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February 23

Left null space and cokernel

According to the article on the four fundamental subspaces, the left null space is the cokernel, but according to the section on left null space, it is dual to the cokernel. Which one is correct, and also, how is the definition $\mathbf {x} ^{T}A^{T}=\mathbf {0} ^{T}$ related to the quotient space definition of the cokernel?--Jasper Deng (talk) 17:30, 23 February 2017 (UTC)

See p. 27 here [1]. I think some wires must have gotten crossed, due to the various notions of "is", and how careful (or not) authors try to be when maintaining distinctions between isomorphic things with similar symbols and names. The way many authors define it, the left null space is a simple subset of the codomain, whereas the cokernel is a quotient space. So, one can say that the left null space is isomorphic to the cokernel, but it's not terribly wrong to say it simply is the cokernel, in anything but the most formal contexts (all we have to do is make a natural identification between elements of the codomain and elements of the quotient space). SemanticMantis (talk) 18:40, 23 February 2017 (UTC)

@SemanticMantis: But certainly being dual to is more specific than just "isomorphic to". In the finite-dimensional case, being isomorphic simply means being of the same dimension over the same field; I'm guessing what's being looked for here is a "natural" isomorphism of sorts.--Jasper Deng (talk) 19:29, 23 February 2017 (UTC)

Yeah, I think that dual statement might just be wrong, but I'm not sure. Do you have access to a university library? This is a case where a real textbook can clear things up far faster than combing through lots of online refs. If it's not wrong, then I think the dual statement must be assuming some different definitions than those used in the other articles. I'm not even sure what kind of "dual" it would be. But I'm very rusty on this stuff, hopefully someone else will chime in. SemanticMantis (talk) 22:54, 23 February 2017 (UTC)

I actually do have access to such a library (in fact I'm in one as I speak). I have in my possession an upper-division linear algebra book, but I don't think that one covers the left null space and cokernel, so I probably will want to dig in the library.--Jasper Deng (talk) 22:57, 23 February 2017 (UTC)

User:Jasper Deng did you find some clarifying definitions? If you do, let me know, and I/we can clarify our articles. SemanticMantis (talk) 18:00, 25 February 2017 (UTC)

I haven't had a chance to look in the library just yet. It may be a little while before I can.--Jasper Deng (talk) 19:55, 25 February 2017 (UTC)

February 25

Rank of an approximation sequence for a given accuracy

How can the rank of an approximation sequence of an irrational number (like e) be determined for a given accuracy with m decimal digits, for instance in the classic rendering of e as :$e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$? What equation must be solved in order to get the rank involved in the approximation? (Thanks)--82.137.9.121 (talk) 01:08, 25 February 2017 (UTC)

$e-(1+n^{-1})^{n}\approx en^{-1}\leq 10^{-m}$

Bo Jacoby (talk) 10:27, 25 February 2017 (UTC).

I got $e-(1+n^{-1})^{n}\approx {\frac {e}{2n}}$, or more specifically $(1+{\frac {1}{n}})^{n}\approx e(1-{\frac {1}{2n}}+{\frac {11}{24n^{2}}}\dots )$. The OP said it was just an example though so I'm not sure that answers the question. Would a link to Numerical analysis help? --RDBury (talk) 15:11, 25 February 2017 (UTC)

I made a mistake. You are correct. Bo Jacoby (talk) 17:28, 25 February 2017 (UTC).

I think you mean Order of approximation rather than rank? Dmcq (talk) 15:45, 25 February 2017 (UTC)

There are good answers, indeed. The next step is to formulate a general case analysis from this specific examples from list_of_representations_of_e#As_the_limit_of_a_sequence for a specific irrational number, e in this case, to any irrational number approximated as a limit of sequence in terms of series acceleration and rate of convergence, that is to get a better approximation to m digits of an irrational at a lower rank of the approximation sequence or at a lower iteration of the approximation. Taking another representation for the same irrational e, for instance replacing n with n!(factorial) what gain in accuracy can be get in the same iteration for a better converging sequence?--82.137.8.83 (talk) 10:54, 26 February 2017 (UTC)

Also what function is more easily to calculate values and apply in the equation obtained, the exponential or the natural logarithm?--82.137.8.83 (talk) 11:02, 26 February 2017 (UTC)

