Wikipedia:Reference desk/Archives/Mathematics/2008 August 5

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August 5

Integral of an Absolute Value

What technique should I use to integrate $\int (|x^{2}-4|,x,0,3)$? I tried u-substitution but the $2x$ does not cancel out.

Thank you!

User:Syst3m3rr0r (talk) 00:41, 5 August 2008 (UTC)

Do you mean $\int _{0}^{3}|x^{2}-4|\,dx$? If so, just split it as $\int _{0}^{2}|x^{2}-4|\,dx+\int _{2}^{3}|x^{2}-4|\,dx$. Algebraist 00:44, 5 August 2008 (UTC)

(edit conflict)Do you mean $\int _{0}^{3}\left|x^{2}-4\right|dx$? I think the only reasonable way is to use the definition $\left|x\right|={\begin{cases}x{\mbox{ if }}x\geq 0\\-x{\mbox{ if }}x<0\end{cases}}$ and from there split the integral into two parts. Confusing Manifestation(Say hi!) 00:48, 5 August 2008 (UTC)

That's weird. Why does your use of \left and \right force the small space before dx? Algebraist 00:53, 5 August 2008 (UTC)

Haven't a clue. Confusing Manifestation(Say hi!) 00:10, 6 August 2008 (UTC)

I think when you do \rigth Tex interprets this as a right side deliminator, which has different spacing than a verticle bar (which I think is normally interpreted as a binary operator). Oded (talk) 01:33, 6 August 2008 (UTC)

Alright, I get it now. Thank you very much. Sorry for the poor notation, I haven't used LaTeX before. —Preceding unsigned comment added by Syst3m3rr0r (talk • contribs) 00:52, 5 August 2008 (UTC)

I messed around trying to see if you could do a substitution and ended up having to do integration by parts repeatedly and getting a power series: which is not quite desirable. The only two ways I know of to antidifferentiate an absolute value are what they’ve done above, or to use the fact that ${\frac {d}{dx}}|x|={\frac {|x|}{x}}$ which is essentially the same thing just more cumbersome but avoids the use of piecewise functions. When combined with the chain rule for more complicated things, it gets very cumbersome quite fast, for instance ${\frac {d}{dx}}|x^{2}-4|={\frac {|x^{2}-4|}{x^{2}-4}}\cdot 2x$ GromXXVII (talk) 11:53, 5 August 2008 (UTC)

You think that's bad, try using the alternative definition $\left|x\right|={\sqrt {x^{2}}}$. Confusing Manifestation(Say hi!) 00:10, 6 August 2008 (UTC)

Doesn't that method fail because the region of integration includes a non-differentiable point? Doesn't integration by parts require the things you're differentiating to be differentiable? --Tango (talk) 00:39, 6 August 2008 (UTC)

I think yes and no. If you were to break it into two(or more) different integrals at the non-differentiable point(s) you’ll get the same expression for both pieces – and so that single expression holds for all other points. However, because of the introduced the denominator to do the antidifferentiation, the expression doesn’t exist at that point(s) anyway. So yes I should have been more careful, but I think it is still valid. GromXXVII (talk) 17:59, 6 August 2008 (UTC)

Permutations

how many different groups of six can i make using numbers 1 through 54 without using the same number twice —Preceding unsigned comment added by 99.207.12.77 (talk) 09:43, 5 August 2008 (UTC)

Do you mean without using the same number twice in each group (for example {[1,2,3,3,4,5];[1,2,3,4,4,5]} would be illegal but {[1,2,3,4,5,6];[1,2,3,5,6,7]} would be legal) or without using the same number twice at all (for example {[1,2,3,4,5,6];[6,7,8,9,10,11]} would be illegal)? --Slashme (talk) 09:51, 5 August 2008 (UTC)

Assuming you mean the first of these (equivalent to the question 'how many different hands of six cards are there from a standard deck of cards with two (different) jokers included), the answer is ${\frac {54!}{6!(54-6!)}}={\frac {54\times 53\times 52\times 51\times 50\times 49}{6\times 5\times 4\times 3\times 2\times 1}}$. See Combination.--130.88.123.142 (talk) 11:40, 5 August 2008 (UTC)

Small order of operations fix: It should be ${\frac {54!}{6!(54-6)!}}$. StuRat (talk) 13:08, 5 August 2008 (UTC)

