Delicious Vegetarian Stuffed Mushroom Recipe

I had some portabello mushrooms to use before they went bad and I looked online for a stuffed mushroom recipe and I found http://whatscooking.us/2008/03/05/lunch-special-stuffed-creamy-portabella-mushrooms/

Which looked delicious so I decided to modify the recipe to make it vegetarian and made out of stuff I had on hand. I was extremely happy with the result so I'm going to share.

Step 1) put a stick of butter, 6 0z of cream cheese 1/4 cup water, 1 tablespoon sugar, and 1 tablespoon grapefruit juice in a sauce pan and heat and stir until it thickens a bit.

Step 2) Add a diced red bell pepper, garlic powder and chili powder to taste to the sauce and let simmer

Step 3) Dice and Sautee an onion in a separate pan. Remove the stems from your portabell0 mushrooms chop them up and add them to the onion and let them cook for a bit.

Step 4) Add the onion and mushrooms to the cream sauce and fill the mushrooms with the mixture.

Step 5) Top the mushrooms with parmesan cheese and bread crumbs.

Step 6) Broil the mushrooms until the bread crumbs are a golden brown.

I was making 4 stuffed portabello's with the above recipe and I had a bit too much sauce but some of my mushrooms were much smaller than the others so your milage may vary.

enjoy

Commencement

This last Friday I finally participated in the commencement ceremonies for the degree I earned a year ago. There was a university wide commencement ceremony which I attended but had to leave early from and a convocation ceremony which I got to sit all the way through. After coming out on the other side of the graduation ceremony I think I would have been better served to simply go to the university wide commencement ceremony and just ignored the existence of the convocation. At the convocation I got to walk across a stage while someone read my name... it wasn't really much to get too excited about. I would rather go to the ceremony where the bulk of the program is made up of speeches instead of disconnected contextless names.

Although it is somewhat nice to have some personal recognition in hindsight it definitely wasn't worth sitting through everyone elses names in order to get my few seconds to cross the stage. Or at least it isn't worth it when I could just as easily get a ceremonial conferral of my degree just by attending the commencement, albeit not a a personal one.

My poker book got delivered to a gynecologist:: or why ups should be spelled phonetically

I ordered "The Mathematics Of Poker" a little while ago and on Wednesday of last week it got delivered.... or at least on Friday when I checked the tracking data online to find out what had become of it I discovered it had been delivered and signed for... but not by me. At 3:00 on the dot Wednesday someone named "Kristie" had signed for my decidedly absent package.

I live in a house in which multiple rooms are rented out independently of each other. I wasn't certain if a Kristie had perhaps moved in recently and for some reason had decided to retain my package for safe keeping in her room instead of in the mantle in the front room. At least before I went about complaining I needed to find out if in fact I lived with a Kristie. After asking a number of my room mates and someone at the coffee shop that shares the plot of land on which this house is built I discovered that there was indeed a new female resident of the house but no one knew her name. Eventually the matter was solved by calling the land lord who informed me that the only woman living here went by the name of Angela (we shall ignore for the moment the fact that Yu-ling has been living here for longer than I have and she is definitely female).

Satisfied that there was no Kristie in the house the question became what address had my package really been delivered to? A little information could be gleaned from the tracking page.

SALT LAKE CITY, UT, US	03/03/2010	3:00 P.M.	DELIVERY
	03/03/2010	1:51 P.M.	A CORRECT STREET NUMBER IS NEEDED FOR DELIVERY. UPS IS ATTEMPTING TO OBTAIN THIS INFORMATION
SALT LAKE CITY, UT, US	03/02/2010	9:16 P.M.	DESTINATION SCAN
	03/02/2010	2:51 P.M.	A CORRECT STREET NUMBER IS NEEDED FOR DELIVERY. UPS IS ATTEMPTING TO OBTAIN THIS INFORMATION
	03/02/2010	6:59 A.M.	OUT FOR DELIVERY
	03/02/2010	4:01 A.M.	ARRIVAL SCAN
SPARKS, NV, US	03/01/2010	7:04 P.M.	DEPARTURE SCAN
	03/01/2010	12:59 P.M.	ORIGIN SCAN
US	03/01/2010	2:19 A.M.	BILLING INFORMATION RECEIVED

A correct street number is needed for delivery? What on earth did that mean? How did UPS manage to lose my address? I decided to go look and see what was on the adjacent streets at 1031 E and I found that in fact there was a house at 1031 one street over and in the other direction there was a hospital. There was no response when I knocked at the 1031 house and I assumed that delivery to the hospital was essentially impossible. Checking online showed that the address of the hospital was 1050 E giving it no commonality with my own.

