User:Husseinshimaljasimdini
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Name:hussein.shimal.jasim.dini Location:baghdad-iraq. Specialty:physicist.university of baghdad. personal contact:jasimshimal@yahoo.com......... this is some personal posts and ideas in math.i hope one day it would be useful.thank you. I`ve been reviewing the list of special functions here and i want to add another special function. i also need your opinions about it. Function no1: Assume we put any real number in the form,(a.b),this new function "L" is going to transform that number to(0.b),for example,if we take the function f(x)=x,then,L(x)=0,if x=integer,and, L(a.b)=0.b,if x=a.b,for example,L(1.2)=0.2,and so on.Obviously the maximum value of "L" never exceeds one.Function no2:this function is about the area we can form by agiven length. Assume we have alength,s, and we want to start generating different areas by closing this length in away we keep all the SUBLENGTHS WE CUT EQUAL.Oviously we will start with atriangle where each sublength=(s\3).The function,A(numbers of sublengths) has aminumium value at the triangle, A(3)=[(s^2)\36](3)^(1\2),and amaximum value at the circle,A(n)=(s^2)\4pi,where,n→infinity. Now if we give up the conditions about the sublengths should be equal,and start with, A(1),
to represent the triangle,A(2),to represent the square and so on,we may put,A(n)=nA(1).thank you
LOGICAL SYSTEMS+NEW PRINCIPLES+ATTEMPT TO SOLVE COLLATZ CONJECTURE.
I have viewed the laws of logic and I think that there is asort of aconfusion about how the logical system works? and how the logical process conducts the conclusion?there is also aconfusion about the limits of the logical laws, specially the mixing between the law of contradiction and the law of noncontradiction. In this post I will be trying to show that all laws of logic are serving two simple prima principles and we will call them "principle of interior contradiction”and”principle of exterior contradiction”. Now let, A, be acase we want to prove,the logical process that followed by any logical or mathematical system to prove A,is 1=(A,exists,end of proof), 2=(A,doesnot exist), we know that the pair(1,2) is called"contradiction law".we have to notice that the whole idea to build aproof start with that pair. now,the pair(1,2)leads to alogical processes depend on aseries of steps of "the noncontradiction law "until we end to astate we cannot step over with anew contradiction or astep of afinal contradiction which is the conclusion,(A,exists,end of proof),the problem is ,whenever the noncontradiction law stops working during the process then we have to create or find anew contradiction other wise the statmment,A,stills open or unsolvable or we get aparadox. We will call the relation between the law of contradiction=pair(1,2)and the law of noncontradiction that works in the logical process as"EXTERIOR CONTRADICTION PRINCIPLE "or the ECP. Now the essential question,(I) is the contradiction law exists?(II)or it doesnot exists?if it exists then we will never be able to proof or know any judgement that the logical statement,A,exists or true or false .if it doesnot exists then why we need the proof in the first place?because we will know for sure any judgement about the statement,A.now we will call the the pair(I,II)”THE INTERIOR CONTRADICTION PRINCIPLE”or the ICP.
The ICP. and the ECP.are the two principles. that all the logical and mathematical systems and processes depend on. In asipmle word,(ICP) →(ECP) →(conclusion).or in another word,(contradiction law) →(prosseses of steps of noncontradiction law) →(final contradiction,or what we call it the conclusion).
Law of identity. P ≡ P The proof, P,exists
1- P ≡ P,end of proof 2- P no ≡ P (1,2) → if (P no ≡ P)→ (P ≡anything ,anything includes, P)→( P ≡ P),end of proof. (1,2) → if (P no ≡ P) →( P ≡nothing) → (P,doesnot exist) →(final contradiction=conclusion).as amatter of facts all logical laws are eventually serving the LIC.
law of the excluded middle. Eventhough , some systems of logic have different but analogous laws, while others reject the law of excluded middle entirely. We will trying here to prove that this law is serving the LIC. The law states, "P ∨¬P" 1-p,exists. 2-p,doesnot exist. (the pair (1,2)=contradiction law).it sems that law of the excluded middle is identical to law of contradiction or another forming to the law of contradiction.
