phi$ be the proposition $(\psi \land \neg \psi )$. By ex falso quodlibet, $\phi \to \neg \phi$.

Conversely, if φ were any proposition such that $\phi \to \neg \phi$ then by conditional exchange, we can conclude $\neg \phi$. Sławomir Biały (talk) 22:10, 7 June 2011 (UTC)

Well actually in that last bit you can't conclude $\neg \phi$ as ψ may be false and false implies false is true. Dmcq (talk) 22:20, 7 June 2011 (UTC)

Sorry, the psi should have been a phi. I've corrected it. Sławomir Biały (talk) 22:33, 7 June 2011 (UTC)

Niels Bohr is quoted for saying: "The negation of a truth is a falsehood, but the negation of a deep truth is another deep truth". I don't know whether this is a truth, a falsehood, or a deep truth. Bo Jacoby (talk) 09:01, 8 June 2011 (UTC).

Tango gave the correct answer as the first post. If you can prove that no consistent system exists then you should restrict the rules of inference. Taemyr (talk) 11:41, 8 June 2011 (UTC)

As I showed above, there is nothing inconsistent about the existence of a proposition $\phi$ for which $\phi \to \neg \phi$. In fact, it's easy to construct such a proposition in standard propositional logic. You might take issue with the rules of inference of propositional calculus (many logicians have), but it's not for want of consistency. Sławomir Biały (talk) 11:54, 8 June 2011 (UTC)

Indeed. We have ${\text{False}}\to P$ for any proposition P because $\neg ({\text{False}}\land \neg P)$. In particular, we therefore have ${\text{False}}\to \neg {\text{False}}$.Gandalf61 (talk) 12:55, 8 June 2011 (UTC)

That's what I meant when I asked for an example. I can't think of a statement that implies its negation. The ant example wasn't quite right. I think that a correct "if A then not A" is inconsistent. If A implies not A then both A and not A are true. That's a contradiction. If we could find an A such that A implies not A then we could imply any statement (by ex falso quodlibet). — Fly by Night (talk) 21:00, 8 June 2011 (UTC)

"If 2 is odd, then 2 is even" is a true statement, using the standard mathematical meaning of "if…then". Since the premise is false, the statement as a whole is true, regardless of whether the conclusion is true or false. See Material conditional. —Bkell (talk) 21:26, 8 June 2011 (UTC)

Heh, whaddya know, "if 2 is odd then 2 is even" is used verbatim in that article as an example. :-) —Bkell (talk)

I think we've got our wires crossed somewhere. "if 2 is odd then 2 is even" is vacuous because 2 is not odd. (Notice that it's equivalent to its own contrapositive.) The point is when you have "if A then not A" holding while at the same time A is true. That's when the problems start. Otherwise, like I said, it's vacuous. — Fly by Night (talk) 21:36, 8 June 2011 (UTC)

Certainly, a true statement of the form "if A then not A" can cause a contradiction, but only if A is true. Above you said you couldn't think of a statement that implies its negation, so I gave you an example of one that does, namely, the statement "2 is odd". You also said, "If A implies not A then both A and not A are true"—that holds only if you additionally assume that A is true. Alone, the statement "A implies not A", even if true, does not inherently cause a contradiction, because it can be the case that A is false. —Bkell (talk) 21:54, 8 June 2011 (UTC)

But that's my hang-up. The statement "2 is odd" does not imply that "2 is even". The statement "2 is odd" is false, so a statement of the form "if 2 is odd then B" does not imply anything. It tells us nothing about the validity of B. Think about it: "if 2 is odd"… well it's not… so I stop. I agree with the last point, but that just says that a vacuous statement does not imply a contradiction. So yes: there are many statements of the form "if A then not A" that do not imply a contradiction; but that's because they are either false or they do not imply anything (i.e. are vacuous). — Fly by Night (talk) 22:48, 8 June 2011 (UTC)

From a strictly logical point of view, vacuous truth is still truth, and "2 is odd" does imply "2 is even", because the statement "if 2 is odd, then 2 is even" is true (vacuously). Your difficulties seem to be based on what you think the English construction "if…then" should mean (in ordinary language, it usually indicates cause and effect) and the idea that a logical implication of the form "A implies B" should tell you something about B (it doesn't, if A is false). Perhaps you should read Material conditional#Philosophical problems with material conditional, which addresses some of these issues. —Bkell (talk) 23:31, 8 June 2011 (UTC)

