Wikipedia:Reference desk/Archives/Mathematics/2012 April 17

= April 17 =

is a single line parallel?
is a single line parallel? I would think, "yes", since at any point it is at a fixed distance of 0. — Preceding unsigned comment added by 80.99.254.208 (talk) 10:50, 17 April 2012 (UTC)


 * Being parallel is not a property that lines posses, but something that holds between lines; so a single, all alone, is not parallel. Or do you mean to ask if it is parallel to itself? Then, essentially, yes; though, it doesn't really give you any new information, so outside of a bigger context, it doesn't really illuminate anything. Was there some specific thing that led you to this, or just generally curious? Phoenixia1177 (talk) 11:06, 17 April 2012 (UTC)
 * Good answer. Another perspective: we could say that all coincident lines are parallel to each other, e.g. if L1 is y_1=mx_1+b, and L2 is y_2=mx_2+b, then they are parallel :) SemanticMantis (talk) 13:16, 17 April 2012 (UTC)
 * That's not correct, I think. When you describe lines with their equations you presumably use THE SAME coordinate system, so there's no need to distinguish some x1 coordinate from some (different?) x2 coordinate, or y1 from y2. However it makes sense to distinguish parameters m and b associated with respective lines. So we would say: lines y = m1×x + b1 and y=m2×x + b2 are parallel (i.e. they are parallel to one another) iff m1 = m2, consequently line y = m×x + b is parallel to itself. --CiaPan (talk) 15:13, 17 April 2012 (UTC)
 * I guess I was inadvertently slipping into a statistics notation (where the different subscripts indicate the same coordinates, but for different samples), basically just to avoid writing the exact same expression twice. The point being that, in some contexts, the "same line" could be arrived at through two different methods, or have two different names, and it would be appropriate to consider the two parallel. SemanticMantis (talk) 19:43, 17 April 2012 (UTC)
 * Wouldn't "identical" be a better adjective? --  ♬  Jack of Oz  ♬ [your turn]  19:59, 23 April 2012 (UTC)

Inequalities (2)
Hi all. I am working with ordered triples (a,b,c), on which I can perform the operations S : (a,b,c) → (c,-b,a) if ca, where we select the ± as appropriate so that |b±2a|<|b| (for those in the know, I am calculating the reduced binary quadratic form equivalent to (a,b,c) but that knowledge is unimportant for my question). We iterate these operations until we finally have (r,s,t) satisfying either 0≤s≤r=t or -ra so we wish to apply T±. Application of T+ produces (-150, -419, 1073) and application of T- produces (-150, 181, 1342), both of which have increased a+|b|. However, this is not currently in reduced form as a is negative. I am sure I have made some simple error but it completely escapes me at the moment. Could someone please have a look and see if they can work out the issue? Thanks. meromorphic  [talk to me]  16:26, 17 April 2012 (UTC)


 * I added "(2)" to the title, to distinguish this from your previous question with the same name. StuRat (talk) 16:46, 17 April 2012 (UTC)


 * Thanks. meromorphic   [talk to me]  18:22, 17 April 2012 (UTC)


 * Never mind, I've resolved the problem. For anyone interested, the BQF I was working with is indefinite, hence the problem. meromorphic   [talk to me]  19:21, 17 April 2012 (UTC)


 * OK, glad you figured it out. I will mark this Q resolved. StuRat (talk) 19:25, 17 April 2012 (UTC)

This Integration problem is killing me!
Alright, this problem is killing me. Basically, I need to find two different areas under the curve $$f(x)=e^x$$ One of the areas is 3 times the area of the other. So..

$$A = \int_0^a e^x dx $$

$$B = \int_0^b e^x dx $$

$$3A = B$$

Express b in terms of a.

I've tried different methods but they all leave me scratching my head!

