Wikipedia:Reference desk/Archives/Mathematics/2010 June 18

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June 18

Order of polynomial / Gaussian Numerical Integration

Is sqrt(x^7) + x^2 a polynomial of order 7 or 3.5? I want to know this because I want to find the minimum number of Gauss points I need to exactly integrate it, and minimum number of Gauss points = ceiling ( (order of polynomial +1) /2 )

122.57.166.246 (talk) 10:08, 18 June 2010 (UTC)

Neither. ${\sqrt {x^{7}}}+x^{2}$ is not a polynomial. This expression is not difficult to integrate, but I don't think you can integrate it exactly with Gaussian quadrature. Gandalf61 (talk) 10:24, 18 June 2010 (UTC)

The exam question is to find the minimum number of Gaussian points needed to integrate it exactly...is there a mistake in the question then? I would just like to know if the expression is considered to have an order of 3.5 or 7. Thanks

122.57.166.246 (talk) 10:54, 18 June 2010 (UTC)

Or it's a trick question. -mattbuck (Talk) 11:27, 18 June 2010 (UTC)

Try substituting x=u^2. Then you have to manage u^7+u^4. -- SGBailey (talk) 12:47, 18 June 2010 (UTC)

Good idea. But don't forget the extra factor of 2u from dx = 2udu. Gandalf61 (talk) 13:05, 18 June 2010 (UTC)

Usually the word degree rather than order is used. Michael Hardy (talk) 16:37, 18 June 2010 (UTC)

Getting help from refdesk for an exam question doesn't seem appropriate. 75.57.243.88 (talk) 19:11, 18 June 2010 (UTC)

For what it's worth, it is a previous exam question. Thanks everyone for their help. 122.57.169.173 (talk) 23:40, 18 June 2010 (UTC)

Hahn fields

What is a "Hahn field"? While Googling the phrase to get information about the Hahn Field Archeological District, I keep finding websites that refer to Hahn fields or Hartman-Hahn fields; here is one for you with JSTOR access — I don't have it, so I can't read it. This is another one, but it's too in-depth for me to understand. I've searched; we don't appear to have an article on the topic. Nyttend (talk) 12:59, 18 June 2010 (UTC)

See Hans Hahn and Hahn series. I don't know who Hartman is. -- 58.147.53.127 (talk) 19:10, 19 June 2010 (UTC)

How do you call this number?

If you take four numbers, which are not all equal, i.e. 4277, and reorder them from bigger to smaller - 7742 - and re-order them from smaller to bigger - 2477 - and subtract this second number from the first, and go on with the same algorithm with the resulting number, you'll always obtained a specific number. I forgot the name, but it is the name of an Indian mathematician. Does anyone know what it is called?--Quest09 (talk) 17:39, 18 June 2010 (UTC)

Kaprekar's constant Dmcq (talk) 17:54, 18 June 2010 (UTC)

Why don't you do those calculation yourself? You would easily find that the number is 6174. Then you'd find an appropriate Wikipedia article. --CiaPan (talk) 17:59, 18 June 2010 (UTC)

No biting there CiaPan, it was a legit question :-( hydnjo (talk) 02:33, 19 June 2010 (UTC)

I don't bite. I didn't know the answer either, so just tried it with a simple four-operation calculator, and got the result witin two minutes, with no advanced maths or even any special efforts. I just started with the proposed 4277 and got following results:
7742–2477 = 5265
6552–2556 = 3996
9963–3699 = 6264
6642–2466 = 4176
7641–1467 = 6174
and saw 6174 has same digits as 4176, so it loops back to the same subtraction it came from, so this is the number, which Quest09 asks for. Then I simply entered it into the Wikipedia's search box and got it: '6174 is known as Kaprekar's constant'. It definitely didn't require lots of work. Or did it?
Finding it ourselves is much faster than waiting for the answer, and with the algorithm given anyone is able to do it. One only needs to try.... --CiaPan (talk) 06:58, 21 June 2010 (UTC)

CiaPan, no I wouldn't have reached the number. Since I didn't know how many repetitions were possible. I'd have gone through it certainly, but how could I recognize it?--Quest09 (talk) 10:32, 19 June 2010 (UTC)

You know you have reached it, because at the next repetition it gives itself again. --84.220.119.125 (talk) 12:19, 19 June 2010 (UTC)

How about trying it out and having look instead of giving up before you start? Dmcq (talk) 10:53, 19 June 2010 (UTC)

