Wikipedia:Reference desk/Archives/Mathematics/2007 February 1

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[edit] February 1

[edit] 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)

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)
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 γ, \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)

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)

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)

[edit] 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)
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)

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)

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)
Generalizing, the sum S of a finite arithmetic series consisting of n terms is S = n/2 (a1 + an). n/2 is 50, a1 is 1, and an is 100. Arithmetic series are things like 1 + 3 + 5 + (and so on). —The preceding unsigned comment was added by 172.146.58.73 (talk) 08:52, 26 February 2007 (UTC).

[edit] 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)
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)
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)
  • (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)

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).

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)
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