# Wikipedia:Reference desk/Archives/Mathematics/2007 November 26

Mathematics desk
< November 25 << Oct | November | Dec >> November 27 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

# November 26

## Multiplication through drawing lines and finding intersections?

I've just seen a YouTube video of doing multiplication (allegedly Chinese, though as one myself I've never seen it) through drawing lines according the number and finding intersections. What is it actually called and is there a mathematical proof on this method? For 2 digit numbers if the original number are AB and CD (each representing a digit), the result seems to be (A×C | A×D + B×C | B×D) , so how does this continues on to more digits? -antilivedT | C | G 06:10, 26 November 2007 (UTC)

It is a nice video! The value of the two-digit number AB is 10·A+B. So the proof is that (10·A+B)·(10·C+D) = 100·A·C+10·(A·D+B·C)+B·D. Bo Jacoby (talk) 07:17, 26 November 2007 (UTC).

The extension to more digits is simple - ABC * DEF = | AD | AE+BD | AF+BE+CD | BF+CE | CF, and so on. This "method" is identical to long multiplication, only more cumbersome - compare
  123
*
321
-----
123
246
369
-----
39483

The digits you see here, obtained by multiplying corresponding digits, are exactly the numbers of intersections in the video. -- Meni Rosenfeld (talk) 16:34, 26 November 2007 (UTC)

The method of the video only uses counting, while the long multiplication requires you to know the little multiplication table. So it is arguable which method is the more cumbersome. Bo Jacoby (talk) 18:03, 26 November 2007 (UTC).

Long multiplication is arugably more advanced than counting intersections, as familiarity with the multiplication table is required. But there is no question as to which is more cumbersome, in the sense of amount of effort required. -- Meni Rosenfeld (talk) 18:26, 26 November 2007 (UTC)
Hmm actually thinking about it make it seem so simple. Thanks guys. --antilivedT | C | G 00:30, 27 November 2007 (UTC)

## LaTeX (again)

What is the escape character for a square bracket? Things like [T]_B sometimes give errors but sometimes no - but I would prefer it if it was fully escaped. 16:02, 26 November 2007 (UTC)

Can you give an example where square brackets give an error? I am not aware of any alternative. -- Meni Rosenfeld (talk) 16:24, 26 November 2007 (UTC)
It throws up on the following:
   \documentclass[a4paper]{report}

\begin{document}

$\begin{array}{rl} [S]_B & =\left(\begin{array}{ccc} 0&1&0\\ 0&0&2\\ 0&0&0 \end{array}\right)\\ [T]_B & =\left(\begin{array}{ccc} 1&1&1\\ 0&-1&-2\\ 0&0&1 \end{array}\right)\\ [ST]_B & =[S]_B[T]_B\\ & =\left(\begin{array}{ccc} 0&1&0\\ 0&0&2\\ 0&0&0 \end{array}\right) \left(\begin{array}{ccc} 1&1&1\\ 0&-1&-2\\ 0&0&1 \end{array}\right)\\ & =\left(\begin{array}{ccc} 0&-1&-2\\ 0&0&2\\ 0&0&0 \end{array}\right) \end{array}$

\end{document}

with the error:
 /<blah>/test.tex:12: Missing number, treated as zero.
T
l.12    [T]
_B &

16:54, 26 November 2007 (UTC)
Putting an extra blank line between the arrays solved the problem for me. A lame fix, but give it a try. –King Bee (τγ) 17:09, 26 November 2007 (UTC)
It works, but does anyone know why? $S$_B doesn't work, for example, it breaks with a different message. 17:25, 26 November 2007 (UTC)
I don't know for sure, but I would guess that it's because LaTeX is expecting an optional argument to one of its commands, and this particular optional argument should be a number. Optional arguments to LaTeX commands come in square brackets, so when it reads the opening bracket it expects an argument, not just more math. Try putting an empty set of curly braces {} in front of it so LaTeX will realize you are not trying to give an argument. Or just use plain TeX. ;-) —Bkell (talk) 18:42, 26 November 2007 (UTC)
Yes, it's the fact that LaTeX is looking for an optional argument there. Braces around the bracketed expression are the best option as {}[B] will cause spacing to be slightly off. Donald Hosek (talk) 19:26, 26 November 2007 (UTC)
Using {[T]}_B (and similarly for the others with []'s) works. kfgauss (talk) 19:00, 26 November 2007 (UTC)

## Most efficient path on a sphere

What is the shortest route between a large number of nodes arranged approximately equally on a sphere (eg, each of the vertices in this image or points on a geodesic)? I've narrowed the options down to 2 which seem most reasonable; a spiral pattern from one pole of the sphere to the other and a polar pattern which goes in circles and crosses each pole many times. Which would be shortest? Also, is there a formula for finding the length of a spiral pattern on a sphere (with each turn equidistant from the next) if I know the radius of the sphere and the number of turns (in other words, if I flew in a satellite from the north pole to the south pole as the Earth rotates underneath me, what length would the path appear to be from an Earth-bound observer)? I've read the articles on Rhumb line and Spherical spiral, but neither was any help. Laïka 20:51, 26 November 2007 (UTC)

