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February 1

complex integration or integration w.r.t. complex measure?

According to Cauchy's theorem the integral of an analytic function along a closed curve is zero. But an analytic function is harmonic and so the integral should be its mean value ( not necessarily zero). This apparent contradiction may have something to do with what I suggested in the title. Who can explain it nicely?DeepthiPP 05:48, 1 February 2007 (UTC)[reply]

As I understand, f(c) is the mean value of an analytic function f(z) on the circumference of any circle with center c. It is an integral with respect to the arc length (real). Look up Gauss' Mean Value Theorem. Cauchy's theorem is about complex integrals. Hope this helps. Twma 07:00, 1 February 2007 (UTC)[reply]

Take a complex integral like

\int _{\gamma }f(z)\,dz

, the dz points along the tangent to the curve. So to go in a closed curve, the

dz

will point in one direction (in the complex plane), and eventually must point as much in the other direction to get back to the starting point. So obviously for example for any closed curve

\gamma

,

\int _{\gamma }\ \,dz=0

; and has nothing to do with any "mean value". In a related way, if you were to integrate an analytic function in a "closed curve" on the real line, you can trivially see that it is always zero, because such a curve consists of a line that goes one way from a to b, and then exactly back the other from b to a, so the integrals along these two paths must be opposites, and when you add them you get zero. --Spoon! 17:04, 1 February 2007 (UTC)[reply]

Spoon! is right, but I'll describe it another way. Suppose the closed arc has length 1, and let t be a variable that measures arc length along the arc from some starting point. If the arc is $z=z(t)$ , then the usual change of variables gives

\int _{\gamma }f(z)\,dz=\int _{0}^{1}f(z(t))z'(t)\,dt.

Thus, Cauchy's theorem is saying (under some conditions on f) that the mean value of $f(z(t))z'(t)$ on the arc is zero. McKay 05:06, 2 February 2007 (UTC)[reply]

The line integral along a closed curve need not be zero unless the vector integrand is ir-rotational. This is the case for analytic functions. Twma 03:03, 4 February 2007 (UTC)[reply]

adding every number from 1 to 100

i heard that there is a formula like adding 1 to the end of 100 so you get 1001 and then dividing by something to go 5050 and i don't know what it is... my math tutor told me but i just don't know could someone help me out.... it's not a formula but a manipulation of the number 100 and then dividing by something...

Thanks, J- http://jdswebservice.com/xmswx

Sums of the form 1+2+3+...+n are Triangular numbers. Dugwiki 18:44, 1 February 2007 (UTC)[reply]

Ah, the arithmetic series and mathematical ingenuity! It is said that this was figured by Carl Friedrich Gauss on his early years, when an evil teacher wanted to keep his students busy for a while. Take a look at arithmetic series, summation and yes, triangular numbers — Kieff | Talk 19:53, 1 February 2007 (UTC)[reply]

The above is right - check the link - but if you want a simple way to work this out what you need to do is add the beginning and end numbers, then the next from beginning and end

eg

100+1
99+2
98+3
97+4
.
.etc
.
53+48
52+49
51+50

Note that each time the result is 101 eg 98+3=101

Also note that there are 50 or these sums

So the result is 50 times 101 = 5050 (that's the answer)

Hopefully that should explain what your teacher was on about.87.102.77.95 19:37, 1 February 2007 (UTC)[reply]

By the way thats 100x(100+1)/2 (there are fifty pairs = 100/2, and each pair adds to 101=100+1. = n(n+1)/2 when you add the number from 1 to n.87.102.77.95 19:45, 1 February 2007 (UTC)[reply]

Statistics/Normal Distribution (moved here from Miscellaneous desk)

The IQs of 600 applicants to a certain college are approximately normally distributed with a mean of 115 and a standard deviation of 12.If the college requires an IQ of at least 95, how many of these students will be rejected on this basis regardless of their other qualification?

First, this would be better suited to the math reference desk, and second, we don't answer homework questions. I'm certain that the correct way to solve this problem can be found in your text book. Dismas|^(talk) 13:21, 1 February 2007 (UTC)[reply]

Right, I could point you at a bunch of articles here but they will likely be more confusing than your text (not a slam to wikipedia). Just so you know, what we do here rather than perhaps do your homework for you is steer you toward references that might clear up the confusion you are having that prevents you from doing it yourself. While that is hard to diagnose "over the internet", if you give us some clue to what part you don't understand then we can help. But this does belong on the math desk. --Justanother 15:03, 1 February 2007 (UTC)[reply]

How many standard deviations below the mean is 95? What Z score does that correspond to in a normal distribution table in your textbook? What portion of the normal distribution lies below and above that Z score? How many students does that proportion imply? (Boy, do I ask a lot of questions, when all you wanted was a simple answer to the homework!). Post your answer and someone might even tell you if you got it right. Edison 19:11, 1 February 2007 (UTC)[reply]

(Question was also posted at science Ref Desk and moved here with replies. Please do not double post.) Edison 16:19, 2 February 2007 (UTC)[reply]

