Wikipedia:Reference desk/Archives/Mathematics/2012 March 6

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

March 6

unwrapped script

here is my unwrapped script:

http://codepad.org/bU68pFDc

it busts! Run sequentially, how many times in a row will it bust, on average? (Before it exits with the success condition.) — Preceding unsigned comment added by 80.99.254.208 (talk) 10:28, 6 March 2012 (UTC)

sorry, it has a mistake (it shouldn't exit but just exit the loop), but you get the idea... — Preceding unsigned comment added by 80.99.254.208 (talk) 10:30, 6 March 2012 (UTC)

So you have a total which starts with 300 and have a 1 in 5 chance of subtracting 25 and 4 in 5 chance of adding 6.375. You bust if the total is less than zero. It looks like your doing a form of Random walk, there are formula there to consider.--Salix (talk): 12:27, 6 March 2012 (UTC)

Right, formulas, or if not formulas at least empirical (monte carlo) testing. This is why I chose this referrence desk... any takers? 78.92.82.6 (talk) 15:18, 6 March 2012 (UTC)

OK lets make it a bit simpler, at each step the value increases by 6.375 and there is a 1 in 5 chance that the value decreases by 31.375. So the value after n trials when there are k successes is 300+6.375*n-31.375*k. To bust we need $300+6.375*n-31.375*k<0$ or $9.563+0.203n<k$. Not we have a binomial distribution so the chance of k successes in n trials is ${n \choose k}p^{k}(1-p)^{n-k}$ call this $b(n,k)$. To work out the chance of ever busting, first work out $p_{1}2=b(12,12)$ the chance of failing on the 12^th time, the first time you could fail. Then it starts to get messy, you can't fail on 13th go but can on 14th, with chance $p_{1}4=b(12,11)*b(2,2)$, i.e. 11 successes in first 12 goes and 2 in the next two.--Salix (talk): 08:21, 7 March 2012 (UTC)

The probability for 'success' is p = 0.2. The mean number of successes after n = X² trials is μ = pn = 0.2X². The standard deviation is σ = √(np(1−p)) = 0.4X. The number of successes is k ≈ μ±σ = 0.2X²±0.4X. The value is 300+6.375n−31.375k ≈ 300+6.375X²−31.375(0.2X²±0.4X) = (300/X+0.1X±12.55)X. The minimum value of 300/X+0.1X is 10.9545. This value occurs for n = X² = 300/0.1 = 3000. It has busted at n=3000 if 10.9545±12.55<0. The probability for this to happen is excellently approximated by the normal distribution. Pr(10.9545±12.55<0) = Pr(±1<−0.872869) = Φ(−0.872869) = 0.191367. Bo Jacoby (talk) 08:07, 13 March 2012 (UTC).

Proof that a series tends to zero

In my lecture notes, my professor has used the following claim:

$\log {n}\sum _{a=1}^{\lfloor n\cdot 2^{-{\sqrt {\log {n}}}}\rfloor }{n \choose a}e^{-(a\log ^{2}n)/3}<0.1$ for n large.

In particular, I guess he's claiming that: $\log {n}\,\sum _{a=1}^{\lfloor n\cdot 2^{-{\sqrt {\log {n}}}}\rfloor }{n \choose a}e^{-(a\log ^{2}n)/3}\to 0$ as $n\to \infty$, if I've understood my notes right. Is there some nice obvious reason why this sum tends to 0? I've tried to bound it so that things work out but every time I keep trying to put the upper bound in things get messy. (That upper bound is $\lfloor n\cdot 2^{-{\sqrt {\log {n}}}}\rfloor$, in case it's hard to read above the sum.)

