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May 31

Taking union and intersection in modulus inequation (or inequalities)

I am confused in solving modulus inequations. I solve them by making different cases (i.e., dividing the whole problem in different parts). In the final answer, sometimes I have to take union of the cases (or parts), sometimes I have to take intersection of the cases. So, my question is - when should I take union and intersection of different cases in solving modulus inequations. 117.247.209.35 (talk) 09:35, 31 May 2014 (UTC)[reply]

Hi. I'm a high-school level mathematician and no more, so if this is a question beyond that, and I in my ignorance didn't realise, please ignore this. But if it is, and as no one better has answered below, then my best advice is to sketch graphs of both functions. The inequalities should become clear when you consider where function f(x) (say) is greater than g(x). Alternatively substitute in values between the critical values (the ones you've found by 'dividing the problem in different parts' ?), and then see for which ones the inequalities hold.86.157.134.7 (talk) 21:04, 31 May 2014 (UTC)[reply]

[This question is about solving equations with absolute value operators, also called modulus, and not modular arithmetic or any of the other meanings of modulus.] Addressing your union vs. intersection question with an example, if we wish to solve |x-1|≥2|x+1|, consider the critical values (as suggested above by 87. ... .7) of x=1 and x=-1, and break up the number line at those points: (-∞,+∞) = (-∞,-1) ∪ [-1,+1] ∪ (+1,+∞). (Note that the choice of which intervals the critical points are placed in is not important, because when f(x)=0 then |f(x)|=-f(x)=+f(x), so we could have equally chosen the partition (-∞,-1] ∪ (-1,+1) ∪ [+1,+∞). The critical points are where the means of simplification via removing the absolute value signs changes, but either form works at the critical point.)

We want all possible solutions to this problem, so the answer will be the union of the solution in the first interval with the solution in the second interval and the solution in the third interval. (Think union because we want them all.) Looking at the first interval, (-∞,-1), |x-1| simplifies to -(x-1) and |x+1| simplifies to -(x+1), so the inequality becomes -(x-1)≥2(-(x+1)) ⇔-x+1≥-2x-2 ⇔x≥-3, the solution of which is the interval [-3,+∞). But the inequality clearly does not hold for arbitrarily large values of x, so the "+∞" part does not belong in our answer. That is because our simplification, and thus this solution, was only valid for the interval (-∞,-1), so we need to take the intersection of the two intervals, (-∞,-1) ∩ (-3,+∞) = [-3,-1). Think intersection because both conditions must happen at once, that is, we only want that part of the solution to the simplified equation where the assumptions on which the simplification were based hold.

The next two intervals are done similarly, and the final answer is the union of all three results. In fewer words:

{x: |x-1|≥2|x+1|} = ( (-∞,-1) ∩ {x: -(x-1)≥2(-(x+1))} ) ∪ ( [-1,+1] ∩ {x: -(x-1)≥2(+(x+1))} ) ∪ ( (+1,+∞) ∩ {x: +(x-1)≥2(+(x+1))} )

= ( (-∞,-1) ∩ {x:-x+1≥-2x-2} ) ∪ ( [-1,+1] ∩ {x:-x+1≥2x+2} )∪( (+1,+∞) ∩ {x:x-1≥2x+2} )

= ( (-∞,-1) ∩ {x:x≥-3} ) ∪ ( [-1,+1] ∩ {x:-1≥3x} ) ∪ ( (+1,+∞) ∩ {x:-3≥x} )

= ( (-∞,-1) ∩ [-3,+∞) ) ∪ ( [-1,+1] ∩ (-∞,-1/3] ) ∪ ( (+1,+∞) ∩ (-∞,-3] )

= [-3,-1) ∪ [-1,-1/3] ∪ ∅

= [-3,-1/3]

Does that address your question? --ToE 11:12, 1 June 2014 (UTC)[reply]

Lottery

Statistically speaking does ones Luck of winning the Lottery increase when there is a mega jackpot rollover? IrishLotto (PLC) (talk) 16:07, 31 May 2014 (UTC)[reply]

By "luck" I assume you mean probability, and no it does not. The chance of winning the jackpot in a lottery doesn't change according to the amount of money it is worth. JaeDyWolf ~ Baka-San (talk) 16:21, 31 May 2014 (UTC)[reply]