THE LOGICAL SPACE. The logical spaces is asystem of operations ,relations,groups,variables,constants and operators where the logical statement exists.logical space is not aring or field.it is agroup of tools that we can use to create either ATRUE logical statement or FALSE logical statement.ofcourse we can shrink or expand the logical space depend on what we have or create of tools.
For example,the statement , 1/x ‹x ,x≠0 .this statement exists in the logical space (1,=,/,x,‹,G)where ,G,is the real numbers› 1.
assume that ,A, is alogical statement exists in the logical space Q, and K,is adifferent logical space 1-A,is true in,Q 2-A,false in,K The pair(1,2) is not acontradiction. This means that we will not be able to prove the existance of,A,is true. in the example above. we either testing all the numbers of,G,which is impossible,to prove the statement, 1/x ‹x, x≠0, or we start with the contradiction 1-A,is true in,Q 2-A,false in,Q (1,2) →assume 1/x ‹x ,x≠0,is true in another space,such as, (1,=,/,›,x,G) →(1/x›x)→(1›x)→(1›1) → contradiction of identity law=final contradiction=conclusion.
We also can find anothers spaces to serve the case. One could say ,1/x ‹x ,x≠0 , in (‹,G)is so obvious and doesnnot need any proof,even if it does,we can prove it by ashorter way like,assume, (1/x ›x)→(1/2›2)→(contradiction).or (1/x =x)→(1/2=2)→contradiction of law identity. The answer is yes,it is the same above but all the mathematical process in the argument is going intuitively while we want it to goes more axiomatically by defining what logical space the statement exists?and if it is atrue? or false?.when we are unable to know the limits where contradiction and noncontradiction lwas work,when we are unable to define the logical statement and in which logical space it exists,then we will get an open problems or paradox.
The important thing here is to notice that the time is not actually amthematical operator we depend on.instead of time we use more well defined operators.for example 1 ≠0 , in the space(G,›),where ,G,is any group.not now not any time,but it is simple to notice that ,(1,could=0),in any space hase the operator,d/dx,i.e,d/dx(1)=0.
THE ONE DIRECTION INFINITY PRINCIPLE,THE STEADY STATEMENT Assume he statement ,(a1,exists)→(a2,exists)→ …….an,exists).the qustion is there astatement,a∞ exists? Obviousely the possiblity of existance of any of the statements,an, is,1,which leads us to say the possibility of existance , a∞,is also 1.we will call, a∞,asteady statement.
THE TWO DIRECTIONS INFINITY.
Assume he statement ,(a1,exists)→(either,a2,exists,or,a3,exists). (a2,exists)→(either,a4,exists,or,a5,exists). (a3,exists)→(either,a6,exists,or a7,exists)……etc. Obviousely the possiblity of existance , a∞,is,1\2.
THE MULTI DIRECTIONS INFINITY.
Assume he statement ,a1,exists→(either,a1,or,a2,or,………. an,exists) Obviousely the possiblity of existance , a∞,is,0.
COLLATZ CONJECTURE AND STEADY STATEMENT. Post:the staedy statement exists in collatz formula and it equals,1. collatz formula, A-F(n)=n/2……if,n≡0(mod2). B-F(n)=3n+1…if,n≡1 (mod2).
Proof.
(some,A→ A),(some,A→ B),BUT(all,B→A,in the space(1,=,2k,/,N),Where,k,apositive integer and,N,is the natural numbers set.by using the one direction infinity principle,there exists asteady statements,k∞,that makes 2k=1,is atrue statement in the space above ,namely,2k. end of proof.
The purpose behinde this post is an attempt to expand integration space to include functions with undesirable points where the function could be undefined.I need help to understand if this post makes sense.The main question here, is this way could be more general than using the measurement set theory to expand the integration?
Assume the function F(x),
F:[a,b]→R We will put the interval[a,b] as acombination of subsets, UGk,
UGk=GNUGQUGQ̀....etc.(U,here stands for combination symbol) which stands for natural set,rational set ,irrational set,….etc.
