Wikipedia:Reference desk/Archives/Mathematics/2007 April 8
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April 8
[edit]Advantages!
[edit]I am trying to find out in which instances would be advantageous to use a mixed number and in which an improper fraction. 72.40.60.1 00:01, 8 April 2007 (UTC)jc
- If what you care about is the magnitude, and you require more precision than offered by rounding to a whole number, then a mixed number is probably more reasonable than an improper fraction. Imagine asking for a 315/8th inch joist, rather than a 393/8" joist. (Of course, 100 cm is even easier.) In this case mixed notation is also traditional, and using traditional notation, even if irrational, has certain advantages when attempting to communicate with people who are used to traditional notation. In mathematics, the convention is to use vulgar fractions for rational numbers (except of course integers), improper or not. So the fourth convergent of the continued fraction for π will be given as 355/113, not as 316/113. Doing arithmetic is easier with vulgar fractions: try to compute 25/14 ÷ 213/21. --LambiamTalk 01:15, 8 April 2007 (UTC)
Big pi notation
[edit]What is the origin of the "big pi" notation for products? Who introduced it and when? Is it related to Gauss's notation for the gamma function (was one derived from the other)? Fredrik Johansson 00:30, 8 April 2007 (UTC)
- I don't know when the notation was introduced, but I could not find a use by Euler and Leibniz. They still wrote a bunch of multiplied factors (or summed terms for a sum), followed by "etc." (in present-day notation "···"). But the introduction of the integral sign by Leibniz clearly paved the way for a summation sign. Since the Latin letter "S" for Summa had already been taken (in a shape nowadays no longer recognized as a letter), the Greek letter "S", or Sigma, was used instead. From there the step to using the Greek letter "P", or Pi, for Productum, is obvious. The large size gives more horizontal room for denoting the limits of summation (subscript and superscript) and matches the traditionally long shape of the long s. --LambiamTalk 01:50, 8 April 2007 (UTC)
- A list of notational "firsts" can be found here. There seems to be a conflict of claims:
- The product symbol (∏) was introduced by Rene Descartes, according to Gullberg.
- Cajori says this symbol was introduced by Gauss in 1812 (vol. 2, page 78).
- The references in question are
- Cajori, Florian. A History of Mathematical Notations. 2 volumes. Lasalle, Illinois: The Open Court Publishing Co., 1928–1929.
- Gullberg, Jan. Mathematics: From the Birth of Numbers. New York: W. W. Norton & Co., 1997.
- Since Descartes flourished long before Gauss, he would have priority; however, since one would expect numerous intervening uses, Gauss seems more likely. --KSmrqT 03:07, 8 April 2007 (UTC)
- A list of notational "firsts" can be found here. There seems to be a conflict of claims:
- In Gauss 1812, Disquisitiones generales circa seriem infinitam ... §18, the notation Π (k, z) is introduced for a two-argument function, and in §20 the notation Π z is defined as "the limit of Π (k, z) for k = ∞". This is the same as the Gamma function shifted one over. Although the two-argument function could have been defined as a product, it is not; the expression used by Gauss is
- This should, in my opinion, not be considered an introduction of a product symbol. It is not in any way different from the introduction, a few sections later, of Ψ for denoting what we now call the Digamma function. The case is not closed. What we really want to see is a use with a running index.
- I also looked at Euler's use of Σ for summation, said to be the first (Institutiones calculi differentialis, 1755). Interestingly, what Euler defines is an "indefinite sum" or "antidifference" operator, just like the integration symbol without bounds functions works as an indefinite integral or antiderivative operator. When the notation is introduced, Euler writes:
- complete with an "indefinite constant of summation". Again no running index. It is funny to see how the printer, at some point apparently having run out of 's, resorts to using 's rotated by a quarter turn. --LambiamTalk 10:27, 8 April 2007 (UTC)
- In Gauss 1812, Disquisitiones generales circa seriem infinitam ... §18, the notation Π (k, z) is introduced for a two-argument function, and in §20 the notation Π z is defined as "the limit of Π (k, z) for k = ∞". This is the same as the Gamma function shifted one over. Although the two-argument function could have been defined as a product, it is not; the expression used by Gauss is
- For those who would like to examine Gauss for themselves, here is a link. (Euler's works are also online.) --KSmrqT 11:07, 8 April 2007 (UTC)
