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June 10

Solution to a nonlinear ODE

The ODE f·f’’ - (f’)^2 = 1 has the simple solution f = cosh (x). Do deformations of the type n·f·f’’ - (f’)^2 = 1 with n an integer also have analytical solutions?XYZsquared (talk) 19:01, 10 June 2009 (UTC) ·

Yes; without entering general theorems, any solution is analytic in the open set where it is not 0. On the other hand, no solution is regular at any point where it vanishes, even if unless you allow complex valued solutions (in the equation, if you try equating the coefficients of the Taylor expansion of a solution $f$ at a zero, you will immediately meet obstructions). Notice also that $\scriptstyle g:=f^{\frac {n-1}{n}}$ puts your equation in normal form; computing you should get

$\scriptstyle g\,^{''}={\frac {n-1}{n^{2}}}\,g^{-{\frac {n+1}{n-1}}}$.

--pma (talk) 19:57, 10 June 2009 (UTC)

thanks, the transformation to g is very useful for my purposes.XYZsquared (talk) 11:55, 11 June 2009 (UTC)

Topology question

Suppose we have two sets, $A_{1}$ and $A_{2}$, both in the metric universe $\mathbb {R} ^{2}$. Each of these sets consists of two discs touching each other at one single point. In $A_{1}$, both discs are open. In $A_{2}$, both are closed. Does the point where they are touching each other have any special property that any other point in their boundary (the distinction between a closed and an open disc) would not have? JIP | Talk 20:07, 10 June 2009 (UTC)

It has the property that deleting it from the boundary disconnects the boundary. Algebraist 20:10, 10 June 2009 (UTC)

Is it actually possible for two open discs to intersect at a single point as in A1?Rich (talk) 03:24, 11 June 2009 (UTC)

No; the intersection of two open sets is open, and an open subset of R^2 is either empty or has the cardinality of the continuum. In fairness to JIP, he didn't say that the point was in the intersection of the two discs. --Trovatore (talk) 03:53, 11 June 2009 (UTC)

thanks75.45.106.99 (talk) 05:53, 11 June 2009 (UTC)

June 11

Pooling related binomial statistics

So, the motivation is this: in an online game I play, when you attack an enemy you get a line-by-line summary of the battle ("You hit for X. Enemy hit for X. You missed." etc.). I know that each enemy has a particular miss chance, p, which is known only to the developers. However, various effects can adjust p by a known amount that differs in each battle, giving a modified miss chance p'. Given any single battle log, I can estimate p' (and by subtracting the modifier, p) and construct a confidence interval around it by counting the number of hits and misses, but generally these are too few to construct a particularly tight CI (usually hits + misses < 20).

If I have a large dataset of the number of enemy hits and misses, and the modifier for each battle, can I pool that data to get a good estimator for p? Confusing Manifestation(Say hi!) 01:05, 11 June 2009 (UTC)

You are saying there are enemies E1,E2,... shooting at you, and you get messages like "enemy hit" but it doesn't tell you which enemy hit you? If you have separate stats for each enemy then you should be able to add them up for all the battles. 207.241.239.70 (talk) 01:49, 11 June 2009 (UTC)

There's only one enemy at a time, and in battle you basically trade blows with them until one of you runs out of health. So for each battle, I have a log that's something like this:

• Enemy miss chance + 20%
• You did x damage to enemy.
• Enemy missed.
• You did x damage to enemy.
• Enemy did x damage to you.

So if there weren't that change to the miss chance each time (I've seen it as low as 0 and as high as 30), I could just add everything up. If the battles went on a bit longer I'd probably be able to justify a normal approximation which would make everything additive, but instead I've basically got a bunch of B(ni, pi') where the ni, like I said, are typically less than 20. Confusing Manifestation(Say hi!) 05:48, 11 June 2009 (UTC)

Hi. Suppose you have a 20% increase of missing, and the probability of hitting is p. And you get H hits and M misses. Then the likelihood function for p is $(p+0.2)^{M}(0.8-p)^{H}$, assuming the hits and misses are independent of one another (also assuming that I've interpreted "+20%" correctly). Then, suppose that the 20% turns into 30% and you have H2 hits and M2 misses. Then we would have $(p+0.2)^{M}(0.8-p)^{H}\cdot (p+0.3)^{M2}(0.7-p)^{H2}$. Maximizing this gives the maximum likelihood estimate. HTH, Robinh (talk) 07:07, 12 June 2009 (UTC)

Aha, maximum likelihood. Of course! Thanks for the help, I'll see if I can follow it all. Confusing Manifestation(Say hi!) 08:26, 14 June 2009 (UTC)

Graphing complex functions online

Does anyone know of a good online resource for graphing complex valued functions defined on the complex plane? I'm looking for something that can make images like File:Complex zeta.jpg, but for simple functions like polynomials and Möbius transformations. Thanks in advance. -GTBacchus^(talk) 13:52, 11 June 2009 (UTC)

