Wikipedia:Reference desk/Mathematics

Wikipedia:Reference desk/headercfg

September 28

Estimating the variance-covariance matrix of a Multinomial function

I wish to estimate the variance-covariance matrix of a multinomial likelihood function containing two parameters.

I think the steps are:

1. Log-transform the likelihood function.

2. Take the second partial derivative of the log-likelihood with respect to each parameter.

3. Insert the parameter point estimates (obtained via maximum likelihood estimation) into the second partial derivatives.

4. Construct a matrix of the resulting 4 values. I believe this would be the Hessian Matrix.

5. Take the negative of that matrix, giving the Information Matrix

6. Invert the Information Matrix.

Could someone please verify that the above steps are correct, or point out which steps are missing or incorrect?

Once I know that I have the steps correct, I want to post an example. Thank you for any thoughts on this matter Mark W. Miller 00:34, 28 September 2007 (UTC)

I believe the steps above are correct after having worked entirely through one example. I did intend to post that example here in the hope others could check my partial derivatives. However, those equations are very long and very messy. In the last day I learned how to use the program MAPLE, with which I was able to check my partial derivatives for myself... in addition to evaluating them numerically on a computer with a different program. I'm not 100% certain I am using the terms "Hessian Matrix" and "Information Matrix" exactly as intended. Nevertheless, my computations do appear to be correct.

Mark W. Miller 04:25, 29 September 2007 (UTC)

So simple yet so hard

Radioactive particles decay. The unobtainium particles have a halflife of 1 second. That is, for each second there is a 50% chance the unobtainium particle will disintegrate.

At t==0 seconds , there are 3 unobtainium particles.

.

At t==1 seconds , there are X unobtainium particles.

.

At t==2 seconds , there are 1 unobtainium particles.

How many unobtainium particles are there at t==1 second?

It's obvious to me that the number of particles at t==1 is a discrete PDF whereby

Pr(X==0) = 0

Pr(X==1) = x1

Pr(X==2) = x2

Pr(X==3) = x3

Pr(X>3) = 0

1 = x1 + x2 + x3

But how to find x1, x2 and x3? The simple answer of x3 = comb(3,0) * p^0 * (1-p)^3 is wrong.( p= 0.5 )

220.239.107.201 13:33, 28 September 2007 (UTC)

You know the half life, so you could try calculating P(t1=X, t2=1), for X=1,2,3 (hint: what is the distribution of the number of decays in the first second? in the second second, given that there are now X at t=1?). Then note that the events described by those three probabilities are your entire conditional sample space, and do the appropriate thing (hint: the three probs calculated will not sum to 1). Baccyak4H (Yak!) 14:46, 28 September 2007 (UTC)

You know that one of the particles survived to 3 seconds, and the other two did not, so the only question is whether each of the other particles decayed during the 1st second or the 2nd second. There are four possibilities, that both decayed in the 1st second, that only the 1st particle decayed, that only the 2nd particle decayed, or that both neither decayed. Do you know how to figure each of those probabilities ? StuRat 19:00, 28 September 2007 (UTC)

I think one of those possibilities should, of course, be that "none decayed in the 1st second"Richard B 23:41, 28 September 2007 (UTC)

Yes, of course, good catch. Correction made. StuRat 01:43, 29 September 2007 (UTC)

A way of tackling this is given by Bayes' theorem. Calling Y the number of particles after two time units (without the observation fixing Y at 1), you can determine P(X = i), 1 ≤ i ≤ 3, P(Y = 1), and P(Y = 1|X = i), 1 ≤ i ≤ 3. Bayes' theorem then allows you to compute P(X = i|Y = 1), 1 ≤ i ≤ 3.  --Lambiam 22:27, 28 September 2007 (UTC)

Most functions are not lines

The article about differential calculus states that "Most functions, however, are not lines." I would like to know:

1. Is this true?
2. If it is true, what does it mean?
3. What's the purpose of the article stating it? A.Z. 21:16, 28 September 2007 (UTC)

Function (mathematics) should help. Friday (talk) 21:21, 28 September 2007 (UTC)

I should explain myself better. I understand "most" to mean more than 50%, and "most functions" to mean "more than 50% of all functions". It seems to me that there's an infinite number of functions that are lines, and an infinite number of functions that are not. I have a hard time figuring what the sentence means. A.Z. 21:42, 28 September 2007 (UTC)

The article meant 'most functions are not straight lines' in the context of the example given - I've changed it to add that additional clarity.

Does that solve the problem or...?87.102.83.163 22:10, 28 September 2007 (UTC)

If you re-read that part - you will see that it is refering to the slope of straight lines. —Preceding unsigned comment added by 87.102.83.163 (talk) 22:13, 28 September 2007 (UTC)

Yes, I took "lines" to mean "straight lines". A.Z. 04:25, 29 September 2007 (UTC)

To compare the sizes of infinite sets, mathematics uses cardinal numbers. The cardinality (size) of the set of functions whose graphs are lines is ${\mathfrak {c}}$, the cardinality of the continuum. The cardinality of the set of all functions (at least in classical mathematics) is 2^{${\mathfrak {c}}$}, which, by Cantor's theorem, is genuinely more than ${\mathfrak {c}}$. The cardinal number 2^{${\mathfrak {c}}$} is also known as $\beth _{2}$, or Beth two. See also Cardinality of the continuum#Sets with cardinality greater than c. —Preceding unsigned comment added by Lambiam (talk • contribs) 22:05, 28 September 2007 (UTC)

Thank you a lot for those links! I had never read about classical and non-classical mathematics. I read your observation about classical mathematics to mean that there's controversy about the topic. Your post obviously doesn't answer my question, but it's very helpful. I would like to be able understand one day why, in classical mathematics, they say that the set of all functions is bigger than the set of functions that are lines. This looks like something hard to prove. A.Z. 04:23, 29 September 2007 (UTC)

Think about picking a random function. This means for every real number, assigning it to a random value. If you were to do this 100 times, how often would the values lie perfectly on a line? What if you did it a million times? It would seem very, very unlikely (practically impossible). This is the intuition behind what the author means when they say that most functions are not lines. J Elliot 22:33, 28 September 2007 (UTC)

Thank you for your reply (really), but we are talking about all functions, not just a hundred of them or one million of them. A.Z. 04:23, 29 September 2007 (UTC)

Common sense works here. A real-valued function takes a real number x and produces a real number y. How likely is it that those xy pairs lie exactly on a line, any line? Is it not intuitively obvious that the likelihood is small? Even if we restrict the functions to be continuous (so that small changes in x produce small changes in y), lines are unlikely. In fact, we could restrict attention to polynomial functions of degree at most three,

$y=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}.\,\!$

For this to be a line, both c₂ and c₃ must be exactly zero. So in real life we're extraordinarily lucky if a function, say in physics or chemistry or ecology, looks like a line.

Why is this worth mentioning? Because we need to deal with such functions, to answer questions that only calculus can handle. --KSmrq^T 22:41, 28 September 2007 (UTC)

c₂ and c₃ will be exactly zero an infinite amount of times, and they won't be exactly zero also an infinite amount of times. Thank you a lot for your post, but that an infinite set can be bigger than another infinite set seems rather counter-intuitive to me. A.Z. 04:23, 29 September 2007 (UTC)

As an example consider the set of polymonials in x; of these polynomials with order 3 or less is a subset, and of those polynomials with c2 and c3 (see above) is a subset. So despite this set having infinite members it is smaller than the sets in which it is nested.83.100.183.116 06:02, 29 September 2007 (UTC)

It is certainly not new that there are different sizes of infinite sets. Lambiam has already mentioned the relevant concepts; I suggest you take a look at cardinality, and perhaps set for some more basic background. -- Meni Rosenfeld (talk) 08:27, 29 September 2007 (UTC)

Defining this in any strict sense becomes very difficult. First what is the set of all functions? Are we talking about just polynomial functions, or following Elliot, for every real number, assigning it to a random value which gives a much larger set including the cantor set. Then you would need to talk about the distribution of functions in a statistical sense. Here pure maths falls down, yes we can talks about the codimension or measure of our set. However it becomes a physical problem, the physics of a situation determines the set of functions and distribution of such. If we are working with two orbiting bodies, then the class of all functions we consider becomes just the conic sections. Its possible to construct a physical situation where most functions in that situation will be straight lines. You can make this at bit more formal by considering a Generic property of the set you are considering. --Salix alba (talk) 08:31, 29 September 2007 (UTC)

If your intuition isn't helping, the solution is to train a better intuition. Consider a stone-age culture whose numbers are "one, two, three, many". That's really how you're approaching infinity: just throwing up your hands and saying "really big". Fair enough, but we can also define and study infinity more carefully, as mathematicians.

Long ago people kept tallies of sheep or bushels of grain or whatever using physical tokens, one token for each item. Our formalism for "counting numbers" reflects this primitive idea, that we can match five pebbles with five sheep or five bushels of grain, so we have an abstract number "five". If one of our sheep goes missing, when we do the match-up we'll have a pebble left over. This leads to a formalism for comparing numbers. In fact, we can define and compare "cardinal numbers", including infinities, in just this way.

Suppose we are given two sets, A and B. A function ƒ: A→B assigns to each element a in set A an element b in set B. For our purposes, we insist that ƒ be an injection, so that if a₁ ≠ a₂, then ƒ(a₁) ≠ ƒ(a₂). That is, we're matching each different element in A with a different element in B. If such a function exists then we say, by definition, that the size ("cardinality") of the set A is less than or equal to the size of the set B, and write |A| ≤ |B|. If we also have an injection from B into A, then A and B are the same size.

From such a humble definition we can draw some fascinating conclusions. We can recognize (or define) an infinite set by the property that it has a proper subset (omitting one or more elements) of the same size. For example, the counting numbers {1,2,3,…} are an infinite set because ƒ(n) = 2n is an injection. Any set of this size or less is called "countable". It turns out that fractions (the rational numbers) are also countable, but that real numbers are not. The set of real numbers is a "larger infinity" than the set of counting numbers.

Our untrained intuition, accustomed only to finite sets, says that a proper subset must always be smaller. Our expanded intuition says, not always. But we can prove that the set of all subsets of a set A, sometimes denoted by 2^A, is always strictly bigger than A itself. Note that we include the empty set as one of the subsets, and that the theorem applies to infinite sets as well as finite sets. Now, as Lambiam has said, we can use this to show that the set of all real-valued functions is strictly larger than the set of functions which are lines.

We can argue that this should be plausible for the following reason. To specify a line function we need precisely two numbers, say the slope m and the y intercept b, defining y = mx+b. To specify a single arbitrary function we need an infinite number of (x,y) pairs, as much data as all the lines put together. Still, plausibility depends on intuition, so we'd prefer proof. Let L = R² be the set of all line functions, 2^L the set of all subsets of L, and let F = R^R be the set of all functions. The theorem assures us that |L| < |2^L|; so if we can prove that there is an injection from 2^L into F (implying |2^L| ≤ |F|), we will have proved that |L| < |F|, which is our goal. I hesitate to propose an exercise I have not attempted myself, but I'm going to leave this one to the reader.

