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February 18

Cubic function

If you go here and enter as the equation ax³ + bx² + cx + d = 0 and tell it to solve for x, you get a long expression as the solution. Can someone show algebraically how you can solve ax³ + bx² + cx + d = 0 for x? Or link to a page that does? Thanks. 70.162.25.53 (talk) 03:39, 18 February 2008 (UTC)[reply]

See cubic function, specifically the Cardano's method section. Be warned, it's not quite as nice as the quadratic formula... 134.173.92.17 (talk) —Preceding comment was added at 03:59, 18 February 2008 (UTC)[reply]

(ec) The Wikipedia article on cubic functions may be of your interest, or at least this subsection of it. Outside Wikipedia, this page from MathWorld is another interesting choice. Pallida Mors 04:06, 18 February 2008 (UTC)[reply]

I intended a reference to the subsection Root finding in that article, but the link doesn't seem to work fine. Nevermind. Pallida Mors 04:19, 18 February 2008 (UTC)[reply]

Logic

Does anyone know where I can find a whole bunch of "Knights and Knaves" problems, with answer/explanations/solutions? Thanks. (Joseph A. Spadaro (talk) 07:51, 18 February 2008 (UTC))[reply]

Here's a bunch of problems with a twist. These problems explore the consequences of adding a third type of person, the philosophers. For more traditional questions, a Google search turns up a few pages. —Bkell (talk) 12:40, 18 February 2008 (UTC)[reply]

Also, check out the books of Raymond Smullyan. He has tons of logic puzzles that all can enjoy. –King Bee ^{(τ • γ)} 13:48, 18 February 2008 (UTC)[reply]

I especially recommend "Alice in Puzzleland" by Raymond Smullyan. 129.67.157.6 (talk) —Preceding comment was added at 15:15, 18 February 2008 (UTC)[reply]

Hi. Thanks to all for the input. I guess I should have made my question a little clearer. So, let me re-phrase it. I am looking for Knights and Knaves problems ... that are accompanied by answers / solutions / explanations. Right now, I only want Knight/Knave questions ... and not just logic puzzles in general. Also, I am only looking for the "simple" type (knights and knaves only), where they do not introduce third-parties (spies, randoms, philosophers, etc.) into the equation. Also, I'd like to find this on the Internet --- as I don't have too much time to go out, order a book, wait for its delivery, etc. At some later point, I will probably go out and get some Smullyan books ... but right now, I don't have that time / luxury. Any suggestions? Any good websites out there? I can't seem to find any, with my searches. Many thanks! (Joseph A. Spadaro (talk) 16:33, 18 February 2008 (UTC))[reply]

The Wikipedia article on Knights and Knaves has a few problems with explanations and solutions. There are some pages with individual knights-and-knaves problems (from one site, [1], [2], [3], [4], [5], [6], [7], [8]). —Bkell (talk) 17:46, 18 February 2008 (UTC)[reply]

Smullyan's book entitled "What is the name of this book?" has a lot of Knights/Knaves puzzles complete with solutions. His books should be available at your local library, so you needn't purchase a thing. –King Bee ^{(τ • γ)} 19:21, 18 February 2008 (UTC)[reply]

Thanks for all the input. Much appreciated. In this day of Internet and digital information, it actually never dawned on me to visit my local library. I actually decided to do that -- after reading King Bee's suggestion -- and indeed they had nine of Smullyan's books. I checked out all nine! Thanks a lot. (Joseph A. Spadaro (talk) 00:59, 24 February 2008 (UTC))[reply]

help with aproof

i need help with how to prove that a number (a) is a zero divisor mod (n) iff gcd(a,n)>1?thank youHusseinshimaljasimdini (talk) 10:52, 18 February 2008 (UTC)[reply]

Think about n/gcd(a,n) - let's call this b. First of all, you can be sure that b is an integer - why ? Secondly, you can be sure that if gcd(a,n) > 1 then b mod(n) is not 0 - why ? Now think about the product ab - what is ab mod(n) ? How does this help you show that a is a zero divisor mod(n) ? Gandalf61 (talk) 12:09, 18 February 2008 (UTC)[reply]

Collecting Terms

i get how to do everything else but i seem to bomb out on collecting the terms forgive me if it is wrong
Simplify: $={\sqrt {3}}-2{\sqrt {2}}+2{\sqrt {3}}+{\sqrt {2}}$
now im not sure but im pretty sure this is wrong $=1{\sqrt {3}}(notsurewhatgoeshere)3{\sqrt {2}}$
—Preceding unsigned comment added by 124.176.206.9 (talk • contribs) 12:28, 18 February 2008 (UTC)[reply]

You have one square root of 3, minus two square roots of 2, plus 2 square roots of 3, plus one square root of 2: so in the end how many square roots of 3 do you have, and how many square roots of 2? —Bkell (talk) 12:31, 18 February 2008 (UTC)[reply]

If it helps you, forget for a moment what

{\sqrt {3}}

means, and just think of it as a symbol, like a circle, perhaps. The same thing for

{\sqrt {2}}

: think of that as a triangle. So you have one circle, minus two triangles, plus two circles, plus one triangle. How many circles do you have, and how many triangles? Just be sure to write "

{\sqrt {2}}

" in your answer instead of a triangle, of course. —Bkell (talk) 12:37, 18 February 2008 (UTC)[reply]

3{\sqrt {3}}+-1{\sqrt {2}}(={\sqrt {2}})

so $3{\sqrt {3}}+{\sqrt {2}}$ ? do you just take the sign infront? so that 12a - 10b + 9c + 5b
=12 - 5b + 9c?

Think about this,

ab+ac+de+df=a(b+c)+d(e+f)

this is just two applications of the Distributive property. If it's not clear why that works, just try going the other way. (Note that algebraic laws are reversible!)A math-wiki (talk) 12:56, 18 February 2008 (UTC)[reply]

You're right: the simplified answer is

3{\sqrt {3}}+(-1){\sqrt {2}}

. Usually −1 times something is just written −something (for example, −1 × 5 = −5), and when "+−" occurs in a mathematical formula it's usually just written "−" [for example, 3 + (−2) = 3 − 2]. So you can write

3{\sqrt {3}}+(-1){\sqrt {2}}

in a simpler way. —Bkell (talk) 13:56, 18 February 2008 (UTC)[reply]

geometry - spheres and unit cubes

seeing as there's quite a few mathematicians in here i thought i'd ask about a geometry problem i've been having. i think it's quite easy but i don't really have a clue.

imagine a unit cube. now imagine a point within this cube (call it p). i want to place a sphere with its centre at p. how would i work out the radius needed for this sphere so that the intersection volume between it and the cube equals a given value (call it v)? AlmostCrimes (talk) 14:21, 18 February 2008 (UTC)[reply]

If the sphere just intersected one side you could find the volume of the Spherical cap cut off, then subtract this from the volume of the sphere. If the sphere intersects more than one side then things get a little trickier. --Salix alba (talk) 15:07, 18 February 2008 (UTC)[reply]

[ec]It's actually not that simple. To do that you will first need some way to calculate the volume of intersection of a cube and a ball, which is pretty messy. Fortunately this latter problem is one I have worked on a while back, so I have thought of a few approaches. Since none of them is perfect, I can help you more only if you provide more details about your application (with special attention to details such as relative and absolute error tolerance, computational resources, platform on which you will do the calculations, etc.). If the question is purely out of curiosity I suggest you just leave it, though I can describe the general ideas. -- Meni Rosenfeld (talk) 15:10, 18 February 2008 (UTC)[reply]

salix alba - Yes sadly, it is quite possible the sphere intersects more than one side, for example if the point was at one of the corners of the cube. the basis of the problem is that I'm investigating various approaches to deciding whether a given colour in an RGB colour space (here represented as a cube like the following RGB#Representations) is "similar" to another colour. the natural approach would be to calculate the distance between the two colours (or points) and to then determine whether this is below a certain threshold. This technique suffers in that not all colours are treated evenly - colours in the centre of the colour space (so at [0.5, 0.5, 0.5], or grey) would be flagged as being similar to a higher percentage of the total colour space than a colour like white would. I was hoping to even this out by dynamically modifying the threshold (analogous to the radius) so that every colour has an equal proportion of the colour space flagged as being similar to it. As i say, this is mostly investigative and it isn't important that I use this exact technique. it's very possible i'm completely overcomplicating things.

for what it's worth though error tolerance isn't too important, neither are computational resources. I'm programming it in MATLAB and basically just seeing what works well. the most important factor is simplicity and implementation time really as I don't want to get too hung up on one technique. If any of your solutions apply I'd be interested in hearing them. I thought briefly about representing both the sphere and cube in a boolean matrix and physically determining the best sphere radius but I was kind of hoping there'd be a simple algebraic method. --80.4.203.142 (talk) 19:44, 18 February 2008 (UTC)[reply]

Well, I can see several problems with this approach. If you're going for a notion of similarity which matches that of a human observer, I don't think using a simple Euclidean metric as a basis will cut it. I don't know a lot about this, but at the very least you should give different weights to the components - Grayscale mentions weights used for a different purpose, but they might be appropriate here. This would then leave you dealing with ellipsoids which is even messier. Second, I'm not sure any sort of "Affirmative action" is necessary - white will have less colors labeled as similar to it, simply because there are less colors genuinely similar to it. You next have symmetry to consider - You mentioned a ball centered around one point, but when doing a comparison there are two points to consider, and the result will depend on which one you choose to calculate the threshold. If you do some symmetric calculation taking both into account, it will be difficult to guarantee that indeed every color has the same portion of the color space labeled similar to it.

As said, solving the problem in you original question is not impossible, but not simple to do with satisfying accuracy and efficiency. Since I gather that these are not crucial, I can offer a simple approximation method for the volume of intersection of the cube and a given ball. Just pick several (something like 100 or 1000) randomly chosen points in the cube, and check whether each is inside the ball. The proportion which are is an estimation for the volume of intersection. You can do this for a ball centered at one point and with radius equal to the distance to the other point, compare it to the volume you want to be labeled similar and make the decision accordingly.

I hope this helps. -- Meni Rosenfeld (talk) 22:25, 18 February 2008 (UTC)[reply]

it's true there are problems with the general approach and while I am aware there are colour spaces that lend themselves more easily to this kind of work I have been asked to look at this RGB cube method specifically. As for the "affirmative action", it seems to me it is necessary. A general distance approach is far better at finding areas that look grey than areas that look white when given the same threshold. Increasing the threshold for white increases the perceived accuracy, which is why I wanted to try this method. Your symmetry point is interesting. I actually hadn't considered that...

Because of the nature of the exercise it isn't important that every technique I try is successful but regardless you've brought me round to not trying the technique, which I believe won't be worth the effort. It helped to have this discussion even if the focus of it changed a little, for which I thank you. --80.4.203.142 (talk) 23:44, 18 February 2008 (UTC)[reply]

Math Riddle

While studying for my math test, I came across this problem:

I am thinking of some polynomial with positive integer coefficients. You can ask me for the value of p[x], where p is the polynomial and x is some positive integer as many times as you want. What is the minimum number of questions you need to ask in order to find out what the polynomial is?