- Did a bit of digging myself, and found one use of for summation in Fourier's 1822 Théorie analytique de la chaleur and one in Cauchy's 1826 Leçons sur les applications de calcul infinitésimal. Both authors assume the reader is unfamiliar with the notation and provide an explanation. Some thirty years later, Riemann in his 1859 Über die Anzahl der Primzahlen unter einer gegebenen Größe and Dedekind in his 1863 Vorlesungen über Zahlentheorie are both using and without further explanation. This indicates that the summation and product notations passed into general usage sometime between 1820 and 1860. Gandalf61 11:34, 8 April 2007 (UTC)
Carpet cleaning
[edit]I s u p p o s e this counts under "Accounting"...I spilled some Pepsi on white carpet, and Mum's making me pay for cleaning it. How much do you think it would cost to have the carpet cleaned?--the ninth bright shiner talk 02:30, 8 April 2007 (UTC)
- The Misc Desk might be more appropriate. If you give us your city and country we might be able to find a price to clean the carpet in one room in your area, though. (They will likely need to clean the entire room so the cleaned spot matches the rest of the carpet.) The minimum charge is $75 at one company in northern Kentucky, just to give you an idea of the prices you are dealing with: [1]. StuRat 04:00, 8 April 2007 (UTC)
- I'm in Kernersville, North Carolina (27284-7844). Something to consider is that the spot in question is in the room of my grandmother, who is crippled and therefore bedridden, and I doubt we'll be able to move her, or the computer desk in the room. This carpet is relatively new, less than half a year old, so maybe cleaning all of the carpet won't be necessary.--the ninth bright shiner talk 04:40, 8 April 2007 (UTC)
- Why not just get a good spot carpet cleaner? Althought the sooner after the spill you do it, the better / easier it is to remove it, in general. --Wirbelwindヴィルヴェルヴィント (talk) 18:20, 8 April 2007 (UTC)
- I agree. Get some good carpet cleaner and do as good a job as possible yourself. Show it to your mum and hopefully she will accept that. You could also apologize and promise to help her out in some other ways to make it up to her (like washing dishes, etc.). StuRat 04:53, 9 April 2007 (UTC)
- Or try (diluted) white vinegar with a dash of mild detergent; spray, let stand, blot gently; repeat as needed; spray with clean water and blot. Do not wet too much or the carpet may wrinkle. --LambiamTalk 05:55, 9 April 2007 (UTC)
- I agree. Get some good carpet cleaner and do as good a job as possible yourself. Show it to your mum and hopefully she will accept that. You could also apologize and promise to help her out in some other ways to make it up to her (like washing dishes, etc.). StuRat 04:53, 9 April 2007 (UTC)
- What on earth has this got to do with mathematics?
- It is like integration over an area. --LambiamTalk 05:55, 9 April 2007 (UTC)
- Not quite, you don't need to be so accurate. Some polygonal region will suffice. —The preceding unsigned comment was added by 58.163.129.56 (talk) 06:40, 9 April 2007 (UTC).
- Well, it would cost money, and that sort of fits into Accounting. A spot cleaner does sound good though, but I think she's against the idea for some reason. I'll ask about it.--the ninth bright shiner talk 14:58, 10 April 2007 (UTC)
- If carpet cleaning relates to Accounting, then I should be able to ask questions about anatomy on the Computing reference desk, questions about food on the Humanities desk, questions about illnesses on the Language reference desk, etc., etc., etc. —The preceding unsigned comment was added by 129.78.208.4 (talk) 00:28, 11 April 2007 (UTC).
- Well, there's no need to get snippy about it. When I posted the question, I was just confused as to where to put it, and when I saw "Accounting," I thought of "Money."--the ninth bright shiner talk 23:30, 11 April 2007 (UTC)
- By that argument, does that mean that you can put anything on the Mathematics reference desk if it involves money? "Which TV should I buy?" "What's the cheapest place to buy a Coke in the San Francisco Bay Area?" Please. Carpet cleaning has very little to do with mathematics.
- Well, there's no need to get snippy about it. When I posted the question, I was just confused as to where to put it, and when I saw "Accounting," I thought of "Money."--the ninth bright shiner talk 23:30, 11 April 2007 (UTC)
- If carpet cleaning relates to Accounting, then I should be able to ask questions about anatomy on the Computing reference desk, questions about food on the Humanities desk, questions about illnesses on the Language reference desk, etc., etc., etc. —The preceding unsigned comment was added by 129.78.208.4 (talk) 00:28, 11 April 2007 (UTC).
- It is like integration over an area. --LambiamTalk 05:55, 9 April 2007 (UTC)
Factorial Problem
[edit]1! + 2! + 3! + 4! ... 97! + 98! + 99! + 100! is divided by 18. What is the remainder?
