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December 28

Integral formula?

Is there a formula just like the derivative formula to integrate functions? 200.63.228.34 (talk) 03:49, 28 December 2007 (UTC)[reply]

No. Also there is no formula which will give the derivative of all functions.--Shahab (talk) 04:44, 28 December 2007 (UTC)[reply]

It's the same concept than with derivatives really.

You got a list of some derivatives/integrals you know (the derivative of exp(x) wrt x is exp(x), that of ln(x) is 1/x...), and then formulas to convert derivatives into others (the derivative of x ln(x) wrt x is the derivative of x times ln(x) plus x times the derivative of ln(x), thus reducing the derivation of x ln(x) to that of x and ln(x), which we know we can do.)

For integrals, similarly, there's a list of "basic" integrals (which is quite extensive) and then formulas to convert to other integrals (integration by parts, substitution, ...).

But there are integrals which we just don't know how to do (and it can be proved it can't be done with a certain set of functions, for example, elementary functions) - so we make up new functions (ie the error function erf, that is, up to a constant factor, the antiderivative of exp(-x²) wrt x).

PS. I think it's true that derivating any combination of elementary functions yields another elementary function, whilst that is not true for integrating, but I'm not totally sure. -- Xedi (talk) 05:12, 28 December 2007 (UTC)[reply]

Yes, that is correct. You have formulae for computing the derivative of a product, ratio, composition, and so on, of functions. This allows you to easily find the derivative of any elementary function, which is itself an elementary function. For integrals, you have much less potent formulae, so in many cases the integral of an elementary function is not elementary, and even if it is, it's not that easy to find it (though methods like the Risch algorithm exist). -- Meni Rosenfeld (talk) 10:36, 28 December 2007 (UTC)[reply]

And let me add something about the term "elementary". It refers to something basic, to be used to build up other objects. So, this is clearly the case of an elementary function. But it should be clear that, whether or not to include a function among the class of elementary functions, was just a matter of choice, mainly due to historical and practical reasons. In a world with more guys from probability theory and parabolic PDE's around (I'm absolutely not complaining about the present world under this respect), we would probably call erf(x) an elementary function too, just like we do with log(x) or sin(x), and learn it at the elementary schools. In any case, what we know about the erf function is not less than what we know about the log function. For these still useful but not-so-widely-used functions, there is the term special functions. What I mean is, there is nothing lacking or peculiar in a given function if it is not a composition of functions of a previously defined class, even if this may be a very interesting and deep mathematical fact. In a more optimistic way the point is not, that integration formulae are not potent enough, but on the contrary, that some constructions in analysis used to solve problems are so powerful that they bring more and more new functions under our knowledge. Maybe there is a kind of moral about all this. You do not belong to my community, you are a function different from who arrived here before you? There's nothing lacking or strange in you, it is just a contingent fact, join us and we will be a richer community etc. We are talking about integration, aren't we ;-) PMajer (talk) 13:09, 28 December 2007 (UTC) (talk) 12:51, 28 December 2007 (UTC)[reply]

Let me add that, strikingly enough, requisites for a function to have a primitive are relatively weak (continuity, etc). In other words, there are "much more" integrable functions than differentiable ones, though many of the first possibly have no closed-form primitives. Pallida Mors 14:24, 28 December 2007 (UTC)[reply]

Elementary functions in this sense is whatever function you can get with a finite sequence of composition form identity, complex numeric constants, addittion, subtraction, multiplication, division, exponential, logarithm, trigonometric and inverse trigonometric functions. There's some problem with these not being defined everywhere and branch cuts and whatever, but basically you can differentiate such functions symbolically in a mechanic way and get elementary functions from them: you just use the chain rule and the known derivatives of the basic building blocks I've listed. The differential function you get does not necessarily have the full domain of the original function, but in typical cases it has most of its domain and it is the differential of that function in most places, so this method is very usable in practice.

On contrary, most elementary functions don't have an integral in elementary form, not even a function that's reasonable close. Well-known examples are

\int (\sin x/x)dx

and

\int (1/\ln x)dx

. On the other hand, it's still useful to know which functions have an integral in elementary form. Rational functions (quotients of two polynomials) always do, for example, and there's an algorithm to integrate them symbolically, though it's not as simple as for differentiation. – b_jonas 20:46, 30 December 2007 (UTC)[reply]

A vocabulary question in matrix theory

Hi, consider the following property, for an n by n square matrix A:=(a_i,j)_{i≤n,j≤n}:

"For any k≤n, the diagonal element a_k,k is equal to the spectral norm (i.e., the operator norm w.r.to the euclidean norm) of the k by k sub-matrix (a_i,j)_{i≤k,j≤k}."

Has this property a name? I guess it is well known. It's kind of diagonal domination (not the usual one).