As I was walking back from the 1031 abode however I saw the UPS carrier making his rounds and hurried to talk to him. I asked if any packages had perhaps given him some trouble on Wednesday and he responded that none had. I told him that I had been expecting a package and that it had been delivered and signed for by a Kristie but that I had no idea who that was. Very fortunately for me he knew who this Kristie was and told me that my package must have gotten delivered to a Doctor Hinson. Why this could possibly have happened neither he nor I had the slightest Idea though he assured me that the package was addressed 1050 E 100 S and to doctor Hinson.

I got online and found out that Doctor Hinson was an OB/GYN though I failed to find the location of her office. I let the matter sit until Monday. On monday my package arrived battered and bruised with a ups label that had apparently gone over the top of a label with my correct name and address partially ripped off. The box had been opened and retaped though the book inside was no worse for wear.

Checking the tracking page now there are two delivery lines one right after the other.

SALT LAKE CITY, UT, US	03/08/2010	3:42 P.M.	DELIVERY
SALT LAKE CITY, UT, US	03/03/2010	3:00 P.M.	DELIVERY

And then immediately after it dated 2 days after I had already received my book

SALT LAKE CITY, UT, US	03/10/2010	8:25 P.M.	A CORRECT STREET NUMBER IS NEEDED FOR DELIVERY. UPS IS ATTEMPTING TO OBTAIN THIS INFORMATION / THE RECEIVER PICKED UP THE PACKAGE THAT WAS BEING HELD FOR THEM

I still have no idea what really happened in this whole mess but I thought the oddity was sufficient to merit sharing. This at the very least justifies me in a long time habit of half jokingly (now somewhat less than half jokingly) pronouncing ups as oops.

Kissing Numbers

The kissing number in a certain number of dimensions is the greatest number of spheres that can be brought to touch (or "kiss") a central sphere if all the spheres are the same size. in one dimension the kissing number is 2. In two dimensions the problem is slightly less trivial but hardly difficult. It will probably not come as much of a surprise that the best arrangement in 2 dimensions is the familiar hexagonal arrangement.



A little thought will show that all the angular space around the central circle is taken up. Since the centers of 3 equally sized spheres mutually in contact with each other form an equilateral triangle.



So the minimum angle between the centers of any two circles touching the central circle is just Pi/3, the angle of an equilateral triangle. The sum of the angles between the centers of all of the circles touching the central circle must add up to a full 2*pi and each of these angles must be at least Pi/3. So the most circles we could possibly get to touch the central circle is 2*Pi/(Pi/3) = 6 circles. Since the hexagonal arrangement actually achieves this maximum we have our proof.

In three dimensions a good solution is found by taking the optimal 2 dimensional hexagonal arrangement of spheres around a central one in a plane but now it is possible to add spheres above and below the plane that touch the central sphere. Three spheres can be placed above and three spheres below the plane of the hexagonaly arranged spheres while still contacting the central sphere. The arrangement gives 6 + 3 + 3 = 12 as a lower bound for the kissing number in three dimensions. The centers of the spheres arrayed around the central one now form the vertices of one of the Archimedean solids, the Cuboctahedron.

Is this the optimal arrangement or is there possibly some more clever arrangement which could fit an extra sphere or two in? Generalizing our tactic for the two dimensional case we can ask what is the minimum angular space taken up by each sphere. A calculation of the solid angle taken up by a sphere touching the central sphere gives. Pi*(2-sqrt(3)) = 0.841 steradians. Dividing the total number of steradians by this value gives 4*pi/(Pi(2-sqrt(3)) = 14.92 So this gives us a definite upper limit of 14 spheres that we can fit around a central sphere.

But this bound ignores the fact that spheres do not fit flush with each other so we can't hope to fill all the angular space around the central sphere. Consider an arrangement of equal sized spheres placed at the vertices of a regular tetrahedron such that all the spheres touch each other. If all the spheres around the central sphere could be made to form such tetrahedral arrangements such that the lines between the centers of all touching exterior spheres form equilateral triangles clearly this arrangement would be optimal. I worked on a proof for a bit but already in three dimensions the problem is quite hard and I am not quite sure how to approach it. Nevertheless such tetrahedral arrangements are not possible in 3 dimensions or in any higher dimension either. The triangular packing of 2 dimensions is unique.

The Robots are gone

So the vast swarm of high school students is finally gone. It did end up turning out that my office was pretty much unusable between the hours of 2:30 and 9:00 and even during the day there came to usually be someone running around. Apparently there were roughly 150 students involved in the project in total though the number that I saw there usually averaged much closer to 30 people at a time.

