Wikipedia:Reference desk/Archives/Mathematics/2015 November 24

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November 24

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Path of steepest descent?

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I'm wondering how, given a differentiable function of two spatial variables, how to compute the steepest paths of descent on its graph. All I know is that such a path must have the gradient as a tangent vector at every point, and thus should be itself a smooth curve. I also know that that means that the x and y components of the derivative of the function (of one parameter) describing the path should be parallel (as a vector) to the gradient. I'd much rather not have to use the calculus of variations.--Jasper Deng (talk) 00:16, 24 November 2015 (UTC)[reply]

I think what you're asking for are essentially orthogonal trajectories. --RDBury (talk) 00:51, 24 November 2015 (UTC)[reply]

Maths fields

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Discrete mathematics is opposed to continuous mathematics, which is basically numerical analysis, which is opposed to symbolic manipulations. How is the relation between the discrete math, and symbolic manipulations? Is there something like an internal opposition in discrete math - symbolic discrete methods vs. numeric discrete methods? Or, is discrete math just symbolic manipulation (or a subset of this latter)?--3dcaddy (talk) 15:12, 24 November 2015 (UTC)[reply]

Er... why should continuous mathematics be identified with numerical analysis? Numerical analysis usually means approximating continuous problems by discrete ones. —Kusma (t·c) 15:23, 24 November 2015 (UTC)[reply]
I have changed the redirect to go to mathematical analysis instead. —Kusma (t·c) 15:30, 24 November 2015 (UTC)[reply]
Yes, that's right. Anyway, the first part of the question is still more confusing than explanatory. So, I am leaving the bottom part and striking it.--3dcaddy (talk) 16:13, 24 November 2015 (UTC)[reply]
I think you may be referring to something like the discrete element method which, despite the name, would fall under numerical analysis and therefore continuous mathematics and not discrete mathematics. Typically, the term discrete mathematics refers to finite (countable sometimes) structures while analysis or continuous mathematics refers to structures involving real numbers in some way. But there are areas that don't fall into either camp and our Areas of mathematics divisions cut across this line. Mathematics Subject Classification has main divisions for discrete math/algebra and analysis, but it also has divisions for foundations, geometry, and applied mathematics. But Lie groups, classified in with other areas of group theory under discrete math/algebra, might more properly be put under analysis. And while most of geometry would fall under continuous mathematics, there is a whole area of discrete geometry that wouldn't. The boundaries between different areas of mathematics are usually fuzzy and often based more on tradition than anything else, so it's difficult to give a precise meaning to the distinction. --RDBury (talk) 18:07, 24 November 2015 (UTC)[reply]
Previous answer is about mathematics aspect of the question. From the computing point of view, there are three main classes of mathematical computation: numerical computation (sometimes improperly called numerical analysis or even scientific computation), which essentially deals with real numbers represented as floating point numbers; symbolic computation, also called computer algebra for which the objects of computation are essentially mathematical expressions; and discrete mathematics computations that deals essentially with finite sets and finite structures. These three classes of methods, although using very different algorithmic methods, are not completely independent. For example, computational geometry makes heavily use of the three kinds of methods.
In your question, you use "symbolic discrete method" and "numeric discrete methods". Personally I find these terms confusing and useless: All computer science may be qualified as "discrete", as, inside computers, everything is expressed in terms of the finite set with two elements. Also, you are wrong when searching oppositions between methods. Many of the best recents achievements were obtained by importing methods of one of the above classes of computation for improving solutions given by other classes. D.Lazard (talk) 13:08, 25 November 2015 (UTC)[reply]

Why isn't complex numbers taught as vectors equivalent?

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Teaching complex number to school kids should not be hard. Complex number addition is just equivalent to vector addition. Complex number multiplication is just treating one complex number as a vector and using the other complex number as a procedure to perform scaling and follow by an angle rotation. This can be done in the argand diagram. When intepreted in the geometrical format there is nothing special about I (imaginary number) which is nothing but a rotation of 90 degrees counter-clockwise in the argand diagram.

It is not just teachers in high school. Look at the wikipedia article on complex numbers, it does not mention that addition of two complex numbers has similar properties to vector addition nor does it mention that multiple of two complex number to be consider a form of scaling and rotation of a complex number (about the origin) in the complex plane. The smart student may figure this out all by themselves but there is no reason not to mention this to all readers of the wikipedia article.