Lets define the subsets,
өi={gk:F(xi)≥gk≥0},
xiЄ[a,b],
Let,(pi) be apartitoning of Gk in [a,b] and (pөi) apartitioning of(өi),
(Mөi=SUPөi), and (mөi=infөi),
(UFөi,pi)= Σi Mөi(xi-xi-1), (UFөi,pi)=upper darboux sum.
(LFөi,pi)= Σi mөi(xi-xi-1),(LFөi,pi)=lowe darboux sum.
(UFөi,pi,Pөi)=inf{UFөi:piPөi,(Pөi) partioning of(өi), (pi) partitioning of (Gk)},
(LFөi,pi,Pөi)=sup{LFөi:piPөi,(Pөi) partioning of(өi),(pi) partitioning (Gk)},
now, we put the integration in the form,
∫Fөi,over,Gk={0,UFөi}={0,LFөi}=the subsets,sөi,
∫F,over,[a,b]=Usөi
for example;
f(x):[0,1]→R,
f(x)=x , xЄ irrational numbers=Q̀, within[0,1],
f(x)=1,xЄ rational numbers=Q, within[0,1],
∫F,over,[o,1]={0,1\2}Q̀,U{0,1}Q={0,1\2}R,U,{1\2,1}Q ,(Q̀,Q AND,R,ARE SUFFIXES HERE AND U STANDS FOR COMBINATION symbol.)
this way enable us to exclude the undesired points from the integration where f(x)may be undefined. 88.116.163.226 (talk) 11:02, 30 January 2008 (UTC)husseinshimaljasim
And this is some feeds back from people on(http://en.wikipedia.org/wiki/Wikiped...sirable_points) What you are talking about seems to be similar to the difference between Riemann integration and Lebesgue integration. In Lebesgue integration, if two functions are identical everywhere except a set of measure zero (very roughly, a set of points that isn't too big), then their integrals will be equal. Thus, for example, a function which is continuous everywhere except x=0, where it is infinite, will have the same integral as the same function, but where at x=0 it takes the value 0, or similar. Since the rationals have measure zero over the reals, you could even have a function which is poorly-behaved over an infinite set of points (as long as it's a "small" infinite), and still be integrable. Confusing Manifestation(Say hi!) 04:37, 31 January 2008 (UTC) well,i dont think so,in Lebesgue integration,the above example would be, μ(0,1\2)inQ`set+μ(0,1)inQset=1\2+0=1\2,iam suggesting to keep integration in form of combination of subsets regardless they were countable or not.wouldnot this be more genral?210.5.236.35 (talk) 14:36, 31 January 2008 (UTC)husseinshimaljasim.thank youHusseinshimaljasimdini (talk) 11:49, 4 March 2008 (UTC)...the purpose of this post is an attempt tp show that real numbers set could be generated intensively,it also could be counted somehow by defining aspecial surjective or injectice function.i think the mathematical constructure of this post need to be fixed by an expert,thats why i need some help here.
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 integer Now if, am+bm=pm+10,pm<10 pm,positive integer Then we define the relationship R, ARB={pm*(10)^(n)}+{(pm+1)*(10)^(n-1)+..... Obviously, R; looks like adding backwards.e.g, 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
i.e, S0=0.123456789101112131415161718192021222324.... ,. In order to generate or count* the real numbers within the interval,e.g. (0,1),
We define the surjective function,F;
F:N→positive irrational numbers subset in(0,1)
Where , F(n)=SnR0.1,
Sn, the set of sequences,
S1=s0 R 0.1,
S2=s1 R 0.1,
Sn=Sn-1R0.1,
etc.
notice that (n-1) is suffix,
post(1)
There are an infinite sequences,s1,s2 that we can make S1RS2 Close enough to any real number.
post(2) if the relation,R,has aseriouse mathematical use , can we solve equations of the form, xRx=s,where,s=0.3,0.5,..etc ,or xR1=10? i mean can the relation,R,be generalized to involve such equations?or even negative numbers?