Thank you for your help, PMajer (talk) 11:22, 28 December 2007 (UTC)[reply]

Are you sure there is any non-diagonal matrix with this property? -- Meni Rosenfeld (talk) 00:09, 30 December 2007 (UTC)[reply]

Oops, sorry I think I've put the wrong question. You are completely wright, very good.

Well, the real question was about this property, of an n by n square matrix A:=(a_i,j)_{i≤n,j≤n}:

"For any k≤n, the diagonal element a_k,k is larger or equal to the spectral norm of the k by (k-1) sub-matrix (a_i,j)_{i≤k,j<k}." PMajer (talk) 09:27, 31 December 2007 (UTC)[reply]

Integration and irrational numbers generatings.

To who may concern.I wish these couple of to be useful.I want to know if they works.I dont have P.HD in math.Iam not aprofessor,that is why I hope to get some help to fix the mathematical constructure of these two theories,if they make sense.

Split into two separate subsection put below for ease of replying. The PDF file referred to is here: MY MATH. IDEAS.pdf. --Lambiam 13:27, 28 December 2007 (UTC)[reply]

Today i published aPDF file on wikipedia where symbols are clear.

Husseinshimaljasimdini (talk) 11:47, 28 December 2007 (UTC)Hussein shimal jasim dini.baghdad-iraq.[reply]

Sorry, it makes no sense to me. Which unsolved problem are you trying to solve? Bo Jacoby (talk) 13:06, 28 December 2007 (UTC)[reply]

I found a counterexample for this guy's prime tester (1331), but I don't understand the other two "theories" and don't have time to look closer at them right now. Can anyone point out something wrong with them? I always feel sorry for people who waste their time on things like this. —Keenan Pepper 18:09, 29 December 2007 (UTC)[reply]

Theory-1: expand integration space

Assume the function F(x), R→ F:[a,b] We put the interval[a,b] as acombination of seb sets, UGk; UGk=GNUGQUGQ’…….etc. N,Q, Q’; are the natural,rational and irrational sets respectively

Lets define the subsets, өi={gk:F(xi)≥gk≥0}, F(xi)Є[a,b] Let pi be apartitoning of Gk in [a,b] and pөi apartitioning of өi, Mөi=SUPөi, mөi=infөi, UFөi,pi= Σi Mөi(xi-xi-1), LFөi,pi= Σi mөi(xi-xi-1),now; UFөi,pi,Pөi =inf{UFөi:piPөi,Pөi partioning ofөi, pi partitioning of Gk}, LFөi,pi,Pөi =sup{LFөi:piPөi,Pөi partioning ofөi, pi partitioning Gk},

∫ Fөi over Gk={0,UFөi}={0,LFөi}=the sets sөi

∫ F over [a,b]=Usөi e.g, R→ F:[0,1] F(x) = 1 , x ЄQ in[0,1], Q = rational numbers set. F(x) = x , x ЄQ’ in[0,1], Q’=irrational numbers set

  ∫ F over[0,1] = {0,1\2}Q’U{0,1}Q

                            ={0,1\2}R U {0,1\2}Q

the point here is as long as we keep the partitioning pi over [a,b].i.e,we dont apply the measurement theory on the subsets of(UGk),we can use the partioining(Pөi)to exclude any undesired points ofөi.

Theory-2: generating irrational numbers

preview the purpose of this theory is an attempt to show that real numbers can be generated or counted randomly intensive with out using Cauchy sequences.

Consider we express the tow positive real numbers ,A&B as,

A=Σam[(10)^(n-m)]

B=Σbm[(10)^(n-m)]

Where,( n,m=0,1.2,……) am,bm,positive integers<10

Now if, am+bm=pm+10,pm<10

pm,positive integer

Then we define the relationship R,

ARB={pmΧ (10)^(n)}+{(pm+1) Χ(10)^(n-1)+……

Obviously, R; looks like adding backwards.For example,

341R283=525

(3+2)=5,(4+8)=12,(1+1+3)=5

Lets now pick up arbitrarily the infinite sequence

S0=Σn\(10)^(n),+Σn\(10)^(n+1) + Σn\(10)^(n+2)+..…

Where n=1to9 ,10to99,100to999 ,…etc.respectively

s0=0.123456789101112131415161718192021.... n, is apositive integer.

In order to generate or count* the real numbers within the interval (0,1), We define ,F; F:N→IR

Where ,

F(n)=SnR0.1

Where, Sn, the set of sequences

S1=s0 R 0.1 S2=s1 R 0.1 . . .

hypothesis There are an infinite sequences,s1,s2 that we can make

S1RS2

Close enough to any real number.

Obviousely inorder to solve the equations, XR0.1=1, x→1.99999… XR1.1=1, x→0.99999… But how to solve xRx=1?