Wikipedia:Reference desk/Archives/Mathematics/2008 September 3

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September 3

hilbart space

what is Hilbert Space in metric space? —Preceding unsigned comment added by Debadhi (talk • contribs) 05:55, 3 September 2008 (UTC)

All Hilbert spaces are metrizable (by definition).

Topology Expert (talk) 10:02, 3 September 2008 (UTC)

for a proof

Is any element of a Hilbert Space complet? —Preceding unsigned comment added by Debadhi (talk • contribs) 06:02, 3 September 2008 (UTC)

Could you please make your question clearer? If you are asking whether any element of a Hilbert space is complete then it is meaningless; an element can not be complete but a space can.

Topology Expert (talk) 09:57, 3 September 2008 (UTC)

Similarly, if I have interpreted your question correctly, all Hilbert spaces are complete by definition. Perhaps you should have a look at Hilbert space for discussion on these spaces.

Topology Expert (talk) 10:03, 3 September 2008 (UTC)

Complicated Maths Question

Here's the original question: The Harvard University Examination Board dispatches copies of an examination question paper yearly to schools in 50,20 or 5 copies. The delivery instructions produced by the board's computer are in the form of x/y/z meaning x packets of 50 copies, y packets of 20 copies and z packet of 5 copies. The following rules are observed during the dispatch: one copy of the question paper is supplied to each candidate each school receives at least 3 extra copies. the no. of question papers sent to schools must be as small as possible, satisfying the two above rules. once the number of question papers is determined, the board dispatches the minimum number of packets to the school.

The question: in 1999, the school receives 11 packets of papers for 1 subject. calculate the no. of copies received in that year.

The question seems to have one rule lacking as the model answer is 385 but mine is 55(all the packets may only contain 5 papers). —Preceding unsigned comment added by Invisiblebug590 (talk • contribs) 07:28, 3 September 2008 (UTC)

Your answer violates the rule that the number of packets must be minimized. If they wanted to send 55 papers, they'd send it as 50+5 (2 packets) not 5+5+5+5+5+5+5+5+5+5+5 (11 packets) --tcsetattr (talk / contribs) 07:45, 3 September 2008 (UTC)

385 may be the least number of copies that require 11 packets if packaged to minimise number of packets, but it it obviously not the only number of copies that the school could have received - another possibility is 550 papers in 11 packets of 50. Unless it specifies "least number of copies", I think the question is ambiguous. Gandalf61 (talk) 10:54, 3 September 2008 (UTC)

You are correct. The full list of answers requiring 11 packets at a minimum are: 385, 415, 430, 445, 460, 475, 490, 505, 520, 550; With 385 being the smallest of the answers. Anythingapplied (talk) 15:27, 3 September 2008 (UTC)

A bit of a strange question...

Take this example problem: If a 10 foot tall ladder leans against a wall, and the base of the ladder is 5 feet away from the wall, how far up the wall does the ladder go? The answer using Pythagoras' theorem is:
$5^{2}+b^{2}=10^{2}\,$
$b={\sqrt {100-25}}=5{\sqrt {3}}\,$
Now the square root gives a positive and a negative answer. In this real world example it is quite obvious that the one we want--the correct answer--is the positive one. My question is with things like this where we must pick between 2 or perhaps more mathematically valid answers is there a more "formal" way of deciding of which one to use. For example, in a case where imagining the problem in our heads is difficult common sense may not be of much help to us. --212.120.246.239 (talk) 15:06, 3 September 2008 (UTC)

It really depends on the context. In a physical problem, there are instances where negative solutions are meaningless, or where non-integer solutions are not acceptable.

Consider a family in which the number of children is X. If two less than the number of children, multiplied by five less than three times the number of children, equals 4, then we've got a quadratic equation: $(x-2)(3x-5)=4$, which is equivalent to $3x^{2}-11x+6=0$. Solving this quadratic equation, we obtain two answers: $x=3$ and $x={\frac {2}{3}}$. The latter answer is clearly inapplicable to the problem.

On the other hand, if we're just solving the equation $(x-2)(3x-5)=4$, with no thought of application, then we'll keep both answers. The key is in your question: "is there a ... way of deciding which one to use?" If there's an application, then it may provide constraints on acceptable solutions. Determining which solutions apply and which do not apply to a particular application is part of the challenge of applied mathematics. I don't know of a general method, simply because applications can vary widely. Sometimes, a solution that seems meaningless in a physical problem will lead to new physical ideas. If memory serves, the existence of Black holes was predicted when some equations "blew up", with an infinite term appearing, which was at first assumed to be extraneous. I know there are other examples of this sort of thing, but that's the one that comes to mind just now.