Is there some clever, concise way to see that this sum obviously tends to 0? I'm happy to use whatever machinery is necessary to prove it, although of course a more direct solution which appeals to few external results would be preferable. Thanks for the help! Otherlobby17 (talk) 13:40, 6 March 2012 (UTC)

Change variables $n=e^{t}$ and note that ${\binom {n}{a}}<e^{at}$ so the sum can be compared to the integral

$t\int _{0}^{\infty }e^{xt-xt^{2}/3}dx$ which tends to zero. Sławomir Biały (talk) 14:32, 6 March 2012 (UTC)

Ah, very nice! I forgot to mention, the logarithms here are in base 2 - that won't cause any problems will it? Otherlobby17 (talk) 14:36, 6 March 2012 (UTC)

No. Just replace all es with 2. Sławomir Biały (talk) 15:03, 6 March 2012 (UTC)

Thanks Sławomir! Otherlobby17 (talk) 18:49, 6 March 2012 (UTC)

what is the problem with this proof that 2=3

Let's start with the hypothetical equation
0 ?= 0
Now let's add 5 to both sides to it
5 ?= 5
let's square the thing
25 ?= 25
and let's take the square root in two different ways
-5 ?= 5
now let's multiply both sides by two
-10 ?= 10
and add 50
40 ?= 60
dividing by ten yields
4 ?= 6
which reduces to
2 ?= 3
Now obviously if two equals three, then three equals four, four equals five, and so on. This seems reasonable to me based on the heap argument, but is it mathematically true? 78.92.82.6 (talk) 15:30, 6 March 2012 (UTC)

Just because there are two square roots for a number does not mean they are equal. --LarryMac | Talk 16:17, 6 March 2012 (UTC)

When you "take the square root two different ways", you're not making a move that preserves equality. That is, unless 1=-1. This is not as absurd as it may sound, for instance 1=-1 in the finite field GF(2), and 1+1=1-1=0. So the "proof" is unsound in the context of real numbers, but is sound for the finite field of two elements, in which 2=3=1+...+1=n=0. SemanticMantis (talk) 16:19, 6 March 2012 (UTC)

If 2=0, then you can't divide by 10 or 2, which were the last two steps OP took. --COVIZAPIBETEFOKY (talk) 18:21, 6 March 2012 (UTC)

Yes, good point. I guess I was caught up with explaining how it could be valid in the proper context, and that the OP's intuition was correct, in that if if 2=3, then m=n, for all m,n>2. SemanticMantis (talk) 20:04, 6 March 2012 (UTC)

In addition, there is only one answer to sqrt(25), which is positive 5. It is true that there are two solutions to x^2 = 25, but you cannot say that both solutions are equal, just because they both make an equation true. — Preceding unsigned comment added by KyuubiSeal (talk • contribs) 20:07, 6 March 2012 (UTC)

Generally, the mistakes in such "proofs" can be found by checking each step separately. In this case it's easy to see that "25=25", but not "-5=5".

Also, as mentioned before, "let's take the square root in two different ways" is an obvious mistake. If you want to make such "proof" into a slightly harder and (perhaps) more entertaining puzzle, first change "25=25" into "(-5)^2=5^2", which is actually true. Then getting to "5=-5" will be less obviously wrong.

By the way, we have an article about such puzzles: "Mathematical fallacy". Still, I wonder if there is any research about their "entertainment value" that could be cited..? --Martynas Patasius (talk) 20:28, 6 March 2012 (UTC)

Taking square root of both sides should maintain quantities at both sides just as the way you started the equation 0=0. because that is the way you defined your statement first. OP did not start the statement by saying -1<1 but rather something equals something, so the equality in the quantities at both sides should be maintained throughout.--NII AFRAH (talk) 22:46, 6 March 2012 (UTC)

No, taking the square root of both side does not "maintain quantities". You can see this because it leads to the obviously incorrect equation -5 = 5. This is because the square root function on positive numbers has two branches - the positive square root and the negative square root. You cannot choose to take one branch on one side of the equation and the other branch on the other side, and still expect to have an equality. An analogy - from my town's train station I can take a train going north to A and a train going south to B, but this does not mean that A and B are the same place. Gandalf61 (talk) 09:11, 7 March 2012 (UTC)