${\frac {54!}{6!(54-6!)}}=25827165$. These answers assume the order of the numbers doesn't matter, for example [1,2,3,4,5,6] and [2,1,6,4,5,3] are not different. Somebody other than the original poster added the headline "Permutations", but order usually does matter in a permutation, and then the answer would be ${\frac {54!}{(54-6)!}}=18595558800$. The original poster used the word "groups". Group (mathematics) has a special meaning which was not intended here. Set would be a better word if the order doesn't matter. PrimeHunter (talk) 13:27, 5 August 2008 (UTC)

Again, the parens seem to be in the wrong place (in the first equation, but correct in the second). I believe that (54-6!) would mean (54-6×5×4×3×2×1) = (54-720) = -666, which, while an interesting number (the anti-antichrist ?), is not what was meant here. I believe it should be written as (54-6)! or 48!. StuRat (talk) 01:26, 6 August 2008 (UTC)

Yes, the "permutations" was me; I should have said "combinations". Very sloppy, sorry ;-) --Slashme (talk) 13:55, 5 August 2008 (UTC)

NP-completeness of 3-colouring

I'm trying to figure out the proof of the NPCness of 3-colouring (this isn't homework). Karp's paper is very terse/cryptic. Garey and Johnson leave it as an exercise. Sipser also leaves it as an exercise, but gives the hint to use an OR-gadget like this:

   *
  / \
 *---*
 |   |
 *   *

OK, but how do I use this gadget? I'm having trouble figuring out which are the inputs and which is the output. Also, graph-colouring isn't "biased" towards any colour but OR-gates are "biased" towards true --- how does one resolve this assymetry? --Taejo|대조 15:32, 5 August 2008 (UTC)

It’s in Algorithms by Johnsonbaugh and Schaefer, although [possibly] different from the approach you have. They reduce 3SAT to 3-colorability by constructing a [fairly complicated] graph which is 3 colorable iff the corresponding expression is satisfiable. GromXXVII (talk) 20:34, 5 August 2008 (UTC)

I don't know how it was originally solved, but I think this will work. The graph has

• nodes labeled T and X with an edge between them;
• a node for each variable;
• a node for the negation of each variable, forming a triangle with the variable and X;
• a node connected to X for each internal wire connecting an OR gate to an OR gate in the CNF circuit; and
• one of those OR-gadgets for each OR gate in the CNF circuit, with the bottom nodes being the inputs and the top node being the output. If the output wire goes to an AND gate then the output node is T.

The OR-gadget doesn't implement an OR gate, but I don't think it matters. The important thing is that the output can't be T unless at least one input is T. It can be F when one input is T, but that will never lead to spurious solutions. In effect you're replacing every $\vee$ in the CNF expression with a variable ranging over $\{\vee ,\wedge \}$, but it's never helpful to set it to $\wedge$. -- BenRG (talk) 21:25, 5 August 2008 (UTC)

Which graphical calculator is best for me?

Hi, I'm considering getting a graphical calculator for my statistics, and would like to know which to get. I want one that allows the greatest number of statistical functions (multiple regression, ANOVA, etc.) and also, just as importantly, has the greatest capabilities for solving equations and systems of equations (eg. being able to solve non-linear simultaneous equations in 2 or more variables, exactly or by approximation). Ease of use is also important, but less so. Are there any I should look at? It's been emotional (talk) 16:42, 5 August 2008 (UTC)

To be able to solve equations exactly you’ll want something with a Computer Algebra System – which essentially puts you at a TI-89/92 or the competing models made by other brands. As for statistics – if you’re going to doing anything beyond a class you might consider a computer package (there are some available for free). Some of the more advanced calculators will do a lot of statistics, but it can be cumbersome especially if you have a lot of data. GromXXVII (talk) 23:10, 5 August 2008 (UTC)

Price support example: utter nonsense, or not?