The thing is, while we can discuss some interesting mathematics concerning sizes of infinities, it's a diversion from the intent of the remark in the article. It would suffice to observe that many functions of practical interest are not line functions. --KSmrq^T 08:56, 29 September 2007 (UTC)

I changed the article according to your last sentence. I am not yet able to understand what you wrote because you used too many words and symbols that I don't understand, and probably because it is complicated. I'm currently unable to have an opinion about whether the following is true:

Suppose we are given two sets, A and B. A function ƒ: A→B assigns to each element a in set A an element b in set B. For our purposes, we insist that ƒ be an injection, so that if a₁ ≠ a₂, then ƒ(a₁) ≠ ƒ(a₂). That is, we're matching each different element in A with a different element in B. If such a function exists then we say, by definition, that the size ("cardinality") of the set A is less than or equal to the size of the set B, and write |A| ≤ |B|. If we also have an injection from B into A, then A and B are the same size.

I should study cardinality and sets, I know, but Wikipedia's article is terrible at teaching me that. Did you prove (or tried to prove) above that "the set of all subsets of a set A, sometimes denoted by 2A, is always strictly bigger than A itself"? I don't see why, "by definition", we say that "the set A is less than or equal to the size of the set B". A.Z. 06:52, 30 September 2007 (UTC)

You can't "see" why something is so by definition. Once it has been agreed to define "positive" as meaning "greater than 0", then henceforth "positive" means "greater than 0", not for any particular deep reason that you can see, but simply by definition. The definition of one set's size being less than or equal than that of another set, or more precisely the mirror relation of being greater than or equal in size, can be found in our article on the cardinal numbers; search for "greater than or equal". If the size of B is greater than or equal to the size of A, then (by definition) the size of A is less than or equal to the size of B. That the set 2^A (not 2A) is larger than the set A is precisely the conclusion of Cantor's theorem. The proof is not particularly complicated.  --Lambiam 18:39, 30 September 2007 (UTC)

Generally when mathematicians say "most" they're referring to measure, not cardinality. It's perfectly correct in most contexts to say that most continuous functions from R to R are not linear, even though the linear functions and the continuous functions have the same cardinality (namely the cardinality of the continuum). Measure theory formalizes the intuitive idea that a "randomly selected" continuous function is unlikely to be linear. -- BenRG 01:01, 30 September 2007 (UTC)

What does this section mean? I can barely understand it. Does it mean that there are mathematicians that don't believe in cardinality? A.Z. 06:52, 30 September 2007 (UTC)

In classical mathematics, one can make unrestricted use of the law of excluded middle, or the equivalent rule of double negative elimination. In particular, to show that some proposition P is true, classical mathematicians can use a proof by contradiction: assume that P is false, derive a contradiction from that assumption, and conclude that therefore P is true. Not so in constructive (or in intuitionistic) mathematics. The proofs for Cantor's results crucially depend on non-constructive proof methods, like showing that A < B and A = B lead to a contradiction, and concluding that therefore A > B. In general, there are several different equivalent ways in which you can define a mathematical concept. Whether "P implies Q" or "(not P) or Q" is used in the definition makes no difference, classically. But constructively, these two are not the same, and two different definitions that classically define the same concept may define different, non-equivalent concepts when interpreted constructively. It is possible to give a constructive definition of cardinality, but in fact it is possible to give different definitions that are not mutually equivalent, and whatever the definition selected, most classical results (such as Cantor's theorem) are no longer valid. Thus the whole edifice of cardinal numbers collapses. This is what Hilbert referred to when, in expressing his resolve to fight intuitionism, he proclaimed: Aus dem Paradies, das Cantor uns geschaffen, soll uns niemand vertreiben können ("No one shall be able to expel us from the paradise that Cantor has created for us").  --Lambiam 19:42, 30 September 2007 (UTC)

September 29

History of Knuth–Morris–Pratt algorithm

(This is a repost of a question I've mentioned months ago on Talk:Knuth–Morris–Pratt algorithm but got no attention there.)

The algorithm described in Item 179 of HAKMEM seems quite similar to the Knuth–Morris–Pratt algorithm, despite that HAKMEM is from 1972 so it predates the publication by KMP. Were variants of the algorithm for fast string matching known long before this publication or am I misinterpreting something and the two algorithms differ in some important point? – b_jonas 15:52, 29 September 2007 (UTC)

The Gosper algorithm is in fact closer to a similar algorithm also based on a finite-state automation, due to Robert Boyer and J Strother Moore (A fast string searching algorithm, Communications of the ACM 20:10 (October 1977) pp. 762–772 [1]). In the final section "Historical Remarks" the authors write: "We have since learned that R.W. Gosper, of Stanford University, simultaneously and independently discovered the delta₁ version of the algorithm (private communication)." I once saw a presentation of a development by stepwise refinement of the various algorithms, in which the different variants of the same basic idea arose through subtly different design choices, but I've forgotten all details, including who presented this when and where. Based on the publication dates, I'd agree that Gosper appears to have scientific priority for the basic idea.  --Lambiam 16:33, 29 September 2007 (UTC)

We actually have an article on the Boyer–Moore string search algorithm.  --Lambiam 16:39, 29 September 2007 (UTC)

I could be wrong, but I don't think the algorithm in HAKMEM is similar to the Boyer-Moore algorithm. – b_jonas 16:53, 29 September 2007 (UTC)

I think you're right. I did not look at the actual algorithm in HAKMEM, but, like KMP, in incremental matching it essentially proceeds along the pattern left-to-right as long as there is a match. It would be interesting to see if Knuth references this in his The Art of Computer Programming (which I don't have access to). Bill Gosper was research assistant of Knuth in the period 1974–1977, so it is hardly conceivable that Knuth was not aware of his algorithm.  --Lambiam 19:35, 29 September 2007 (UTC)

Clearly we see common ideas among the algorithms. From the ACM page we can see that Boyer and Moore cite both HAKMEM and the 1974 Stanford technical report that became the 1977 Knuth–Morris–Pratt paper; Moore's web page says he and Bob Boyer invented their algorithm around 1975. Looking at the second edition of TAOCP Volume III: Sorting and Searching, I find no mention of string searching or the Boyer–Moore algorithm or even Gosper. I believe the first PhD thesis on computational complexity was in 1975 by Knuth's student Bob Sedgewick (founding Chair of the Department of Computer Science at Princeton), so the whole idea of analyzing and proving complexity bounds was in its infancy as these algorithms were being developed. The HAKMEM item strives for efficiency, but predates the complexity era. Physically, these authors were in close proximity (Xerox PARC was on Stanford land, and SRI International was still the "Stanford Research Institute"), and the institutions had close ties. But why are we discussing this on the mathematics reference desk? The computing desk would seem more appropriate. --KSmrq^T 22:24, 29 September 2007 (UTC)

Thanks for all comments. Just let me comment that KMP is part of the material that shall be in TAOCP volume 5. – b_jonas 08:27, 30 September 2007 (UTC)

September 30

N-th power Legendre Symbol of an integer mod integer (Java)

I was wondering if there was java code somewhere on the internet that deals with this problem. I looked everywhere but could not find any.

Suppose you have a number n mod p, where n is an integer and p is prime, then if Legendre Symbol is = 1 there is a solution such that x^2 = n mod p, if Legendre Symbol is = 0 then n has a factor in common with p, if Legendre Symbol is -1 there there is no solution for X such that X^2=n mod p.

This problem is trivial with the case of a prime mod and a power of 2 because of the Euler's criterion.

The terms that will help with this question are: Jacobi Symbol,Quadratic Residue,Euler's criterion, Legendre Symbol,P-adics,ect... Number theory stuff basically....

Here is what I wanted the program to do: Given (not specifically prime) integer A, B, C This function outputs integers {-1,0,1}, -1 is false, 0 is N/A, 1 is true

Checks that A mod C is congruent to X^B mod C, but does "not" give the value of X. The variable "B" is not just 2 but can be any prime integer.

It would be really helpful if you can provide be to some source code to something that does something similar. 24.250.139.137 14:11, 30 September 2007 (UTC)ForgotMyLogin —Preceding unsigned comment added by 24.250.139.137 (talk) 14:08, 30 September 2007 (UTC)

It's homework, but I need help!

I have a test on Monday and I'm doing my review for it. I'm not able to go in to the class for tutorials befopre the test so I need some help with solving a couple problems I don't know how to do.

1. (3) Write $f(x)=x^{3}-6x^{2}+12x-5$ in the form $f(x)=(x-k)q(x)+r$ when $k=3$
2. (4) Write the quadratic function, $f(x)=2x^{2}+8x+17$, in standard form. THen find the vertex, axis of symmetry, and x-intercepts. (on this problem I think I remember soemthign about (-h,k) or something)
3. (19) A rectangular field has a perimeter of 306 feet and is to have an area at least 5670 square feet. Within what bounds must the length, x, of the rectangular field lie?
4. (26) Solve: $x^{2}-5x-24<0$ (do I jsut factor and the answer is $(x+3)(x-8)$?)
5. (27) Solve: $4.1x^{2}-7.1<-3.1x$
6. (20) Find the domain and range of the function. $q(x)=(3x)/(x+8)$
7. (21) Find the domain of the function. $f(w)=sqrt(25-w^{2})$ (i don't know how to do the square root symbol, but sqrt is square root)

I can't explain how much it helps for anyone on this desk to help me with these problems. Myt grade i nthe class rides on this test and if I get less than an 80 on the test, I will be barred from any extracurricular activities. These are only 7 of the 50 problems so I pretty much got it down but these are the hardest ones for me. Thank so SO MUCH!! edit: didn't sign in before savign the page: schyler 17:16, 30 September 2007 (UTC)

Do your own homework. The reference desk will not give you answers for your homework, although we will try to help you out if there is a specific part of your homework you do not understand. Make an effort to show that you have tried solving it first. —David Eppstein 17:18, 30 September 2007 (UTC)

Schyler, you're more likely to get help if you put your questions in a more abstract form; that at least shows us that you understand what kind of problem you're dealing with. If you've got this far in your course without being able to that, what do you expect us to do? —Tamfang 19:46, 30 September 2007 (UTC)

[Personal attacks removed] schyler 17:19, 30 September 2007 (UTC)

Please be WP:CIVIL. —David Eppstein 17:20, 30 September 2007 (UTC)

Oh, I wouldn't have removed the entire whine. —Tamfang 19:46, 30 September 2007 (UTC)

I've restored the questions themselves (trimmed down in presentation) for context. I see what might have provoked you, Schyler, but profanity and namecalling are unacceptable behavior.

You need to understand both the rules of the page and the attitude of the people who help here. Many of us are teachers, and are more than willing to help you understand a problem and how to solve it. We will not solve homework problems for you, and don't appreciate being asked.

Your plea lies in a gray area, but I believe most of us would agree with David Eppstein that it asks for too much and demonstrates too little. Here's why.

Question (26) is well presented, because it shows us that you have thought about the problem and are left with a specific uncertainty, which you tell us. But for all the others, what can we say? We don't know what has been taught so far in your class, we don't know what puzzles you about the questions, we don't know what (if anything) you have tried.

And your heartrending saga about being barred from extracurricular activities falls on battle-toughened hearts. What would a soccer coach say to a player who did not work hard at practice but then expected to be in the game? Any good teacher will try to help a student who demonstates commitment; so far, that's not you. --KSmrq^T 20:57, 30 September 2007 (UTC)

Since you have made an effort on number 26, I will help you solve it:

(26) Solve: $x^{2}-5x-24<0$ (do I jsut factor and the answer is $(x+3)(x-8)$?

