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October 2

Is there any short-cut method for finding the derivative or integral of a function?

Finding limit of a function may be lengthy in some cases, but those limits can simply be solved using L'Hôpital's rule. In fact, almost every question related to limits can be solved easily using L'Hôpital's rule. I am searching for some short-cut methods (like L'Hôpital's rule) for finding the derivative or integral of a function. You know finding integral or derivative of a function in some cases is really very lengthy. So, I need a method (or rule) by which I could find derivative or integral of a given function in shortest time possible. Thank you for you response! Concepts of Physics (talk) 15:34, 2 October 2013 (UTC)[reply]

Graphical methods might work if you only need approximate numeric values within a certain range. That is, graph it, then find the slope, area under the curve, etc. StuRat (talk) 15:38, 2 October 2013 (UTC)[reply]

See our articles on automatic differentiation and symbolic integration. In general, integration is more difficult than differentiation. Gandalf61 (talk) 15:44, 2 October 2013 (UTC)[reply]

This may not be what you are looking for, but Wolfram Alpha and programs like Maple (software) and Mathematica do them. Bubba73 ^{You talkin' to me?} 15:49, 2 October 2013 (UTC)[reply]

In fact, almost every question related to limits can be solved easily using L'Hôpital's rule. ^{[citation needed]}, I think. In any case, application of L'Hôpital's rule involves finding the derivatives of two functions, so it's hardly a 'shortcut' compared to differentiation. AndrewWTaylor (talk) 16:15, 2 October 2013 (UTC)[reply]

I think the OP meant that just as L'Hopital's Rule is a shortcut method for finding limits, so also method "?????" is a shortcut for differentiating. Duoduoduo (talk) 17:43, 2 October 2013 (UTC)[reply]

You think that "every question related to limits can be solved easily using L'Hôpital's rule" because in your course you have been given exercises that are to be solved using L'Hôpital's rule. This doesn't apply generally.

If you want a symbolic expression for a derivative, the "shortcut" is to use Differentiation rules rather than the limit definition. I don't believe there is a shorter way.

For integration, while there are some systematic methods, in practice it is more of an art, with many integration methods to try. -- Meni Rosenfeld (talk) 08:00, 3 October 2013 (UTC)[reply]

If you need to evaluate a high order derivative in a point, then the fastest method is to use the Taylor expansion around that point. To evaluate a Taylor expansion to order N, the required number of computations is of order of some power of Log(N), so to find the trillionth derivative of a complicated function in some point only requires some dozens of computations. Count Iblis (talk) 14:06, 3 October 2013 (UTC)[reply]

I'm curious — how does one derive a Taylor series that efficiently? --Tardis (talk) 21:12, 6 October 2013 (UTC)[reply]

Definite integrals can sometimes be immediately evaluated from the series expansion of the integrand e.g. using Ramanujan's master theorem or some suitable generalization of this based on Umbral calculus. These methods sometimes allow you to immediately write down the result of an integral that Wolfram Alpha or Mathematica cannot evaluate. Count Iblis (talk) 14:40, 3 October 2013 (UTC)[reply]

October 3

PPM?

How much μg Iodine there is in 30ppm Iodine (per 100 g) Iodized salt? Thanks. Ben-Natan (talk) 04:53, 3 October 2013 (UTC)[reply]

30 ppm is 30 μg per g, so 3000 μg in 100 g.31.54.112.70 (talk) 11:54, 3 October 2013 (UTC)[reply]

This assumes that the parts per million is by mass, as opposed to by volume or count of molecules. StuRat (talk) 11:57, 3 October 2013 (UTC)[reply]

True. Typically for solids, ppm, is a concentration based on mass and for liquids, it is a concentration based on volume. Iodised salt indicates the concentrations, at least for those sources, are in terms of mass. --Mark viking (talk) 21:09, 3 October 2013 (UTC)[reply]

Interesting. This could possibly cause a strange result where a 1 ppm solid becomes 2 ppm, when dissolved in a liquid. StuRat (talk) 01:16, 6 October 2013 (UTC)[reply]

Proof involving binomial

Help please! Can anyone explain the step at 7:10 involving Binomial? Its the only part I don't understand and it's doing my head in!

http://www.youtube.com/watch?v=XHQ0OzqTjd0&feature=c4-overview-vl&list=PL32446FDD4DA932C9 — Preceding unsigned comment added by 31.48.32.249 (talk) 20:50, 3 October 2013 (UTC)[reply]

