User talk:Kevin Baas/Archive1

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- For example, x3 + x, at x = 2, equals 10; the inverse function equals 2, when x = 10. If one tries to find the derivative of f(x)&minus;1, at x = 10, one might note that f ' &minus;1(10) = 1 / f ' [f&minus;1(10)] = 1 / f ' (2); and since f ' is 3x2 + 1; then, f ' &minus;1(10) = 1 / [3(2)2 + 1] = 1 / 13. Indeed, f ' (2) = 13; thus f ' &minus;1 should be 1 / 13.

Is there something wrong with this? Pizza Puzzle

I didn't ask if there was something wrong with the other edits. I asked if there was something wrong with the above paragraph. Pizza Puzzle

Ahahhaa- just as long as you recognize el presidente for what he is, we should get along just fine. Pizza Puzzle

Hi there! I love your presentation on fractional calculus. Could you give me some comments on topology, group theory and ring theory?

There is something about Wikipedian mathematicians that worries me. It seems they are overconcentrated on the mathematical contents that they simply ignore the related history. By suggest a standard format, maybe we can "correct" this bias.

I suggest we can start a WikiProject on "Styles of Mathematics Articles". What do you think? Wshun

Yes, I have the same feeling. Someone may move our discussion to meta page soon, I suppose. :) wshun 20:35, 4 Aug 2003 (UTC)

Hi Kevin. Good to see some engagement with my recent addition to the Logistic map article. However, I stand by my original text. I think that it is in fact true, but I take what may be your point -- it is not entirely clear on the whole story.

What I had written was meant to say that if we had perfect knowledge of the system and its initial state, then in principle we could know every future state -- I mean this in precisely the sense that it is deterministic. Of course, if we were measuring a system or using a digital computer to calculate the future states, then we would have imperfect knowledge or we would fail in practice to accurately calculate future states. But, perfect is, well, perfect, and immeasurability has nothing to do with it.

I'll make the change to what I meant, and try to incorporate your edit as well to make it clearer. Ben Cairns 02:27, 10 Dec 2003 (UTC)

In response to your reply on my talk page, I think we agree philosophically about the impossibility of obtaining perfect knowledge about something, but that does not make the concept of perfect knowledge meaningless. I think that the word "perfect" as I used it would be understood by most of the Wikipedia's readers as meaning what I intended it to mean, and moreover the contrasting point about the impossibility of perfect knowledge in practice has been made in the article. However, in response to your comments, I'll have a go at defining "perfect" more clearly. --Ben Cairns 23:54, 10 Dec 2003 (UTC)

Thanks again for your comments. You have an interesting view on 'perfect' -- and I'm happy we can agree to disagree. Ben Cairns 22:18, 11 Dec 2003 (UTC)

Do you honestly think that "physics does not admit real variables?" Then tell me why all the laws of physics are stated with calculus. Tell me further how we can have dimensions if the universe only contains the set of computable numbers, which has a dimension less than one. Tell me further how fractals are at all possible. Tell me further why a turing machine can never be trully artificially intelligent, and yet we are intelligent. Tell me further how Kurt G&ouml;del's incompleteness theorem proved that no formal symbolic system is complete, and that, as you suppose, using only the set of computable numbers, and thus a formal symbolic system, we can have a complete description of the universe. -Kevin Baas 2003.12.20


 * 1) I didn't write "physics does not admit real variables."
 * 2) If you are complaining about the hypercomputation article, it's enough that we can't measure real variables to arbitrary accuracy for analog computers to reduce to Turing machines.
 * 3) I would be very interested to see a proof that a Turing machine cannot be artificially intelligent.
 * 4) I think you are grossly misunderstanding the real-world consequences of G&ouml;del's First Incompleteness Theorem.  We don't have a complete (in the logical sense) symbolic description of the universe now&mdash;why would it be surprising that we can never have one?
 * 5) Regarding the Super-Turing computation redirect: there still doesn't seem to be a clear enough distinction between Super-Turing computation and the previous hypercomputation article to justify two separate articles.  Perhaps you could concentrate on improving the hypercomputation article rather than making your own parallel article.

Populus 14:05, 20 Dec 2003 (UTC)


 * Hi Kevin, I put that claim in, and I'll defend it. Real numbers form a field, as do computable numbers.  As far as I am aware, it is an open question as to whether all of the physics, which physicists habitually model using the reals, can in fact be modelled with computable numbers alone.  In that case, one would say that physics does not admit uncomputable reals (and thus, it does not admit real variables in general). -- pde 08:05, 23 Dec 2003 (UTC)

Two responses, Kevin. Firstly, I never said that physics does not admit real variables. I was just defending the conditional, which I placed on the hypercomputation page, "if physics admits real variables". You were implying that it is absurd to counterance the fact that it doesn't.

