February 10

LOGICAL SYSTEMS+NEW PRINCIPLES+ATTEMPT TO SOLVE COLLATZ CONJECTURE.

First i would like to say that am honoured to share my thoughts with great people in here who always provide help.I will not say iam right or wrong.I hope this post will be aspark to good mindes.

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.Husseinshimaljasimdini (talk) 10:17, 10 February 2008 (UTC)husseinshimaljasimdini[reply]

You should talk to User:BenCawaling. He also claims to have a proof of the Collatz conjecture. —Keenan Pepper 17:22, 10 February 2008 (UTC)[reply]

I don't think your system will work. You should probably test-drive this idea before using it on something new. Try to prove and disprove things that are already known. Try to prove, for instance, that there are only finitely many prime numbers, or that there are infinitely many positive integers less than 10,000. If you succeed, it might make it clearer what's wrong. Black Carrot (talk) 01:15, 11 February 2008 (UTC)[reply]

Hello, please help me! I have the number 2008. The sum of any natural numbers shall give the number 2008. But the multiplicative inverse of these numbers shall give 1. Which numbers must be taken? --85.178.52.149 (talk) 10:19, 10 February 2008 (UTC)[reply]

OK, just so I have this clear, you want numbers x such that ${\displaystyle \sum \limits _{i}x_{i}=2008}$ and ${\displaystyle \sum \limits _{i}{\frac {1}{x_{i))}=1}$, is this correct? If so, there's no way to work it out without using trial and error - n equations can only ever determine at most n variables, so if we treat the ${\displaystyle x_{i))$ as the variables, then we have two equations and so could only work it out for i = 1 or 2. Do you have more information? -mattbuck (Talk) 10:25, 10 February 2008 (UTC)[reply]

E.g.:

• 3 + 44 + 44 + 526 + ... = 2008
• 1/3 + 2/44 + 1/526 + ... = 1

Everything clear? --85.178.52.149 (talk) 10:30, 10 February 2008 (UTC)[reply]

Yes, but as I said, it's not possible to solve it explicitly - you need to just do trial and error. -mattbuck (Talk) 10:49, 10 February 2008 (UTC)[reply]

I will give you (or someone else) a barnstar if you will get out the answer. Is the price worth enough? ;) --85.178.52.149 (talk) 10:55, 10 February 2008 (UTC)[reply]

If I give you the answer then you learn nothing. If I give you a hint and you follow it and work out the answer for yourself then you learn something. So here is a hint:

• If we take a repetitions of the number b then their sum is ab and the sum of their reciprocals is a/b.
• So if we can make b equal to a then the sum of the reciprocals will be 1. But this will only work if the sum ab is a square number. For example, 9 = 3x3 = 3+3+3 and 1/3 + 1/3 + 1/3 = 1.
• 2008 isn't a square number, so we can't use this idea directly. But suppose we can make b equal to 2a. Then a/b = 1/2 and ab = 2a². In other words, if the sum is twice a square then we can partition it so that the sum of the reciprocals is 1/2. For example 18 = 2x3^2 = 6+6+6 and 1/6 + 1/6 + 1/6 = 1/2.
• So if we could find two numbers each of which is twice a square and which sum to 2008, then we could partition each of these numbers into a set of smaller numbers whose reciprocals sum to 1/2. Then when we take all the smaller numbers together, they will sum to 2008 and their reciprocals will sum to 1. For example 26 = 18+8 = (6+6+6) + (4+4).
• If 2008 is the sum of two numbers that are twice squares, then 1004 would be the sum of two squares. Unfortunately, 1004 isn't the sum of two squares, so this doesn't work either.
• But, going back to our original line of thought, suppose we make b equal to four times a. Then a/b = 1/4 and ab=4a². So if we can make 2008 equal to the sum of four numbers, each of which is four times a square, then we are done.
• To find four numbers, each of which is four times a square, which sum to 2008 then we need to find four squares whose sum is 502. And Lagrange's four-square theorem says that any integer can be expressed as the sum of at most four squares, so this approach is looking hopeful.