I am unable to figure out the answer. Since it doesnt tell you how many terms or the order of the polynomial, it's hard to find a starting spot. Can anyone help? 99.240.177.206 (talk) 18:37, 18 February 2008 (UTC)[reply]

I remember that one. The answer is 2. Does that help? Black Carrot (talk) 18:45, 18 February 2008 (UTC)[reply]

It helps me, at least. Nice problem. Algebraist 19:45, 18 February 2008 (UTC)[reply]

If I understand Black Carrot correctly, the value you choose for x in your second question depends upon the answer you receive to your first question - yes ? Gandalf61 (talk) 20:06, 18 February 2008 (UTC)[reply]

It would have to, polynomials aren't completely determined by their values at two points. There must be some clever trick based on the answer to the first questions... I haven't worked it out yet, but give me some time... --Tango (talk) 20:08, 18 February 2008 (UTC)[reply]

You don't just have 2 points, you have 2 points plus the knowledge that the coefficients are positive integers. I just got it! Oh that is clever. --tcsetattr (talk / contribs) 20:40, 18 February 2008 (UTC)[reply]

...which was the point I was trying subtly (perhaps too subtly) to make. Answer to first question gives you information about coefficients which you use in second question. Gandalf61 (talk) 22:42, 18 February 2008 (UTC)[reply]

Haha, wow, that's pretty cool. —Bkell (talk) 21:23, 18 February 2008 (UTC)[reply]

Hm..suppose if one chose p(0) and the answer turned out to be 4 or 5. Obviously, this would indicate the the constant at the end is either a 4 or 5. Where would you go from there?Acceptable (talk) 20:39, 18 February 2008 (UTC)[reply]

Asking for p(0) is a bad start (aside from the fact that it wasn't technically allowed by the problem statement). --tcsetattr (talk / contribs) 20:55, 18 February 2008 (UTC)[reply]

Funnily enough, 0 (if you take it to be positive) is the only first question that doesn't work. Suppose you got f(0)=a (say), and used n for your second question. If you got f(n)=n^2+a, then f could be nx+a or x^2+a. Algebraist 21:50, 18 February 2008 (UTC)[reply]

Finally, I think I've got it! That shouldn't have taken me that long... --Tango (talk) 22:29, 18 February 2008 (UTC)[reply]

Just in case the OP needs another hint - how would you solve this if you also knew that all coefficients are less than 10? -- Meni Rosenfeld (talk) 22:32, 18 February 2008 (UTC)[reply]

It finally clicked when I twigged that it was only positive coefficients. -mattbuck (Talk) 23:02, 18 February 2008 (UTC)[reply]

I think the smallest number of questions you need to ask is (the degree of the polynomial + 1).--George (talk) 22:43, 18 February 2008 (UTC)[reply]

I wonder if there's a finite bound for more general polynomials, like with positive rational coefficients. Black Carrot (talk) 22:47, 18 February 2008 (UTC)[reply]

To George - that'd be right for polynomials with no restrictions on the coefficients, but you can take advantage of their positive integerosity to jump straight to the answer. Black Carrot (talk) 22:50, 18 February 2008 (UTC)[reply]

Do you have any specific technique for doing this or is this just a general statement ?--George (talk) 22:56, 18 February 2008 (UTC)[reply]

To the OP's riddle? Yeah, I've got the answer. Give it a little while, and reflect on Meni's hint. BTW, for a polynomial in n variables, with positive integer variables and coefficients, it could be done within n+1 questions. Black Carrot (talk) 23:00, 18 February 2008 (UTC)[reply]

I got it ,thanks. --George (talk) 23:53, 18 February 2008 (UTC)[reply]

I think 2 questions ought to work for real non-negative coefficients - that they're integers I wouldn't think was strictly necessary. -mattbuck (Talk) 23:02, 18 February 2008 (UTC)[reply]

Really? It doesn't seem like that would be enough. That is, for any two inputs you choose, it seems like there would be at least one pair of polynomials with the same output. Black Carrot (talk) 23:06, 18 February 2008 (UTC)[reply]

If you choose any two points, there are an infinite number of polynomials that go through them. But, restricted to non-negative coefficients, if you choose the 2nd point based on the first one, it ought to be possible to find the order. -mattbuck (Talk) 23:12, 18 February 2008 (UTC)[reply]

Thinking about Black Carrot's extended question, I can see that if you only know that all (non-zero) coefficients are real and positive, then with two questions you can find the maximum degree of the polynomial - say it is m (so at most m+1 non-zero coefficients). But don't you still need a further m-1 questions to find the values of the coefficients (so that altogether you have m+1 simultaneous equations) ? Key difference from the positive integer case is that there is no longer a lower bound on the magnitude of a non-zero coefficient. Gandalf61 (talk) 00:00, 19 February 2008 (UTC)[reply]

Generally with polynomials, rational coefficients and integer coefficients work in exactly the same way (you just multiply through by the lowest common multiple of all the denominators), so I imagine the case of positive rationals is doable (you may end up needing to additional fact that the polynomial is monic in order to get a unique answer, I'd have to try). The case of positive reals probably isn't, though. The key fact that p(1)<10 (or whatever) gives you is that 10 doesn't divide any of the coefficients, in the real case that fact is no longer true (it's not true for rationals, either, but you can probably fudge it). I haven't actually tried either case, this is just intuition and could well turn out to be wrong. --Tango (talk) 12:50, 19 February 2008 (UTC)[reply]

But how do you determine the lowest common multiple (or even any common multiple) of the denomoinators ? I think the positive rational coefficients case is the same as the positive real coefficients case - it only takes a finite number of questions, you don't know in advance how many, but after two questions you know how many further questions you need. Gandalf61 (talk) 15:08, 19 February 2008 (UTC)[reply]

That comment was just to demonstrate why polynomials over the integers and polynomials over the rationals are closely related. Obviously, you can't directly convert in this case, but there might be some trick that allows you to use the same basic method. I haven't looked for one, so I don't know. --Tango (talk) 16:44, 19 February 2008 (UTC)[reply]

For those still struggling, I suggest using p(1) as the first question. It's not necessary, but thinking about what I could learn from p(1) was what led me to the solution. And don't forget the coefficients are all positive! --tcsetattr (talk / contribs) 23:21, 18 February 2008 (UTC)[reply]

Sorry, I still am not getting it. The question does not state how many terms there are. So suppose I realize that none of the coefficients are greater than 10, I still dont know anything about how many terms or how large the coeff of each term is.

Let's take an example: suppose the equation was f(x)= 4x^3 + 3x^2 + 7x + 6. If I plug in p(1), i get p(1)=20. What do I ask next? Thanks. 99.240.177.206 (talk) 04:29, 19 February 2008 (UTC)[reply]

OK, that's enough. Full spoiler: suppose you had chosen that function and I asked you for p(1) and you said p(1)=20. I know the polynomial is A+Bx+Cx^2+Dx^3+... so what do I get when x=1? A+B+C+D+... in other words the sum of the coefficients. I don't know how many coefficients there are, but I know they are all positive integers and their sum is 20. If a set of positive numbers adds up to 20, then none of them individually can be greater than 20. The next question I'd ask is for p(100). I could ask for p(21) instead - any number greater than 20 will do - but I'm choosing 100 because it might make the procedure more obvious.

Your answer will be p(100) = 4(100^3) + 3(100^2) + 7(100) + 6 = 4030706. See how the coefficients 4,3,7,6 revealed themselves? If I had asked for p(21) you would have said 38520 which would make the procedure a little more mysterious-looking, but the secret is I would then convert 38520 into base 21, and the result would be: 4376₂₁!

Now we can go back to the first step and see how p(1) doesn't have to be used. If I asked for p(2) first, you'd say 64. In this version, 64 is not the sum of the coefficients, but it's still an upper bound on them individually. If any one of your coefficients was greater than 64, it would contribute a term of (something greater than 64)*(some power of 2) to your answer, so your answer would have been greater than 64 also. Knowing this upper bound, I could ask for p(65), get the answer 1111636, and convert it to base 65, where it is: 4376. --tcsetattr (talk / contribs) 04:44, 19 February 2008 (UTC)[reply]

And of course, my point was that if all coefficients are less than 10, you don't need two questions - you just ask what is

p(10)

. -- Meni Rosenfeld (talk) 09:49, 19 February 2008 (UTC)[reply]

This is nice. Ask for $p(1000000)$ and the problem is solved in one question for all the cases where the coefficients are less than a million. You do not have to require that the integer coefficients are positive, just that they are not negative. What if the integer coefficients may be negative too, how do you solve that? Bo Jacoby (talk) 15:33, 19 February 2008 (UTC).[reply]

If the coefficients can be arbitrary integers, then infinitely many questions are required (consider the case when every question gets the answer '0'). If they're arbitrary positive reals, then finitely many questions suffice (as Gandalf pointed out above) but I'm not sure if you can get a fixed finite bound or not. Algebraist 16:25, 19 February 2008 (UTC)[reply]

You can't have infinitely many questions all giving 0, since the polynomial will always be of finite degree (otherwise it's not a polynomial). --Tango (talk) 16:44, 19 February 2008 (UTC)[reply]

Well, I suppose there's the 0 polynomial... it would take an infinite number of questions to spot that. Any non-0 polynomial would be worked out with finite (but unbounded) steps, though, I'd think. --Tango (talk) 16:52, 19 February 2008 (UTC)[reply]

No. However (finitely) many times I give you the answer zero, you won't know if it's the zero polynomial or just some other poly that happens to have roots at every point you've asked about. Algebraist 18:35, 19 February 2008 (UTC)[reply]

That's not quite what I said. I said that any non-zero polynomial could be worked out in finitely many steps, not that finitely many steps is enough to tell if a polynomial is non-zero. There are only finitely many roots to any polynomial (at most, it's degree), so after one more question than that (which is still a finite number), you'll know it's non-zero. Whether you can work out exactly what it is or not, I'm not sure - repeated roots would be an issue, but you would know it's non-zero. --Tango (talk) 20:39, 19 February 2008 (UTC)[reply]

That (while true) is not much use to solve the problem as stated. Sure, for any polynomial there is a question-asking strategy that proves the poly is non-zero in finitely many steps, but what we want is a strategy that for any polynomial will prove it is non-zero in finitely many steps. We can't base our strategy on an answer we don't know. Algebraist 22:34, 19 February 2008 (UTC)[reply]

How about asking for p(0), p(1), p(-1), p(2), p(-2), etc. until you get a non-zero answer? If the polynomial is non-zero, you will get a non-zero answer after a finite number of questions. If it is the 0-polynomial, then you'll never get an answer, but you're asking for a proof that it's non-zero, not a proof that it is zero. --Tango (talk) 12:08, 20 February 2008 (UTC)[reply]

Some day I'll learn not to trust maths I do drunk. Algebraist 16:34, 21 February 2008 (UTC)[reply]

Set of convergence of a sequence of measurable functions

This is an exercise from Lang's Real and Functional Analysis, Chapter VI: Let $\{f_{n}\}$ be a sequence of measurable functions. Show that the set of points for which $\{f_{n}\}$ converges is a measurable set.

