Testwiki:Reference desk/Archives/Mathematics/2012 November 5
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November 5
Teach integration
Hi. I realize that mathematical integration contains dozens of human variants of the How-wise, including the Quadrature methods, Archimedes' quadruplets and likewise. For example, I may need to integrate a piecewise function, and require to explanify the various modulus on f(x) to derive the antiderivative in two, three, n and n-1.5 dimensions. My brain works best in acute visual calculus, and so to predict a subspace or manifold I must first visualize, and then derive, otherwise my dopamine level approaches 0 as t approches 1. Since I will need to use integrative methods for processing the universe within the next three weeks, I would great-infinitely appreciate some explanatory guidance unto this matter. As it stands, I have absolutely no background in integration, and only limited edusophical background in calculus. My understanding of complex erudite logic is far superior to my thought-attention span in numerical and differential mathematics. Some history of integration across the lands of the English, the Arabs and the Chinese would also serve interest. You may recall that I asked a question in Science many years ago on the flow of water into a geo-sedimentary basin of lower height, yet this is likely a common application of integral calculus. I find the quadrature of the Parabola highly interesting, and the proposition of self-same continuity across x as the scale rotates about the x-y plane is also. Please understand the meaning of this post, and give any resources based on my understanding or lack thereof of the subject that which I would highly very like to learn about regarding. Thank you! Verily. ~AH1 (discuss!) 19:54, 5 November 2012 (UTC)
- The phrasing of this question is so clever that I can't figure out what the question is. I get that it has something to do with integration, but what exactly? Looie496 (talk) 22:29, 5 November 2012 (UTC)
- Yes, unfortunately he has integrated so many different fields into his Q that we can't differentiate between them. In the hopes that my answer won't seem derivative, let me offer this calculus: "If you want to be able to visualize integration, consider that 'the area under a curve is the integral of the equation of that curve' ". So, for example, what's the area between the X-axis and curve y=x2, between x = 1 and x = 2 ? The integral is C + x3/3. So, ignoring the constant C (which is the area below the X-axis, in this case), and evaluating between x = 1 and x = 2, we get 23/3 - 13/3 = 8/3 - 1/3 = 7/3. Plot the curve out (on graph paper, if you must), and see if that isn't correct. StuRat (talk) 03:01, 6 November 2012 (UTC)
- You may recall that I asked a question in Science many years ago on the flow ..... Er, in a word, no. -- Jack of Oz [Talk] 03:03, 6 November 2012 (UTC)
- I thought you guys had infinite memory. At that time I wanted to know the equation for calculating how quickly water fills a depression of lower height, yet now I know there are many complex non-linear effects such as Darcy's laws that make an approximation difficult - I had assumed that the forward surface speed at the entry point was equal to the speed of sound in air. Now, I want to know whether this might be a possible application of integration, and how much integrative knowledge I'd need to know in order to derive an equation for it. ~AH1 (discuss!) 18:47, 6 November 2012 (UTC)
The Ramanujan master theorem. Suppose f(x) is analytic in a neighborhood of zero and . Then we have:
where is defined as follows. For integer values it is given in terms of the series expansion coefficients of f(x):
This is then analytically continued to the complex plane. Count Iblis (talk) 19:20, 7 November 2012 (UTC)
localization
in the article about localization it reads:
- The ring Z/nZ where n is composite is not an integral domain. When n is a prime power it is a finite local ring, and its elements are either units or nilpotent. This implies it can be localized only to a zero ring. But when n can be factorised as ab with a and b coprime and greater than 1, then Z/nZ is by the Chinese remainder theorem isomorphic to Z/aZ × Z/bZ. If we take S to consist only of (1,0) and 1 = (1,1), then the corresponding localization is Z/aZ.
I don't agree with the statement that for n = prime power, the localization must be a zero ring. For example:
Z/pZ where p is prime is a field. So any element a!=1 in Z/pZ* generates Z/pZ*.
Take for example: Z/5Z and a = 2, then S:=<2> = {2,4,8=3,6=1} = Z/5Z* but then S^-1 Z/5Z ~= Z/5Z (because the field of fraction of a field is the same field)
--helohe (talk) 21:36, 5 November 2012 (UTC)
- I agree; it should somehow be rephrased. As you said, if you localize a field with respect to all nonzero elements, you end up with the same field but not a zero ring. I suppose it should say a zero ring or the same field. -- Taku (talk) 12:35, 9 November 2012 (UTC)
- I assume non-trivial localization was intended.--80.109.106.49 (talk) 22:23, 9 November 2012 (UTC)