The factoring was a good first step. Next, to get an answer less than 0 (or, in other words, negative) you must make one factor negative and one positive. This is because a positive times a positive is positive, and a negative times a negative is also positive. We can only get a negative value by multiplying a positive by a negative. So, let's make a chart:

 x  x+3  x-8
--  ---  ---
 9   12    1
 8   11    0
 7   10   -1
 6    9   -2
 5    8   -3
 4    7   -4
 3    6   -5
 2    5   -6
 1    4   -7
 0    3   -8
-1    2   -9
-2    1  -10
-3    0  -11
-4   -1  -12
-5   -2  -13
-6   -3  -14

Now, you can see from this chart that we only have one positive and one negative factor if x is between -3 and 8. Do exactly -3 and 8 work ? Let's try them in the original equation:

   x^2 - 5x    - 24 < 0 
(-3)^2 - 5(-3) - 24 < 0 
     9 - (-15) - 24 < 0 
     9 +   15  - 24 < 0 
                  0 < 0 

x^2 - 5x   - 24 < 0 
8^2 - 5(8) - 24 < 0 
 64 - 40   - 24 < 0 
              0 < 0

So no, the values of -3 and 8 would only work if the original equation was: x^2 - 5x - 24 ≤ 0. Thus, we get an answer of -3 < x < 8, instead of -3 ≤ x ≤ 8. Now let's try a value that should work, x = 0:

x^2 - 5x   - 24 < 0 
0^2 - 5(0) - 24 < 0 
  0 - 0    - 24 < 0 
            -24 < 0

It does work, good. Now try two that shouldn't work (9 and -4):

x^2 - 5x   - 24 < 0 
9^2 - 5(9) - 24 < 0 
 81 - 45   - 24 < 0 
             12 < 0

   x^2 - 5x    - 24 < 0 
(-4)^2 - 5(-4) - 24 < 0 
    16 - (-20) - 24 < 0 
    16 +   20  - 24 < 0 
                 12 < 0

They don't work, good. So, we have verified our answer. A much quicker way to solve this would be with a graphing calculator, but I'm guessing you haven't gotten that far yet. StuRat 23:29, 30 September 2007 (UTC)

I'm feeling generous, so will also show you the setup for problem 19:

(19) A rectangular field has a perimeter of 306 feet and is to have an area at least 5670 square feet. Within what bounds must the length, x, of the rectangular field lie?

The formula for the perimeter of a rectangle is twice the height plus twice the length equals the perimeter. Since they said to use x for the length, let's use y for the height, to get:

2x + 2y = 306

Now solve for y:

     2y = 306 - 2x

      y = 153 -  x

The formula for the area of a rectangle is height times length, or xy, in our case. This gives us:

xy ≥ 5670

Now plug in the value for y we calculated in terms of x previously:

x(153 - x) ≥ 5670

Now solve for x:

153x - x^2 ≥ 5670

-x^2 + 153x - 5670 ≥ 0

(-x + 63)(x - 90) ≥ 0

(63 - x)(x - 90) ≥ 0

Now you have it just like the previous problem I showed you, except that you are looking for a positive or zero value on the left. This means the two factors must either both be positive (or zero) or both negative (or zero). See if you can solve it from there. Hint: You don't need to chart every integer, just look at those around x = 63 and x = 90:

 x 63-x  x-90
-- ----  ----
91
90
89
 .
 .
 .
64
63
62

Let me know if you need more help with this problem.StuRat 00:24, 1 October 2007 (UTC)

Some hints for problems that I don't think have been covered by anyone else yet: (3) Notice that writing "f = aq + r" is saying that "f equals a times q, plus r", which is equivalent to "f divided by a gives q, remainder r". So, you should be able to perform polynomial long division to get the answer, or, as I suspect you would have been taught about, the polynomial remainder theorem. (4) I assume from the context that "standard form" is y = a(x-h)^2 + k. Can you write it like that? Does that give you anything that can be used to find the vertex, etc.? (19), (26) and (27) are all of the same form, but arranged differently. In each case, you can write it as (quadratic function) > 0, and solve in a fashion similar to StuRat's hints above.

Again, more specifics on how far you got with each question will help us know how much we can help you. Confusing Manifestation 23:40, 1 October 2007 (UTC)

to the power of negative?

What's ten to the power of negative one, two , three etc. and how do you figure it out? (Wookiemaster 17:29, 30 September 2007 (UTC))

10^-1 = 1/10, 10^-2 = 1/100, 10^-3 = 1/1000, and so on. Exponent#Negative_integer_exponents has the full rule. The reason negative exponents are defined this way is so the formulas in Exponent#Identities_and_properties hold true. --Ben Kovitz 17:37, 30 September 2007 (UTC)

Thank you very much but can I get the answer in decimals(Wookiemaster 17:40, 30 September 2007 (UTC))

"What's ten to the power of negative one, two , three etc." 0.1 0.01 0.001 24.250.139.137 17:47, 30 September 2007 (UTC)ForgotMyLogin

Thank you very much but can I get the answer in decimals? In the old days, you do long division. Nowadays kids just punch it into their calculators and get a decimal answer. Does this answer your question? It would be much much better if you understand the basic concept of division (mathematics). 202.168.50.40 23:36, 30 September 2007 (UTC)

In the old days you punched it into your pocket calculator. Nowadays kids just go to Wikipedia. —Blotwell 05:22, 1 October 2007 (UTC)

Or (for those with really big pockets) Google AndrewWTaylor 08:02, 1 October 2007 (UTC)

Dividing decimals?

I know that when dividing decimals you have to move the decimal to the right as many times as you have to for it to become a whole number.Ex. (4.56 divided into 0.8) becomes (45.6 divided into 8) My question is "when we get the answer do we put the decimal back where it was or do we leave it wher we put it? (Wookiemaster 17:53, 30 September 2007 (UTC))

I haven't heard of that method of dividing decimals. If I understand the method correctly, then you would leave the decimal where you put it. Better, though, would be to try both ways yourself on a few examples where you already know the answer. Then the right way will become clear. Ben Kovitz 18:48, 30 September 2007 (UTC)

Let us try mathematics rather than folklore. The number 45.6 is 10×4.56, yes? Likewise 8 =10×0.8. Now compare ^0.8⁄_4.56 to what you propose:

${\frac {10\times 0.8}{10\times 4.56}}={\frac {8}{45.6}}.\,\!$

This works, but I, too, find it a puzzling and unnecessary distraction. --KSmrq^T 20:10, 30 September 2007 (UTC)

What is the density of triangular-free numbers?

What I mean is: What's the density of natural numbers that fail to be divisible by 3,6,10,15,21,..n(n+1)/2..? (I left out divisibility by the triangular number 1, by necessity, just as it must be left out for squarefree numbers) It's well known that the square-frees have density 6/pi^2, so I would find it fascinating to know, if a density formula for triangular-frees exists, whether it explicitly involves pi. Thanks, Rich Peterson —Preceding unsigned comment added by 130.86.14.86 (talk) 21:34, 30 September 2007 (UTC)

I don't know, but the sequence itself is in the Online Encyclopedia of Integer Sequences as A112886. —David Eppstein 21:53, 30 September 2007 (UTC)

I ran some code in R to find the triangular-free numbers up to 100,000 and show their distribution, and the density does seem to converge to something resembling 2/3 (in fact, there are 66,666 triangular-free numbers in the range). Of course this doesn't prove anything, but it does look like a promising result, especially since it's such a fast convergence (see below graphs, where the first is on normal axes and the second is semilog). Confusing Manifestation 06:30, 2 October 2007 (UTC)

• 
  The density of triangular-free numbers up to 100,000. Appears to converge to two-thirds.
• 
  The convergence is fast, even on semilog axes.

But doesn't the density have to be less than 2/3 because we eliminate straight away 1/3 of integers which are divisble by 3 ? This also eliminates numbers divisible by 6, but after that we also have to eliminate those numbers divisible by 10. This gives an upper limit on the density of (2/3)x(9/10) or 0.6. Gandalf61 15:50, 2 October 2007 (UTC)

I've looked at this again, and I only get 55,622 triangular free numbers in the first 100,000 numbers. I looked at it in blocks of 5000 integers, and the lowest was 2774 (twice), highest was 2788. The density in each of the 20 blocks of 5000 seemed fairly constant - with some noise. As Gandalf points out, there are 66,667 numbers not divisible by 3 in the first 100,000, so putting in the fact that 10 and 20 are triangular-free, must mean that the answer is less than 66,666 as you mentioned. Richard B 17:24, 2 October 2007 (UTC)

OK, I obviously used a dumb choice of method that gave me completely wrong answers (actually, it was pretty much knocking off only the multiples of 3), but if you look above I've uploaded new, apparently more correct versions, which still seem to get close to some kind of value around 0.56. I also changed it to cover values up to 10,000,000, and took off some values from the start which skewed the axes. Confusing Manifestation 02:17, 3 October 2007 (UTC)

October 1

Math

I would like to know what are the instruments of addition? (A list of instruments) —Preceding unsigned comment added by 71.252.34.246 (talk) 01:39, 1 October 2007 (UTC)

Clarinets of summation? Strad 02:18, 1 October 2007 (UTC)

Organs of increase? -- JackofOz 03:40, 1 October 2007 (UTC)

Maybe he means the cymbals.

+,OR,||,...--Mostargue 04:50, 1 October 2007 (UTC)

Perhaps he means devices used to perform addition calculations, like an abacus, adding machine, or calculator. StuRat 14:51, 1 October 2007 (UTC)

What's the biggest set in use?

Okay, we have the finite sets, then the integers, then the reals, the the complex, but what after that?--Mostargue 05:03, 1 October 2007 (UTC).

The set of irrational complex numbers. 24.250.139.137 06:25, 1 October 2007 (UTC)ForgotMyLogin

The previous response is nonsense. See cardinal number --- there's a whole theory about this sort of thing (and there is no biggest set). If you want an explicit example of something larger, I should first note that the cardinality (think "size") of the complex numbers is the same as that of the reals. For an actual example of something larger, see power set; the power set of the reals is larger than the set of reals. (Also, I should note that the reals have the same cardinality as the power set of the integers.) As for biggest set "in use," that's presumably subjective and I have no idea, but the references I gave should get you started.Kfgauss —Preceding signed but undated comment was added at 06:31, 1 October 2007 (UTC)

It's difficult to answer your question, because the mathematical objects you name don't really fit into a sequence. Most of the sets you describe are sets of numbers, which fit into a hierarchy:

1. The natural numbers
2. The integers
3. The rational numbers
4. The real numbers
5. The complex numbers

Each step in this hierarchy is obtained from the last by making the set closed under certain arithmetic operations:

1. The integers are obtained from the natural numbers by closing under subtraction
2. The rational numbers are obtained from the integers by closing under division
3. The real numbers are obtained from the rational numbers by closing under the process of taking limits
4. The complex numbers are obtained from the reals by closing under square roots.

In some sense, the process of adding more numbers is complete at this point. In fact, the fundamental theorem of algebra states that any polynomial equation over the complex numbers has a solution. On the other hand, various generalization of the complex numbers have been defined, including the quaternions and Clifford algebras. Matrices can also be viewed as a generalization of the concept of number to multiple dimensions.