Hi. The R programming language is free and includes methods for plotting complex numbers in the elliptic package, which deals with elliptic functions. The examples in the online help files include Möbius transforms (I think; I'll add some if not). HTH, Robinh (talk) 06:57, 12 June 2009 (UTC)

Sadly, I'm looking for something I can use on a computer where I haven't got admin rights to install anything. At home, I've got a good graphing program, but I'm obliged to work in a lab at school where I'm stuck with no ability to install software. Thanks, though. -GTBacchus^(talk) 15:11, 12 June 2009 (UTC)

If you have a Hex editor and understand how to use it to edit a BMP file then you have what you need to put together an image by specifying the colour of each pixel. I suggest trying this with a very small image that you can store as a .bmp file using the MS PAINT program found in every Windows OS. The PAINT program is convenient for drawing axes and text. Then you have a lot of pixels to paint... That will be hard work using only the hex editor so good luck. BTW memory sticks are inconspicuous and can easily carry software but you didn't hear that from us. Cuddlyable3 (talk) 17:19, 12 June 2009 (UTC)

I know how to write code, and I've done it, but if that's the hurdle, then this ain't gonna happen. At home, I just type in: $f(z)=z^{1+i}$, and boom: I get a graph where hue represents argument and saturation represents modulus. It would be cool if someone made an applet that did that, too. It only seems slightly more complicated than I'm motivated. -GTBacchus^(talk) 20:07, 12 June 2009 (UTC)

Proof that the axiom of choice implies the well-ordering principle?

Proven math provides a proof of the fact in the title, but I can't for the life of me find a key to the notation used there. Can someone point me to such a key, or even save me the trouble of wading through all that heavy notation, and spell out the proof right here? Also, is there a reason that the proof cannot be found on wikipedia?

I appreciate it. --69.91.95.139 (talk) 20:43, 11 June 2009 (UTC)

Are you familiar with ordinal numbers? The simplest way to prove the well-ordering principle from the axiom of choice is via transfinite recursion over the ordinals. Algebraist 20:51, 11 June 2009 (UTC)

Yes, though I should probably hasten to add that I only know about them from what I've read here on wikipedia. Can you explain how one goes about proving that, for any set X, there exists a sufficiently large ordinal number α such that the members of X can be enumerated by all the ordinal numbers less than or equal to α? --69.91.95.139 (talk) 21:26, 11 June 2009 (UTC)

That is the well-ordering principle. You prove it by transfinite recursion: when you get to the ordinal β, choose an element x of X which you haven't already mapped anything less than β to and map β to x. Of course, you need the axiom of choice to make all those choices. The process must terminate at some ordinal α less than the Hartogs number of X. Algebraist 21:34, 11 June 2009 (UTC)

Thanks for the link. I myself was worried no such upper bound would exist. The proof in the Hartogs number was very illuminating. JackSchmidt (talk) 23:30, 11 June 2009 (UTC)

Ok, can I try a different angle here? Because unless there's something I'm missing with this transfinite recursion, I'm still not sure I accept the existence of sufficiently large ordinal numbers (though I promise I will as soon as I accept the well-ordering principle).

Suppose we are setting out to develop a well-ordering ≤ on the set X, using the choice function f on the power set of X. Let Y be a variable which we will use to keep track of our progress. To begin with, we define Y = X. At each step, we take the element of Y f(Y). We define x ≤ f(Y) for all x in X but not in Y, and similarly f(Y) ≤ x for all x in Y (and yes, I realize that just one of these definitions is sufficient; the other is redundant). Finally, define the new Y to be the old Y minus the chosen element f(old Y). Repeat the process. Now, in order to prove the well-ordering principle, we need only show that this process will somehow eventually end, with Y = ∅. What guarantee is there of that happening? --69.91.95.139 (talk) 22:08, 11 June 2009 (UTC)

The guarantee is the existence of Hartogs numbers. Algebraist 22:12, 11 June 2009 (UTC)

At a slightly more conceptual level, the way these things work is that if the process never terminated, it would mean you could get a (class) injection of the ordinals into a set. Then you pull it back and see that there has to be a set containing all the ordinals. But that can't be, because of the Burali-Forti paradox. --Trovatore (talk) 22:18, 11 June 2009 (UTC)

Ah, I got it. I should have guessed there was a proof by contradiction just for that purpose.

Many thanks, --69.91.95.139 (talk) 23:27, 11 June 2009 (UTC)

There's a proof in Erich Kamke's book on set theory, reprinted by Dover. Michael Hardy (talk) 21:14, 11 June 2009 (UTC)

June 12

A question of sociopathic probability...

The last week or so, Swedish media has reported on a horrible murder, meticulously planned and executed by two 16-year-old kids against one of their classmates. Much of the reporting has emphasized that while it is exceedingly rare for two kids that young to plan and commit cold-blooded murder, what makes this case absolutely unique in recent Swedish crime-history is that one of the culprits is a girl.