Presumably this involves finding some collection of obviously measurable sets that can be unioned/intersected together to end up with the set in question, but I haven't been able to find such a collection. The only information available is the basic properties of σ-algebras and measurable functions.

[As a side issue, does anyone have experience with or thoughts on this book? I can't say I find it easy going, but struggling with it seems to be good practice and the more time I spend with it the more I appreciate it. I got it originally for the development of integration, which is the best I've seen anywhere--just beautiful! If only there were a solutions manual... — merge 23:33, 18 February 2008 (UTC)][reply]

One way of doing this is to first show that sup_n{f_n} is measurable. Dually the inf of countably many measurable functions is measurable, so the lim sup and the lim inf of the f_n are both measurable. Then the set of x for which f_n(x) converges is the set of points at which two measurable functions are equal, and is thus measurable. Slight care might be needed in dealing with infinity. Algebraist 23:51, 18 February 2008 (UTC)[reply]

Thanks! This idea had not occurred to me. However as it turns out the properties of lim sup and lim inf you refer to are dealt with in the next exercise! So I'm thinking there's a way to do this one that doesn't use them. Also, I forgot to mention that the

f_{n}

take values in a Banach space, so one can't take the sup directly. — merge 12:52, 19 February 2008 (UTC)[reply]

In that case, you could use the fact that f_n(x) converges exactly if f_n(x) is Cauchy. The latter statement is just

\forall {\epsilon }\exists {N}\forall {n,m}(n,m\geq N\rightarrow |f_{n}(x)-f_{m}(x)|<\epsilon )

So the set of x for which f_n(x) converges is just

\bigcap _{\epsilon }\bigcup _{N}\bigcap _{n\geq N}\bigcap _{m\geq N}(f_{n}-f_{m})^{-1}(B_{\epsilon })

where

B_{\epsilon }

is the open ball of radius epsilon about 0 (I'm assuming this is measurable) and all the intersections and unions are countable (in particular we can restrict to positive rational epsilon). Algebraist 16:22, 19 February 2008 (UTC)[reply]

Wonderful. Someone else suggested the same idea in conversation and it works perfectly. I think this is "the" solution. Thank you! — merge 21:23, 19 February 2008 (UTC)[reply]

Googol

Why is one googol not $10^{100}$ ? The Wikipedia article states it as being $10^{100.0784}$ . Why is this? Thanks, Zrs 12 (talk) 23:50, 18 February 2008 (UTC)[reply]

One googol is by definition 10¹⁰⁰, as correctly stated in the article. The article also correctly states that 70! is approximately 10^100.0784. I don't see any source of confusion here. Algebraist 23:55, 18 February 2008 (UTC)[reply]

Perhaps OP did not understand Order of magnitude, and just skipped over those words to read "One googol is the same as 70!." --PalaceGuard008 (Talk) 23:57, 18 February 2008 (UTC)[reply]

Haha! How dumb of me! Gah, I must start reading more carefully. Zrs 12 (talk) 00:38, 19 February 2008 (UTC)[reply]

February 19

Singular Value Decomposition & Hermitian Matrices

This is a question posed in my class and the teacher himself could not figure it out either. The question is that given the singular value decomposition of a mxm square matrix $A=U\Sigma V^{*}$ where the * represents the complex conjugate transpose of a matrix, is there anything that we can say about the eigenvalue decomposition (diagonalization) of the 2mx2m Hermitian matrix $B={\begin{bmatrix}0&&A^{*}\\A&&0\end{bmatrix}}$ ? Can B's eigenvalue decomposition be written in terms of $U,\Sigma ,$ and $V$ ? I have also tried a few numerical examples on Matlab and it appears to me that the two are completely unrelated. Any help would be appreciated.A Real Kaiser (talk) 06:10, 19 February 2008 (UTC)[reply]

If

A=U\Sigma V^{*}

, then

U^{*}AV=\Sigma

. As

\Sigma

is self-adjoint, we get

V^{*}A^{*}U=(U^{*}AV)^{*}=\Sigma

as well. Now compute:

{\begin{pmatrix}0&U^{*}\\V^{*}&0\end{pmatrix}}{\begin{pmatrix}0&A^{*}\\A&0\end{pmatrix}}{\begin{pmatrix}V&0\\0&U\end{pmatrix}}={\begin{pmatrix}U^{*}AV&0\\0&V^{*}A^{*}U\end{pmatrix}}={\begin{pmatrix}\Sigma &0\\0&\Sigma \end{pmatrix}}.

Moving the unitaries (one can check that those block matrices on the left and right of your

B

are unitary) to the right gives the singular value decomposition. You have the absolute values of the eigenvalues at this point, but I'm not sure exactly what you mean by eigenvalue decomposition. Do you mean diagonalizing B? J Elliot (talk) 07:13, 19 February 2008 (UTC)[reply]

The eigenvalue decomposition goes similarly. For motivation, find the eigenvalue decomposition of [0,1;1,0] (taking A to be the 1x1 identity matrix). A=USV*, so A*=VSU*, AV=US, A*U=VS, so [0,A*;A,0][V;U] = [VS;US], so [V;U] is an "eigenvector" with eigenvalues S, and similarly [0,A*;A,0][V;-U]=[-VS;US], so [V;-U] is an "eigenvector" with eigenvalues -S. Since our matrix is hermitian, we want our eigenbasis to be unitary, so we'll divide the eigenvectors by their overall norm, 1/sqrt(2). Putting it all together gives:

{\begin{bmatrix}0&A^{*}\\A&0\end{bmatrix}}=\left({\frac {1}{\sqrt {2}}}{\begin{bmatrix}V&V\\U&-U\end{bmatrix}}\right)\cdot {\begin{bmatrix}\Sigma &0\\0&-\Sigma \end{bmatrix}}\cdot \left({\frac {1}{\sqrt {2}}}{\begin{bmatrix}V^{*}&U^{*}\\V^{*}&-U^{*}\end{bmatrix}}\right)

So the eigenvalues of B are plus or minus the singular values of A. JackSchmidt (talk) 15:53, 19 February 2008 (UTC)[reply]

Thanks guys, that makes a lot more sense. But I still have a follow-up question. You have shown that [V;U] and [-V;U] are eigenvectors with respect to S and -S but how do we know that those are the only eigenvalues? What if S^2 of -2S^5 also turn out to be eigenvalues of our matrix B? How can we conclude that S and -S are the ONLY eigenvalues?A Real Kaiser (talk) 23:53, 19 February 2008 (UTC)[reply]

Sorry, I spoke too informally. U and V are actually m x m matrices, and S is a diagonal matrix with m values. [V,V;U,-U] has full rank (because it is unitary), so is actually 2m independent eigenvectors for B, and S,-S gives the 2m eigenvalues. The informal language was just to indicate how block matrices can simplify things. JackSchmidt (talk) 00:58, 20 February 2008 (UTC)[reply]

fitting a conic

I want to fit a general parabola (of unknown size and orientation) roughly to a given set of points in the plane. My first thought was to seek the coefficients that minimize the sum of the squares of $A{x_{i}}^{2}+2{\sqrt {AC}}x_{i}y_{i}+C{y_{i}}^{2}+Dx_{i}+Ey_{i}+1$ ; to make the problem more linear, I then sought to settle for a general conic, $A{x_{i}}^{2}+2Bx_{i}y_{i}+C{y_{i}}^{2}+Dx_{i}+Ey_{i}+1$ ; but then it occurs to me that this penalizes those curves that go near the origin.

My next idea is to consider the family of cones tangent to some plane $z=z_{0}\neq 0$ ; I'm not sure what to minimize.

Anyone know a better way? —Tamfang (talk) 06:24, 19 February 2008 (UTC)[reply]

Minimize the sum of squares of

A{x_{i}}^{2}+2Bx_{i}y_{i}+C{y_{i}}^{2}+Dx_{i}+Ey_{i}+F

using some other normalizing condition than

F=1\,

(which excludes curves through the origin). Bo Jacoby (talk) 07:50, 19 February 2008 (UTC).[reply]

You could try an iterative approach. Define for brevity

F(x,y)=A{x_{i}}^{2}+2Bx_{i}y_{i}+C{y_{i}}^{2}+Dx_{i}+Ey_{i}+1

and

F_{i}=F(x_{i},y_{i}).\,

Given estimates for the coefficients A, B, etcetera, you can determine the distance e_i of each point (x_i,y_i) to the curve determined by F(x,y) = 0. If we give weight w_i to the term F_i² in a weighted sum of squares, we want the weighted square to come out like e_i², which suggests setting

w_{i}={\frac {e_{i}^{2}}{F_{i}^{2}}}

as the weights for a next iteration.

Instead of determining the values e_i exactly, which is computationally expensive, you can approximate it by using the linear approximation

F(x+\Delta x,y+\Delta y)\approx F(x,y)+{\frac {\partial }{\partial x}}F(x,y)\cdot \Delta x+{\frac {\partial }{\partial y}}F(x,y)\cdot \Delta y.

Applied to the point (x_i,y_i), we write this as

F(x_{i}+\Delta x,y_{i}+\Delta y)\approx F_{i}+{\frac {\partial }{\partial x}}F_{i}\cdot \Delta x+{\frac {\partial }{\partial y}}F_{i}\cdot \Delta y.

The least value for (Δx)² + (Δy)² for which the right-hand side can vanish, which provides an estimate for e_i², is then given by

(\Delta x)^{2}+(\Delta y)^{2}={\frac {F_{i}^{2}}{{\frac {\partial }{\partial x}}F_{i}^{2}+{\frac {\partial }{\partial y}}F_{i}^{2}}}\,.

So for the weights for the next iteration, you can use then

w_{i}=({\frac {\partial }{\partial x}}F_{i}^{2}+{\frac {\partial }{\partial y}}F_{i}^{2})^{-1}\,.

