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May 8

Integration Problem

I've tried evaluating $\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\cdots \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}{\frac {1}{1-x_{1}x_{2}\cdots x_{n-1}x_{n}}}\,dx_{1}\,dx_{2}\cdots \,dx_{n-1}\,dx_{n}$ for values of n=1, 2, and 3, and they came out nicely to $-\infty$ , ζ(2), and ζ(3) respectively. But I know that you can't take infinite integrals, from the article on Multiple integral. Does this integral actually converge to 1 as n approaches infinity? This is NOT my homework; this problem came up while my friend and I were giving ourselves some calculus problems to do. Thanks. —Preceding unsigned comment added by 70.111.161.98 (talk) 02:28, 8 May 2008 (UTC)[reply]

The integral is indeed equal to

\zeta (n)

for any n. Since

\zeta (n)

converges to 1 as n goes to

\infty

, so does the integral. -- Meni Rosenfeld (talk) 09:42, 8 May 2008 (UTC)[reply]

JC-1 H2 Maths: Partial Fractions headache

I am revising for a test. I can do most of the partial fractions questions, until this one:

8x^2+4x+1/(x^2+1)(2x-1)

I have to split the big fraction into two small fractions.

The normal method is to substitute values of x so one of the functions in the denominator = 0, but I cannot make x^2+1=0, because then it will be x^2= -1 and squares cannot be negative. (We did not learn imaginary numbers.)

What to do? —Preceding unsigned comment added by 166.121.36.232 (talk) 05:18, 8 May 2008 (UTC)[reply]

When you have an irreducible polynomial of degree 2 or greater in the denominator (i.e. when one of the factors on the bottom is more than linear and can't be broken down), you have to do a couple of things:

First, you have to note that in the partial fraction decomposition, the fraction with the non-linear denominator will have a numerator of degree one fewer than the denominator. That is, if you have a quadratic factor, as in this case, its part in the fraction will not just be C/(x^2+1), but (Bx+C)/(x^2+1), which you need to take into account when finding the decomposition. (If you have a repeated factor, e.g. (x+3)^2, you need a decomposition that includes A/(x+3) and A/(x+3)^2.)

Secondly, to actually find the coefficients in the decomposition, you can either do it the hard way, or the usually-not-quite-so-hard way. In both cases, you start with an equation such as ${\frac {1}{(x^{2}+1)(2x-1)}}={\frac {A}{2x-1}}+{\frac {Bx+C}{x^{2}+1}}$ . Then, you multiply both sides by the denominator, giving $1=A(x^{2}+1)+(Bx+C)(2x-1)$ . In the hard way, you then expand it all out, and equate coefficients (giving you n simultaneous equations to solve, where n is the degree of the original denominator). In the usually-not-quite-so-hard way, you notice that the equality holds for all values of x, and choose smart ones to simplify your calculations. The simplest values are those that make one of the terms zero - in this case, x = 1/2 works nicely to give you A. After that, it's up to you to choose nice ones. Some suggestions include x = 0, 1 and -1, since they normally result in things you can work with easily. Also remember that once you've found one of the coefficients, you can use its value in further calculations rather than trying to eliminate it. Confusing Manifestation(Say hi!) 07:02, 8 May 2008 (UTC)[reply]

That’s something I’ve never quite understood about the “usually-not-quite-so-hard-way”, in the original inequality you couldn’t plug in x=1/2 because it was in a denominator. Why does the equation, after multiplying by the denominators hold true for 1/2? Further, why does using that point give you any information about A,B,C? The best I’ve been able to get at is that because the new equation is continuous, but can’t quite string out the logic as to exactly how that implies that the A,B,C found are the same as the original. GromXXVII (talk) 11:07, 8 May 2008 (UTC)[reply]

You've answered your own question. The equality (before multiplying, and consequently, afterward) holds for any

x\neq {\tfrac {1}{2}}

. After multiplying, the equation is continuous, so from being true for any

x\neq {\tfrac {1}{2}}

we can deduce it being true for the limit point

{\tfrac {1}{2}}

. It gives you information about A, B and C because not for all values of these the equality will hold, thus the fact that it holds tells you something about them. -- Meni Rosenfeld (talk) 11:19, 8 May 2008 (UTC)[reply]

If you move onto the complex plane, you can talk about analyticity and the fact that you create a removable singularity (i.e. a discontinuity that you can replace by a limit value and turn it into a point of continuity). Confusing Manifestation(Say hi!) 04:55, 9 May 2008 (UTC)[reply]

And, building up on that, while the original rational functions can't be said to be equal at 1/2, their residues at that point certainly are. -- Meni Rosenfeld (talk) 15:03, 9 May 2008 (UTC)[reply]

Another integration problem

Is there a solution for:

S=\int _{a}^{b}x^{x}\,dx.

Or is there any antiderivative for that matter? /81.233.41.161 (talk) 15:57, 8 May 2008 (UTC)[reply]

No, the function does not have an elementary antiderivative. For some particular values of a and b you can say something intelligent about the definite integral, e.g. sophomore's dream. -- Meni Rosenfeld (talk) 16:38, 8 May 2008 (UTC)[reply]

No antiderivative

Hi. I've been playing with some integrals, and I was working on one which, after a change of variable, turned into $\int \cos(x^{2})\,dx$ , which I don't think has any elementary anti-derivative. I say that because my copy of James Stewart's Calculus identifies $\sin(x^{2})$ as a function for which we have no anti-derivative, so it must also be true for $\cos(x)=\sin(x+{\frac {\pi }{2}})$ . I have two questions: First what's the best way to tell when you've found one of these functions without any elementary anti-derivative? Secondly, I know that some functions with no anti-derivative, such as $e^{-x^{2}}$ , have got special functions (the Error function) defined to be their anti-derivative. Are such functions listed somewhere? Is there one for my $\cos(x^{2})$ function? How can one tell?

Thanks in advance for any insights anyone can offer. -GTBacchus^(talk) 17:52, 8 May 2008 (UTC)[reply]

I doubt there is any general procedure to determine if a function has an elementary antiderivative. For practical purposes, the best course of action will be to feed the function to a CAS and see if it is able to express it. In your case, you should take a look at Fresnel integral. -- Meni Rosenfeld (talk) 18:19, 8 May 2008 (UTC)[reply]

I think Meni's advice on feeding a CAS is usually the best method of finding out whether some function has an elementary primitive.
There are so many non-elementary antiderivatives that it would be hard to build a useful list. In any case, the article List of special functions and eponyms contains references to many of these non-elementary functions (as well as to many other ones). The article on special functions may also be of your interest. Here is also an external site. Pallida Mors 18:41, 8 May 2008 (UTC)[reply]

Thanks very much! -GTBacchus^(talk) 23:48, 8 May 2008 (UTC)[reply]

There is in fact such a general procedure: the Risch algorithm – although it may be disputed if this procedure is truly an algorithm in the formal sense, as it needs an oracle that tells whether a given expression equals 0. In any case, the procedure is too complicated for manual use. --Lambiam 09:39, 9 May 2008 (UTC)[reply]

Oh, for some reason I was under the impression that it wouldn't terminate for functions with no elementary antiderivative. -- Meni Rosenfeld (talk) 11:03, 9 May 2008 (UTC)[reply]

One more thing - you don't even need to have a CAS available. You can simply go to the Wolfram Integrator. -- Meni Rosenfeld (talk) 19:28, 10 May 2008 (UTC)[reply]

Proof Generator?

