Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

May 28

Smooth approximation of a hypercube

I'm working on an optimization problem where the solution must fall on the unit hypercube. Is there a simple smooth approximation to this constraint? It must be differentiable, and being twice differentiable would give me more flexibility in solving the problem. The simplest solution I can think of is to use a sphere, then project it to the nearest face and use gradient descent along that face to fine-tune the result. However, if there is a reasonably simple approximation that is more cube-shaped it may get me a good enough result faster because it doesn't cut off the corners of the cube as much. Katie R (talk) 12:20, 28 May 2014 (UTC)

You could use a $p$-norm for large $p$. The unit sphere is $\sum _{i=1}^{n}x_{i}^{2}=1$. If you change that to $\sum _{i=1}^{n}x_{i}^{p}=1$, you'll get a better approximation to the hypercube as $p$ increases; indeed, the limit is the unit hypercube.--80.109.80.78 (talk) 12:35, 28 May 2014 (UTC)

[ec] $\sum _{i=1}^{n}|x_{i}|^{\alpha }=1$ approximates the surface of the cube $\forall i(-1\leq x_{i}\leq 1)$. With higher $\alpha$ the approximation is better but it's less smooth. -- Meni Rosenfeld (talk) 12:37, 28 May 2014 (UTC)

Thanks! That sounds perfect - quick to evaluate, simple derivatives, and it will be easy to work with analytically when I start working on analyzing the convergence of the algorithm. Katie R (talk) 13:08, 28 May 2014 (UTC)

You will have to restrict α to even integers to keep it analytic. —Quondum 17:20, 28 May 2014 (UTC)

The points on the surface of the unit hypercube satify the equation

$\prod _{i}(x_{i}^{2}-1)=0$

The points on the surface of the unit hypersphere satify the equation

$(\sum _{i}x_{i}^{2})-1=0$

So the equation

$(1-p)\prod _{i}(x_{i}^{2}-1)+p((\sum _{i}x_{i}^{2})-1)=0$

is a hypercube for p=0 and a hypersphere for p=1, and for 0<p<1 it is a surface in between. Bo Jacoby (talk) 22:17, 28 May 2014 (UTC).

That's not correct. There are many more points satisfying the first equation than just the surface of the hypercube. It seems to me the last equation is only satisfied by $x=0$ or $y=0$. -- Meni Rosenfeld (talk) 22:35, 28 May 2014 (UTC)

Right - as long as one element has an absolute value of 1, the others are unbounded and can fall anywhere on the plane that face defines. I'm very happy with the unit ball in a p-norm solution. It's very simple, and so are the Jacobian and Hessian, which is important because I may need to evaluate it several thousand times a second. It's also convex which makes it useful for global optimization. Katie R (talk) 02:31, 29 May 2014 (UTC)

We agree that the points on the surface of the unit hypersphere do satify the equation (for p=0) and that inside points do not. It is not a problem that some outside points do. Choose 0≤x₂²≤1 , . . . , 0≤x_n²≤1 and compute x₁² from the linear equation.

$(1-p)(1-x_{1}^{2})\prod _{i=2}^{n}(1-x_{i}^{2})+p(1-x_{1}^{2}-\sum _{i=2}^{n}x_{i}^{2})=0$

It is simpler than the p-norm solution for p>n. Bo Jacoby (talk) 05:35, 29 May 2014 (UTC).

It does matter that points outside the cube satisfy the equation, when you're trying to create a new equation with a convex combination of the two equations. Did you actually try to plot your last equation? I did in the 2D case and got gibberish. And as I said, it is only satisfied for $x=0$ or $y=0$ (fixed above) when $0<p<1$. -- Meni Rosenfeld (talk) 11:41, 29 May 2014 (UTC)

Yes I did actually plot my last equation by this J program:

  'dot; pensize 4' plot(,j.)(,-)(,+)j./%:(,:-.%[:-.0.9&*)?100#0

Explanation: (?100#0) creates 100 random numbers 0≤x²≤1. (,:-.%[:-.0.9&*) appends the corresponding y² values for p=0.1. (-.x) is 1-x. (%) is division. (%:) is square root. (j./x,:y) creates x+iy. (,+) appends x-iy. (,-) appends -(x+iy). (,j.) appends i(x+iy). The program displays 800 dots showing a square with round corners.