Since the value being inverted can become arbitrarily small and even vanish, you must exercise caution not to make this numerically unstable, and put limits on the size of the weights. --Lambiam 08:49, 21 February 2008 (UTC)[reply]

Rationalising Surds

I was going over some of my notes and i tried this one but my answer isnt the same as in the book.. im not sure what im doing wrong
Rationalise the denominator
${\frac {1{}}{{\sqrt {5}}-{\sqrt {7}}}}$
$={\frac {1{}}{{\sqrt {5}}-{\sqrt {7}}}}$ $*{\frac {{\sqrt {5}}+{\sqrt {7}}}{{\sqrt {5}}+{\sqrt {7}}}}$
$={\frac {{\sqrt {5}}+{\sqrt {7}}}{-2}}$
Kingpomba (talk) 11:26, 19 February 2008 (UTC)[reply]

Looks fine to me and quick calculator check confirms your answer. Could also be written as

{\frac {-{\sqrt {5}}-{\sqrt {7}}}{2}}

or

-\left({\frac {{\sqrt {5}}+{\sqrt {7}}}{2}}\right)

. What does your book say ? Gandalf61 (talk) 11:47, 19 February 2008 (UTC)[reply]

it says: ${\frac {1{}}{2}}$ $({\sqrt {5}}+{\sqrt {7}})$ (hmm i got a different answer on paper and i understand how the 1/2 thing works i guess typing it out on wikipedia helped me solve it, well cheers anyway =] Kingpomba (talk) 11:57, 19 February 2008 (UTC).[reply]

Are you sure that's what it says? There should be a minus sign in front, shouldn't there? --Tango (talk) 12:54, 19 February 2008 (UTC)[reply]

Here's a quick sanity check: 5 < 7, so sqrt(5) < sqrt(7) (by the properties of the square root), and hence sqrt(5) - sqrt(7) < 0. Thus the denominator of the original fraction is negative, meaning the whole fraction is negative. Confusing Manifestation(Say hi!) 23:47, 19 February 2008 (UTC)[reply]

February 20

zeros of a periodic function

Is there any way to find an $x$ which solves the equation

${\frac {1}{2}}+{\frac {1}{4}}\left(\cos(ax)+\cos(bx)\right)=0$

for any arbitrary $a$ and $b$ ? I know that some value of x will make both $ax$ and $bx$ equal to an odd multiple of $\pi$ but can't think of a way to find this value. I know that the answer may lie in recasting the equation in the form

${\frac {1}{2}}+{\frac {1}{2}}\left(\cos({\frac {a-b}{2}}x)\cos({\frac {a+b}{2}}x)\right)=0$

and in this case know that some value of x makes this product of cosine terms equal to negative one, but again can't think of how to find this x value. Can this be solved analytically or must I resort to numerical calculation (which would be very unsatisfying). Thank you very much for your assistance. —Preceding unsigned comment added by 128.223.131.21 (talk) 00:21, 20 February 2008 (UTC)[reply]

If the combined function is periodic, then a/b is a rational number, so after rescaling x, you can assume a and b are relatively prime integers. Then I believe x=pi is a solution if both a,b are odd, and there is no solution if one of a or b is even. JackSchmidt (talk) 01:16, 20 February 2008 (UTC)[reply]

To explain JackSchmidt's answer a bit: since (for

x\in \mathbb {R}

)

|\cos x|\leq 1

, the only way to have

\cos x+\cos y=-2

is to have

\cos x=\cos y=-1

. So

ax=(2n+1)\pi ,bx=(2m+1)\pi

, so

{\frac {a}{b}}={\frac {2n+1}{2m+1}}

and the simplest form of

{\frac {a}{b}}

must have both numbers odd. When that condition is satisfied, there is some smallest integer

k\ni \{ak,bk\}\subset \mathbb {I}

; k is the common denominator for a and b. Any

x=(2l+1)k\pi

with integer l solves the equation. --Tardis (talk) 16:20, 20 February 2008 (UTC)[reply]

Donald in Mathmagic Land

Two questions about the above-named 1959 Walt Disney film. (Question 1) In the film, a character incorrectly states the value of pi. Does anyone know how a Disney film that was designed to be a mathematics educational film for children (and presumably had math consultants working on it) could possibly contain such an error? How would that error slip by unnoticed? (Question 2) With regard to that error, someone stated (in a blog) that it was not really an error, since "the value of pi has changed so much since 1959". That cannot possibly be true, can it? I am sure that we have become enlightened about the accurate values of many digits of pi over the course of thousands of years of history ... but not since 1959, correct? Any ideas? Thanks. (Joseph A. Spadaro (talk) 04:10, 20 February 2008 (UTC))[reply]

Pi is an irrational number meaning that its' decimal representation will have and infinite amount of digits. Given the fact that we can never calculate an infinite amount of digits it is the case that new digits have been calculated since 1959. Of course this does not mean the the value of pi has changed, just the precision with which one can write it. Perhaps this is what 'someone' was reffering too? Any errors in the calculation of digits of pi since 1959 would, i suspect, occur at least several thousand (being conservative here) digits out so it would in fact be an error. I have no input on how the error actually occured. —Preceding unsigned comment added by 74.242.224.136 (talk) 04:55, 20 February 2008 (UTC)[reply]

The error could very well have been caused by something as simple as the voice actor getting lost while reading the digits. —Bkell (talk) 05:00, 20 February 2008 (UTC)[reply]

Thanks. Let me follow-up. I should have made this clearer. (Though, remember, we are dealing with a Disney children's video.) First ... the value of pi was recited to, perhaps, 15 or 20 decimal places ... certainly, no more than 20. So, since 1959, nothing has "changed" in that early chunk of the number - correct? Second ... regarding the voice actor flub. Certainly, that is a possibility. I guess my point was ... is not that an "inexcusable" error (that would call for a retake ... or, at the very least, a dubbed edit in post-production), given that it is an education video, and a math one to boot ( ... as opposed to, say, some character misstating the digits of pi on an inconsequential TV episode of Seinfeld) ...? Thanks. (Joseph A. Spadaro (talk) 05:16, 20 February 2008 (UTC))[reply]

Pi was known to 527 digits by the end of the 19th century, and to 2037 digits by 1949 (a decade earlier than the film was made). Our article on Donald in Mathmagic Land says that the bird only recites "the first few digits" of pi. The error was not due to pi not being known accurately enough in 1959. MrRedact (talk) 05:25, 20 February 2008 (UTC)[reply]

Probably the producers didn't consider it important. If the voice actor flubbed, either no one noticed or no one thought it was worth it to rerecord the line just to be pedantic. The character reciting pi is basically a throw-away gag anyway. They got the first few digits right (out to maybe eight or ten places, if I remember correctly), and that's certainly more than anyone really needs. The purpose of the film is not to teach children the digits of pi but to introduce them to some general mathematical concepts. —Bkell (talk) 06:12, 20 February 2008 (UTC)[reply]

Yes, I agree with all that is said above. However, as an education video --- I am not sure that being "pedantic" is a bad thing ... quite the contrary, I'd assume. I will also assume that Disney's multi-zillion dollar fortune could easily afford any "expensive" retakes. Finally, mathematicians are nothing, if not precise. I can't imagine a mathematician letting that glaring error slide. A film producer, yes ... a mathematician, nah. (Joseph A. Spadaro (talk) 06:56, 20 February 2008 (UTC))[reply]

I agree completely. This is just a symptom of a much more general problem. Very few people care about spreading misinformation, and very many then wonder how come people are so misinformed. -- Meni Rosenfeld (talk) 10:45, 20 February 2008 (UTC)[reply]

What a succinct -- yet woefully accurate -- synopsis, Meni. Bravo! (Joseph A. Spadaro (talk) 16:20, 20 February 2008 (UTC))[reply]

As Bertrand Russell appears to have written: 'PEDANT—A man who likes his statements to be true.' Algebraist 10:55, 21 February 2008 (UTC)[reply]

Thanks for all the input. Much appreciated. (Joseph A. Spadaro (talk) 01:05, 24 February 2008 (UTC))[reply]

Predicate logic

I have to show, using a transformational proof (logical equivalence and rules of inference) that the following two premises are a contradiction:

1.

\neg \exists x[((Q(x)\land R(x))\lor (Q(x)\land \exists yP(y))]

2.

\exists x[((Q(x)\land R(x))\lor (Q(x)\land \forall yP(y))]

Now these two statements look quite similar because they were actually two more complex statements I tried to "whittle down" into a similar form in order to see where I could take it. They almost contradict one another apart from the $\exists yP(y)$ part in 1. versus the $\forall yP(y)$ part in 2.

Can you offer me a suggestion for manipulating these statements further to show a clear contradiction? My first line of thought would be to show that the conjunction, $\land$ , of the two statements is false but I get bogged down in symbols. Perhaps there is a more elegant way using inference rules, I'm not sure. Damien Karras (talk) 13:39, 20 February 2008 (UTC)[reply]

Start by consulting De Morgan's laws, including the ones for quantifiers, if you're not familiar with them. Then you should be able to make use of the fact that

(\forall x\,A(x))\land (\exists x\,B(x))\rightarrow \exists x\,A(x)\land B(x)

. Also, it may not be strictly necessary, but consider what it means to say that

\exists x\,A(y)

. --Tardis (talk) 16:34, 20 February 2008 (UTC)[reply]

The two premises are contradictory if (and only if) (2) implies the negation of (1). You will have to use somehow, directly or indirectly, a rule that says that

\forall yP(y)

implies

\exists yP(y)

. If you have a logic in which the domain of predicate P can be empty, so that this implication does not hold, while the domain of the predicates Q and R, from which variable x assumes its values, is not necessarily empty, the two premises are not contradictory: take for Q the constant always-true predicate and for R the always-false predicate; the premises then simplify to

\neg \exists yP(y)

and

\forall yP(y)

. I don't know the precise rules you are allowed to use, but the logical connectives

\land \,\!

and

\lor \,\!

, as well as existential quantification, are all "monotonic" with respect to implication, so that

\exists x[((Q(x)\land R(x))\lor (Q(x)\land \forall yP(y))]\quad \rightarrow \quad \exists x[((Q(x)\land R(x))\lor (Q(x)\land \exists yP(y))]

is a consequence of

\forall yP(y)\quad \rightarrow \quad \exists yP(y).

--Lambiam 09:58, 21 February 2008 (UTC)[reply]

Thanks both, and let it be known that P, Q, and R share the same domain. Lambiam, I assume what you're saying is that if, essentially, if I prove

\forall yP(y)\rightarrow \exists yP(y)

I can show the two statements are a clear contradiction? Damien Karras (talk) 13:58, 21 February 2008 (UTC)[reply]

I still don't understand what I have to do. If

\neg \exists xQ(x)

in one statement and

\exists xQ(x)

in another, doesn't that show that the two sides are dependent on

\exists yP(y)

and

\forall yP(y)

. I just don't get this at all. If

\forall yP(y)

is true, then so is the existential quantification. If it is false then the existential quant. is still true? "It is not the case that, for ALL y, P(y) is true - however there is a y for which P(y) IS true." Damien Karras (talk) 15:31, 21 February 2008 (UTC)[reply]

Sorry, but as I wrote, I don't know the precise rules you are allowed to use, so I am not sure what the steps are that you can take. If the domain of

P\,\!

is not empty, and

\forall yP(y)

is true, then so is

\exists yP(y)

. The latter can be true while the former is false, but it is also possible that both are false. This is summed up in the statement that

\forall yP(y)\rightarrow \exists yP(y)

, which is true for nonempty domains. The contradiction should then follow from this. --Lambiam 20:02, 24 February 2008 (UTC)[reply]

There are no countably infinite σ-algebras

Hoping someone can help me spot the holes in my argument. ;)

Suppose ${\mathcal {S}}$ is a countably infinite σ-algebra over the set X. If we can show that ${\mathcal {S}}$ contains an infinite sequence of sets $E_{1}\subset E_{2}\subset \ldots$ or $F_{1}\supset F_{2}\supset \ldots$ , then the sequence of differences $E_{1},(E_{2}-E_{1}),\ldots$ or $\left(F_{1}-\cup _{n>1}F_{n}\right),\left(F_{2}-\cup _{n>2}F_{n}\right),\ldots$ will be a countably infinite collection of pairwise disjoint sets $\{D_{1},D_{2},\ldots \}$ , and the map $\{n_{1},n_{2},\ldots \}\mapsto \cup _{k\geq 1}D_{n_{k}}$ will be an injection from the power set of N into ${\mathcal {S}}$ .