I think a very similar question has been asked (quite some time) before, and I'll try to dig it up later if no-one bears me to it. This is a version of the Travelling salesman problem, which is notoriously difficult to solve if you want the really shortest route; so you'll have to settle for something less ambitious: a reasonably short route. One (fairly complicated) thing you could do is use a space-filling curve on the sphere. For example, you can deform a cube with a Hilbert curve on each of the six faces to a sphere, or do the same with an icosahedron with a triangle-filling curve on each face. The curves can be glued together to form a single sphere-filling curve (even a loop). The curve defines a continuous surjective mapping C from, say, the interval [0, 1) to the points on the surface of the sphere. Using an inverse of C, you can assign a value in the range [0, 1) to each node; sort them according to that value and visit them in that order.  --Lambiam 21:27, 26 November 2007 (UTC)
To answer the further questions: a spiral pattern is almost certainly better than the polar overpasses if you do the spiral right – there are many kinds of spirals, some of which have infinite length. You'd want a spherical analogon of something resembling an Archimedean spiral. Very roughly, if the windings of the spiral are a distance h apart, and it has a length L, it covers an area of about Lh – with a bit of overlap of the swaths near the poles. If it covers the whole sphere with an area of 4πr2, where r is the radius of the sphere, we find that L is approximately 4πr2/h. But for the route through given nodes in a spiral pattern, you'll have to deviate from the smooth spiral to reach the actual positions of the nodes. While I don't know the optimal value of h, it should be of the same order of magnitude as the average distance between a node and its nearest neighbour.  --Lambiam 21:38, 26 November 2007 (UTC)
Well, if the cities are equally spaced, this is a very specific instance of the travelling salesman problem which need not be NP-complete, I think, which would mean that finding the shortest route may be tractable. What can you tell us about the layout of the nodes? Phaunt (talk) 23:20, 26 November 2007 (UTC)
The wireframe vertices problem is easy. An optimal route using the (longitude, latitude) notation on a 5 segment sphere is as follows. Start at the south pole (0,-2). Move to the north pole in a straight line (0,-1), (0,0), (0,1), (0,2). Move south one ring adjacent to the point you came from (1,1). Move in a circle around the ring stopping just before you reach a visited point (2,1), (3,1), (4,1). Move south one ring (4,0). Move in a reverse circle along that ring again stopping before you reach a visited point (3,0), (2,0), (1,0). Continue forward-backward pattern until you reach the south pole again.
Now why is this optimal? As wee look for a complete tour through all vertices, it does not matter at all where we start. We chose the south pole. As the north pole must be visited at least one full cirlce of vertical edges must be on the path. Each point on the sphere has to be visited once and for each point the adjacent horizontal edges are not longer than the vertical edges. Our path visits each node only once and uses the minimum possible amount of vertical edges. So it is optimal.
I just invented that proof right now, and its 2:00 in the morning here, so if it's garbage; sorry. ^^ This is off course if you want to move *along* the wireframe. Otherwise it might still be optimal, but I am not sure. —Preceding unsigned comment added by 84.187.110.92 (talk) 01:10, 27 November 2007 (UTC)

## Limits

Hi, im currently in my first year of a Mathematics degree and at the moment i'm studying limits. I've been given an exercise sheet to do but i am having trouble with one of the questions.

Consider the series: Σ(n≥1) (n+1)¼ - n¼

By using a suitable comparison test investigate whether S converges or diverges.

It looks like a telescoping question but i have to use a comparison test. I'm not too fond of the comparison tests, I've tried a few tests but i keep finding when the test series is bigger than this, the test series diverges and converges whens its smaller telling me nothing.

I understand the answer can't just be posted on here but could somebody give me a pointer in the right direction, thanks.

