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Wikipedia:Reference desk/Archives/Mathematics/2010 August 29

Contents 1 August 30 1.1 Graph question 1.2 Magic Numbers 2 September 1 2.1 Rotation matrix: counterclockwise vs. clockwise 2.2 Convert procedure to formula? 2.3 Another number theory problem 2.4 Name of a curve 3 September 2 3.1 find scalar values a, b and c 3.2 Numerical analysis notation 3.3 Drawing 45-45-90 triangles on spheres 4 September 3 4.1 Standard deviation 4.2 Random variables 4.3 Formula images 4.4 Homogeneous polynomials 5 September 4

August 30

Graph question

What is the function for a graph that starts at x=a, y=k, which begins to rise slowly then much more steeply until nearly perpendicular to the x axis, then cuts off at x', and resumes in a mirror image at x'', descending until it reaches y=k again? (Apologies for using non-standard notation, I'm afraid I don't know any better. (The shape I'm envisioning is sort of like the elevation of a cooling tower at a power plant.) I suppose it's a Hyperbola, but I looked at that article and it's too much information for me to wade through.--82.113.121.52 (talk) 18:39, 30 August 2010 (UTC)

Just cutting off at suitable (that is, defining it only when ) would do something like that — choose a and b to get the right amount of "slowly/nearly vertical" and the right amount of separation between the two branches. --Tardis (talk) 21:43, 30 August 2010 (UTC)

Taking seems to be a nice choice. Cooling tower says that they are hyperboloid, so the cross section is a hyperbola: in this case, it's , which becomes completely vertical at . Replace x and y with and to scale/shift to taste. --Tardis (talk) 21:56, 30 August 2010 (UTC)

Dear Dr. Who, thank you! Reason I was asking was, I read an article about a speculative theory that attempts to explain why gravity is so much weaker a force than, say, electromagnetism. According to the theory, if our universe is envisioned as a two-dimensional sheet lying flat then other universes are stacked on both sides like pancakes and gravity dissipates by becoming spread out among universes. I was trying to picture in my mind how universes could interact gravity-wise without them slamming into each other and hit on the idea of a graph that rises from nothingness asymptotically and, after an interval, reverses back to nothing. The graph is made by one function so the two halves are in fact a single entity but there is a break in between them. Hmm. On second thought, maybe part 1 and part 2 should switch places. A few more, if I may: (1) Why are so few of us mathematically talented? (2) I am still heartbroken about Rose Tyler, she should still be your companion! (3) Do you ever tire of stupid questions from earthlings?--82.113.106.31 (talk) 01:34, 31 August 2010 (UTC)

I'm not he, of course, but the Psychic paper tells me that he read this message in 2045:

1. That's what's so brilliant! You can't become a talented ballet dancer or a talented werewolf, but anyone can learn mathematics because it's all written down!
2. Aren't we all — but think of the other you over there to whose world she was added.
3. Apparently not: I keep coming back for more.
4. ...Four things: just "the Doctor."

--Tardis (talk) 02:44, 1 September 2010 (UTC)

Magic Numbers

Can someone explain why there are "magic numbers", and particularly why they work.

I was thinking of the number: 142857. When multiplied by 7 and multipiles thereof produce unusual results, or indeed of any non-multipiles-of-seven.

When multiplied by 55 and then 7, look at the result! Why? MacOfJesus (talk) 19:21, 30 August 2010 (UTC)

There's an article on this at Cyclic number. 85.226.205.150 (talk) 20:02, 30 August 2010 (UTC)

1/7 has the decimal value 0.142857142857142..., so 142857 is just below 1000000/7. Multiplying by 7 gives a whole number just below 1000000. Similarly, 1/13 is 0.076923076..., so 76923 shows the same result when multiplied by 13.→86.132.161.214 (talk) 20:13, 30 August 2010 (UTC)

Thank you. MacOfJesus (talk) 21:11, 30 August 2010 (UTC)

The Cyclic number article mentions a number of restrictions:

This restriction also excludes such trivial cases as:

1. repeated digits, i.e.: 555
2. repeated cyclic numbers, i.e.: 142857142857
3. single digits preceded by zeros, i.e.: "005"

I understand the first two, but how is #3 problematic? 005 * 2 = 010, so it doesn't seem to be cyclic, trivially or not. -- ToE^T 00:14, 31 August 2010 (UTC)

Is it because "Zero" "0" is used by us so differently? "Zero" can mean "Zilch" or that there is no value in that space, such as the difference between "ten" and "one", a zero!? MacOfJesus (talk) 11:37, 31 August 2010 (UTC)

