Talk:Broyden–Fletcher–Goldfarb–Shanno algorithm

rank of matrix v. rank of tensor

What's the deal with the rank 2 matrix claim? Do we mean rank 2 tensor? It seems to me that if you take U = u v^T, then U is a matrix with rank of one (i.e. all rows/cols are linearly dependent). In other words, I think there is potential confusion between matrix rank and tensor rank. Birge 22:51, 28 March 2006 (UTC)

Relation to steepest decent (gradient descent)

Could someone explain what the relationship between BFGS and gradient descent is? I could (very well) be wrong, but isn't it pretty much doing an approximate newton step? My notes say that when the Hessian is approximated by an identity matrix then BFGS will take a gradient descent step. Perhaps there's another relationship that I don't get, but it confused me to read that BFGS is "derived" from gradient descent.129.59.43.172 23:45, 3 May 2006 (UTC)

too technical tag

this tag was first added when the article was just a stub; it was just one paragraph long. the article has considerably grown since then, and the stub tag was long ago removed. if the person who added the tag wants to add it back, please do but leave some suggestions on what details need to be better explained to a general audience. Lunch 03:58, 24 September 2006 (UTC)

under-explained math

I'm totally confused as to what the variables are in the equations. If they are explained in other topics (like Newton's Method, etc.), then links should be made.

I am particularly confused by step 4 of the algorithm. B is a matrix, y and s are vectors, correct? Then the first term on the right is a matrix, but the second is a scalar (i.e., quotient of two vector dot products). The third term is another scalar. So, what am I missing?