Yesterday, I was having a conversation about this case with a friend of mine, and the conversation made me regret I didn't listen more closely when my high school teacher tried to teach us probability theory. My friend was asking how a person could do something like this, why some basic human instinct didn't stop them during their weeks of planning, to which I responded that both of them are probably psycopaths. My friend then asked "Yeah, but what are the chances that two people in the same classroom are psycopaths?" I thought that was a very good question, and one I'd like you fine people's help with answering (for the record, my answer at the time was "I have no idea, but surely in the hundreds of thousands of Swedish classrooms, it's not so strange that there are two psycopaths in at least a few of them").

Wikipedia states that it is estimated that about 1% of people are psycopaths (it seems to be a good average from the other sources I've looked at). So what is the probability that in a group of 30 people, two of them have that particular affliction? And how do you make that calculation? 195.58.125.52 (talk) 08:27, 12 June 2009 (UTC)

If one has a group of 30 people sampled from a particular population, and in that population it is estimated that 1% of people have some property X, then the probability that 2 people have property X in the sample of 30 can be computed as follows:

I am assuming that you want to know the probability that exactly 2 people have property X and not at least 2 or more than 2. To compute this probability one uses the binomial distribution. Practically, you want to know the probability of at least 2 people having property X in the sample of 30. In this case, we compute the probability that either 1 person, or 0 people have property X, and subtract the sum off from 1. This probability is:

1 - [(0.99)³⁰ + (0.99)²⁹ x (0.01) x 30]

= 0.03614799831

Thus, 0.03614799831 is the probability that at least 2 people in the given sample of 30 have property X. If you can not follow the above, I am sure that other Wikipedians can explain but the relevant article is binomial distribution. In percentage, there is an approximately 3.6% chance that at least two people in the class are psycopaths. --PST 09:31, 12 June 2009 (UTC)

That was both A) very understandable and B) incredibly scary (seriously, 3.6 percent???). Now that you said it, I had a flashback to learning about the binomial distribution using dice and coins. Thanks! 195.58.125.52 (talk) 10:37, 12 June 2009 (UTC)

This calculation assumes that the probability of a person being afflicted is independent of the class in which he studies. However, it is reasonable to assume that it actually depends on factors like geographic location, socioeconomic status, field of study and so on. So the quoted 1% figure is actually a mixture of classes where the probability is higher and those where it is lower. This can make the probability (of at least 2 psychopaths in a given class) even higher. -- Meni Rosenfeld (talk) 11:00, 12 June 2009 (UTC)

Just a note that that sort of calculation applies when the property X is exactly specified beforehand. If instead we were to ask "what the chance that at least two people share some property" (where the details of the property are left unspecified), we have to calculate it slightly differently. This is explained more in depth at the Birthday problem article. -- 128.104.112.114 (talk) 17:50, 12 June 2009 (UTC)

The way of computing the probability given here assumes you're talking about just one specified classroom. Michael Hardy (talk) 01:51, 13 June 2009 (UTC)

There are also different types of psychopathy. Not all psychopaths have the ability to meticulously plan and execute a murder. (Igny (talk) 17:33, 14 June 2009 (UTC))

And there is also an unlikely event of two psychopaths recognizing and trusting each other. However if there is only a chance of $5\cdot 10^{-5}$ for a particular pupil to be just such a psychopath, the probability of two such psychopaths in at least 1 class out of 1 million classes (of ~30 pupils per class) worldwide will exceed 50%.

crazy hard math question

How many unique ways can you arrange the tiles on a standard Scrabble board?

The board is 15x15 tiles and uses the following numbers of letters: A - 9 B - 2 C - 2 D - 4 E - 12 F - 2 G - 3 H - 2 I - 9 J - 1 K - 1 L - 4 M - 2 N - 6 O - 8 P - 2 Q - 1 R - 6 S - 4 T - 6 U - 4 V - 2 W - 2 X - 1 Y - 2 Z - 1 Blank - 2

The letters do not have to be in an order that makes sense. Not all tiles need to be used at once. E in spot 1,1 with the remaining spaces empty is valid (though a second setup that looks identical but uses a different E tile is not counted a second time). I am not looking for an exact answer, just an order of magnitude. 65.121.141.34 (talk) 14:45, 12 June 2009 (UTC)

Ok, you've got 100 letter tiles, and 225 places to put them. So, if we ignore the markings on tiles, the number of subsets of the board that can be covered would be $\sum _{n=0}^{100}{225 \choose n}$. We know that $\sum _{n=0}^{225}{225 \choose n}=2^{225}$, and only going as far as 100 would be roughly half of that, by symmetry (a bit less, because 100 < 112, where the actual middle is). Thus, I'd say the number of subsets of the board that can be covered is on the order of $2^{224}$. Next, you have to deal w/ the markings on the tiles... -GTBacchus^(talk) 15:20, 12 June 2009 (UTC)

If I haven't screwed up the calculation, you have 3 × 10¹⁸¹ possibilities with all tiles used. Might give an impression of a lower bound on the total. —JAO • T • C 15:26, 12 June 2009 (UTC)

