Wikipedia:Reference desk/Archives/Mathematics/2006 December 27

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December 27

hardcore problem

I came across this while on the net today. If it's not a trick question (which would be annoying), it has to have a little maths involved. I have no idea where to start on this.

"FICTIONAL:

The Ipswich Massacre occurred in January 1967. A similar massacre occurred the day before where the 13 survivors called themselves "The Lucky Thirteen." The survivors of the Ipswich Massacre decided to copy this trend and call themselves their own number with "The Lucky" at the start. In Ipswich, on that day, the survivors watched outside through a window into a hall with a capacity of 100 people. The hall was supposed to be packed with 96 hostages but by the time some of the hostages were shoved in, the medical examiner and 3 guards realized they were short of a few. They were short of the survivors, of course, who were waiting outside.

The guards searched outside, but could not find the missing hostages. So they gave in and began to unceremoniously shoot the hostages that were inside the hall. The medical examiner wrote in his diary that night: "about 11% of them were being popped off every minute, and 6 minutes into the shootings, about 43 of them were dead."

Give the two possible names for the group of survivors."

does anyone have a clue?

I can't say the puzzle appeals to me; too many words for too little mathematics, and a horrific reminder of the worst in human behavior. The first paragraph merely gives an upper bound on the number of survivors (which is what we are to find), namely no more than 96 (and presumably far fewer). The second paragraph suggests that 66% (about 2/3) of the number of unlucky hostages is 43, which should allow us to calculate the full number. If we subtract the unluckies from 96, we may determine the lucky number. I'm guessing the correct name reminds us of The Lucky 13. --KSmrq^T 06:52, 27 December 2006 (UTC)

It's also unclear whether the 11% is always 11% of the total, or only 11% of those alive at the time. Can't this be rewritten in some less violent fashion, like kids eating donuts ? StuRat 12:48, 27 December 2006 (UTC)

The use of "about" as in "about 43" suggests that it could actually also be 42 or 44, which however lead to other answers. Same for "about 11%"; and if you combine the two, you get a rather large uncertainty. There is further an apparent but unstated and unrealistic assumption that the rate of doughnut consumption (shouldn't that be broccoli? doughnuts are high carb) is and will remain completely steady. All together the story is not only disgusting but also contrived, and as a puzzle unconvincing and utterly boring.  --Lambiam^Talk 13:48, 27 December 2006 (UTC)

An infinitely-long rod

Well, I will soon if I can find a bespoke tailor. But at any rate... Say I had a rod, a metre long, of perfectly circular cross-section. I then find a rod two metres long, and attach it to the end of my first rod, leaving a perfect two-dimensional seam showing where the first rod ends and the second one begins. Then I find a four-metre long rod and stick it on the end, and soon enough I have an infinitely-long rod with seams starting a metre from the end and doubling in distance between themselves each seam. So far, so good. I now get a second set of rods and do exactly the same thing. But I then get a half-metre long rod and stick it to the front end, then a quarter-metre long rod, an eighth, a sixteenth, etc. I know that there are differing sizes of infinity, but I'm unclear on the maths, which is why I'm asking this. As far as I know the second rod is longer than the first rod. Now I get a third set of rods and make a construction exactly the same as the first rod. I then paint over the seams, leaving the one where the first rod joined the second, and from there halve the distance from the first seam to the front end and draw a perfect 2D line around it, I then halve the distance from the second line to the front end and draw a line there, and so on. I then go back to my first seam and find the place that is, from the first seam, double the distance than it is from the front end to the first seam and draw a line around that. Quadruple the distance, and so on. It is now seemingly exactly the same as the second rod. But how long is it? Vitriol 01:09, 27 December 2006 (UTC)

My mind hurts from thinking about your rod. 202.168.50.40 02:29, 27 December 2006 (UTC)

See, logically it'd be the same length as the first rod, but it would seem to be the same as the second now. Vitriol 02:34, 27 December 2006 (UTC)

All three of them are infinitely long. Formally, we can express this as saying that the length of a rod is greater than every real number. There is no meaningful way to compare infinite quantities like these in our standard number system, so it is not meaningful to discuss which of the rods is longer than any other. Maelin (Talk | Contribs) 03:48, 27 December 2006 (UTC)

