Wikipedia:Reference desk/Mathematics

Reference Desk

Science Mathematics Computing/IT Humanities Language Entertainment Miscellaneous Archives
How to ask a question Search first. It's quicker, because you can find the answer in our online encyclopedia instead of waiting for a volunteer to respond. Search Wikipedia using the searchbox. A web search could help too. Common questions about Wikipedia itself, such as how to cite Wikipedia and who owns Wikipedia, are answered in Wikipedia:FAQ.
Sign your question. Type ~~~~ at its end. Be specific. Explain your question in detail if necessary, addressing exactly what you'd like answered. For information that changes from country to country (or from state to state), such as legal, fiscal or institutional matters, please specify the jurisdiction you're interested in. Include both a title and a question. The title (top box) should specify the topic of your question. The complete details should be in the bottom box. Do your own homework. If you need help with a specific part or concept of your homework, feel free to ask, but please don't post entire homework questions and expect us to give you the answers. Be patient. Questions are answered by other users, and a user who can answer may not be reading the page immediately. A complete answer to your question may be developed over a period of up to seven days.	Do not include your e-mail address. Questions aren't normally answered by e-mail. Be aware that the content on Wikipedia is extensively copied to many websites; making your e-mail address public here may make it very public throughout the Internet. Edit your question for more discussion. Click the [edit] link on right side of its header line. Please do not start multiple sections about the same topic. Archived questions If you cannot find your question on the reference desks, please see the Archives. Unanswered questions If you find that your question has been archived before being answered, you may copy your question from the Archives into a new section on the reference desk. Do not request medical or legal advice. Ask a doctor or lawyer instead.
After reading the above, you may ask a new question by clicking here. Your question will be added at the bottom of the page.
How to answer a question Be thorough. Please provide as much of the answer as you are able to. Be concise, not terse. Please write in a clear and easily understood manner. Keep your answer within the scope of the question as stated.	Link to articles which may have further information relevant to the question. Be polite to users, especially ones new to Wikipedia. A little fun is fine, but don't be rude. The reference desk is not a soapbox. Please avoid debating about politics, religion, or other sensitive issues.

February 19

The length of a curve when the metric is not positive semidefinite

What is the length of a curve when the metric is not positive semidefinite? I suppose it's

$L=\int _{a}^{b}{\sqrt {\left|\sum _{i,j}a_{ij}{dx^{i} \over dt}{dx^{j} \over dt}\right|\,}}dt\$

but I can't find any source for this definition. The Infidel 11:45, 19 February 2006 (UTC)

It may depend on your application, but I don't see that the length is well-defined unless the context allows you to make a meaningful choice of branch of the square root function in a consistent way. For example, if the rootand is positive or negative consistently over the whole curve ("spacelike" or "timelike" throughout, in Minkowski space terminology) you're sorted (though the length comes out imaginary in one case). —Blotwell 06:44, 20 February 2006 (UTC)

Is this a mathematical definition or is it just what physic students usually do? I know that in engineering, if a length turns out negative they often flip sign without comment (yet many of the houses do not collapse ;-). The Infidel 19:27, 20 February 2006 (UTC)

find the inverse function

Inverse function <=== found it! Dmharvey 15:57, 19 February 2006 (UTC)

LMAO nice one -zappa 16:42, 22 February 2006 (UTC)

Pythagorean-like problem

Find the sum of all $0\leq a\leq 100$ such that

${\frac {1}{a^{2}}}+{\frac {1}{b^{2}}}={\frac {1}{c^{2}}}$

where

$a,b,c\in \mathbb {Z}$

---

I am wondering if there is a way of solving the problem above without enumerating all possible Pythagorean triples in the range, since this problem was taken out of a math competition.

Thanks.

206.172.66.43 17:34, 19 February 2006 (UTC)

Are you thinking this is the same as the Pythagorean problem?

$a^{2}+b^{2}=c^{2}\,\!$

On the contrary, the left side is

${\frac {a^{2}+b^{2}}{a^{2}b^{2}}},$

and we can reciprocate both sides and clear the denominator to get

$a^{2}b^{2}=c^{2}(a^{2}+b^{2})\,\!.$

Clearly it is true that a²+b² must be a perfect square, but the problem is more subtle than that. --KSmrq^T 20:51, 19 February 2006 (UTC)

It's precisely Pythagorean triples $(a,b,d)$ satisfying $d\mid ab$. Let $(a,b,d)=t(a',b',d')$ with (pairwise) coprime $(a',b',d')$. It follows that $d'\mid t$, hence it is essentially sufficient to consider relatively coprime Pythagorean triples $(a',b',d')$ satisfying $a'd'\leq 100$, which are $(3,4,5)$ and $(5,12,13)$.--gwaihir 21:30, 19 February 2006 (UTC)

gwaihir's clue was the final piece of the puzzle I needed to nail this problem. In fact I had hit upon the idea of $d\mid ab$ as a criteria for generating the triples for the original equation. But I was unable to prove that multiples of basic Pythagorean triples where $a'd'\leq 100$ were all that's needed. Essentially, my mental block was on the fact that either the numbers in a Pythagorean triple are all coprime with each other or all three share the same common factor. I kept on trying to find the case where two number out of the three had a common factor that was not divisible by the third, turned out this is impossible, because for a Pythagorean triple $(a,b,c)$:

Assume that $c$ has no common factor with $a$ or $b$, but $a$ and $b$ have a greatest common factor, $d$, that is greater than 1. Then

$c^{2}=a^{2}+b^{2}=(a'd)^{2}+(b'd)^{2}=a'^{2}d^{2}+b'^{2}d^{2}=d^{2}(a'^{2}+b'^{2})$

Clearly, $c$ is also divisible by $d'$, which is a contradiction with the original assumption that $c$ has no common factor with $a$ or $b$.

Assume that $a$ has no common factor with $b$ or $c$, but $b$ and $c$ have a greatest common factor, $d$, that is greater than 1. Then

$c^{2}=a^{2}+b^{2}\Rightarrow a^{2}=c^{2}-b^{2}=(c'd)^{2}+(b'd)^{2}=c'^{2}d^{2}+b'^{2}d^{2}=d^{2}(c'^{2}+b'^{2})$

Clearly, $a$ is also divisible by $d'$, which is a contradiction with the original assumption that $a$ has no common factor with $b$ or $c$.

Symmetrically, it could be shown that it is impossible for $a$ and $c$ to share a greater than 1 common factor that is not divisible by $b$.

Hence, for Pythagorean triples, either all three numbers have a greater-than-one common factor or they are pairwise coprime, that means $gcd(a,b)=1$ and $gcd(b,c)=1$ and $gcd(c,a)=1$ all at the same time! There are no other situations possible in terms of common factors.

With this proof, along with gwaihir's clue, it is not hard to figure out the following possible values for $a$:

15, 30, 45, 60, 75, 90	for the (3, 4, 5) case
65	for the (5, 12, 13) case

Thus, the answer to the original question (the sum of all such $a$) is: 350.

Hurrrrrrrrrrrah, thank you so much for your help, gwaihir and everyone else!

129.97.252.63 05:03, 21 February 2006 (UTC)

My addition of the above numbers gives 380, but more importantly, you forgot the (4,3,5) case. Minor details about counting: For a = 60, there are two solutions: should 60 be counted twice? If yes, what about negative b's and c's?--gwaihir 10:41, 21 February 2006 (UTC)

Waaaaaaaaaaaaah, I'm stupid, I'm stupid. 129.97.252.63 21:04, 21 February 2006 (UTC)

---

OK, using gwaihirs ideas, if a,b have a common factor p we can reduce it as $p^{4}a'^{2}b'^{2}=c^{2}(p^{2}a'^{2}+p^{2}b'^{2})$ with c = pc', a'p=a, b'p=b to $a'^{2}b'^{2}=c'^{2}(a'^{2}+b'^{2})$, so we need only to consider coprime a, b.

If p is a prime factor with p^n dividing a (but not b, as they are coprime), the $c^{2}b^{2}=0(mod\,p^{2n})$ so p^n|c as $b^{2}\neq 0(mod\,p)$. This holds for every p^n| a, so a divides c, there exists u such that au =c. We can rewrite $a^{2}b^{2}=a^{2}u^{2}(a^{2}+b^{2})$ so $b^{2}=u^{2}(a^{2}+b^{2})$ and as all numbers are integers with $u\geq 1$, we have b^2 > b^2 as a contradiction, no solutions exist, the sum is zero.

The Infidel 22:36, 19 February 2006 (UTC)

(15, 20, 12) is a solution.--gwaihir 23:00, 19 February 2006 (UTC)

Yep. JackofOz 23:25, 19 February 2006 (UTC)

Ouch, you'r right. I've lost solutions in the first step when reducing by a common factor. c =pc' is only on possibility, the other is p| a'^2+b'^2. The Infidel 18:59, 20 February 2006 (UTC)

Statistics: probability

I knew I shouldn't have decided to major in a degree that requires a statistics class. I acutally have two questions.

1. If given a variable X with a given, discrete sample space S={0, 1, 2, 3, 4, 5}, with accompanying probabilities of each value occurring, how can you verify that that is a legitimate discrete distribution?
2. If given P(A), P(B), and P(A and B), how do you figure P(B not A)?

Thanks, Hermione1980 18:32, 19 February 2006 (UTC)

First: don't look at our article about probability (no joke)

Second: think of the samples as boxes labled 0 to 5 with different weights that add up to 1.