I agree with your calculation (well, up to a factor of two, and it's hardly worth chasing that when the numbers are of this magnitude). Algebraist 15:33, 12 June 2009 (UTC)

Same here wrt the result, but I think the exact answer is still important, if only to show that it indeed can be found. Providing it is free, so what I got is

3456981978217014632075525114928779982118337773021077837923560875149664042972031452007170111...
...5929753351999852423974761359826877638987511529912396851416083377852241674240000000000000000

This is, again, assuming that all tiles need to be used. If not, I think it's around $10^{190}$ but I'll run an exact calculation tomorrow (if nobody beats me to it). -- Meni Rosenfeld (talk) 15:56, 12 June 2009 (UTC)

Ok, if my calculations are correct (a huge if), the answer to the original question is

7722117868441671924993195080976319850952236039054771466584888315955839702441216546371222929...
...7029197060990587627600742490611073623755475870190820272102817641617988346426872233220965236

Which is roughly $7.72\cdot 10^{181}$. -- Meni Rosenfeld (talk) 19:12, 13 June 2009 (UTC)

Differentiating the Jacobian determinant, need some help!

Hi everybody, I'm trying to solve this problem and need some hint indeed:

File:Jacobian001.JPG

where the hints are:

File:Jacobian002.JPG

and,

File:Jacobian003.JPG

thanks in advance! --Re444 (talk) 17:19, 12 June 2009 (UTC)

I do not think that we are permitted to answer questions from you unless you have provided sufficient evidence that you have attempted the question yourself. Could you please cite the book from which this question originates? It is then possible that we may be able to answer the question for you should you provide us the required evidence. --PST 22:46, 12 June 2009 (UTC)

It appears to be from "Introduction to Tensor Calculus and Continuum Mechanics" By J. H. Heinbockel [1]. I think it is ok to supply hints in this situation, just not answer the whole question. 207.241.239.70 (talk) 00:55, 13 June 2009 (UTC)

The referenced book is true : "Introduction to Tensor Calculus and Continuum Mechanics" By J. H. Heinbockel (a free e-book). And I'm a self-learner of that (not a homework then). --Re444 (talk) 19:23, 13 June 2009 (UTC)

Where is the book online for free? I'd like to try reading it. Thanks. 67.122.209.126 (talk) 03:02, 14 June 2009 (UTC)

80% of the book is available free, here (at the end of the page) —Preceding unsigned comment added by Re444 (talk • contribs) 08:16, 14 June 2009 (UTC)

June 13

Paradox of the twots

Natural Numbers are positive integers, N={1,2,3,4,5,...}

A twot is a natural number which is the product of two unique primes.

For example:
6 is a twot because 2 * 3 = 6
10 is a twot because 2 * 5 = 10
14 is a twot because 2 * 7 = 14
15 is a twot because 3 * 5 = 15

Now as a twot is the product of two unique prime numbers, one of the two prime numbers must smaller than the other. Therefore we can call the smaller prime number the base prime and classify twots by their base primes

For base prime two
1st twot is 2 * 3 = 6
2nd twot is 2 * 5 = 10
3rd twot is 2 * 7 = 14
and so on

For base prime three
1st twot is 3 * 5 = 15
2nd twot is 3 * 7 = 21
3rd twot is 3 * 11 = 33
and so on

Because the number of prime numbers is infinity, we can match every twot of base prime two to a natural number. Therefore there are as many twots of base prime two as there are natural numbers.

Therefore we can say there are more twots than there are natural numbers because all the natural numbers can be matched to twots of base prime two and we still have twots of base prime three and other base primes leftover.

Yet, we know not all natural numbers are twots, in fact the set of twots is a subset of the set of Natural Numbers.

So the paradox of the twots. There are more twots than there are Natural Numbers and yet at the same time there are less twots than there are Natural Numbers.

PS: Originally, I did not call it by the name twot but by different name. Later I was told that the name I have chosen was an unmentionable word in the English language, so I had rename it to twot. 122.107.207.98 (talk) 06:16, 13 June 2009 (UTC)

Visit Hilbert's hotel for an analysis of the paradox. Bo Jacoby (talk) 07:29, 13 June 2009 (UTC).

There is no paradox. By showing there is a bijection between the set of natural numbers and one of its proper subsets, you have simply proved that the set of natural numbers is a Dedekind-infinite set. A simpler example is to consider the obvious bijection between the set of natural numbers and its proper subset of even numbers. Gandalf61 (talk) 08:16, 13 June 2009 (UTC)

(ec)The error lies in the statement " Therefore there are as many twots of base prime two as there are natural numbers." You are basically saying that these two infinities are equal just because they both are infinities. Infinity is not a number, and hence does not follow the rules of usual arithmetic. By the above logic i can devise hundreds of new paradoxes :

• Infinity + Infinity = Infinity
  
• But since infinity is equal to infinity, infinity - infinity is 0
  
• Therefore Infinity = 0.
  