Is there a program that can generate mathematical proofs if given an equation? If so, where is it? 75.170.42.250 (talk) 22:26, 8 May 2008 (UTC)[reply]

See Automated theorem proving. Algebraist 22:43, 8 May 2008 (UTC)[reply]

What do you mean by "given an equation"? You can't prove an equation, you can just prove that the equation is implied by some premise. --Tango (talk) 22:51, 8 May 2008 (UTC)[reply]

The same can be said about most statements that we prove. Eg. when we prove that the sum of angles in a triangle is 180 degrees we are "just" proving that this is implied by the axioms of euclidian geometry. Taemyr (talk) 23:17, 8 May 2008 (UTC)[reply]

You can word it as "If a, b and c are the angles in a triangle, then a+b+c=180", you can then prove that equation given the stated premise (obviously, you need definitions of the terms, and that's where the axioms of Euclidean geometry come in). Just "x²+5x-4=0" is not something which can be proven or disproven, it's just an equation. It's true for certain values of x and not true for others. For "prove" to be meaningful, the statement has to be always true. (The closest you can get to proving equations is proving identities, which are a subset of equations.) --Tango (talk) 15:28, 9 May 2008 (UTC)[reply]

I can’t seem to find the wikipage for it, but it sounds to me like the difference between a statement and an open statement in formal logic. Most equations are open statements. Some are trivially statements though, such as x+(1-x)=1, or 1 = 2. GromXXVII (talk) 20:10, 9 May 2008 (UTC)[reply]

I've heard some good things about Isabelle (theorem prover). I don't think any currently available automatic prover is very easy to use - you would probably need to know quite a bit about formal logic, as well as the syntax of the program. Even then, they might not be able to solve anything you give them. -- Meni Rosenfeld (talk) 23:15, 8 May 2008 (UTC)[reply]

You might enjoy reading about the Robbins problem ("are Robbins algebras equivalent to Boolean algebras?"). This problem, proposed in 1933, was first solved by the automated theorem prover EQP in 1996. [1] JohnAspinall (talk) 14:42, 9 May 2008 (UTC)[reply]

May 9

Indeterminate relational operator

Is there an official mathematical relational operator that can be used when the equality - or lack of it - of two expressions is not known a priori? For instance, to determine which is the greater of ${\frac {3}{8}}$ and ${\frac {4}{9}}$ one might write,

${\frac {3}{8}}<>{\frac {4}{9}}$ and then by the ususal operations,

$27<>32\,\!$

Thus proving that ${\frac {4}{9}}$ is the greater. However, the symbol <> is not a good choice as this could be construed as "less than or greater than" (ie not equal to) rather than the meaning I intended. SpinningSpark 13:49, 9 May 2008 (UTC)[reply]

OT: 4*8=32^{[citation needed]} Algebraist 13:56, 9 May 2008 (UTC) oops! thankyouSpinningSpark 15:09, 9 May 2008 (UTC)[reply]

I don't think there is anything standard for this. However, operators like

\pm

are generally taken to have a consistent instantiation - for example,

1\pm 2\pm 3

means

1+2+3

oder

1-2-3

, but not

1+2-3

. Analogously,

2\gtrless 3\iff 4\gtrless 6

could be taken to mean "

2>3\iff 4>6

and

2<3\iff 4<6

". But this is non-standard. You can also just not write anything - put the two numbers in front of each other, and leave it blank until you figure out the correct relation. -- Meni Rosenfeld (talk) 14:18, 9 May 2008 (UTC)[reply]

Leaving it blank does not really fit the bill as the idea is not so much to determine the correct relation but rather to be able to manipulate the equation while at the same time showing that the inequality status is not known. The

\gtrless

looks good though -what is its usual meaning? If it is "greater or less than" then it is undesirably excluding "equal to". SpinningSpark 15:09, 9 May 2008 (UTC)[reply]

Uhh, whatever it is you're doing, it's wrong. 4/9 exceeds 3/8. –King Bee ^{(τ • γ)} 15:03, 9 May 2008 (UTC)[reply]

I know - Algebraist already pointed that out above, but was kind enough to write it small. I have now corrected the original question. SpinningSpark 15:09, 9 May 2008 (UTC)[reply]

Too small for my eyes, apparently. Apologies. –King Bee ^{(τ • γ)} 15:14, 9 May 2008 (UTC)[reply]

I don't know of any standard way of writing that. Normally, you would work out which way round it was before writing it - writing it down is then just a proof, rather than the actual working. The actual working you can do on rough paper and it doesn't make any difference what notation you use as long as you understand it. --Tango (talk) 15:31, 9 May 2008 (UTC)[reply]

For my own notes I use

{\stackrel {?}{=}}

or "vs.". The former emphasizes equality vs. non-equality more than I wish it did but I haven't found any better notation. Eric. 144.32.89.104 (talk) 17:15, 9 May 2008 (UTC)[reply]

One way that would avoid the need for such a notation would be to write it as:

{\begin{aligned}{\frac {3}{8}}={\frac {3\cdot 9}{8\cdot 9}}={\frac {27}{63}}\\{\frac {4}{9}}={\frac {4\cdot 8}{9\cdot 8}}={\frac {32}{63}}\\\implies {\frac {3}{8}}<{\frac {4}{9}}\end{aligned}}

--Tango (talk) 18:03, 9 May 2008 (UTC)[reply]

For what it's worth, TeX has the operator \gtreqless:

a\gtreqless b.

The very first meaning that comes to my mind though is that there are instances for which term a is greater than b, some other instances for which

a=b

, and finally some cases for which

a<b

. Pallida Mors 18:16, 9 May 2008 (UTC)[reply]

(Edit conflict) I've used similar for other operators. For example

{\overset {?}{<}}

. Note for this instance though you could probably just circumvent the issue by subtracting one from the other and going from there:

{\frac {3}{8}}-{\frac {4}{9}}={\frac {3\cdot 9}{8\cdot 9}}-{\frac {8\cdot 4}{8\cdot 9}}={\frac {27-32}{8\cdot 9}}=-{\frac {5}{8\cdot 9}}<0

\therefore {\frac {3}{8}}<{\frac {4}{9}}

Ahem. 8*9. We so stink at arithmetic. Too many abstract operators swimming around in our heads I guess. --Prestidigitator (talk) 18:19, 9 May 2008 (UTC)[reply]

I am deeply regretting having given an example (and not just because I got the answer wrong). The question was not to find an alternative way of solving that particular trivial problem. The example was given just to make clear the meaning of the operator I required a symbol for. I guess that

\gtreqless

is the closest I am going to get. Thanks everyone. SpinningSpark 18:30, 9 May 2008 (UTC)[reply]

All the methods given are precisely equivalent, they differ only in notation, which is what you were asking about. --Tango (talk) 19:49, 9 May 2008 (UTC)[reply]

It is exactly your misunderstanding that led to my regret at phrasing the question that way. I know the methods are equivalant but my question was not one of notation or method in general, but for the symbol for a specific operator. SpinningSpark 21:23, 9 May 2008 (UTC)[reply]

It doesn’t seem like he misunderstood – he specifically stated that the notation is what you were asking about.

Anyways, I’m personally biased toward blanks, filling in the correct information when you know it. Such as

{\frac {3}{8}}

___

{\frac {4}{9}}

I also like the question mark relations if there are only two options you might care about: the relation or it’s negation GromXXVII (talk) 23:15, 9 May 2008 (UTC)[reply]

Yeah, I always use the question mark over the less-than symbol. I doubt there's any official symbol, other than the standard variable symbols for relations (R, etc). Black Carrot (talk) 02:11, 10 May 2008 (UTC)[reply]

"Notation" is a superset of "symbols", thus "not... notation... but... the symbol" doesn't make much sense. Surely you know that this "operator" is not a real operator\relation, but rather a handwavish way to remind you what is the goal of your calculation. The reason people pointed out alternative approaches to solving your toy example is to demonstrate why the sought symbol isn't really necessary. -- Meni Rosenfeld (talk) 17:32, 10 May 2008 (UTC)[reply]

May 10

Nodes of Vibrating Plate

I've seen this video [2] and others like it many times, but I can't find what shape those curves are - a maathematical description of them. Can anyone point me in the right direction? Black Carrot (talk) 02:32, 10 May 2008 (UTC)[reply]

Those are Chladni plate interference surfaces

Look here:http://ozviz.wasp.uwa.edu.au/~pbourke/surfaces_curves/chladni/

They have a *LOT* of applications in the design of wood saw blades to reduce noise. 71.193.2.115 (talk) 14:10, 10 May 2008 (UTC)[reply]

Parabolas and Lines

Hello. One of my students (I am a mathetmatics tutor) has a problem which I am unable to help them with. They need to do the following:

Given the parabola, y=X^2

write an equation of the line through a point P (a,a^2) with slope m.
Write a quadratic equation involving x and m that must be satisfied by all points of intersection of any line through point P and the parabola.
Find the slope of a tangent line m, in terms of a. (Hint: The tangent line to a parabola intersects the graph only once.)

—Preceding unsigned comment added by 64.26.98.90 (talk) 20:39, 10 May 2008 (UTC)[reply]

fixed format of original question -- Meni Rosenfeld (talk) 20:45, 10 May 2008 (UTC)[reply]

1 is a standard "find a line with a given slope through a point" exercise. For 2 - you have found the equation of a general line through P. Now you just need to equate it to the parabola. For 3 - use the hint. -- Meni Rosenfeld (talk) 20:45, 10 May 2008 (UTC)[reply]

Seems to me that calculus could be used to find the answer to question 3. Zrs 12 (talk) 21:45, 10 May 2008 (UTC)[reply]

No need for that - you already have a way of finding the points of intersection of a line through P with the curve, the tangent to the curve at P is the line that intersects the curve at only P (with multiplicity two). That's easy enough to find (I won't say more). --Tango (talk) 22:01, 10 May 2008 (UTC)[reply]

Presumably, the question was posed to a class which hasn't studied calculus yet, or was supposed to demonstrate that there are some things you could do even without calculus. I think it does the latter rather nicely. -- Meni Rosenfeld (talk) 22:35, 10 May 2008 (UTC)[reply]

I agree - it's a nice question. --Tango (talk) 23:55, 10 May 2008 (UTC)[reply]

Number of unique outcomes for 5 dice

How many unique outcomes are there if 5 dice are rolled simultaneously?