Third: play around with subsets of these, looking at the total weight

Forth: look at the article measure theory. A probability measure is just a measure where the total set of samples has measure 1.

The Infidel 19:59, 19 February 2006 (UTC)

For question 2, the inclusion-exclusion principle article might interest you. —Blotwell 06:19, 20 February 2006 (UTC)

For the first: all probablilities must be nonnegative and their sum must be exactly 1. For the second: P(B and not A) = P(B) - P(A and B) or something like that. – b_jonas 17:19, 23 February 2006 (UTC)

February 20

Powerball

Powerball says that the odds of matching just the last number, the Powerball, is 1 in 68.96. The article also says that the Powerball goes from the number 1 to 42. So shouldn't the chance of matching the powerball be around 1 in 42? What explains the discrepancy? zafiroblue05 | Talk 00:57, 20 February 2006 (UTC)

• That's the odds for matching the powerball only. So it's 1 in 42 minus the chance of matching one or more of the white numbers. Perhaps this should be pointed out on Powerball, since it's asked often enough that it shows up on the Powerball FAQ. --jpgordon∇∆∇∆ 03:04, 20 February 2006 (UTC)

Trigonometry question

I want to write a program for a robot. One thing I need is something that allows my robot to move to any given location from its current location. Now, I know how to calculate the distance (difference between x and y coordinates each squared and added then the square root of that number) but I don't know how I would calculate the angle. Taking the arctan of the change in y over x won't always work because arctan (-1/1) /= arctan (1/-1). Is there any way around this?

Of course there's a way around it: just add 180 degrees if x is negative. Many programming languages have a special function set aside for this, for example atan2() in C, but that's just a convenience. —Keenan Pepper 02:50, 20 February 2006 (UTC)

FORTRAN, and I suspect many other languages, also have an ATAN2 function. StuRat 05:52, 20 February 2006 (UTC)

In fact, I needed that atan2 function in BASIC once so I implemented it. Here's the code.

   FUNCTION Atan2 (Y, X)
   
   hpi = 1.570796326794897#
   pi = 2 * hpi
   
   IF ABS(Y) < ABS(X) THEN
    IF X > 0 THEN
     Atan2 = ATN(Y / X)
    ELSE
     Atan2 = ATN(Y / X) + pi
    END IF
   ELSE
    IF Y > 0 THEN
     Atan2 = ATN(-X / Y) + hpi
    ELSE
     Atan2 = ATN(-X / Y) - hpi
    END IF
   END IF
   
   END FUNCTION

– b_jonas

Non-integer numeral-system bases

In some Wikipedia articles I've seen the statement that the most "efficient" (in some unstated way) number base is e (2.718...). This raises some questions:

1. Is it reasonable to discuss a non-integer base?
2. Have such bases been studied?
3. Presuming this means position-implies-power system of notation, doesn't that mean that under an irrational base (like e), no rational number would have a finite expression?
4. For instance, in base e an integer would have to be approximated (to as many e-places as desired). Wouldn't counting in base e be rather problematic?
5. In what way would the base e be "efficient"?

Thanks. -R. S. Shaw 04:30, 20 February 2006 (UTC)

I have a slight feeling that the "base" they're talking about is not the base as in the base you write your numbers in (base 2, base 10), but as the base of an exponential (the e in $e^{x}$, the 10 in $10^{100}$). If you use e as the base of an exponential, it's neater to differentiate for example: ${\frac {d}{dx}}e^{x}=e^{x}$. enochlau (talk) 05:40, 20 February 2006 (UTC)

No, they're talking about e as a positional number system base, as in decimal, binary, and e-ary. One example is Computer numbering formats#Why binary?. -R. S. Shaw 10:02, 20 February 2006 (UTC)

To answer #3, no they wouldn't, because e is irrational transcendental. Now, positional notation numeral systems work on exponentials. The number 352 in base 10 is $3\times 10^{2}+5\times 10^{1}+2\times 10^{0}$. 110 in binary (base 2) is $1\times 2^{2}+1\times 2^{1}+0\times 2^{0}=6_{base\ 10}$. In order to write numbers on base e, you'd use the digits 0, 1, and 2, and you'd have to work out the exponents to match the value you're trying to represent. Now, I'm not aware of how well this would work, but considering all the nice properties e has as an exponential base, and since numerical bases work based on the exponent principle, seems reasonable to think that this base would have some very interesting properties. I can't tell which, though, so I'll leave that to someone else :P. Also, we should create base e eventually. ☢ Ҡi∊ff⌇↯ 06:06, 20 February 2006 (UTC)

Right, the integers 0, 1, and 2 can be expressed finitely. But >2 would not. For instance, 3 would equal 1 × e^1 + 0 × e^0 + 0 × e^-1 + 2 × e^-2 + ... to several e-places, 3 (base 10) = 10.0200112000... (base e). -R. S. Shaw 10:02, 20 February 2006 (UTC)

I think non-terminating representations for integers is due to e being transcendental, not irrational. Integers can have terminating representations in irrational bases - see golden ratio base for example. Gandalf61 15:31, 20 February 2006 (UTC)

Good call. Fixed my statement. ☢ Ҡi∊ff⌇↯ 18:13, 20 February 2006 (UTC)

... or indeed an even better example is base sqrt(2), in which every integer has a finite representation that is created by inserting 0s between the digits of its binary representation so 2 is 100, 3 is 101, 4 is 10000 etc. Gandalf61 17:28, 20 February 2006 (UTC)

Any algebraic number has an associated base in which all integers terminate. There is also a "standard form" which allows only certain sequences of digits, for example the golden ratio base in which more than one 1 in a row is forbidden. —Keenan Pepper 17:58, 20 February 2006 (UTC)

(The forbidden sequences are determined by the polynomial of which the algebraic number is a root. The golden mean is a root of the polynomial $x^{2}-x-1$, so "011" can always be replaced by "100". —Keenan Pepper 18:00, 20 February 2006 (UTC))

Donald Knuth discusses such topics in TAOCP Vol. 2:Seminumerical Algorithms. (See the reference at the end of the positional notation article.) A helpful article can be found online in American Scientist. See also The ternary calculating machine of Thomas Fowler. --KSmrq^T 19:31, 20 February 2006 (UTC)

Thanks for the link to the Hayes essay in American Scientist; it is very helpful wrt question 5. (The "efficiency" is an optimum of a measure of little or no practical significance.) The golden ratio base article answers 1 & 2. Q 3 has been clarified (most integers have only infinite expressions in transcendental bases). And yes, counting in base e would be horrible; the "carry" in the addition 2 + 1 would not only affect the position to the left of the units position, but also an infinite number of places to the right. -R. S. Shaw 22:22, 21 February 2006 (UTC)

Yes such systems are studied, usually in the form of $\sum _{n}q_{n}t^{n}$ where $q_{n}$ is a rational number and t is not. For example, Diophantine analysis and Galois theory when t is an algebraic number. I've seen even crazier things; I wrote rational zeta series on one such crazy system. The point is not so much to "count" in base e, but to study the filligree of how rationals "fit totether" with real numbers. I say "filligree" because the "fit together" part is very fractal-ish with things like crazy near-identities, and etc. Borwein, for example, found a very simple approximation for pi, which is accurate to the first 50 billion digits before it breaks down (!!!!) So take that 22/7! Another, simpler example: over on talk:real number theres a never-ending discussion of 1.00000....=0.99999... These kinds of "accidental" degeneracies between number systems are a lot more interesting in non-integer bases, and you get a much better idea of what the "true meaning" of 1.00000....=0.99999... (and its not what most people say it is). linas 07:06, 22 February 2006 (UTC)

Where can I read more about this Borwein approximation? By the way, there's also $(1+9^{-4^{7\times 6}})^{3^{2^{85}}}$ which gives e to 18457734525360901453873570 decimal places. [1] Fredrik Johansson 07:46, 22 February 2006 (UTC)

That approximation for e is just substituing a very large number in $e=\lim _{n\to \infty }{(1+{\frac {1}{n}})^{n}}$. Nothing special about it, you could just as well have written $(1+999^{-999^{999}})^{999^{999^{999}}}$ for an even better approximation. When you finish doing the actual calculation, I'm not sure the universe will be still there... -- Meni Rosenfeld (talk) 09:00, 22 February 2006 (UTC)

Well, there is one thing special about it: it is almost pandigital (has all decimal algarisms once - except for zero) ☢ Ҡi∊ff⌇↯ 07:09, 23 February 2006 (UTC)

Where would it have gone to? JackofOz 06:20, 23 February 2006 (UTC)

Kieff: Hey, you're right, that is interesting. Alas, This doesn't have anything to do with e really.