All of these have the same length, so far as I can tell. I don't think you can theorize a physical object with a greater or lesser infinite length, but it is possible to create infinitely large sets of different "size". The easiest place to start in wikipedia is probably Cantor's theorem, but there are more general (but more terminology-dependant) articles like Aleph number or Cardinality. - Rainwarrior 05:37, 27 December 2006 (UTC)

You might want to read Hilbert's paradox of the Grand Hotel, an amusing discrete infinity paradox of which the simpler part is analogous to yours. Using interval notation, let (a, b) stand for all real numbers between a and b. Is the length of, say, (3, 8) the same as that of (7, 12)? Yes: the difference between them is just a matter of shifting up or down the real number line by a distance of 4. But then the length of (0, ∞) should be the same as that of (1, ∞), even though the first interval is a proper subset of the first.  --Lambiam^Talk 09:17, 27 December 2006 (UTC)

Guess what - your rod is actually -1 metres long ! Why ? Because if you take the 1 metre rod away your first rod is now 2 metres long, your second rod is now 4 metres long etc. so each rod is twice as long as the rod in the corresponding position in the original arrangement. So if the total length of the rods was originally L metres, it is now 2L metres. So we have L-1=2L, and so L = -1 metres. Of course this is absurd - it just shows that you cannot treat "infinite lengths" (or, more properly, "unbounded limits") as if they were finite quantities. It was this sort of "paradoxical" behaviour that led 19th century mathematicians such as Cauchy and Bolzano to introduce the modern concepts of limit (mathematics) and convergence, and put mathematical analysis on a rigorous foundation. Gandalf61 11:29, 27 December 2006 (UTC)

Not all the rods would have an infinite length, and those which did would approach infinity at a different rate. One that doesn't have an infinite length, and instead converges on a finite length, is the 1 + 1/2 + 1/4 + 1/8 ... series. The quickest way to determine the limit here is to use a chart:

$\sum _{n=0}^{\infty }{\frac {1}{2^{n}}}$

 n  value      sum    decimal sum
==  ======  ========= ===========
 0   1          1       1.000
 1  1/2        3/2      1.500
 2  1/4        7/4      1.750
 3  1/8       15/8      1.875
 4  1/16      31/16     1.934
 5  1/32      63/32     1.969
 6  1/64     127/64     1.984
 7  1/128    255/128    1.992
 8  1/256    511/256    1.996
 9  1/512   1023/512    1.998
10  1/1024  2047/1024   1.999

This makes it pretty clear that we are approaching 2. Of course, this isn't a proof, more formal methods are needed for that. StuRat 12:33, 27 December 2006 (UTC)

You have misread the question. The second rod is constructed by first making a rod like the first, and then adding to it the 1/2, 1/4, 1/8 and so on. So it is also infinite.

About Gandalf's comment, 1+2+4+8+... is indeed not equal to -1 according to the usual definition of infinite sums of real numbers, but that's not the same as saying that it is absurd (it is correct, for example, in 2-adic numbers, and it is correct in reals if a more liberal definition of infinite sums is used). Similarly, there is at least one good reason to have 1+2+3+4+5+... = -1/12. -- Meni Rosenfeld (talk) 15:53, 27 December 2006 (UTC)

triangles

Kindly give me a hint to solve this problem:

The distance between two parallel lines is 1. A point 'A' lies between the lines at distance 'a' from one of them.Find the length of a side of an equilateral triangle ABC,the vertex B of which lies on one of the parallel lines and the vertex C on the other.

Thank you.

First, draw a diagram, keeping in mind that the interior angles of an equilateral triangle are each 60 degrees:

-------------------*----------------
  |    |       ? * *
  |   1-a      q   *
  |    |     *     *
  1   ---- * 60    r
  |    |     *     *
  |    a       p   *
  |    |  THETA  * *
-------------------*----------------

Then, determine the angles of intersection with the sides of the triangle and the parallel lines. Can you find the angle marked with the question mark as a function of angle THETA ? Next, can you find the lengths of p, q, and r as a function of THETA and a ? After that, you only have to set those expressions equal to each other and solve. Show us some work and let us know where you get stuck. StuRat 12:25, 27 December 2006 (UTC)

Giving hints is an art; a good hint depends on insight into the problem, but also into the problem solver. We see the problem, but do not know what background you may bring, so our hints must be crude.