1 2 3 4 5 and 2 3 4 5 1 should be considered the same outcome, since they have the same combination of numbers.

May 11

Permutation notation

I just want to check this is correct, I'm not 100% sure.

$X_{7}$ = {1, 2, 3, .. , 7}

So, for instance, the permutation notation (127) can be broken down as f(1) = 2, f(2) = 7 and f(7) = 1. I assume this means then: (2, 7, 3, 4, 5, 6, 1) if you write it in full.

Does it therefore mean to say (127) * (12) where * is function composition is the following:

(12) = (2, 1, 3, 4, 5, 6, 7)

then,

(127) = (7, 2, 3, 4, 5, 6, 1)

which could be written (17) with regards to the original permutation? 86.146.141.180 (talk) 10:31, 11 May 2008 (UTC)[reply]

Yes, this is correct. Note, however, that the order of multiplication is not universally agreed. That is, some writers will take (127)*(12) to mean (12) first and (127) second, like you did, but others will do it the other way around. -- Meni Rosenfeld (talk) 10:57, 11 May 2008 (UTC)[reply]

Note that the second line should read (1 2 7)(1 2)=(7, 2, 3, 4, 5, 6, 1)=(1 7), because you’ve performed the operation. That is, it is not (1 2 7) still.GromXXVII (talk) 11:42, 11 May 2008 (UTC)[reply]

Equations of lines

I've read through the book but its not very helpful..

Question 1

Find the equation of line with the gradient -6 that passes through the point (1,5)
Now it says use the formula Y-Y1=M(X-X1) [Y1 and X1 Being the points supplied, M being the Gradient(AKA Slope).

So Logically:
Y-5=-6(X-1)
Y-5=-6x+6 [Expand]
Y-5+5=-6x+6+5 [Make Y the subject]
Y=-6x+11

However the book says 6x+Y=11 Why? (Perhaps it is for this reason, a little earlier in the book i read something about putting it in gradient intercept form or general form, which ever you prefer)

y=-6x+11

is exactly the same as

6x+y=11

. The first form focuses on finding y when x is known, the second focuses on the symmetry between x and y. -- Meni Rosenfeld (talk) 10:48, 11 May 2008 (UTC)[reply]

y=-6x+11 is the same as 6x+y=11, so either can be used according to taste. The second form has the merit that the line can be drawn immediately, as the x and y intercepts are 11/6 and 11. …86.146.174.17 (talk) 10:51, 11 May 2008 (UTC)[reply]

So its some form of fancy shuffling but why does -6 become positive 6? —Preceding unsigned comment added by 60.230.6.43 (talk) 10:56, 11 May 2008 (UTC)[reply]

Because you move it from one side of the equation to the other? If

a=b-c

then

a+c=b

, that's the definition of subtraction. -- Meni Rosenfeld (talk) 10:59, 11 May 2008 (UTC)[reply]

Because 6x is added to both sides, in the same way that you added 5 to both sides of the equation y-5=-6x+6.…86.146.174.17 (talk) 11:00, 11 May 2008 (UTC)[reply]

Question 2

Find the equation of the line that passes through (3,4) and (-2,8)
$Y-Y_{1}=M(X-X_{1})$ where $M={\frac {y_{2}-y_{1}}{X_{2}-X_{1}}}$
So ${\frac {8-4}{-2-3}}={\frac {4}{-5}}$
$Y-4={\frac {4}{-5}}(X-3)$
$Y-4={\frac {4}{-5}}X+2{\frac {2}{5}}$
$Y-4+4={\frac {4}{-5}}X+2{\frac {2}{5}}+4$
$Y={\frac {4}{-5}}X+4{\frac {4}{5}}$

This is also not correct. Any ideas?

The problem is in the last step:

2{\frac {2}{5}}+4=6{\frac {2}{5}}

, not

4{\frac {4}{5}}

. Then there is some room for rewriting the equation in a nicer looking way; your book probably uses

4x+5y=32

.-- Meni Rosenfeld (talk) 10:51, 11 May 2008 (UTC)[reply]

Where do all those numbers come from60.230.6.43 (talk) 11:07, 11 May 2008 (UTC)[reply]

And in both questions, you use upper and lower case x and y indiscriminately - it's best to be totally consistent in notation.…86.146.174.17 (talk) 10:56, 11 May 2008 (UTC)[reply]

y-4={\frac {-4}{5}}(x-3)

\Rightarrow 5(y-4)=-4(x-3)

\Rightarrow 5y-20=-4x+12

I'll let the questioner take it from there. Gandalf61 (talk) 11:36, 11 May 2008 (UTC)[reply]

squares crossed by trhe diagonal

the number of squares that are crossed by the diagonal of a rectangular field covered by m by n squares each of unit squares where m and n are relatively prime is given by ( m+n ) - 1 ... could you please explain why is it so 117.200.1.99 (talk) 15:34, 11 May 2008 (UTC)[reply]

Start at one corner and go along the diagonal counting how many times you cross a line. Bear in mind that, since m and n are coprime, you'll never cross at the corner of a square. Now relate the number of line crossings with the number of squares crossed. --Tango (talk) 15:57, 11 May 2008 (UTC)[reply]

Also note that the answer makes sense because you must go through at least m squares horizontally, and n squares vertically, so you could reduce the question to asking how many of them count toward both the horizontal and vertical components. GromXXVII (talk) 16:04, 11 May 2008 (UTC)[reply]

Once you have covered the coprime case, generalise to the case where m and n are not coprime. Gandalf61 (talk) 18:25, 11 May 2008 (UTC)[reply]

Math question

How do I find the stationary points of a function with two variables???

f(x,y) = 2(x^3) +x(y^2) + 5(x^2) +y^2 Mr Beans Backside (talk) 17:29, 11 May 2008 (UTC)[reply]

See the article on stationary point, which will then lead you to gradient. --Kinu ^t/_c 18:28, 11 May 2008 (UTC)[reply]

Trigonometry

An isosceles triangle has base angles which measure 48 degrees. the length of the base is 16 cm. find the length of the altitude of the triangle to the nearest tenth. Mr Beans Backside (talk) 17:30, 11 May 2008 (UTC)[reply]

As indicated at the top of this page: Do your own homework. The reference desk will not give you answers for your homework, although we will try to help you out if there is a specific part of your homework you do not understand. Make an effort to show that you have tried solving it first. So tell us what you can do on your own first. --Kinu ^t/_c 18:25, 11 May 2008 (UTC)[reply]

I assure you this is not homework, as I am too cool for school. But seriously, its a puzzle in my local newspaper testing intellect. Obviously I've failed by asking here, but I hoped if I saw the correct answer I might be able to figure out the correct method. Mr Beans Backside (talk) 18:37, 11 May 2008 (UTC)[reply]

Well, a) what is the angle between the altitude and the base, and b) what does the altitude do to the length of the base? Answering those two questions is the key. Try drawing a diagram and going from there. --Kinu ^t/_c 18:48, 11 May 2008 (UTC)[reply]

Split it in half to get two right triangles. Think about just one of the two. The base will be 24 rather than 48 since the base has been split in half. Michael Hardy (talk) 23:00, 11 May 2008 (UTC)[reply]

The 48 was the angle, not length. Do you mean to say it "will be 8 rather than 16"? --Tango (talk) 23:06, 11 May 2008 (UTC)[reply]

What does it mean when a frequency distribution is uniform?