February 26

Infantry squares field of fire

I was watching the film Waterloo last night. Really impressive battle scenes, nothing done with CGI, thousands of soviet soldiers used as extras apparently. I was wondering about the fields of fire from the British infantry squares which were formed when they were under cavalry attack. The squares are portrayed as being very close together and some seem to be directly firing on each other. See this still from the film. Is there a mathematical solution to the best orientation of the squares to avoid "friendly fire"? The solution is easy for a single line of soldiers, a line of diamond orientated squares like: ◇◇◇◇. However, in the film (and probably in reality as well) the squares extended to a depth of three or four lines. This is much more tricky to solve. Arbitrarily reducing the density of squares can produce a solution, but the density of squares has to be taken as fixed by the density of the troops while they were in-line formation. I don't believe any solution is possible for an arbitrary number of lines and infinite range of weapons, so the range of the rifles has to be taken as a parameter. Alternating rows of diamonds and squares works quite well,

◇   ◇   ◇   ◇
   ▢   ▢   ▢
◇   ◇   ◇   ◇

Rows of diamonds works even better for up to three rows if they can be spaced wide enough to allow the squares in the two rows behind to fire between them. That is a reasonable spacing on the assumption that the in-line length of the formation is 4× the side of the square.

◇    ◇    ◇    ◇
◇    ◇    ◇    ◇
◇    ◇    ◇    ◇

Is there a better solution? I guess this counts as a kind of packing problem. SpinningSpark 13:32, 26 February 2017 (UTC)

I guess the first question would be what do you mean by "better"? Are you're trying to pack the soldiers as densely as possible, do you want to maximize the area covered, or do you want to maximize the chances of winning a battle based on some model of combat? As you mentioned above, the range of the weapons has to be considered, but there are two parts to that. First, what is the effective range to hit your target, and second, how far can a shot be expected to travel and possibly hit an ally if it misses the enemy. Another factor would be how disciplined your soldiers are, since it's one thing to tell them to only shoot straight ahead, but it's another to have them actually do it when there's a armed horseman coming at them at an angle. Not having engaged in early 19th century warfare myself, I's sure there are many other realities of the situation that I simply can't imagine.

But if you reduce the problem to it's simplest possible form you might phrase it in terms of chess pieces. Namely, how many rooks and bishops can be placed on an nxn board without attacking one another? For rooks only the answer is trivially n. For bishops only it seems you can do a bit better - 4 on a 3x3 board, 6 on a 4x4 board and in fact the answer is known to be 2n-2, see [2]. --RDBury (talk) 01:45, 27 February 2017 (UTC)

One way to pack in more firepower without having them shoot each other is to fire from different heights. The front row would lie down, the 2nd row kneel, and the third row stand. Thus, the back rows fired over the heads of those in front. If they are prepared to fire to the rear, as well, that means 6 rows of soldiers, and perhaps more in the middle in support roles, such as replenishing ammunition, caring for the wounded, replacing fallen soldiers, etc. And if you have a natural elevation difference, such as a hill, even more rows can fire over the heads of those in front. StuRat (talk) 04:39, 27 February 2017 (UTC)

• Not a mathematics answer, but my memories of The Mask of Command by John Keegan suggest me a hypothesis to the "friendly fire" question. (One big thesis of that book is that the increasing firearm range between Napoleon and the Civil War lead to a radically different style of command, because generals on the battlefield would be at a much higher risk of being targeted by sniper rifle in the latter than in the former.)

See Napoleonic_weaponry_and_warfare#Firearms: rifles only started to appear during the Napoleonic wars and were apparently reserved for semi-elite marksmen. The common soldier had a musket without rifling; from the various linked articles I gather that the effective range of non-rifled firearms was about 100-200 meters, after which the bullet deviated too much.

If friendly-fire casualties beyond 200m were negligible because almost all bullets would run in the ground or slow down too much by friction, separating the infantry squares by that distance would guarantee they could always shoot without any risk of hitting one another. Tigraan^{Click here to contact me} 12:41, 27 February 2017 (UTC)

We should also look at it strictly from a cost v. reward basis. The naive answer here is that if shooting in one direction is more likely to hit an enemy than your own forces, it's worth it. However, this would only work if both sides had equivalent forces. A more sophisticated formulation would be to say, if your side takes 1/10th the casualties of the enemy, due to superior equipment, tactics, training, numbers, etc., then you only want to fire if you are 10 times as likely to hit the enemy as your own forces. If course, this is only a rule of thumb, and specific situations will change that logic. For example, if an enemy is approaching with a bomb that could wipe out your entire infantry square, you'd better shoot at him, even if your own forces are directly behind him. StuRat (talk) 15:06, 27 February 2017 (UTC)