Let ${\mathcal {S}}$ be partially ordered under inclusion. We may assume that ${\mathcal {S}}$ contains no infinite ascending chain $E_{1}\subset E_{2}\subset \ldots$ , for if it does this gives us what we wanted.

The set ${\mathcal {S}}_{1}={\mathcal {S}}-\{\emptyset ,X\}$ has a maximal element F₁ by Zorn's lemma. It follows that the only elements of ${\mathcal {S}}_{1}-\{F_{1},X-F_{1}\}$ are subsets of F₁ and their complements; in particular this set contains infinitely many subsets of F₁. The collection ${\mathcal {S}}_{2}$ of these subsets cannot contain an infinite ascending chain, so has a maximal element F₂. It follows that the only elements of ${\mathcal {S}}_{2}-\{F_{2},F_{1}-F_{2}\}$ are subsets of F₂ and their complements relative to F₁. Proceeding in this way yields an infinite descending chain $F_{1}\supset F_{2}\supset \ldots$ . — merge 16:36, 20 February 2008 (UTC)[reply]

I think you have confused maximal element with greatest element. -- Meni Rosenfeld (talk) 19:46, 20 February 2008 (UTC)[reply]

Meni's right -- your last "it follows that" is wrong. The result is right, though. Hint: You don't need Zorn's lemma; the hypothesis that the sigma-algebra is countably infinite already gives you a bijection between the sigma-algebra and the natural numbers. Do induction along that. --Trovatore (talk) 20:08, 20 February 2008 (UTC)[reply]

I had confused myself, although not about that--sorry! I'll definitely try what you said, but I've also tried to fix the above--any better? — merge 21:08, 20 February 2008 (UTC)[reply]

Revised version at σ-algebra redux below. — merge 13:26, 21 February 2008 (UTC)[reply]

Summing 1^1 + 2^2 +3^3 ..... + 1000^1000

I'm trying to improve my mathematics skills by doing the problems at Project Euler, but have hit a stumbling block with problem 48: (http://projecteuler.net/index.php?section=problems&id=48

I need to find the last 10 digits of 1^1 + 2^2 +3^3 ..... + 1000^1000

I've read the wikipedia page on Geometric_progression , but I want to do $\sum _{n=1}^{1000}n^{n}$ rather than $\sum _{n=1}^{1000}r^{n}$ (i.e. I don't have a constant value r)

I'm not so much looking for an answer, just what techniques or terminology I should be reading up on. Pointers welcomed!

Lordelph (talk) 16:56, 20 February 2008 (UTC)[reply]

The key point is that you're looking for the last 10 digits, not the whole number (which is enormous). Think about which numbers contribute what to the last ten digits. --Tango (talk) 17:14, 20 February 2008 (UTC)[reply]

I don't think there is a mathematical shortcut for this. But Project Euler is as much about programming as it is about mathematics, so you could take a brute force approach. To calculate the last 10 digits of 999^999, for example, start with 999; multiply it by 999; just keep the last 10 digits; repeat another 997 times and you are done. You only need to work with interim results of up to 13 digits, so its not even a bignum problem. I guess you end up doing around half a million integer multiplications altogether, but I don't think that will take very long. In the time it's taken me to write this, someone has probably already programmed it. Gandalf61 (talk) 17:27, 20 February 2008 (UTC)[reply]

Actually even the ultimate brute force approach of computing the whole 3001-digit sum will be more than fast enough on a modern computer. My 2004 laptop took a negligible fraction of a second to evaluate sum [n^n | n <- [1..1000]] in GHCi. They should've gone up to 1000000^1000000. -- BenRG (talk) 19:24, 20 February 2008 (UTC)[reply]

I've seen similar problems where there has been a shortcut, I can't spot one for this problem, though. It's a trivial programming exercise, though, so I can't see why they would have set it. I can see a few possible ways to optimise the code, but there's really no need (for example, 999^999=(990+9)^999, expand that out using the binomial expansion and each term will have a 990^n in it, which has all 0's for the last ten digits for all but the first 10 terms, so you can ignore all but those terms - probably not much quicker for 999^999 (you need to calculate the coefficients, which won't help), but could be for 999999^999999). --Tango (talk) 19:52, 20 February 2008 (UTC)[reply]

Many thanks for pointers, problem solved Lordelph (talk) 20:37, 20 February 2008 (UTC)[reply]

On a slow computational platform, this can be somewhat sped up as follows. The recursive algorithms for exponentiation by squaring works in any ring, including the ring Z/nZ of modulo arithmetic. So in computing nⁿ, all computations can be done modulo 10¹⁰. And if n is a multiple of 10, nⁿ reduces to 0 modulo 10¹⁰, so 10% of the terms can be skipped. —Preceding unsigned comment added by Lambiam (talk • contribs) 11:13, 21 February 2008 (UTC)[reply]

Resistance

These are more questions of physics than maths but considering the overlap between the two, and how this ref desk is much more specific than the science one, I'll ask them here.

I know that the formula for the overall resistance of two resistors in parallel is ${\frac {1}{R_{t}}}={\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}$ . So if you have two fifty ohm resistors in parallel, the total resistance is 25 ohms. What I don't understand is how the total resistance can be less than that of either resistor; it doesn't make any sense to me.

Also, if you have the below setup but R1 is just a wire, how would you calculate the total resistance of the circuit, ignoring internal resistance?

Many thanks, 92.1.70.98 (talk) 18:07, 20 February 2008 (UTC)[reply]

It's actually very simple intuitively. Resistance measures how hard it is for electrons to move from one place to another. If there are two resistors in parallel, the electrons have 2 ways to choose from, thus travel is easier. The analogy is a mass of people trying to cross a narrow corridor - they will have a much easier time if there are two parallel corridors.

Any resistor network can be solved using Ohm's law (

V=IR

) and Kirchhoff's current law (the sum of currents at every point is 0). This one is simpler because it's just two resistors in a series, where the second is composed of two resistors in parallel. Thus the equivalent resistance is

R_{3}+{\frac {R_{1}R_{2}}{R_{1}+R_{2}}}

.

There are many Wikipedia articles from which you can learn more about this - Resistor might be a good start. -- Meni Rosenfeld (talk) 18:35, 20 February 2008 (UTC)[reply]

Well, Meni beat me to it. But here it is. First of all, the total resistance decreases when you are talking about two resistors in PARALLEL not in series. In series, you get exactly what you think that you should get. The total resistance is just the sum. In parallel, you can think of it like this, you have two branches, so the current going in splits up in the branch so the energy loss in each branch is smaller than what it should have been. But then the current adds up again when the branches meet. In series, the entire current has to go through both of the resistors, so the loss is greater. As for the picture, first you calculate the combined resistance of R1 ad R2 as resistors in parallel (call it R12) which gives you

R_{12}={\frac {R_{1}R_{2}}{R_{1}+R_{2}}}

and then you simply treat R3 as a resistor in series with the combined resistor R12 and you get that the total resistance of the circuit is

R=R_{3}+R_{12}=R_{3}+{\frac {R_{1}R_{2}}{R_{1}+R_{2}}}

.A Real Kaiser (talk) 18:39, 20 February 2008 (UTC)[reply]

OK the explanation for why the resistance is less in parallel is helpful but I think you both missed the bit about R1 being just a wire, not a resistor (it was the best picture I could find lying around on WIkipedia). Using the equation for resistors in parallel, it would make the resistance of the parallel section 0 (assuming that a wire has no resistance). I would interpret this as meaning that all current goes through the wire and none goes through the resistor. Is that right? Thanks 92.1.70.98 (talk) 18:45, 20 February 2008 (UTC)[reply]

That would be an idealised situation. Nothing really has 0 resistance, so R₁ would be very small, but it would be non-zero. This would mean that the total resistance over R1 and R2 was almost 0, since R2's contribution would be negligible. -mattbuck (Talk) 19:24, 20 February 2008 (UTC)[reply]

Still, in the limit where

R_{1}\to 0

, there will be no current going through R2 (this has nothing to do with interpretation, it's just a calculation). Assuming we want the current to go through R2, the extra wire is a short circuit. -- Meni Rosenfeld (talk) 19:32, 20 February 2008 (UTC)[reply]

Random calculus question

A friend of mine asked me to help him solve the following question, and I'm not a math major by any means. So, any help would be appreciated. :) Zidel333 (talk) 19:40, 20 February 2008 (UTC)[reply]

[1+1/N] ^ (N+1/2), prove that it is always increasing from zero to infinity.

It will help to try to prove something that is true: that function is always decreasing for

N\in (0,\infty )

. Its limit at large N is e, as can be seen immediately from that article. The normal procedure is to evaluate the function's derivative and show that it is negative for all positive N. However, the algebra involved is non-trivial; perhaps some information is missing? --Tardis (talk) 20:07, 20 February 2008 (UTC)[reply]

I'm sorry, that's all the info I have from my friend. If he tells me more, I'll add it to the discussion. Thanks for your help so far, far better than what I could have come up with. :) Zidel333 (talk) 20:12, 20 February 2008 (UTC)[reply]

As it's "N" not "x", I would assume N is ranging only over the integers, so taking the derivative isn't necessary. You can either show that f(n+1)-f(n)<0 or f(n+1)/f(n)<1 for all n. The latter is probably easier. --Tango (talk) 22:22, 20 February 2008 (UTC)[reply]

Hint: Perhaps rewriting the sequence as exp[(N+1/2)*log(1+1/N)] can help you (sorry, I'm not comfortable with LaTeX). --Taraborn (talk) 10:33, 21 February 2008 (UTC)[reply]

Sum

How would you work out, for j and k ranging over the integers, the sum of (j+k)^-2? Black Carrot (talk) 20:36, 20 February 2008 (UTC)[reply]

Well, the term for j=k=0 is going to be a problem :-) --Trovatore (talk) 20:50, 20 February 2008 (UTC)[reply]

But assuming you really meant the positive integers, the sum diverges. For a fixed j, the sum over all k>0 of (j+k)⁻² is Ω(1/j). --Trovatore (talk) 20:53, 20 February 2008 (UTC)[reply]

(Edit conflict) Assuming you mean positive integers, since all integers doesn't make sense, as Trovatore points out, observe that for a fixed j, and varying k, j+k varies over all the integers, so

\Sigma _{j,k\in N}(j+k)^{-2}=\Sigma _{j=1}^{n}(\Sigma _{k=1}^{n}k^{-2})

Now, the inner sum is clearly greater than 0, and summing an infinite number of non-zero constants gives you infinity (or, strictly speaking "doesn't converge"). So, unless I've missed something, the series doesn't converge. --Tango (talk) 21:00, 20 February 2008 (UTC)[reply]