I don't buy that. The article explicitly addresses the way leading zeros are handled. -- ToE^T 13:05, 31 August 2010 (UTC)

Anything like "00005" is trivially cyclic because the cyclic permutations are "50000", "05000", "00500", "00050", and "00005". All of these are multiples of 5. (The fact that 005 * 2 = 10 doesn't mean it's non-cyclic. Cyclic means the permutations are all multiples, not that all multiples are permutations.) Staecker (talk) 13:24, 31 August 2010 (UTC)

But according to the article, cyclic numbers specifically deal with "successive multiples". -- ToE^T 13:30, 31 August 2010 (UTC)

FWIW, the three restrictions were in place when the article burst forth fully formed from the prolific keyboard of 198.99.123.63. -- ToE^T 13:38, 31 August 2010 (UTC)

Ah OK- I missed the bit about "successive". Reading more carefully now... In the bit you quoted above, "This restriction" refers to the "successive multiples" condition. So those three categories are examples of numbers which would trivially be cyclic were it not for the "successive multiples" condition. So "0005" trivially satisfies all properties of cyclic numbers except for the "successive multiples" condition, as you said. I'll try to clarify this a bit in the article. Staecker (talk) 11:57, 1 September 2010 (UTC)

I too noted the "This restriction" part, and was going to address that after figuring out #3. I disagree with your interpretation and your edit. The "consecutive multiples" condition dose not lead to restrictions #1 and #2 -- they still need to be explicitly excluded as trivial cases. I suspect that you were close to the mark when you mentioned that #3 would need to be exclude as trivial cases of numbers whose permutations are (non-consecutive) multiples. Perhaps one of the sources used that definition of cyclic number. Anyone here have a copy of TPDoCaIN to see if they define it differently than the Wolfram page does? -- ToE^T 00:54, 2 September 2010 (UTC)

Well, repeated digits (#1) do give multiples when you make the permutations: "555" has permutations "555" and "555", each of which is 555 * 1; so, were it not for the "successive" condition, this would be cyclic. For (#2), the example "142857142857" does indeed give multiples when you make permutations, but they are not successive (e.g. 428571428571 = 142857142857 * 4). So this one too would be cyclic were it not for the "successive" multiples. Staecker (talk) 11:22, 3 September 2010 (UTC)

Staecker meant to type 142857142857 * 3 in the example above. -- ToE^T 16:36, 3 September 2010 (UTC)

With #2:

142857142857 × 1 = 142857142857

142857142857 × 2 = 285714285714

142857142857 × 3 = 428571428571

142857142857 × 4 = 571428571428

142857142857 × 5 = 714285714285

142857142857 × 6 = 857142857142

which are successive multiples (and are all six distinct cyclic permutations that exist for that twelve digit number). With #1, 555 only has one distinct cyclic permutation and the single multiple 1 is trivially successive. I'm not trying to pick nits here, but this is why such trivial cases are explicitly excluded. -- ToE^T 16:36, 3 September 2010 (UTC)

Thanks to all. MacOfJesus (talk) 17:11, 1 September 2010 (UTC)

Wikipedia:Reference desk/Archives/Mathematics/2010 August 31

September 1

Rotation matrix: counterclockwise vs. clockwise

Hello, I am trying to teach myself algebraic matrices. I seem to understand them, and am getting right answers, except in one particular case. For the life of me, I can't understand why the matrix for a 90-degree counterclockwise rotation is what it is.

We are told (for example at Rotation matrix) that the 90-degree counterclockwise rotation matrix is

but to me that looks like a clockwise rotation. Consider the illustration at Matrix (mathematics)#Interpretation_as_a_parallelogram, which represents the transform of the unit square by the matrix

If we plug in the numbers from the above counterclockwise rotation matrix, then the (a,b) vertex — originating at (1,0) as a corner of the unit square — is transformed to (0, –1). No? Meanwhile the (c,d) vertex, originating at (0,1), is transformed to (1,0). To me that produces a clockwise rotation: The unit square has been rotated such that it is now below the x-axis but still to the right of the y-axis. What gives?

I get the same result when I multiply the row vectors of the unit square [0 0], [1 0], [1 1], [0 1] (also from Matrix (mathematics)#Interpretation_as_a_parallelogram) by the 90-degree counterclockwise rotation matrix above. This seems to produce the vertices [0 0], [0 –1], [1 –1], [1 0] — which again describes a clockwise rotation below the x-axis.