And these are the feedbacks form others and my responses. Originally Posted by HallsofIvy The fundamental problem is that your whole concept is flawed. The operations you define can only produce a countable set of numbers and the set of all real numbers is not countable. Or do you refuse to accept that? You define "the surjective function,F;
F:N→positive irrational numbers subset in(0,1)" which is, of course, impossible since that would imply the set of irrational numbers between 0 and 1 was countable. Originally Posted by husseinshimal i dont get it,you said that The operations i defined can only produce a countable set of numbers and we know the set is countable if there exists acountable subset belong to the original set.
From ramsey2879 The numbers that both HallsofIvy and I see here are only the rational numbers, but real numbers include the irrational numbers, so your set is incomplete. While you can make rational numbers close to irrational numbers they are still only the rational numbers and not the irrational numbers. Moreover for every rational number you say is close to an irrational number you can create an infinite number of irrational numbers still closer, so which of these irrational numbers does your counting number actually count? Originally Posted by husseinshimal we know the set is countable if there exists acountable subset belong to the original set Originally Posted by HallsofIvy No, that's non-sense. A set if countable it there exist a countable set containing it, not the other way around! Any set of real numbers, complex numbers, etc. contains the natural numbers- that does prove that any such set is countable Originally Posted by husseinshimal . first let me express my great apreciation for your being patient with me thank you all guys.I am just confused here alittle bit.i am not saying that real number set is countable.all what iam trying to understand is this,if the arbitrary sequence,S0,that i gave represents one irrational number in(0,1) and if the realtion,R,that i defined is not mathematically flawed.and if the definition of counable set which is,(G, is countable, i.e. there exists an injective function ,F:G→N. Either, G, is empty or there exists a surjective function,F:N→G) is right.then dont you think that what i was trying to do might be right? all what iam saying is ,F(1)=S0R0.1=0.223456789101112....f(2)=S0R0.2=S1R0 .1=0.323456....,i.e,F,is asurjective function.but ,F,keeps all the irrational numbers within(0,1) as long as it just shifts the digits to the left as it defined in the relation,R,the question here guys,and iam sure that you are the experts,is , if there is no logical or mathematical objection about this then dont you think it is worth to study it?thank you From ramsey2879 If your function creates unique irrational numbers from a set of k "ordered" non-negative integers where k is a constant then I would say that your set is countable because all such sets of k integers can be sorted first by the sum of the k integers, then by ascending order, for instance from left to right of the k integers. Thus if k = 4, 1 = F(0,0,0,0), 2 =F(0,0,0,1), 3 = F(0,0,1,0), 4 = F(0,1,0,0), 5 = F(1,0,0,0), 6 = F(0,0,0,2), 7 = F(0,0,1,1), etc. When k itself is infinite this is not the case.
And these are another feedbacks from other people on (http://en.wikipedia.org/wiki/Wikiped...al_numbers_set) . He's describing an adjustment of addition. Instead of carrying to the left, you carry to the right. So, 16(+)4=10.1, instead of 20. The decimal system remains the same, and is still intended to represent the real numbers. He describes a sequence of numbers in which each is the previous "plus" 0.1, and says that the "sums" of pairs of consecutive members of the sequence are dense in an interval. In the case of the number he started with, the unit interval. He says he's now trying to solve equations involving this operation, including x(+)0.1=1 and x(+)x=1. He doesn't seem to completely understand the words he's using, but he's still got an interesting question. Off the top of my head, 0.99999...