This is nothing but a global elitist conspiracy to prevent the understanding of the true nature of complex numbers from the dumb masses. Instead of allowing them to understand complex numbers, the students are forced to memorize complex procedures (pun not intended) and endure deliberate obfuscation. Where as what the students need to know is what the complex numbers are and then the procedures for dealing with them which can be programmed into a smart phone similar to calculators which can perform addition , subtraction , multiplication and division. I can't even remember when I last did complex multiplication and complex division by hand. My point is that the conspiracy to prevent the understanding of complex numbers is prevents literally billions of people from using complex numbers in their daily lives. — Preceding unsigned comment added by 175.45.116.59 (talk) 23:33, 24 November 2015 (UTC) 175.45.116.59 (talk) 23:02, 24 November 2015 (UTC)[reply]

Note: The redirect Argand diagram -> Complex plane is tagged {{R with possibilities}}, and while "Argand diagrams" is bolded in the lede, they are not defined there and are not mentioned in the article's body. -- ToE 23:28, 24 November 2015 (UTC)[reply]

As part of the global elitist conspiracy why would people pay us anything but peanuts if they understood how easy it all is ;-) Actually if you do look at the complex number article you'll see vector addition in the diagram beside the description of addition. The corresponding diagram showing multiplication is equivalent to adding the angles and scaling is in the section about polar representation. That was a bad oversight leaving that in, ah well too late I'd have people complaining if I removed it. Dmcq (talk) 00:02, 25 November 2015 (UTC)[reply]
But aren't complex numbers already part of the standard school curriculum? Sławomir
Biały
00:31, 25 November 2015 (UTC)[reply]


Please do elaborate on what use billions of people would find for complex numbers in their daily lives. That would have given me a more convincing answer when I used to teach College Algebra and the students wanted to know why they needed to learn about complex numbers. The actual answer I gave, "you should learn about complex numbers because they're really cool", is one I stand by, but I'm not sure they totally bought it. And maybe they shouldn't have, because my mental reservation was, "... they're really cool, but if you take College Algebra and then stop, you're never actually going to find out why they're cool". --Trovatore (talk) 00:44, 25 November 2015 (UTC) [reply]
I think the IP's point, the incivilities aside, is that most people are not taught that complex numbers can also be thought of as vectors in R2. However, that is quite an advanced treatment for high school and below... I'm not even sure if the connection really can be done without going further than vectors. In particular, division of complex numbers does not have a straightforward vector counterpart. (I'm also about 1/3-sure that the IP is also just trolling.)--Jasper Deng (talk) 03:23, 25 November 2015 (UTC)[reply]
Division in complex number is easy. If A times B equal C. And this means, a vector A is scaled and rotated (via B) to obtain the result C. So division just means, given the resultant vector C and the "scaling and rotation" B, find the original vector A. 175.45.116.59 (talk) 03:35, 25 November 2015 (UTC)[reply]
This isn't a result that is readily apparent to most high school students, for it requires the notion of the argument of a complex number. The closest high school usually gets to that is de Moivre's formula (even then, only calculus classes delve into its derivation from Maclaurin series), hence why I don't consider it straightforward from the standpoint of a high school student. I also think that the concept of a bivector and geometric algebra is useful when talking this, since the imaginary unit has a very straightforward representation as a bivector.
Whether this should be taught, I'm not so sure. You're assuming that high school students intrinsically want to use advanced mathematics in their everyday life. You must remember that complex numbers are introduced as roots of otherwise irreducible polynomials, and that only a small subset go on to the point where complex numbers become useful in this geometric sense. I also would highly caution you against making highly general statements such as a "global conspiracy". I know plenty of highly rigorous high schools which have honors classes dealing with this (for students who are truly interested).--Jasper Deng (talk) 04:31, 25 November 2015 (UTC)[reply]

Choose a zero point, 0, and a unit point, 1. Similar triangles have proportional sides. If triangle (0,1,a) is equiangular to triangle (0,b,x), then x/b=a/1, or x=a·b. That's all. Bo Jacoby (talk) 12:22, 25 November 2015 (UTC).[reply]

  • I do not think teaching complex numbers as vectors will make them simpler to understand. On contrary it over-complicate everything. In addition, the vector interpretation of them is not practically useful. Ruslik_Zero 13:45, 26 November 2015 (UTC)[reply]