What am I misunderstanding about this interpretation of the rotation matrix? It is particularly vexing since I've had success with every basic type of transform, except for rotations greater or less than 180 degrees. Help is greatly appreciated. -Jordgette (talk) 00:16, 1 September 2010 (UTC)

My guess is that the mistake is in your Matrix multiplication. Using your matrix A, . Perhaps you were calculating ? The convention is to use column vectors and (therefore) multiply with the matrix on the left and the vector on the right. Given A and v, , so reversing the multiplication (and transposing v to make the multiplication work then) is equivalent to transposing A, which in your case makes it equal to -A, so of course it generates the negative of the vector that you want (which is equivalent to a further rotation by 180° and changes 90° to -90°). --Tardis (talk) 02:13, 1 September 2010 (UTC)

I was multiplying a row vector on the left and the matrix on the right. That would be my mistake. Thank you for taking the time. -Jordgette (talk) 07:04, 1 September 2010 (UTC)

Convert procedure to formula?

How do I convert something like: if(x>50) y=(30+x), else y=(x/2), to an equation of the form y = f(x)?

Is this a standard 'area' of mathematics, if so what is it called, and where can I go online to learn it?

I have also seen precedures that involve for/while loops that I know should be representable as a single mathemtical function and shouldn't need interation. But I have no clue how to build a function from the procedure. Can anyone point me in the right direction? Thanks.--Dacium (talk) 03:27, 1 September 2010 (UTC)

• To answer the first part of your question, it looks like what you want is a Piecewise function... the value of x (i.e., where it is in the domain) determines which piece of the function you're using to evaluate the dependent variable:
  .
  Essentially it's saying the same thing that you've expressed as an "if-else" or another programming function that involves selecting cases based on the input. --Kinu ^t/_c 03:50, 1 September 2010 (UTC)

Selecting cases is avoided by using Iverson brackets: y=[x>50]·(30+x)+[x≤50]·(x/2). For-loops are avoided by using J. Bo Jacoby (talk) 05:58, 1 September 2010 (UTC).

You could also use Recursion to avoid for loops.

For some for loops it may be posible to replace the loop by a simple formula without recursion. For instance sum =0; for i=1 to n: sum=sum+i; is an Arithmetic progression and could be replace by the simple function . Similar results could be obtained for other loops but I'm not aware of a general procedure, the article Series (mathematics) might be of help.--Salix (talk): 08:45, 1 September 2010 (UTC)

Corresponding J expressions may be written n=.10 +/1+i.n NB. brute force summation without for loop 55 n*(n+1)%2 NB. computing the sum 55 Salix alba ment to write . Bo Jacoby (talk) 10:08, 1 September 2010 (UTC)

Another number theory problem

I have two problems:

• If p is an odd prime, show that every prime divisor of must be of the form 2pk+1 for some natural number k.
• If are the Fibonacci numbers then show that Euclid's algorithm takes n steps to determine gcd. I have been racking my brain on this one (the first one I have no idea about), using induction etc but to no avail. -Shahab (talk) 13:05, 1 September 2010 (UTC)

The second one's pretty trivial: the Euclidean algorithm constructs a sequence of numbers, whose first two elements are f_n+2 and f_n+1. What is the third element?—Emil J. 13:11, 1 September 2010 (UTC)

As for the first one: you want to show that , or in other words, . You can work mod 2 and mod p separately, as p is odd. The rest is Fermat's little theorem.—Emil J. 13:19, 1 September 2010 (UTC)

Thanks Emil for part 2. But for the 1st part I dont understand as to how you concluded that I wished to show or equivalently for some k. Could you be a little more explicit. Thanks-Shahab (talk) 14:05, 1 September 2010 (UTC)

Sorry, I misread the question. So, we have a prime divisor q | 2^p − 1, and we want to show that . It's clear mod 2, so we only need it mod p. Now, consider the multiplicative order k of 2 modulo q. The assumption gives k | p, and Fermat's little theorem gives k | q − 1. Conclude p | q − 1.—Emil J. 14:16, 1 September 2010 (UTC)

Thanks. I have another problem, (sorry for posting so many, I am appearing for an exam after a long long time). If gcd(a,b,c)lcm[a,b,c]=abc then show that (a,b)=(b,c)=(c,a)=1.-Shahab (talk) 21:33, 1 September 2010 (UTC)

Hi Shahab. For any prime let the nonnegative integers be the exponents of the greatest powers of that divide respectively . Then , and from this you have to deduce that at most one out of is positive (meaning that divides at most one out of ). You may think w.l.o.g. that of course.--pma 00:38, 2 September 2010 (UTC)

Thanks pma. I hope you are doing well.