The Kaprekar's_constant article and the D. R. Kaprekar#Kaprekar constant section both state that this property (all non-trivial orbits leading to a single fixed-point) is unique in the decimal system to 3 and 4 digit numbers, but the reference given only states that this "appears" to be the case "only for three and four digit numbers", and examples are given through ten digits. Has this assumption been proven? -- 58.147.53.127 (talk) 13:50, 19 June 2010 (UTC)

No, this is not unique to 3 or 4 digit numbers in base 10. The Kaprekar mapping has one or more non-zero fixed points for almost any number of digits. The reference you gave has a table (near the end) of fixed points or "kernels" of the mapping up to 10 digits - for example, with 6 digits we have:

$182738\rightarrow 763533\rightarrow 431766\rightarrow 631764\rightarrow 631764\rightarrow \dots$

OEIS Sequence A099009 lists fixed points up to 14 digits long, and also gives patterns for generating 3 different infinite sequences of fixed points with increasing numbers of digits. One of these patterns generates a fixed point for any even number of digits ≥ 4; another generates a fixed point for any odd number of digits ≥ 9; and the third generates a fixed point for any number of digits that is a multiple of 3. So the only digit lengths for which a non-zero fixed point does not exist in base 10 are 2, 5 and 7. The Kaprekar Routine page at MathWorld lists fixed points for other bases as well. Gandalf61 (talk) 08:55, 20 June 2010 (UTC)

Thank you; that is very interesting. My question is, however, about the unreferenced assertion that three and four digit numbers are unique (in base 10) for having a single fixed point to which all non-trivial orbits lead. -- 58.147.53.145 (talk) 12:09, 20 June 2010 (UTC)

Oh, yes, I see. Well, looking at the much longer list of fixed points here, we can see other families of fixed points which provide at least two fixed points for each even length ≥ 8 and each odd length ≥ 17. So the only digit lengths with a single fixed point are 3, 4, 11 and 13 (6, 9 and 15 have a second fixed point because they are multiples of 3). And a quick spot check of 11 and 13 shows that there are longer loops for these digit lengths. So 3 and 4 are indeed the only digit lengths for which the Kaprekar mapping sends all numbers eventually to a single fixed point.

A related question is whether are other digit lengths for which the mapping has only fixed points (even though not unique) and no longer loops. I doubt this, as I expect there are extendible patterns for generating loops for any digit length, in the same way as there are for fixed points - but it would make an interesting investigation. Gandalf61 (talk) 13:38, 20 June 2010 (UTC)

Excellent. Thank you. -- 58.147.53.145 (talk) 06:12, 21 June 2010 (UTC)

Mathematicians by country

Which countries provides the world with the best mathematicians?--Quest09 (talk) 17:41, 18 June 2010 (UTC)

Before anyone attempts to answer this question I'd like to ask Quest09 to define what they mean by "best". Zunaid 18:38, 18 June 2010 (UTC)

And what they mean by "provide". The country which provides the best (for pretty much any definition of best) mathematicians with jobs and money for their research has got to be the US. If the meaning is where the "best" mathematicians are born, for example, now that's a rather different question.—Emil J. 18:48, 18 June 2010 (UTC)

On a quick look at [1] from The MacTutor History of Mathematics archive I see that Scotland has had almost as many as the US and Ireland is getting up there with China and India, and England has the most by far. Dunno why but I get this niggling feeling that it might be just a teensy weensy bit biased ;-) Dmcq (talk) 19:04, 18 June 2010 (UTC)

My experience has always been that the Russians, and more generally former USSR countries turn out the most brilliant mathematicians. The problem is that they have never been recognised in the past because much of their work was secretive and even if it hadn't have been; they wouldn't have been given any credit in the West because of political pressures. I think the Russian education system has a lot to do with it; IMHO it is vastly superior to the American education system (and indeed many of the European systems too). The late, great V. I. Arnold was very critical of the American system (he didn't like the simplification of the curriculum and the lowering of standards required to accommodate ill-prepared students, he made the criticism that they could graduate with a jazz history course, but not necessarily with a history of algebra course ). •• Fly by Night (talk) 10:44, 19 June 2010 (UTC)

I think saying that Soviet mathematicians "have never been recognised in the past" or "given any credit" in the west is a bit of an exaggeration. Izvestiya: Mathematics was translating Soviet mathematics into English from 1937 1967, Sbornik: Mathematics from 1967, Russian Mathematical Surveys from 1971. There were many Soviet mathematicians known by their reputation and their work to western colleagues. Tinfoilcat (talk) 12:31, 22 June 2010 (UTC)