Hi, basically my teacher said that frequency distributions can either be symmetricaql or skewed, have one peak or many pearks, and or be uniform? I don't really understand what she means by uniform so can somebody explain it to me simply please? Thanks very much. 79.77.201.36 (talk) 21:46, 11 May 2008 (UTC)[reply]

Uniform means the same everywhere, so the graph would just be a horizontal line. --Tango (talk) 22:51, 11 May 2008 (UTC)[reply]

Specifically, the graph of the probability density function is the same height everywhere. Michael Hardy (talk) 22:58, 11 May 2008 (UTC)[reply]

We have articles on uniform distribution (continuous) and uniform distribution (discrete). Gandalf61 (talk) 08:37, 12 May 2008 (UTC)[reply]

May 12

Odds of winning in Lotto 6/49

Per 6/49, the chances of winning the jackpot is roughly 1 in 14 million. However, lotto 6/49 consists of matching 6 numbers from a box of 49 numbers. So therefore, shouldn't the odds of winning the jackpot be 1 in (69 x 68 x 67 x 66 x 65 x 64) or 1 in 86311779840? Where have I gone wrong? Acceptable (talk) 00:50, 12 May 2008 (UTC)[reply]

First, you morphed 49 into 69 for no apparent reason. Then you failed to make the distinction between combinations and permutations. There are indeed 49*48*47*46*45*44=10068347520 different possible drawings, but they're not all considered distinct from each other. If your ticket is "1,2,3,4,5,6", you will win whether the drawing is "1,2,3,4,5,6" or "6,5,4,3,2,1" or "3,5,6,2,4,1"... There are 6*5*4*3*2*1=720 such orderings, so you will win 720 out of 10068347520 drawings instead of just 1 out of 10068347520. The number in the article, 13983816, is 10068347520/720. —Preceding unsigned comment added by Tcsetattr (talk • contribs) 02:57, 12 May 2008 (UTC)[reply]

_{_{(Edit conflict)}}Your approach would be correct if you needed to have the numbers in the same order, but that is not the case.

Take a simpler example, choose 3 numbers from a total of 10, suppose the winning combination is 1 2 3, then you have in total 3! = 6 ways of choosing the combination 1 2 3, not just only one, which means the odds are 1 in (10 x 9 x 8)/6 = 1 in 120, instead of 1 in 720 if the order mattered and 1 2 3 was different from 2 3 1.

Similarly, for the 6/49 lotto, the odds are

{\frac {49!}{(49-6)!}}\cdot {\frac {1}{6!}}

, where ! is the factorial sign.

So the formula is the same as yours except it accounts for the different ways you have of getting the winning combination, as

{\frac {49!}{(49-6)!}}\cdot {\frac {1}{6!}}={\frac {49\cdot 48\cdot 47\cdot 46\cdot 45\cdot 44}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}}=13,983,816

.

Hope that helps. -- Xedi (talk) 03:08, 12 May 2008 (UTC)[reply]

Oh ok, haha I don't know why i used 69 and 68's instead of 49 and 48's. Thanks. Acceptable (talk) 22:28, 12 May 2008 (UTC)[reply]

Re: Trigonometric functions

The article at http://en.wikipedia.org/wiki/Trigonometric_function states, "Trigonometric functions are commonly defined as ratios of two sides of a right triangle containing the angle, and can equivalently be defined as the lengths of various line segments from a unit circle. More modern definitions express them as infinite series or as solutions of certain differential equations, allowing their extension to arbitrary positive and negative values and even to complex numbers."

Almost!

1. The example of the triangle shown to the right side is NOT the right triangle which is the trigonometric triangle. The triangle shown linearly (and incorrectly) demonstrates ratios only for the sin, sec and tan and displays them all wrong. That it is not the trigonometric triangle is evident in that a SECOND illustration is necessary to linearly display those ratios for the cos, cot and csc as linear expression of functions. NOT ONE TRIGONOMETRIC TRIANGLE, RATHER TWO ERRORS.

2. Modern definitions, and therefore the rote of modern mathematics, fail to recognize two very interesting and distinct functions which are quite present in the trigonometric triangle. Those functions are the dav and codav. They are so present in the trigonometric triangle they cannot be denied. One just needs to first achieve an understanding of the Trigonometric Triangle.

To do so see Google Videos

Search term: trigonometric triangle trilogy

or view (to make it easy) at these Google links:

(self-aggrandizing spam links removed)

Kindest regards, Keith Davie (talk) 02:14, 12 May 2008 (UTC)[reply]

Yes, clearly, we are all going to change our understanding of mathematics, which has existed for centuries, because of YouTube videos uploaded by you. Please do not troll the reference desk. --Kinu ^t/_c 02:29, 12 May 2008 (UTC)[reply]

Dear god, that was boring. And disappointing. I'd been hoping for some insanity-fueled entertainment, maybe a rant about timecube or something, but it's not. It's just describing a way to fit all six of the ordinary trig functions into a single diagram as the lengths of line segments. BTW, a XZ is a function of the angle. The angle isn't a function of XZ. Black Carrot (talk) 08:36, 12 May 2008 (UTC)[reply]

C'mon guys, that was a bit harsh. I don't think he deserved to have his head bitten off for attempting to discuss pedagogical techniques for trigonometry. Assume good faith, even here on the refdesks. Maelin (Talk | Contribs) 14:24, 12 May 2008 (UTC)[reply]

Perhaps you are right in that the gun was jumped, but it is easy to assume shenanigans when someone comes in with some links to videos they created, talks about a function named after themselves, and proceeds to imply that the foundations of trigonometry are all wrong. Perhaps had the poster worded things a little more... diplomatically... i.e., asking for feedback on his methods rather than pointing out "errors" in our ways, it would yield to providing more constructive commentary. --Kinu ^t/_c 22:50, 12 May 2008 (UTC)[reply]

Having now viewed the video in its entirety, I have a couple of comments. Firstly, don't feel too discouraged by the above responses. Any self-motivated explorations of mathematics are a worthy goal. Now, a few comments about the actual video

Your understanding of what constitutes a mathematical "function" appears a little off - you seem to require that functions be "unique" over the domain, hence your dismissal of the function yielding the length of XY since it was equal for complementary angles. This property, called injectivity, is an often-desirable but by no means essential property of functions. Remember, even sine and cosine aren't unique if you consider their domain to be all 360° of a full circle.
Whilst your construction of the triangle certainly yields valid definitions of the trigonometric functions, it does not do so in a particularly motivating sense. The construction seems somewhat arbitrary and contrived. Teaching school students trigonometry is difficult enough, and that's when the function definitions lend themselves intuitively to simple applications. A construction like this one would serve as merely an extra layer of complexity in an already challenging new area for young students.
Your claims that mathematics has "failed to recognise" the functions yielding the lengths of WZ, VX and XY are somewhat unfounded. They are easy to define, as given the angle θ, we can prove that WZ = OZ - OW = sec(θ) - cos(θ), that VX = 0X - OV = cosec(θ) - sin(θ), and that XZ = XY + YZ = cotan(θ) + tan(θ). What mathematics (or, more accurately, mathematicians) hasn't done is give these functions special names, probably because they are not especially useful or interesting.

Trigonometry offers many opportunities for "Ooh, neat!" moments where you discover that various identities can be proven geometrically. Good luck in your trigonometric adventures! Maelin (Talk | Contribs) 15:10, 12 May 2008 (UTC)[reply]

For what it's worth, the author of this topic and I have had a discussion about this at my talk page... feel free to digest. --Kinu ^t/_c 03:23, 13 May 2008 (UTC)[reply]

You know, after looking at the history of the article, I now fully understand the first part of Mr. Davie's comment... which brings up an interesting point: why was this image done away with for the current two-graph system of defining the ratios? (Perhaps this is a discussion best suited for the article's talk page, but I thought I'd pose it here for starters.) --Kinu ^t/_c 00:08, 13 May 2008 (UTC)[reply]

After having read most of the above, davie, your ideas are certainly worthy of discussion but your presentation is apt to offend those with significant mathematical knowledge. I to have had to learn how to pose my thoughts in a manner that avoids me coming off like I am the god of math or what ever. I find the best tool is to pose one's arguments a question (after all this desk is for questions not rants). As for those who quickly concluded that his post was trolling, I think the lesson here is don't assume something is true just it fits the bill, look a little deeper into whether or not his arguments make any sense, if not then such a post would be motivated, but if they do seem to have some merit, I would say one should address it as any other question. A math-wiki (talk) 06:15, 13 May 2008 (UTC)[reply]

Amusing. Videos posted on YouTube.

"your dismissal of the function yielding the length of XY"

Prejudice is an ugly bedmate.

The functions which above are described as simple subtractions and additions in the linear geometric perform exponentially when studied through 91 degrees (0 through 90), hence sin(sq)/cos and cos(sq)/sin are more appropriate descriptions.

Regarding an 'extra layer of complexity", there is far more elegance in the Trigonometric Triangle of my video than either two triangles or a first quadrant angle with a triangle in the fourth quadrant, both products of Wiki. Imagine using one triangle to measure the angle. Oh my, too complex for Wiki.