The example in the Price support article makes no sense to me. It seems to be comparing things that can't logically be compared. That's a vague complaint, but one specific issue I have with it is that the government wouldn't just throw away the 200 units of goods it bought for $1200. Is there a good reason for neglecting the value of those goods, or is the example just nonsense? 128.165.101.105 (talk) 19:02, 5 August 2008 (UTC)

I can't comment on the maths in the example but the government is quite likely to store the surplus in a huge warehouses until it is about to go off, and then sell it in third world countries at a price that undercuts local producers and puts them out of business. Who this benefits I'm not at all sure. Dmcq (talk) 19:28, 5 August 2008 (UTC)

In the case of agriculture at least I think they normally pay not to produce as opposed to acting as a retailer. GromXXVII (talk) 20:23, 5 August 2008 (UTC)

I don’t think the value of the goods is neglected, the example makes note about how the cost to the consumers is actually 450+1200, the 1200 being the cost to keep the goods off the market. GromXXVII (talk) 20:23, 5 August 2008 (UTC)

The US government has bought and/or destroyed perfectly good food crops to prop up prices. This seems illogical, however, since they're unwilling to allow competition and capitalism to work, this is the only way to keep prices high during periods of overproduction. StuRat (talk) 01:19, 6 August 2008 (UTC)

My question was actually supposed to about the logical consistency of the argument. At the end it just says "For this reason, price supports are considered inefficient.". I don't see where that conclusion comes from at all. There's all this arithmetic going on that doesn't make sense, and doesn't support the conclusion.

Also, even if I believe that price supports are bad, that doesn't change the fact that the article doesn't mathematically prove that. 128.165.101.105 (talk) 14:10, 6 August 2008 (UTC)

I don’t think the article is claiming they’re bad, but rather saying their inefficient meaning that the benefit is monetarily less than the cost. That is, the direct cost to the consumers is only 450, and the benefit to producers is 550. But the indirect cost to consumers adds another 1200, which together with the previous 450 is larger than 550. Also note I’m not an economist, I’m just trying to interpret the what the article is saying GromXXVII (talk) 15:04, 6 August 2008 (UTC)

I think inefficient may refer to Pareto efficiency, its inefficient because somebody is worse off. It wouldn't be my first objective. Dmcq (talk) 18:32, 6 August 2008 (UTC)

Sorry it could be Pareto efficient because the producers get more money, I believe the article is saying it isn't efficient because the consumers could give the money directly to the producers so the producers go the same amount but the consumers lost less. Dmcq (talk) 18:43, 6 August 2008 (UTC)

In other words, it's more efficient for the government to just pay farmers to produce less, rather than buying the surplus and throwing it away. Algebraist 18:48, 6 August 2008 (UTC)

This would certainly be a saving because the government could pay less to the producers but the producers would still get the same amount because they don't have to pay to produce the surplus. You have to wonder what the whole point of keeping the price high is though. If it is to support the producers through lean periods it would be better just to subsidize them in such years - or even just have a basic subsidy anyway. I believe this is the way the EU is going with agriculture. A bit of a pity really as set aside protected lots of wild space. Dmcq (talk) 20:22, 6 August 2008 (UTC)

Ray-tracing in more exotic non-euclidian spaces?

I've seen ray-traced views of elliptic space and hyperbolic space, but I was wondering about some of the stranger non-Euclidian spaces, like projective space and taxicab space, if there were a way to draw ray-traced images in those spaces. --Zemyla^t 21:58, 5 August 2008 (UTC)

Well, to do ray tracing you need at least some notion of straight line and a way of embedding the viewer in the space. If you want directional lighting and reflection and refraction and such you need some notion of angle also. A Riemannian manifold gives you all of those automatically, and you can find ray-traced images made with more complicated Riemannian manifolds than the elliptic and hyperbolic planes, like these movies. The projective space RP³ with the usual geometry is just a model of elliptic geometry. I'm not sure what would count as an angle or even a straight line in the taxicab space. -- BenRG (talk) 23:07, 5 August 2008 (UTC)

A natural definition of 'straight line' is 'shortest path' (or something similar). For taxicab geometry, the shortest path between two points is massively non-unique, so I doubt you can get very far. Algebraist 23:14, 5 August 2008 (UTC)

Taxi-cab might be possible - assuming there are opaque objects in a transparent field. Then replacing the rays with a 'project plane' eg a X x Y array of points. Then there are two options:

a. Each path is equally likely (and contributes) - therefor the 'illumination' of a point in the array by an object is proportional the fraction of paths un-obscured.

b. If there is a unblocked path to the object then it is seen (black & white)

because multiple 'objects' would be visible from a given 'screen point' some sort of summable transparent representation of the 'projection' would be needed eg The screen point intensity is proportional to the sum of the 'colours' of each point visible.

It's not really a good geometry to do ray tracing in.87.102.5.5 (talk) 02:14, 6 August 2008 (UTC)

For hyperbolic geometry specifically, see negative curves and other blog entries on the same blog.