Probably you committed an elementary error. Bo Jacoby (talk) 22:10, 29 May 2014 (UTC).

Ok, one issue is that in your original comment you had a sign error. This was corrected in your second comment and gives a more reasonable shape. However, the point remains that there are also points outside the cube that satisfy the equation. -- Meni Rosenfeld (talk) 09:11, 30 May 2014 (UTC)

The equi-p-norm shape is called a superellipse or superellipsoid by the way (or more specifically, a squircle). --catslash (talk) 12:39, 30 May 2014 (UTC)

Looks like squircle is the special case where p=4. Too bad, because I was hoping my high-dimensional version could be called a squircloid. :-P Katie R (talk) 15:54, 30 May 2014 (UTC)

Using only one branch of, say, a hyperbola is not prohibited. It is interesting how reluctance prevented Meni Rosenfeld from catching the idea. Bo Jacoby (talk) 11:32, 1 June 2014 (UTC).

Algebra brain-fade.

I'm having trouble with a bit of algebra:

If:

 A = B x

 x = ???

(B is in the range 0..1 and x>1 if that helps at all). 65.111.112.94 (talk) 20:54, 28 May 2014 (UTC)

$x=\log _{B}A={\frac {\log A}{\log B}}$. See Logarithm. -- Meni Rosenfeld (talk) 21:13, 28 May 2014 (UTC)

Logarithm turns multiplications into additions and exponential into multiplications.

So

A = B ^x

becomes

Log(A) = Log(B ^x )

Log(A) = Log(B) * x

So you can easily solve for x. 202.177.218.59 (talk) 01:05, 30 May 2014 (UTC)

May 30

Math joke, riddle, or puzzle

I am trying to recall a specific math joke, riddle, or puzzle. It was basically a "play on words", if I remember correctly. The punch line (or, rather, the humor) was based on the reader needing to understand the binary number system, which uses just two values (0 and 1). I believe that it was just one simple sentence, such as might appear on a T-shirt or a bumper sticker or such. The saying was something like this: "There are only two kinds of people in this world; those who understand binary and those who don't." (or something like that). But, within that sentence, there was some play on words; I believe the sentence used some binary value of 0 or 1 or some combination to provide the joke/humor. Or perhaps it was something along these lines: "There are only 10 kinds of people in this world; those who understand binary and those who don't." Does this ring a bell with anyone? Does anyone recall the specific statement? Thanks. Joseph A. Spadaro (talk) 05:13, 30 May 2014 (UTC)

See Mathematical joke#Jokes with numeral bases. —Quondum 05:27, 30 May 2014 (UTC)

That's it! Thanks! I guess my memory was pretty good, after all. I got the quote exactly correct. Thanks. Joseph A. Spadaro (talk) 15:47, 31 May 2014 (UTC)

There are only three kinds of people in this world; those who can count and those who can't. Bo Jacoby (talk) 12:11, 1 June 2014 (UTC).

Good one! Joseph A. Spadaro (talk) 14:56, 1 June 2014 (UTC)

base eight math

In an episode of Stargate SG1 Jack O'Neill has his brain made smarter by aliens. During this time he starts doing "base eight math". While I know that "base eight math" is a real thing used in computers, is it something a smart person would use to write out math equations on a blackboard? SupremeThor (talk) 21:11, 30 May 2014 (UTC)

Algebra is the same in any base, of course, except for the different representation of numerical constants. Arithmetic can be done in any base, and the reason that nearly everyone finds base ten easier is that they have had lots of practice in that base. Someone who has been taught in base eight from an early age would be just as competent in that base, and might struggle with base ten if they had not practised decimal arithmetic. Perhaps the writers of Stargate just wanted to indicate that Jack's brain had been "taken over"? Using base eight is no indication of "smartness". Dbfirs 06:17, 31 May 2014 (UTC)

But being able to instantly switch over and do math in different bases in your head is a sign of intelligence.