I'm not really looking for a practical military solution. If I was, I wouldn't have asked the question at this board. Perhaps I have not been clear enough. The problem is to devise a packing that maximises the perpendicular distance from the square edges to any other square while maintaining the overall density of troops equal to the in-line formation. Arbitrarily setting the squares 200 metres apart is therefore going outside the problem boundaries (and I believe it is also militarily impractical). The line density of troops in-line and in square edges should be assumed equal (or at least proportional). According to Line (formation) a line consisted of two to five ranks with three being most common, and, according to Infantry square, Wellington's squares at Waterloo were 500 men in four ranks in squares less than 20 metres on a side. A distance between lines is also needed. I have assumed this to be just enough to allow squares to form and have taken all lines to have the same number of ranks. SpinningSpark 15:17, 27 February 2017 (UTC)

March 1

Activity network question

Suppose that there are two activities emanating from the source node, A and B. There is one activity C, that depends on A alone; one, D, that depends on B alone; and one, E, that depends on A and B. It isn't obvious to me how to use a dummy in this case, so what do I do?--Leon (talk) 19:50, 1 March 2017 (UTC)

Can you add some context, and say what your objective is? (I'm familiar with dummy variables, but only in the context of regression.) Maybe you could give a similar example in which you can show how a dummy is used. Loraof (talk) 17:34, 2 March 2017 (UTC)

An ASCII text diagram may help:

                  E
                  ^
                  |
                  ^
                 / \
           C <- A   B -> D

StuRat (talk) 18:16, 2 March 2017 (UTC)

• Thanks, but that just restates what he said. I'd like to know what the OP's objective is and how he considers that dummies can be used in this context. Loraof (talk) 18:41, 2 March 2017 (UTC)

• Correct. I was replying to the OP, not you, as shown by my indent level. BTW, I fixed your indent level to show that you were replying to me. StuRat (talk) 18:45, 2 March 2017 (UTC)

• Sorry. Thanks. Loraof (talk) 21:55, 2 March 2017 (UTC)

I think in the particular precedence chart style the OP is referring to (there are many), each task is represented by the edge of a directed graph and each node represents a possible stage of completion. Say node 0 is nothing done, 1 is A done, 2 is B done, 3 is A and B done, 4 is A and C done, 5 is B and D done, and 6 is A, B and E done, with other nodes to be filled in as more tasks are added to make a complete diagram. Then A connects 0 to 1, B connects 0 to 2, C connects 1 to 4, D connects 2 to 5, and E connect 3 to 6. So far there is nothing to show 3 is connected to 1 and 2, so you need two dummy tasks, say X to connect 1 and 3, and Y to connect 2 and 3. Not my area of of expertise (the whole field is pretty woolly if you ask me), but hopefully that helps. --RDBury (talk) 03:50, 3 March 2017 (UTC)

March 2

March Madness Auction

I've been invited to participate in an upcoming auction for the NCAA basketball tournament, in lieu of the typical fill-out-a-bracket, with rules roughly as follows:

• Each team (final 64 only; play-in games are ignored) goes up for auction in a random order
• Depending on the seed of the team, there's a minimum bid price (ranging from $1 to $20, but I don't know the specific distribution)
• Each team must be sold
• Payouts are based on the games that each team you own wins, as a percentage of the total pool of money spend buying teams, as follows: 1% for 1st round win; 1.5% for second; 2% for 3rd, 3% for 4th, 4% for 5th and 8% for winning the championship.
• The total amount of the pool (and the number of players, though I think that's irrelevant) is, of course, unknown in advance.

With data for the likelihood that each team will reach each round in the tournament (which I can get from 538 or the like), I can calculate the expected percentage of the total pool that each team will win. What I don't know how to do is translate that info into an appropriate bid amount, since I won't know until the end what the total pool is.

To finally get to asking a question, any idea how I can get a plausible estimate for the "correct" amount to bid on each team? I could always just decide in advance what I think the total pool amount will be, but that's unlikely to be inaccurate and would lead to systematic over or under bidding, so I think I need some process (hopefully quick, since this'll be happening live) that could quickly update my total pool estimate (and thus my break even bid amount for each team) as each team is purchased. --Strohbot (talk) 23:03, 2 March 2017 (UTC)

EDIT: Should have googled first, but it turns out this is a thing with a name: Calcutta auction. So I guess my real question is, any information available for a bidding strategy for a Calcutta auction? --Strohbot (talk) 23:12, 2 March 2017 (UTC)

You should be able to estimate the number of total bidders there will be by the number who bid on earlier auctions. However, we get into game theory here, in that others may also take a wait-and-see approach to gauge the amount of bidders, too. So, it could end up like an Ebay auction, where it's pointless to bid until the last few seconds, as doing so gives away too much info. StuRat (talk) 04:00, 3 March 2017 (UTC)