And if one starts counting at angle no. 1, like the trig tables tell us, at zero degrees (it is not to be feared), when one arrives at 360 degrees that is angle 361. Good grief, I am far to radical to be taken seriously, claiming there are 361 degrees in a complete circle. Have a fun time with that at my expense guys. :)

Kindest regards Keith Davie (talk) 01:14, 14 May 2008 (UTC) Keith Davie[reply]

I defend you when others assume that you are just another Timecube madman, and offer you constructive criticism about your ideas and try to help correct your misunderstandings about mathematical conventions, and you mock me? Your approach to mathematical communication with others needs some real work. Maelin (Talk | Contribs) 02:09, 14 May 2008 (UTC)[reply]

Mathematics-Quadratic Equations

The solutions to a particular equation can consist of different kinds of real numbers,for example rational numbers,or irrational;they can be equal roots or unequal,positive,negative or zero.There can be even no real solutions at all.All of this depends on the values of a,b and c in a given equation. The quantity under the square root(delta) controls what kind of roots there are for a given equation.For example if this quantity is negative then there are no rea roots.

QUESTION- What are the conditions that apply to delta, so that the solutions will fit into these categories-real,unreal,rational,irrational,aqual,unequal? —Preceding unsigned comment added by 196.207.40.212 (talk) 14:23, 12 May 2008 (UTC)[reply]

The quantity you call delta is the discriminant of the polynomial. That articles answers all your questions, except the question of (given that the roots are real) when the roots are rational. Assuming the coefficients of the polynomial are rational, the roots are also rational iff the discriminat is the square of some rational number. Algebraist 14:27, 12 May 2008 (UTC)[reply]

Left/right division

This is a question about notation. Specifically I’m working in a noncommutative ring with nonequivalent left and right division algorithms. The question is if there is any accepted notation for saying that “a left divides c” or “b right divides c”. Clearly, “a|c” won’t work because it’s ambiguous if a doesn’t both left and right divide c. GromXXVII (talk) 15:15, 12 May 2008 (UTC)[reply]

I don't know if there's a standard notation or not, but if I had to make one up on the spot, I'd go with

a|_{L}b

and

a|_{R}b

. --Tango (talk) 15:47, 12 May 2008 (UTC)[reply]

If you want to say that a divides b on the left then you can just write a|b on the left.A Real Kaiser (talk) 05:16, 13 May 2008 (UTC)[reply]

See quasigroup, or the matlab documentation for a reasonable use of / and \. JackSchmidt (talk) 05:24, 13 May 2008 (UTC)[reply]

If you just mean a if a left multiple of b, then just say a in Rb. One sided ideals are made for this. JackSchmidt (talk) 05:25, 13 May 2008 (UTC)[reply]

Connection between the definition of the unit sphere and the real-projective plane

What is the connection between the definitions of the unit sphere and the real-projective plane? I was told they were connected. I have seen how the geodesics on the sphere correspond to lines in the real-projective plane, but when you just look at definitions of spaces I get all confused.. Also, how is it that the real-projective plane can be regarded as a disc? I can't find an explanation of this anywhere! —Preceding unsigned comment added by 152.78.120.74 (talk) 15:26, 12 May 2008 (UTC)[reply]

~~Perhaps Stereographic projection is what you are looking for?~~

~~The real projective plane cannot be regarded as a disc unless you puncture it.~~ -- Meni Rosenfeld (talk) 15:56, 12 May 2008 (UTC)[reply]

A punctured real projective plane is not a disc, it's a Möbius band. kfgauss (talk) 20:44, 12 May 2008 (UTC)[reply]

Ouch. The heavenly gates to topological wisdom were obviously closed to me at the time I made this post. -- Meni Rosenfeld (talk) 18:45, 13 May 2008 (UTC)[reply]

I have an exam in projective geometry in a little under 3 weeks, and I'm not all the confident about passing it, so take this with a large pinch of salt, but, as I understand it, the complex projective line is homeomorphic to the sphere, but the real projective plane isn't (C and R² are homeomorphic, but when you go to the projective versions, they're no longer the same - C just gets a single point at infinity, R² gets a whole projective line at infinity. It also isn't a disc... If memory serves, the real projective plane is a sphere with a disc removed and replaced by a Moebius band. --Tango (talk) 15:55, 12 May 2008 (UTC)[reply]

Perhaps you will find Real projective plane useful. -- Meni Rosenfeld (talk) 15:59, 12 May 2008 (UTC)[reply]

(after edit conflict) The real projective plane is the space of lines in R³ that pass through the origin. Since each of these lines intersects the unit sphere in just two antipodal points, the real projective plane has the same toplogy as a unit sphere in which opposite pairs of points are counted as identical i.e. the quotient of the unit sphere by the antipodal map. You can also think of the real project plane as being the compactification of the real plane R² formed by adding a "projective line at infinity" to R² - think about how the lines through the origin in R³ intersect with the plane z=1.

Don't confuse this construction with the one-point compactification of R², the "complex projective line" or Riemann sphere, which is topologically equivalent to the unit sphere without identification of opposite points. Gandalf61 (talk) 16:02, 12 May 2008 (UTC)[reply]

One can "think of" the real projective plane as a disk, but with each point on the boundary of the disk identified with the boundary point opposite from it (i.e. identify antipodal boundary points). The way to see this is to think of RP^2 as the unit sphere with antipodal points identified, as Gandalf61 discusses, and then throw out the (open) bottom hemisphere, leaving only the (closed) top hemisphere. This is fine because each point in the (open) bottom hemisphere has an antipodal point in the (closed) top hemisphere. When we identify antipodal points to get RP^2, what we get is the top hemisphere (which is topologically a disk), but with antipodal boundary points identified. kfgauss (talk) 20:21, 12 May 2008 (UTC)[reply]

Non-degenerate conic in the real-projective plane

I have exams coming up in two weeks and this is a past paper question, i'm worried something similar will come up again this year.. can someone please explain? If you had 5 points in the real-projective plane, no three of which are collinear, how would you prove that there is a unique non-degenerate conic passing through each of the points? 152.78.120.74 (talk) 15:40, 12 May 2008 (UTC)[reply]

Here's an outline of one way to prove this:

1) Show that you can make a (linear) change of variables so that three of your points are (1,0,0), (0,1,0), and (0,0,1).

2) Write down the general equation for a conic, ax^2 + by^2 + cz^2 + dxy + exz + fyz = 0, and impose the condition that the above three points solve your equation. Write down the general equation for a conic through those three points.

3) Show that, given two more points A and B such that no three of these two and the above three are collinear, (a) none of the three coordinates of A or B are zero (so we may normalize z to be 1 and work in the plane with these two points), (b) if we do normalize z to be 1 and work in the plane, A and B are not collinear with the origin. (One can draw more conclusions, but we won't need them.)

4) Working in the plane, let A = (x1,y1) and B = (x2,y2) and write down the two equations for these two to be solutions of the equation you found in (2). Show that (up to multiplying the coefficients by a scalar) there is a unique choice of a through e (from (2)) such that A, B, and the three points from (1) are solutions. You'll need to use what we learned in (3). Hint: If you know some linear algebra, this step shouldn't be messy at all. Try to turn it into a question about showing that the kernel of a certain matrix is 1-dimensional.

This solves the problem, as the conic we find is necessarily nondegenerate. For completeness, you may wish to argue this, i.e. 5) Given 5 points on a degenerate conic, at least 3 are collinear. Hope this has helped. kfgauss (talk) 20:05, 12 May 2008 (UTC)[reply]

Combinatorial problem

I am trying to figure out how many permutations I can have of 4 values, where the values are integers from 0 to 100 inclusive. The four values sum to 100 and the position is important. I have figured out that if I have 1, 2, or 3 values, the answer is 1, 101, and 5050 (101 choose 2). How can I do the next step for the 4 value case? I figure this is a nested binomial coeffienct or choose 2 problem. (I realize a good upper bound is 101^4 ~ 10^8. I have calculated 2.5 x 10^5 as a close solution, but I'd like to be exact if possible.)