If you're more curious about sets than numbers, the first step is indeed finite set, followed by infinite set (such as the integers or rationals), followed by uncountable set (such as the real or complex numbers). See also cardinality and the other links provided by Kfgauss. It is important to realize that not every set in mathematics is a set of numbers: as has already been mentioned, the simplest way to make larger sets is to consider sets of sets (e.g. power sets). Jim 06:47, 1 October 2007 (UTC)

If you think my first answer is nonsense then prepare for this one: The largest set is defined recursively as the superset of itself. So just think of any set, take the superset of it and you will get the largest set if you keep applying this definition an infinity of times. 24.250.139.137 08:17, 1 October 2007 (UTC)ForgotMyLogin

Befor anyone can understand your response, I think you have to explain what you mean by "take the superset". You are clearly not using superset in the standard sense of the term, because (a) a superset of A is any set that contains A as a subset - how do you distinguish the superset ? (b) with infinite sets, a set and one of its (proper) supersets can have the same cardinality - for example, the set of even integers has the same cardinality as the set of integers - so how do you ensure that your operation of "take the superset" produces a superset that has a greater cardinality ? Gandalf61 08:55, 1 October 2007 (UTC)

Perhaps anon has meant power set, or the union of a set with its power set? Even if so, the proposed method is flawed, since you can continue after taking the union of ω applications, and anon hasn't specified any other interpretation of "applying an infinity of times". Of course, in ZFC set theory there is no largest set; at best, there is the collection of all sets, which is the largest class but is not a set.-- Meni Rosenfeld (talk) 09:24, 1 October 2007 (UTC)

Hi. Yes I am talking about cardinality. A finite set has a smaller cardinality than an infinite set, right? The only thing I had to think twice about was the jump from reals to complex. The set of complex numbers, is it a higher cardinality of the set of real numbers? --Mostargue 18:50, 3 October 2007 (UTC)

No, the cardinality is the same. Think of a complex number in terms of its real and imaginary parts, and interleave the decimal digits of those to get the decimal expansion of a real number. Well, roughly speaking anyway. You might have to do a little work to make sure the 999... problem doesn't cause any ambiguity.

This is the result that made Cantor write "Je le vois, mais je ne le crois pas". --Trovatore 19:03, 3 October 2007 (UTC)

I think I get it, but I'm not quite sure. Is there any formal proof that complex numbers and real numbers have the same cardinality?--Mostargue 21:49, 3 October 2007 (UTC)

Yes, there is. Trovatore has indicated one method; the 'little work' referred to is not particularly difficult, and can be made easier by use of the Schroder-Bernstein theorem. Algebraist 16:44, 4 October 2007 (UTC)

Coupling from the Past

OK, so random Wiki-browsing led me from Markov Chains (which I understand) to MC Monté Carlo (which I kinda get) to Coupling from the Past, which I get, a little. I get enough to wonder about one thing.

In this paper, they rightly make the point that the first place where all the different chains coalesce might not be able to take a certain state, even if that state has a presence in the stationary distribution (e.g. if that state is only "fed" directly from a previous state, where all the chains would have had to have coalesced one time interval ealier). But they then say that if, rather than looking at the first time T after 0 when the chains coalesce, you look at the first time M such that STARTING at -M the chains coalesce at 0, THEN you get an accurate representation of the stationary distribution.

Is this because, when you're running "from the past", adding the new random seeds (r_t) to the start of the chain means that the later chain is just "messed up" somehow, whereas adding the seeds at the end of the chain doesn't have this "messing up" effect?

So yes. This is all pretty muddled. I don't usually come across maths I don't understand! I hope that, despite this subject not having a wiki article, someone might be able to help me out here ^^

Rawling4851 15:47, 1 October 2007 (UTC)

Your link doesn't seem to work for me, and I don't quite get what you mean by "messing up", but have you checked out the original Propp & Wilson paper available here?

The basic trick in the proof that the output of the algorithm is from the stationary distribution of the chain, is that once you have coalesced the n-step maps, you can conceptually extend the chain further backward, as if you had run the chain forward for a large fixed number of steps. Then you can let the number of steps approach infinity, and you can do this without requiring any extra work in the algorithm.

On the other hand, if you run multiple copies of the chain forward until they coalesce, the same extension trick does not work - you would have to actually run the chains forward for a large, fixed number of steps in order to approach the stationary distribution. 84.239.133.38 19:16, 1 October 2007 (UTC)

Ah, I Think I'm a little out of my depth. Better keep taking my Probability units, eh? Cheers Rawling4851 07:57, 2 October 2007 (UTC)

Hey, if you are interested, it's worth your while to take another look. It is a cool algorithm and the proof is surprisingly simple. 84.239.133.38 17:22, 2 October 2007 (UTC)

Rydeberg formula question

According to Rydberg formula, this is the equation:

${\frac {1}{\lambda }}=R_{\mathrm {H} }\left({\frac {1}{n_{1}^{2}}}-{\frac {1}{n_{2}^{2}}}\right)$

Now, I was asked to solve for "n₂". So this is what I did:

${\cfrac {1/\lambda }{R_{\mathrm {H} }}}=\left({\frac {1}{n_{1}^{2}}}-{\frac {1}{n_{2}^{2}}}\right)$

${\cfrac {1/\lambda }{R_{\mathrm {H} }}}-{\frac {1}{n_{1}^{2}}}=-{\frac {1}{n_{2}^{2}}}$

$-{\cfrac {1/\lambda }{R_{\mathrm {H} }}}+{\frac {1}{n_{1}^{2}}}={\frac {1}{n_{2}^{2}}}$

$-{\cfrac {R_{\mathrm {H} }}{1/\lambda }}+{\frac {n_{1}^{2}}{1}}=n_{2}^{2}$

${\sqrt {-{\cfrac {R_{\mathrm {H} }}{1/\lambda }}+n_{1}^{2}}}=n_{2}$

But then I'm still wrong. What did I do wrong?? --Jeevies 18:47, 1 October 2007 (UTC)

$-{\cfrac {1/\lambda }{R_{\mathrm {H} }}}+{\frac {1}{n_{1}^{2}}}={\frac {1}{n_{2}^{2}}}$

$-{\cfrac {R_{\mathrm {H} }}{1/\lambda }}+{\frac {n_{1}^{2}}{1}}=n_{2}^{2}$

The incorrect step is between these two lines. It looks like you tried to take the inverse of both sides, but the inverse of A + B is not 1/A + 1/B. —Keenan Pepper 18:52, 1 October 2007 (UTC)

Then what is? --Jeevies 19:10, 1 October 2007 (UTC)

The inverse of A + B is 1 / (A + B). That probably doesn't help solve the problem though. --LarryMac | Talk 19:21, 1 October 2007 (UTC)

I hate to tell you this, and I don't know how to say it without sounding patronizing, but by the time you are assigned such problems as solving the Rydberg formula for hydrogen for one of the two numbers, you ought to be well beyond the stage in which a student makes such errors. It seems to me that you'd be well advised to invest some time in brushing up on elementary algebra.  --Lambiam 22:40, 1 October 2007 (UTC)

>_< [[Chilling Effect|Chilling.] It took me a while to get that whole thing to "parse" right! Anyways, I looked through a formula sheet, but I'm still confused. How do I do this? Please? --Jeevies 23:06, 3 October 2007 (UTC)

Help teaching Scientific Notation?

I am currently doing field experience in a Middle School in New York State, and one of the 7th grade students is having trouble with scientific notation. It seems the heart of the problem is that he has not learned the faster method of multiplying by ten, since he uses the classic

   1262
   x 10
   -------

to solve what should be a simple mental calculation. Unfortunately, I am at a loss as to how to help this student. Does anyone have any ideas on how I can teach the method of adding zero/moving the decimal, other than standard math classroom methods? Some concrete examples would be very helpful as well, since I can't think of any. US Sports examples would be especially helpful, as I am know next to nothing about sports.

209.51.73.60 21:58, 1 October 2007 (UTC)

Can't you tell the student that in a multiplication you are allowed to shift final digits 0 from one number to the other: 7 × 800 = 70 × 80 = 700 × 8?  --Lambiam 22:24, 1 October 2007 (UTC)

This kind of learning gap can be usually be addressed by means of visualization. Building on the example provided by Lambiam, you might suggest that 70 × 80 means "exactly the same thing" as 700 × 8 because the rules of multiplication allow the zeros to "jump" back and forth.

Eventually, you will have to introduce that 700 × 8.0 and 7000 × .8 and 70000 × .08 also mean "exactly the same thing" but that the "jump" rules are a little different when dealing with decimals, because you have to remember about the "dot". Then you just give the student a few iterations of applying the "jump" rules on their own.

Some math teachers wince at this kind of method, because it is rather crude, imprecise, inelegant and informal (for example, this assumes Base10), but many times a student's eyes will light up because they will be able to picture the symbolic manipulations more easily, and it will be less "abstract" to them. Sometimes this is far more helpful than trying to talk about the numbers in the context of sports scores or applied mathematics, because the student is not struggling with practical application of the knowledge, but with the fundamental customs of symbolic knowledge representation as have been handed down to us by fiat. dr.ef.tymac 02:08, 2 October 2007 (UTC)

I humbly suggest consulting the experienced teachers at The Math Forum.

Like all personalized teaching, it's a matter of probing the student's existing mental model and helping the student construct a more effective revised and expanded model. You've made a start, and it looks like several concepts are missing.

Better understanding of multiplication will be crucial. At most a dozen exercises of multiplying by powers of 10, plus a tiny amount of discussion, should solidify the idea of shifting decimal points. Ideally, the student understands the theory of position notation, but I'm guessing not.

Next we need to bring in the commutative law, so that we can go from (a×10^b)×(c×10^d) to (a×c)×(10^b×10^d). We fervently hope this is already solidly familiar, and merely reinforce its existence and applicability.

The next step is to discuss adding exponents; this is an important concept that must not be taken as "obvious". Thus we gain the ability to reduce 10^b×10^d to 10^b+d. Be sure to cover negative exponents.

Finally we come to the hardest part. Having introduced the freedom to slide decimal points around by adjusting exponents, we now must confront the need, in addition and subtraction, to align decimal points. Our theoretical basis is the distributive law. That is, given (a×10^b)+(c×10^b), where the exponent in both summands is b, we can write (a+c)×10^b. Without great emphasis and a solid mental model a student will be irresistibly tempted to skip the alignment step, with disastrous consequences.

Most likely we will have already discussed a normal position for the decimal point already. Since our calculations will often leave the decimal point somewhere else, we must finally remember to "normalize" the result.

At each stage we may wish to have a few exercises/examples to display a pattern, then discussion of theory to point out the pattern and its importance and use, then adequate exercise to build competence and confidence.

I haven't offered specific exercises to use (again, try The Math Forum); I do hope the framework is helpful. --KSmrq^T 05:09, 2 October 2007 (UTC)

Response to KSmrq

Like all personalized teaching, it's a matter of probing the student's existing mental model and helping the student construct a more effective revised and expanded model

Ideally, the student understands the theory of position notation, but I'm guessing not.

I humbly suggest consulting the experienced teachers at the Math Forum.