(+)0.1=1, assuming normal rules of carrying to infinity, and there's no solution to the other question because the first digit will have to be 5 (0.5), and the second will have to be a digit n so that 2n+1=10m for some m, which is impossible. The article P-adics might be of interest. Black Carrot (talk) 20:42, 27 January 2008 (UTC) Corrections. The first digit could be zero, so either x=0 or the same problem. Also, I meant 1.9999... for the other, which I see he has. Problem, though. If 1.99999....=2, and 2(+)0.1=2.1, and 1.999...(+)0.1=1, then 1=2.1, which is an odd system. May have to be careful with the infinite decimals. Black Carrot (talk) 20:46, 27 January 2008 (UTC) I don't know what I was thinking. 50(+)50=1, of course. Black Carrot (talk) 20:57, 27 January 2008 (UTC). January 2008 (UTC) Well done to Black Carrot for successfully interpreting the original post. Seems to me that the binary operation it describes is base-dependent - for example, in decimal, 2(+)2=4, but in binary 2(+)2 = 102 (+) 102 = 1. This means it is can have little or no serious mathematical use or interest, in my opinion. Gandalf61 (talk) 13:28, 28 January 2008 (UTC) by reviewing the general mapping,R,where,ARB={pm*(10)^(n)}+{(pm+1)*(10)^(n-1)+..... it seems that binary operation is irrelevant.my main consern here is,can this relation with the set of sequences obove be applied to generate or count uniformly the irrational numbers set independently of considering cardinals concept or cauchy sequences?also can this relation be generalized to solve equations like xR1=10? OR xRx=(-1)?or xR1=(-1)? 210.5.236.35 (talk) 14:04, 28 January 2008 (UTC)husseinshimaljasimdini. also i mean here, can the mathemetical constructure of R,be fixed to be an equlevance relation involves negative numbers?210.5.236.35 (talk) 14:14, 28 January 2008 (UTC)Husseinshimaljasimdini Negative numbers would be tricky. One way to define the negative of a number, say -a, is as the number such that aR(-a)=0. So, -1 would be the number x such that 1Rx=0, which would make it equal to 9.99999... . I still recommend reading about the P-adics. For xR1=10, x=19.999999... . For xRx=(-1)=9.9999..., x=(-50)=59.99999... . For xR1=(-1)=9.99999..., x=(-2)=8.99999... . I don't understand what you want to do with the sequences S1 and S2 etc. By "generate or count uniformly" do you mean generate randomly? Black Carrot (talk) 04:47, 30 January 2008 (UTC) Thank you very much Black carrot for your advice.as amatter of fact the department of mathematics in my college has ran out of experts, specially after the war and when i asked them the were making fun of me because iam physicist,thats why i am annoying you guys here whith my questions.85.17.231.25 (talk) 09:06, 30 January 2008 (UTC)husseinshimaljasimdini I'm glad to help. I think I've figured out what you're trying to say. You have a sequence of numbers dense in an interval. In other words, given a point in the interval, you can approximate that point more and more closely with a subsequence of the original sequence. You want to use the indexes of the subsequence to keep track of the given irrational number. Is that right? Black Carrot (talk) 17:19, 30 January 2008 (UTC) yes black carrot.but the part of countablity conserns me.you know very well that ,G,is countable set if the sub sets that form ,G,are countable,now by reviwing the general concepts of countable set, we know that,K,is countable,if there exists an injective function F:K→N,either,K, is empty or there exists a surjective function,F:N→K.in the example that i gave above ,dont you think that ,F:N→(0,1),where,F=(SnR0.1),is asurjective?i also think F:(0,1)→N,is aninjective .(N,is the natural numbers set).if such function exists,does this make the positive irrational numbers set countable?because we know it is not.88.116.163.226 (talk) 12:31, 31 January 2008 (UTC)husseinshimaljasimdini .Husseinshimaljasimdini (talk) 11:52, 4 March 2008 (UTC)................................I have to questions here,is the function,f(x)=1^x,afixed point function?what is the value of g(x)if ,x=an irrational number like,2^1/2?thank you.Husseinshimaljasimdini (talk) 13:36, 10 March 2008 (UTC) Exponentiation over the complex numbers is inherently a multivalued function. In some cases we can choose a nice branch and it will be single-valued; in other cases we cannot. For 1x the obvious choice of branch is f(x) = 1 which is a constant function. For ( − 1)x there is no such obvious choice. Its values are (-1)^x=[e(2k+1)pi*i]^x=e(2k+1)x*pi*i=cos[(2k+1)*x]+isin[(2k+1)x] .......................Husseinshimaljasimdini (talk) 11:00, 13 March 2008 (UTC)................................................how to solve equation like,sin(x)=exp(x)?AND can we put x=log[sin(x)]?thank youHusseinshimaljasimdini (talk) 11:18, 4 April 2008 (UTC)