A few things we can say with ease. If a is ¹⁄₂, then the side length is 1, for BC is perpendicular to both the B and C lines. If a is 0, then the side length is that of an equilateral triangle with height 1, namely ²⁄_√3 (approximately 1.1547), for AB lies on the B line. And 1−a should give the same side length as a. We now have some sanity checks for any proposed solution.

As for how to find a solution, the fact that the problem is stated in terms of a rather than an angle suggests that we look for an end result that is algebraic, not trigonometric. However, we may find it convenient to use trigonometric reasoning. Let r be the desired side length. Referring to StuRat's elegant ASCII diagram, we observe that ^a⁄_r is cosine of θ. And since the angle inside each corner of the triangle is 60°, we also know that ¹⁄_r is cosine of θ+60°. Using a few trigonometric identities we can get an expression for r in terms of c = cos θ = ^a⁄_r. The final solution will involve a square root. --KSmrq^T 21:51, 27 December 2006 (UTC)

Assign transport cost based on weight and volume

I am trying to spread the cost of transport across a given load (cost of transport per box), and am trying to find out if there is a standard formula for this calculation.

I have come across the articles:

Freight rate http://en.wikipedia.org/wiki/Freight_rate

Dimensional weight http://en.wikipedia.org/wiki/Dimensional_weight

But was not able to find what I need. Here are a few examples:

Lets say that it costs 100€ to send a package from A to B

The load consists of:

A) 1 pallet of 20 boxes, therefore it cost:

100€ / 20 boxes = 5€ per box to move from A to B

B) 1 pallet (1 m³/100 Kg) containing: 10 boxes item X with a volume of 0,4 m³ and weight of 40 Kg

20 boxes item Y with a volume of 0,6 m³ and weight of 60 Kg

The cost of this movement that needs to be allocated to each item would be:

Total	Item X	Item Y
Weight	40% (40 kg)	60% (60 kg)
Volume	40% (0,4m3)	60% (0,6m3)
Total cost (100€)	40€	60€

But what happens when the numbers are not so easy/obvious? How do we contemplate the fact that a trucks volume may be 40 m³ but due to weight restrictions we could only fill half? (and vice versa).

I am guessing the final formula must include the following:

Total weight available
Total volume available
Total weight used
Total volume used
Total weight specific item
Total volume specific item

Any help would be greatly appreciated.

One way to approach this is the following. Determine the cost Cost(X) if you only ship the X items, Cost(Y) for only shipping the Y items, and Cost(X+Y) for shipping both. Presumably Cost(X+Y) is less than Cost(X)+Cost(Y); if not, just ship separately. Otherwise, assign the respective shares in the combined cost as follows:

Share(X) = Cost(X)/(Cost(X)+Cost(Y)) × Cost(X+Y)

Share(Y) = Cost(Y)/(Cost(X)+Cost(Y)) × Cost(X+Y)

Then Share(X)+Share(Y) = Cost(X+Y). This method passes several tests of being fair and reasonable. For example, if the cost is proportional to volume, the allocation is the cost of the contribution of each to the volume. Likewise for weight. In the case that Cost(X+Y) = Cost(X)+Cost(Y), you simply get back Cost(X) and Cost(Y). If Cost(Y) = 0, all is allocated to X. It can easily be generalized to more kinds of items.

A disadvantage is that you have to compute transport cost three times. In general that is not a simple formula; it works for example with weight classes or bulk discounts.  --Lambiam^Talk 14:49, 27 December 2006 (UTC)

Don't you know that any problem involving translating boxes is NP complete and there's no way to solve it? Don't even try. :) – b_jonas 19:25, 27 December 2006 (UTC)

I did not know that that also applied to calculating the fare. The USPS Postage Rate Calculator works rather quickly also for packages, but perhaps they use a supercomputer.  --Lambiam^Talk 20:06, 27 December 2006 (UTC)

What's the difference between analysis and calculus?

Both articles don't seem to solve the problem. Moreover, to add even more confusion, some articles such as vector calculus says it can also be called vector analysis. Thanks.

