Wikipedia:Reference desk/Archives/Mathematics/2009 March 25

= March 25 =

Using two results to generate a "combined" result
Hi - I have been asked a question I can't quite get my head round.... Each day at work I report two results, A and B. A is calculated using the formula A = L/R, and B is calculated using B = D/(R+T). I have now been asked to report these as a "combined result". This seems a little nonsensical to me (believe me, it is), but in order to keep the powers that be happy I need to report this new result. The question is: how do I combine the results in a way that makes some mathematical sense? Thanks for your help! sparkl!sm hey! 09:22, 25 March 2009 (UTC) The results I report relate loading cages of stock onto lorries in a depot. Result A is the percentage of cages that did not get loaded onto a lorry, where L represents the number of cages that did not get loaded (lost), and R represents the total number of cages delivered to store (retail). Result B is the percentage of cages that were deleted from the depot computer system before they reached the loading bays, where D = no. of deleted cages, R again represents the total number of cages delivered to store, and T represents the number of cages delivered elsewhere (trunk). Is there a sensible way to combine the numbers? Thanks again! sparkl!sm hey! 10:08, 25 March 2009 (UTC)
 * You'll have to tell us a good deal more about what's going on and what everything means to get a useful answer. Algebraist 09:31, 25 March 2009 (UTC)
 * OK, I tried to keep it simple, but I will attempt to explain.
 * A number of ideas:
 * 1. Figure out a cost for each and report a total cost.
 * 2. Report the maximum of the two
 * 3. Toss a coin and report one depending on heads or tails
 * 4. Report a random number and see if they notice
 * 5. Report the exact same number every day and see if they notice.
 * 6. Report 2L.3R.5D.7T this loses absolutely no information.
 * 7. Alternatively just interlace the decimal digits of A and B - this is easy to reverse back to A and B. Dmcq (talk) 12:40, 25 March 2009 (UTC)
 * Thanks, I think I'll go with number 4 ;) sparkl!sm hey! 13:34, 25 March 2009 (UTC)

Easy! Consider A to be the length of the X-axis and B to be the length of the Y-axis.

Report back C=Sqrt(A^2 + B^2). 122.107.207.98 (talk) 11:46, 26 March 2009 (UTC)


 * You could use (L+D)/(R+T), which would equal the percentage of boxes that disappear at any point in the process. I don't think there's any way to calculate that based only on A and B, though. Black Carrot (talk) 22:34, 26 March 2009 (UTC)
 * There is not, as you suspect. Algebraist 10:16, 27 March 2009 (UTC)


 * I'd have tried going for the costing version myself. What's the point of reporting something if there is no profit or loss involved? And if two things are to be combined which are not really comparable otherwise what better way to combine them? Dmcq (talk) 10:10, 27 March 2009 (UTC)

(L+D)/(R+T) is what we went for in the end, the rationale being, as Black Carrot suggested, that it does give an indication of the success of the overall process. Thanks everyone, the powers that be are satisfied with the nonsense results that they have dreamed up... sparkl!sm hey! 10:53, 31 March 2009 (UTC)

Necklace with pendante catenary shape
Hi,

I can work through the derivation for the shape of a catenary with no problem. However, what I want to do is find the shape for a catenary with an additional downwards force in the centre - e.g. like a necklace with a pendant. Would the chain on one side of the pendant still follow a catenery shape? I thought it would, with boundary conditions given by the length of the chain and the additional force, but I can't seem to adjust the parameters in the standard catenery expression to give an asymettric shape.

I am ultimately working towards 'inverting' this to find the optimum shape for an arch bridge which supports its self-weight plus a point load in the centre. Any help greatly appreciated! LHMike (talk) 10:48, 25 March 2009 (UTC)


 * Yes it'll be two parts of a catenary stuck together. Each little bit of the chain only knows whats beside it so the equations at any point except the centre are exactly the same. For any point in the catenary there is a force along the catenary supporting the centre bit. The upward component of that force will be supporting the weight of the catenary from that point to the bottom so is proportional to half the length of the catenary at that height. So you need to find the length of catenary that equals the weight of your pendant, chop that part out of the catenary and join the two halves together. By the way Gaudi is reputed to have designed the Sagrada Família by hanging weights to nets that way. Dmcq (talk) 12:21, 25 March 2009 (UTC)


 * Ah! Yes, chop out the length equal to the weight of the pendant. Fantastic, thanks, now to Maple! LHMike (talk) 14:20, 25 March 2009 (UTC)

Show xsin(1/x) is integrable rigorously?
Hi there guys - pretty much wondering what the title says - how would one show rigorously that $$xsin\frac{1}{x}$$ with f(0)=0 is integrable? I've been working with Riemann integrals but I'm not following the theory brilliantly so if anyone could give me a little help on this that'd be great. I have that a function is Riemann integrable if its upper and lower integrals are equal, or alternatively if for any $$\epsilon$$ there exists some dissection $$D$$ with $$U(f,D)-L(f,D)< \epsilon$$, but I don't really know how to start proving that in this example case...

