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April 11

Name for the set of all solutions

In a constraint satisfaction problem, is there a name for the set of all solutions? Black Carrot 06:59, 11 April 2007 (UTC)[reply]

If "the set of solutions" is too long, you can use "the solution set". --Lambiam ^Talk 12:40, 11 April 2007 (UTC)[reply]

Volume

I need to know what A4³ is for my design project. I worked it out to be something along the lines of 230 billion billion so I know I'm wrong. Can someone help? —The preceding unsigned comment was added by Nebuchandezzar (talk • contribs) 10:28, 11 April 2007 (UTC).[reply]

Could you explain what you mean by that expression? Our article A4 lists many possible meanings, but none that you'd try raising to the cube. --Lambiam ^Talk 12:43, 11 April 2007 (UTC)[reply]

I mean the area of an A4 piece of paper as a volume. i.e. (210 × 297)³

Huh? Isn't V = lwh? Splintercellguy 15:08, 11 April 2007 (UTC)[reply]

What you are trying to do is somewhat meaningless. The units in your representation of the A4 paper size are millimeters, so it is 210 mm × 297 mm. This corresponds to an area of 62370 mm², which is – not coincidentally - almost the same as 1/16th of a square meter: 62500 mm². If you use other units, for example inches, you get completely different numbers. Then A4 is 8.27 inch by 11.69 inch, giving an area of 96.67 square inch. Area has the dimension of L², where L stands for "length". If you raise that to the cube, you get a dimension of L⁶; for example, 242620354053000 mm⁶ (where the number of millimeters-to-the-sixth is closer to 240 million million). But volume has a dimension of L³. Trying to view one as the other is a bit like asking "How much does a kilometer weigh?", or "What is the speed of a calorie?". --Lambiam ^Talk 16:21, 11 April 2007 (UTC)[reply]

Assuming the OP meant to calculate the volume of the paper, Splintercellguy gave the right formula - you just need to multiply the area by the ~~width~~ thickness. -- Meni Rosenfeld (talk) 16:30, 11 April 2007 (UTC)[reply]

I think Meni meant to say the paper thickness. StuRat 17:13, 11 April 2007 (UTC)[reply]

Definitely. -- Meni Rosenfeld (talk) 17:42, 11 April 2007 (UTC)[reply]

Perhaps you mean the volume of a cube made of 6 sheets of A4 paper ? Since A4 paper isn't square, you would need to trim off the longer edge to make it square. This, and some tape, would allow you to create a 210 x 210 x 210 cube. You do the multiplication. If any rectangular prism will do, then you can make a 210 x 210 x 297 prism. Again, you do the math. StuRat 17:01, 11 April 2007 (UTC)[reply]

Also, if you don't mind sides being made of two sheets of paper taped together, you could make a 210 x 297 x 297 prism with 7 sheets of paper or a 297 x 297 x 297 cube with 8 sheets. You could, of course, make any sized cube or prism by continuing to tape more pieces of paper together for each side. StuRat 17:10, 11 April 2007 (UTC)[reply]

Volume of the area.... Well, um. If you need the volume of one piece of A4 paper, you know the width and length, else it wouldn't be an A4 piece of paper. Then you just need to find the thickness. Since the thickness is probably too hard to measure, you can measure the thickness of a larger number of paper, like 100, or 500, then divide it by the number of papers. So if you have 100 sheets of A4 paper, and the thickness of those 100 sheets is 1 cm, then each sheet is 0.1 cm, or 0.1 mm. Then you would get the volume of that sheet of paper by multiplying that number by 210 mm and 297 mm. Make sure you use the same units. Then the resulting volume will be in mm^3. You can convert that into something else if needed. --Wirbelwindヴィルヴェルヴィント (talk) 23:16, 11 April 2007 (UTC)[reply]

I believe User:Wirbelwind meant 0.01 cm. StuRat 02:40, 12 April 2007 (UTC)[reply]

what is the fastest way to test prime aprime number?

is there any fast way to test agiven priem number?is there any fast way to test an arbitrary number and find out if this number is aprime number or not?80.255.40.168 12:05, 11 April 2007 (UTC)mdm[reply]

See our article Primality test. --Lambiam ^Talk 12:45, 11 April 2007 (UTC)[reply]

You actually asked the same question just a week ago here, and also the day before here. Did you not like the answers given? --Lambiam ^Talk 12:51, 11 April 2007 (UTC)[reply]