I think that is enough of a hint. I'll let you take it from there. (Incidentally, there are many solutions to this problem - I have just found 6 or 7 different solutions using this method). Gandalf61 (talk) 14:55, 10 February 2008 (UTC)[reply]

At first, thank you! --- I have now this: 20^2 + 10^2 + 2*2^2 = 502. But it doesn't work because i got two times the same number. Can you help me again? --85.178.31.13 (talk) 17:16, 10 February 2008 (UTC)[reply]

I think you mean 20^2 + 10^2 + 2*1^2 = 502. Having 1^2 appear twice is fine - this still leads to a solution to the original problem. Gandalf61 (talk) 17:23, 10 February 2008 (UTC)[reply]

No, it doesn't work. 2*1^2 is not agreeable because 1^1 is 1, the reciprocal is 1, too! So the solution is wrong (I had written 2*2^2, because I wanted to get 508. This was a mistake by me). 2*2^2 would also be a problem: 2*4 means 4*(2*4) at the end. But the reciprocals would be then 2 (and this is surely wrong). --85.178.31.13 (talk) 18:11, 10 February 2008 (UTC)[reply]

You don't want to take the reciprocal of 1, you want to take the reciprocal of 4*1. --Tango (talk) 18:54, 10 February 2008 (UTC)[reply]

?? The reciprocal must be at the end only 1. --85.178.31.13 (talk) 19:32, 10 February 2008 (UTC)[reply]

Yes, and you'll end up with 1/4 four times, which adds up to 1. Read through Gandalf61's hint again, it might take a few tries to get your head round it, but it does work. --Tango (talk) 20:22, 10 February 2008 (UTC)[reply]

Please guys, can you give me just one solution? Just one for the understanding. --85.178.31.116 (talk) 14:25, 12 February 2008 (UTC)[reply]

But you almost have the solution yourself. Hint: 400+100+1+1=502, not 2008. So you don't add with 1. Taemyr (talk) 15:18, 12 February 2008 (UTC)[reply]

I came up with a different solution, in which all the integers are powers of 2. Note the following sets of numbers, for which the recipricals add up to 1 (and the sum of the numbers is in parentheses):

2, 2 (4)

2, 4, 4, (10)

2, 4, 8, 8 (22)

2, 4, 8, 16, 16 (46)

2, 4, 8, 16, 32, 32 (94)

Also note that you can take one of these sequences, and replace any number with two copies of twice that number, which increases the sum by three times the number you removed. So, for example, if I take that last sequence, and replace the 8 with two 16s, I get:

2, 4, 16, 16, 16, 32, 32 (118)

And the sum from the previous sequence was increased by 3*8=24.

So, for example, if I wanted to get a sum of 1000, I would start with the nearest I could get using only the last number twice:

2, 4, 8, 16, 32, 64, 128, 256, 256 (766)

Now I have to increase the sum by 1000-766=234. Which means picking numbers to replace that add up to 234/3=78. And 64+8+4+2=78, so we remove each of those numbers, and replace them with two copies of their doubles, giving us

4, 4, 8, 8, 16, 16, 16, 32, 128, 128, 128, 256, 256 (1000).

This method, with this particular set of initial sequences only works when the desired sum is congruent to 4 mod 6, but fortunately 2008 is such a number. Other sequences may be able to give other numbers, however. Chuck (talk) 22:22, 14 February 2008 (UTC)[reply]

Can you get the circumference if you only have the diagonals e and f (nothing else is given)? Could you write down the formula to get this? --85.178.52.149 (talk) 10:46, 10 February 2008 (UTC)[reply]

I wouldn't have thought so. There are 3 variables in a parallelogram - the height, the overhang, and the length of the long side. Three pieces of information cannot be uniquely determined with only two equations. -mattbuck (Talk) 12:18, 10 February 2008 (UTC)[reply]