Thanks for any help,

Otherlobby17 (talk)Otherlobby17 —Preceding undated comment added 10:49, 25 March 2009 (UTC).


 * Here is a hint: for a given t, think of the region of integration in two parts, the part where |x| >= t and the part where |x|<t. How does the integral behave in each of the two parts?  How large is the contribution of the "messy" part to the total, if t is very small? 75.62.6.87 (talk) 11:05, 25 March 2009 (UTC)


 * It's a continuous function, what's the matter? On an interval [a,b] you always have $$\scriptstyle U(f,D)-L(f,D)\leq(b-a)\omega(|D|)$$, where $$\omega$$ is a modulus of continuity of f and |D| is the maximum distance between consecutive points of D (the so called "modulus" or "norm" of the "partition" D). Here f is 1/2 Hoelder continuous, that is, it has a modulus of continuity of the form $$\omega(t):=Ct^{1/2} $$.  --PMajer (talk) 13:33, 25 March 2009 (UTC)
 * This approach easily shows that all continuous functions are Riemann/Darboux integrable, once one has the Heine–Cantor theorem that continuous functions on closed bounded intervals are uniformly continuous. Algebraist 14:03, 25 March 2009 (UTC)


 * So let me add that if you are still "not following the theory brilliantly", the example of your f(x) is a nice exercise to understand better the general facts mentioned by Algebraist. Notice that $$\scriptstyle U(f,D)-L(f,D)\leq (b-a)\omega(|D|)$$ is quite a simple and elementary inequality, and that you can compute explicitly the constant C in the modulus of continuity of your f (you need the Heine-Cantor theorem only in the generality of the above mentioned statement about all continuous function, of course).--pma 16:02, 25 March 2009 (UTC)

A function is Riemann-integrable on a bounded interval if and only if it's continuous almost everywhere in the interval. The function you give is continuous. Even if the concept of "almost everywhere" has not been mentioned (as it might well not be, when the Riemann integral is treated) it would probably be mentioned that if a function is everywhere continuous then it's Riemann integrable. Michael Hardy (talk) 02:16, 26 March 2009 (UTC)
 * You forgot the boundedness condition: f from [a,b] to R is Riemann-integrable iff f is bounded and continuous almost everywhere. Algebraist 02:29, 26 March 2009 (UTC)
 * In this case, it doesn't matter, though - a continuous function on a closed interval is always bounded, so Michael's last statement (the important one) is accurate (as long as you take "interval" to mean "closed interval", anyway). --Tango (talk) 05:04, 26 March 2009 (UTC)
 * That's OK, as a further information; however let's recall that the OP asked for a rigorous proof of the Riemann integrability of $$\scriptstyle x\sin(\frac{1}{x})$$, simple and elementary, as (s) he is a beginner. From this point of view, the best answer is maybe the first one. Let's go back to it for a moment. Let D be a partition of the interval [0,1] and t a point of D, to be specified later. The contribution to $$U(f,D)-L(f,D)$$ on the interval [0,t] is of course bounded by $$2t$$ because $$f$$ satisfies $$\scriptstyle|f(x)|\leq|x|$$ for all $$x$$. Our function is Lipschitz on the interval [t,1] with constant $$\scriptstyle 1+\frac{1}{t}$$ (just because this is a bound for $$\scriptstyle|f\,'(x)|$$ there). So the contribution to $$U(f,D)-L(f,D)$$ on [t,1] is bounded by $$(1+1/t)(1-t)|D|$$. Now choose $$t:=|D|^{1/2}$$ (that we may assume to belong to D because in any case we can add it); summing up we have
 * $$\scriptstyle U(f,D)-L(f,D)\leq 3|D|^{1/2}$$.
 * Note that we could have obtained this more directly by the Hölder continuity of f, but the Hölder continuity is maybe less direct to get than the local lipschitz. pma(talk) 21:29, 26 March 2009 (UTC)