You see? You can't call two infinities the same, just because they are infinities. Lim_x->inf(x) is definitely smaller than Lim_x->inf(x²) Rkr1991 (talk) 08:22, 13 June 2009 (UTC)

Rkr1991, I am afraid your comment is incorrect. By showing a bijection from the set of 2-twots to the set of natural numbers, the OP has correctly demonstrated that there are as many 2-twots as natural numbers. The rest is questionable as well - $\lim _{x\to +\infty }x=\lim _{x\to +\infty }(x^{2})$ in most common constructions. -- Meni Rosenfeld (talk) 17:47, 13 June 2009 (UTC)

I would not call it a paradox, at least not if you have a well-defined way to speak of when two sets have the same number of elements. For example, if I have the set {1,2,3}, I know that there are 3 elements in this set. If I have the set {a,b,c}, again there are three elements. "Without having a counting system," how do we know this? We know this because of the association:

1 -> a

2 -> b

3 -> c

In fact, no matter whether we have three element sets {a,b,c}, {d,e,f}, {5,100,93045134314}, this "association method" allows us two decide whether the two sets have the same number of elements. However, as always in mathematics, this method must be formalized so we can check its validity in applications. I think the articles on countable set and bijection formalize this. --PST 08:25, 13 June 2009 (UTC)

This idea is originally due to Galileo (who used square numbers instead of nonsquare semiprimes). We have an article Galileo's paradox. Algebraist 11:13, 13 June 2009 (UTC)

Galileo called attention to this in the 17th century and Cantor addressed many similar problems in the 19th century.

Usage note: you should say "two distinct primes" if that's what you mean. "There is a unique even prime number" means there is exactly one of those. "The prime factors of 15 are distinct; whereas those of 18 are not distinct" is the right way to use that word. Michael Hardy (talk) 16:13, 14 June 2009 (UTC)

Twin prime version of Copeland-Erdos constant

Can anyone prove the constant created by concatenating prime p that p+2 is also a prime

0.35111729...

is irrational?

Assume that there are infinitely many twin primes.

Thank you for your help.Motomuku (talk) 13:38, 13 June 2009 (UTC)

At the first glance, I'd say you need more information about the distribution of twins primes to conclude. Note that the statement about the irrationality of that Copeland-Erdős constant is just a consequence of a rough form of the prime number theorem, e.g. the number of primes with n digits diverges as n increases (therefore if by contradition that decimal expansion were periodic, some primes would be repeated more times in the list). Of course, any strictly increasing sequence of positive integer numbers which is 10^o(n) gives rise to a non-periodic decimal expansion as well. --pma (talk) 20:54, 13 June 2009 (UTC)

Well 1,11,111,1111, etc or the prime in them which probably are infinite stuck together give a rational number. However for the original problem p and p+2 would eventually have repeating sequences in them - and if one had the repeats the other wouldn't. Perhaps the OP was wondering about trancendental numbers which are rather a bit harder? Dmcq (talk) 23:41, 13 June 2009 (UTC)

But he wants to put only "p" not "p+2" in the list... So the simplest reason for the irrationality seems that the sequence be 10^o(n). --pma (talk) 23:59, 13 June 2009 (UTC)

Oops yes, quite right. I should have looked at his example more carefully. Dmcq (talk) 00:13, 14 June 2009 (UTC)

There is a bit of a problem in that even if there are infinitely many twin primes they may not be distributed as the Hardy–Littlewood conjecture on the Twin prime conjecture page says. It is possible, though extremely unlikely!, that above a certain level all the p of the type you want are of the form of a Repunit like 11111. Dmcq (talk) 09:06, 14 June 2009 (UTC)

On the other hand, if the integers (primes or not) of the sequence are eventually numbers of the form 11..1, the sequence has an exponential growth; that's why I was talking of 10^o(n). However I must add that this way of making a decimal expansion out of an integer sequence by concatenation of digits gives me a sort of feeling of unhappiness... Isn't it just a way of restating some properties of an integer sequence in a quite involute and complicated way? I suspect that the previous constant with the primes has been introduced just as an exercise or a curiosity. In that case, I wonder what is the point of making variants with other sequences. Recreational mathematics suffers repetitions and generalizations. --pma (talk) 19:52, 14 June 2009 (UTC)

That's interesting, it hadn't occurred to me before, I guess a repunit in any base must be a repunit in any other base, barring bits being chopped off te beginning or end. Dmcq (talk) 11:56, 14 June 2009 (UTC)

Um sorry the Goormaghtigh conjecture says I'm very very wrong, I must have a good look at that. Dmcq (talk) 12:00, 14 June 2009 (UTC)

Need help with radical equation questions.

Question One: (y/y-1)^2 = 6(y/y-1)+7

Question Two: 3x(x^2 + 2x)^1/2 - 2(x^2 +2x)^3/2 = 0

Your help is much appreciated! Thanks! —Preceding unsigned comment added by 76.78.133.108 (talk) 18:11, 13 June 2009 (UTC)

For the first question: solve for (y/y-1) by factoring. Then expand and solve for y.