All of these are excellent points, and in fact reinforce my original emphasis on (the possibility) of using fanciful visualizations as an option. Because it tends to be the last method considered by mathematicians, and the first method to make sense to young people.

Concrete and simplified example:

To clarify what I am talking about, here's a very simple example. The question is:

 "When dealing with inequalities, why do you flip the inequality sign when   
 multiplying by a negative number?" 

The following links all show different mathforum.org responses to this "stumbling block" for students, with a summary description of the answer methodology enclosed in parenthesis:

• [2] ;; (algebraic description )
• [3] ;; (verbal description [based on textbook definition] )
• [4] ;; (visualized description [number line] )
• [5] ;; (visualized description [balanced scales with "anti-gravity"] )
• [6] ;; (visualized description [number line and balanced scales])

All of the answers here are legitimate, but note the last answer. The respondent clearly indicates there are different ways to think about this, by saying:

   The answer depends on whether you need to convince yourself or a 
   mathematician. Here's how I like to convince kids ...

This answer is quite unique, because it is the only one out of the bunch that acknowledges there are different ways to "convince" someone of this basic principle of arithmetic. Each kind of "convincing" employs a different methodology. Some students simply are not "convinced" by algebraic descriptions or an explanation of the distributive law, even though these are the "modalities of choice" for a trained mathematician, even though these are (ostensibly) the principles that you are intending to convey to your students.

All of this just re-emphasizes the point already made: it's a matter of probing the student's existing mental model -- and sometimes that mental model does not respond well to the symbolic and theoretic apparatus of arithmetic and algebra. dr.ef.tymac 15:34, 2 October 2007 (UTC)

October 2

"Super" composites

Hi,
For no particular reason I recently thought to myself what would be the opposite of a prime number. Well I suppose the conventional answer would be a composite number. But what I then thought was what would really be an opposite would be a number which is divisible by all numbers below it - ie it has every number as a factor. Yes that is a bit of an abstract concept but I was wondering if such a thing had been described anywhere and if it had a particular name? An interesting thing about this "ideal" number would be that it would be impossible to calculate since there are infinite primes and by it's definition it would need to be divisible by all primes. And hence it could only ever be defined within a limited set.

Anyway I had a shot at calculating these numbers in progressive sets and this is what I got for the first 25:
1, 1
2, 2
3, 6
4, 12
5, 60
6, 60
7, 420
8, 840
9, 2520
10, 2520
11, 27,720
12, 27,720
13, 360,360 (interesting no.!)
14, 360,360
15, 360,360
16, 720,720
17, 12,252,240
18, 12,252,240
19, 232,792,560
20, 232,792,560
21, 232,792,560
22, 232,792,560
23, 5,354,228,880
24, 5,354,228,880
25, 26,771,144,400
Where the LHS shows the number at which all integers below it are factors of the RHS number. (ie 2520 is divisible by all numbers from 1-10) Now I know this in now way constitutes a formula or anything but I noticed the following whilst generating this list: you need to multiply "the number" each time you reach a prime; each time you reach a square number you have to multiply by the lowest prime factor of the square root of this number (eg for 16 you needed to multiply the previous figure by 2 which is the lowest prime factor of 4 and when I reached 25 I needed to multiply by 5); I haven't generated enough numbers yet (probably should have gone to 27) but judging from 8 it looks the same "square rule" applies for cubic numbers...
Anyway would appreciate any comments regarding this! --Fir0002 03:29, 2 October 2007 (UTC)

The closest thing I can think of is the Primorial, but that's not quite the same. It is a number that is the product of all primes below a given number. - Rainwarrior 03:39, 2 October 2007 (UTC)

Oh, there's a link in that article to Highly composite number, which is getting a lot closer to your idea. - Rainwarrior 03:40, 2 October 2007 (UTC)

And continuing on Superior highly composite number is even closer. - Rainwarrior 03:43, 2 October 2007 (UTC)

Thanks for your links Rainwarrior, nearly what I was after. But they were useful anyway coz it lead me to Perfect number which is pretty interesting too. --Fir0002 03:57, 2 October 2007 (UTC)

A simple way to define this sequence is that its nth term is the Least common multiple of the integers 1 through n. It's in the On-Line Encyclopedia of Integer Sequences at [7]. --tcsetattr (talk / contribs) 03:49, 2 October 2007 (UTC)

Thanks a lot tcsetattr!! --Fir0002 03:57, 2 October 2007 (UTC)

Note that this sequence has very little to do with your primary idea of "a number which is divisible by all numbers below it". This is clearly impossible (for numbers greater than 2), since n would never be divisible by $n-1$. -- Meni Rosenfeld (talk) 08:52, 2 October 2007 (UTC)

That's true but I already acknowledged that such a number would be impossible due to the infinite number of primes --Fir0002 10:35, 2 October 2007 (UTC)

On a similar note, n! has the greatest number of divisors (namely, 2^n-1) out of numbers less than n!. —Preceding unsigned comment added by 122.148.69.121 (talk) 11:02, 2 October 2007 (UTC)

No. n! has 2^n-1 divisors for n = 1 to 5, but less for all n > 5. For example, 6! has 30 divisors. And numbers with more divisors than all smaller numbers are called highly composite numbers. The only factorials among them are n! for n = 0 to 7. PrimeHunter 12:07, 2 October 2007 (UTC)

Did anyone else notice that the digital number of every super composite after 9 (including 9 too) is 9? Xhin Give Back Our Membership! 21:45, 2 October 2007 (UTC) —Preceding unsigned comment added by Xhin (talk • contribs)

Wait, disregard that, I forgot that the digital number of any number that's multipled by nine is nine :( Xhin Give Back Our Membership! 21:46, 2 October 2007 (UTC) —Preceding unsigned comment added by Xhin (talk • contribs)

You must be referring to digital root. PrimeHunter 00:59, 3 October 2007 (UTC)

Mathematics problem!Need help please!

The product and quotient rules give you the ability to symbolically calculate the derivative of a wide range of functions. However, many calculators and almost every computer algebra system (CAS) can do this work for you. Discuss why you should learn these basic rules anyway. —Preceding unsigned comment added by Nelsonkhoo (talk • contribs) 05:23, 2 October 2007 (UTC)

The most basic answer is that it's impractical to reach for a calculator every time you want to take a derivative. Students who fail to learn the product rule or quotient rule will slow at differentiating, and therefore slow at solving calculus problems. They will also find themselves unable to understand integration by parts, they will have difficulty solving differential equations using integrating factors, they will be surprised by the form of the multivariable chain rule, and so forth. Learning mathematics involves mastering each stage before moving on to the next, and students who don't learn these rules will find that future mathematics makes less sense than it would otherwise.

In addition, the idea behind the product rule is one of the most important ideas in calculus: when multiple quantities are changing, calculate the change due to each effect separately and then add. The small errors accrued by neglecting the interaction between different changes can safely be ignored. This idea is almost as important as the product rule itself, and any good student of calculus ought to be able to explain why the product rule works, in addition to merely having it memorized. Jim 06:28, 2 October 2007 (UTC)

It's not as if learning these rules is difficult. Besides, isn't it irresistible to know that, to first order, multiplication resembles addition? For, let λ be a nonzero "infinitesimal" quantity such that λ² = 0; then (1+λa)(1+λb) = 1+λ(a+b).

Actually, we needn't invent infinitesimals; we can use 2×2 matrices.

${\begin{aligned}\Lambda &={\begin{bmatrix}0&1\\0&0\end{bmatrix}}&\quad \Lambda ^{2}&={\begin{bmatrix}0&0\\0&0\end{bmatrix}}\\I&={\begin{bmatrix}1&0\\0&1\end{bmatrix}}&\quad I^{2}&=I\\(I+a\Lambda )(I+b\Lambda )&={\begin{bmatrix}1&a\\0&1\end{bmatrix}}{\begin{bmatrix}1&b\\0&1\end{bmatrix}}\\&={\begin{bmatrix}1&a+b\\0&1\end{bmatrix}}\\&=I+(a+b)\Lambda \end{aligned}}$

This fact is important to the use of homogeneous coordinates for, say, computer graphics; the geometric operation called translation is essentially addition, yet we can use matrix multiplication to do it. --KSmrq^T 07:12, 2 October 2007 (UTC)

One might equally ask, why learn to do basic arithmetic when calculators can do it; or why think about your homework when you can get people on WP:RD to do it for you.. AndrewWTaylor 10:18, 2 October 2007 (UTC)

Also, one can derive more complicated theorems from these rules that a calculator couldn't handle. – b_jonas 09:01, 3 October 2007 (UTC)

Simple: Because there are problems that not even Computer Algebra Systems can solve. Of course, you could write a patch of code for the CAS so that it can solve your problem, but then we come full circle: you need to be able to understand how to solve the base case (much more difficult) in order to implement it. Ultimately, you're going to have to do some math on your own.--Mostargue 05:42, 4 October 2007 (UTC)

Axiom of Choice

Can someone please clear my doubts regarding this equivalence of axiom of choice: Prove that (i) implies (ii) where:

(i) For every nonempty set whose elements are non empty sets there exists a choice function.

(ii) If $\{a_{i}\}$ is a family of nonempty sets indexed by a nonempty set I, then there exists a family $\{x_{i}\}$ with $i\in I$ such that $x_{i}\in a_{i}$ for each $i\in I$.

Here is the proof which is in my book:

Let A be a collection of disjoint sets. We have $A\subset P(\cup _{i\in I}a_{i})$. (P is the power set). By (i) there exists a choice function f on $P(\cup _{i\in I}a_{i})$. Let b be the image of A. Pick an element a $\in$ A, f(a) $\in a\cap b$ since f(a) $\in$ a. Let y $\in$ b where y $\neq$ f(a) thus we have y = f(a') where a' $\neq$ a, and thus y $\in$ a'. Since a and a' are disjoint y $\notin$ a. Thus the only element of b $\cap$ a is f(a).

My doubts are:

• Why do we need to prove this? Won't the choice function f give us the family {f(i)} anyway.
• How is the existence of A guaranteed?
• Can't we take A={a_i} and work through the proof.
• How does b $\cap$ a being f(a) guarantee that (ii) holds.

Thanks a lot. Cheers--Shahab 06:08, 2 October 2007 (UTC)

It seems that the main difference is between a family of sets (as in (ii)) and a set of sets (as in (i)). I feel this proof is very confusingly worded. Here's an attempt at a clearer version; perhaps it will be helpful. I've tried to stay as close to the structure of your proof as I could.

Let A = {a_i} be a collection of disjoint sets, and let U be the union $\cup _{i\in I}a_{i}$ (for universe, or perhaps union). Then since each a_i can be thought of as a subset of the union U, we can say that each a_i is an element of P(U); this lets us think of A as a subset of P(U). There exists a choice function f on P(U); by definition, this maps a subset $S\subset U$ to some element $f(S)\in U$ with the property that the element f(S) must lie within the set S: $f(S)\in S\subset U$.

Now let B be the image of A under f; since A is a subset of P(U), B will be a subset of U: B = {f(a_i)|a_i in A}. That is, for each of our original sets a_i, B contains the "chosen element" f(a_i). At this point we are done, since B is the desired family, but we don't know it yet -- we need to check that each f(a_i) is in a_i where it should be. This is the B $\cap$ a = f(a) that you asked about. That equation says that B contains exactly one element from a_i, and that element is f(a_i) -- this says exactly that B is the family that (ii) claims exists. The proof is a simple idea (I will use a₁ and a₂ for simplicity, but of course your index set I may not be the integers).