The previous equation (the 6th on that slide), is the Cauchy product and has an error in it. If you listen very carefully (the sound is bad), the speaker mumbles an apology for omitting a factorial sign. It should read

1=\sum _{n=0}^{\infty }\sum _{\mu =0}^{n}{\frac {\mathrm {\beta } _{\mu }}{\mu !}}{\frac {1}{(n-\mu +1)!}}s^{n}

--catslash (talk) 22:23, 3 October 2013 (UTC)[reply]

Ah i see now! Thanks! That makes so much more sense! Cheers mate! Your a lifesaver! — Preceding unsigned comment added by 81.155.129.116 (talk) 15:02, 4 October 2013 (UTC)[reply]

October 4

Absolute value (algebra)

In this article, I don't understand what the "No-Solution Case" section is trying to say. It looks like it is trying to say that the absolute value is never negative, but I could be wrong. At any rate, it seems out of place with the rest of the article; the talk page doesn't seem much visited, so I figured I'd ask here. I don't want to remove/edit something if I've missed something- maybe it's something commonly seen in textbooks that needs to be reworded? Thanks:-)Phoenixia1177 (talk) 05:23, 4 October 2013 (UTC)[reply]

It might be an attempt at adding what you just said, but it is borderline incoherent as written. I've reverted it. --Kinu ^t/_c 05:42, 4 October 2013 (UTC)[reply]

Thanks. I was figuring that'd be the best course, just wanted to make sure I wasn't glossing over/forgetting something.Phoenixia1177 (talk) 05:55, 4 October 2013 (UTC)[reply]

Possible columns in an integral matrix

This probably has a very easy answer and I'm just being dense. It is clear that a necessary and sufficient condition for a pair of non-zero integers to appear as a column in some invertible 2 x 2 integral matrix (edit: whose inverse is integral) is that their greatest common divisor is 1. Does this property hold for n-tuples of non-zero integers when considering invertible n x n integral matrices for larger n? It's clearly still a necessary condition. I feel if I hadn't forgotten all my basic module theory, I'd be able to answer this pretty quickly! Thanks, Icthyos (talk) 17:19, 4 October 2013 (UTC)[reply]

Unless I'm missing something, the premise is not true. Consider:

 1  2
 0  4

That's invertible, isn't it? Looie496 (talk) 17:58, 4 October 2013 (UTC)[reply]

Ack, sorry, I only ever work with integral matrices, so forgot to specify that I require the inverse to be integral too (i.e. the determinant must be 1 or -1). Icthyos (talk) 18:06, 4 October 2013 (UTC)[reply]

(ec)To answer the revised question, see Extended Euclidean algorithm for the 2 by 2 case. Two integers a and b are relatively iff there exist integers x and y so that ax+by=1. But ax+by is the determinant of

{\begin{bmatrix}a&-y\\b&x\end{bmatrix}}.

There is a similar theorem that says if a, b, c, ... n are relatively prime (as a group, not necessarily in pairs), iff there exist p, q, r, ... z so that ap+bq+cz+...+nz=1. But this isn't in the form of a determinant so I think something a bit stronger is needed, such as:

A vector (a, b, c, ... n) of integers is a column of a square integer matrix of determinant 1 iff the gcd of a, b, c, ... n is 1.

I've sketched out a proof of this following the method use in the Euclidean algorithm, but it's rather long to post here, plus I assume the result is easier to obtain using more sophisticated methods. --RDBury (talk) 19:40, 5 October 2013 (UTC)[reply]

It seems to me that gcd(all elements of the column)=1 should be sufficient. Consider the column (a b c)^T for given a, b, c. Find d, e, f, g, h, and i such that the matrix

a d g

b e h

c f i

has determinant = 1. If my unchecked algebra is correct, setting this determinant to unity and solving for d gives

d={\frac {a(ei-fh)+b(fg)-c(eg)-1}{bi-ch}}.

Now setting the denominator of this equal to 1 is the same as the problem in one lower dimension of finding i, h that make

b h

c i

have a unit determinant. Since this can be done, we can make our denominator equal to 1; then use any integer e, f, g in the numerator and use the equation to get the resulting integer d.

