April 26

I need positive integers x,y satisfying ${\displaystyle xy+4x\leq 500}$ that minimizes the function ${\displaystyle {\frac {1}{xy}}(0.9+0.1y)}$. Money is tight (talk) 08:37, 26 April 2012 (UTC)[reply]

Ignore for the moment the restriction that x and y are integers, and let x and y be any real numbers. If we have (x,y) in the interior of the feasible region i.e. such that ${\displaystyle xy+4x<500}$, then we can move towards the boundary by increasing y while keeping x constant, which reduces the value of ${\displaystyle {\frac {1}{xy}}(0.9+0.1y)}$. This suggests that you first look for a real valued (x,y) on the boundary ${\displaystyle xy+4x=500}$ that minimises ${\displaystyle {\frac {1}{xy}}(0.9+0.1y)}$. Then if x and y in this solution are not integers, examine the integer value points that are closest to this (x,y). Gandalf61 (talk) 09:36, 26 April 2012 (UTC)[reply]

${\displaystyle z={\frac {1}{xy}}(0.9+0.1y)}$ is minimized. ${\displaystyle dz=0}$.

${\displaystyle y(y+9)dx+9xdy=0}$.

The equation ${\displaystyle x(y+4)=500}$ is differentiated to give

${\displaystyle (y+4)dx+xdy=0}$.

Eliminating dy:

${\displaystyle (y(y+9)-9(y+4))dx=0}$.

So ${\displaystyle y^{2}-36=0}$. Substituting the positive solution ${\displaystyle y=6}$ into ${\displaystyle x(y+4)=500}$ gives ${\displaystyle x=50}$. This solution ${\displaystyle (x,y)=(50,6)}$ happens to consist of integers. Bo Jacoby (talk) 10:21, 26 April 2012 (UTC).[reply]

Thanks for your help! Money is tight (talk) 07:21, 27 April 2012 (UTC)[reply]

I just read the article about functional derivatives, and while I think I understand the idea now, I am confused about the introduction. It states: In mathematics and theoretical physics, the functional derivative is a generalization of the gradient. While the latter differentiates with respect to a vector with discrete components, the former differentiates with respect to a continuous function.

Would it be more correct to use the term directional derivative instead of gradient? Both a directional derivative and a functional derivative are expressed as real numbers, the first in case of a finite number of variables and the second for an infinite number so to say. A gradient on the other hand is expressed as a vector.

Thanks,

M.

--111.251.197.34 (talk) 09:59, 26 April 2012 (UTC) — Preceding unsigned comment added by 111.251.197.34 (talk) 09:58, 26 April 2012 (UTC)[reply]

It used to say directional derivative. I have changed it back. Sławomir Biały (talk) 11:58, 26 April 2012 (UTC)[reply]

I'm pretty sure the gradient analogy is correct. The inner product is important: ${\displaystyle \left\langle {\frac {\delta F[\varphi (x)]}{\delta \varphi (x)}},f(x)\right\rangle =\left.{\frac {d}{d\epsilon }}F[\varphi +\epsilon f]\right|_{\epsilon =0}}$ corresponds to ${\displaystyle \left\langle \nabla F({\vec {\varphi }}),{\vec {f}}\right\rangle =\left.{\frac {d}{d\epsilon }}F\left[{\vec {\varphi }}+\epsilon {\vec {f}}\right]\right|_{\epsilon =0}}$. In fact, the article then goes on to say that f having unit norm produces directional derivatives. (To the OP:) The functional derivative is a vector in the vector space of functions. The continuity and/or smoothness of the functional F and of the test functions f is crucial to this being well-defined by allowing the entire behavior (in a neighborhood) to be defined in terms of "a few" variables (here, one functional derivative that is independent of f). --Tardis (talk) 13:38, 26 April 2012 (UTC)[reply]

You need an inner product to define what a gradient is, but you don't need to know anything about inner products to take directional derivatives (or functional derivatives, for that matter). So talking about gradients in the general context of differentiation in function spaces is not such a good idea. —Kusma (t·c) 08:56, 27 April 2012 (UTC)[reply]

Let X be a hausdorff space, and ${\displaystyle a,b}$ two distinct points in it. Is always there a continuous function ${\displaystyle f:X\to \mathbb {R} }$ such that ${\displaystyle f(a)\neq f(b)}$? — Preceding unsigned comment added by 84.228.237.103 (talk) 14:57, 26 April 2012 (UTC)[reply]

I can't think of a counterexample off the top of my head, but I"m pretty sure there isn't always such a function. What you state is a special case of Urysohn's lemma. --COVIZAPIBETEFOKY (talk) 17:39, 26 April 2012 (UTC)[reply]

Apparently a space with this property is a completely Hausdorff space, and PlantMath has an example of a Hausdorff space that is not completely Hausdorff. 60.234.242.206 (talk) 23:37, 26 April 2012 (UTC)[reply]

I'm assuming one of our learned mathmeticians will be familiar with the game of craps. (I'll be happy to explain if not.) A "fire bet" requires the shooter to make all six points (4,5,6,8,9,10) and returns between 999 and 1999 to 1. Smaller payouts are given for making four or five points. I have yet to find a casino floorperson or pit boss who can tell me the true odds of a player doing this. Googling "fire bet odds" has taken me to sites which proclaim the house edge to be between 20 and 25%. While this is a huge house edge, I can't help but think it's even bigger than that. Joefromrandb (talk) 19:09, 26 April 2012 (UTC)[reply]