JackOfOz: Well, being that I'm not a phsycist, I don't know where the universe will be in $10^{10^{25}}$ years. I'm not sure physicists know that either :-) -- Meni Rosenfeld (talk) 12:09, 23 February 2006 (UTC)

Like, are there any other places? JackofOz 10:35, 24 February 2006 (UTC)

Trajectory of a projectile with air drag

Despite the heading, this isn't a question in physics since I'm actually interested in the solution of a specific (type of) set of differential equations. Suppose a projectile is moving in the x-y plane, experiencing a constant gravitational acceleration of magnitude g in the negative direction of the y axis, and air drag proportional to the square of its velocity. This gives rise to the equations:

${\frac {d^{2}x}{dt^{2}}}=-\alpha {\frac {dx}{dt}}{\sqrt {\left({\frac {dx}{dt}}\right)^{2}+\left({\frac {dy}{dt}}\right)^{2}}}$

${\frac {d^{2}y}{dt^{2}}}=-\alpha {\frac {dy}{dt}}{\sqrt {\left({\frac {dx}{dt}}\right)^{2}+\left({\frac {dy}{dt}}\right)^{2}}}-g$

As far as I know, these equations have no (non-trivial) elementary solution. However, does anyone know a way to represent its solution in some other way, perhaps using non-elementary functions or power series? Since this motion has a singularity point, I think there should be some sort of power series involving t, the time since the point of singularity, and $\theta$, the angle to which the direction of velocity converges in the singularity. Perhaps an expansion in $\left({\frac {t-1}{t+1}}\right)$? How about a way to express a solution in terms of the boundary conditions of location and velocity at a specific time? Any help in the matter would be appreciated. -- Meni Rosenfeld (talk) 16:12, 20 February 2006 (UTC)

OK, what do we know:

1. |x'| is decreasing and eventually reaches 0, where it stays.
2. Once (and while) If x'=0, the solution for y can be represented as one of the following:

$y(t)={\begin{cases}y_{0}+{\frac {\log \cos {\sqrt {\alpha g}}(t_{0}-t)}{\alpha }}&t_{0}-{\frac {\pi }{2{\sqrt {\alpha g}}}}<t\leq t_{0}\\y_{0}-{\frac {\log \operatorname {cosh} \,{\sqrt {\alpha g}}(t-t_{0})}{\alpha }}&t\geq t_{0}\end{cases}}$

$y(t)=y_{0}-{\sqrt {\frac {g}{\alpha }}}t$

$y(t)=y_{0}-{\frac {\log \operatorname {sinh} \,{\sqrt {\alpha g}}(t-t_{0})}{\alpha }}$ where $t>t_{0}+{\frac {\operatorname {arcsinh} 1}{\sqrt {\alpha g}}}$

I'm not sure what you're looking for in the interesting region. Arthur Rubin | (talk) 02:21, 24 February 2006 (UTC)

I don't agree that |x'| reaches 0. It does converge to 0, and the solution in the case x'=0 is indeed simple, but alas, I am looking for an exact solution, so this does not solve the problem. As I said, what I am looking for is a way to represent the exact solution in the entire region, in terms of either non-elementary functions, or an infinite series (preferrably a rapidly converging one). -- Meni Rosenfeld (talk) 06:21, 24 February 2006 (UTC)

Sorry, my mistake. That's still the exact solution for |x'| = 0, and I think I can produce the first few terms of series as t approaches +infinity and the singularity from above, but I don't know about exact solutions. We'll see how the first few terms work out. Arthur Rubin | (talk) 08:13, 24 February 2006 (UTC)

If you know an algorithm that can produce arbitarily many terms of the exact power series, that would be great. I doubt there is a closed form for the nth term, but I could be mistaken. What series do you have in mind? Laurent? -- Meni Rosenfeld (talk) 08:21, 24 February 2006 (UTC)

Approach 1: Expand around the singularity

$x(t)=a_{-1}\log t+a_{0}+a_{1}t+a_{2}t^{2}+\ldots$

$y(t)=b_{-1}\log t+b_{0}+b_{1}t+b_{2}t^{2}+\ldots$

which can be solved to find

$x(t)={\frac {\cos \theta }{\alpha }}\log t+x_{0}+g{\frac {\sin 2\theta }{24}}t^{2}-\alpha g^{2}{\frac {5\cos \theta +4\cos 3\theta }{1440}}t^{4}+\ldots$

$y(t)={\frac {\sin \theta }{\alpha }}\log t+y_{0}-g{\frac {5+\cos 2\theta }{24}}t^{2}-\alpha g^{2}{\frac {3\sin \theta +\sin 3\theta }{360}}t^{4}+\ldots$

Approach 2: Convert to polar coordinates:

$(x',y')=\,(r\cos \phi ,r\sin \phi )$

leaving the equations as

$r'=\,-\alpha r^{2}-g\sin \phi$

$\phi '=-{\frac {g\cos \phi }{r}}$

Approach 3: Expand around t=Infinity.

I can't find the proper form of the expression for the series.

Found it! Write:

$\gamma ={\sqrt {\alpha g}}$

$x(t)=a_{0}(t)+a_{1}(t)e^{-\gamma t}+a_{2}(t)e^{-2\gamma t}+\ldots$

$y(t)=b_{0}(t)+b_{1}(t)e^{-\gamma t}+b_{2}(t)e^{-2\gamma t}+\ldots$

where the a_i are polynomials in t. The first few terms, writing

$\tau =\,\gamma t$

are

$x(t)=x_{0}+{\frac {A}{\alpha }}e^{-\tau }+{\frac {A^{3}}{36\alpha }}(2+12B-\tau )e^{-3\tau }+\ldots$

$y(t)=(y_{0}-{\frac {\tau }{\alpha }})+{\frac {A^{2}}{4\alpha }}(4B-\tau )e^{-2\tau }+{\frac {A^{4}}{256\alpha }}(1+16B+256B^{2}-(4+64B)\tau +8\tau ^{2})e^{-4\tau }+\ldots$

How the parameters x₀, y₀, A, and B propagate under shifting time, and how this may relate to other characteristics of the solution, are left as an exercize for the reader. Note also that if A=0, it makes sense for B to become infinite such that A^{2 B is constant, producing the solution above with x'=0.}

Arthur Rubin | (talk) 18:43, 24 February 2006 (UTC) and 20:13, 24 February 2006 (UTC)

February 21

ovaloid : definition (finite projective geometry)

There seems to be little talk about ovals , ovoids and ovaloids on this page. Now before I can help make some pages, I would like to clear something up :

an ovaloid is a set of points, no three collinear, in PG(3,q) that is maximal with this property. Now, when one says maximal, do they mean :no set with this property and more points can be found, or do they mean : it cannpot be extended to a bigger set such that no three are collinear.

There is a subtle difference in definition there am I not right?

Thanks,

evilbu

You are certainly right that there is a difference in definition, although I am too ignorant of finite geometry to say whether the two definitions do not happen to be equivalent in this particular case. However, the word maximal (which makes sense with respect to any partial order) would generally be understood to be with respect to the inclusion order on subsets, so from the sentence you gave I would expect the second of your definitions to be correct. If I were writing the sentence and meant the first definition I would have written something like has maximal cardinality. But again, I'm not a finite geometer so no absolute guarantees. —Blotwell 02:25, 23 February 2006 (UTC)

one billion seconds

How many minutes is one billion seconds?

Google: http://www.google.com/search?q=one+billion+seconds+in+minutes (divide by 60) - Fredrik Johansson 18:03, 21 February 2006 (UTC)

Hey, maybe you could write a musical about it. You could call it Mortgage. --Trovatore 18:13, 21 February 2006 (UTC)

There are 60 secs in a min. Try arithmetics, which records elementary properties of certain operations on numerals : 100/6 = 16.6. 1000/6 = 166.66 ... Or try Google for the same operation. --DLL 22:08, 21 February 2006 (UTC)

In other words, divide one billion by 60 to get your answer. StuRat 03:13, 22 February 2006 (UTC)

Very crudley: about 31 years. gelo 07:09, 22 February 2006 (UTC)

Or, considering 365 days = 1 year, 31 years 259 days 1 hour 46 minutes and 40 seconds. Disregard the months because those are too innacurate. ☢ Ҡi∊ff⌇↯ 07:28, 22 February 2006 (UTC)

Gee, so many replies and not even one straight answer :-) It's 16,666,666 minutes and 2 thirds of a minute (40 seconds). -- Meni Rosenfeld (talk) 09:16, 22 February 2006 (UTC)

If you care about years, it may help to know that π seconds is a nanocentury.--gwaihir 11:20, 22 February 2006 (UTC)

You meant "is roughly a nanocentury". There's a 0.5% difference. -- Meni Rosenfeld (talk) 13:21, 22 February 2006 (UTC)

There's roughly a 0.5% difference. SCNR--gwaihir 10:41, 23 February 2006 (UTC)

You're right, there's roughly 12% difference between 0.5% and the difference. -- Meni Rosenfeld (talk) 12:33, 23 February 2006 (UTC)

:-) --gwaihir 13:01, 23 February 2006 (UTC)

How to calculate confidence intervals?

I have normally distributed observations x_i, with i from 1 to N, with a known mean m and standard deviation s, and I would like to compute the upper and lower values of the confidence interval at a parameter p (typically 0.05 for 95% confidence intervals.) What are the formulas for the upper and lower bounds? --James S. 20:30, 21 February 2006 (UTC)

I found http://mathworld.wolfram.com/ConfidenceInterval.html but I don't like the integrals, and what is equation diamond? Apparently the inverse error function is required. Gnuplot has:

     inverf(x)         inverse error function of x

--James S. 22:00, 21 February 2006 (UTC)

The simplest answer is to get a good book on statistics, because they almost always have values for the common confidence intervals (you may have to look at the tables for Student's t-distribution in the row labelled n = infinity). Confusing Manifestation 03:57, 22 February 2006 (UTC)

The question was how to calculate, not how to look up in a table. --171.65.82.76 07:36, 22 February 2006 (UTC)

In which case the unfortunate answer is that you have to use the inverse error function, which is not calculable in terms of elementary functions. Thus, your choices are to look up a table, use an inbuilt function, or use the behaviour of erf to derive an approximation such as a Taylor series. (Incidentally, I think equation "diamond" is meant to be the last part of equation 5.) Confusing Manifestation 16:12, 22 February 2006 (UTC)

Any general statistical package worth its salt (including free ones such as R) will let you calculate the inverse error function mentioned above for any p, as will most spreadsheets (e.g. see the NORMSINV function in MS Excel). -- Avenue 12:07, 23 February 2006 (UTC)

Perl5's Statistics::Distributions module has source code for

 $u=Statistics::Distributions::udistr (.05);
 print "u-crit (95th percentile = 0.05 sig_level) = $u\n";

...from which I paraphase this...