So it looks to me like you could generate a proof by induction for successively larger problem sizes, the sufficiency proof for each being dependent on the sufficiency in the problem of one dimension lower. Duoduoduo (talk) 19:13, 5 October 2013 (UTC)[reply]

Actually this assumes that some pair of a, b, and c is relatively prime, but in the case 100, 45, 72, the numbers are relatively prime as a group but not in pairs. However

{\begin{bmatrix}100&11&7\\45&5&3\\72&8&5\end{bmatrix}}.

has determinant 1, so having a pair of relatively prime entries isn't needed. --RDBury (talk) 19:51, 5 October 2013 (UTC)[reply]

Okay, so my post above shows that a sufficient condition for the 3×3 case is that at least one pair of the entries be coprime. So the question remains: is the alternative condition gcd(a, b, c) = 1 also sufficient? I.e., it works in the case (100, 45, 72), but does it work in all such cases? Duoduoduo (talk) 22:19, 5 October 2013 (UTC)[reply]

Here's a proof that's a bit better than the one I was talking about above. I'll do the dim 2 => dim 3 case which easily generalizes to dim n => dim n+1. We are given a,b,c with gcd(a,b,c)=1. Let d=gcd(b,c). then gcd(b/d, c/d) = 1 and there is a matrix

{\begin{bmatrix}b/d&p\\c/d&q\end{bmatrix}}.

with determinant 1. gcd(a,d)=1 so there is a second matrix

{\begin{bmatrix}a&r\\d&s\end{bmatrix}}.

with determinant 1. Then the product

{\begin{bmatrix}1&&\\&b/d&p\\&c/d&q\end{bmatrix}}{\begin{bmatrix}a&r&\\d&s&\\&&1\end{bmatrix}}.

has determinant 1 and the first column is (a, b, c).

I believe this can be generalized as follows: Let R be a p.i.d. and let F be a free, finitely generated R-module. If L is a submodule of F so that F/L is torsion free then L is a direct summand of F. This is an application of the theory of modules over a p.i.d. --RDBury (talk) 00:17, 6 October 2013 (UTC)[reply]

Ah-hah! Very clever. Thank you both for your replies. Icthyos (talk) 11:08, 6 October 2013 (UTC)[reply]

October 6

I understand Complex Numbers

Complex number used to be really hard. But I understand it now. It's is just ScaleRot. Scaling and Rotation.
4+3I is just PolarCoor( 5 , 36.86 deg ) or ScaleRot( 5 , 36.86 deg )
So the square of (4+3i) is just scale the unit number (1) by 5 twice and rotate by 36.86 degrees twice.
1*5*5 = 25
0 degree + 36.86 degree + 36.86 degree = 73.72 degree

So the answer is just ScaleRot( 25 , 73.72 degree ) or PolarCoor( 25 , 73.72 degree ) or 7.008 + 23.997i

Why does the maths teacher at school makes it seems so difficult??? 220.239.51.150 (talk) 02:51, 6 October 2013 (UTC)[reply]

We don't know how your teacher taught it, or why this way seemed difficult to you.

But while it's true that multiplying by a complex number scales and rotates, I wouldn't say that complex numbers are just that, it's deeper than that. Perhaps the teacher was trying to present a more complete view of complex numbers. -- Meni Rosenfeld (talk) 06:26, 6 October 2013 (UTC)[reply]

Is it deeper than that? Complex numbers is just a convenient way of expressing the 2 by 2 matrix group whose Lie algebra is given by the generators

I_{2}

(1 in complex numbers) and

\sigma _{z}

(i in complex numbers) which, as pointed out above, is exactly the same group of continuous scaling and rotation operations in 2D. They are one and the same thing. — Preceding unsigned comment added by 109.144.255.136 (talk) 14:40, 6 October 2013 (UTC)[reply]

That captures only the multiplicative structure. They also have an additive structure and various sorts of topological/differential structure, and these interact in deep ways. --Trovatore (talk) 22:49, 6 October 2013 (UTC)[reply]

Symbol in the center of a commutative diagram?