Also, I am thinking about a case with a larger number like 23. Where I have 23 integer values that sum to 100 inclusive. This would be a massive nested problem. Thanks for your help. --Rajah (talk) 22:22, 12 May 2008 (UTC)[reply]

You are looking for Stars and bars (probability), in particular theorem 2, with

k=100

and

n=4,\ 23

. The answer is

{\binom {n+k-1}{n-1}}

. The sequence goes 1, 101, 5151 (not 5050 like you wrote), 176851, 4598126,..., and for 23 you have 941388310113310598422026. -- Meni Rosenfeld (talk) 22:40, 12 May 2008 (UTC)[reply]

Awesome, thanks! --Rajah (talk) 22:49, 12 May 2008 (UTC)[reply]

nets for making geometric solids

Hi. These are not specific homework questions that I have to figure out, nor are they things I have to figure out by myself using some kind of formula, as I am asking for a way to measure these and not the answer to any questions or specific measurements (and am seeing if my formulas are correct). These questions are about making nets for constructing a shell of a geometric solid. First question I have is about the net of a right circular cone. I am to make a net so that the edges so not overlap and to only use tabs for connection. The net for a cone consists of a circle and a shape that is almost like a geograhpic isoceles triangle, where the congruent edges are straight but the edge connecting to the circle is an arc. I understand that the arc's length is equal to the circumfrence of the circle, but how do I draw the arc to the right curvature? I have the height of the cone as well as the diametre of the circle. If you extend the arc, what shape is it and how large is the resulting shape? I'm thinking that it will be a circle, with the centre of the circle at the vertex of the cone's net, as the distance from the vertex to the bottom of the shape which connects to the circle should remain the same, as this is a right cone and the distance on the curved face of the cone between the vertex and the circular edge should be the same. Am I correct, or is there a different simple formula for this? Thus, is the formula for the distances between the vertex and the round edge on the triangle-like portion of the cone's net [math] sqrt(r²+h²) [/math]? Sorry, I'm not very good with LaTeX syntax and r means the radius of the circle and h means the height of the cone. Ok, next I am to do a chevron-based pyramid. What I mean by that is, the base is a chevron shape, which is kind of like, in this case, an isoceles triangle with another triangle cut out of it at the only edge not congruent to the other two edges, that is also isoceles. So, the end result (chevron) should be a quadrilateral with three acute angles, and one reflex angle, and the reflex angle, were is subtracted from 360 to make an acute angle, would be greater than the angle that is pointing in the same direction and is not the two congruent acute angles. Basicly, I hope you know what a chevron is. The vertex for the pyramid is directly above the reflex angle, and the resulting pyramid should have 5 faces, 8 edges, and 5 verticies. However, that's not my question. In order to make a net, will I need to use any trigonometric functions (sine, cosine, tangent) to calculate the lengths of the edges for each of the 5 faces for the net, other than the basic Pathegorean theorem? I know the distance between the two congruent angles on the chevron (the width), the length between the other acute angle on the chevron and the reflex angle on the chevron (the length₁), the distance between the reflex angle and the middle of the width line, which forms the edge of the nonexistant triangle in the chevron that does not share the edges with the real shape (the length₂) [length₁ and length₂ should form a straight line with each other, as well as the height, which is the distance between the reflex angle and the top vertex of the pyramid. My other question is, is there a way to calculate whether or not a net with 6 congruent squares will succesfully form a cube if folded, without actually folding it? Something like x edges to seal, x edges to fold, etc. Thanks. ~A H 1^(TCU) 22:34, 12 May 2008 (UTC)[reply]

Right-Angled Cone: You're right, it'll be a circle for the base and a piece of a circle for the rest, and the radius of the partial circle is equal to the distance from the vertex to an edge of the base, sqrt(r²+h²).

Chevron: You can do it all with the Pythagorean theorem. Just draw a picture of the base chevron, including the width and height lines, and start filling in numbers until you've labeled every line. Once you've got the base, three of the vertical edges are hypoteneuses of right triangles, with the fourth being their shared side.

Cube: The number of edges to fold is always five, and the number to seal is seven, and the perimeter is therefore fourteen. That doesn't narrow it down enough, though. Try this. There can be no more than four squares in a row, and there should never be a c-shape anywhere in it. Black Carrot (talk) 02:13, 13 May 2008 (UTC)[reply]

May 13

Unique Factorization Domains

Suppose I have a ring extension of Z, say Z[w].

How can I tell whether or not Z[w] is a unique factorization domain?
What if w happens to be a root of 1? Are there tables that tell me the answer?

mike40033 (talk) 01:05, 13 May 2008 (UTC)[reply]

I'm learning about this too at the moment, you might be interested in the articles about the Stark-Heegner theorem and the Cyclotomic fields. -- Xedi (talk) 01:21, 13 May 2008 (UTC)[reply]

Gerschgorin Theorem and Taussky's Result

Okay, so in my numerical analysis class, studying for the finals, I came across the Gerschgorin criterion. One of the exercises is the proof of the criterion which is split up into three parts. I have proved the first two parts but the third one (which I learned later is called the Taussky's theorem), I can't get. Here is everything I have done:

Let $C=(c_{i,j})$ be an arbitrary $d\times d$ singular complex matrix.
Then there exists a nonzero $x$ such that $Cx=0$ .
Choose $l\in \{1,2,...,d\}$ such that $|x_{l}|=\max |x_{j}|$ .
This maximal element is of course, nonzero.
By considering the $l$ th row of $Cx$ , I have already proven that
$|c_{l,l}|\leq \sum _{j=1,j\neq l}^{d}|c_{l,j}|$ .

Next, taking a dxd matrix B and letting $\lambda$ be an arbitrary eigenvalue of B.
Substituting, $C=B-\lambda I$ , I have also shown that $\lambda$ is in a Gerschgorin disc $S_{l}$ where
$S_{l}=\{z\in \mathbb {C} :|z-b_{i,i}|\leq \sum _{j=1,j\neq i}^{d}|b_{i,j}|\}$
And since, $\lambda$ was an arbitrary eigenvalue of B, the entire spectrum of B is contained the union of all the Gerschgorin discs.

Now here is the problem which I can't get. Let us suppose that the matrix B is irreducible and the previous inequality holds as an equality, i.e.
$|c_{l,l}|=\sum _{j=1,j\neq l}^{d}|c_{l,j}|$
for some $l$ between 1 and d. I need to show that this equality implies that
$|c_{k,k}|=\sum _{j=1,j\neq l}^{d}|c_{k,j}|$ for all k between 1 and d.
This would imply that if an eigenvalue is on the boundary of one Gerschgorin disc, then that eigenvalue will be on the boundary of all Gerschgorin discs. Any ideas on how to prove it? I have looked in several books but everyone just states it and calls it a famous result but no one proves it.A Real Kaiser (talk) 05:51, 13 May 2008 (UTC)[reply]

Proceed as in your proof of the first step and get to the point

|c_{l,l}||x_{l}|\leq \sum _{j=1,j\neq l}^{d}|c_{l,j}||x_{j}|

. Now if we have the equality you mention, what does that mean about those

x_{j}

's for which

c_{l,j}\neq 0

? Now use that irreducibility is equivalent to strong connectivity of the directed graph with d vertices with an edge from vertex i to vertex j if and only if

c_{i,j}\neq 0

. kfgauss (talk) 07:16, 13 May 2008 (UTC)[reply]

Proving Integrability

On a completely different note, I am trying to prove some properties for this function.
Let $f:[0,1]\rightarrow [0,1]$ be defined as $f(x)={\begin{cases}[1/x]^{-1},&{\mbox{if }}x\in (0,1]\\0,&{\mbox{if }}x=0\end{cases}}$
where [x] is the truncation function (i.e. [x]=the integer part of x). Notice that at each unit fraction, the function has a jump discontinuity and in between, the function is a constant. We have already had some discussion about this function but now I am trying to prove that this function is Riemann integrable. Here are some facts that I proved:

Fact I

$f(x)$ is a non-decreasing function
Proof:
Let $x>y$

$\Rightarrow 1/x<1/y$

$\Rightarrow [1/x]\leq [1/y]$

$\Rightarrow [1/x]^{-1}\geq [1/y]^{-1}$

$\Rightarrow f(x)\geq f(y)$ .

Fact II

$f(x)$ is continuous at $x=0$ from the right.

Proof:Let $\epsilon >0$ be given and then let $\delta =\epsilon /2$ .

$x-0<\delta$

$\Rightarrow x<\delta$

$\Rightarrow x<\epsilon /2$

$\Rightarrow 1/x>2/\epsilon$

$\Rightarrow [1/x]\geq 2/\epsilon$ (for $\epsilon$ sufficiently small)

$\Rightarrow [1/x]^{-1}\leq \epsilon /2<\epsilon$

$\Rightarrow f(x)<\epsilon$

$\Rightarrow f(x)-0<\epsilon$

$\therefore \lim _{x\rightarrow 0^{+}}f(x)=f(0^{+})=f(0)=0$

Fact III

$f(x)$ is Riemann integrable. This will be shown by proving that for every $\epsilon >0$ , we can find a partition p of [0,1] such that the upper sum U(f,p) minus the lower sum L(f,p) using p, is smaller than $\epsilon$ .