It depends. If you win you will win a bigger prize, but, because it is likely that more people will have taken part, your chance of winning may be less. If the payout ever becomes larger than the takings it can even make financial sense to take part. Thincat (talk) 17:33, 31 May 2014 (UTC)[reply]

The concept you are after is called expectation - the probabilities of winning something times the amount you get in each case, minus the cost of entry. Certainly when it was launched, the British National Lottery became much better value to play during rollover weeks, and even more so during double rollovers. It's certainly conceivable that a strategy of only playing during rollovers would be more profitable - or at least vastly less loss-making - than a strategy of playing every week. But the actual likelihood of breaking even depends in part on whether it's only the top jackpot that goes up during rollovers, or whether some of the smaller pots do too. If you're still only 1/13,838,916 likely to win the jackpot, and that accounts for the entire increase in the expectation, you're still overwhelmingly likely to lose money. AlexTiefling (talk) 21:31, 31 May 2014 (UTC)[reply]

The expectation is less than one may think due to another factor: splitting of the jackpot between multiple winners. At a roll-over, there tend to be more tickets bought, thus increasing this effect. —Quondum 20:04, 1 June 2014 (UTC)[reply]

Colored fractals

I formerly had a roommate who was doing his maths Ph.D. in topology, and he introduced me to the concept of the Julia set — basically because he (correctly) thought I'd enjoy the pretty pictures. Why are they just almost always depicted in lots of different colors? The same is true (and thus my question also applies) to the Mandelbrot set; most images in that article are multicolored. Nyttend (talk) 20:25, 31 May 2014 (UTC)[reply]

I hope this is not to technical, but the different colors indicate how quickly iterations of the function on that that point reaches the distance that you can be sure that it isn't in the set. Bubba73 ^{You talkin' to me?} 21:09, 31 May 2014 (UTC)[reply]

The colours are assigned with the express purpose of illustrating the values in an interesting way. There are many different colour schemes that can be used. Some produce pretty pictures and others look rather plain. I once spent many happy hours choosing colour schemes and sub-regions that brought out the beauty of the representation (or so I thought). Dbfirs 21:38, 31 May 2014 (UTC)[reply]

I'd say the correct answer to the "why" question would have to be: "Because it is pretty that way." —Quondum 04:30, 1 June 2014 (UTC)[reply]

Bubba, it might be too technical; let me try to paraphrase, and please tell me whether I got it right. Different colors indicate different regions of the drawing, and each region comprises areas in which the iterations are basically reaching their limit (mathematics), in which they're growing so slowly that you know that they'll never grow out. Paraphrase #2: one color shows the areas that basically stop growing after x iterations, another the areas that stop growing after y iterations, etc. Nyttend (talk) 12:45, 1 June 2014 (UTC)[reply]

I'm more familiar with the similar Mandelbrot set, and that is what I had in mind. If you look at the first image there, the black area is in the set. There is no easy way to know if a point is in the set. You pick a point and take the first equation there and iterate it. If the absolute value of that ever gets above a threshold (I think it is 1.0) then it is not in the set. Count the number of iterations it took to exceed that threshold and color that point according to the scheme of your choosing (e.g. 1-2 is one color, 3-5 is another, 5-10 another, etc). Repeat that for all of the points in the image. Bubba73 ^{You talkin' to me?} 16:52, 1 June 2014 (UTC)[reply]

So the colour (in the case of the Mandelbrot colouring) indicates how quickly the sequence diverges to infinity. Using the absolute value criterion (and 1 is not large enough: I believe one must use |z|>2 as the criterion to ensure this divergence) does not lead to a particularly pleasing set of colour boundaries. There are other schemes that produce more natural, smooth colour gradings. —Quondum 17:13, 1 June 2014 (UTC)[reply]

It has been a while, but I think you are right about |z|>2. When I wrote a program for it in the late 80s, I used the number of iterations to reach that threshold - of course, other schemes can be used. Bubba73 ^{You talkin' to me?} 18:00, 1 June 2014 (UTC)[reply]

Markov chain with weighted nodes

I apologize if this is somehow described in the Markov chain article. I have a Markov chain where each node has a predefined weight (as well as probabilities to move to other nodes), and a sink. How can I calculate the expected sum of the nodes visited when I simulate the system from the source to the sink? 70.190.182.236 (talk) 22:31, 31 May 2014 (UTC)[reply]