Correction: For fixed j, varying k, j+k varies over all integers greater than j, I was thinking of k varying over all integers, rather than just positive ones... it doesn't change the conclusion, though - the series still diverges. --Tango (talk) 22:19, 20 February 2008 (UTC)[reply]

The statement that "summing an infinite number of non-zero constants gives you infinity" is very often not true, even assuming you're limiting yourself to positive constants. For example,

\Sigma _{j=1}^{\infty }2^{-j}=1

. See the Taylor series for e^x or tan x as a couple other examples. MrRedact (talk) 01:50, 21 February 2008 (UTC)[reply]

None of those examples are constants. --Tango (talk) 12:08, 21 February 2008 (UTC)[reply]

I guess it depends on what you mean by a "constant". My first example is the sum of 1/2, 1/4, 1/8, 1/16, 1/32, etc., all of which are positive constants, and whose sum is finite (1). I see now that you're talking about an infinite sum of identical positive terms, i.e., an infinite sum of terms that don't depend on the index of summation, in which case, yeah, that would always diverge. MrRedact (talk) 17:35, 21 February 2008 (UTC)[reply]

Yeah, I should probably have said "an infinite number of copies of a non-zero constant". I'm pretty sure my conclusion was correct, I just didn't explain to too well, sorry! --Tango (talk) 18:42, 21 February 2008 (UTC)[reply]

Here's another way to do it. Let i=j+k. Now, you have 1 case where i=2 (j=1, k=1); 2 cases where i=3 (j=1, k=2; and j=2, k=1), and in general i-1 cases for a given i. Thus we can express the desired sum as

\sum _{i=2}^{\infty }{\frac {i-1}{i^{2}}}

, which is equal to

\sum _{i=2}^{\infty }{\frac {1}{i}}-\sum _{i=2}^{\infty }{\frac {1}{i^{2}}}

. The first sum diverges and the second converges, so the overall sum diverges. Chuck (talk) 18:53, 21 February 2008 (UTC)[reply]

Integral Question

Hi I've got an integral question which i have the answer for but i don't understand why it works.
The question is: If f(x) is continuous on [0; ¼], show that: $\lim _{n\rightarrow \infty }\int _{0}^{\pi }|sin\ nx|f(x)\ dx={\frac {2}{\pi }}\int _{0}^{\pi }f(x)\ dx$
I've got it into this form: $\lim _{n\rightarrow \infty }\sum _{k=1}^{n}\int _{\frac {(k-1)\pi }{n}}^{\frac {k\pi }{n}}|sin\ nx|fdx$
The next step i have written is: $\sum _{k=1}^{n}f(\xi nk)\int _{\frac {(k-1)\pi }{n}}^{\frac {k\pi }{n}}|sin\ nx|dx$
But i don't understand how to get to that step. It doesnt help that my lecturer makes the odd mistake now and again. Can somebody explain how to get between these steps and how to complete the rest of the question. Thanks for any help. 212.140.139.225 (talk) 23:00, 20 February 2008 (UTC)[reply]

In the one-but-last formula, write f(x) instead of just f. I think the last formula should be something like

\lim _{n\rightarrow \infty }\sum _{k=1}^{n}f(\xi _{k})\int _{\frac {(k-1)\pi }{n}}^{\frac {k\pi }{n}}|sin\ nx|dx,~\mathrm {where} ~{\tfrac {(k-1)\pi }{n}}<\xi _{k}<{\tfrac {k\pi }{n}},

which is justified by the first mean value theorem for integration. --Lambiam 11:39, 21 February 2008 (UTC)[reply]

February 21

Quick question about log convexity

I've read the article on power series, and it mentions that for more than one variable, the domain of convergence is a log-convex set instead of an interval.

But what is a log-convex set here ? I mean, how do we take the log of a set ? Surely the set can contain negative values for example...

Thanks. -- Xedi (talk) 02:13, 21 February 2008 (UTC)[reply]

Sorry, I have no idea what they could mean by log-convexity of a set. The multi-variable stuff was added in June 2004, and the log-convex comment was made in Oct 2004. It does not appear to have been touched since then. I'll ask the author about it.

The domain of convergence is a union of polydisks, but I think perhaps it need not be a polydisk. Some reasonably elementary notes are at Adam Coffman's website, specifically his research notes on Notes on series in several variables. It includes the polydisk result, as well as the continuity, and differentiability of functions defined by a series, and includes the nice "recentering" result familiar for one complex variable allowing analytic continuation. JackSchmidt (talk) 04:25, 21 February 2008 (UTC)[reply]

I've added an example and a little more explanation to the article. Terry (talk) 23:07, 21 February 2008 (UTC)[reply]

Thanks. I suppose when the "center" is (a₁,a₂,...) then we consider (log|x₁-a₁|,log|x₂-a₂|,...) ? —Preceding unsigned comment added by Sam Derbyshire (talk • contribs) 13:08, 22 February 2008 (UTC)[reply]

Natural Logs

I'm being asked to evaluate natural logs and I'm getting very confused. One problem I'm having particular trouble with is lne(-x/3)

Could anyone please help explain to me how to solve problems like these and how to determine how to get started?

Thanks, anon. —Preceding unsigned comment added by 66.76.125.76 (talk) 03:53, 21 February 2008 (UTC)[reply]

What is lne(-x/3) ?

Is it Log[ Exp[-x/3] ] ?

or is it Log[-x/3] where Log is Log based e.

202.168.50.40 (talk) 04:34, 21 February 2008 (UTC)[reply]

ln is usually a shorthand for log_e --PalaceGuard008 (Talk) 05:07, 21 February 2008 (UTC)[reply]

y = log(-x/3)
e^y = -x/3
x = -3e^y

--wj32 ^t/c 05:16, 21 February 2008 (UTC)[reply]

If the question is about

, what can you say in general about

? --Lambiam 12:05, 21 February 2008 (UTC)[reply]

is math the same as logic?

Is math the same as logic, or why not.

In other words, is there math that isn't logical, but just evaluates truths, for example "experimenting" with numbers to find out "facts" but not using logic to show that these must be necessarily true. After all, the other sciences don't rely on their findings to be NECESSARILLY true, just that they happen to be true... so which is math? —Preceding unsigned comment added by 79.122.42.134 (talk) 10:12, 21 February 2008 (UTC)[reply]

You might be interested in logicism and experimental mathematics. Algebraist 10:43, 21 February 2008 (UTC)[reply]

You might also be interested in David Hilbert's project to formalize mathematics.--droptone (talk) 12:45, 21 February 2008 (UTC)[reply]

Experimental mathematics uses numerical methods to suggest conjectures which are then formally proved (or perhaps disproved if they turn out to be just a numerical coincidence). All mathematical results are proved using the techniques of logic, so, yes, they are necessarily true. But this does not mean that mathematics is the same as logic - logic does not determine what a mathemtician sets out to prove, or why a particular result is considered to be beautiful, deep or interesting. Logic is to mathematics as calligraphy is to the Rubaiyat of Omar Khayyam. Gandalf61 (talk) 14:04, 21 February 2008 (UTC)[reply]

Moment

Suppose you have a ladder leaning against a wall, which makes an angle θ with the ground, and a man, who can be treated as a particle, stands a distance l up the ladder. Now his weight will obviously be acting downwards and so working to work out the moment produced, you would have to resolve his weight vector to find the component of it that acts perpendicular to the ladder.

My question is, if you used Pythagoras or trig to determine his perpendicular distance from the ladder's point of contact with the ground to the line of action of the weight vector and then multiplied that by his weight, would that give you the correct value of the moment? Cheers in advance 92.3.49.42 (talk) 12:45, 21 February 2008 (UTC)[reply]

Unless I've misunderstood, yes. What I tend to do is extend the "arrow" of the force such that the perpendicular goes through the point you are trying to resolve against, then it's a simple case of force times distance from point. x42bn6 Talk Mess 13:02, 21 February 2008 (UTC)[reply]

Yes 92.3.49.42, both those methods are valid ways of calculating the moment around the ladder's point of contact with the ground, and both will yield the same answer. --mglg_(talk) 17:18, 21 February 2008 (UTC)[reply]

σ-algebra redux

A revised version of There are no countably infinite σ-algebras, hopefully closer to being correct?

Lemma. If ${\mathcal {A}}$ is an infinite algebra over a set X, partially ordered by inclusion, then ${\mathcal {A}}$ contains an infinite ascending chain $a_{1}<a_{2}<\ldots$ or an infinite descending chain $b_{1}>b_{2}>\ldots$ .

Proof: Let ${\mathcal {A}}_{1}={\mathcal {A}}-\{X\}$ . If ${\mathcal {A}}_{1}$ has a chain with no upper bound, then that gives us what we wanted. Otherwise ${\mathcal {A}}_{1}$ has a maximal element b₁. If $x\in {\mathcal {A}}_{1}$ and $x\not \leq b_{1}$ then $b_{1}<x\cup b_{1}$ so $x\cup b_{1}=X$ by maximality, and therefore $X-x\leq b_{1}$ . That means there are infinitely many subsets of b₁ in ${\mathcal {A}}_{1}$ .

Let ${\mathcal {A}}_{2}$ be that collection of subsets, minus b₁ itself. Then ${\mathcal {A}}_{2}$ has a maximal element b₂, and so again contains infinitely many subsets of b₂. Inductively we obtain an infinite descending chain $b_{1}>b_{2}>\ldots$ .

Corollary. Every infinite algebra contains an infinite collection of pairwise disjoint sets.

We just take the sequence of differences of elements of the chain.

Corollary. There are no countably infinite σ-algebras.

An infinite σ-algebra is an infinite algebra, and by the above has an infinite subcollection $\{D_{1},D_{2},\ldots \}$ of pairwise disjoint sets. The map $\{n_{1},n_{2},\ldots \}\mapsto \cup _{k\geq 1}D_{n_{k}}$ is then an injection from the power set of N into the σ-algebra.

— merge 13:58, 21 February 2008 (UTC)[reply]

That seems to work, but as Trovatore commented above, you don't need the Axiom of Choice (you may not care about AC, of course, but I do). Since you have no infinite ascending chains, you don't need Zorn to find a maximal element (the top element of any chain is maximal). If you start with a bijection between your (putative) countably infinite sigma-algebra and N and, whenever you have to choose an element, always choose the one of least index (in terms of this bijection), the proof becomes wholly choice-free. Algebraist 16:30, 21 February 2008 (UTC)[reply]

Many thanks for the review. You're right of course that there's no need to invoke the ghost of Max Zorn to get maximal elements there! I'm not too concerned about using AC, but I do like to improve my awareness of when and how it's being used and when it is or isn't necessary, so your comments are much appreciated. Although there are other ways to solve the original problem, the results for infinite algebras in general seem more useful than the fact that there are no countably infinite σ-algebras, and it's nice that once you have them the statement about σ-algebras becomes trivial. — merge 17:19, 21 February 2008 (UTC)[reply]

(slightly modified the above) It's worth noting also that for the results on infinite algebras we aren't assuming countability, so for these I believe AC is required (?). — merge 13:00, 22 February 2008 (UTC)[reply]

I think the argument goes through if you just know that every infinite set has a countably infinite subset, which does need a little choice to prove. But you don't need full AC to prove it -- for example, it follows from the axiom of countable choice, which is consistent (for example) with every set of reals being Lebesgue measurable. --Trovatore (talk) 22:23, 22 February 2008 (UTC)[reply]

Differential equation gives weird result!