In other cases, simply checking whether our solution satisfies the original problem will lead to some solution being thrown out. Consider: ${\frac {x+1}{x-2}}=x+9+{\frac {3}{x-2}}$. If we multiply both sides by $x-2$, we obtain another quadratic equation: $x+1=x^{2}+7x-15$. We can solve this to obtain $x=-8$ or $x=2$, but the latter solution clearly does not work in the original problem. (We could have caught this earlier by being more careful, and noting that $x\neq 2$ is implied by the original equation.)

I realize this isn't quite a succinct answer, but I hope it helps illuminate your question somewhat. -GTBacchus^(talk) 16:02, 3 September 2008 (UTC)

Whenever you use mathematics for a real world problem, you have to come up with a mathematical model that is basically a way of converting your real world concepts into mathematical concepts. In your example you are modelling lengths as non-negative real numbers. The fact that they are non-negative is part of the model, so acts as a constraint to the mathematics. There is no absolute way of working out what constraints you will need when modelling any given situation, coming up with a model is the first, any often hardest, part of solving the problem (once you have a good model all that's left is often basic arithmetic). --Tango (talk) 18:00, 3 September 2008 (UTC)

• (ec) You ask 'how FAR UP the wall does the ladder go?', so you define a vertical axis, along which you measure the 'height', and the axis is oriented upwards. The negative result then means simply a point below the floor. An algebraic equation, which you used to express the problem, does not contain a condition of 'only positive ordinate allowed', so the Pythagorean formula describes two triangles: the one between the ladder, the floor and the wall, and the other one, which is the reflection of it in respect to the floor. Eventually you get two solutions, defining the upper and the lower vertex of the upper and the lower triangle, respectively. --CiaPan (talk) 18:11, 3 September 2008 (UTC)

• In 'formal' mathematics the values which satisfy the equation are the ones that are in the problem's domain. Going from a real world problem to the mathematics involves deciding the domain of the variables as well as the equations that apply to them. For number of children the domain would be the non-negative integers (except in statistics of course!). If you want only heights and don't allow negative heights then you can either have a condition that the height is greater than or equal to zero and use the domain of real numbers (but of course negative ones won't satisfy the condition), or else you could use a domain consisting only of the non-negative reals. Dmcq (talk) 18:46, 3 September 2008 (UTC)

• The quadratic equation is a tool you can use to find out the answer you want. But the tool does not know the context of the question, so it just spits out all the possible answers that could, in some context, satisfy it. An analogy is this question: How do I get from A to B? Well, the options include walking, crawling, driving a car, swimming, flying by plane, taking a bus, train, taxi etc etc etc. All those answers could apply to specific As and Bs. But if your A was your kitchen and your B was your bedroom, then only one of these options (walking) would actually apply, and you'd know which one it was. (Ok, maybe crawling could be an option, too, but not the others). -- JackofOz (talk) 01:26, 5 September 2008 (UTC)

Peanut shaped

Is there a name for that geometric shape? If so, what is it?--Honeymane_{Heghlu meH QaQ jajvam} 22:07, 3 September 2008 (UTC)

As far as I know, there isn't a specific name for a "peanut shaped" object. If you draw a Cassini oval with b/a slightly greater than 1 - say between 1.1 and 1.2 - then you get a two-lobed curve that resembles the cross-section of a peanut. Gandalf61 (talk) 11:02, 4 September 2008 (UTC)

Really? I was under the impression that there was names for everything. Oh well.--Honeymane_{Heghlu meH QaQ jajvam} 20:48, 4 September 2008 (UTC)

This Prolate spheroid certainly looks pretty much like a peanut to me. I got linked to the article from Rugby ball because, well, I find it comical that some people refer to rugby-players as peanut-huggers. 194.221.133.226 (talk) 11:18, 5 September 2008 (UTC)

I was thinking more about the outer shell of the peanut, the part that looks sort of like dumbbells.--Honeymane_{Heghlu meH QaQ jajvam} 16:02, 5 September 2008 (UTC)

I think a mathematician would most likely call it "peanut shaped". Playing around with Greco-Roman roots I came up with "arachiform", for which a Google search actually turns up a few uses alongside such terms as "obpyriform". But I can't deduce from those few references whether "arachiform" refers to the shape of a peanut shell or of the nut inside or neither. -- BenRG (talk) 18:44, 5 September 2008 (UTC)