Second, if we continue to have no resolution to the question &mdash; if none of our best physical theories require uncomputable reals, then Occam's razor might actually suggest that it is wise to work with the hypothesis that they are non-physical. Occam's razor might be the kind of logical operator you're looking for. But note, I don't know enough about the research on this point to hold an opinion either way. -- pde 12:44, 23 Dec 2003 (UTC)

Kevin Bass wrote:

''Okay, I'm supposing you read some paper by some young student that spoke of this "the universe is a turing machine" theory. I have to say, PDE, you're very gullible. Open your eyes! Look around you! Mathematics is a tool. "The Matrix" is a movie. etc. etc. Physics is not reality. Physics is a very piecemeal and makeshift description. It cannot "prove" anything. Perhaps you need to study a little philosophy of science. Start with Husserl's phenomenology. Sir Francis Bacon might be a good resource, as well. -Kevin Baas''


 * No Kevin, I didn't get that argument from "some young student" &mdash; although I did write it up for a philosophy of mind class (and I think I may have persuaded my initially anti-strong-CTT lecturer, too :P). And I'm not sure what makes you think I know nothing about epistemology.  Of course mathematical physics is not reality.  It's just an attempt to model or simulate reality.  If the theories get good enough, then perhaps they can aspire to having captured all of the things which make reality what it is.  In the case of Turing completeness, the mathematical conditions for membership (any non-infinite system of state transitions) are so general that it's entirely possible that the universe meets them.


 * Should I be suggesting some reading in exchange? With your combination of confusion and arrogance (please note, one of those on its own is not so bad), I'm not sure that it would do much good.  To be brutally honest, I wonder whether your net contribution to Wikipedia is helpful. -- pde


 * And I'm sorry the, fourier transform is principally used in signal processing. It is not "used" in pure mathematics, because pure mathematics doesn't "use" anything, rather it rejects all meaning and works in absurd logical manipulations

The Fourier transform has many applications to things other than signal processing, in pure mathematics, in physics, in engineering, in cryptology, in number theory, in combinatorics, in probability theory, in statistics, and in many other fields. See Dym & McKean's book to read about many of these. Michael Hardy 01:20, 17 Jan 2004 (UTC)


 * Pay no attention to the gibberish below.

I must say, you're very rude. My answer was respectful and responsive; yours neglected the question, which was: what is the vector space involved? Michael Hardy 01:44, 17 Jan 2004 (UTC)


 * perhaps I need only to remind you that the Poisson distribution is a continuous distribution, not a discrete distribution.

This is incorrect; the Poisson distribution is discrete, supported on the countably infinite set { 0, 1, 2, 3, ... }. Perhaps you meant the parameter space that parametrizes the family of all Poisson distributions is continuous. That much would be true. But each Poisson distribution is a discrete probability distribution. See also the fractional probability discussion page, where I have posted some comments on this. Michael Hardy 02:12, 17 Jan 2004 (UTC)

Mr. Baas, you have been very rude on a number of occasions on talk pages, and that which you call my rudeness was in fact merely my description of the facts, plus my opinion that you often write unclearly.

You are confused about the Poisson distribution. Any Poisson distribution is a discrete probability distribution; the family of all Poisson distributions is parametrized by one non-negative real parameter. Your reference to "events as they occur in continuous time" suggests that you are confusing the Poisson distribution with the Poisson process. Indeed, we are talking about the distribution, not the events themselves. The support of this probability distribution is the set { 0, 1, 2, 3, ... } of non-negative integers; therefore, it is necessarily a discrete distribution.

You write that "It is impossible to simulate processes that involve poission distributions to perfect accuracy on a turing machine." But the same is true of all probability distributions, including a single coin-toss. You wrote: "Turing machines can perform any discrete mathematical operation to perfect accuracy." But they cannot simulate a coin toss. And you forget that the probability that a biased coin comes up heads can be a noncomputable irrational number, and that in no way diminishes the fact that the number of "heads" that appear -- either zero or one -- is a random variable whose probability distribution is discrete.

In your manner of writing about mathematics, you appear to consistently make matters more complicated than they really are.

That I do not follow conventions dogmatically is proved by the nature of some of my contributions to Neil Weiss's new book on probability, in which I argued at length in favor of some unconventional nomeclature that he ultimately adopted. I follow conventions in order to understand others and to be understood by others.

Your statement that "One of [two parameters] is continuous, therefore, the distribution itself is continuous." is utter nonsense; even the simple coin-toss random variable is parametrized by a continuous parameter.