Ok...maybe you can do something with their values; e is 7, f is 9. Maybe now (all sides are natural numbers)? --85.178.52.149 (talk) 12:28, 10 February 2008 (UTC)[reply]

Well, if we take h to be height, a to be the length of the long side and b to be the overhang:

${\displaystyle e={\sqrt {h^{2}+(a-b)^{2))))$ and ${\displaystyle f={\sqrt {h^{2}+(a+b)^{2))))$ which gives ${\displaystyle 4ab=f^{2}-e^{2))$, but again, you can't get further unless you assume both a and b are integers, in which case you would at least get a finite number of possibilities which you could then work out, remembering that ${\displaystyle {\sqrt {b^{2}+h^{2))))$ must be an integer. -mattbuck (Talk) 13:06, 10 February 2008 (UTC)[reply]

You'll need 1 more information since you can draw a number of parallelograms with 7 and 9 as the length of diagonals. —Preceding unsigned comment added by Ftbhrygvn (talk • contribs) 15:27, 10 February 2008 (UTC)[reply]

What is it about the universe that makes logic work. I don't see why it needs to.

Also, does logic work everywhere, or are there some places (maybe the bermuda triangle?) where logic doesn't hold, and something that would of necessity follow, elsewhere, there it does not??? —Preceding unsigned comment added by 79.122.117.186 (talk) 11:34, 10 February 2008 (UTC)[reply]

I think this is a prime example of the possible application of the anthropic principle - put simply, if the universe weren't logical, we wouldn't be asking the question. -mattbuck (Talk) 12:16, 10 February 2008 (UTC)[reply]

It depends on what you mean by "logic". If you're asking why the world seems to make as much sense as it does, I guess we just got lucky. Not extremely lucky, considering how much of it nobody understands even now, but still. If you're asking why a particularly popular logical system works so well, it's because it was designed by some very smart philosophers to mimic the real world (which, as I pointed out, usually makes sense) and has been improved over many years. If there was anything wrong with it, someone would have noticed by now. In fact, flaws have been pointed out every once in awhile, and the system has been improved more than once. You've heard of it because it works, and it works because people have fixed it until it does. It applies, by definition, to everything it applies to, and nothing else. The particular system I linked to, for instance, only applies to statements that are either completely true or completely false. Black Carrot (talk) 00:57, 11 February 2008 (UTC)[reply]

I'm curious about the function

${\displaystyle w(n)=\int _{-\infty }^{\infty }(1-(1-\alpha )^{n}-\alpha ^{n})\,dx}$ ,

where ${\displaystyle \,\alpha =\Phi (x)}$, i.e. the cumulative standard normal distribution, and n is an integer greater than one. I suspect that the usage of α without any indication that it is a function of x may be non-standard (at least, it had me confused). However, I wanted to reproduce it exactly as it appears in the paper in which I found it: Tippet, LHC. On the Extreme Individuals and the Range of Samples Taken From a Normal Population. Biometrika vol 17, 364-387, 1925. It is my understanding that this function gives the expected value of the range of a sample of size n, taken from a standard normal distribution. In the field of Statistical process control, the function w is known as d2.

These are my questions:

• For n = 2, it is easy to show that the integral is numerically equal to 2/sqrt(π) within machine precision, and I feel reasonably certain that 2/sqrt(π) is indeed the exact value. I would like to know how one determines whether this is the case. As I am neither a statistician nor a mathematician, I would need the details spelled out.
• Can this integral be expressed in terms of simple functions for values of n greater than 2? If so, how?
• Is my suspicion avove, that the notation today would be considered non-standard, correct? If so, what would standard notation be?

Thanks. --NorwegianBlue ^talk 14:45, 9 February 2008 (UTC)[reply]

I haven't done much probability, so I'm not really sure what's going on there, but it does seem odd to me to be integrating the CDF - normally you integrate the density function to get the CDF, so I'm pretty much lost. As for the standard notation, wouldn't that just be:

${\displaystyle w(n)=\int _{-\infty }^{\infty }(1-(1-\Phi (x))^{n}-\Phi (x)^{n})\,dx}$?