The IQs of 600 applicants to a certain college are approximately normally distributed with a mean of 115 and a standard deviation of 12.If the college requires an IQ of at least 95,how many of these students will be rejected on this basis regardless of their other qualification,,,,, —The preceding unsigned comment was added by 203.128.5.84 (talk) 11:42, 2 February 2007 (UTC).[reply]

The number of applicants rejected will be equal to the fourth root of the number of femtoseconds you took to write this question, integrated with respect to x over a range from minus infinity to the time it would take you to do your own homework. — QuantumEleven 13:18, 2 February 2007 (UTC)[reply]

If you did your own statistics homework, you might be able to estimate where in the roughly Gaussian distribution of class grades yours is likely to fall if you keep up this behavior. -- mattb @ 2007-02-02T13:28Z

February 2

pentiminoes

Can someone show me how to fit 12 diffrent penteminoes into a six by ten rectangle in 5 diffrent ways?

You might want to try and look at the article on pentaminoes. The question sounds like homework, so I'm not going to help you further. Oskar 07:16, 2 February 2007 (UTC)[reply]

Well, the proper article would be pentomino. — Kieff | Talk 15:51, 2 February 2007 (UTC)[reply]

Here's one possiblity as a starter:

   ***@@@@@**
   *%::::%%:*
   *%%%:%%::*
   @@*%@%*:@*
   @***@**:@@
   @@*@@@**@@

– b_jonas 19:04, 2 February 2007 (UTC)[reply]

Poincaré conjecture

The Poincaré conjecture says that if a 3space is enough like a 3sphere, then it is like a 3sphere. Um, so what does that mean? If a Poincaré-violating manifold (PVM) exists, what properties might it have that distinguish it from a 3sphere? —Tamfang 08:14, 2 February 2007 (UTC)[reply]

The conditions that make a 3-dimensional space "close enough" to a 3-sphere for the Poincaré conjecture to apply are:

The space must be a manifold - if you look at a small enough neighbourhood of any point, it will resemble Euclidean 3-space.
The space must be simply connected - given two points in the space, any path between these points can be continuously transformed within the space into any other path between the same two points.
The space must be a compact space - roughly speaking, it has a finite size and does not extend to infinity in any direction.
The space must be without boundary - roughly speaking, it doesn't have any edges where you can suddenly fall out of the space.

A Poincaré-violating manifold would have to posess all of these properties, yet still fail to be topologically equivalent to a 3-sphere. I'm not actually sure what topological property might allow you to easily distinguish such a manifold from a 3-sphere. Gandalf61 11:33, 2 February 2007 (UTC)[reply]

seeking hyperbolic cookbook

It appears that the algebraically easiest way to do stuff in hyperbolic geometry is with the hyperboloid model, in which each isometry apparently can be expressed with a matrix. But which matrix? I've read a couple of books on h.g. and plenty of websites, but found nothing that goes beyond "it can be done." Where do I find a book (or website) that will tell me:

Given a line, what matrix expresses a unit translation along that line?
Given a plane, what matrix expresses a reflexion in that plane?

...and like that. I might be able to work it out for myself, given free time and a better night's sleep than I've had in recent years! —Tamfang 08:47, 2 February 2007 (UTC)[reply]

If you want to experiment with the hyperbolic plane, a nice place to start is

Stillwell, John (1992). Geometry of Surfaces. Springer-Verlag. ISBN 978-0-387-97743-0.

One reason we use so many different models is because each has its virtues. Also remember that all isometries can be composed using reflections (reflexions) only. In the hyperboloid model, a hyperbolic line corresponds to a plane of the ambient space, and any isometry must map the surface to itself. --KSmrq^T 18:23, 3 February 2007 (UTC)[reply]

Does the Stillwell book contain answers to the specific questions I ask, or only principles that I already know as in your paragraph? —Tamfang 00:41, 4 February 2007 (UTC)[reply]

triangle angle bisectors2

From Wikipedia:Reference_desk/Mathematics#triangle angle bisectors

Somebody asked "is there an expression to get each side of the triangle from given values of the angle bisectors?…86.132.237.140 21:07, 29 January 2007 (UTC)"

I've got this equation

(C²-ab)(a+b)²/c = (B²-ac)(a+c)²/b = (A²-bc)(c+b)²/a

Where capitals are bisector lengths, lower case are side lengths..

Now I could solve this.. But given that I usually fail to manipulate equations through more than two steps without making a mistake could someone check that this is correct.. Or, is there a better way to do this? Thanks87.102.4.6 11:28, 2 February 2007 (UTC)[reply]

I've just read an article in the American Mathematical Monthly of Jan 94 which shows that, given 3 arbitrary values, a unique triangle exists which has these as angle bisectors. As the easily-obtained equations giving the bisectors separately in terms of the sides were not manipulated into a vice versa form, I rather assume that this is not possible, however.86.132.238.155 15:44, 2 February 2007 (UTC)[reply]

Statistics/Normal Distribution

In a mathematics examination the average grade was 82 and the standard deviation was 5.All students with grades from 88 to 94 received a grade of B.If the grades are approximately normally distributed and 8 students received a B grade,how many students took the examination"""" 12:03, 2 February 2007 — Preceding unsigned comment added by 203.128.5.84 (talk • contribs)