Solution

 x = -ln(4 * significance_level * (1 - significance_level));
 
 critval = sqrt( x * ( 1.57079628800
               + x * ( 0.03706987906
               + x * (-0.8364353589E-03
               + x * (-0.2250947176E-03
               + x * ( 0.6841218299E-05
               + x * ( 0.5824238515E-05
               + x * (-0.1045274970E-05
               + x * ( 0.8360937017E-07
               + x * (-0.3231081277E-08
               + x * ( 0.3657763036E-10
               + x *   0.6936233982E-12)))))))))));
 
 if (significance_level > 0.5) then {critval = -critval};
 
 ci_top = mean + (standard_deviation/sqrt(N_obs)) * critval/2;
 ci_bot = mean - (standard_deviation/sqrt(N_obs)) * critval/2;

Where significance_level is, e.g., 0.05 for 95% confidence.

--James S. 07:30, 24 February 2006 (UTC)

February 22

pseudo-BCH formulas

Suppose we have a smooth function f(.) - is it possible to write this operator A = f(X) B f(-X) for operators B and X in terms of a bunch of commutators, like the Campbell Baker Hausdoff formulas? --HappyCamper 05:05, 22 February 2006 (UTC)

I'm not sure what to do with such a question. BCH refers to Lie algebras, tangent spaces at the identity of a Lie group along with the operator sometimes called Lie bracket. How is this supposed to relate to your situation? --KSmrq^T 01:52, 23 February 2006 (UTC)

It's sort of awkward for me to describe, because I'm trying to venture into a new area unknown to me previously...let me try explaining this way:

$e^{-X}Ae^{X}=A+\left[A,X\right]+{\frac {1}{2}}\left[\left[A,X\right],X\right]+\cdots$ - this I know holds. Now, the other day, I read that it can be rewritten as

$e^{-X}Ae^{X}=Ae^{\left.X\right]}$

I am not sure whether the ] refers to "left commutation" or "right commutation", but regardless, it seems to make sense structurally. Now, is it true that

$e^{-X}Ae^{X}=Ae^{\left.X\right]}=e^{\left[-X\right.}A$?

Since there really isn't anything special about the exponential, if I replace the exponential with some sort of arbitrary function, does

$f(-X)Af(X)=Af(\left.X\right])=f(\left[-X\right.)A$ hold? If so, can I write the resulting infinite commutator series with the same structural form as the first line above, but with just different coefficients? --HappyCamper 09:51, 23 February 2006 (UTC)

Actually, the exponential is special. It maps a vector from the Lie algebra to an element of the Lie group, and has important properties in doing so. --KSmrq^T 03:59, 25 February 2006 (UTC)

Oh, I see. So, it would not be possible to replace the exponential with say, a q-exponential, and see what happens? --HappyCamper 18:20, 25 February 2006 (UTC)

Gödel's theorem of incompleteness.

What are the philosophical implications of this theorem?...If there are any...I'm not sure, I'm just wondering.--Cosmic girl 19:29, 22 February 2006 (UTC)

Did you read Gödel's incompleteness theorem? (Completely?) Dmharvey 19:33, 22 February 2006 (UTC)

Yes, I've read it completely! :S..but from what I understood, even if my understanding of it is consistent, I will nevcer find a proof that it is...haha (lame joke).--Cosmic girl 21:16, 22 February 2006 (UTC)

How did I know you would come up with this?

It implies that "There are more things in heaven and earth, Mylady, Than are dreamt of in your philosophy". ;-)

Skip that Gödel stuff for now and go directly to the Turing machine and the Halting problem The Infidel 19:58, 22 February 2006 (UTC)

I thought so, Hamlet...did you really know I was gonna ask this? hahaha...creepy.--Cosmic girl 21:18, 22 February 2006 (UTC)

Yes, I suspected it. The question was already a seed in your mind when you asked about other kinds of maths. Next logical step would be Chomsky. The other branch of your thoughts would take you to the questions about existance. Indian philosophy of rebirth (what of our beeing is (thought to be) reborn) could be of interest here. The Infidel 17:50, 23 February 2006 (UTC)

Why chomksy?...isn't he just a lingüist who talks about politics?...what are his thoughts on 'truth' for example, or on 'god'...I thought Chomsky was only a communist linguüist who has strong opinions when it comes to politics...but I've never encountered any original or interesting philosophical propositions of his...and also...what does karma and hindi philosophy have to do with the theorem? I don't see the conection...I thougt this theorem was about undertinty or something like that.--Cosmic girl 19:18, 23 February 2006 (UTC)

Chomsky because of Chomsky hierarchy in formal languages. (I don't care what he said about politics. Read the newspapers to know the world is crazy. Read history books to know it's always been.)

The second was not so much about karma but about reincarnation, and not about Gödel but about your other question: does a mathematical entity exist for itself and if so, in what sense does it exist (Your wording was different.)

Closely connected to the question of existance is the question of what it is that exists. Hindu philosophers have spent a lot of thought about what it is that (they think is) reborn. Into a different species. In a different time. Deprived of memories. But still in a weird way the same. You don't have to believe all this, but the huge variety of sophisticated distinctions of self is very interesting. (See also Poincaré's reccurrence theorem.)

The Infidel 21:16, 25 February 2006 (UTC)

I asked something about this on my early Wikipedia days. I was wondering if I could use the theorem against Intelligent Design\Theism... I think it makes sense. :P ☢ Ҡi∊ff⌇↯ 21:11, 22 February 2006 (UTC)

How can it be used against ID and Theism?--Cosmic girl 21:18, 22 February 2006 (UTC)

Agree, Kieff, but this is mathematics. There was once a big farce about philosophers using science concepts to create a smoke screen. Analogy helps but it is not falsifiable. --DLL 21:20, 22 February 2006 (UTC)

Well, it's hard to refute an argument you haven't actually given, but I would be surprised if there were any very convincing argument against theism from the Gödel theorems. By the way Gödel himself is generally thought to have believed in God, though the record is less than clear. --Trovatore 21:28, 22 February 2006 (UTC)

By the way, no one has really said much about Cosmic Girl's original question. The theorems are of great philosophical importance in the philosophy of mathematics; applications to philosophy outside phil of math are much dicier.

Even within philosophy of mathematics, while almost everyone agrees the theorems are important, precisely in what way they're important remains contested. My take on it would be something like this: They made it more difficult to give a "formalist" or "logicist" account of mathematics and why it works. Since they didn't make it any more difficult to give a Platonist/realist account, they somewhat enhanced the position of the Platonist viewpoint with respect to its competitors (though without in any way helping to explain "where" or "how" mathematical entities exist independently of our reasoning about them). --Trovatore 22:17, 22 February 2006 (UTC)

I'm kind of lost now...what does the theorem have to say about whether mathematical entities exist independently of our reasoning about them?..does it have to say anything in the 1st place? I don't want to sound patronizing because I'm really stupid when it comes to math... but I believe that there's no possible way of knowing without a doubt, EVER, that mathematical entities exist appart from our reasoning about them.--Cosmic girl 02:01, 23 February 2006 (UTC)

The theorem has nothing at all to say about whether mathematical objects exist; that's not the subject matter of the theorem. What I said was, the theorem makes alternative accounts more difficult. In particular, it screws up Hilbert's program, which was the big formalist thrust pre-Gödel, pretty badly.

My personal view is that wanting to know things about mathematics "without a doubt" is the wrong goal. I'll grant that it's a time-honored one, going back to the Greeks, but I still think it's wrong. We don't know mathematical truths without a doubt, even if we've proved them; the assumptions that we used to prove them could be wrong, or there could be a mistake in the proof. We have more certainty than exists in, say, biology, but it's a difference of degree rather than kind. --Trovatore 03:44, 23 February 2006 (UTC)

What did Gödel find so unconfortable about this that he went crazy? I'm not disturbed by this theorem, maybe I'm not getting something...haha... ( I hope I'm not punished by some power from beyond for this joke).oh and also...what does the mystery of the Aleph have to do with this?.--Cosmic girl 20:45, 23 February 2006 (UTC)

At the time, the mathematicians were getting really worked up about the nature of maths, and they were asking themselves "just how much can we prove with a simple set of axioms?". Unfortunately, GIT says the answer is "no", and mathematicians have already found a whole bunch of things that they would like to prove true (or false, as it may be), but know they never will, but which they sometimes don't feel comfortable stating as axioms because it's impossible to know whether the theorem itself or its contradiction is true (because neither is provable). The article itself mentions a few examples, but the ones that come to mind are the continuum hypothesis (which is probably the "mystery of the aleph" you're talking about) and the axiom of choice - at a rough guess there are probably comparable numbers of papers written that assume the axiom of choice as those that try to avoid it entirely. Confusing Manifestation 02:08, 24 February 2006 (UTC)

Are you sure mathematicians have 'accepted' there are things they can never prove true or false? like what things? and can they 'prove' anything true in the first place? if so, what can they prove?.--Cosmic girl 17:39, 24 February 2006 (UTC)

Several points I have to quarrel with in the above. First, we don't have any theorems of the form "such and such can never be proved or disproved", only of the form "from this specified set of axioms, such and such can never be proved or disproved". So for example we know that the continuum hypothesis, CH, can never be proved or disproved from ZFC, granting that ZFC is consistent. It does not follow that we'll never know whether it's true, only that we'll never know a proof or disproof from ZFC. A current line of work by W. Hugh Woodin provides an argument, convincing to many, that CH is false (which was also what Gödel believed).