Solving all quintics by Algebraic means

I have submitted my paper on General Reducibility and Solvability of all polynomial equations to a maths journal. Thank you all for your comments made on it and i am sure by June this year it will be published. To answer Gandalf61 question on x^5 - x + 1 = 0, it becomes very simple and more interesting to give out solutions which have troubled mathematicians for many years. I will make sure to provide the solutions to all quintics and above after my paper has been published in the journal to this forum (network). Please be expecting it in the late June of this year.--NII AFRAH (talk) 22:29, 6 March 2012 (UTC)

Which math journal ? StuRat (talk) 22:37, 6 March 2012 (UTC)

Please see the Abel–Ruffini theorem. A polynomial can be solved with radicals if and only if its Galois group over the rationals is solvable. There are quintics, such as x⁵ - x + 1 = 0, which have Galois group S₅, and this isn't a solvable group. Of course you can still solve quintics and higher degree polynomials numerically, and there are many techniques for doing this, but expressing the roots exactly in terms of radicals is proven not possible in general. Rckrone (talk) 02:00, 7 March 2012 (UTC)

Perhaps the OP is going on a trip like Hardy who is said to have told a conference he was going to present a proof of the Riemann hypothesis before going on a sea trip, on the basis God wouldn't let him die with people thinking he had a proof. Dmcq (talk) 12:02, 7 March 2012 (UTC)

I understand Rckrone, but i want to ask him if he believes that there should be a mathematical method to express the equation x⁵ - x + 1 = 0 to what i am calling a standard equation t⁵+bt³+ct²+dt+e=0 where x = g(t), solve for t and hence for x as Leonhard Euler believed? this is why i am saying there are severe problems with the correctness of all of Galois theory and Abel–Ruffini theorem. i want everyone to follow this post since some theories in algebra need severe corrections. --NII AFRAH (talk) 14:09, 7 March 2012 (UTC)

As a general question, given that your above equation doesn't have a quartic term, is there a difference in the solvability of quintics in general vs. quintics with zero quartic terms?Naraht (talk) 14:45, 7 March 2012 (UTC)

If you are talking about suppressing the term in t⁴, then that is trivial. If we have a general monic quintic:

$x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+e=0$

then the sum of the roots is -a. Since we know there are five roots, a simple linear transformtion

$t=x+{\frac {a}{5}}$

should give us a quintic in t whose roots sum to 0, and hence it will have no term in t⁴. Indeed, we have

$x^{5}=(t-{\frac {a}{5}})^{5}=t^{5}-at^{4}+{\text{ terms in }}t^{3},t^{2}\dots$

$ax^{4}=a(t-{\frac {a}{5}})^{4}=at^{4}+{\text{ terms in }}t^{3},t^{2}\dots$

and we can see that the terms in t⁴ cancel. I don't see how this helps you produce an algebraic solution of the general quintic. Gandalf61 (talk) 14:59, 7 March 2012 (UTC)

I believe we should just wait till June like the poster said. If it has been submitted for publication what's the point of speculating or arguing about something which has fooled Abel and Galois and anyone else who has looked at it in the last two hundred years without some preprint copy to look at? Dmcq (talk) 15:21, 7 March 2012 (UTC)

Fair enough. But I am worried this might be like one of those films that don't quite live up to their trailers. Gandalf61 (talk) 15:31, 7 March 2012 (UTC)

OK, so for any general polynomial of degree d > 1, we can generate a polynomial with the d-1 term suppressed. If I remember that sort of suppression was the first step in solving the cubics, but I was curious as to whether he was saying something about the solvable quintics. Given what you showed, the answer to my question is no. Thank youNaraht (talk) 15:42, 7 March 2012 (UTC)

I'd also like to note that there *are* areas of mathematics that I find the idea of a someone who doesn't do mathematics for a living extending the subject. I'd put Tiling in that area for example, possibly some of knot theory, and semi-regular higher dimension polytopes (possibly even someone who has mystical feelings about them). Those are places where the solutions tend to be at "Yes, I agree with everything that's out there currently and here is something more" rather than "A chunk of the foundation of a branch of current mathematics that has been accepted by those who do Mathematics for a living for the last century is wrong".Naraht (talk) 15:57, 7 March 2012 (UTC)