It's not particularly unusual not to use a notation that doesn't explicitly state a dependence, although when you're only dealing with functions one variable, I can't see much point in not being explicit. --Tango (talk) 13:40, 10 February 2008 (UTC)[reply]

Thanks! Well, the function works, and there's no doubt that it's the CDF that is to be integrated. Below is a very crude implementation in R. The standard normal density in R is dnorm, while the distribution function is pnorm:

     d2 = function(n, distribution=pnorm, lolim=-12, hilim=12)
     {
         dx = 0.01
         steps = seq(lolim, hilim, by=dx)
         alpha=distribution(steps)
         rectangles = (1 - (1-alpha)^n - alpha^n)*dx
         return(sum(rectangles))
     }

And here's the output when testing it:

     > format(d2(2), digits=15)
     [1] "1.12837916709551"
     > format(2/sqrt(pi), digits=15)
     [1] "1.12837916709551"

--NorwegianBlue ^talk 14:08, 10 February 2008 (UTC)[reply]

A little more experimenting shows that d2(3)=3/sqrt(pi). However d2(4) = 2.058751, while 4/sqrt(pi) = 2.256758. Can anyone see a pattern? --NorwegianBlue ^talk 00:32, 11 February 2008 (UTC)[reply]

I can't offer much, but Mathematica confirms that ${\displaystyle w(2)={\tfrac {2}{\sqrt {\pi ))))$ and ${\displaystyle w(3)={\tfrac {3}{\sqrt {\pi ))))$. It is also worth noting that ${\displaystyle w(1)=0}$. It doesn't readily succeed in calculating ${\displaystyle w(4)}$ symbolically. The numerical value is 2.058750746..., for which Plouffe's inverter gives no match. So I doubt there is any significantly simpler way to express it. -- Meni Rosenfeld (talk) 09:05, 11 February 2008 (UTC)[reply]

Thanks a lot, Meni. --NorwegianBlue ^talk 13:02, 11 February 2008 (UTC)[reply]

sorry, I doin't mean to be incendiary, but I wonder if anyone has heard of it being more difficult for a non-Jew to become a mathematician for whatever reason? —Preceding unsigned comment added by 79.122.117.186 (talk) 11:42, 10 February 2008 (UTC)[reply]

You what? I can't say I've known anyone ever ask about religion here in the UK. Where exactly are you asking about? -mattbuck (Talk) 12:13, 10 February 2008 (UTC)[reply]

united states. —Preceding unsigned comment added by 79.122.117.186 (talk) 12:35, 10 February 2008 (UTC)[reply]

Why do you ask? What makes you think there would be any problem? Is there even a higher proportion of Jews in the US Mathematical community than in the general population? --Tango (talk) 13:31, 10 February 2008 (UTC)[reply]

Yes, there was a period of about 20 years when only Jews were allowed to become mathematicians. Fortunately the Supreme Court overturned the law. Strad (talk) 16:39, 10 February 2008 (UTC)[reply]

(My joke removed.. apologies for those who missed it - still wanting attention though,, )87.102.79.203 (talk) 19:34, 11 February 2008 (UTC)[reply]

This is the most nonsensical statement I have ever heard. Phils 16:51, 10 February 2008 (UTC)[reply]

Also consider the fact, that only Roman Catholics may become professonal popes. The VEOB (Vatican Equal Opportunity Board) appears to be rather conservative, when it comes to gender, race and creed. But then again, I only ever applied for a missionary position...