I'm not entirely clear on what the bet is. He has to make all six points before crapping out? What if his own point comes up before he's made all six or craps? Is he allowed to keep going to make the remaining points, or does he lose the fire bet and establish a new point on the next roll? --Trovatore (talk) 19:25, 26 April 2012 (UTC)[reply]

The shooter must establish, and make, all six points. Joefromrandb (talk) 19:42, 26 April 2012 (UTC)[reply]

When does he lose the bet? When he loses the dice? --Trovatore (talk) 19:46, 26 April 2012 (UTC)[reply]

Correct. If the shooter sevens out before establishing, and making, "4", "5", "6", "8", "9", and "10" individually, the bet is lost. Joefromrandb (talk) 19:48, 26 April 2012 (UTC)[reply]

Please do explain. It may be difficult to find a symbolic expression but easy to find an approximation by simulation. I simulated what I understood from reading our article but my results are inconsistent with the odds you state, so I must have misunderstood. It will help if you explain exactly what are the conditions. -- Meni Rosenfeld (talk) 19:48, 26 April 2012 (UTC)[reply]

See Craps#Player bets for the conditions. The essence is

As different individual points are made by the shooter, they will be marked on the craps layout with a fire symbol. The first three points will not pay out on the fire bet, but the fourth, fifth and sixth will pay out at increasing odds. The fourth point pays at 24-to-1, the fifth point pays at 249-to-1 and the 6th point pays at 999-to-1.

Duoduoduo (talk) 20:19, 26 April 2012 (UTC)[reply]

Correct. Also, instead of "24-1, 249-1, 999-1", some casinos pay "9-1, 199-1, 1999-1". Joefromrandb (talk) 20:25, 26 April 2012 (UTC)[reply]

Also, at least one casino I know pays 1.5-1 for three points. Joefromrandb (talk) 20:28, 26 April 2012 (UTC)[reply]

If the fourth and fifth points are both achieved, am I right that they don't both pay out, but only the last one does? Duoduoduo (talk) 21:04, 26 April 2012 (UTC)[reply]

Correct. Only the highest-reached level is paid. Joefromrandb (talk) 21:10, 26 April 2012 (UTC)[reply]

(copied from User_talk:Meni_Rosenfeld)

I'm putting this here because of multiple edit conflicts. Feel free to copy it to the ref desk. The fire bet is only available on the shooter's first come out roll. The numbers "2", "3", "7", "11", and "12" are irrevelant for the fire bet, so we'll disregard those. The shooter rolls until s/he throws a "4", "5", "6", "8", "9", or "10". Let's say the shooter rolls a "6". This number is the "point". The shooter continues to roll until a "7" is thrown and s/he loses, or a "6" is thrown, in which case the point is made and one-sixth of the fire bet is complete. The shooter must continue in this fashion, establishing a point, and making each point without rolling a "7" first. Any "7" thrown on a come out roll does not affect the fire bet. If the shooter establishes, and makes a "6", and then establishes and makes another "6", the second "6" is irrevelant to the fire bet. "4", "5", "6", "8", "9", and "10" must each be established (thrown on the come out roll), then made (thrown after the come out roll, before a "7" is thrown). Any "7" thrown between the establishing and making of a point renders the bet lost. Please let me know if I need to clarify anything else. Joefromrandb (talk) 20:08, 26 April 2012 (UTC)[reply]

Ok, I think I get now the requirements and I have a model to calculate the exact probability. It's possible I made a mistake but what I get for all 6 points is 3700403899126040038831518494284887738125 / 22780863797678919004236184338193605974839452 or roughly 0.000162435 which is 1:6155 odds. My mathematica code in case anyone wants to try to verify is:

 op[4] = 3; op[5] = 4; op[6] = 5; op[8] = 5; op[9] = 4; op[10] = 3;
 prob[X_] := Product[p = op[X[[i]]]/24; 
  q = Sum[(op[X[[j]]]*op[X[[j]]])/(op[X[[j]]] + 6), {j, 1, i - 1}]/24;
  r = 1 - p - q;
  p/(p + r)*op[X[[i]]]/(op[X[[i]]] + 6), {i, 1, Length[X]}]
 Total[Map[prob, Permutations[{4, 5, 6, 8, 9, 10}]]] 

By using permutations of length 5 and 4, I get for 5 points 16299535440740507819343049336436045 / 9043400932320159658679162374214129124 which is 1:554, and for 4 points 230470444882074000130253 / 21741374240278642986445695 which is 1:93. -- Meni Rosenfeld (talk) 21:11, 26 April 2012 (UTC)[reply]

Very impressive. Thank you! Joefromrandb (talk) 21:23, 26 April 2012 (UTC)[reply]

You can ignore 2, 3, and 12? I thought you immediately lost your bet if you throw one of those on your first roll? --Trovatore (talk) 21:56, 26 April 2012 (UTC)[reply]

If you bet "pass" you do indeed lose with a "2", "3", or "12" on the come out roll. However, the shooter keeps the dice, and continues shooting until a point is established. Then the shooter continues until the point is made, and a new come out roll begins, or a "7" is rolled, at which time the shooter loses and passes the dice. Just as "7" and "11" are winners on the come out roll, "2", "3", and "12" are losers. These do not affect the fire bet. Joefromrandb (talk) 22:26, 26 April 2012 (UTC)[reply]