As for the axiom of choice, most mathematicians consider it unproblematic these days, though there are still reasons to prefer choice-free proofs when you can find them.

None of this addresses Cosmic girl's question as to why Gödel lost his marbles. I would not assume that had anything to do with his work in mathematical logic. Maybe he'd have gone crazy faster if he hadn't done logic. --Trovatore 03:41, 24 February 2006 (UTC)

wow...I'm so lost now, so you are saying that Gödel meant that say axiom A and axiom B can't prove such and such? what I meant by the question was that IF the theorem meant that we can't know if axiom A and axiom B are 'true' in the first place...well of course they are true, because they are axioms...but I thought that the theorem meant that while axioms can appear to be automatically true to us, they may not be so. and the rest of the stuff I don't get. :| --Cosmic girl 17:37, 24 February 2006 (UTC)

To be honest I don't really understand what you're saying above. No, the theorems don't mean that we can't know whether the axioms are true. No, it's not an "of course" that axioms are true; axioms are automatically provable, not automatically true. You can have a deductive system in which some of the things you can prove are in fact false. One of the repercussions of the incompleteness theorems (please note there are two of them! you keep using the singular) is that truth cannot be identified with provability in the sense that, say, Hilbert would have wanted. There are still those who identify truth with provability, but they have to give up properties of truth that most of us are not so ready to part with.

But you keep asking what the theorems mean. What they mean, literally, is what they actually say, which is a completely concrete claim about what happens in formal deductive systems. I think you should first try to understand the theorems at that level, before trying to understand their importance for philosophy. That will give you a much better nonsense filter, which you badly need in this arena, because there's so much nonsense written on this topic. --Trovatore 21:15, 24 February 2006 (UTC)

You are right, sorry for all the nonsense...=P.--Cosmic girl 03:11, 25 February 2006 (UTC)

A formal system is build with axioms ; no question of truth in an axiom, only practical results.

A consistent system contains no contradictions.

• GIT1 : every consistent system allows to construct - using its axioms and statements built from them – at least a true statement which is not provable (using the same). Thus it is incomplete.

The above unprovable statement may be provable in a richer system, and so on.

A formal theory including basic arithmetical truths and also certain truths about formal provability is “rich enough” to enter the second contest : Zermelo-Fraenkel set theory with the axiom of choice (ZFC) is a good example, as it allows the construction of the ordinal numbers.

• GIT2 : a “rich enough” formal system including a statement of its own consistency is inconsistent.

ZFC does not try and admits its own incompleteness.

Consequences for philosophy : few. For the philo of maths : some problems are undecidable. However, few people play with them ... but they make one of the paths that help maths to evolve. See the articles above and, Cosmic, another question, plz. --DLL 22:04, 24 February 2006 (UTC)

Thank you, your answer was the simplest one,and I will read the articles above.--Cosmic girl 03:11, 25 February 2006 (UTC)

February 23

Econometrics question: Autocorrelation

Hi everyone... am having trouble with a specific aspect of first-order autocorrelation.

The setup is y_t=B*y_(t-1)+u_t ; u_t=p*u_(t-1)+e_t ... y_t and u_t are covariance-stationary processes.

I'm supposed to eventually find cov(y_(t-1), u_t) and I've started off by breaking the covariance into cov(y_t/B , u_t) and cov (-u_t/B , u_t). I'm not sure this is the right approach, am becoming convinced it probably isn't. Anyways, continuing from there I've got the second covariance (with the u's) solved, but the first one, no matter how I break it down, still comes back to something in the form of cov(y_t , u_t).

Any ideas of how to get around this? Thanks very much for any thoughts. rabbit84.92.181.246 00:44, 23 February 2006 (UTC)

repeating decimals

is .999... repeating to infinity equal to 1?

• Yep! See Recurring decimal#The case of 0.99999.... --jpgordon∇∆∇∆ 07:44, 23 February 2006 (UTC)

And also Proof that 0.999... equals 1, and for some additional background real number. -- Meni Rosenfeld (talk) 12:16, 23 February 2006 (UTC)

And if you want your head to explode, see Talk:Proof that 0.999... equals 1, where people identified only by IP addresses try to bring that fact into dispute, and manage to use "phd" as an insult. Confusing Manifestation 12:24, 23 February 2006 (UTC)

You're right, that argument is hilarious. Sad to see that the anon posting there, clearly has some mathematical knowledge, and yet can mix up facts so badly. All the more reason to read real number - The more one knows about where real numbers come from, the less he is likely to make ridiculous claims. -- Meni Rosenfeld (talk) 12:46, 23 February 2006 (UTC)

It's one of the funniest I've ever seen -- all six or so archives as well. Is there a special place for mathematical WP:BJAODN? --jpgordon∇∆∇∆ 03:33, 24 February 2006 (UTC)

What's sadder still is the endless stream of misguided "helpers" who can't see that they're being trolled. --KSmrq^T 18:34, 25 February 2006 (UTC)

Quadratic formula

In the quadratic equation article, it says an alternate form of the quadratic formula is:

$x_{1,2}={\frac {2c}{-b\pm {\sqrt {b^{2}-4ac\ }}}}$

Am I correct in saying that this would not work where c is equal to zero, e.g. in the case "x² - 5x = 0"? That formula would return zero, but the actual answer would be 0 or 5. --210.246.30.87 07:35, 23 February 2006 (UTC)

Congratulations! You have uncovered a longstanding instance of error, delusion, or vandalism. The formula is totally bogus. It went in via this edit (later merged en bloc into quadratic equation). The amazing thing is that this has survived unchallenged for almost two years. -R. S. Shaw 09:49, 23 February 2006 (UTC)

It's not vandalism at all - the formula is perfectly valid. It is used whenever you need to worry about numerical errors from accumulating if you use the so-called "normal" or "canonical" expression for the quadratic formula. --HappyCamper 09:59, 23 February 2006 (UTC)

The expression arises from resolving the radical (is this the correct term for the procedure, as it seems to be an awful mouthful). Here, for example, we have

$x_{+,-}={\frac {-b\pm {\sqrt {b^{2}-4ac\ }}}{2a}}={\frac {\left(-b\pm {\sqrt {b^{2}-4ac\ }}\right)\left(-b\mp {\sqrt {b^{2}-4ac\ }}\right)}{2a\left(-b\mp {\sqrt {b^{2}-4ac\ }}\right)}}={\frac {4ac}{2a\left(-b\mp {\sqrt {b^{2}-4ac}}\right)}}$

Finally, note how some terms magically cancel out on the top and bottom...we get the celebrated (at least in numerical methods and analysis circles)

$x_{+,-}={\frac {2c}{-b\pm {\sqrt {b^{2}-4ac\ }}}}$

of course, valid when a is not zero, and when c is not zero. So, to answer the original question, yes, you are correct to say that the formula fails when c is zero. When c is zero, the conjugate term vanishes:

$\left(-b\mp {\sqrt {b^{2}-4ac\ }}\right)$

Sort of subtle, but do you see how? This should give you a hint to figure out when the alterative formula would be advantageous over the other one...Let me update the article :-) --HappyCamper 10:12, 23 February 2006 (UTC)

But you're not entirely correct to say that the formula incorrectly gives zero when $c=0$. When you take the minus sign it correctly gives 0, because 0 is one of the roots as it always is when $c=0$. When you take the plus sign to find the other root, the denominator becomes zero so the formula gives 0/0, which you recognize as meaning "Now would be a great time to use an alternative formula". So yes, the formula fails, but it doesn't lie to you and tell you an incorrect 0. —Blotwell 12:09, 23 February 2006 (UTC)

In numerical analysis the question of stability of quadratic root-finding is an early cautionary example, illustrating the dangers of naively pulling a formula out of a mathematics text. Assuming a is nonzero and d =b²−4ac is positive, typical code might be

if (b > 0)

 then q := -0.5 * (b + sqrt(d))

 else q := -0.5 * (b - sqrt(d))

r1 := q / a

r2 := c / q

A copysign function could be used instead of the test, for languages that adequately support IEEE floating point. --KSmrq^T 18:52, 25 February 2006 (UTC)

3-torus

This is not a factual question, it is a request. Can anyone generate an animation of the 3-torus? I was thinking something similar to the pentatope one you can see at the right, where each frame is a 3D slice of it. ☢ Ҡi∊ff⌇↯ 10:20, 23 February 2006 (UTC)

Grr...edit conflict....um...this request sort of begs the question: What does a 5d torus look like? Does it even exist? Can 5-D objects have holes in them? Are holes only found in 3-D objects? I wonder if any topologists are here on Wikipedia? Hm... :-) :-) :-) --HappyCamper 10:27, 23 February 2006 (UTC)

Can't say that I know much topology, but...

• I see no reason why 5-D objects shouldn't have holes.
• The torus article, I believe, supplies enough information to decide how does this 3-torus look like, for those able to understand it (myself not included).
• The pentatope is 4-D, making it reasonable to represent it with a 3D animation. But how would you represent a 5-D object? Looks tricky to me.
• Um... whom did you have an edit conflict with? Yourself? I understand, it happens to me all the time.