I think i have to ask Gandalf61 this fine question that did he believe x⁵ - x + 1 = 0 can be rewritten in the form t⁵+bt³+ct²+dt+e=0 by reduction? as for solving t⁵+bt³+ct²+dt+e=0, it is included in my paper for publication.so lets wait till june as i said earlier. --NII AFRAH (talk) 16:39, 7 March 2012 (UTC)

As has been said above, x⁵ - x + 1 is already in that form. Set b = 0, c = 0, d = -1, e = 1. Gandalf61 (talk) 17:02, 7 March 2012 (UTC)

Gandalf61 and everybody else can see that x⁵ - x + 1 = 0 can be written as t⁵+0t³+0t²+(-1)t+1=0 with t=x, b=c=0, d=-1, e=1. The problem is that you haven't posted and clearly don't have a solution to this equation in terms of radicals. Here is a numerical approximation to test your alleged solution: x = -1.16730397826141868425604589985484218072056037152548903914008. PrimeHunter (talk) 17:03, 7 March 2012 (UTC)

Gandalf61 and everybody should use the substitution x=5/(4+t) to reduce the equation x⁵ - x + 1 = 0 to a form t⁵+at⁴+bt³+ct²+e=0 before transforming it by the substitution t = g(y) to get y⁵+by³+cy²+dy+e=0. Please wait till i bring the solution of y⁵+by³+cy²+dy+e=0 after whole paper has been published. note that there are important mathematical theorems which can be found in my paper and PrimeHunter don't worry, you will be the first to have the solution of all quintics and above. we are seeing it to be mystery but it is real; we have answered the problems that have existed for so many years. — Preceding unsigned comment added by NII AFRAH (talk • contribs) 18:01, 7 March 2012 (UTC)

When the journal rejects your paper, please do publish it yourself and give us a link. I could do with a laugh. --Tango (talk) 20:25, 7 March 2012 (UTC)

Dear Mr. Afrah,

You are obviously a gifted young person with a passion for mathematics, and for this you should be congratulated and nurtured.

However, I honestly think you got some bad advice here awhile back, to try to finalize your work and submit it for peer review. It would have been better of the reference desk to recommend that you spend a year learning group theory and then Galois theory so that you can understand the proof of the Abel-Rufinni theorem. These techniques underlie much of modern number theory, and there are real mathematical challenges that are within reach of a determined student. Unfortunately, members of this forum encouraged you to pursue your own course, which is fruitless. You will be disappointed with the results. The paper will be rejected from any serious journal. You should abandon this endeavor now, and nurture your mathematical talents with hard work. If you are interested in number theory, a good starting point for a research track is Bushnell and Heniart's "The Local Langlands Conjecture for GL(2)". If you get enough background to read this (e.g. Knapp "Basic algebra", Knapp "Advanced algebra", and a book on class field theory e.g., the one by Milne which is free), then you will be on your way to a real research track, together with real validation from your peers. This course that you are pursuing now will only discourage you. It will be hurtful and counterproductive. You will never get the recognition you think you deserve from your peers. Dmcq's sarcastic treatment of you is a taste. The world will have lost another young gifted person for lack of good guidance. I hope you take this advice and abandon your present course. By careful studying of the texts I have suggested, you can easily learn enough to apply somewhere for a PhD. When you are a professor of mathematics, then you can look back on your old work and see if it is still worth publishing. If so, you will be taken more seriously by your peers. Sincerely, Sławomir Biały (talk) 21:37, 7 March 2012 (UTC)

Well I hope they take your advice even if they didn't take any the last time they were here and persisted. And you are right, I shouldn't be sarcastic. Anyway on the actual books line the first book I bought myself on it was Edgar Dehn 'Algebraic Equations' and I thought it was very good but a more modern work like Ian Stewarts's 'Galois Theory' might be better - he's a good writer. Dmcq (talk) 22:02, 7 March 2012 (UTC)