Proof: For a given $\epsilon >0$ , let $\epsilon _{1}=\epsilon /4$ , and let N=the largest integer less than or equal to 1/ $\epsilon _{1}$ . This gives us that 1/N will be the next unit fraction. Hence x=1/N will be the next point (on the right) after $x=\epsilon _{1}$ where $f(x)$ will have a jump discontinuity. Let $\epsilon _{2}=\epsilon _{1}/2N$ and let the partition be

$P=\{0,\epsilon _{1},{\frac {1}{N}}-\epsilon _{2},{\frac {1}{N}}+\epsilon _{2},{\frac {1}{N-1}}-\epsilon _{2},{\frac {1}{N-1}}+\epsilon _{2},...,{\frac {1}{3}}-\epsilon _{2},{\frac {1}{3}}+\epsilon _{2},{\frac {1}{2}}-\epsilon _{2},{\frac {1}{2}}+\epsilon _{2},1\}$

This is a perfectly valid partition, because the division by 2N ensures that none of the points go out of order. The intervals will not overlap. The point of the first fact was to make the Riemann evaluation easier. Since the function is nondecreasing, in order evaluate the area of a Riemann rectangle, for the upper sum, we simply take the value of the function on the right endpoint, and for the lower sum, we simply take the value of the function at the left endpoint. Both facts also show (for my teacher) that the function does NOT become unbounded as x goes to zero. In fact, if we extend the function to the interval [-1,1], the limit from the left as x goes to zero also exists so I can show that the function is actually continuous at x=0. But, that is just something extra.

Now when the difference between the upper sum and the lower sum is taken, the rectangles which fall on the flat region, with the base being of the form $x={\frac {1}{n}}+\epsilon _{2}$ to $x={\frac {1}{n-1}}-\epsilon _{2}$ cancel out because the upper sum is the same as the lower sum. The same thing happens with the last rectangle. The only rectangles that remain are the columns of width $2\epsilon _{2}$ which have the jump discontinuity in the middle.

So,
$U(f,p)-L(f,p)={\frac {1}{N}}\epsilon _{1}+\left({\frac {1}{N-1}}-{\frac {1}{N}}\right)2\epsilon _{2}+\left({\frac {1}{N-2}}-{\frac {1}{N-1}}\right)2\epsilon _{2}+\cdots +\left({\frac {1}{3}}-{\frac {1}{4}}\right)2\epsilon _{2}+\left({\frac {1}{2}}-{\frac {1}{3}}\right)2\epsilon _{2}+\left(1-{\frac {1}{2}}\right)2\epsilon _{2}$

$U(f,p)-L(f,p)={\frac {1}{N}}\epsilon _{1}+\left({\frac {1}{(N-1)(N)}}\right)2\epsilon _{2}+\left({\frac {1}{(N-2)(N-1)}}\right)2\epsilon _{2}+\cdots +\left({\frac {1}{12}}\right)2\epsilon _{2}+\left({\frac {1}{6}}\right)2\epsilon _{2}+\left({\frac {1}{2}}\right)2\epsilon _{2}$

Since height of each rectangle is less than or equal to one, and the number of terms with $2\epsilon _{2}$ is precisely N-1, we have that

$U(f,p)-L(f,p)\leq \epsilon _{1}+\underbrace {2\epsilon _{2}+2\epsilon _{2}+\cdots +2\epsilon _{2}+2\epsilon _{2}+2\epsilon _{2}} _{N-1}$

$U(f,p)-L(f,p)\leq \epsilon _{1}+(N-1)(2\epsilon _{2})\leq \epsilon _{1}+N(2\epsilon _{2})=\epsilon _{1}+N\left({\frac {2\epsilon _{1}}{2N}}\right)=\epsilon _{1}+\epsilon _{1}=\epsilon /2<\epsilon$ .

Therefore, the function $f(x)$ is Riemann integrable on the interval [0,1]. The reason, I put all of this up is that I just want to run it by a couple of people here to make sure that I have not made a stupid mistake and to see if this is "rigorous" enough. I will be presenting this to him (and maybe we can finally settle this argument once and for all about this function being Riemann integrable) and he is very particular about rigor. So, do you guys think that am I missing out some particular detail which I should include or is this good enough for a rigorous Mathematical argument? Any comments and suggestion will be welcome about the proof of all three facts. The third one is the most important one, of course. In addition, in the proof of the second fact, in line three, it is obvious that the line is true for a sufficiently small epsilon but how can I write that better? I don't really like it the way I have it but I need it there. Thanks everyone!A Real Kaiser (talk) 09:33, 13 May 2008 (UTC)[reply]

In case you don't know, any monotone bounded function on a bounded real interval is Riemann integrable. This is easy to prove (at least assuming Riemann's integrability criterion, which you seem to be doing): pick an equal width partition of mesh

\delta

, and the difference between the upper and lower Darboux sums is

(f(b)-f(a))\delta

. Your proof of III may be a special case of this argument; I don't know, as I haven't read it. Algebraist 10:59, 13 May 2008 (UTC)[reply]

Your proof of II is flawed: it is not the case that x < epsilon/2 implies floor(1/x) is at least 2/epsilon for sufficiently small epsilon. You can conclude, however, that floor(1/x) is at least 2/epsilon - 1, which is at least 3/2epsilon (say) for epsilon small enough. Algebraist 11:05, 13 May 2008 (UTC)[reply]

Revised Proof of Fact II

Algebraist, you are absolutely right. I just said the opposite of what is true. So here is the revised proof.

Proof:Let $\epsilon >0$ be given and then let $\delta =\epsilon /2$ .

$x-0<\delta$

$\Rightarrow x<\delta$

$\Rightarrow x<\epsilon /2$

$\Rightarrow 1/x>2/\epsilon$

$\Rightarrow {\frac {1}{x}}-1>{\frac {2}{\epsilon }}-1$

$\Rightarrow [1/x]>{\frac {1}{x}}-1>{\frac {2}{\epsilon }}-1={\frac {2-\epsilon }{\epsilon }}$

$\Rightarrow [1/x]^{-1}<{\frac {\epsilon }{2-\epsilon }}<\epsilon$ because $\forall \epsilon <1$ we have that $2-\epsilon >1$

$\Rightarrow f(x)<\epsilon$

$\Rightarrow f(x)-0<\epsilon$

$\therefore \lim _{x\rightarrow 0^{+}}f(x)=f(0^{+})=f(0)=0$

Fact IV

Here is another thing I thought of. I will try to prove that $2x\geq f(x)\geq x$ for all $x\in [0,1]$ .

Proof:
First, let {x}=x-[x] be the fractional part of a real number x.

Now we have that for all $x\in \mathbb {R} ,\exists ![x]$ and $\{x\}$ such that $x=[x]+\{x\}$ and $0\leq \{x\}<1$ .

Now let us consider all $y\in \mathbb {R}$ such that $y\geq 1$ .

We now have that $y=[y]+\{y\}<[y]+[y]=2[y]$ .

$\Rightarrow y<2[y]\Rightarrow y/2<[y]$

$\Rightarrow {\frac {1}{2x}}<[1/x]$ for all $x\in [0,1]$

and using the fact that $[y]\leq y$ for all $y\Rightarrow {\frac {1}{2x}}<[1/x]\leq {\frac {1}{x}}$

$\Rightarrow 2x>[1/x]^{-1}\geq x$

$\Rightarrow 2x>f(x)\geq x$

and then using the squeeze theorem, I can show that the limit of f(x) from the right as x goes to zero is indeed zero and this function is most definitely not unbounded in [0,1]. I think that after these four facts combined, there is no question that f(x) is bounded and Riemann integrable in [0,1].

Fact V

In fact, we can also find the exact value of the integral as follows:

Since, the function as a jump discontinuity at every unit fraction, with 1/2 being the first one from the left as x goes to zero, and in between two consecutive unit fractions, the function is constant

(i.e. $f(x)={\frac {1}{n}}{\text{ for }}x\in \left({\frac {1}{n+1}},{\frac {1}{n}}\right]{\text{ for all }}n\in \mathbb {N}$ ), we can just find the area underneath the curve by drawing rectangles

with length ${\frac {1}{n}}-{\frac {1}{n+1}}$ with height ${\frac {1}{n}}$ and then add up the areas of all such rectangles.