Mathematical Transforms

I am currently taking an engineering course on Signals and Systems,where we spend a big deal of our time studying many types of transforms ( e.g. Laplace Transform,Fourier transform, Z-transform,Hilbert transform,etc...). - We study these transforms to solve a wide range of engineering problems which would be very hard and tedious to solve without utilizing a transform.For example, it is very easy to solve ordinary diffrential equations and convolution integrals using the laplace transform. - My professor pledged that if someone in class submitted a new transform with some helpful properties,he would put a full grade on the course for that student. - So,can somebody please refer me to some pages where i can learn about the mathematics behind transforms and their theory? - And under what branch of mathematics do they fall after all? - - Thanks in Advance - Hisham1987 13:07, 11 April 2007 (UTC)[reply]

For a list of transforms you can use as a starting point for your investigations, see List of transforms. You can also investigate Category: Transforms. Your best bet of finding something new might be in the area of wavelet transforms. The common idea is to find a family of functions that span a vector space of interesting functions you want to approximate. Usually the vector space is infinite-dimensional, but you approximate the functions by a finite linear combination of functions from the family. For the last, you typically pick an orthonormal basis. Good luck. --Lambiam ^Talk 15:36, 11 April 2007 (UTC)[reply]

Existence of a radical of a Lie algebra

Hello, a radical of a lie algebra is the unique maximal solvable ideal of that Lie algebra, containing all other solvable ideals. I need to prove existence of such a radical. Now my book explains why the sum of two solvable ideals is again a solvable ideal. But then it seems the problem is already solved...but is it?? It's not because the property holds for two solvable ideals that it holds for all of them (quite likely infinitely many?)Thank you,Evilbu 20:53, 11 April 2007 (UTC)[reply]

I'm not quite sure why you think the problem is solved by the sum property. You have to show essentially three things, of which one comes for free. First, you must show that there exists at least one maximal solvable ideal. This kind of thing is usually done by somehow giving a construction for it, even if the "construction" is not terribly constructive, and showing (a) that the construction gives a solvable ideal, and (b) that it is maximal in the collection of solvable ideals. But you must also show something stronger, namely (b') that it contains all solvable ideals, and then maximality follows, so you can skip (b). Finally, you must show (c) uniqueness, so assume that you have two maximal solvable ideals, and show that they coincide. This last step is trivial. --Lambiam ^Talk 22:10, 11 April 2007 (UTC)[reply]

The sum property immediately shows that a maximal solvable ideal contains every other solvable ideal (which also gives uniqueness, of course). Thus, only existence of a maximal solvable ideal remains. This is trivial for finite-dimensional Lie algebras, and I suspect that the general case uses Zorn's lemma. However, I do not see why the union of an ascending chain of solvable ideals should be solvable again.--80.136.175.200 22:22, 11 April 2007 (UTC)[reply]

Actually, I think there is no radical in general. Let

{\mathfrak {b}}_{n}

be the Lie algebra of upper triangular n×n-matrices. Then, every

\bigoplus _{k=1}^{n}{\mathfrak {b}}_{k}

is solvable, but

\bigoplus _{k=1}^{\infty }{\mathfrak {b}}_{k}

is not because it contains direct factors having arbitrarily long derived series.--80.136.168.151 13:12, 12 April 2007 (UTC)[reply]

Triangulation

Hi. This is technically a follow-up to my last question, if you could remember it. When I am trying to calculate the lenths of the sides of a triangle according to the diagram in Triangulation, and the way I tried to calculate it last time, I have a scientific calculator. Which mode do I use? My calculator only has DEG, RAD, and GRA(D) . Which one do I use? They all give different answers. I couldn't really find an answer to this on wikipedia. Also, is the angle A in the diagram in degreesor radians, or does it not matter? Could someone help? Thanks. – A stroHur ricane 00 1^{(Talk+Contribs+Ubx)(+sign here+How's my editing?)} 23:59, 11 April 2007 (UTC)[reply]

Pick a mode:

(1) degrees

(2) radians

(3) gradians

And stick to it. They should give you exactly the same answer. If you can demonstrate that they do not then publish it here and we shall point out to you what the problem is.202.168.50.40 01:31, 12 April 2007 (UTC)[reply]

Bear in mind that some things you might take for granted (depending on background) depend on the units you're using. A right angle, for example, is 90 degrees, π/2 radians, or 100 gradians. Make sure that all your quantities are in the right units. Tesseran 05:03, 12 April 2007 (UTC)[reply]