If you don't like this treatment of implication and vacuous truth, there are different logical systems that may appeal better to your intuition, like relevance logic. These are not the standard logical system used in most mathematics, however. —Bkell (talk) 23:45, 8 June 2011 (UTC)

It might happen that a universally quantified statement is a true statement about the mathematical world, but that when it is actually instantiated, it produces a vacuous implication at certain values of the quantified variable. For instance,

$\forall x((x{\text{is even}})\to \neg (2x^{2}-1{\text{is even}})).$

is clearly a true sentence. But instantiating at $x=1$ gives

$(1{\text{is even}})\to \neg (1{\text{is even}}).$

-- Sławomir Biały (talk) 01:26, 9 June 2011 (UTC)

I don't have a "difficulty". Just because I don't agree with your point of view does not mean that I have a difficulty understanding. — Fly by Night (talk) 11:11, 9 June 2011 (UTC)

I'm sorry, I didn't mean to imply that you had difficulty understanding. My use of the word "difficulty" was in response to your use of the word "hang-up", which, according to wikt:hang-up, means "an emotional difficulty or a psychological inhibition". I interpreted that to mean that you were having difficulty reconciling mathematical ideas with your intuition, which is common for everyone from time to time. —Bkell (talk) 11:29, 9 June 2011 (UTC)

Ha ha ha :-) You're not far off there. I do have many emotional difficulties, and psychological inhibitions; but I try to carry on regardless. The medication helps a lot too. — Fly by Night (talk) 14:18, 9 June 2011 (UTC)

To Slavomir. OP did not ask for the existence of a statement that implies it's opposite. He asked for the implications of a proof that *any* statement implies it's opposite. So not only do you have ${\text{False}}\to \neg {\text{False}}$, which is not a problem, you also have $\neg {\text{False}}\to \neg \neg {\text{False}}$, which is a bit more of a problem.Taemyr (talk) 08:45, 10 June 2011 (UTC)

How to average ranks?

As an example, there are several listings of the best 100 (or whatever) novels of the 20th. century. A particular novel is not going to have the same rank in every list. To get the "average" rank of each novel, is there any better method than merely taking the mean average of its ranks? Thanks 92.28.240.238 (talk) 20:24, 5 June 2011 (UTC)

Average rank is not a very good way of doing it unfortunately. The main problems are that firstly it is very bad at dealing with incomplete orders and secondly in general the rank can vary more in the middle rather than the beginning and end so it is not combining things with similar ranges. I tried doing a quick search for merging rank order and combining rank order but didn't find the one I was looking for which is iterative and assigns a real number value rather than an integer order. Other ways of doing it involve setting up a matrix of wins/losses and getting the overall order from that. Dmcq (talk) 23:59, 5 June 2011 (UTC)

Instead of attempting to order them, you can group them. For example, assume you have 200 books that appear in 10 different top 100 listings. You can place that data in a matrix and use singular value decomposition on the matrix. Then (assuming the books are on the left vertical column of your matrix) the first table will have 5 columns - one per book. Replace all negative values with 0 and all positive values with 1. Each book will have an identifier like 01101. All books with the same identifier are in the same group. With little work, you can sort the groups into the most popular, somewhat popular, and less popular groups. Further, SVD allows you to fill in a rather good estimate for missing data. So, if a book doesn't appear on a particular top 100 list, you can estimate where it is on the list: 101, 110, 250... -- kainaw™ 00:06, 6 June 2011 (UTC)

Just looking again at the type of stuff you want to order I think you can probably get a quick and fairly good order by averaging the logarithm of the order including something like 1+log(100) for all the missing ones. Rather empirical but far better than just averaging the order without getting the log. Dmcq (talk) 00:27, 6 June 2011 (UTC)

If I were doing this, I would probably just use the median of the reported ranks. The main problem with using the mean is that it can be badly affected by outliers -- a single ranking that widely differs from the others can significantly change the result. The median does not suffer from that problem, and is simple to compute. Looie496 (talk) 00:42, 6 June 2011 (UTC)

Isnt there any maths/stats theory which would indicate the best method to use? 2.97.212.124 (talk) 12:17, 6 June 2011 (UTC)

Well there's lots of studies of different algorithms which are based on different assumptions! I found something mentioning what I was looking for above in http://deepblue.lib.umich.edu/bitstream/2027.42/66929/2/10.1177_001316447303300104.pdf , I'm sure there's much better now with people using Monte-Carlo methods since they've oodles of performance to waste on computers. Dmcq (talk) 13:41, 6 June 2011 (UTC)