$\int _{0}^{1}f(x)dx=\sum _{n=1}^{\infty }\left(\left({\frac {1}{n}}-{\frac {1}{n+1}}\right){\frac {1}{n}}\right)=\sum _{n=1}^{\infty }\left({\frac {1}{n^{2}}}-{\frac {1}{n(n+1)}}\right)$

$=\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}-\sum _{n=1}^{\infty }{\frac {1}{n(n+1)}}={\frac {\pi ^{2}}{6}}-\sum _{n=1}^{\infty }\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)={\frac {\pi ^{2}}{6}}-1.$

The first sum is just from the Riemann Zeta function (the values are known) and the second sum is a telescoping series. We can split up the series because it is absolutely convergent. So, what do you guys think? Should I change something or is this enough? Any more stupid mistakes in what I wrote or my reasoning? Come on guys, don't be shy. I just need your input and some constructive criticism.A Real Kaiser (talk) 23:23, 13 May 2008 (UTC)[reply]

Sum

"Let ABCDEFGH be an eight digit number where A,B,C,D,E,F,G,H are the digits 1,2,3,4,5,6,7,8 in some order. For example, A=4,B=1,C=3,D=2,E=5,F=6,G=7,H=8 gives 41325678. FInd other eight digit numbers in which the three digit number ABC is divisible by 7, the three-digit number BCD is divisible by 6, CDE is divisible by 5, DEF is divisible by 4, EFG is divisible by 3 and FGH is divisible by 2."

I know that E is 5 but don't konw the others. Can someone pls give me some hints as the maths exam is round the corner(to be exact, tomorrow)? tks. —Preceding unsigned comment added by Invisiblebug590 (talk • contribs) 10:26, 13 May 2008 (UTC)[reply]

How many other numbers does it want? It looks like there are quite a lot. I suggest starting at the end and working back. FGH being divisible by 2 tells you something about H, but doesn't restrict F or G, for example. You can then work your way up narrowing down your options bit by bit. --Tango (talk) 10:42, 13 May 2008 (UTC)[reply]

The "easy" way to solve this is a brute force computer program. You only have 8! = 40320 permutations to evaluate. My program has determined that there are five solutions to your question. Note that BCD, DEF and FGH are all divisible by 2, so that uses up all but one of your even mumbers for D, F & H. BCD and EFG are both divisible by 3, so B+C+D+E+F+G is also divisible by 3. Since the sum 1..8 = 36 then A+H will also be divisible by 3 which reduces the possibilities a lot. -- SGBailey (talk) 11:10, 13 May 2008 (UTC)[reply]

If it's any help, the way I found a solution to this was to observe that there are only two choices for F, one of which leads to a unique choice for G. Fixing F and G at these values allows everything else to be worked out without too much difficulty. Algebraist 11:13, 13 May 2008 (UTC)[reply]

E=5, DEF is divisible by 4 therefore EF=52 or 56. EFG is divisible by 3 therefore of the div 3s (522, 525, 528, 561, 564, 567) we can try 528, 561, 564, 567. -- SGBailey (talk) 22:22, 13 May 2008 (UTC)[reply]

The above was a reply to a comment that got itself deleted, Hence the apparent non-sequitur. -- SGBailey (talk) 22:25, 13 May 2008 (UTC)[reply]

The Answer is 73485612. FIgured it out in 15 min. This is how you do it (not very mathematic but it works):

the last 4 digits (EFGH) are even, so H is 2,4,6, or 8. You know in order to use all 8 numbers the last 5 will be every-other even odd even odd even (keep that in mind), so G will be either 1,3,5,or 7. Even next. So depending on what H is, F isnt that. E isthe same but with odd. You start by figuring out the last 4 first because to know the DEF, EF needs to be divisible my 4, and EFG needs to be divisible by 3. so if you make a little chart using this info. you have over 120 possiblities now. Now figure out the EFG set that E is 5, F is even and G makes the sum of the three a multiple of 3. You get 561H, or ABCD561H ( H is 2,4, or 8). Now you have 73248 left to choose from. This part is easier. D will be even, ABC will add to a multiple of 7 and BCD adds to 3 or a maultiple of 3. Looking at this you realize that the only 3 numbers that will give you a multiple of 7 is 734 (14). Now you just eliminated the one that ended with 4. 2 options left. D is 2 or 8. you see what will give you x3. so 7+4+8, 7+3+2, 3+4+8, 7+4+2, 7+3+2, 3+4+2. 348 works. Now you have your answer. A3485612 or 73485612. Its a lot easier to do than to hear.--Xtothe3rd (talk) 01:45, 14 May 2008 (UTC)[reply]

Logic circuit design

Please provide symbol/math tags in TeX/HTML for logic gates like NOT, AND, OR, XOR,... Currently, users are uploading image files to depict circuit diagrams. A XML version would be immediately useful beyond Wiki.Anwar (talk) 14:35, 13 May 2008 (UTC)[reply]

Feature requests should go to [3], this page is for asking maths questions. --Tango (talk) 18:21, 13 May 2008 (UTC)[reply]

Homeless Statistics in China

Hi, I am an academic student currently researching on China's demographic. I am having a difficult time locating the most current information concerning "the homeless percentage in China overall and by region or province and cities vs. rural." I would really appreciate it if anyone can provide me with the most up to date data or point me to the right direction on where to search.

Thank you for your time and help. Your prompt reply will be greatly appreciated.

Sincerely, username: darjeelinguru

Hi. This is really a question for the Humanities Reference Desk. This seems like difficult information to come by. You might start by looking at the page Demographics of the People's Republic of China, which has some related information, but not what you ask for specifically. Then you might look at the references and links at the bottom of that page. Best of luck.

Also, to sign your name automatically, type four tildes in a row (i.e. ~ four times in a row). kfgauss (talk) 18:29, 13 May 2008 (UTC)[reply]

We certainly don't make it any easier to choose a reference desk section for such questions by including the ambiguous term "statistics" as a sub-topic for the mathematics section. -- Meni Rosenfeld (talk) 18:40, 13 May 2008 (UTC)[reply]

Mathcad functions

Using the Mathcad statistical functions how would I convert percentile to a z-score and vice versa? 71.100.14.205 (talk) 16:32, 13 May 2008 (UTC)[reply]

Pi

When was pi first used in England how was it measured? —Preceding unsigned comment added by 82.71.27.169 (talk) 18:26, 13 May 2008 (UTC)[reply]

I couldn't be sure, and I am sorry if I am stating the obvious, but have you looked at Pi?

Numb3rs

An episode of Numb3rs involved a (false) proof for the Riemann hypothesis that was used as a ransom with the intent of breaking encryption. Did they get it mixed up with the P = NP problem, or are they that closely related? — DanielLC 23:33, 13 May 2008 (UTC)[reply]

It is suggested that a proof for the Riemann hypothesis may be a key factor in understanding and breaking the pattern of prime numbers, what would in turn speed up our best factorization algorithms of very large numbers. These are frequently used in cryptography for public keys. I guess that's their whole point. — Kieff | Talk 23:42, 13 May 2008 (UTC)[reply]

Is the idea that it's still exponential time, but faster than normal, or did they simultaneously (almost) prove that P = NP without even mentioning it? Also, it would have to be something in the way it was proved, or they'd just assume the Riemann hypothesis to be true and then just see if it works. In might not be formal, but if you can break encryptions, who cares. — DanielLC 23:59, 13 May 2008 (UTC)[reply]

I think they said something non-specific about the method of the proof giving an easy way to factor numbers. See [4] by Chris Caldwell and the links [5] and [6] in one of the replies. PrimeHunter (talk) 00:07, 14 May 2008 (UTC)[reply]

There's no evidence that factoring is NP-complete, and most complexity theorists think it isn't, so factoring being in P doesn't imply P=NP. I don't think it would even affect most theorists' faith in P≠NP very much; factoring is considered much easier than known NP-complete problems (though a polytime algorithm would still be very surprising). -- BenRG (talk) 01:06, 14 May 2008 (UTC)[reply]

May 14

what is the diffrent with prism and pyramid?

--24.78.51.208 (talk) 00:17, 14 May 2008 (UTC)[reply]

Maybe you can find what you want in prism (geometry) and pyramid (geometry). PrimeHunter (talk) 01:55, 14 May 2008 (UTC)[reply]

regression toward the mean

The eugenics article says: [Galton] reasoned that, since many human societies sought to protect the underprivileged and weak, those societies were at odds with the natural selection responsible for extinction of the weakest; and only by changing these social policies could society be saved from a "reversion towards mediocrity," a phrase he first coined in statistics and which later changed to the now common "regression towards the mean."[18] Isn't this a completely different concept from "regression towards the mean"? My understanding of regression is that it is symmetric with respect to both time and direction. So although extreme data points will regress to the mean, the total number of extreme data points does not increase or decrease on average. --MagneticFlux (talk) 03:00, 14 May 2008 (UTC)[reply]