Hi there, I've been given differential equation $dy/dx=e^{-3x}/y^{3}$

I separated variables and integrated, and I eventually came up with the general result $y=(-{4e^{-3x}/3}+D)^{1/4}$ where D is 4C, my initial constant of int.

But... the function is always going to be negative, and I have to take the fourth root of it! Have I messed it up, or am I ok so long as D makes the stuff under the fourth root positive??

Thank you! Psymun (talk) 15:20, 21 February 2008 (UTC)[reply]

Your working is correct. Could you not have a result that includes complex numbers? -mattbuck (Talk) 15:42, 21 February 2008 (UTC)[reply]

'Spose! I didn't see it coming in this course... although seeing as I did them in the last maths course I possibly should have! Thanks for reassurance! Psymun (talk) 15:47, 21 February 2008 (UTC)[reply]

No need to consider complex numbers. The function just isn't defined for

x<{\frac {1}{3}}\log \left({\frac {4}{3D}}\right)

. This isn't any different from solving

2y'y=1\;\!

and getting

{\sqrt {x+C}}

for a result. -- Meni Rosenfeld (talk) 16:09, 21 February 2008 (UTC)[reply]

Your general solution should be

y^{4}=-{\frac {4}{3}}e^{-3x}+C

. If you're restricting yourself to the real numbers, this is equivalent to

y^{2}={\sqrt {-{\frac {4}{3}}e^{-3x}+C}}

. But if you solve for y by taking the fourth root like you did, you're excluding negative values for y, so your solution isn't the general solution any more. —Bkell (talk) 05:20, 22 February 2008 (UTC)[reply]

Is there anything that HAS TO be true but isn't?

Has mathematics ever proved anything, so that it has to be true, but in fact it isn't? For example, I'm thinking of something like proving that something can be done within some constraints, but in fact all the variations have been tried by computer and none of them are successful. So there HAS TO be a way, but there isn't. —Preceding unsigned comment added by 79.122.42.134 (talk) 15:32, 21 February 2008 (UTC)[reply]

No, because if it could be proved true but all attempts to give examples fail, then you did it wrong. -mattbuck (Talk) 15:38, 21 February 2008 (UTC)[reply]

(edit conflict) No. This state of affairs is impossible based on the any commonly accepted concept of a mathematical proof. If you modify your statement to "Has anything ever been accepted as proven by the mathematical community...", then I do not know. Taemyr (talk) 15:47, 21 February 2008 (UTC)[reply]

There are of course things that seem to have been logically proven but aren't true. See paradox, and especially Zeno's paradox. DJ Clayworth (talk) 15:51, 21 February 2008 (UTC)[reply]

Do you guys mean to tell me that the universe is 100% consistent 100% of the time, and anything that follows is in fact thus, with no exceptions, ever? I find that a little hard to swallow... —Preceding unsigned comment added by 79.122.42.134 (talk) 16:15, 21 February 2008 (UTC)[reply]

We're not talking about the universe, we're talking about mathematics. Mathematics has not been proven to be consistent (and in a certain sense can't be), but is generally believed to be so. But yes, in the universe as in mathematics, the result of a valid deduction from true premises is true (in the jargon, valid deductions are truth-preserving). Algebraist 16:23, 21 February 2008 (UTC)[reply]

Indeed. Don't confuse reality with mathematics. And even in mathematics it is possible to have a valid proof that a statement is true and an equally valid proof that the same statement is false (simplest case is that you start with both P is true and P is false as axioms). Unfortunately, that result indicates that you are working in a system that is inconsistent, and so not very interesting or useful. Gandalf61 (talk) 16:28, 21 February 2008 (UTC)[reply]

A number of logicians, such as Graham Priest, argue that inconsistent systems can be both interesting and useful. See paraconsistent logic for example. -- Dominus (talk) 00:29, 22 February 2008 (UTC)[reply]

Two possibly relevant Einstein quotes [9]:

One reason why mathematics enjoys special esteem, above all other sciences, is that its laws are absolutely certain and indisputable, while those of other sciences are to some extent debatable and in constant danger of being overthrown by newly discovered facts.
As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.

--mglg_(talk) 17:10, 21 February 2008 (UTC)[reply]

Two blog entries that might be relevant are [10] and [11]. – b_jonas 17:46, 21 February 2008 (UTC)[reply]

I think it's possible for a mathematical theory to prove something exists, but for it to be provably impossible to find an example (using a different mathematical theory) without either being inconsistent. For example, you might be able to prove that there is an even number above two that isn't the sum of two primes using one theory, and prove that you could never find an example by using a hypercomputer and the original theory with allowance of infinite steps (in other words, your example, but with an infinite number of variations and an infinitely fast computer). — Daniel 01:38, 22 February 2008 (UTC)[reply]

Indeed. For example, ZFC proves the existence of a well-ordering of the real numbers R, but (assuming ZFC is consistent) there is no formula that provably well-orders R. Algebraist 21:00, 22 February 2008 (UTC)[reply]

There can't be any rules necessary for basic mathematical reasoning that don't work... i.e. disproven by counterexample. Because then they simply wouldn't be rules! Interestingly enough I've been reading about aspects of mathematics which can be logically seen to be true but not provable with the commonly accepted "axioms" or rules of mathematics (Roger Penrose - The Emperor's New Mind, Oxford University Press). See Gödel's incompleteness theorems! Psymun (talk) 22:56, 22 February 2008 (UTC)[reply]

Vandermonde determinant and polynomials

Hi, im having trouble doing this problem from a book on galois theory. Its from the start of the book (Galois theory - Ian Stewart) so wont have much to do with real galois theory, mostly just algebra realated to polynomials. The problem is:

Two polynomials $f,g$ over $\mathbb {C}$ define the same function if and only if they have the same coefficients.

If $a_{1},a_{2}\cdots ,a_{n}$ are distinct complex numbers, then the Vandermonde determinant is defined as $D=\left|{\begin{array}{rrrrr}1&1&1&\cdots &1\\a_{1}&a_{2}&a_{2}&\cdots &a_{n}\\a_{1}^{2}&a_{3}^{2}&a_{3}^{2}&\cdots &a_{n}^{2}\\\vdots &\vdots &\vdots &\ddots \\a_{1}^{n-1}&a_{2}^{n-1}&a_{3}^{n-1}&\cdots &a_{n}^{n-1}\end{array}}\right|$ Then there is a hint which doesnt really help me:

Consider the $a_{j}$ as independent determinants over $\mathbb {C}$ . Then D is a polynomial in the $a_{j}$ , of total degree $0+1+2+\cdots +(n-1)={\frac {1}{2}}n(n-1)$ . Moreover D vanishes whenever $a_{j}=a_{k}$ for some $k\neq j$ as it has 2 identical rows. Therefore D is divisible by $a_{j}-a_{k}$ hence it is divisible by $\prod _{j<k}(a_{j}-a_{k})$ now compare degrees.

Thats as far as i got really, any help would be great, thanks. —Preceding unsigned comment added by 137.205.93.126 (talk) 15:50, 21 February 2008 (UTC)[reply]

Huh? I'm confused. The statement 'Two polynomials

f,g

define the same function if and only if they have the same coefficients (up to multiplication by a constant)' is simply false (multiplication by a constant changes the function, different polynomials can define the same function over finite fields), and I can't see any obvious true statement it could be a misprint of. Moreover it seems to have little to do with the Vandermonde matrix. The one standard problem about the Vandermonde determinant is to show that it is non-zero if the a_i are distinct, and that seems to be what the hint is leading you towards proving. Algebraist 16:18, 21 February 2008 (UTC)[reply]

Sorry i didn't explain it well. I seemed to have added the up to multiplication bit for no reason as now i think about it it is wrong, ive taken it out of the question and rewote it exactly as it appears in the book. In the book it gives a different proof using calculus, then goes on to say "For a purely algebraic proof see [this problem]". I had the following idea: consider

{\underline {x}}=(x^{0},x^{1},x^{2},\cdots ,x^{n-1})

, the equation

V{\underline {x}}={\underline {0}}

(where V is the vandermonde matrix) has a solution if and only if

det(v)=D=0

so showing the vandermonde determinant is non zero if the coefficients in it are distinct, that is the polynomials in the vector solution of

V{\underline {x}}={\underline {0}}

have different coefficients. Is this in any way correct? Also if you could give me a little help showing the Vandermonde determinant is 0, that would be helpful too. Sorry if my wording is off, im having alot of trouble understanding this. Thanks. — Preceding unsigned comment added by 137.205.93.126 (talk • contribs)

Unfortunately, I was taught Vandermonde determinants by someone that went far too fast, so I don't fully understand them. I have one general tip, though: A matrix has zero determinant if the columns (or, equally rows) aren't linearly independent - in particular, if one row/column is a scalar multiple of another (that's not necessary, but it is sufficient). --Tango (talk) 18:39, 21 February 2008 (UTC)[reply]

Looked in my copy of Stewart's Galois Theory (2nd edition). The only problem I can find that mentions the Vandermonde determinant is Exercise 18.4 - but that is in the penultimate chapter of the book, not near the beginning, and it isn't worded like your question. Is your question from a different edition ? Gandalf61 (talk) 20:17, 21 February 2008 (UTC)[reply]

Mine is third edition. it appears on page 28, exercise 2.5*. Nothing about it is mentioned in the preface to the third edition though. —Preceding unsigned comment added by 137.205.93.126 (talk) 22:29, 21 February 2008 (UTC)[reply]

Okay. I'm still not clear just what the problem is asking. Is it just asking you to show that

D=\pm \prod _{j<k}(a_{j}-a_{k})

? Gandalf61 (talk) 12:09, 22 February 2008 (UTC)[reply]

The start of the question wants you to show that D is nonzero. Then [somehow] it wants you to use that to prove that "Two polynomials

f,g

over

\mathbb {C}

define the same function if and only if they have the same coefficients."—Preceding unsigned comment added by 137.205.93.126 (talk) 22:29, 21 February 2008 (UTC)[reply]

Hmmm. The sentence "Two polynomials

f,g

over

\mathbb {C}

define the same function if and only if they have the same coefficients" seems to me to be a definition rather than the objective of the problem. And you can't show that D is unconditionally nonzero, because D is zero if the a_i are not distinct. If you are not clear about what the question is asking, then you start with a big handicap when trying to solve it. Gandalf61 (talk) 15:37, 22 February 2008 (UTC)[reply]

I don't think that's a definition. Conceivably it might be possible to have two different polynomials (i.e., polynomials with different coefficients) that happen to give the same value when evaluated at any

x\in \mathbb {C}

. Stating that two polynomials define the same function if and only if they have the same coefficients is equivalent to stating that this never happens, which would require proof. —Bkell (talk) 17:18, 22 February 2008 (UTC)[reply]