I resisted posting over and over again about the inconvenience. I don't mind working in the library but there is a certain indignity in not being able to use my own office. Of course as everyone would tell me I was more than welcome to come and use my desk but when there are dozens of people coming and going and working together and discussing things and when those people are furthermore high school students... I think it will not be hard to convince you that the library was a much more conducive work/study area.

I really found it very irksome that the administration didn't deem this intrusion of my work space significant enough to merit a simple e-mail warning me of the swarm in advance. I first learned of the plan to share my office with these students when someone came by to install a box with the key to my office right next to my door. I wondered if perhaps my presence in this office had somehow gotten overlooked and on some official roster somewhere I was really supposed to be in one of the more populated grad student offices.

But after the last students had left and the copious amounts of garbage, and carpet, and plywood, and electronics, and dirty dishes, etc had been cleaned out of my office I got a letter from the department chair in my box.

Of course my first response was that this was some sort of official departmental action and therefore bad news. When I opened it I discovered it had a gift card in it. It actually turned out to be something of an apology.





Tim Anderton
Department of Physics and Astronomy
115 S 1400 E #201
University of Utah
Salt Lake City, Utah 84112

Dear Tim,

I would like to thank you for your accommodating the West High Robotics team during the past six weeks as they built and tested their robot in the James Fletcher Building. The group was very appreciative of the use of the grad student offices and the lab next door as this facilitated the construction and testing of the robot. Last year, the labs and office spaces for the robot build were located in Chemistry, but the mechanical construction took place in the Department of Physics and Astronomy machine shop.

I had not envisioned the amount of time the team would spend in the Department, and the number of people and/or computers that eventually were crowded into your small office. I know the imposition was rather severe at times. On behalf of the Physics Department and West High Robotics, I have enclosed a small gift card to the University of Utah bookstore. I sincerely appreciate your patience in the midst of all the chaos!

with Best Regards,
Dave Kieda
Chair, Department of Physics and Astronomy
Professor of Physics
University of Utah
Salt Lake City, Utah 84112




I am of somewhat mixed feelings about this. After the robotics team had gone I was happy to have shared my office with them. True it was quite a pain at times but at the same time I definitely approve of the project and think it is a good idea for them to have access to such a space in the department. The thing I most would have wanted is to have been asked permission or, failing that, at least been warned. On the other hand of course if I had been warned then I very probably would not have received this gift and apology. Especially since I would probably have ok'd the the use of the office anyway perhaps this is preferable. I can't help but wonder though if perhaps this is a manifestation of the fact that is often easier to ask forgiveness than permission. Ah, well C'est la vie.

P.S. On the off chance that Dave Kieda is reading this, Thanks for the letter and the card! I appreciate it despite my griping about my lowly status within the department.

ISS flyover

The International Space Station (ISS) will be visible over salt lake for about 7 minutes for the next few nights.

http://www.spaceweather.com/flybys/search_results_printable.php?zip=84112

its pretty fun to watch at least if you are inclined to be impressed by stuff like that, which I definitely am.

Recession a Speedo advertising ploy?

The red is the number of jobs created/lost during the last months of the Bush administration and the blue is the jobs created/lost during the first part of the Obama administration. Is it just a coincidence that the graph looks like a red and blue speedo with the crotch at the inauguration? I think not...

That being of course either because the stimulus plan worked or because speedo just ran the most expensive ad campaign in history.

the original graph can be found here

Mouse activity: Surfing and chatting

I was surfing the net using stumble upon and I stumbled upon this little program which logs your mouse position. The lines are mouse paths and the blobs are places where the mouse sat for a while.

a best k term approximation problem

Here is an interesting little question.

For all dictionaries of M vectors chosen from a N dimensional vector space what is the minimum value of the maximum relative error in the best k term approximation for any vector in the N dimensional space.

The term "dictionary" in this usage comes from signal processing (or at least that is where I got it from). A "dictionary" in this case is just any set of vectors which we can use to express other vectors in terms of. Generally in the case where the vectors are not linearly dependent on one another we call such a collection of vectors a "basis" a dictionary is the same except we remove the restriction that the vectors be linearly independent.

For a particular dictionary of vectors the best k term approximation of a vector X is the vector whose difference from X has the smallest value which can be constructed as a linear combination of k vectors chosen from our dictionary. For instance take the example of the plane and our dictionary is the usual basis {(0,1), (1,0)} then the best 1 term approximation to the vector X = (x, y) is (x, 0) if x > y and (0, y) otherwise. Obviously no other dictionary with only 2 vectors in it will do a better job making 1 term approximations. Also obviously the best 2(or more) term approximation of any vector is just that vector itself since our dictionary vectors span the space of all vectors in the plane.