Method 1: $$3A = B$$

$$B = 3 ( \int_0^a e^x dx) $$

$$\int_0^b e^x dx = 3 (\int_0^a e^x dx) $$

$$e^b - e^0 = 3(e^a - e^0) $$

$$e^b - 1 = 3e^a - 3 $$

$$e^b = 3e^a - 2 $$

$$b = \ln(3e^a - 2)$$

Method 2:

$$ 3A = B $$

$$ B - A = 2A = \int_a^b e^x dx $$

$$ 2A = e^b - e^a $$

$$ 2(e^a - e^0) = e^b - e^a $$

$$ 2(e^a - 1) = e^b - e^a $$

$$ 3(e^a) - 2 = e^b $$

$$ \ln(3e^a - 2) = b $$

Both of my methods converge at the same place. Yet, is this the right answer? It seems too inelegant of a relation between the two terms. I've been trying to find examples for it. For example, I know that:

$$\int_0^{\ln 2} e^x dx = 1 $$ and that

$$\int_0^{\ln 4} e^x dx = 3 $$

But if I have $$ a = \ln 2 $$, I end up with:

$$ \ln(3e^{\ln 2} - 2) $$, which simplifies to $$ \ln 4 $$. So it's true for one example. But how can I prove that it holds true for all a? Or have I already done that? Help!--99.179.20.157 (talk) 16:56, 17 April 2012 (UTC)
 * Presumably you know the evaluation part of the FTC, and that ex is its own antiderivative? That combined with the solution you found gives you what you want. Rschwieb (talk) 17:48, 17 April 2012 (UTC)


 * Work out A and B, set B = 3A and then solve for b. You know that A = ea – 1 while B = eb – 1. You're also told that B = 3A and so eb – 1 = 3ea – 3. It follows that b = ln(3ea – 2). — Fly by Night  (  talk  )  14:53, 18 April 2012 (UTC)

Why is the autocorrelation vector of an Nx1 vector (2N-1)x1 long?
Shouldn't it give me a vector of Nx1? I'm a biology student trying to use autocorrelation to extract signals from fruit fly tracks, and I'm trying to plot autocorrelation versus time, given some signal like average speed over time. 137.54.36.97 (talk) 23:18, 17 April 2012 (UTC)
 * ... But that's what autocorrelation produces. It's a result of the definition of the algorithm that computes autocorrelation for a discrete signal.  You can read our autocorrelation article, which outlines the math, and includes information about autocorrelation of sampled signals.  As you may know, the output of autocorrelation of a real signal has a special property: it is symmetric, which means that (just about) half the information is redundant.  As an algorithm designer, you can choose to selectively discard either half of the output, knowing that it provides no additional information.  That will get you a "Nx1" signal.  Nimur (talk) 23:51, 17 April 2012 (UTC)


 * I'm sorry, but the premise of the question is not correct. The autocorrelation of a scalar time series is (2N-1)x1, where N is the number of offset time bins used for the calculation.  The autocorrelation of a time-varying vector yields a matrix.  Also it is not possible to see the autocorrelation change over time -- the autocorrelation is itself summed over time. Looie496 (talk) 00:09, 18 April 2012 (UTC)
 * Well, unless you repeat the autocorrelation on frames of data, as is done in, e.g., SONAR and RADAR signal processing. This allows you to evaluate slow changes to repeated signals.  For example, in a simple RADAR, you can autocorrelate the signal chirp at the pulse repetition rate; so you're autocorrelating a time-varying signal over several different, longer time-intervals.  After re-reading our original question, I think the OP may be misunderstanding and/or mis-using autocorrelation.  Nimur (talk) 02:19, 18 April 2012 (UTC)

Seventeen or Bust project
Does anyone know anything about the status of the Seventeen or Bust project? Their website has been down for several days. I am a member of the project, but now my CPUs are starting to run out of work and can't get more from the website. Bubba73 You talkin' to me? 23:53, 17 April 2012 (UTC)


 * It is back up. Bubba73 You talkin' to me? 01:31, 18 April 2012 (UTC)