PS: Re Strad´s comment: Me thinks your instrument has been tuned by the late Beethoven (one of his many places of abode is just next door but one to the block I live in). --62.47.136.108 (talk) 18:10, 10 February 2008 (UTC)[reply]

From my experience the mathematical community is particularly indifferent to questions of race/religion. Its how good at maths they are which is the most important thing. There are quite a number of notable jewish mathematicians see List of Jewish American mathematicians.--Salix alba (talk) 20:08, 10 February 2008 (UTC)[reply]

The question was about racism towards *non*-Jews. I agree with the rest of your point, though - my Maths department barely has two people from any one country! I once got all the appropriate flags off commons for a poster I was making about the department (as part of a summer job) - it took quite a while! --Tango (talk) 20:12, 10 February 2008 (UTC)[reply]

Maybe the question should be do Jewish get an easier ride due to the large list of very notable jewish mathematicians? BTW I find the whole suggestion of this topic rediculous.--Dacium (talk) 22:20, 10 February 2008 (UTC)[reply]

my question (not the subject, the actual body of the question) asks "is it more difficult for non-Jews" -- how is this not the same as asking "do Jewish get an easier ride"???? I mean, if it's easier for Jews then it means it is harder for non-Jews. You can't say "It's not any harder to be a NASCAR driver if you're a woman, it's just easier if you're a man..." That's a completely nonsensical sentence!!!! —Preceding unsigned comment added by 212.51.122.22 (talk) 22:41, 10 February 2008 (UTC)[reply]

As far as I know, there's no bias against non-Jews. In fact, the only bias I've ever seen mention of until today was in one of Feynman's several autobiographies, where it claimed that Jews had a harder time because they were unpleasant to work with. With the sole exception, naturally, of Feynman himself. Black Carrot (talk) 01:18, 11 February 2008 (UTC)[reply]

I have two large matrices A and B, and I have to find the inverse of A + c*B, for many different values of c. Is there a more efficient way of doing this rather than finding the inverse of the sum A + c*B every time. Thanks, deeptrivia (talk) 19:46, 10 February 2008 (UTC)[reply]

Well, you can do it in terms of c and plug the values in at the end. It might end up being quite a mess (for n-by-n matrices, you'll potentially be dealing with (n-1)-degree polynomials in c divided by n-degree polynomials in c - if A and B are nice, some of it might cancel out, though). I don't know any formula for directly finding the inverse of a sum of matrices - matrix addition doesn't behave very nicely - but that doesn't mean there isn't one out there somewhere. If you know any nice properties of A and B, it might make things easier. The best option with anything to do with matrix inverses is just to do it on a computer, really. --Tango (talk) 20:20, 10 February 2008 (UTC)[reply]

I don't know if this is in any way helpful, but (A + cB)⁻¹ = (I + cA⁻¹B)⁻¹A⁻¹, and for sufficiently small c you can use (I + cM)⁻¹ = I − cM + c²M² − c³M³ + ... .  --Lambiam 20:51, 10 February 2008 (UTC)[reply]

Does that first formula have a name? I'm curious to see the derivation. The binomial approximation is only useful if an approximate answer is sufficient, which depends on what deeptrivia is trying to do - plugging it all into Mathematica or Maple seems like an easier approach if you need more than the first order term, anyway. --Tango (talk) 21:00, 10 February 2008 (UTC)[reply]

Such CASes are no match for C code optimized for this purpose, which could be what deeptrivia is after. If performance is really crucial and c is assumed to be small, a power expansion may be the best way to go. Of course, you don't want to calculate it naively - a good way to calculate ${\displaystyle (I+X)^{-1))$ is to iterate ${\displaystyle Y_{n+1}=I-XY_{n))$, and an even better way is probably to calculate ${\displaystyle (I+X)^{-1}=(I-X)(I+X^{2})(I+X^{4})(I+X^{8})\cdots }$. -- Meni Rosenfeld (talk) 08:52, 11 February 2008 (UTC)[reply]

If by "many", we're talking thousands, maybe. It's easier just to plug the general case into a CAS, get a general answer, and plug the values for c in at the end - substituting in a value for a variable is a pretty quick function on any platform, it's just evaluating n^2 rational functions. --Tango (talk) 11:05, 11 February 2008 (UTC)[reply]