-- Meni Rosenfeld (talk) 12:29, 23 February 2006 (UTC)

Uh... The 3-Torus is a 4-D object, so an animation would be adequate. I can imagine a 4-D sphere, but a 4-D torus clogs my imagination. ☢ Ҡi∊ff⌇↯ 13:06, 23 February 2006 (UTC)

Ah, I see. So, an animation of the 3-torus would start as a sphere, that would split into two and then join again? ☢ Ҡi∊ff⌇↯ 14:22, 23 February 2006 (UTC)

In general the n-Torus is just S^n-1 X Sⁿ, where Sⁿ is the n-dimensional sphere. (Basically, you take a (n-1)-sphere and rotate is around an axis... think of 3D case, where you rotate a circle around).

All the more reason for me to believe it is a 5-D object... But like I said, I don't really know what I am talking about. -- Meni Rosenfeld (talk) 13:43, 23 February 2006 (UTC)

Ummm..... I think the statement "n-Torus is just S^n-1 X Sⁿ, where Sⁿ is the n-dimensional sphere" is incorrect. In the case n = 3, this would mean that the 3-torus is $S^{2}\times S^{3}$, which is incorrect because the latter space is simply connected whereas the 3-torus is not simply connected (in fact its fundamental group is isomorphic to the sum of three copies of Z). Also it's not entirely clear to me that "the 3-torus is a 4-D object". I can see easily how to embed it in five dimensions: in fact since it's just $S^{1}\times S^{1}\times S^{1}$, we already know that the first $S^{1}\times S^{1}$ can be embedded in three dimensions (that's just an ordinary 2-torus), and then the last $S^{1}$ can be embedded in two dimensions (since it's just an ordinary circle), giving five dimensions altogether. How do you squeeze it into four dimensions? I'm not saying it can't be done; I'm just not a geometer and can't see immediately how it would work. Dmharvey 18:11, 23 February 2006 (UTC)

OK, I think I see now. The statement should be replaced by "the n-torus is just $S^{1}\times T^{n-1}$", where $T^{n}$ is the n-torus. Then you can embed the n-torus in n+1 dimensions as follows. Draw the circle $S^{1}$ in your first two dimensions. We're going to "drag an n-1 torus around the circle". So we have one spare dimension (in the direction of the radius of the circle) plus n-1 dimensions perpendicular to our first two. By induction the n-1 torus that we are dragging requires n dimensions, which is (n-1) + the spare one, so we have enough room. So: a 3-torus needs 4 dimensions, a 4-torus needs five dimensions, etc. That was clear as mud. Sorry. I guess the way to think about it is that in the ordinary 2-torus case, we're dragging a 1-torus (which also happens to be a 1-sphere) around a circle. The correct generalisation is to drag a 2-torus around, not a 2-sphere. Perhaps this helps to visualise the thing, which was the original question :-) Dmharvey 18:28, 23 February 2006 (UTC)

OK, I messed up there... what I meant to say was that the n-torus was $S^{1}\times S^{n-1}$. Mind you, that was still not correct. Tompw 18:49, 23 February 2006 (UTC)

So, what a 3-torus would look like as an animation then? ☢ Ҡi∊ff⌇↯ 08:20, 24 February 2006 (UTC)

I think I can work out what it looks like from one particular "direction", and maybe someone skilled in animation will be able to make an animation for you (I have no idea how to do that). Let's call the dimensions A, B, C, and D. Our main circle (call it M) will be in dimensions A and B. Let R be the radial direction of that circle, so we're going to rotate a 2-torus T which lives in dimensions R+C+D around our circle. Now, I haven't yet specified exactly how to embed T in those three dimensions; there are a few choices, and I think the versions of the 3-torus you get will look kind of different. I'm going to choose the embedding where the axis passing through the "hole" in the middle of the 2-torus T runs along dimension C. Now let's slice the 4-dimensional object we have created. I'm going to take a slice in the A+B+D plane. What this corresponds to is taking a slice of T perpendicular to the C axis and rotating it around the circle M. Such a slice of T looks like two circles with the same centre but different radii.

So here's what the animation looks like: let t (time) run from say 0 to 1. (t is moving along the C axis.) When 0 < t < 1, it looks like an annulus rotated around the circle; or in other words it looks like a "thin 2-torus" contained "inside" a "fat 2-torus". At time t = 1/2 the fat torus has maximal radius and the thin torus has minimal radius. At times t = 0 and t = 1 the two tori coalesce into a single 2-torus, before vanishing completely.

I'm sure other embeddings give interesting results too, I think that's the easiest one to visualise. In the animation you might want to make the outer one transparent and the inner one a different colour so you can see what's going on. Dmharvey 13:27, 24 February 2006 (UTC)

You can represent any n-D object in 3-D or 2-D projections, knowing that you miss a lot ; quasicrystals are a 3-D projection of a 6-D grid. An animation here would be gorgeous. --DLL 21:30, 24 February 2006 (UTC)

A topological 3-torus is trivial to illustrate. Show a point moving inside a solid unit cube. When the point hits a face while moving away from the origin, it teleports to the opposite face, moving toward the origin. For a geometric 3-torus, a specific geometry must be chosen. This can be rather entertaining. Consider the familiar 2-torus, commonly seen as the surface of a donut, depicted in 3D space. It has a geometric form in 4D with the fascinating property that it is "flat", as measured by Riemannian curvature of the surface. (A web search for 'flat torus' finds some nice images.) A comparable option for the 3-torus might map (u,v,w) to the 6D point (cos u,sin u,cos v,sin v,cos w,sin w)/√3; but then a series of projections to 3D or 2D must be chosen. --KSmrq^T 06:07, 26 February 2006 (UTC)

Sub-groups of general linear group

Does there exsist any non-trival subgroup of the (real) General linear group that is not a sub-group of the Orthogonal group? Further, does there exsist such a sub-group that doesn't have the Orthogonal group as a sub-group?

(Alternatively: is there a geometry which is an Affine geometry (which deals with invarients under GL_n), but not a Euclidean geometry (which deals with invarients under O_n)? Further, is there such a geometry which does not "contain" Euclidean geometry?) Tompw 13:40, 23 February 2006 (UTC)

There sure is. For example, take the scalar multiples of the identity matrix by integer powers of 2. This is a group because the product of any two scalar multiples of the identity matrix is another scalar multiple of the identity matrix, and the product of any powers of 2 is another power of 2. None of its elements are orthogonal except the identity matrix itself. —Keenan Pepper 14:16, 23 February 2006 (UTC)

Thanks for that one... any others? (Ignoring those in a similar vein) Tompw 18:43, 23 February 2006 (UTC)

There is plenty to be sure. The symplectic group, the diagonal subgroup, the complex general linear group embedded in the real general linear group, and the group of upper triangular matrices are a few that immediately come to mind. -- Fropuff 20:11, 23 February 2006 (UTC)

Estimation methods

In doing a mock maths paper, one question asked me to estimate the sum of six four-digit numbers by a)the front-end method and b)the "cluster" method. None of us have any idea of what these terms meant, and after some google search(wikipedia search on both terms failed, returning results on topic of statistics), the results shows that the front-end method is to truncate the number to have only two most significant figures, then add them up as usual.(notice in case some number has three digits only as in my example, like 729, it takes 700, not 720). But unfortunately, I still can't find out what the "cluster" method is. Any helps would be appreciated. Thanks. --Lemontea 15:18, 23 February 2006 (UTC)

See if this helps. It's the first Google hit from arithmetic estimation cluster. hydnjo talk 21:17, 23 February 2006 (UTC)

Thanks! It helped a lots. Some follow-up questions: a)From the webpage you provided, does that means both methods are not strictly algorithmic--e.g. it's up to the user to determine which number the group cluster to, how many sig. fig. the front-end method should use? b)If we do front-end method all the way to completion(added all digits), then it is no longer an estimation method, is that right? Finally, it seems wikipedia didn't give any mention of these things, maybe we can add them to the corresponding entries. --Lemontea 02:59, 24 February 2006 (UTC)

It seems that "clustering" is a special case say, "estimate the total weight of 24 containers that nominally weigh 10 pounds each (9.8, 10.1, 10.3, 9.9 ....)", so a reasonable estimate would then be 240 pounds. And yes, if you "front-end" to completion then it is no longer an estimate (neglecting mistakes). The idea behind estimating is to quickly get a ballpark answer to the degree of precision that you want but can also be utilized to get the correct answer via mental arithmetic and amaze your friends.  ;-) This could well be the substance of an article on say, Estimating arithmetic. hydnjo talk 05:29, 24 February 2006 (UTC)

Statical Mechanics

(Transferred from the French Reference Desk)

I know that planck's:

${\mathcal {\,}}E=h\nu (e^{h\nu /kT}-1)^{-1}\,$

mulitiplied by 1/n then do another parts of statics. Then do N*n backing to the original formulas

Same or not between these two cases?

note:for 3D

${\mathcal {\,}}E=3h\nu (e^{h\nu /kT}-1)^{-1}\,$

.

.

.

.

.

.

${\mathcal {\,}}E=(1/3)*3h\nu (e^{h\nu /kT}-1)^{-1}\,$

........then others

Between

${\mathcal {\,}}E=h\nu (e^{h\nu /kT}-1)^{-1}\,$

.......then others in Math.....

.

.

.

.

.

.

The same or Not?

• I see that they are not the same. Because of hv/ (e{hv/ kT}-1) do not need to set somewhat coeffiecients, even in this case.

And setting somewhat numbers does not respect Planck.