The expected value of the path from a node is the node's value plus the expected value of the paths from each successor of the node (weighted according to the transition probabilities, of course). Writing this for all nodes simultaneously will produce a linear system

(I-P)v=w

to solve (which will have a solution only if the stationary distribution (which is the cokernel of the matrix for the linear system) is orthogonal to the node weight vector; for a single sink this means it must have 0 weight). --Tardis (talk) 10:10, 1 June 2014 (UTC)[reply]

To clarify, by "is the cokernel" I mean that the set of stationary distributions spans the cokernel; each contributes a constraint on the weights so that each absorbing communicating class must separately contribute no weight (since it would do so infinitely). Since

I-P

is singular, some special solution technique may be required (e.g., for numerical reasons): the obvious approach is to remove the equations (rows) and terms (columns) for the absorbing classes since they cannot contribute to any expected values. In the simple one-sink case, this means removing a row of all 0s (including in the vector!) and a column not all 0s. --Tardis (talk) 13:59, 1 June 2014 (UTC)[reply]

How many napkins does McDonald's pay $1 for?

If that varies, I'm referring to a McD's in the US, and Kansas if we need to get that specific. Wonder what supplier they get it from, how much they charge, and how much profit the supplier gets, if you know about this issue that deeply.

I'm wondering if I buy a $1 menu item for $1.08 post-tax, how many napkins I'd have to take from the dispenser to cancel their net profit difference, and even cancel the entire gross profit.

Their food is still appetizing, but their business ethics are morally questionable. That's why if I'm to eat from there, I'd like to do so in a way that doesn't let them profit off of me. Thanks. --2602:30A:2EE6:8600:9DC0:990F:7175:706A (talk) 22:57, 31 May 2014 (UTC)[reply]

Basically, their net profit from the $1 is all that's relevant. Everything else is overhead (business) or fixed costs (I can't tell the difference), which aren't relevant — they're going to pay the same amount in electricity, the same amount in wages, etc. whether or not you buy anything, so the profit is $1 minus the costs directly related to producing the $1 item. Nyttend (talk) 03:07, 1 June 2014 (UTC)[reply]

Nyttend - I think you partly missed the OP's point. Napkins don't go off, but they are consumed at a rate depending on customer choice. If the OP spends $1.08 on an item, eats that item, but also takes 50 cents worth of napkins, there's a fair chance that McDonalds won't really make a profit on that transaction. How many napkins are there in 50 cents worth?

The OP mentioned "their net profit difference" and "the entire gross profit". My point is that they're functionally the same here, since if McDonald's makes 50¢ on the item, they make 50¢ on the whole process — aside from the cost of the item itself, they have identical costs whether or not the item is bought. Since the core question was "how many napkins can I steal in order to keep them from making anything", we must discover both McDonald's cost per napkin and their profit per $1 item. Nyttend (talk) 04:03, 1 June 2014 (UTC)[reply]

Yes, and the OP's question can be paraphrased as "What is McDonald's cost per napkin?", so he's halfway there. Profit per $1 item is also needed to do the calculation. HiLo48 (talk) 04:07, 1 June 2014 (UTC)[reply]

And profit per $1 item is different depending on the item. Soft drinks are almost all profit, with fries being next after that (this is why they are always pushing fries and drinks). Anything with meat is lower profit. If they thought people would go there just for fries and drinks, if they stopped selling meat, then they would do just that. StuRat (talk) 17:50, 1 June 2014 (UTC)[reply]

Has anyone figured out if I just bought dollar-menu burgers, how many napkins I'd need to take to cancel their net profit? Next, how about dollar-menu fries and then dollar-menu drinks? Thanks, guys. --2602:30A:2EE6:8600:C52C:E22F:1249:FECC (talk) 20:29, 1 June 2014 (UTC)[reply]

Although I don't understand why you don't just eat somewhere else that you don't have ethical concerns with, calling their 1-800 number once per meal to complain about their labor practices would likely cost them a lot more than a stack of napkins and would let you get your message to them. Katie R (talk) 13:31, 2 June 2014 (UTC)[reply]

Also note that if people start taking too many free napkins, they will just stop providing unlimited free napkins, and give a set number away with each meal, instead. Thus you will fail to bankrupt the business, and make things less convenient for the patrons, as well. StuRat (talk) 13:48, 4 June 2014 (UTC)[reply]