Wikipedia:Reference desk/Archives/Mathematics/2007 April 20

= April 20 =

The integral of cosz / z(z^2 +8) dz?
Let C denote the positively oriented boundary of the square whos sides lie on the lines (x= + / - 2 and y = + / - 2)

Evaluate integralc cosz / z(z^2 +8)

My problem with this question is that up until now I could solve the questions by using the Cauchy Integral f(z0) = 1/(2πi) Integralc f(z) dz / z - z0

Because the denominator just had a z - something so it was easy to use. Now this doesn't have that so I don't even know if this formula is correct. Which formula should I use?

This is just a question from the text and I have the answer, πi/4 but that hasn't helped me.. :) Thanks


 * Hint 1: $$z^2+8$$ can be factored into two linear terms like $$(z-a)(z-b)$$.
 * Hing 2: What goes in "the denominator" depends only on what you want to put there. You don't have to put everything that is in your other denominator in there at once. --Spoon! 05:47, 20 April 2007 (UTC)


 * Just to be clear, is the following the integral in question?
 * $$ \oint_{C} \frac{\cos z}{z (z^2 + 8)} \, dz $$
 * If so, consider the partial fraction expansion of
 * $$ \frac{1}{z (z^2 + 8)}, $$
 * especially as it relates to poles and residues. --KSmrqT 06:42, 20 April 2007 (UTC)


 * KSmrq's suggestion regarding partial fractions is an excellent approach to problems like this one in general; in the particular example you have, howeve.

Inter-rater agreement: Are there different alternatives for calculation of the confidence interval for kappa?
We have developed a new method for blood typing that is particular useful in certain clinical situations when conventional blood typing is difficult. For 5 antigens we have found 100% concordance between our new method and the conventional method in 151 blood samples that have been typed by both methods. These results are part of a manuscript that we have recently submitted for an international Hematology journal. One of the referees asked us which statistics we had used to assess concordance between the two methods. Initially we thought that this is an easy problem to solve. We intended to calculate the kappa coefficient with 95% confidence interval (CI) for each antigen and then, for all 5 antigens, to calculate the pooled kappa value with 95% CI. Because there was 100% concordance it is obvious that kappa = 1. To our surprise we found that the standard error of kappa in our case was 0, but by scrutinizing the formula for the standard error of kappa (Altman. Practical Statistics for Medical Research. Chapman & Hall, London 1991) it is evident that the standard error of kappa always equals 0 when there is 100% concordance. Intuitively, we found this difficult to understand: We would be much more confident in our new blood typing method if we have shown 100% concordance with the conventional method in a very large number of samples (for instance 830) samples as compared to a small number of samples (for instance 12). Hence, we expected that the confidence interval would more narrow if had tested a large number of samples as compared to a small number of samples. To us it seems as the 95% CI is not very useful for kappa values close to 1. If we assess the inter-rater agreement using the MedCalc statistical package on a data set of 50 samples in which there is concordance for 49 samples and disagreement for 1 observation, we obtain the following results: Kappa = 0.960, 95% CI from 0.882 to 1.038. This CI does not make sense because kappa cannot exceed 1.

Our question is as follows: Are there other ways of estimating the 95% confidence interval for kappa that are more suitable for kappa values of 1 or close to 1?

129.240.43.46 09:13, 20 April 2007 (UTC)


 * The usual standard error formulae for most statistics (including kappa) are appropriate for large samples, but are based on approximations that can be highly inaccurate when the sample size is low or the distribution of the data is extreme, and especially when both occur at once. A simple example is the usual formula for the standard error of a proportion, $$ \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$, which is 0 for $$\hat{p}=0$$ or $$\hat{p}=1$$ regardless of whether the sample size is 1 or 1000. There are algorithms that solve the small sample aspect of the problem, known as exact tests, which can also help mitigate extreme data distributions. Statistical packages such as SPSS or the SAS System can calculate exact tests and confidence intervals for kappa (see relevant SAS command here, and further background here), if you want to try that route. Another approach might be some form of Bayesian analysis; I recall that Alan Agresti has shown this can at least solve the problem of an extreme proportion I mentioned above. This would probably require some statistical expertise, though. -- Avenue 12:12, 20 April 2007 (UTC)

Freaky triangles - Where did those two squares come from???
I was reading this pop-science magazine at my dentist the other day, and they have these brain-teasers in the back. I can usually figure these things out pretty easily, but one of them stunned me, and I haven't been able to figure it out. It's a picture of a triangle composed of smaller shapes (four other triangles and two sort-of stubbed tetris pieces) that are rearranged in a different way to form the exact same triangle, but this time there are two additional squares. I drew the picture and uploaded it, this is how it looks like:



Please, for the sake of my sanity, can someone please explain where the hell those two blue squares come from? This is killing me! --Oskar 14:27, 20 April 2007 (UTC)


 * Niether of the 2 big figures is a triangle. The 2 "sides" pointing upwards are not straight lines. Got it now? Zain Ebrahim 14:38, 20 April 2007 (UTC)
 * Well, now I feel profoundly stupid. Of course they aren't straight lines! I can't believe I missed that! I see it now. Thank you! --Oskar 14:44, 20 April 2007 (UTC)


 * Longer explanation, in case anyone is still puzzled - if you subtract the first "triangle" from the second, you find that you are left with two long thin parallelograms. If we place the origin at the bottom left hand corner, the co-ordinates of the vertices of the left-hand parallelogram are (0,0), (3,7), (5,12), (2,5), and the second is symmetrically placed on the right hand side. The area of each of these parallograms is 1 unit (i.e. (3x5)-(2x7)), so the area of the second "triangle" exceeds that of the first by 2 units, which accounts for the two squares. Gandalf61 15:06, 20 April 2007 (UTC)


 * Quite a classic puzzle, a nice explanation is here. It seems that Fibonacci numbers come into play when selecting the lengths of sides. I'm sure this puzzle had a name, but it escapes me. --Salix alba (talk) 18:06, 20 April 2007 (UTC)

This is a variant of the Missing square puzzle. Root4(one) 18:16, 20 April 2007 (UTC)


 * I note the article also calls it Curry's paradox.Root4(one) 18:24, 20 April 2007 (UTC)


 * I seem to remember that this, or similar, was one of Sam Loyd's puzzles.…81.159.11.43 10:43, 21 April 2007 (UTC)

Well, I will never understand the numbers, but I sat a straight edge along both "upward pointing" sides of the triangles, and they look like straight lines to me. Do I just need new glasses, or a new straight edge? Bielle 10:17, 26 April 2007 (UTC)