unsolved math mysteries?
I'm wondering if anyone could name any well-known unsolved math mysteries. I don't mean conjectures like Goldbach's (i.e. propositions which most mathematicians think are true, but are missing proofs, even though some of these eventually turn out to be false). By a mystery, I mean something that is knowable and interesting but there is significant uncertainty about what the answer will actually be when it is found. Thanks for any examples. 75.62.6.87 (talk) 11:00, 25 March 2009 (UTC)
 * Unsolved problems in mathematics is the place to be. Livewireo (talk) 14:45, 25 March 2009 (UTC)
 * Unfortunately, most of those articles don't indicate whether they're things that everyone believes (like Goldbach's) or things that there's uncertainty about. I believe an example of the latter is the invariant subspace problem for Hilbert spaces. Algebraist 14:50, 25 March 2009 (UTC)
 * Whether the Euler–Mascheroni constant is irrational is a good one, probably? SetaLyas (talk) 15:22, 25 March 2009 (UTC)
 * Well, there's the problem. As far as I know, everyone thinks it must be irrational (and indeed transcendental), but Wikipedia articles don't tend to give information on this kind of thing, and it might be hard to find reliable sources. Algebraist 15:26, 25 March 2009 (UTC)
 * I would add the continuum hypothesis, though this does require you to liberalize your notion of what it means to "find" an answer. I think it's fair to say that, among set theorists who think it has a well-defined answer, the majority think it's false.  But there is a significant countercurrent represented by Matthew Foreman. --Trovatore (talk) 19:46, 25 March 2009 (UTC)
 * Would you mind taking a stab at what proportion of set theorists do think CH has a well-defined truth value? Algebraist 20:33, 25 March 2009 (UTC)
 * Hmm, that's difficult. Mostly they don't like to commit themselves on the point in public.  It's fairly clear that Woodin thinks so, and Moschovakis; beyond that I can't really say. --Trovatore (talk) 20:38, 25 March 2009 (UTC)
 * Thanks. Algebraist 23:27, 25 March 2009 (UTC)
 * That's an unsolved problem in the philosophy of mathematics, rather than mathematics itself, really, isn't it? --Tango (talk) 23:41, 25 March 2009 (UTC)
 * Actually, no, I'd call it an unsoved problem in mathematics, but one whose solution requires doing some philosophy. Some mathematicians like to pretend that philosophy of mathematics is an add-on, something you don't really have to worry about if you don't want to.  In my view that's a superficial approach. --Trovatore (talk) 00:06, 26 March 2009 (UTC)
 * Yes... I've always noticed this mysterious reserve of set theorists. Maybe they know something? Sometimes I'd almost say they like to show they don't like to commit themselves. It is strange though... --pma (talk) 23:42, 25 March 2009 (UTC)

Independence number of 2^n as a graph.
I became curious about this: what is the maximum cardinality of a subset S of the set $$\mathbf{2}^n$$ of all binary strings of length n, such that for all x and y in S one has $$\scriptstyle(x_2,x_3,..,x_n) \neq (y_1,y_2,..,y_{n-1})$$? We may formulate it as a graph problem: if $$G_n$$ is the graph with vertex set $$\mathbf{2}^n$$ and (x,y) is an edge iff $$\scriptstyle(x_2,x_3,..,x_n)=(y_1,y_2,..,y_{n-1})$$, what is the independence number $$\alpha(G_n)$$? I'd be happy even with asymptotics. I made some hand computations for small cases, but of course the complexity grows very fast. I think something should be known though. But what's the name of these graphs? I'd say "shift graphs", but seems that the name has already another use. This has a group picture of the graphs for n=0,..5. --pma 15:37, 25 March 2009 (UTC)
 * Have you tried searching for your small-case results on the OEIS? Algebraist 15:42, 25 March 2009 (UTC)
 * Well, they are too less, just the first 4 or 5, and now I am not even sure about them; but you are definitely right that this is one thing to do, after computing some more --pma 19:17, 25 March 2009 (UTC)
 * Maybe De Bruijn graph is what you want? 12:57, 26 March 2009 (UTC) —Preceding unsigned comment added by 212.112.160.94 (talk) 212.112.160.94 (talk) 16:02, 26 March 2009 (UTC)
 * Wow! and look at this: ..thanks to you both, a very efficient pair of answers --pma (talk) 21:46, 26 March 2009 (UTC)