As far as I am aware, calculus is just a name for the elementary parts of analysis. -- Meni Rosenfeld (talk) 16:38, 27 December 2006 (UTC)

They're the same, really. – b_jonas 18:57, 27 December 2006 (UTC)

Many topics in mathematics are called "Calculus of XYZ" or "XYZ calculus". There are such things as Propositional calculus, Predicate calculus, Lambda calculus, Superposition calculus, Combinator calculus, Proof calculus, Process calculus, Relational calculus, Umbral calculus, Duration calculus, Region Connection Calculus, Situation calculus, Mueller calculus, Refinement Calculus, Schubert calculus, Kirby calculus, and many more. The general idea of a calculus is that you have a formal language of terms (essentially a term algebra), plus rules to rewrite terms (often as in equational logic) using mainly formal criteria (that is, based on the composition of the term and not on its meaning). Such rules are also called "calculation rules". The point is that these terms also have a meaningful interpretation, and that the calculation rules stand for mathematically meaningful operations. One such set of rules is given by the "Integral and Differential Calculus". Because of its dominant position in mathematics, that calculus is often simply referred to as The Calculus.

The area in which The Calculus operates and where the meaning of the interpreted terms belongs, is Analysis. In theory you can do Analysis without Calculus, finding limits of difference quotients the hard way, but you'd be completely handicapped. It is worse than trying to do division using Roman numerals and Egyptian fractions instead of Algorism – by itself also a calculus. Most calculi (for example SKI combinator calculus) have nothing to do with Analysis.  --Lambiam^Talk 19:25, 27 December 2006 (UTC)

The word "calculus" can be traced back to the Ancient Greek "χάλιξ" (khalix), a pebble used for calculations. (The same word is used in internal medicine for the deposits of a kidney stone, and in dentistry for tartar deposits.) It came to English through Latin, which was used in the time of Newton and Leibniz as the language for scholarly publications. As Lambiam illustrates, it is a popular term in many areas of mathematics.

For a few centuries the rules for derivatives and integrals had no solid formal basis, though they (mostly) worked well. Eventually mathematics evolved to the point where more rigor was needed, for several reasons. Often "analysis" distinguishes the more modern rigorous study from the older traditions. Even more specifically, it often refers to complex analysis, a rich area unknown to early calculus. For example, when we talk about a holomorphic function, we are in the domain of analysis, not calculus. Likewise, when we talk about Lebesgue measure, even though we may use it for integration, we are speaking analysis.

A student majoring in mechanical engineering would probably take a course in calculus; one majoring in mathematics would (eventually) take a course in analysis. --KSmrq^T 23:40, 27 December 2006 (UTC)

KSmrq, do you have any references for the relevance of "χάλιξ"? I've only found one source that derives 'calx'/'calculus' from it, (an Anglo-Saxon dictionary), though I admit that if the words are related it is probably a loan, since Greek 'χ' doesn't generally correspond to Latin 'c'. But that's in the sense of 'limestone': what I'm asking is whether there is any evidence that the Greek word was used for numbering. --ColinFine 00:47, 30 December 2006 (UTC)

I'm surprised you didn't find it, since this seems to be the standard etymology. If you have access to the Oxford English Dictionary, generally considered a definitive source for word origins, try that. (I believe most universities make an online version available for their students.) Otherwise, I recommend The American Heritage Dictionary of the English Language, in its online version. Usually it gives a history with the word, but in this case we must chase back through several words, from calculus to calculate to calx, where we eventually find

• Middle English, from Latin, lime, limestone, pebble, from Greek khalix, pebble.

If we know how to transliterate to the Ancient Greek spelling, we're done; otherwise we may consult the Perseus Project and its authoritative LSJ (Liddle–Scott–Jones) lexicon, where we find χάλιξ (and variations) uses in classic texts of antiquity. For example, we find this word used in Thucydides, The Peloponnesian War book 1, chapter 93, section 5, which in English translation reads (in part)

• Between the walls thus formed there was neither rubble nor mortar, …

Here "rubble" is the same word as "pebble". From this source we see no hint of the connection with mathematics, which seems to be provided most clearly by the Latin use of the word calculus, found at the definition for calculate:

• Late Latin calculāre, calculāt-, from Latin calculus, small stone used in reckoning, diminutive of calx, calc-, small stone for gaming; see calx.

In other words, the Greek refers to pebbles in the role of rubble, but the Latin specializes to calculating. That's a long answer to a short question! Hope it helps. --KSmrq^T 11:07, 30 December 2006 (UTC)

Alec Stewart

(question moved to Wikipedia:Reference_desk/Miscellaneous#Alec Stewart --hydnjo talk 21:42, 27 December 2006 (UTC))