Now for my favorite base 8 joke: "Why do computer programmers confuse Halloween with Christmas ? ... Because OCT31 = DEC25." StuRat (talk) 06:29, 31 May 2014 (UTC)

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)

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)

[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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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

If 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)

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)

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)

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)

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)

k=ab+ac+bc , max k not possible?

Is there a maximum k where k is not expressible as ab+bc+ac where a, b and c are positive integers?Naraht (talk) 03:48, 1 June 2014 (UTC)

Yes (sequence A025052 in the OEIS). The largest value is 462, or possibly one larger value, it seems. —Quondum 16:54, 1 June 2014 (UTC)

Interesting!! If we could find one more term, that would disprove the GRH. (That's a big IF). Bubba73 ^{You talkin' to me?} 18:03, 1 June 2014 (UTC)

Infinity

How do you KNOW that numbers are infinite ? What is the largest number science has discovered to date ? — Preceding unsigned comment added by Ankoosh97 (talk • contribs) 10:37, 1 June 2014 (UTC)

Have you read infinity? Do that and come back. Plasmic Physics (talk) 12:45, 1 June 2014 (UTC)

It depends on what you mean by numbers. If you think of integers or reals, then none of them are infinite. It just exists infinitely many of them. YohanN7 (talk) 17:11, 1 June 2014 (UTC)

These infinities can be quantified by another type of number, the infinite cardinal numbers.

mathematics

quadratic equations problem — Preceding unsigned comment added by 123.238.158.211 (talk) 12:18, 1 June 2014 (UTC)

I'm not sure, but I suspect we'll need just a teensy bit more information. -- Jack of Oz ^{[pleasantries]} 13:15, 1 June 2014 (UTC)

42. Plasmic Physics (talk) 13:53, 1 June 2014 (UTC)

Are you looking for how to solve quadratic equations? You may have been introduced to a method where you "factorized" the expression, leaving you with a pair of brackets. You may have been introduced to a method called Completing the square. You may have been introduced to a method using the quadratic formula. If you can be more specific about what you remember, and what you are trying to do, we can help. Otherwise, BBC Bitesize might help you get started. 86.146.28.105 (talk) 19:18, 1 June 2014 (UTC)

Negative numbers

Are negative numbers "smaller" than positive numbers? Could a negative number ever be bigger than a positive number? Diechaddie (talk) 13:01, 1 June 2014 (UTC)

I assume you refer to magnitude when you say "smaller". If that is the case, then there is no difference. Plasmic Physics (talk) 13:16, 1 June 2014 (UTC)

Size can mean different things. It could mean the magnitude of a quantity (so the absolute value) in which case, yes there are negative numbers that are bigger than positive numbers. For instance, −10 is larger in magnitude than +2. On the other hand "size" might refer to the natural ordering on the real numbers. Every negative number is less than every positive number. I would argue that the former notion of "size", "bigger", and "smaller" is the one that more commonly is used in conversation (for instance, we talk about "bigger debt" meaning a more negative balance). Although obviously it depends on the context. Sławomir Biały (talk) 13:50, 1 June 2014 (UTC)

It depends on how you use them. For example, if you use positive to mean how much money you owe the bank, then negative numbers are what they owe you. In that case, -$10 means they owe more than you would at +$2.

If, on the other hand, it's something that can't reverse direction, like elevation above sea level, you could think of +2 (2 meters above sea level) as higher than -10 (10 meters below sea level). However, -10 is farther from sea level, so it's still "bigger" in that sense.

There are even some strange cases where all negative values are bigger than all positives, such as stellar magnitudes, where the negative values are brighter. (I see no real reason why they had to do it this way, they just did.) StuRat (talk) 17:44, 1 June 2014 (UTC)