Yes, n^2 rational functions, each of which having ${\displaystyle O(n)}$ terms, totaling an ${\displaystyle O(n^{3})}$ calculation for each new c, not to mention the expensive preprocessing. A calculation along the lines suggested by Lambiam is, if I'm not mistaken, ${\displaystyle O\left(n^{2.376}\log {\frac {\log \epsilon }{\log {\|cA^{-1}B\|))}\right)}$ (where ε is the tolerable error) per c. We can hybridate the approaches - invert symbolically, and then expand each term in a way that will facilitate the calculation. It all depends on how many cs we want (for a huge number, preprocessing cost is irrelevant), the size of the matrices and the error tolerance. -- Meni Rosenfeld (talk) 12:03, 11 February 2008 (UTC)[reply]

I don't know if it has a name, but here is a derivation:

(A + cB)⁻¹ = (I(A + cB))⁻¹

                  = (AA⁻¹(A + cB))⁻¹

                  = (A(A⁻¹A + cA⁻¹B))⁻¹

                  = (A(I + cA⁻¹B))⁻¹

                  = (I + cA⁻¹B)⁻¹A⁻¹.

 --Lambiam 02:14, 11 February 2008 (UTC)[reply]

This is a Neumann series. Nevermind, he was referring to the first formula. Phils 02:05, 13 February 2008 (UTC)[reply]

Thanks a lot for your responses. I have a question about ${\displaystyle (I+X)^{-1}=(I-X)(I+X^{2})(I+X^{4})(I+X^{8})\cdots }$. What conditions are sufficient for this formula to converge after a certain number of terms. X is a pretty large symmetric complex matrix. Regards, deeptrivia (talk) 22:58, 11 February 2008 (UTC)[reply]

${\displaystyle \|X\|<1}$ is sufficient, and I think it is also necessary. -- Meni Rosenfeld (talk) 23:48, 11 February 2008 (UTC)[reply]

Sorry if I've asked this before, but is {2, 2} a set or a mistake. I know it's a legitmate mmultiset... —Preceding unsigned comment added by 212.51.122.22 (talk) 20:49, 10 February 2008 (UTC)[reply]

It is a permissible notation for the same set as {2}. So it is a set, yes, with exactly one element.  --Lambiam 20:52, 10 February 2008 (UTC)[reply]

So is the answer to "How many ways can you type a set if you only have braces commas spaces and the number 2" infinite, since you can type {}, {2}, {2,2}, {2,2,2} etc and in all these cases you're typing a set??? (And the answer to that question is not, say, 2, namely {} and {2}, the other typings not being instancse of typing a set...) —Preceding unsigned comment added by 212.51.122.22 (talk) 22:39, 10 February 2008 (UTC)[reply]

{2,2} is notation for a set - the set containing the number "2". It's important to distinguish between the object and the notation. {2} and {2,2} are clearly different notations, but they refer to the same set. It's also important to note that 2,{2},((2)),{2,{2)) etc. are all distinct mathematical objects (in fact, they are all sets using the most common model of natural numbers where 2={0,1}=((},(())}). --Tango (talk) 23:21, 10 February 2008 (UTC)[reply]

You don't even have to get into trickery with sets of sets to get different sets using the symbols listed: sets such as {2, 22, 222} are permitted by the condition that "you only have braces, commas, spaces, and the number 2". -- AJR | Talk 01:23, 11 February 2008 (UTC)[reply]

I disagree, there is a difference between the number 2 and the digit 2. Taemyr (talk) 09:30, 11 February 2008 (UTC)[reply]

Agreed. The question explicitly said "number 2". The fact that in the base ten positional notation for numbers, the same shape is drawn when denoting other numbers is irrelevant. --Tango (talk) 10:59, 11 February 2008 (UTC)[reply]

Talk:Stochastic matrix/Question 3