Wikipedia:Reference desk/Archives/Mathematics/2011 November 12

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November 12

I think my book is missing a page...

Dear Wikipedians:

I am learning curvature from the book Riemannian Geometry, a Beginners Guide by Dr. F. Morgan. However, I am pretty sure that a page is missing. I think the missing page is page 5. The reason I believe this is because the content at the end of page 4 (I believe), which is the first page of chapter 2 (with a big number 2 and the chapter title at the beginning of the page), transitions too jarringly into the contents found at the start of page 6, below Figure 2.1. However, there is no way that I can verify it as I do not possess the original library copy of the book.

So I am wondering if one of you can verify for me that that is indeed the original sequence progression of pages in the original library version of the book?

Thanks,

174.88.35.195 (talk) 02:09, 12 November 2011 (UTC)

I have a vague feeling that the fact that the page numbers seem to go 2,3,4,6,7,8 might be a hint of something here. But I can't quite lay my finger on it...  ;-) --Stephan Schulz (talk) 13:01, 13 November 2011 (UTC)

I suspect that you might just have solved the riddle there. Can anyone provide a scan or a photo of page 5 for the OP (or a link to a complete scan)? Dbfirs 17:13, 13 November 2011 (UTC)

Hi I am the OP. Thanks so much for all your help. In addition to waiting for the scan/photo, I shall personally visit some libraries in my vicinity in an effort to locate this book and the missing page it contains. 174.88.35.172 (talk) 02:22, 14 November 2011 (UTC)

extending an orthornormal set of vectors

I was reading about the singular value decomposition of a matrix and it said that given an orthonormal set of vectors {u1, ... ,ur} you had to extend it to an orthonormal basis of R^n. What does this mean and how do you do it? Widener (talk) 06:20, 12 November 2011 (UTC)

If r=n then U={u1, ... ,ur} spans R^n so U is a basis of R^n and you are done. If r<n, then there is a vector v in R^n that is not in the sub-space spanned by U. Subtract from v its projections along each of the orthonormal vectors in U. In other words, create

$w=v-\sum _{i=1}^{r}(v.u_{i})u_{i}$

Note that w is not the zero vector because v is not in the subspace spanned by U. Now we can calculate the dot product of w with each of the vectors in U:

$w.u_{j}=v.u_{j}-\sum _{i=1}^{r}(v.u_{i})(u_{i}.u_{j})=v.u_{j}-v.u_{j}=0$

because the vectors in U are orthonormal so the only term left in the sum is the jth term, and u_j.u_j = 1. So w is orthogonal to every vector in U. Now just divide w by its length to normalise it, and you have a vector that is normal and orthogonal to each vector in U. So U + w/|w| is an orthonormal set, now with r+1 vectors. Rinse, repeat and continue until you reach an orthonormal set with n vectors. Gandalf61 (talk) 09:37, 12 November 2011 (UTC)

This is also known as the Gram–Schmidt process. Also, Gandalf didn't dwell on the "what does this mean" part. A basis of R^n, as you surely know, is a linearly independent spanning set of vectors. "Orthonormal" means the vectors are orthogonal to each other and all have norm 1. "Extending to a basis" means finding a basis which includes the vectors you started with. -- Meni Rosenfeld (talk) 16:44, 12 November 2011 (UTC)

Higher dimensional equivalent of Johnson solids?

I'm looking for help with finding a list of the higher dimension analog of the Johnson solids, which are the Solids where all faces are regular polygons but are not vertex transitive. ("The corners aren't all the same"). There is sort of a split in higher dimensions that you don't have in 3 dimensions as to whether all of the polytopes that make up the higher dimensional equivalents have to be regular as well. As an example in 4 dimensions, the hyper pyramids with the Octahedron and Icosahedron bases are entirely made of regular polyhedra (the base plus tetrahedra), but the hyper pyramids with Cube and Dodecahedron bases are not made up entirely of regular polyhedra (because they have square and pentagonal pyramids) but those solids are made up of regular polygons. Any ideas where to look?Naraht (talk) 13:33, 12 November 2011 (UTC)

Mimesis

Somebody has edited Mimesis (mathematics) so that where it once said "The term geometric integration denotes a very similar philosophy", it now says "The term geometric integration denotes the same philosophy". Is this vandalism? Since I generally ignore mathematics and I haven't a scrap of the background knowledge the article expects, I can't tell.  Card Zero  (talk) 15:15, 12 November 2011 (UTC)

No that is not vandalism. It is a good faith edit. It may be controversial though. You may discuss it on the discussion page. Bo Jacoby (talk) 20:12, 12 November 2011 (UTC).