By definition of B, f(a₁) is in B, and since f is a choice function f(a₁) is in a₁. We want to show that f(a₁) is the only element of B that lies in a₁. So take some other element y of B. Since the elements of B were defined to be the f(a_i), it must be that y = f(a_i) for some i -- say for simplicity that y = f(a₂). Then since f is a choice function, we must have that y is an element of the set a₂. But a₁ and a₂ are disjoint, so y cannot be in a₁; thus f(a₁) is unique, as desired.

Finally, the existence of A is not in question; you are given a collection A to start with. You're correct that you could take A = {a_i} and work through the proof that way -- in fact, this is what the original proof is doing. Tesseran 07:12, 2 October 2007 (UTC)

Thanks. But how is A given to us. The only thing that is given is the family {a_i} which consists of nonempty sets not non empty disjoint sets. But why can't we simply use {f(i)} in place of {x_i}. The meaning of the choice function guarantees that f(i) is in a_i. --Shahab 07:48, 2 October 2007 (UTC)

Pls Help me on my maths

Wiki, Pls I need wide information on the following mathmatisc topics Logic, Binomial theorems, permutation & Combination —Preceding unsigned comment added by 62.56.133.78 (talk) 12:14, 2 October 2007 (UTC)

Try logic, binomial theorem, permutation and combination. Algebraist 13:23, 2 October 2007 (UTC)

If you have more specific questions, feel free to ask them here. Algebraist 13:23, 2 October 2007 (UTC)

Stats n00b question

Would the quartiles of the standard normal deviation simply be the values for z = 0.25 and z = 0.75? Probably not. I'm confused.--Sonjaaa 19:15, 2 October 2007 (UTC)

No, because z=0 defines the median. The quartiles will lie above and below that. You're looking for the points where 25% of the area under the bell curve. You can find a table in your textbook and also here [8]. Donald Hosek 20:11, 2 October 2007 (UTC)

In which textbook?  --Lambiam 09:35, 3 October 2007 (UTC)

The quartiles of a probability distribution are those values of z for which F(z) = 0.25, 0.50 (the median itself is also a quartile) and 0.75, where F denotes the cumulative distribution function of the distribution. The cumulative distribution function of the standard normal distribution is usually denoted by Φ. Because Φ(−0.6745) = 0.2500... and Φ(0.6745) = 0.7500..., the first and third quartiles are (about) at z = −0.6745 and z = 0.6745. They are each other's opposite because the standard normal distribution is symmetric about 0.  --Lambiam 09:35, 3 October 2007 (UTC)

women

there are only 3 types of women in this world, and the types can be plugged into the equation: y=(a+b)x. if someone knows the answer, this would help me immensly. --Etro —Preceding unsigned comment added by 151.196.108.226 (talk) 20:40, 2 October 2007 (UTC)

The function is actually |a-b|x, A = the amount she wants there to be equality, B = the amount she likes being a woman. X = the amount of anger she has over not being able to be both. One way or another, you end up with a negative number, although if one trait overpowers the other, that number will be enormous, and you consequentially will be screwed (figuratively, yes, but literally, no). Xhin Give Back Our Membership! 21:41, 2 October 2007 (UTC) —Preceding unsigned comment added by Xhin (talk • contribs)

What is a way to represent the infinity symbol with a non-unicode keybord?

that is, how can I represent infinity as a symbol or a couple of symbols with my keyboard, instead of using the obtuse 1/0 ? Much help appreciated ! Xhin Give Back Our Membership! 21:35, 2 October 2007 (UTC)

"oo" is commonly used. —Tamfang 21:59, 2 October 2007 (UTC)

In personal communications and informal settings, many mathematicians use \infty, from the TeX code for $\infty$. Tesseran 23:00, 2 October 2007 (UTC)

Wow, a Unicode keyboard. Wouldn't that be something! —Bromskloss 07:32, 3 October 2007 (UTC)

A Unicode keyboard with one key per character would be … large: Chinese characters, and Korean characters, and Arabic characters, and on and on — each with one key. The Chinese do not use such a keyboard even for "simplified Chinese".

Thus we enter the wonderful world of character encodings.

• For Web pages we can use a numeric entity, a string of ASCII characters of a special form, to obtain any Unicode codepoint. The encoding for the usual infinity character is "&#x221E;", producing "∞" (with the exact appearance depending on the typeface).
• This particular character also has an HTML named entity, "&infin;", making it easier to remember.
• Many browsers and text editors support UTF-8, so if an instance of the character can be found somewhere, copy and paste will work. As it happens, Wikipedia stores its pages using this encoding, which is a default Web standard, so "∞" works fine.
• The MathML standard for Web mathematics markup defines a long list of named entities; don't expect these to work with plain HTML markup.

Your options depend on your needs. For text email, the double-small-o works well; it is usually instantly recognized with no explanation. --KSmrq^T 08:57, 3 October 2007 (UTC)

In HTML one can use the character entities ∞ or ∞ for ∞. Works here on Wikipedia too. —Ilmari Karonen (talk) 08:35, 3 October 2007 (UTC)

October 3

TI-89 Question

I want to solve an equation but restrict the domain to integers. ex. Solve(2^x+1=3^y-1,{x,y}), but i want to restrict it. How would I do that? xxpor ^yo!|_{see what i've done} 02:33, 3 October 2007 (UTC)

You wouldn't. You can't. You'd need a very, very advanced calculator to solve diophontine (sp?) equations, seeing as we can't always find a method for solving them -- that's a number theory thing, not simple math. Sorry! Gscshoyru 02:39, 3 October 2007 (UTC)

I wish to point out that 2^x and 3^y are not polynomials. This makes it even harder to solve. Looking at the equation

2^x + 1 == 3^y - 1

2^x + 2 == 3^y

Since x and y are integers. We can try y=0,1,2,3,4,...

and look for 3^y that are even since "2^x + 2" is even.

Find all 3^y that are even and "3^y - 2" that are of the form 2^x

But of course this is just brute force and not elegant. I must be stupid because I'm having trouble finding integer values of y such that 3^y is even. How odd! 202.168.50.40 03:47, 3 October 2007 (UTC)

In case the original questioner misses the point, the answer is that there are no solutions to 2^x+1 = 3^y-1 in positive integers because 2^x+1 is odd and 3^y-1 is even. No calculator required. Gandalf61 09:26, 3 October 2007 (UTC)

If you don't require the integers to be positive, however, there is one solution.  --Lambiam 09:42, 3 October 2007 (UTC)

lol @ Lambiam's approach. I love loopholes like that!

• $2^{x}+1=3^{y}-1\!$

--Mostargue 18:29, 3 October 2007 (UTC)

Also, see Diophantine equation.--Mostargue 18:34, 3 October 2007 (UTC)

Haha, I mixed up, its $x^{2}+1=x^{3}-1$. but there must be someway in general restrict the domain to integers. xxpor ^yo!|_{see what i've done} 18:36, 3 October 2007 (UTC)

Again, see diophantine equations. In the case that you've given, though, with only one variable, there are exactly three complex solutions (possibly real, possibly rational, possibly integer). With two variables, the problem becomes much more complicated. Although $y^{2}+1=x^{3}-1$ can be easily re-written in a standard form as an elliptic curve and solved using those methods. Donald Hosek 20:16, 3 October 2007 (UTC)

• $x^{2}+1=x^{3}-1$

Collect like terms:

• $x^{3}-x^{2}-2=0$

From here, you can use synthetic division to find the roots.

Just from looking at it right now, I can tell there is an odd number of roots between 1 and 2. So, either you have 2 more chances of finding getting integer roots, or none. Good luck ^_^.--Mostargue 05:36, 4 October 2007 (UTC)

Actually, that's not quite right. If the single-variable equation has any integer solutions, they can be any of $\{-2,-1,1,2\}$. The simplest approach is not to use synthetic division though. It's far easier to just plug each of these four possible values into the polynomial $x^{3}-x^{2}-2$ and see if they work out (here's a hint, they don't). I think the most likely scenario is that the OP was really looking for integral solutions to the elliptical curve $y^{2}=x^{3}-2$. Donald Hosek 21:24, 4 October 2007 (UTC)

Linear System Elimination Method

I can use the elimination method to solve equations, but I can't understand why it works, which really bothers me when I use it. My algebra book explains absolutely nothing about why anything is true (which is extremely frustrating!). Could someone please give me an explanation of why this works. Thanks in advance. Imaninjapirate^{talk to me} 03:24, 3 October 2007 (UTC)

Gaussian elimination, which I believe is the method being discussed, relies on the fact that the solution to a system of linear equations is not changed under any of the following operations:

1. Multiplying both sides of one equation by a non-zero number. (e.g. 2x = 2 has the same solution as 4x = 4).
2. Swapping two of the equations around. (e.g. 3x + 2y = 5, x - y = 1 has the same solution as x - y = 1, 3x + 2y = 5).
3. Adding a multiple of one of the equations to another. (A little trickier, but if you just think in terms of adding two equations together, then 3x + 2y = 5, x - y = 1 has the same solution as 3x + 2y = 5, (x - y) + (3x + 2y) = 4x + y = 5 + 1 = 6).

Gaussian elimination is then just a systematic method of applying these three operations to simplify the equations. Confusing Manifestation 03:56, 3 October 2007 (UTC)

We can think of a system of linear equations in many ways, each providing insights and benefits.

• When we have a single linear equation in a single variable, we can always write it as ax = b. That is, a constant multiplied times a variable equals a constant. Assume a is not zero (important!); then it has a reciprocal (a multiplicative inverse) denoted by ¹⁄_a or a⁻¹. If we multiply both sides of an equation by the same quantity, the truth of the equality is preserved. Since a⁻¹a = 1 (by definition), we easily obtain x = a⁻¹b, or x = ^b⁄_a as a solution.
• In advanced mathematics, we can use matrices to follow an almost identical approach with a system of n equations in n variables, writing Ax = b. The definition of "multiplicative inverse" becomes richer, so we need to be more careful to decide if A⁻¹ exists. For large systems, elimination is almost always approached using matrices, though explicit use of an inverse is rare.

But let's look at two linear equations in two variables. Generically, the form is

${\begin{alignedat}{3}a_{11}x_{1}&+&a_{12}x_{2}&=b_{1},\\a_{21}x_{1}&+&a_{22}x_{2}&=b_{2}.\end{alignedat}}$

As before, we can multiply either equation by a constant (on both sides). We can also preserve equality by adding equal quantities to both sides. That sounds simple enough, but consider that the left side of the first equation equals its right side. Thus we can add the left side of the first equation to the left side of the second, and the right side of the first equation to the right side of the second. We call this "adding the two equations".

Assume that a₁₁ is not zero (again, important). Then it has an inverse, and we may multiply by it.

$a_{11}x_{1}+a_{12}x_{2}=b_{1}\longrightarrow x_{1}+a'_{12}x_{2}=b'_{1}\,\!$

Save this for future reference; but now multiply the new equation by −a₂₁.