Set theory paradox
Is the cardinality of the set containing exactly the finite natural numbers finite or infinite? Lucas Brown (talk) 18:35, 25 March 2009 (UTC)
 * Infinite. Where's the paradox? --Trovatore (talk) 18:45, 25 March 2009 (UTC)

...and similarly the set of all countable ordinals is uncountable. I don't think things like that are generally considered paradoxical. Michael Hardy (talk) 02:10, 26 March 2009 (UTC)

What would an infinite natural number be anyway? —130.237.45.207 (talk) 13:53, 26 March 2009 (UTC)
 * I would think Supernatural numbers are the closest thing. —Preceding unsigned comment added by 99.255.228.5 (talk) 14:07, 26 March 2009 (UTC)
 * Non-standard arithmetic provides another reasonable answer. The ordinals and cardinals both generalize the natural numbers, so I suppose they could be said to contain 'infinite natural numbers' as well, though I've not encountered such a usage. Algebraist 14:19, 26 March 2009 (UTC)

The paradox is that they are the finite naturals - so the cardinality is the largest element of the set, which is a finite number. However, there is no limit to the finite naturals - so the cardinality is infinite. So something finite is infinite - paradox! Lucas Brown (talk) 17:20, 27 March 2009 (UTC)
 * But why would we think that the cardinality of a set of numbers is equal to the largest number in the set? {5} does not have cardinality 5. Algebraist 17:43, 27 March 2009 (UTC)

Because this set is the set of all finite natural numbers. Therefore, when the set is ordered so that the ith element is greater than the (i+1)th, the ith element in fact equals i. It should be painfully obvious by now that the cardinality of this particular set is equal to the largest element of the set. Lucas Brown (talk) 19:27, 27 March 2009 (UTC)
 * However there is no largest natural number and thus no largest element of the set of natural numbers. &mdash; Carl (CBM · talk) 03:08, 28 March 2009 (UTC)
 * If I understand, the paradoxical fact referred in the OP is, that an ordered set may not posses a maximum element. Maybe this can be called a paradox, in the etymological sense of something "contrary to the common opinion" --it is at least contrary to the intuition of who is acquainted to treat only finite collections of objects. I don't see how it could be called a set theory paradox, as the language of set theory, already at a rather naive level, is exacty made to describe naturally situations like this. In fact, many mathematical objects that are pathologies and look paradoxical in a context, are a perfectly natural object of study in another one (e.g. the square root of 2, or the Cantor function, or fractals, or the Moebius band &c........) --pma (talk) 18:41, 28 March 2009 (UTC)

Random number generation
You see the function on calculators to produce a random number, how is this generated? Surely there must be a logarithm to work out which number to select, which in itself means the number would not be random. —Cyclonenim (talk · contribs · email) 20:53, 25 March 2009 (UTC)


 * I think you mean algorithm, not logarithm. See the article Pseudo-random number generator. Aenar (talk) 21:02, 25 March 2009 (UTC)


 * But you are correct that "random numbers" generated on a calculator or computer aren't truly random, they are only pseudo-random. To get truly random numbers, one must rely on natural events, such as the decay of radioactive atoms.  It's possible that even these events aren't truly random, but, in any event, they follow a pattern too complex for us to understand, so they are unpredictable (by us), which is what's important in a random number. StuRat (talk) 15:06, 26 March 2009 (UTC)


 * A calculator that offers a (pseudo)random number with no particular quality constraint likely gets it from one of the register structures shown at LFSR.Cuddlyable3 (talk) 20:19, 27 March 2009 (UTC)


 * The usual source of physical randomness in calculators is the crystal or RC oscillator that clocks the calculator's CPU. When you physically push the "random" button, you do that an essentially random number of microseconds after you first power up the calculator.  The internal clock counts those microseconds, giving at least a dozen or so reasonably good random bits.  It can also capture the timestamps of all your other button presses to get a long random seed (see Fortuna (PRNG) for a cryptographic construction of this sort).  The calculator then puts the entropy through some numerical PRNG, but I would usually expect it to be a non-cryptographic PRNG, so I'd avoid using the output for cryptographic purposes, if that's what you were thinking.  75.62.6.87 (talk) 21:36, 28 March 2009 (UTC)


 * Unless perhaps the calculator asks for a button press or two before each random number, so that the user gives it entropy through keystroke timing.