In fact if you take a moment to reflect you will quickly see the answer to the general question if k >= N is 0. This is clear since if k >= N then M >= k >= N so we must be able to choose a basis of our N dimensional space in which case the best k term approximation of a vector X is just X itself. Slightly more interesting but with just as obvious a solution is the case where M < N. If we have fewer than N vectors to choose from then it doesn't matter which particular vectors we put in our dictionary just so long as they are all linearly independent of one another. Then we can build a M dimensional subspace of the N dimensional one and any M dimensional subspace is "as good" as any other one. Of course actually it is important to choose linearly independent vectors only if we also care about the average relative error of the best k term approximation as well as its maximum value. Since we can't even span the space with our M terms when we consider any vector orthogonal to our M dimensional subspace we get a relative error of 1. (relative error is the number |X-Y|/|X| where X is the vector to be approximated and Y is the approximating vector)

Lets return to the plane and examine the first interesting case N = 2 M = 3 k = 1. A first thought is to chose three vectors such that they form an equilateral triangle. With this choice no vector will ever be more than 30 degrees different than its best approximation by one of the 3 vectors. Since adding in -1 times the three vectors gives 6 equally distributed vectors yielding 360/6 = 60 degrees between vectors meaning we can be at most 30 degrees off from one of them. This gives a maximum relative error of sin(30)=1/2. The observation that we should equally distribute the vectors as much as possible solves the problem for odd M but obviously such solutions do not hold for even numbers since then half of our vectors would be just -1 times other vectors leaving us with a very obviously non-optimal dictionary.

A slight modification of our proceedure remedies this problem. We simply equidistribute the vectors not through the angles 0 to 2*Pi but from 0 to Pi treating the first vector as occupying both the positions at 0 and at Pi. This works for both odd and even numbers of vectors in M and it is not hard to see that this is in fact the optimal solution. Since if we consider the case k>=2 for n=2 we can obviously always exactly match any vector so this is a complete solution of the problem for 2 dimensions.

One might expect for there to be similar unique and hopefully even simple and structured solutions for higher dimensions. On this I cannot pretend to know what the solution to the problem is like but I am certain that the 3 dimensional and higher cases of the problem whatever the solution is it isn't simple. For the 2d case finding the most equidistributed set of M vectors was not hard. There is a unique(up to a rotation) such configuration for every M. But for 3 dimensions and higher this is no longer the case. Though if we have say 6 vectors a good bet for the optimal solution is to put them at the centers of half the faces of a dodecahedron. Very probably the platonic solids each correspond to a provably optimal and unique(again up to rotation) solution/s of the problem for specific numbers of vectors M.

Clearly though these solutions do not provide the general solution. Since the solution of the general case in exact terms would certainly be at best very difficult and at worst practically impossible here we move into the realm of heuristics for a bit. The most obvious first approximation to picking equidistributed vectors in some vector space is simply to choose vectors off the unit n-sphere with a completely uniform probability density.

If M>>N then this is probably a good enough approximation to work and a dictionary of random vectors picked from N dimensional space will give us close to the minimum possible value of the maximum error. Just how much larger than N M would have to be in order for this to be sufficiently good is certainly not clear.

What if we were to choose a first vector at random and then choose subsequent vectors by picking the vector which has the smallest maximum magnitude dot product with a vector already in our dictionary with ties being broken randomly. Clearly this would first result in the building of an orthogonal basis of the space. Once a complete orthogonal basis has resulted we would begin picking vectors from a complimentary basis and so on down the line. Such a construction is tempting until you realize that such a prescription doesn't give the right answer even for the simple 2D case.

On the other hand a probabilistic version of the same algorithm is perhaps worth considering. Consider the same algorithm but now modified so that every time we have chosen M vectors we randomly delete half of them and then again choose deterministically and again randomly remove vectors and continue this process for some large number of steps.

A man of 18 Letters (maybe 52)

I just got my bachelor's degree diplomas from the university in the mail. I do admit I now have the urge to go out and get some frames that will fit them and hang them up on my wall. Also that would help make it so that I don't lose them. (Note to self begin using better filing system than bedroom floor provides)

They are identical to one another except one has the word "Mathematics" in the middle and the other one has the word "Physics" in the middle. I guess now I'm a man of letters 18 to be exact 7 in physics and 11 in mathematics. Of course you could include the letters in "Bachelor of Science" which gives an extra 17 letters for each degree almost tripling my total count to 52.