Wikipedia:Reference desk/Archives/Mathematics/2008 July 7

= July 7 =

reduction in training time
I found a process which results in the least number of queries required to identify an element.

Recently I decided to automate the identification process to make it even faster using a single hidden layer FF/BP neural network. Accidentally I used the data set that was not optimized for training and the optimized data set for testing. When I discovered my error I replaced the optimized data test set and experienced no problem.

Out of curiosity I decided to train and test the network using the optimized data set and to my delight discovered that optimization consistently reduced the number of training epochs by almost half. Data set after data set, test after test gives the same results.

While I know that an optimized data set will reduce the number of queries necessary for manual identification I had no idea that optimization of the data set would result in fewer epochs necessary for convergence. This discovery makes artificial neural networks seem even more biological. Any explanations? Mimus polyglottos (talk) 02:27, 7 July 2008 (UTC)

Statistics: Pearson's chi squared vs Fisher's exact test
This link provides a PDF file with a single page of output from SPSS. In the first cross-tab, you see that the data was apparently unsuitable for Pearson's (33% of cells have values less than 5) but it goes through anyway and does not bother to calculate Fisher's exact test. On the second analysis, the data are suitable for Pearson's chi squared, but the program goes ahead and calculates Fisher's exact test-statistic. How is this sensible? Seans Potato Business 11:12, 7 July 2008 (UTC)

Cosets
Hi, I've got a group theory question, from an algebra book I'm working through. The question says: "If G is a group, and H is a subgroup of G, then prove that the number of left cosets of H is the same as the number of right cosets."

If H is finite, whether G is finite or infinite, it's easy. I'm having trouble with the case where H is infinite. An infinite subgroup can certainly have finitely many cosets, such as 5Z as a subgroup of Z (with addition), or it could have infinitely many, such as Q as a subgroup of R (with, say, addition).

Can someone give me a hint? I feel as though I should set up some bijection from left cosets to right cosets, but I can't figure one out. Am I barking up the wrong tree? -GTBacchus(talk) 20:22, 7 July 2008 (UTC)


 * You just need something 1-1,onto that reverses multiplication and sends subgroups to subgroups. For instance, (ab)^-1 = b^-1 a^-1, but H^-1 = H.  This implies that (aH)^-1 = H(a^-1).  You can check this is a well defined 1-1 correspondence. JackSchmidt (talk) 20:27, 7 July 2008 (UTC)
 * Yep, that makes perfect sense, and I feel kind of silly for not seeing it myself. I was trying conjugation, and I needed to think simpler. Thanks very much! -GTBacchus(talk) 21:42, 7 July 2008 (UTC)


 * Thanks GTBacchus for your feedback, you're a rarity around here. Most never comment one way or the other - like who knows if they've even read our responses. Check us out, you may be of help to another questioner :-) -hydnjo talk 01:52, 8 July 2008 (UTC)


 * Hi. As a mathematics student and teacher, I find this reference desk to be one of the most useful resources on the internet. I have used the desk from both sides - both asking and answering - although I've been online so infrequently of late that I haven't been around. When I start a Ph.D. program in late August, you may be seeing more of me around here. -GTBacchus(talk) 18:20, 10 July 2008 (UTC)