For the second one, factor out (x^2+2x)^1/2. The remaining part is a quadratic, and trivial to solve by factoring. --COVIZAPIBETEFOKY (talk) 18:25, 13 June 2009 (UTC)

June 14

alzebra quary of 9 standard

question  : x⁴+1+x² —Preceding unsigned comment added by V neeta (talk • contribs) 01:00, 14 June 2009 (UTC)

That's not a question, it's a context-free string of symbols. What do you want us to do with them? Algebraist 01:23, 14 June 2009 (UTC)

The fact that this question is in the context of 9th grade algebra and that he has not explicitly given the question suggests that the OP wants us to simplify the given algebraic expression. However, it is somewhat clear that the expression cannot be simplified further, unless of course you consider factoring x² from x⁴ + x² to be a "simplification." In this case, therefore, the question appears to be a "trick question" in that no simplification is possible, although in the given context of questions, an algebraic expression in any other question could be simplified. Of course, I may have interpreted the question completely incorrectly. --PST 04:21, 14 June 2009 (UTC)

The polynomial can be viewed as a difference of squares, namely (x⁴ + 2x² + 1) − x², leading to the factorization (x² − x + 1)(x² + x + 1). This may be useful as an intermediate step on the way to the roots, if that's what the OP wants. Michael Slone (talk) 04:55, 14 June 2009 (UTC)

Or it could be that he wants to solve the equation x⁴ + 1 + x² = 0. In that case, I'd start with u = x², solve the quadratic equation in u, and then solve the two separate quadratic equations in x. Alternatively, solving the two quadratic equations in x suggested by Michael Slone's posting would give the same result.

Probably a textbook said something like "Factor the following polynomials", then had a numbered list of problems, and somewhere among them there appeared x⁴ + 1 + x². Then the poster neglected the words written at the top and posted only the polynomial. The words are where the actual question was. The poster failed to realize that. In other words, the poster had no question about mathematics, but was unsuccessfully attempting to transmit the textbook's question to us.

Just a guess.

That such absurd lunatic behavior is routine shows one aspect of what's wrong with pretending to require people to learn math when they don't want to do so and spend the whole course kicking and screaming. Michael Hardy (talk) 16:09, 14 June 2009 (UTC)

I don't completely agree; I think in most cases it is just a problem of lack of correct mathematical language, and sometimes, just a problem of English for non-native speakers (I suspect the present case could be one). In fact, while the "Assume good faith" guideline is generally followed, I fear we can't say the same of "Assume no total stupidity" (even between answers, sometimes!). For instance, we often see people asking for "solution of a polynomial", which is a nonsense taken literally, yet we don't need to presume idiocy. Sometimes in these cases we eventually see that the questioner did know what he said and even understood the answer. Here, as PST points out, the context of 9th grade algebra makes it clear that the request is a factorization. If a person of mean intelligence has only seen polynomials as objects to be factorized, he may well assume that it is implicitly understood that you can't do other, and that the meaning of an polynomial be "factorize me", as "3 x 2" means "multiply 3 by 2". To make an example, if someone offers me a banana, I am authomatically lead to take and eat it, and I don't even wonder if he may mean I should put it in my ears.

Note the approach by Algebraist: just a short answer: your question is not properly formulated / please restate your question. This way the poster will not be intimidated, and we can also understand better the situation. The subsequent defence of the questioner by PST was maybe not necessary --possibly refrained a bit the reply. However, I do not have solutions... maybe sometimes we should wait for the OP's reply, when he's been requested for clarifications. If we only could, the best would maybe something like in Brazil: as soon as somebody puts an unclear question, a group of Regulars implodes in his room, take him and in the case furnishes him with the needed mathematics. --pma (talk) 21:20, 14 June 2009 (UTC)

I agree with pma above. Although the OP may have phrased his question in an "incompetent style", one should be aware that this does not give an indication of his mathematical knowledge. On the other hand, it could well be the case that the OP knows English reasonably well but cannot use mathematical terminology appropriately. In this case, I still feel that we should not attack the OP because I am sure that most of us were at this stage at some point. In particular, when one first learns a concept, appropriately employing correct terminology is difficult for the average.

I guessed that the OP wanted a "simplification" of some sort, performed by adding/subtracting like terms. Usually, in my experience, I would expect such questions not to be given in the 9th grade for usually such students do not have a good understanding of algebra. I could be wrong, however.