Yes, you can rewrite the equation as x = log(sin(x)), as long as you add the constraint that sin(x) > 0. From the graphs of sin(x) and exp(x) it is clear that there are an infinite number of solutions, all less than 0. I very much doubt that there is a closed form solution - I think the best you could hope for would be a power series. Gandalf61 (talk) 13:02, 4 April 2008 (UTC) A better method might be to write sin as , which would at least give solutions without resorting to logs of trig functions. That being said, it's still pretty much impossible to solve analytically I'd think. -mattbuck (Talk) 21:07, 4 April 2008 (UTC) Approximate exp(x)−sin(x) by p(x)=1+x2/2+x3/3+x4/24+x6/720+x7/2520 and solve the algebraic equation p(x)=0 by the Durand-Kerner method. Bo Jacoby (talk) 15:51, 5 April 2008 (UTC). If you're going to approximate, it's much better to use Newton's method. Use negative multiples of π as seeds. -- Meni Rosenfeld (talk) 16:54, 5 April 2008 (UTC) I respectfully disagree. Newton's method is likely not to converge. Smart choises of seed are needed. Nonreal complex roots are often not found. Bo Jacoby (talk) 17:54, 5 April 2008 (UTC). So? I wasn't talking about the general case (though I could), but rather for this particular problem. If we are only interested in real roots, I have already given the needed smart seeds. The Durand-Kerner may be good for polynomials, but this is not a polynomial, and approximating it as one is likely to cause problems and should only be used as a last resort. In this case, using the polynomial you have given, the first root is -3.01303. The actual root is -3.18306. Just guessing the solution to be -π is already better. -- Meni Rosenfeld (talk) 18:21, 5 April 2008 (UTC) The formulation "how to solve equation like,sin(x)=exp(x)" is referring to the general case. Bo Jacoby (talk) 07:08, 6 April 2008 (UTC). That's true. Your suggestion does generalize a bit better than "the roots are clearly close to negative multiples of π", but I maintain my position that it is only applicable if we are willing to ignore roots distant from the point of expansion, to find a few spurious roots, and to either have low accuracy or refine it with a general iterative method (such as Newton's). -- Meni Rosenfeld (talk) 14:38, 6 April 2008 (UTC) For any entire function f(x), and for any precision ε>0, and for any radius R>0, there exists a degree N>0 such that the Taylor polynomial approximation pn(x) to f(x) for any n>N and for any |x|<R satifies |f(x)−pn(x)|<ε. So a general method to find zeroes of a transcendent entire function f(x) within a circle |x|<R is to compute the zeroes of polynomial approximations. Newton's method finds one root at a time, and it is hard to know if all roots inside some circle have been found. That's why I prefer the Durand-Kerner method which computes all the roots of a polynomial. Bo Jacoby (talk) 15:22, 6 April 2008 (UTC).......................................................................................................Iam not sure but I think I have developed asimple technique to solve equations like,e^x=sin(x).I need your opinions if this technique works.Now,if,f(x)=x,at x=x0,then, f(f(f(…….f(x0))=x0.This technique depends on choosing apoint around the interval where, g(x)=h(x).It doesnot matter which point we choose as long as it is inside the interval.for example, Cos(x)=x,if we start with any point within the interval (0,pi\2),say,0,then we will get, Cos(cos(cos(…….cos(0))=0.73908….it does not matter wich point we choose within (0,pi\2),we will always get the same result.As for the equation,e^x=sin(x),we put, [Inv(sin(e^x)=x],as long as sin(x)has aperiod of(pi).If we start with(-pi),then we should get for the first approximation,[ Inv(sin(e^-pi)=0.04322…=x1].and for the second approximation,[Inv(sin(e^-(pi+x1)=x2]….and so on until we be close enough and the point would be(-pi-xn),where sin(-pi-xn)=e^(-pi-xn).We also could apply the same way at the zeros of sin(x) and find all of the other solutions.thank youHusseinshimaljasimdini (talk) 11:30, 22 April 2008 (UTC