Name of a curve

Say A is an arbitrary smooth curve, and P is some point on that curve. I draw a line normal to A through P, and I mark off point Q on the line such that PQ is some constant distance. I then trace the locus of Q as P moves along A, thus forming a new curve B. Is there a name for the relationship between curves A and B? I vaguely had in mind "evolute", but looking that up I see it means something completely different.—Preceding unsigned comment added by 86.135.28.150 (talk) 13:59, 1 September 2010 (UTC)

B is more or less a Parallel curve of A.—Emil J. 14:07, 1 September 2010 (UTC)

Yes, it is indeed a parallel curve. I have added this ref (the first) to Parallel curve. DVdm (talk) 14:21, 1 September 2010 (UTC)

Thanks (duh). 86.184.27.6 (talk) 17:14, 1 September 2010 (UTC)

September 2

find scalar values a, b and c

Given , find scalar values a, b, c (NOT ALL ZERO) for which
I get these 4 simultaneous equations in 3 unknowns:




What do I do now? Wikinv (talk) 01:41, 2 September 2010 (UTC)

Check your initial work. I believe one of the four equations is wrong. Next, take a look at Simultaneous equations for various techniques for solving these. One approach is to rearrange one of the equations to isolate one of the unknowns (i.e., ) and then substitute the results into the next equation. After the second iteration, you should have a value for one of the unknowns. Repeat the process until you've solved them all. -- Tom N (tcncv) talk/contrib 02:09, 2 September 2010 (UTC)

After attempting to solve it myself, I found that there's a gotcha in those equations. The problem has a solution, but that solution may not be unique. I assume this is homework, so I will not give you too obvious a hint. Write back if you need additional help. -- Tom N (tcncv) talk/contrib 02:33, 2 September 2010 (UTC)

Fixed the equation. I was aware that there are multiple solutions, indeed that is why I don't know how to solve it.--Wikinv (talk) 07:05, 2 September 2010 (UTC)

Although, having fixed the equation, the solution becomes quite easy by inspection, but is there an analytic way of solving it?--Wikinv (talk) 07:09, 2 September 2010 (UTC)

Just realised that the second and third equations are in fact exactly the same. Furthermore, it is evident that there is an infinite number of solutions (that's why it was so easy to solve by inspection!)--Wikinv (talk) 07:14, 2 September 2010 (UTC)

Observe that if (a,b,c) is a nonzero solution to the equation where is any square matrix, and k is a nonzero number, then (ka,kb,kc) is a nonzero solution too. Bo Jacoby (talk) 05:15, 2 September 2010 (UTC).

In general, the equation has a one-dimensional space of solutions (a, b, c) for any 2x2 matrix . To see this, let

Then

but

so

so

or (as Bo says) any multiple of this. Gandalf61 (talk) 08:54, 2 September 2010 (UTC)

Alternatively, just take the coefficients of the Characteristic polynomial. —Preceding unsigned comment added by 203.97.79.114 (talk) 09:11, 2 September 2010 (UTC)

See Gaussian elimination. Properly used, it works also for systems of linear equations that have no solutions, multiple solutions, andor redundant equations. I don't know if our articles cover the details involved, but if not, any introductory linear algebra book will. -- Meni Rosenfeld (talk) 09:17, 2 September 2010 (UTC)

Numerical analysis notation

I was reviewing my numerical analysis textbook, and stumbled across a notation I didn't remember, concerning propagation of errors. For background, the question is, assuming that is a differentiable function, and that is an approximation of , what is an upper bound on the uncertainty ? Now, the mean value theorem directly gives that there is a between and such that . We know that , but we do not know an upper bound on .

Now, what the book does here is that it replaces the unknown with the known . The idea, presumably, is that since is small, then so is . The book tacitly makes this assumption, mind you, without requiring a single fact about —not even that it be continuous! Of course, it is not then necessarily true that , and the book does not claim so, but replaces the sign with a sign, telling the reader to read it as "less than or approximately equal to"—I would say it seems reasonable to interpret this as "unequality that almost certainly holds unless the function is too irregular". It feels kind of sloppy though. Is it just my book or is this common notation? Is this a case of "since we're dealing with applications, we don't have to care about badly-behaving functions"? 85.226.206.114 (talk) 07:57, 2 September 2010 (UTC)

Trying to find an estimate for gets you back to the original problem but with derivative of the function. So this estimate will be in terms of and if you try it estimate this it will be in terms of etc. But each time you're multiplying the estimate by which is assumed to be small. So I think the interpretation of is "Less than equal up to a quantity that is small compared to the other quantities." You can construct examples where the derivatives get very large or infinite so that the error term is still large even with the factors, but these aren't encountered often in practice. The notation is a bit vague perhaps but if you want more precise notation try a different numerical analysis book, there is no shortage of them.--RDBury (talk) 15:05, 2 September 2010 (UTC)