On the other hand, the OP is probably not going to return to this help desk (and has probably not seen this discussion) so I do not really see much point in continuing. --PST 00:52, 15 June 2009 (UTC)

The comments of pma and Point-set topologist seem to make a certain amount of sense until I go back and look at what the original poster actually wrote. It's understandable that incorrectly used terminology appears on this page, but this poster used no terminology at all and behaved in the manner suggested by my guess. But maybe all sorts of weird people that I don't know about are out there.... Michael Hardy (talk) 22:51, 15 June 2009 (UTC)

For the OP, now you've seen how that one was solved you can try factorizing x⁴+4 Dmcq (talk) 12:16, 15 June 2009 (UTC)

June 15

Predicting Score in a T20 match

Hi! I am trying to build a mathematical model to predict the number of runs that would be scored by a team if we know the number of runs and the wickets fallen after a few overs. I have made an initial empirical formula:

Let n_x be number of runs scored in x overs and let the number of wickets fallen be w, if x ≥ 10 then

n₂₀ ≈ n_x ( 1 + 2 log (x/w²)) + (10 – (x/2)) log(n_x x/w(20-x))

Can you help me in improving this model? shanu 04:30, 15 June 2009 (UTC) — Preceding unsigned comment added by Rohit..god does not exist‎ (talk • contribs) I fixed the wiki-formatting of the original post. Abecedare (talk) 05:24, 15 June 2009 (UTC)

It would be best to start by plotting n₂₀ versus x, n_x and w, using the database of all previous match scores. That way one can have a general idea of the shape of the curves (surfaces) and also compare the performance of various proposed formulae.

For those who have no idea of what is being discussed: See Twenty20, which is a form of cricket. Abecedare (talk) 05:17, 15 June 2009 (UTC)

In case you don't know this already, be aware that the Duckworth-Lewis method is the canonical way of doing this. Algebraist 07:56, 15 June 2009 (UTC)

Olympiad question

In my wanderings, I came upon a paper for the 1981 International Mathematics Olympiad. It contained quite an interesting question:

Suppose that f(x,y) satisfies:

$(1)\qquad f(0,y)=y+1$

$(2)\qquad f(x+1,0)=f(x,1)$

$(3)\qquad f(x+1,y+1)=f(x,f(x+1,y))$

Hence determine f(4,1981)

I've tried to solve it. The direction I choose has the potential for success, but would take many hours to complete. I figured that, beginning with

$f(4,1981)=f(3,f(4,1980))\,\!$

and continuing thus to find

$f(4,1980)=f(3,f(4,1979))\,\!$

and then

$f(4,1979)=f(3,f(4,1978))\,\!$

would result in a sequence ending with f(4,0). I therefore concluded that f(4,0) was critical, and determined that it produced 7 13. But that doesn't help much, because we then need to plug in 7 13 for

$f(4,1)=f(3,f(4,0))=f(3,13)\,\!$,

and then complete resultant plug ins for y in f(4,y) all the way from 1 through 1981. Obviously, this would take an unfeasible period without mechanised assistance: the Olympiad participants were given 4 hrs 30mins for 6 such problems. I concluded that my approach was "the long way round" and that a more elegant solution probably expending half or a single page is to be found here. Can anyone provide some insight? Cheers, —Anonymous Dissident^Talk 08:03, 15 June 2009 (UTC)

This is actually the Ackermann Function and so there should be something on that link that should provide some help. It is very odd for them to ask a question about such a well-known function and the more recent Olympiads would be very unlikely to contain such a question. --Anthonymorris (talk) 08:12, 15 June 2009 (UTC)

Indeed - the solution is in Ackermann function. But if you want to tackle the problem from first principles, I would start out by finding a recurrence relation for f(m, n) for small fixed values of m. So:

$f(1,n+1)=f(0,f(1,n))=f(1,n)+1{\text{ for }}n\geq 0$

$f(1,0)=f(0,1)=2$

$\Rightarrow f(1,n)=n+2$

then we can proceed to f(2,n):

$f(2,n+1)=f(1,f(2,n))=f(2,n)+2{\text{ for }}n\geq 0$

$f(2,0)=f(1,1)=3$

$\Rightarrow f(2,n)=2n+3$

and so on. Two more steps and you can find an explicit expression for f(4,n) (incidentally, I think your value for f(4,0) is incorrect - it should be 13). Then you can determine an explicit expression for f(4,1981) - you won't be able to write out all of its digits, but it is close to a (very large) power of 2. Gandalf61 (talk) 08:46, 15 June 2009 (UTC)

I agree, doing it again. I'll try your method now. —Anonymous Dissident^Talk 09:19, 15 June 2009 (UTC)

The problem is that those in the competition were not permitted a calculator. How could they have computed the output without one? —Anonymous Dissident^Talk 09:28, 15 June 2009 (UTC)

Actually, it is not so odd for them to ask a question on a well-known concept. For instance, on a recent Tournament of the towns paper, a question was asked on the concept of combinations. In fact, this question was relatively simple if the solver knew some basic identities related to the concept. Furthermore, this question was one of the "tougher questions" given more points than some of the other questions on the paper. On the other hand, the international mathematics olympiad is much more reputable, and it does strike as surprising that a question on such a well-known concept should be asked. --PST 09:11, 15 June 2009 (UTC)

Conic section-Simultaneity in relativity

This is a question about conic section related to simultaneity in special relativity. If the tilt of the plane is determined, can the conic sections always parallel-shifted in the same direction to match another conic section of the same cone but and a plane which is parallel to, and has the same distance to the apex of the cone as, the first plane? Like sushi (talk) 12:49, 15 June 2009 (UTC)

I divide every value in a table with the smallest value in the same table. What have I done.