212.140.139.225 (talk) 21:28, 26 November 2007 (UTC)Pete

Use the fact that a4−b4 = (a−b)(a+b)(a2+b2), with a = (n+1)1/4 and b = n1/4.  --Lambiam 21:42, 26 November 2007 (UTC)
Doesn't this just show lim (n+1)¼ - n¼ = 0, i know that if it doesn't converge to 0 then the series diverges but the series doesn't necessarily coverge if it does tend to 0. Thats what i think is the case, please correct me if i'm wrong, i don't fully understand the topic. Also the question wants me to use a comparison test. Thanks for your quick response. 212.140.139.225 (talk) 21:59, 26 November 2007 (UTC)
First, you have noted correctly that this is a telescoping series; I understand you are required not to use this fact, but just so you know where you are going, what does it tell you - does the series converge or diverge?
Now, try writing this formula as ${\displaystyle a-b={\frac {a^{4}-b^{4}}{(a+b)(a^{2}+b^{2})}}}$. What is ${\displaystyle a-b}$? What is ${\displaystyle a^{4}-b^{4}}$? How large can ${\displaystyle (a+b)(a^{2}+b^{2})}$ be? What does this tell you about the terms of your series? Can you show that the terms are larger\smaller than some known series which diverges\converges? -- Meni Rosenfeld (talk) 22:32, 26 November 2007 (UTC)
Ok this makes sense to me now. This is the 2nd part of a question and the first part just asks to show that the individual term of the series converges to 1 which i have done using the equation you've both posted. So i assumed you were both describing the first part. Am i right in saying what you were talking about is comparing (n+1)¼ - n¼ but in another form ie using the equation above to another converging or diverging series. Thans again both of you. 212.140.139.225 (talk) 22:57, 26 November 2007 (UTC)
The terms converge to 0, not to 1. And yes, you need to compare the terms of the series with those of a series which you know whether it converges. -- Meni Rosenfeld (talk) 23:01, 26 November 2007 (UTC)
It's not often very helpful, but try writing out the first few terms of this particular series and simplifying. --Tardis (talk) 23:24, 26 November 2007 (UTC)
I read several times that you needed to use a comparison test, and I still forgot. However, my suggestion may help you see which way to try comparing if you don't already know! --Tardis (talk) 23:26, 26 November 2007 (UTC)
I think we've convered that in the "telescoping series" part. -- Meni Rosenfeld (talk) 23:57, 26 November 2007 (UTC)

You can try the integral test and the integral test will tell you that this series diverges (because the integral itself diverges).130.166.172.130 (talk) 01:50, 27 November 2007 (UTC)

Thanks for all of the replys. I've got the question in the form ${\displaystyle {\frac {1}{(n+1)^{3/4}+n^{1/2}(n+1)^{1/4}+n^{1/4}(n+1)^{1/2}+n^{3/4}}}}$ . From here I've said that each component in the denominator is < n for large n so the whole denomintor is less than 4n. So the above equation is bigger than 1/4n. So comparing it to ${\displaystyle \Sigma {\frac {1}{4n}}}$, since this diverges the original equation diverges. Is this valid? I'm not sure about saying each component of the denominator is less than n. It's not true for low values but for large values it is true. Thanks again. 212.140.139.225 (talk) 21:13, 27 November 2007 (UTC)

This is correct, but the "less than n for large n" is indeed something the grader might quibble about, especially since this is an introductory course. Perhaps it would be better to say that each term is less than ${\displaystyle n+1}$ for every ${\displaystyle n\geq 1}$, hence when comparing it to ${\displaystyle \sum {\frac {1}{4n+4}}}$, which clearly diverges (or if it is not considered clear, it can be proven easily), we see that this diverges as well. -- Meni Rosenfeld (talk) 21:38, 27 November 2007 (UTC)
Thanks for your reply, using n+1 instead of n seems a pretty obvious thing to do but i seem to miss these smaller steps for some reason. You say ${\displaystyle \sum {\frac {1}{4n+4}}}$ can be proved to diverge easily, the way i have done it is to say its bigger than ${\displaystyle {\frac {1}{4n}}}$ and since that diverges the other must, is there an easier way i've missed? 212.140.139.225 (talk) 22:22, 27 November 2007 (UTC)
${\displaystyle {\frac {1}{4n+4}}}$ is smaller than ${\displaystyle {\frac {1}{4n}}}$, so this is not correct. How to prove it depends on what theorems you are allowed to use. Since you are summing only for ${\displaystyle n\geq 1}$, you can say that ${\displaystyle {\frac {1}{4n+4}}\geq {\frac {1}{8n}}}$ and use the fact that ${\displaystyle \sum {\frac {1}{8n}}}$ diverges. -- Meni Rosenfeld (talk) 22:39, 27 November 2007 (UTC)
Thanks again, i'm finding it hard to get to grips with the limits section, but i've been told the only way to get better is by doing more examples.

I have another question from a while ago which i haven't found an answer for. Is ${\displaystyle x\in \Re /\left\{\pm 1\right\}=x\in (-\infty ,-1)\cup (-1,1)\cup (1,\infty )}$? 212.140.139.225 (talk) 23:22, 27 November 2007 (UTC)

Yes, however I wouldn't use the equals sign that way; I'd use the biconditional: ${\displaystyle \Leftrightarrow }$. The left-hand side says you have all real numbers except for -1 and 1; the right hand side says you have all numbers smaller than (but not including) -1, all numbers between (but not including) -1 and 1, and all numbers greater than (but not including) 1. Phaunt (talk) 23:45, 27 November 2007 (UTC)
A minor point: you want ${\displaystyle \Re \setminus \{\pm 1\}}$ or (more commonly in my experience) ${\displaystyle \mathbb {R} \setminus \{\pm 1\}}$. / normally means some sort of quotient, while \ is reserved for set difference. Algebraist 00:06, 28 November 2007 (UTC)
An alternative way to say the same thing is to compare the sets directly, in which case the equals sign is appropriate: ${\displaystyle \mathbb {R} \setminus \{\pm 1\}=(-\infty ,-1)\cup (-1,1)\cup (1,\infty )}$. —Bkell (talk) 00:23, 28 November 2007 (UTC)