For a normal distribution centered on 82 you need to know the fraction of the distribution (a value between 0 and 1 - in this case less than 1) falling between 88 and 94 when the deviation is 5. You can get the equation of the distribution when the average is 82 and the deviation is 5. Then you need to integrate the equation between 88 and 94 (or find the fraction using look up tables..) this gives you the fraction of the students getting a B. Then divide 8 by that fraction to get the total number of students...87.102.4.6 12:44, 2 February 2007 (UTC)[reply]

Note that (from Variance) the "the square root of the variance, called the standard deviation" - therefor your variance is 5x5 = 25.87.102.4.6 12:50, 2 February 2007 (UTC)[reply]

Therfor you need to integrate the below equation with respect to x between 88 and 94, with μ=82, σ=5, σ²=25.(I can't think of a simpler way - but there usually is..)87.102.4.6 12:54, 2 February 2007 (UTC)[reply]

${\frac {1}{\sigma {\sqrt {2\pi }}}}\;\exp \left(-{\frac {\left(x-\mu \right)^{2}}{2\sigma ^{2}}}\right)\!$

The form of the OP hardly suggests that the questioner is going to find integration a meaningful idea, given the essential triviality of the problem. I don't think it unreasonable to tell him/her that the key will be to use Normal tables with z=1.2 and 2.4—86.132.238.155 15:56, 2 February 2007 (UTC)[reply]

Hyperbolic geometry

I too have a simple question about hyperbolic geometry.. What lines are considered to be 'straight' in hyperbolic geometry say of a Hyperboloid surface? It seem to me that for spherical geometry great circles are the equivalent of straight lines - since they represent the longest distance.. For a hyperboloid (of one sheet)

{x^{2} \over a^{2}}+{y^{2} \over b^{2}}-{z^{2} \over c^{2}}=1

(hyperboloid of one sheet),

are lines in the plane x/y=constant condsidered to be the equivalent of straight lines (and also parallel)(think yes)?

What about the circles formed by intersection with z=constant - are two circles formed by z=a,z=b considered to be parallel - and are they considered to be straight lines(thinks not)?

I'm using the term 'straight line' to mean the shortest distance between two points on the surface - I looked at the page and understood I think the parallel concept - but didn't get how to find a line that is the equivalent of a straight line in conventional geometry.

If you can make an answer simple I would appreciate it.87.102.4.6 12:15, 2 February 2007 (UTC)[reply]

First, it's not a hyperboloid of one sheet, it's one sheet (chosen arbitrarily) of a hyperboloid of two sheets.

?? Are you sure about this - this image shows only one sheet - that's the equation I gave? Image:HyperboloidOfOneSheet.PNG 87.102.4.6 16:20, 2 February 2007 (UTC)[reply]

True, that's the equation you gave, and I ought to have looked closer at it, rather than assuming that you had in mind the hyperboloid model of the hyperbolic plane. Since I can't tell you anything about geodesics on the hyperboloid of one sheet (other than to agree with your guess about both questions above), I'd delete my remarks below but that's bad form. —Tamfang 19:49, 2 February 2007 (UTC)[reply]

That's ok. thanks.83.100.183.48 20:05, 2 February 2007 (UTC)[reply]

In this model, hyperbolic straight lines are represented by planes passing through the origin, ax+by+cz=0, or (if you prefer) by the intersections of such planes with the hyperboloid. Note that this also describes great circles on a sphere!

(By the way, great circles are "straight" not because they are the longest possible "lines" on the sphere but because the shortest curve between two points on a sphere is a segment of a great circle.)

The lines that you describe are a valid subset of these, with c=0, but they're not parallel: they all intersect at x=y=0. A parallel set would "intersect" in a point on the circle (a line on the cone)

x^{2}+y^{2}=z^{2}

. —Tamfang 16:06, 2 February 2007 (UTC)[reply]

Given that x=y=0 doesn't have a point on the hyperboloid I described would you reconsider your last point - to be honest you've confused me a bit - are you talking about the same thing - look at the picture..87.102.4.6 16:34, 2 February 2007 (UTC)[reply]

I am not aware of any model of hyperbolic geometry based on the hyperboloid of one sheet. As Tamfang says, the model that is usually called the "hyperboloid model" is based on one sheet of a hyperboloid of two sheets. Other standard models of hyperbolic geometry are the Klein model, the Poincaré disc model, the Poincaré half-plane model and the pseudosphere. To define a model based on the hyperboloid of one sheet, you would have to define a family of curves within the surface that meet the axioms of "straight lines" in hyperbolic geometry. The curves x/y=constant within the surface could be a sub-set of this family, but there are not enough of them - there is only one curve through each point, and most pairs of points within the surface are not connected by one of these curves. The circles z=constant are certainly not candidates for "straight lines" because they are finite whereas all "straight lines" in hyperbolic geometry can be extended to infinity (as in Euclidean geometry). Gandalf61 11:07, 3 February 2007 (UTC)[reply]

Thanks.87.102.9.55 11:44, 3 February 2007 (UTC)[reply]

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