$x_{1}+a'_{12}x_{2}=b'_{1}\longrightarrow -a_{21}x_{1}+a''_{12}x_{2}=b''_{1}\,\!$

We need not assume a₂₁ is nonzero, though we're wasting effort if it is. Because our next step is to add this revision of the first equation to the second. When we do, we eliminate x₁; hence the name of the method.

$a_{21}x_{1}+a_{22}x_{2}=b_{2}\longrightarrow a'_{22}x_{2}=b'_{2}\,\!$

But now our second equation has the simple form of a single linear equation in a single variable, which we already know how to solve so long as a′₂₂ is nonzero.

$a'_{22}x_{2}=b'_{2}\longrightarrow x_{2}=b''_{2}\,\!$

Having solved for x₂, we next multiply both sides of this final form of the second equation by −a′₁₂.

$x_{2}=b''_{2}\longrightarrow -a'_{12}x_{2}=b'''_{2}\,\!$

The final step is easy to anticipate; we add this equation to the saved version of the first equation.

$x_{1}+a'_{12}x_{2}=b'_{1}\longrightarrow x_{1}=b'''_{1}\,\!$

Et viola! The first equation displays the solution for x₁, as the second equation does for x₂.

Thoughout, we used only three procedures:

• Given a nonzero coefficient, find its multiplicative inverse.
• Multiply both sides of an equation by a constant.
• Add one equation to another.

For safety, we must include another step:

• Order the equations so that the first equation has the largest coefficient for x₁ (in magnitude).

The methodical approach here used for two equations in two variables extends readily to n equations in n variables. In matrix form, we call it LU decomposition with partial pivoting. (To "pivot" is to reorder the equations.) It is quite likely the single most important, and most heavily used, algorithm in all of applied mathematics. --KSmrq^T 10:20, 3 October 2007 (UTC)

sum of sum of sum... of factors

You start off with a number such as 138. Using the definition given in Perfect number, the sum of its factors is 150. The sum of the factors of 150 is 222. If you continue taking the sum of the factors, you eventually (after 168 times) get a prime number which has a sum of 1. My question is, are there any starting numbers which will never lead to 1 and continue increasing infinitely? I am excluding perfect numbers, numbers which lead to perfect numbers (25 leads to 6), and numbers such as 220 and 284 which lead to each other. 68.231.151.161 06:04, 3 October 2007 (UTC)

There are periodic sequences, such as Amicable numbers. What you are describing is an Aliquot sequence, and it is unknown whether or not there are infinite sequences (that do not repeat). - Rainwarrior 06:23, 3 October 2007 (UTC)

Wow thanks... that's exactly what I was looking for. 68.231.151.161 06:42, 3 October 2007 (UTC)

According to MathWorld, the longest periodic aliquot sequence known has period 28 and smallest member 14316. It is also interesting to note that there are fairly small numbers (like 276) whose aliquot sequences are not completely known, because they go above the range of practical factorisation. Gandalf61 09:10, 3 October 2007 (UTC)

competition

Has USA announced any competition which involves solving of 10 tough problems?If so please inform me of that. 59.96.27.63 11:50, 3 October 2007 (UTC)

There is a list of seven tough problems known as the Millenium Prize Problems, set by the Clay Mathematics Institute. One of the six has been solved. See also Unsolved problems in mathematics.  --Lambiam 12:21, 3 October 2007 (UTC)

If you're talking about national versions of the International Mathematical Olympiad, take a look at USAMO. Black Carrot 07:56, 4 October 2007 (UTC)

And at the university level, there's the Putnam. That's 12 questions, though. J Elliot 16:06, 4 October 2007 (UTC)

vivinia's theorem

I also wanted to know about vivinia's theorem. 59.96.27.63 11:50, 3 October 2007 (UTC)

Could it be that you are looking for Viviani's theorem?  --Lambiam 12:25, 3 October 2007 (UTC)

Sin2A problem

Can you prove sin2A=2sinAcosA geometrically? 59.96.27.63 11:50, 3 October 2007 (UTC)

Sure. Take an isosceles triangle with "top" angle equal to 2A. The area of the triangle, using the formula area = ½bh in which b = base and h = height, can be computed two ways, depending on which side is taken as the base. Equating the two gives you the desired result.  --Lambiam 12:16, 3 October 2007 (UTC)

You can prove most trig identities geometrically. Just rewrite the trig function as its geometric definition. Let a,b,c be the sides of a right triangle.

• $\sin A={\frac {b}{c}}$
• $\cos A={\frac {a}{c}}$
• Thus:$2\sin {A}\cos {A}=2{\frac {ab}{c^{2}}}$

Remember that because it is a right triangle, $c^{2}=a^{2}+b^{2}$ See if you can figure out the rest =)--Mostargue 17:54, 3 October 2007 (UTC)

Perhaps you, 59.96.27.63, are interested in a book called "Proofs Without Words" (well, they are actually two books). There are proofs for what you just asked and many, many more. --Taraborn 20:17, 3 October 2007 (UTC)

Comparing complex numbers

Do the inequality symbols <, >, ≤, and ≥ have any meaning under complex numbers? My first guess was to take the absolute value of the complex numbers and compare them, but that's performing a function on them, which may change the equality (x > y does not imply f(x) > f(y), which definitely doesn't hold true for the absolute value function ).--Mostargue 18:59, 3 October 2007 (UTC)

No, you can't use < and > on the complex numbers, though you can of course on their moduluses.--PaulTaylor 19:09, 3 October 2007 (UTC)

There is no standard meaning. You could introduce your own ordering, for example by defining:

u + iv ≤ x + iy  ↔  u < x OR (u = x AND v ≤ y).

Then 1 < i < 2. This is essentially the lexicographic ordering on a pair of numbers. However, for complex numbers this is not a particularly useful relation for a working mathematician. In particular, it would destroy the symmetry between i and −i: if P[i] is the conclusion of a theorem involving i, then P[−i] is also a true proposition. For example, take P[j] := (1+j)(2–j) = (3+j). Then P[i] is the proposition (1+i)(2–i) = (3+i), which is true, and therefore P[−i], the proposition (1−i)(2+i) = (3−i), is also true. This would be broken by taking P[j] := 0 < j, which is true if you substitute i for j, but false when you do this with −i.  --Lambiam 20:44, 3 October 2007 (UTC)

The most useful ordering I know of is the partial ordering by absolute value. Black Carrot 07:54, 4 October 2007 (UTC)

You can of course use the inequalities separately on the real and imaginary parts of a complex number as you would normally87.102.94.194 15:48, 4 October 2007 (UTC)

You can, naturally, but given the lack of relationship with the algebraic/analytic structure of the complex numbers, I can't immediately see why you would want to. In particular, as Lambiam pointed out above, the symmetry between i and -i means any claim that -i<i (or vice versa) is unlikely to be useful. Algebraist 16:29, 4 October 2007 (UTC)

Volume of a "bowl"

How would you find the volume of a portion of a sphere that has been "sliced" in two? Obviously, if the cut is down the center, the volume of each resulting figure will be .5 that of the sphere, but what if the cut is, say, 1/3 along the diameter of the sphere? 70.146.164.160 21:00, 3 October 2007 (UTC)

See our article on spherical caps. Gandalf61 21:37, 3 October 2007 (UTC)

Which gives the answer, but not necessarily enough information. This sounds like a Calc II homework problem type thing (although it's early in the semester for this question. The calculus approach would be to use an integral to find the result so you're looking at $\int _{a}^{b}\pi f(x)^{2}dx$ where $f(x)$ gives the radius of a slice of the sphere at vertical position x. I'll leave the determination of $f(x)$, a and b as exercises for the reader. Donald Hosek 23:17, 3 October 2007 (UTC)

In other words you need to be able to find volumes using integration.. If you don't know how to do this now would be a good time to ask..87.102.94.194 15:46, 4 October 2007 (UTC)

There are numerous pages that cover this - here is one if you need it - hopefully it should give enough info to get you started http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/volume.html - you can also integrate using angles as well as length - both methods are solvable.87.102.94.194 15:46, 4 October 2007 (UTC)

October 4

NOT A QUESTION

Earlier this week I asked questions on a homework assignment and one user deleted the questions and told me to do my own homework. I became mad 'cussed out' the user. I want to aplogize to anyone I may offended, especially David Eppstein, whom I have sent a message on thier talk page. If anone was wodering, I ended up getting an 88 on the test. If soem Wikipedian believes this needs to be deleted because it's not what the desk is for, then do it, I jsut didn't see a more practical way to apologize. schyler 00:35, 4 October 2007 (UTC)

It's okay, we forgive you. I think this is a good case of WP:IAR to let you post this here.--Mostargue 00:53, 4 October 2007 (UTC)

Thanks for getting back to us - most don't bother. And, 88 ain't bad! - hydnjo talk 02:12, 4 October 2007 (UTC)

No problem schyler, although it's definitely good to avoid insults and name-calling, you have every right to object if you feel someone was being less-than-helpful with their answer. Sometimes people worry more about "enforcement" than about helping people figure out how to properly use Wikipedia and get familiar with all the "rules". This is not to "blame" anyone, just to say we can all do our part to help avoid misunderstandings.

In the future, if you want to make a post like this, (a general-purpose comment instead of a question), feel free to post it at Wikipedia talk:Reference desk. Just like with Wikipedia articles, the reference desk also has a talk page. (See also, Wikipedia:Talk page guidelines and Wikipedia:No personal attacks for more information, both of those have "quick summaries" so you don't have to read through the whole thing if you're busy).

Thanks again for your original question and for your apology. Regards. dr.ef.tymac 06:30, 4 October 2007 (UTC)

The original outburst appeared here, directed at David, so apologies both here and personally to David are fine.

It is not just "good" to avoid profanity and attacks, it is mandatory. Bans can be imposed if civility is flagrantly ignored.

That said, I understand that tests can be stressful. Shoe on the other foot, David is a university professor with years of experience dealing with students, which can also try the patience. And surely he has other stresses in his life as well.

In the end, your willingness to acknowledge that you behaved badly, and to apologize to all concerned, is a testament to your character. It suggests humility and consideration and courage. Thank you; and congratulations on your test showing. --KSmrq^T 09:14, 4 October 2007 (UTC)

Indeterminate form

I know that in early calculus, 0/0, (undefined)/(undefined), and (undefined)-(undefined) are indeterminate forms. Are there any other indeterminate forms that are suitable for newbies (to calculus)? --Kushal^{Click me!} write to me 03:43, 4 October 2007 (UTC)

0^0, 0^∞, 1^∞.--Mostargue 05:20, 4 October 2007 (UTC)

And $0\cdot \infty$, and $\infty -\infty$. – b_jonas 08:43, 4 October 2007 (UTC)

Also ∞/∞ and ∞/0. And (−∞)^L, in which L can be basically any limit, is in general not well defined. Actually, if you want to be exhaustive, −∞ and +∞ should be treated separately.  --Lambiam 12:27, 4 October 2007 (UTC)

Is 0^∞ really indeterminate ? Surely

$\lim _{x\rightarrow 0}(\lim _{n\rightarrow \infty }(x^{n}))=0$

$\lim _{n\rightarrow \infty }(\lim _{x\rightarrow 0}(x^{n}))=0$

so 0^∞ = 0 whichever way you look at it (and so it is different from 0⁰ and 1^∞, which certainly are indeterminate). Gandalf61 12:41, 4 October 2007 (UTC)

Even more general, if $x_{n},y_{n}$ are sequences of positive numbers such that $x_{n}\to 0$ and $y_{n}\to \infty$, then $x_{n}\leq 1$ and $y_{n}\geq 1$ for almost all n, so $0\leq x_{n}^{y_{n}}\leq x_{n}\to 0,$ so all relevant limits are 0, not just the iterated limits considered by Gandalf61. Kusma (talk) 13:42, 4 October 2007 (UTC)

Although there is a slight problem: lim_x→∞−exp(−x) = 0, lim_x→∞exp(x) = ∞, yet I'd say that lim_x→∞(−exp(−x))^exp(x) is undefined rather than 0.  --Lambiam 17:43, 4 October 2007 (UTC)

Win some, lose some.--Mostargue 13:50, 4 October 2007 (UTC)

Interesting, (although I do not understand all of it.) Thanks guys! (this thread is still open I guess) --Kushal^{Click me!} write to me 17:29, 4 October 2007 (UTC)

partial derivative

Is ðy/ðx=1/(ðx/ðy). —Preceding unsigned comment added by 218.248.2.51 (talk) 05:52, 4 October 2007 (UTC)

Generally, no. See Partial derivative.

