## 598 – Upcoming talk: Jodi Mead

October 28, 2009

Jodi Mead, Wed. November 4, 2:40-3:30 pm, MG 120.

Non-smooth Solutions to Least Squares Problems

In an attempt to overcome the ill-posedness or ill-conditioning of inverse problems, regularization methods are implemented by introducing assumptions on the solution.  Common regularization methods include total variation, L-curve, Generalized Cross Validation (GCV), and the discrepancy principle. It is generally accepted that all of these approaches except total variation unnecessarily smooth solutions, mainly because the regularization operator is in $L^2$. Alternatively, statistical approaches to ill-posed problems typically involve specifying a priori information about the parameters in the form of Bayesian inference. These approaches can be more accurate than typical regularization methods because the regularization term is weighted with a matrix rather than a constant. The drawback is that the matrix weight requires information that is typically not available or is expensive to calculate.

The $\chi^2$ method developed by the author and colleagues can be viewed as a regularization method that uses statistical information to find matrices to weight the regularization term.  We will demonstrate that unique and simple $L^2$ solutions found by this method do not unnecessarily smooth solutions when the regularization term is accurately weighted with a diagonal matrix.

## 502 – Cancellation laws

October 28, 2009

Two homework problems. The first one is easier, so you can consider the second one to be extra credit. A proof of these results can be found in different places, for example, the paper Division by three, by Conway and Doyle. (Please don’t look at the paper while working on the homework, of course.) Unfortunately, the paper could use a serious trimming and editing, so I cannot really recommend it, but the proof is carefully written there.

1. Without using the axiom of choice, show that if $A$ and $B$ are sets, and $|A\times 2|=|B\times 2|,$ then $|A|=|B|.$
2. Same as 1., but now with $3$ instead of $2.$