There is a hell of a lot of information on this topic. You are dealing with three main issues: synonymy, polysemy, and sparsity. In your specific case, synonymy and polysemy are not too bad, but there is the issue of how books are listed. For example, what if one list has both the hardcover and softcover version of a book listed? What if one list places an entire series as a single book while another list breaks each independent book out? Your main issue is sparsity. Many books will not be on all lists. So, assume you use median. A book is, for some reason, on one and only one list at rank 25. It didn't even make the top 100 list for any of the lists but one. You end up ranking it 25 because that is the book's median rank. You need to figure out how to handle sparsity such that it doesn't inhibit the result you want to get. -- kainaw™ 14:16, 6 June 2011 (UTC)

Before you get hat far you've of course got to ask yourself how the various lists of novels or whatever were compiled. Were they done by a wide poll or seeing library returns or by a journalist asking a few friends? Was it done by some teenagers or some mothers group? So you have the problems that some might have much more weight than others and some might be much more compatible with your purposes than others. Dmcq (talk) 15:56, 6 June 2011 (UTC)

This is pretty similar to the questions investigated in social choice theory—you're trying to take several rankings and produce one overall ranking. This is difficult to do well. For example, you would probably like the method you use to have nice properties, such as these:

• If every list ranks The Grapes of Wrath higher than A Clockwork Orange, then the overall ranking should also rank The Grapes of Wrath higher than A Clockwork Orange.
• If the various lists were different, but the relative rankings of The Grapes of Wrath and A Clockwork Orange do not change (in other words, every list that previously ranked The Grapes of Wrath higher than A Clockwork Orange still does so, and vice versa), then the relative rankings of The Grapes of Wrath and A Clockwork Orange should remain unchanged in the overall ranking.
• No one list determines the overall ranking; every list is taken into account.

Unfortunately, it is impossible to satisfy all of these criteria simultaneously; see Arrow's impossibility theorem. —Bkell (talk) 16:49, 6 June 2011 (UTC)

Yep there's no sound mathematical method with even very simple assumptions. It's up to people to make choices on what they find works for them. You just have to make an admission like Google who now say their page ranks reflect their judgement of what will be most relevant for a user's query rather than just that it is determined by an algorithm. Dmcq (talk) 10:04, 7 June 2011 (UTC)

Converting between the forms of equations for an ellipse

I know that the standard form for an ellipse is (x-h)^2/a^2+(y-k)^2/b^2=1 or (x-h)^2/b^2+(y-k)^2/a^2=1 depending on the major axis and the general form is ax^2+cy^2+dx+ey=f. I want to be able to convert between these two forms with a simple set of equations. Can anyone help me? --Melab±1 ☎ 21:01, 5 June 2011 (UTC)

It doesn't help that you've used the constant "a" in both forms, making it impossible to get an exact correspondence. If you take the general form as cx^2+dy^2+ex+fy=g, just expand the first form and match coefficients of x^2, y^2, x, y and the constant term.→86.132.165.117 (talk) 21:40, 5 June 2011 (UTC)

To go from the former to latter, just expand as 86.132 suggests. To go the other way, you complete the square for each variable. SemanticMantis (talk) 20:35, 6 June 2011 (UTC)

• An ellipse in the plane is given by F = ax² + by² + 2hxy + 2gx + 2fy + c = 0 where a, b, c, f, g and h are real numbers and h² – ac < 0. (If h² – ac = 0 you have a parabola and if h² – ac > 0 you have a hyperbola.) To calculate the centre of the ellipse (or more generally any conic) you have to solve ∂F/∂x = ∂F/∂y = 0 with respect to x and y. Once you know the centre, say x = p and y = q, you translate the centre to the origin by a substitution x = x + p and y = y + q. After that, you'll see that F(x+p,y+q) = Ax² + 2Bxy + Cy² + D, where A, B, C and D are real numbers (in fact they are functions of the old a, b, c, f, g and h). To calculate the axes of the conic you need to consider the matrix

$\left[{\begin{array}{cc}A&B\\B&C\end{array}}\right],$

which is the matrix of the quadratic form Ax² + 2Bxy + Cy². The eigenvectors of that matrix give the axes of the conic. Finally, you need to calculate an orthogonal change of basis matrix that takes the axes of the conic onto the coordinate axes. It needs to be orthogonal so that you don't stretch the conic in anyway. This is just a rotation about the origin. Then your conic will be in a normal form, e.g. (x/α)² + (y/β)² = 1. I know it might sound complicated, but all that we have done is slide the conic and then rotate it to put it into normal form. — Fly by Night (talk) 18:13, 9 June 2011 (UTC)