Books can't always substitute a good teacher. without feedback, an erroneous interpretation may persist, Usually it's something we learned early on, an elementary fact we take for granted. Wrong interpretations can give correct results, and when they finally lead to a contradiction, you don't doubt that what you have known for years. Can't think of a specific example right now, but it has happened to me more than once, in chemistry, physics, maths electronics etc.. 84.197.178.75 (talk) 05:56, 8 March 2012 (UTC)

We only advised writing the paper once it became clear that we weren't being listened to with other advice. Sometimes people only learn by making their own mistakes. --Tango (talk) 11:53, 8 March 2012 (UTC)

I will always thank my creator for revealing these two powerful equations to me ( vic-emmeous and reneagbed equations). equations which A-R and Galois did not see before coming out with their false theorems.equations which are changing false mathematical theorems accepted in Algebra for so many years now and Sławomir Biały and others believed strongly that they are true without even making an attempt to confer to my paper.I have read particularly on Abel and Galois because my discovery is related to their field of research and found out that all what they are saying are imaginations(assumptions)besides my paper and theirs are talking different things altogether. what i am expecting them to do is to put more morale in this issue because for all you know Galois and others knew something of that sort is there and that is why they searched and searched and concluded impossible because they did not get it. that does not mean the world should accept findings from Abel, Galois and few others only because for all you know Wrong interpretations can give correct results.so please have a second look.--NII AFRAH (talk) 08:04, 8 March 2012 (UTC)

One of the most wonderful things about mathematics is that we don't have to trust those that came before us. Galois, Abel and others certainly weren't infallible and they could have made mistakes in their proofs of their theorems about quintics. However, we can check those proofs ourselves. While I doubt I could ever have come up with Galois theory myself, I was able to take a course in it as an undergrad and learn more than enough about the field to be able to follow the proof myself. That means I know it is impossible to find a general solution to a quintic in radicals, not because Galois says so, but because I've proven it myself. Therefore, I know that whatever you have come up with must be incorrect because it contradicts something that I have proven. I don't accept the findings of Galois and Abel, I accept my own findings. --Tango (talk) 11:53, 8 March 2012 (UTC)

I'm afraid I am all too fallible and quite liable to think I have a proof when I don't, so I trust the loads of people before me who have checked it more than I do myself. Dmcq (talk) 14:10, 8 March 2012 (UTC)

Well, yes, I agree, but I still find theorems more convincing when I've been able to go through the proof and understand it myself. Whether you are trusting yourself or the thousands of other mathematicians that have gone through it, you certainly aren't just trying the person that originally came up with it. --Tango (talk) 16:26, 8 March 2012 (UTC)

All I have to say is "I will wait till June to show the world my earth-shaking findings on Algebra". Thanks.--NII AFRAH (talk) 18:10, 8 March 2012 (UTC)

I guess someone has to cite the instruction from the header: "The reference desk does not answer requests for opinions or predictions about future events. Do not start a debate; please seek an internet forum instead.". Now, do you have an actual question? And if it is "Will anyone believe if a total stranger (who is not a certified professional) will claim to have done something that has been proved to be impossible, but avoid giving any evidence of the feat?", then the answer is "No.". Or "No, for almost all human beings.", if you will ask nicely. --Martynas Patasius (talk) 20:46, 8 March 2012 (UTC)

Considering you can't be bothered to tell us which journal (so we could theoretically keep an eye out for it) or tease us with a solution to $x^{5}-x+1=0$ (without even necessarily telling us how you got it), I'm not holding my breath. --Kinu ^t/_c 23:18, 8 March 2012 (UTC)

It may be worth pointing out that what the theorem says is that there is no solution to these equations by radicals. The OP claims only to be able to solve them by algebraic means, not by radicals. Depending on what one means by that phrase, that might be possible or even sort of trivial (just introduce a new function symbol for a distinguished root to a polynomial, say). --Trovatore (talk) 23:37, 8 March 2012 (UTC)