How would one go around solving such a problem? What concepts are involved? Are there any online calculators that can brute force it? But more interestingly, how can it be solved without doing so? --Proficient 06:07, 8 April 2007 (UTC)
- This is much easier than you might think. Factor 18 as 3×6. Now recall that 6! is 6×5×4×3×2×1, and every higher factorial is a multiple of this one; what major simplification does this allow? --KSmrqT 06:23, 8 April 2007 (UTC)
- Easiest way I can think of for solving it is that it's equal to 100 * 1 + 99 * 2 + 98 * 3 + 97 * 4 + ... + 2 * 99. But that doesn't seem any easier to put into a calculator. --Wirbelwindヴィルヴェルヴィント (talk) 18:19, 8 April 2007 (UTC)
- Perhaps you wanted to say that it's equal to 1*(1 + 2*(1 + 3*(1 + 4*(1 + ... 98*(1 + 99*(1 + 100))...)))) – b_jonas 19:41, 8 April 2007 (UTC)
- Easiest way I can think of for solving it is that it's equal to 100 * 1 + 99 * 2 + 98 * 3 + 97 * 4 + ... + 2 * 99. But that doesn't seem any easier to put into a calculator. --Wirbelwindヴィルヴェルヴィント (talk) 18:19, 8 April 2007 (UTC)
- KSmrq's point is that anything that has a factor of 18 does not contribute to the problem. You can dramatically reduce what you have to calculate. Root4(one) 18:42, 8 April 2007 (UTC)
- To answer your less interesting question, yes, it can be brute forced, after all 100! is only 158 digits long. 1! + 2! + 3! + ... + 100! =
- 94 269001 683709 979260 859834 124473 539872 070722 613982 672442 938359 305624 678223 479506 023400 294093 599136 466986 609124 347432 647622 826870 038220 556442 336528 920420 940313
- Dividing that by 18, the remainder you get is 9. – b_jonas 19:41, 8 April 2007 (UTC)
- I see now. Thanks for your help. --Proficient 22:44, 8 April 2007 (UTC)
- Great. So to wrap up, instead of summing factorials from 1 to 100, we need only sum from 1 to 5. And using the Horner's rule idea, we can compute this efficiently as
- to get 153. If we like, we can use one last shortcut, because 18 = 9×2: casting out nines gives 0 — so the remainder upon division by 18 is either 0 or 9, and we have an odd number — so 9 is the only possibility. A little simpler than brute force, yes?
- We got lucky with 18; if we had to work with 19 (which is a prime), we would have to sum factorials up to 18. Still, that's much nicer than summing to 100.
- To introduce some higher mathematics into the discussion, we stand on the shoulders of Gauss, taking advantage of a homomorphism from the ring of integers, Z, to the ring of integers modulo n, Zn. (In the given example, n is 18.) And if n has coprime factors p and q, say, then we may use Zp×Zq and the Chinese remainder theorem. In simpler language, the homomorphism tells us that the remainder of a sum is congruent to the sum of the remainders, and likewise for a product. That is why we could replace all terms in the sum from 6! and above by their remainder of zero, and thus ignore them.
- If we had n = 20, we could work with the factors 4 and 5, computing
- to get remainders of 1 and 3, then reconstruct the full answer as described under Chinese remainder theorem. According to the algorithm we could note that (−1)×4+1×5 = 1, so the algorithm would sum 1×5+3×(−4) to get −7, equivalent to 20−7, namely 13. (Of course, for a small number like 20, we can simply compare the list (1,5,9,13,17) with the list (3,8,13,18) — each consistent with its remainder — to find the only common member, 13.)
- As luck would have it, we can end this exploration with a recent news item, involving the "exceptional" Lie group named E8. A team of mathematicians worked for four years to describe the structure of this complicated object, and discovered along the way that they would be dealing with huge quantities of very large numbers. At first it appeared that the final calculations would be too big for their available computer to handle. What saved them was the realization that they could use this very idea of homomorphic images and the Chinese remainder theorem to replace one large computation — which would not fit — with several smaller ones — which would fit. (Specifically, they used 251, 253, 255, and 256, with product 4 145 475 840.) To give a sense of scale: "The final answer is contained in a pair of binary files of respective sizes 14 gigabytes and 60 gigabytes." The story of this accomplishment is inspirational. --KSmrqT 11:00, 9 April 2007 (UTC)
- Great. So to wrap up, instead of summing factorials from 1 to 100, we need only sum from 1 to 5. And using the Horner's rule idea, we can compute this efficiently as
significant or nonsignificant
[edit]The study examined moderators of the relationship b/w trait anxiety and information received by patients postmyocardial infraction. When information received was regressed on the product of trait anxiety and gender, the beta was -.10 (p=.42) and the R squared change was .009 F= .66 and p= .42. —The preceding unsigned comment was added by Rtuscan (talk • contribs) 23:28, 8 April 2007 (UTC).
- Besides for not knowing what you're talking about, I have no idea what you're asking. --Wirbelwindヴィルヴェルヴィント (talk) 23:43, 8 April 2007 (UTC)
- Maybe the question is: what does this mean? It should be postmyocardial infarction, not infraction, but even then we too are lost when it comes to explaining what it means to multiply anxiety by gender. --LambiamTalk 02:27, 9 April 2007 (UTC)
- The variable "F" above presumably stands for the value of a test statistic that has an F-distribution. Here is an online table of upper critical values. --LambiamTalk 02:55, 9 April 2007 (UTC)
- Can your heart be arrested for commiting a serious infraction ? :-) StuRat 04:48, 9 April 2007 (UTC)