In the article Banach bundle, all the commutative diagrams have what look to be small copyright symbols in the centers of them. Is this meant to convey that the diagrams commute? If so, is this notation standard? It is new to me. --Mark viking (talk) 03:46, 6 October 2013 (UTC)[reply]

While not a copyright sign, this [1] mentions putting a symbol in the center to indicate the diagram commutes. However, I didn't find this notation anywhere else- and I don't think it's that good of an idea- so it may be unrelated, but it's the only like thing I could find. A lot of the things I work on, and read, involve heavy use of category theory, and I don't think I've ever seen diagrams marked like that. At the very least, if it means something, it should be explained in the article, maybe we can track down who put the diagrams in?Phoenixia1177 (talk) 05:58, 6 October 2013 (UTC)[reply]

It was User:Sullivan.t.j, and I've left a note on his/her talk page. Rojomoke (talk) 10:55, 6 October 2013 (UTC)[reply]

Yes, the c-in-a-circle denotes commutativity, not copyright. I think that there mathematically needs to be some indication that a diagram commutes, whether that's done (a) globally, by saying "all diagrams on this page / in this article commute", (b) locally in the text, by saying "the following diagram commutes", or (c) locally in the diagram itself, as done by the "copyright sign". I'm not particularly wedded to any one way of achieving that aim, and I've seen all three used in practice. Sullivan.t.j (talk) 16:16, 6 October 2013 (UTC)[reply]

I'm used to a little circle, not a copyright sign. That's what Serge Lang uses, if I recall correctly. I think this is less confusing, because the copyright sign might be interpreted as an assertion of copyright. (But only a little less confusing, as it still has to be explained, and anyway in practice one never draws a non-commutative diagram.) --Trovatore (talk) 21:55, 6 October 2013 (UTC)[reply]

I would like to thank you all for explaining the notation and giving insight about its context. The discussion is ongoing, but my particular questions are well resolved. --Mark viking (talk) 22:44, 6 October 2013 (UTC)[reply]

Hopefully nothing went amiss on this wind energy assignment...

Without access to a tutor or the instructor this late at night, I figured that the only way to have the assignment checked was by placing it here. The rules of the RD asks to do our own homework. I already did. I'd just now like for someone to check whether they're all correct, or if I forgot to plug something into a formula here and there, or whether I used the right formula. I do feel iffy on the airplane problem (Problem #3.)

Problems 1-4; P. 45

1. The blade of a wind turbine experiences a force of 100 lbs. when the wind speed is 10 ft/sec. To what value does the force change if under the same conditions the wind speed increases to 15 ft/sec?

a. First force = 100 lbs., First wind speed = 10 ft / sec, Second wind speed = 15 ft / sec
b. 15 ft/sec = Force of 10 ft/sec * (15/10)^2
c. 15 ft/sec = (100)(15/10)^2 = (100)(2.25)
d. = 225 lbs of force at the wind speed of 15 ft/sec

2. If the lift and drag coefficients in an airfoil are 0.998 and 0.05, respectively, determine by how many times the lift force is larger than the drag force in a blade with that airfoil profile.

a. Lift coefficient = .998, Drag coefficient = .05
b. .998 / .05
c. 19.96
d. The lift force is 19.96x larger than the drag force in a blade with this airfoil profile.

3. If a small 3200-lb airplane has two wings with the airfoil profile in problem 2, and the drag force on each wing at a given wind speed is 100 lbs., what is the net lift force on the plane?

a. Drag coefficient = .05, Lift coefficient = .998
b. Lift coefficient is 19.96x larger than the Drag coefficient
c. 100 lbs * 19.96 = 1996 lbs lift force
d. 1996 lbs lift force – 3200 lbs plane weight = 1204 lbs shortfall (Net lift force = -1204 lbs)

4. If a turbine has three blades and the center of pressure (where the lift and drag forces apply) of each blade is 10 ft from the axis of the turbine and the lift force on each blade is 100 lbs., how much is the torque that rotates the turbine?

a. Lift force = 100 lbs; distance from center of gravity on each blade = 10 feet. (Blade lengths = ~20 feet each)
b. 100 lbs * 10 ft = 1000 lb-ft of torque for each blade.
c. 1000 lb-ft of torque per blade * 3 blades = 3000
d. 3000 lb-ft of total torque rotates the turbine

But hopefully everything checks out. --2602:30A:2EE6:8600:9C3C:99E3:B76F:BDBA (talk) 05:30, 6 October 2013 (UTC)[reply]

In 3, I think you have done the lift force calculation for one wing - but we are told the plane has two wings ... Gandalf61 (talk) 07:31, 7 October 2013 (UTC)[reply]

Data distribution

Assuming that data is normally distributed, what percentage of that data would be expected to lie within three standard deviations of the mean (i.e., the mean +/– three standard deviations)? Thanks. Joseph A. Spadaro (talk) 17:04, 6 October 2013 (UTC)[reply]

See Normal distribution#Standard deviation and tolerance intervals. Duoduoduo (talk) 17:59, 6 October 2013 (UTC)[reply]