The official method, is to be the 2nd I wrote before.--HydrogenSu 23 février 2006 à 16:34 (CET)

February 24

Lagrangian dual of a logarithmic barrier problem

Yes, this is a homework question. Yes, I've put some thought into it myself. I just need some help getting over the last little bit.

Here's the primal problem:

minimize $c^{T}x-43\sum _{i=1}^{m}\ln s_{i}$

subject to $Ax+s=b$, $s>0$

Ok, so first I make the Lagrangian:

$L(x,s,u)=c^{T}x-43\sum _{i=1}^{m}\ln s_{i}+u^{T}(Ax+s-b)$

and I can rearrange this so that x & s are separate:

$L(x,s,u)=-u^{T}b+(c^{T}+u^{T}A)x+u^{T}s-43\sum _{i=1}^{m}\ln s_{i}$

and now I find the dual function

$L^{*}(u)=-u^{T}b+\min _{x}[(c^{T}+u^{T}A)x]+\min _{s>0}(u^{T}s-43\sum _{i=1}^{m}\ln s_{i})$

Now, since x is unrestricted, and the dual function can't go off unbounded below, I know that the dual problem must have the constraint $c^{T}+u^{T}A=0$ which means I can reduce my dual function to

$L^{*}(u)=-u^{T}b+\min _{s>0}(u^{T}s-43\sum _{i=1}^{m}\ln s_{i})$

And... now I'm stuck. What, if anything, do I do with the part that's minimized over s? Is there another constraint in the dual problem, or am I stuck with this, or what? moink 02:08, 24 February 2006 (UTC)

Mathematics for the masses

Does anybody know of any puzzles or concepts in mathematics that are:

• Entertaining and educational
• Short
• Easy (People tune out at the sign of a pronumeral)

I work at a Science Centre, and am looking for ideas,

Thanks, --Alexs letterbox 07:56, 24 February 2006 (UTC)

Sure, there are lots. Can't think of any right now, but... Perhaps list of paradoxes is a good place to start looking for all sorts of curiosities. And I think maybe platonic solids match your description. I'll let you know if I think of anything else... -- Meni Rosenfeld (talk) 08:06, 24 February 2006 (UTC)

Take a look at recreational mathematics. :D ☢ Ҡi∊ff⌇↯ 08:18, 24 February 2006 (UTC)

Try cut-the-knot? —Blotwell 10:25, 24 February 2006 (UTC)

I like "guess the number of jelly beans in the jar". It touches on geometry, sampling, and probability (if asked to also estimate the number of each color). And after the contest you all get to eat the jelly beans ! StuRat 11:18, 24 February 2006 (UTC)

Sudoku? Proto||type 12:51, 24 February 2006 (UTC)

We might need an article on pronumeral — does it differ from variable? Gdr 16:28, 24 February 2006 (UTC)

I saw some beautiful graphical proofs of - 1. sum of cubes, 2.pythogorean theorem, etc...which they didnt teach me at school. I found it very nice and inspiring for thinking problems in a very graphical PoV. I could search it up again if u want. -- Rohit 18:07, 24 February 2006 (UTC)

How about fractals? The pictures always look stunning and impressive. --HappyCamper 20:11, 24 February 2006 (UTC)

Sample size

Without having to relearn statistics, how large of a sample would be required to determine, with a normal degree of certainty, valid results about the distribution of topics in Wikipedia (so from a total population of 1 million)? Rmhermen 19:41, 24 February 2006 (UTC)

You'd first have to do a preliminary examination to determine how to determine that, since you don't know what kind of distribution to expect. You'd also have to define the problem in greater detail. --BluePlatypus 20:12, 24 February 2006 (UTC)

How many topics are you thinking of splitting it up into? Is there any overlap? If not, it sounds like the chi-squared test is what you'd use. If so, the only way I know of to find out what sample size you need is to input the data you expect, and try it with progressively larger sample numbers until you get what you want, then take a survey somewhat larger than that in case you guessed wrong on the expected values. Black Carrot 21:30, 24 February 2006 (UTC)

1100 is common number used as a minimum, as it gives about a 3% margin of error over a 90% confidence interval, meaning the results will be within 3% of the actual number 90% of the time. Hopefully a statistician here can show the calculations for this and add details. StuRat 22:45, 25 February 2006 (UTC)

February 25

Pythagorian Triples

Is there a way to prove that there are an infinite number of pythagorian triple families (like 3,4,5 is the family in which 6,8,10 and 9,12,15 belong)? I assume this would involve the proof that there are an infinite number of prime numbers, but you also have to prove that they are in the right proportions. — Ilyanep (Talk) 04:18, 25 February 2006 (UTC)

It's also probably an indirect proof — Ilyanep (Talk) 04:19, 25 February 2006 (UTC)

See Pythagorean triple#Generating Pythagorean triples. It shouldn't be too hard to prove that the triple is primitive if m and n are coprime and one of them is even, and then you're done. —Keenan Pepper 04:31, 25 February 2006 (UTC)

I added a sentence to that article which makes the proof trivial. Since all pythagorean primes are represented by complex numbers of the form $(m+in)^{2}$, one needs only to prove that there are infinite number of complex arguments of such numbers (which is also trivial because their arguments are twice the argument of $m+in$.  Grue  16:36, 25 February 2006 (UTC)

Take any odd integer greater than 1 - call it a. Square it, so you get a². Now express a², which is also odd, as the sum of two integers whose difference is 1 - call these b and c (note that b and c will be coprime). Then

$c^{2}-b^{2}=(c+b)(c-b)=c+b=a^{2}$

So (a,b,c) is a Pythagorean triple - for example, starting with a=13 you get (13,84,85). So each odd integer > 1 gives you a distinct Pythagorean triple. This does not generate all Pythagorean triples, but it does generate an infinite number. Gandalf61 16:44, 25 February 2006 (UTC)

You can also go this more "complicated" method. Here's a sketch. Suppose $x^{2}+y^{2}=z^{2}$. Now divide both sides by $z^{2}$, and set a = x/z, b = y/z. The equation then becomes $a^{2}+b^{2}=1$, which is precisely a unit square. Suppose you have a line that goes from the origin to the circle. If the line has a rational slope, the point of intersection must also be a pair of rational points. So, now, you try and parameterize the original variables in terms of the slope, and this will give you the family of all the possible solutions, and you can show that there are an infinite number of them. --HappyCamper 18:26, 25 February 2006 (UTC)

In my universe, a²+b²=1 is precisely a unit circle. Furthermore, the line with slope 1, which I'm sure is a rational slope, intersects the circle at (¹⁄_√2,¹⁄_√2), which I'm sure is not a rational point. --KSmrq^T 19:04, 25 February 2006 (UTC)

Right. HappyCamper might have been misremembering the following fact. First pick an initial point P on the circle (say (0, 1)), and then consider any line with rational slope passing through P. Except for the tangent line, every such line intersects the circle in exactly one other point whose coordinates are rational; and conversely, if you take a point Q on the circle with rational coordinates, then the line PQ has rational slope. Effectively this is an example of a rational map between the projective line and the circle. Dmharvey 19:22, 25 February 2006 (UTC)

Oh yes, thanks for correcting that critical mistake of mine. A while ago, I used this method to find the all the rational solutions to x^2 + xy + y^2 = 1, which of course, is related to triangles which have angles of 120 degrees instead of 90 degrees. (The 3-5-7 triangle is an example) --HappyCamper 20:50, 25 February 2006 (UTC)

Which nicely brings us to the generation of Pythagorean triples via a rational parameterization of a unit circle. Let the fixed point be (−1,0). Then points on the circle are given by x = (1−t²)/(1+t²), y = 2t/(1+t²). Here t is, indeed, the slope; and clearly a rational slope gives a rational point. More directly, for any rational t we have a Pythagorean triple obtained from 1−t², 2t, 1+t² by clearing denominators. Conversely, a rational point, (x,y), guarantees a rational slope, t = y/(1+x). It suffices to restrict attention to 0 ≤ t ≤ 1, the first quadrant. --KSmrq^T 23:45, 25 February 2006 (UTC)

Golden Times on a clock face

Does anyone have a solution to the following question that would be accessible to an average high school student? "If a rectangle is drawn inside the circle touching where the hands of a clock are pointing, at how many times will the rectangle be 'golden'?" [Don]

Visit http://en.wikipedia.org/wiki/10:08 for clarification. There is an image at http://en.wikipedia.org/wiki/Talk:10:08 [Don]

Well, the answer is either 22 or 44 times in 12 hours, depending on whether you also allow the hands to point to the short side of the embedded rectangle. Suppose x is the fraction of the circle taken up by the long side of the embedded golden rectangle. Then the times are:

${\frac {12}{11}}(n+x),{\frac {12}{11}}(n+1-x)$ and optionally ${\frac {12}{11}}(n+{\frac {1}{2}}+x),{\frac {12}{11}}(n+{\frac {1}{2}}-x)$ hours since midnight or noon, where n is an integer in the range from 0 to 10. Arthur Rubin | (talk) 17:27, 25 February 2006 (UTC)

Using "SohCahToa" trigonometry, I found that the central angle that intercepts the long side of the embedded golden rectangle is approximately 116.565 degrees. That makes your "x" = 116.565/360 = 0.32379. If we let n=9, the (12/11)(9+x) formula gives 10:10:17.07, one instance of "golden time" that I found earlier by a different means. However, I confess that I do not understand why your formulas work. Please explain how you derived them. Thanks! [Don]

What sort of a function is X^X?

A variable to a constant is a parabola, a constant to a variable is exponential, but what is a variable to a variable?