All right, thanks! 85.226.205.5 (talk) 08:42, 3 September 2010 (UTC)

Drawing 45-45-90 triangles on spheres

Can you draw a 45-45-90 triangle on a sphere so all three sides have whole-number lengths? 20.137.18.50 (talk) 16:31, 2 September 2010 (UTC)

You cannot draw a 45–45–90 triangle on a sphere at all. The sum of angles of any triangle on a sphere is strictly more than 180°, see Spherical geometry.—Emil J. 16:36, 2 September 2010 (UTC)

September 3

Standard deviation

Hi all! In physics we're doing a bit of stats and I noticed in the standard deviation formula they divide by N-1 rather than just N. I asked my teacher and he said he didn't get it either, and look it up on Wikipedia or something like that, so here I am. I tried looking at your articles Standard deviation and Bessel's correction, but that didn't really help because I don't have a university-level stats background :/ Can someone who does explain why you divide by N-1, in simpler terms? I'm OK with (and even expect you to) dumb the concept down a little --cc —Preceding unsigned comment added by 76.229.208.208 (talk) 01:58, 3 September 2010 (UTC)

As I understand it, the N-1 come in because you are trying to estimate the actual standard deviation based on sample data. If you put N in the denominator it turns out that the estimate will, on average, be too low. So a correction factor is built into the formula so that the estimate will average to the actual value if the experiment is repeated many times. When the correction factor is added it works out the same as using N-1 in the denominator instead of N. It has been noted here before though, if your sample is small enough that it actually makes a difference then your sample size is too small.--RDBury (talk) 03:46, 3 September 2010 (UTC)

See the Wikipedia article on unbiased estimator, which has the explanation you're looking for. --173.49.14.153 (talk) 04:20, 3 September 2010 (UTC)

If you knew the population (actual) mean rather than estimating it and used that to get the squared differences then N would be correct. However using the sample (estimated) mean makes the sum of the squared differences slightly smaller. In fact the sum of the squared differences from the population mean is equal to the sum of the squares of the differences from the sample mean plus N times the square of the difference between the population mean and the sample mean. This itself gives you an estimate of the probable difference between the population and sample mean so the workings out in the article is just using this to get an estimate of the sum of squared differences from the population mean. A finickety point is that it is only the expression without the square root that is unbiased, the estimated standard deviation from taking the square root is biased but I would worry even less about that than using N instead of N-1 in the denominator. Dmcq (talk) 07:57, 3 September 2010 (UTC)

Random variables

Hello mathematicians! Can you please help me solve this. It's not homework, it's actually work work. Say is the amount of money I make per "event" and is the number of events per year. Let's also say that has a lognormal distribution and is a poisson distribution (the parameters for can be estimated from some data and let's assume that the parameter for is known).

A) Then the total money I make from these events in one year is . Is there an analytic distribution function for ?

B) Will the following monte-carlo methods work to determine a distribution for :

1) sample a random value from , say , then sample values of and add them up - repeat this many times; or

2) sample a random value from , say , and sample a random value of , say , and then use - and repeat this many times.

What is the difference between these two methods? What other possible numerical methods can I use to determine ? Thanks very much. --Mudupie (talk) 17:32, 3 September 2010 (UTC)

Formula images

In every maths page on wikipedia I notice the formulae are images not text. How do you create these? On Mac? Thanks for any replies.86.147.12.111 (talk) 18:05, 3 September 2010 (UTC)

See Help:Displaying a formula. —Bkell (talk) 18:27, 3 September 2010 (UTC)

Thank you86.147.12.111 (talk) 19:42, 3 September 2010 (UTC)

Homogeneous polynomials

The symmetric degree 4 homogeneous polynomial in two variables: x⁴ + x³y + x²y² + xy³ + y⁴ can be written (x⁵−y⁵)(x−y)⁻¹ for x≠y. What is the analogous expression for the symmetric degree 4 homogeneous polynomial in 3 variables: x⁴ + x³y + x³z + x²y² + x²yz + x²z² + xy³ + xy²z + xyz² + xz³ + y⁴ + y³z + y²z² + yz³ + z⁴ ? Bo Jacoby (talk) 22:28, 3 September 2010 (UTC).

First, just to be consistent with the terminology, these are called the Complete homogeneous symmetric polynomials. The expression you're looking for follows from the properties of Schur polynomials.

which turns out to be the complete symmetric polynomial. Here Δ is the product of the differences (x−y)(x−z)(y−z).--RDBury (talk) 04:33, 4 September 2010 (UTC)

Thank you very much! Bo Jacoby (talk) 06:10, 4 September 2010 (UTC).

September 4