Perfect. Thanks! Joseph A. Spadaro (talk) 21:05, 6 October 2013 (UTC)[reply]

Diffusion Model

Hello. I have a diffusion equation modelled by the Gaussian function. $f(x)={\frac {Ae^{-{\frac {x^{2}}{4Dt}}}}{8({\sqrt {\pi Dt}})^{3}}}$ A and D are constants. I want to find the distance from the source where the density of particles is the greatest. Given the time when the density is greatest, why do I take the partial derivative with respect to t instead of x? Thanks in advance. — Preceding unsigned comment added by Mayfare (talk • contribs) 22:45, 6 October 2013 (UTC)[reply]

The gaussian function is a solution to a diffusion equation without source, so your question is confusing. Bo Jacoby (talk) 18:30, 7 October 2013 (UTC).[reply]

October 8

Is there a set of functions that fulfil a certain condition

If I have a linear function in ONE variable, say 4j, for j an integer and j ≥ 1, then 4j = 4, 8, 12, 16 etc.

So if I want to get the values that 4j does NOT represent, I can use the set of functions 4j+n, where n = 1, 2, or 3.

But what if I have a linear function in TWO variables, say 4jk+j+k, for j and k integers, and j and k ≥ 1.

For j = 1 and k = 1, 2, 3, 4, etc., 4jk+j+k = 6, 11, 16, 21, etc.

For j = 2 and k = 1, 2, 3, 4, etc., 4jk+j+k = 11, 20, 29, 38, etc.

For j = 3 and k = 1, 2, 3, 4, etc., 4jk+j+k = 16, 29, 42, 55, etc.

Question: Is there a set of functions which give the values that 4jk+j+k does NOT represent? Bh12 (talk) 23:42, 8 October 2013 (UTC)[reply]

If j or k were allowed to be zero you get all the numbers, so any problem has to do with multiplying so you can think prime numbers. That isn't the answer directly but if you change your expression to ((4j+1)(4k+1)-1)/4 = n then you're asking which values of 4n+1 can't be expressed as (4j+1)(4k+1) when j and k are 1 or more. Dmcq (talk) 07:17, 9 October 2013 (UTC)[reply]

Yes, but is there a function or a set of functions that gives those values of n which cannot be expressed as ((4j+1)(4k+1)-1)/?Bh12 (talk) 14:37, 9 October 2013 (UTC)[reply]

October 9

US default on 17th Oct

Does the probability of "US Default on 17th Oct" exists? And if so what is the proper Mathematical way of calculating it? 202.177.218.59 (talk) 01:53, 9 October 2013 (UTC)[reply]

Yes, as argued here all probabilities are quantum mechanical in nature (even what looks like obvious counter examples like betting on digits of pi turn out not to be so upon a precise analysis). So, the probability is in principle well defined by the laws of physics and has a quantum mechanical origin. Count Iblis (talk) 15:04, 9 October 2013 (UTC)[reply]

Common factors of a sum

This might be a terribly difficult question, given how greatest common divisors behave. Suppose I have two odd integers, A and D, and two even integers, B and C, such that: gcd(A,B,C,D) = 1, but gcd(A,B) ≠ 1 and gcd(C,D) ≠ 1. Does there exist some integer K such that gcd(C + 2KA, D + 2KB) = 1? Searching for a counterexample has proved fruitless - after dozens of attempts, the gcd of the sums has been 1 for K = 1 or K = -1. It seems like there should always be such a K, but I have very little idea of how to show it. Any suggestions are greatly appreciated! Icthyos (talk) 14:02, 9 October 2013 (UTC)[reply]

Let A=3, D=5, B=6, C=10, which fulfil all of your conditions regarding A, B, C, and D.

Let K=2, so that C+2KA=22 and D+2KB=29, and gcd(C + 2KA, D + 2KB) = 1.

(Should K be relatively prime to C and D? Perhaps that helps.)Bh12 (talk) 16:01, 9 October 2013 (UTC)[reply]

Your example certainly works, but I should clarify: I'm wondering if it's possible for there to be A, B, C, D satisfying the above conditions such that there is no integer K for which gcd(C + 2KA, D + 2KB) = 1. For all the examples of A, B, C, D I pick, I can always find some K for which this gcd is 1, but I'm having a hard time proving that there always exists such a K in general. Finding examples is the easy part! Icthyos (talk) 16:15, 9 October 2013 (UTC)[reply]