It is neither of those. X^X is equivalent to e^{X ln(X)}, but I don't know a name for a class of functions that contains that. —Keenan Pepper 17:22, 25 February 2006 (UTC)

I've never really thought of it before, but perhaps Transcendental function? Beautiful beautiful topic: Differential Galois theory. --HappyCamper 18:36, 25 February 2006 (UTC)

Also, note that the function y = x^c is only a parabola if c=2. In general, if c is a constant integer, the function is called a polynomial. ×Meegs 03:11, 26 February 2006 (UTC)

Since we're on this topic, how do we find the lowest point of $f(x)=x^{x}$? Derivating it we get $f'(x)=x^{x}\ln x+x^{x}=e^{x\ln x}\ln x+e^{x\ln x}=e^{x\ln x}(\ln x+1)$, but I can't go any further. ☢ Ҡi∊ff⌇↯ 05:12, 26 February 2006 (UTC)

Psh, I guess I was too tired when I tried this yesterday.

$e^{x\ln x}(\ln x+1)=0$

This will happen when either $e^{x\ln x}$ or $\ln x+1$ are zero, but this only happens for $\ln x+1$. So we have $\ln x=-1$, which gives us $x=e^{-1}$. Bah! ☢ Ҡi∊ff⌇↯ 05:22, 26 February 2006 (UTC)

Define g(x) = ln (f(x)), where f(x) = x^x. Then, since the logarithm is a monotonically increasing, you can just find the minimum of g(x), which in turn will let you find the minimum of f(x). The answer should be "reasonably nice number". Sort of a nifty trick which is very useful in statistics and detection theory. Not to mention statistical thermodynamics. --HappyCamper 05:23, 26 February 2006 (UTC)

Nice trick. Thanks for that! ☢ Ҡi∊ff⌇↯ 06:28, 26 February 2006 (UTC)

Yes, that trick comes in very handy. Let's say you have a nasty exponentiated multivariable beast. What happens, is that very often, when you take the derivative with respect to a certain variable, a lot of the other irrelevant variables just disappear! --HappyCamper 15:02, 26 February 2006 (UTC)

$e^{x\ln x}(\ln x+1)=x^{x}(\ln x+1)$. Which is strictly increasing and isn't real for x < 0 and undefined at x = 0. —Ruud 05:24, 26 February 2006 (UTC)

I just got to $e^{x\ln x}$ because I thought that'd be important for the solution, but in the end leaving it as x^x would be the same. Bleh. ☢ Ҡi∊ff⌇↯ 06:27, 26 February 2006 (UTC)

On second thought, it seems to be quite an interesting function when x < 0. —Ruud 05:45, 26 February 2006 (UTC)

Agreed. I've been trying to make a 3D plot of it for a while but I don't have any program to do so. ☢ Ҡi∊ff⌇↯ 06:27, 26 February 2006 (UTC)

Looks like an exponentially damped corkscrew for x<0.

-lethe ^{talk +} 07:08, 26 February 2006 (UTC)

Higher Maths Formula

Hi there, I searched for one of the numerous parabola formulas: $y=k(x+a)(x+b)$ formula, but got no hits; I also searched the parabola article. Can anyone give me the specific article for this formula? Thanks, ^KILO-_LIMA 22:47, 25 February 2006 (UTC)

Assuming k,a,b are constants and x is a variable, then it's a form of Quadratic equation. --BluePlatypus 22:59, 25 February 2006 (UTC)

It's called a merkwaardig product in Dutch but I wouldn't know how to translate that into English. Maybe irreducible polynomial is of any help? Cheers, —Ruud 23:01, 25 February 2006 (UTC)

I've heard it being called "Factored form", but I don't think there is a Wikipedia article on it, and no section on it in the quadratic equation article. --Borbrav 01:51, 26 February 2006 (UTC)

The article on polynomials calls it 'factored form'. It's not specific to quadratics. --BluePlatypus 03:17, 26 February 2006 (UTC)

Yes, I have heard it referred-to as the "factored form" as well. Among other things, this form makes it easy to find the roots (the "x-intercepts") of a function. The first section of the article Factoring, entitled "Factoring a Quadratic Equation" should help you in dealing with parabolas. ×Meegs 03:23, 26 February 2006 (UTC)

February 26

Mechanical aproximations of functions

I've been interested on mechanisms to aproximate certain functions in the physical world. For example, you can draw a circle by fixing a point somewhere and rotating a fixed length thing around. An ellipse can be drawn with a string with ends fixed on the foci. Sure, easy, because they can be defined like that anyway. But, I'm looking for methods for a few particular functions, not sure if they are even possible, but it's worth asking...

1. sin x
2. x²
3. e^x
4. ln x
5. 1/x
6. (e^θ,θ)
7. the catenary

Note that I'm not looking for straight geometric algorithms, like using a ruler\straightedge and compass, or careful measures (except on the case of certain, necessary proportions), but mechanical devices that actually "plot" these curves in their natural movement. Also, they must be table-top gadget things, so saying I could throw a sphere on a plane to plot x² isn't really what I want. Additionally, I'd like to know if there's any interesting, physical method to approximate e, something akin to Buffon's needle?

Well, that'll be all. :P ☢ Ҡi∊ff⌇↯ 13:25, 26 February 2006 (UTC)

Okay, at least for sinx, if you agree to some level of sophistication, you can create a mechanical device which when you rotate it, it moves a long sheet of paper and simultaneously moves a pen in an essentially harmonic motion, thus plotting a sine wave on the paper (I hope you understood what I meant, couldn't find the best words). As for some other curves, and without using such gizmos, I have an idea which I haven't really thought out through, but it just might be crazy enough to work: Suppose you wrap 2 strings around a wheel in opposite directions - so that when the wheel rolls, it releases one string and pulls the other. If the strings are kept in tension, this should keep the difference of lengths of the strings constant. So, by wrapping them around properly placed pulleys\nails, and attaching them to a pen, you can create a curve where the difference of distances between 2 points is constant - Which I believe is a hyperbola, essentially plotting 1/x. With the proper setting, I think this idea can allow plotting a curve where any given linear combination of distances between any number of arbitary points is constant - Opening many possiblities, and with some tweaking perhaps including a parabola. What do you think? -- Meni Rosenfeld (talk) 14:23, 26 February 2006 (UTC)

For the sine, the problem with the rotating wheel is to extract the movement in one axis without distortion. I've been thinking of a few devices, but they end up with either a cycloid or some weird shape I can't name. ☢ Ҡi∊ff⌇↯ 14:53, 26 February 2006 (UTC)

The catenary's an easy one: put a pen on the rim of a circle and roll the circle I'll have to think about the others. Confusing Manifestation 14:31, 26 February 2006 (UTC)

You're mixing it up with the cycloid. Funnily I, too, mistook it for the cycloid at first, but followed the link to make sure. Besides, I'm not sure your idea is easy to implement in practice. -- Meni Rosenfeld (talk) 14:34, 26 February 2006 (UTC)

Yes, sorry, I just checked the article to confirm what I thought. I suppose to make a catenary you could try to find some way to make a "string" that can be frozen in place somehow ... Confusing Manifestation 14:41, 26 February 2006 (UTC)

Nah, not what I'm looking for :P ☢ Ҡi∊ff⌇↯ 14:53, 26 February 2006 (UTC)

x^2 = 2^x

There are three solutions for $x^{2}=2^{x}$, x = 2, x = 4 and the other one is negative. I have absolutely no idea how to find this, and I've tried everything I know. How do we solve things like these? ☢ Ҡi∊ff⌇↯ 14:47, 26 February 2006 (UTC)

Did you mean "how to find that these are the only roots" or "how to find the negative root"? If the former, that's easy by analysing the functions. If the latter, this is an essentially non-algebraic equation so the root can't be given explicitly in terms of elementary functions, but it should be possible to represent it with non-elementary functions (I can try to work that out if you'd like), and you can approximate it with any numerical root-finding algorithm (it is roughly -0.76666469596212309311). Hope that's what you meant... -- Meni Rosenfeld (talk) 14:58, 26 February 2006 (UTC)

Oh, so I was trying in vain... Nice to know. ☢ Ҡi∊ff⌇↯ 15:39, 26 February 2006 (UTC)

You may be looking for Lambert's W function. —Ruud 15:21, 26 February 2006 (UTC)

Yeah, that does it. Thanks! ☢ Ҡi∊ff⌇↯ 15:39, 26 February 2006 (UTC)

This is (sequence A073084 in the OEIS) - Fredrik Johansson 15:24, 26 February 2006 (UTC)

I wonder if anything can be said by letting x = a + bi ... when you do this, you get two sets of equations by equating real and imaginary parts

${\frac {a^{2}-b^{2}}{2^{a}}}=\sin(b\ln 2)$

${\frac {2ab}{2^{a}}}=\cos(b\ln 2)$

Dividing the first by the second, we get

${\frac {2ab}{a^{2}-b^{2}}}=\tan(b\ln 2)$ - and ah! This can be written as a quadratic equation

$a^{2}\tan(b\ln 2)-2ab-b^{2}\tan(b\ln 2)\;\,$ and I'm sure something can be said about the admissibility of these roots, the conditions on a and b, et cetera... --HappyCamper 15:45, 26 February 2006 (UTC)

A new form of logic?

Goedel's incompleteness theorem only applies to first order logic, right? If so, would that mean that it might be possible to create a different form of logic that is sound, complete, etc. AND is capable of doing what Goedel's incompleteness theorem says that first order logic is incapable of?