And, moreover, as I said, there is just one such parameter, and it is continuous. You seem to have in mind that the family of Poisson distributions is parametrized by two parameters, and one is discrete! I know of no discrete parameter used to parametrize the family of Poisson distributions. Conventionally, this family is parametrized by one parameter &lambda; (I do not insist on that particular letter) and the probability mass function is given by


 * $$P(X=x)={\lambda^x e^{-\lambda} \over x!},$$

where x &isin; { 0, 1, 2, 3, ... }.

"Continuity" of a parameter space in no way implies that a distribution belonging to the family is discrete.

You would both communicate and understand better if you if you learned conventional language instead of exhibiting a holier-than-thou contemptuousness of it.

You would both communicate and understand better if you were not so often so belligerent. Michael Hardy 01:52, 25 Feb 2004 (UTC)


 * I agree that there is turbelence and a million other processes involved in the flipping of a real coin, and that it would be quite inconceivable for a computer to simulate all of these processes. But that's boringly obvious.  Why even mention it?


 * Perhaps "calculate" the probability would be a term that communicates my point better? The probability is a computable number.  These probabilities can be manipulated and composed every which way, without involving random numbers, in what may be called a "simulation".  The end result is a discrete probability distribution which is an exact solution, not an outcome or event.


 * However, a computer cannot produce an exact solution if the probability model involves a poission distribution.


 * If the coin is biased and the resulting probability is an uncomputable number, this implies that it was biased by an uncomputable number. But nondimensionalizing solves this problem.  Essentially, we're not really concerned with "numbers" when we're talking about a computer.  We're talking, rather, about finite states.  So long as the problem can be coded by a finite number of states, it is discrete, and can be operated on by a turing machine.


 * The poission distribution is often represented in the form:


 * $$P(x,\lambda)={\lambda^x e^{-\lambda} \over x!}$$


 * A special case of the poission distribution is the exponential distribution, where the x parameter is 1:


 * $$P(\lambda, t)={\lambda e^{-\lambda t}}$$


 * - Kevin Baas 17:05, 25 Feb 2004 (UTC)


 * You are mistaken on several counts. (And you still give just one parameter &lambda;.)  You are wrong to call the exponential distribution a special case of the Poisson distribution.  The Poisson distribution is a discrete distribution assigning a probability to each nonnegative integer.  The exponential distribution is a continuous probability distribution that assigns a positive probability to every interval in the half-line (0, &infin;).  To say the exponential distribution is a special case of the Poisson distribution would mean that every exponential distribution is a Poisson distribution but not every Poisson distribution is an exponential distribution.  That is false; the exponential distribution is not a Poisson distribution.  As I said, you are confusing two different things with each other: (1) Poisson distributions, and (2) (more complicated) Poisson processes.  A 1-dimensional Poisson process involves discrete Poisson distributions and continuous exponential distributions.  The distribution of the number of "arrivals" in a given time interval is a discrete probability distribution; it is a Poisson distribution.  The distribution of the waiting time until the next "arrival" after a given time is a continuous probability distribution; it is an exponential distribution, not a Poisson distribution.  Every time I've taught probability I've warned students not to confuse these two things with each other, and almost always some of them do.  So be more careful. Michael Hardy 21:24, 25 Feb 2004 (UTC)


 * Heh, you guys argue a lot.

I agree with Michael -- what he is saying is standard nomenclature. If either one of you was wrong, though, there's no need to throw words like "two shits" around to prove your point. :-)

The distribution just describes the probability P(X=x) of an event X=x happening. Usually the value of the variable is not considered a parameter (at least I have never seen anyone talk about it that way). Certainly the final value of the distribution function depends on its parameters and on x, but "parameters" usually refers to the other values that characterize that particular distribution, e.g. for the binomial distribution you have n and p (the number of tosses, the probability of a success per toss).

Maybe that's the source of your debates? I guess you can refer to x as being the "argument" to the function, instead of a parameter of the function.

Finally, as to your examples, you can have various distributions depending on the situation is. As you mentioned, you can make a Markov process out of the poisson distribution (by making the Mean parameter, i'll call it $$\lambda$$, vary proportionally with the time elapsed, so it becomes $$\lambda$$t). Then the probability of the first success is exponentially distributed. That's a different distribution and it does depend on two parameters. (It's a special case of the Gamma distribution, if you want to say that.) The poisson distribution is one of the only distributions that has one parameter (the mean), because it describes pretty simple things. And yes, it is discrete.

PS: Aren't computable numbers those numbers that can be computed in a finite time? If so, I don't see why the probability of a binomial distribuion with parameters n and p is possibly non-computable. (This is to Michael, I guess.)

- Greg Magarshak 12:32, 26 Feb 2004 (UTC)