• ${\frac {\partial y}{\partial x}}={\frac {\partial }{\partial x}}y$
• ${\frac {\partial x}{\partial y}}={\frac {\partial }{\partial y}}x$

Basically, it's because ${\frac {\partial }{\partial x}}$ and ${\frac {\partial }{\partial y}}$ are operators. --Mostargue 06:02, 4 October 2007 (UTC)

Also see Differential operator.--Mostargue 06:06, 4 October 2007 (UTC)

Rather than "no", I'd say that both expressions are not meaningful at the same time. If the expression ∂y/∂x is meaningful, then y is an expression depending on several variables, including x. In that case x is not an expression depending on several variables, including y, so then the expression ∂x/∂y is not meaningful. (If "several" can include the case of just one dependent variable, so that you can write just dy/dx, and the relationship is such that x can be considered as functionally depending on y equally well, then indeed dy/dx = 1/(dx/dy) with some caveats such as that thou shalt not divide by zero.)  --Lambiam 06:35, 4 October 2007 (UTC)

Lamb-"Both Guns Blazing"-Iam. More questions I answer, the more of my own questions are answered.--Mostargue 06:49, 4 October 2007 (UTC)

outcircles?

I've just looked up incircle and was surprised to see the definition given for excircles. What is the circle called that goes through the three corners of a triangle? Do we have an article on it? -- SGBailey 06:24, 4 October 2007 (UTC)

Circumscribed circle.--Mostargue 06:29, 4 October 2007 (UTC)

complex numbers.

what does a term with "i" represent in an equation.as "i"represent squareroot of -1,a term with it (may be sin or cos )represents what part of physical value when squareroot of -1 dosent exist or is imaginary,n still most of forms uses i,just as in equn of electromagnetic radiations or exponential form of fourier series a term with i is mandatory. —Preceding unsigned comment added by 202.141.149.165 (talk) 13:36, 4 October 2007 (UTC)

Not actually sure what you mean. Lots of letters have implicit meanings and are avoided when it can cause ambiguity (sometimes engineers use j instead of i because i has uses in engineering; sometimes people use p, q and r as indices rather than i, j and k because of vector concerns, and so on).

In polar form, a complex number z with modulus r and argument t has real and complex components in the form r cos t and r sin t. It's just a property of a number, like sign. x42bn6 Talk Mess 14:33, 4 October 2007 (UTC)

I believe the question is asking what is physically represented by the imaginary term of a complex number. As with any number, the answer is "it depends on the application". I can no more say in general terms what "3i" represents than what "3" represents. You may find the applications section of our complex number article useful. Additionally, please consider phrasing your questions more carefully -- inconsistent spelling, spacing, and punctuation does make it more difficult to parse your request (the same holds for below, where I'm really not sure what's being asked). — Lomn 14:52, 4 October 2007 (UTC)

"i" usually represents the square root of minus one as you say, however when using vectors, "i" may represent a vector parallel to the x-axis eg from the origin the vector 4i+5j+10k gives the point (4,5,10).87.102.94.194 15:15, 4 October 2007 (UTC)

fourier series.

a fourier form is used 2 represent a time domain signal to frequency domain signal,n orthogonal signals r used,just as in trignometric form of fourier signal "what does each sine term correspond 2 for say (a simple NRZ signal used for digital communication)". —Preceding unsigned comment added by 202.141.149.165 (talk) 13:49, 4 October 2007 (UTC)

In general fourier analysis can convert one function from a signal in terms of time to a signal as a sum of sinusioudals eg sines and cosines - I assume you are thinking specifically of fourier series.

For this to be possible the function must be periodic ie it must repeat itself over a certain constant time frame

As such a NRZ signal can not be expressed as a fourier series - as it is not a periodic function.

You might want to look at square wave which is a function that can be expressed as a fourier series.

If you want know about methods that make a NRZ signal expressable as a fourier series please say so.87.102.94.194 15:39, 4 October 2007 (UTC)

Actually, the Fourier Transform does not need to operate on a periodic function, there are aperiodic forms of it, but this is mostly only useful for theoretical purposes. More practically, a periodic form of the transform can be applied to chunks of a signal at a time, and can be used for all sorts of things, e.g. filters, convolution, etc. (see Fourier analysis#Applications in signal processing for a few ideas). Specifically for a "non-return-to-zero" signal I don't know if there is much application in terms of the content of the signal; at least, I can't think of any reason (in terms of information) that I would want to alter or analyze that kind of signal in the ways that a FT could. - Rainwarrior 16:49, 4 October 2007 (UTC)

trial method equation solvings

a^3 + 2a+5 how can i solve this within 5 secs —Preceding unsigned comment added by 59.93.243.198 (talk) 17:24, 4 October 2007 (UTC)

Not by asking the question here. It isn't quite clear to me what you mean b y "solving" a³ + 2a + 5; that expression is not an equation. Are you asking for the zeros of the function f(a) = a³ + 2a + 5 (or, equivalently, the solutions of the cubic equation a³ + 2a + 5 = 0)? Our article on cubic equations gives an algebraic method for solving such equations; if you consistently manage to do that within 5 seconds, you should contact the Guinness people.  --Lambiam 17:55, 4 October 2007 (UTC)

women

there are three types of women in this world, what are they. i begged my proffesor for a clue as to what these types are pertaining to. he said to me, that it deals only with phisical appearence. there are 3 categories of body types that only women fall into, only three. i am desperate for the answer please help. he then said he asked his quantitative research analysis class the question, and only one person was able to find the answer, and he was very impressed. i am the only student he has given this clue to, and it would be very much appreciated to receive the answer... thanks!

---

Etro —Preceding unsigned comment added by 151.196.108.226 (talk) 17:55, 4 October 2007 (UTC)

This has already been discussed on the Miscellaneous RefDesk as well as above on this desk; there is no need to post a third time. — Lomn 20:12, 4 October 2007 (UTC)

Algebra question

x×.8=4032 —Preceding unsigned comment added by 71.225.135.106 (talk) 19:03, 4 October 2007 (UTC)

Are you serious? First off, a × b is the same as b × a, so you can rewrite the left-hand side of your equation as .8 × x. And .8 is another notation for 0.8, which is the same as 8/10. Then, if you have an equation of the form a × x = b in which a is not 0, you can divide both sides by a to arrive at the solution x = b / a.  --Lambiam 20:37, 4 October 2007 (UTC)

Traveling Salesman computation

I expect i'm asking a question which Wikipedia already has an article for, but i'll ask anyway as i can't find one;

(also, should this be in computing or mathematics?)

With Traveling Salesman computation i was thinking about reducing the time of computation by counting the length of an in-construction path as we go (with each added edge) and if the length of path is greater than the current best upper band at that time abandoning the unfinished path - But is it quicker to do it this way (assuming time to calculate length of path increases with increase to the path size) and removing 'over' paths as we go than computing the length only when the path is complete and deciding if the path is shorter - Also at with how many vertices would one system become faster than the other surely as the number of vertices increases so does the calculation time of the first method but so does the possible gain of not having to construct full paths?

Maybe i should start Differentiating?

Thanks!

-Benbread 19:18, 4 October 2007 (UTC)

I am not fully sure I understand your proposed method, but it sounds like branch and bound. If it is possible to answer the more specific questions at all (some techniques improve the time for some inputs but make it worse for other inputs), this requires a more precise description of the proposed algorithms.  --Lambiam 20:24, 4 October 2007 (UTC)

SHA hash websites

Hi. There are probably different links for SHA-1, SHA-2, SHA-512, etc. The problem is, what exactly are those links? The Template:User committed identity gives the link to SHA-1, and the article gives the links to a few, but where are the links to all the SHA websites, say for example where is the link for calculating SHA-512? Can someone list me the links of those websites, and where you got it from? I don't want links to sites other than those used to directly calculate a string into a hash code, so I don't want any sites with info about the SHA websites, I want the actual links themselves. This question was originally from Talk:SHA hash functions. Thanks. ~AH1^(TCU) 20:43, 4 October 2007 (UTC)

The first return of a Googling of SHA-512 hash calculator is a rather fine site that calculates MD5, SHA-1, SHA-256, SHA-384, and SHA-512 hashes; it has worked well for me. Joe 04:06, 5 October 2007 (UTC)

That's the same one listed at SHA hash functions#Online Hash Calculators. Template:User committed identity suggests the other one because being implemented in Javascript, it doesn't involve sending your super secret string to some random person's web server. (The obvious drawback of requiring you to run Javascript is ignored.) The serious answer has to be: forget web forms. Use the sha256sum program (and other similarly named ones) from coreutils. --tcsetattr (talk / contribs) 04:50, 5 October 2007 (UTC)

y = x^x

Does the function y = x^x have any significance? Does it have a name? I've always found it a particularly intriguing function in the abstract, especially for –1 ≤ x ≤ 1. (Then there are the related functions y = (–x)^x, y = x^(–x), and y = (–x)^(–x).) Thanks. — Michael J 05:21, 5 October 2007 (UTC)

It doesn't have a name as far as I know. But you'll perhaps be interested in the sophomore's dream. - Fredrik Johansson 05:38, 5 October 2007 (UTC)

Interesting. It's a little over my head, so I'll have to take some time with it, Fredrik. — Michael J 05:52, 5 October 2007 (UTC)

Logic

I have to prove the following tautology:

$(P\to Q)\land (Q\to R)\to (P\to R)\equiv T$

Here's my logic:

$(\neg P\lor Q)\land (\neg Q\lor R)\to (\neg P\lor R)$

$\neg (\neg P\lor Q)\lor \neg (\neg Q\lor R)\lor (\neg P\lor R)$

$(P\land \neg Q)\lor (Q\land \neg R)\lor (\neg P\lor R)$

Can't see how to get out of that last statement. I'm thinking that maybe I cocked it up at the second line where I flipped the AND to OR. But $\neg (P\land Q)\equiv \neg P\lor Q$, right? —Preceding unsigned comment added by 91.84.143.82 (talk) 07:31, 5 October 2007 (UTC)