Hi all.

Here comes a table with 2 columns and 3 rows.

200 400

600 800

500 700

Now I divide every value in this table with the smallest value, which happens to be 200.

1 2

3 4

2.5 3.5

Here comes my question: What is this operation called? Does it even have a specific name?

I was thinking that perhaps I have "normalized every value in the table to the smallest value", but I'm not sure. Is normalize the correct answer?

Thx. Contributions/194.237.142.21 (talk) 14:15, 15 June 2009 (UTC)

Yes, "normalize" is the first word that comes to mind. Contributions/67.122.209.126 (talk) 15:40, 15 June 2009 (UTC)

Differential equation

Is it possible to find a general solution for $a{\ddot {y}}y+b{\dot {y}}^{2}+cy^{2}+d=0$? Contributions/76.67.79.151 (talk) 19:15, 15 June 2009 (UTC)

Here is a first information; do you have a particular request? Assuming a≠0, you can normalize and fix a=1. For all ξ, y₀≠0, y₁ the Cauchy problem with initial data y(ξ)=y₀ and y'(ξ)=y₁ has a unique local solution, which has a power series expansion at ξ, tha is $\scriptstyle \;y(x)=\sum _{k=0}^{\infty }y_{k}(x-\xi )^{k}$, where the coefficients (y_k) depend on y₀ and y₁ (not on ξ). You can determine the power series equating the coefficients in the differential equation. You will get a recursion formula for the y_k, possibly too complicated to put in a closed form. Of course, if y₀=0 then also y₁ is fixed by the equation for you have by₁²+ d=0. Notice also the transformation z:=y^b+1, (with a=1 as before) that puts your equation in the form z"=R(y).--pma (talk) 20:54, 15 June 2009 (UTC)

Intro to nonlinear differential and integral equations by Harold Davis (ch8 especially) might be a good read for the OP, this equation is of type 12 in the 1st appendix Silverfish70 (talk) 21:11, 15 June 2009 (UTC)

Formula for Sine based on just angle

If I'm using a circle with radius 1, what is the formula to get the sine from just the angle? Not anything like sin x=oppositie/hypotenuse. I'm looking for the way it is done like with just the angle. --Melab±1 ☎ 20:15, 15 June 2009 (UTC)

this? Trigonometric functions#Series definitions --pma (talk) 21:00, 15 June 2009 (UTC)

See also:

• Exact trigonometric constants for how sin x can be calculated for some angles using trigonometric identities
• Generating trigonometric tables for various numerical approaches to computing sin x for any angle.

Note tat some of the formulas and results are given for angles expressed in radians rather than degrees. Abecedare (talk) 21:12, 15 June 2009 (UTC)

Growth rate of functions

Is there an increasing function (of one real number, obviously) that grows faster than any polynomial, but slower than any exponential? Is there a proof that there is no such function? deranged bulbasaur 20:38, 15 June 2009 (UTC)

Subexponential growth for example $e^{\sqrt {x}}$ (also see Sub-exponential time) (Igny (talk) 20:44, 15 June 2009 (UTC))

Thanks. Actually, that answer was pretty obvious. I guess I should have thought about it more. On a related note, why are these growth rates not really talked about? One frequently hears talk in computer science that would seem to indicate that either a polynomial algorithm can be found, or one is stuck with an exponential worst-case. Is it simply that sub-exponential solutions are rare? Also, shouldn't these time complexities generate their own complexity classes with their own complete problems and such? My trite error was caused by the fact that apparently this is just never mentioned, at least in the relatively introductory materials I have read on the topic. deranged bulbasaur 23:21, 15 June 2009 (UTC)

Integer factorization is a well studied problem where the current best algorithm is subexponential. Taemyr (talk) 00:06, 16 June 2009 (UTC)

What is interesting however, is that there is a function which is faster than $x^{\alpha }$ for any $\alpha >0$, but slower than $e^{x^{\beta }}$ for any $\beta >0$. Can you construct one? (Igny (talk) 02:37, 16 June 2009 (UTC))

June 16

rule finding

I remember an old math teacher taught me a method, given numbers n1, n2, etc. , to determine a function f(x) for which f(1) = n1, f(2) = n2, etc., but I forget how. I recall that it had something to do with subracting the first two terms with each other, then subtracting this difference with the 3 number, and so on until you got 0, but I'm not sure about this. There were also a lot of factorials involved. Does anyone know what I'm talking about? —Preceding unsigned comment added by 74.15.138.134 (talk) 02:30, 16 June 2009 (UTC)

Area of a circle

If a dog is tied to the corner of a shed 6m in width and 3m in length by a rope that is 5m, what is the area within the dog's reach? I know the answer is 62 m squared, but i can find no way to get it.