Well, if you have to prove this, you look at f-g, which is zero everywhere, and so must be the zero polynomial, so f and g must have identical coefficients. But this statement about identical polynomials has no obvious connection with the Vandermonde determinant (as Agebraist has already said above). I suspect it is part of the previous problem that has been copied in error. Stewart often gives exercises that are lists of statements which the student has to mark as "true" or "false"; this statement feels like part of one of these "true or false" exercises. Of course, if someone has the 3rd edition of Galois Theory to hand, they could check this, and maybe also clarify the whole problem and resolve the original posters confusion. Gandalf61 (talk) 17:51, 22 February 2008 (UTC)[reply]

The statement is clearly equivalent to the statement "if a polynomial is zero everywhere, it is the zero polynomial", the intention may be to prove that (it's not always true, it's a property special to polynomials over fields of characteristic 0, so it's far from a trivial statement - it shouldn't be all that hard to prove, however). I do seem to remember a connection between the Vandemonde determinant and checking if functions are equal, but as I said before, those lectures didn't make a lot of sense. --Tango (talk) 19:50, 22 February 2008 (UTC)[reply]

The characteristic is irrelevant, actually. The usual proof is that a polynomial of degree n has at most n roots; this works over any infinite integral domain. Algebraist 20:10, 22 February 2008 (UTC)[reply]

Over a finite field, the number of solutions can be less than n simply because there are less than n elements in the field. Consider

x^{2}+x\in {\mathbb {F}}_{2}[x]

. --Tango (talk) 20:58, 22 February 2008 (UTC)[reply]

But we are not working in a finite field. We are told that f and g define the same function over C. Therefore f-g is zero everywhere in C. Therefore f-g is a polynomial with an infinite number of roots. Therefore by FTA f-g is the zero polynomial. Maybe the point of introducing the Vandermonde determinant is to prove this without using FTA. Gandalf61 (talk) 22:50, 22 February 2008 (UTC)[reply]

I now have the book in my hand. The problem is to show that the Vandermonde determinant is nonzero if the a_i are distinct complex numbers. The hint is as given by the OP. The next question is to use the Vandermonde determinant to show that if a polynomial f(t) over C vanishes at all points of C then it is the zero polynomial. The hint 'Substitute t = 1, 2, 3 ... and solve the remaining system of linear equations of the coefficients' is given. Algebraist 20:18, 22 February 2008 (UTC)[reply]

(Outdent) Gandalf61: You don't need the FToA to prove that a degree n polynomial has at most n roots. You can do that (over any integral domain) with the factor theorem and induction. The FT tells us that (counting multiplicity) the poly has exactly n roots. Algebraist 14:41, 23 February 2008 (UTC)[reply]

Fastest algorithm to find very smooth numbers close to some number

What is the fastest algorithm that could find smooth integers very near some specified integer. In other words which algorithm has the lowest ration of computation/(smoothness of a found number * smoothness of distance of that smooth number from a specified integer). 128.227.1.24 (talk) 20:30, 21 February 2008 (UTC)Mathguy[reply]

How would you compare the following two approximations of 123456789:

2·3⁹·5⁵ = 123018750
2⁷·3⁹·7² = 123451776

Which is better, and why? Do these two approximations have smoothnesses of, respectively, 5 and 7? If not, how do you measure the smoothness of a number. And what do you mean by "smoothness of distance"? —Preceding unsigned comment added by Lambiam (talk • contribs) 22:52, 22 February 2008 (UTC)[reply]

I would prefer the first one because its distance to N has a smaller largest factor. I was looking for a better/faster way to find Integers such that (Number-Integer)*(Integer) are smooth. The fastest method I tried so far is sieving. Dynamic programming subset sum solution is much slower than sieving. —Preceding unsigned comment added by 24.250.129.216 (talk) 00:33, 23 February 2008 (UTC)[reply]

How smooth is "smooth"? What's the highest prime factor you'll allow? --Tango (talk) 00:37, 23 February 2008 (UTC)[reply]

February 22

Transfinite number

What's the name of a transfinite number that can have any finite number subtracted from or added to it, with the result being different than the original number? For example, ω wouldn't be it because there's no ω-1. ℵ₀ wouldn't be it because ℵ₀+1=ℵ₀. — Daniel 02:08, 22 February 2008 (UTC)[reply]

The term "transfinite number" refers specifically to cardinals and ordinals, neither of which have that property. You may be thinking of the nonstandard integers. Black Carrot (talk) 02:21, 22 February 2008 (UTC)[reply]

Indeed, cardinals and ordinals are extensions of the natural numbers, not the integers, so subtraction isn't particularly meaningful (except for undoing addition). --Tango (talk) 11:49, 22 February 2008 (UTC)[reply]

Any nonstandard model of arithmetic has numbers with this property. The OP might also be interested in the surreal numbers. Algebraist 12:41, 22 February 2008 (UTC)[reply]

A non-diagram solution?

Is there a way to solve the following question without drawing a diagram, ie using combinatorials, permutations, etc. only? "How many ways are there to seat two students in a row of four desks so that there's at least one empty desk between them?" Imagine Reason (talk) 20:02, 22 February 2008 (UTC)[reply]

Well, I just solved it without drawing a diagram. There are two ways with exactly one space between them (the other empty desk can be on the left or the right) plus one way with exactly two. Algebraist 20:08, 22 February 2008 (UTC)[reply]

I drew a diagram in my head. Does that count? (I got either three or six, depending on whether you count switching the students.) Maybe a bigger example would help. Let's say we have 10 people and 100 desks. How many ways are there to arrange the people in the desks so that any two people have at least one empty desk between them? This is indeed a standard combinatorics problem, and can be solved without drawing diagrams. It does require imagining the situation in your head, but no more. Black Carrot (talk) 20:23, 22 February 2008 (UTC)[reply]

Let's say we're only worried about which desks have a person in them and which don't. There are 10 people, so there must be at least 9 desks empty - one between each pair of people. In particular, each person but the one on the right end must have an empty desk to their right. So, the question is really, how do we arrange 10 people (each attached at their right hip to an empty desk) in 91 desks? Black Carrot (talk) 20:40, 22 February 2008 (UTC)[reply]

Obviously order matters in this case, so the answer to the original question is six. I'm not sure it's a simple combinatorial question, however, because the one person who doesn't have an empty desk attached to the hip cannot be placed directly to the left of another student, but I don't see a simple way of preventing that from happening. Imagine Reason (talk) 01:55, 23 February 2008 (UTC)[reply]

All have a hip attachment, without exception. Add an extra empty desk to compensate, so we have for example 10 students with attachments + 91 loose desks = 10 people + 101 desks total. Since the rightmost desk is always empty, there is a bijection with the lawful arrangements in the original 100-desk formulation. --Lambiam 23:57, 23 February 2008 (UTC)[reply]

You can decide which seats will contain students first (for example, by ignoring order), and then afterward decide which order to assign the students to the chosen seats. —Bkell (talk) 06:27, 23 February 2008 (UTC)[reply]

Yes, but there's no way of doing that without drawing diagrams in your head, right? Imagine Reason (talk) 18:17, 23 February 2008 (UTC)[reply]

Depends on how you think, I guess. What's wrong with diagrams? The diagram is a way to help you translate the problem into mathematics which you can then solve. The fact that you used a diagram to get there doesn't make the maths any less mathematical. --Tango (talk) 21:54, 23 February 2008 (UTC)[reply]

Changing coins

How many ways can you change a five cent coin (nickel) into one cent coins and/or two cent coins?.

The answer is of course 3 ways. (1+1+1+1+1,1+1+1+2,1+2+2)

But I cannot figure out how many ways there are of changing $100 dollar note (10000 cents) into one cent coins and/or two cent coins. Exhaustive counting is way too exhaustive for me. The only thing I could think of is writing a computer program but I dont think that is how mathematicians in the 19th century would solve this problem. 122.107.226.136 (talk) 22:43, 22 February 2008 (UTC)[reply]

Hint: Once you choose how many 2-cent coins to use, the number of 1-cent coins is determined. -- Meni Rosenfeld (talk) 22:53, 22 February 2008 (UTC)[reply]

I figure out how to do that. If you have n of n+1 cents (

n\in

even numbers), then you have

{\frac {n}{2}}+1

ways to do it. In the question, you have 10000 cents, so you have 5001 ways. Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 02:49, 23 February 2008 (UTC)[reply]

Thanks. Now how many ways are there to change $100 dollar note (10000 cents) into one cent and/or two cent and/or five cent coins. This is even more difficult than just one cent and two cent coins. 122.107.226.136 (talk) 03:43, 23 February 2008 (UTC)[reply]

Hint: Once you choose how many 5-cent coins to use, the number of ways of changing the remainder into 1-cent and 2-cent coins is determined. Michael Slone (talk) 03:53, 23 February 2008 (UTC)[reply]

More ODE madness

Last question of my assignment and I can't do it at all!

The differential equation is ${\frac {\mathrm {d} y}{\mathrm {d} t}}=y(y-1)$ with initial condition $y(0)=2$ .

I've been taught solving by separating variables and also the integrating factor method, but I can't see how to apply either here! One attempt got me 1 = -2 which was most disconcerting! I'm really looking for hints just to get me started.

Thanks all Psymun (talk) 23:40, 22 February 2008 (UTC)[reply]

What goes wrong with separating the variables and integrating? At worst, it's integration by parts, and there is probably a better method I'm not seeing (in my defence, it is nearly midnight here). --Tango (talk) 23:54, 22 February 2008 (UTC)[reply]

It looks like partial fractions to me. --Tardis (talk) 23:57, 22 February 2008 (UTC)[reply]

Oh, yeah... Never did like partial fractions! --Tango (talk) 00:14, 23 February 2008 (UTC)[reply]

Incidentally, I just checked the method I used (not int. by parts, that was a terrible idea!) by using partial fractions, and it turns out I actually it right - I used a change of variables y->1/y->1-1/y. --Tango (talk) 00:17, 23 February 2008 (UTC)[reply]

Right! That's odd, I haven't been taught it, but it won't do any harm for me to look it up. Why did you change the varibles, Tango? Psymun (talk) 10:47, 23 February 2008 (UTC)[reply]

Changing variables is a standard method for integrating and is useful in a wide range of cases. It's not really the best way for this problem, partial fractions is better, but I use change of variables more often so that's what I'm more comfortable with. As for how I chose what to change them to - lucky (educated) guess! --Tango (talk) 14:04, 23 February 2008 (UTC)[reply]

February 23

Finding basic limits

How does one find this limit:

\lim _{x\to 9}{\frac {x-9}{{\sqrt {x}}-3}}

When I enter it into my graphing calculator, I can trace the rational curve and the answer should be somewhere between 5.8 and 6.1. But how do I find the exact value? Thanks. Acceptable (talk) 03:12, 23 February 2008 (UTC)[reply]

Hint:

(a-b)(a+b)=a^{2}-b^{2}

. PrimeHunter (talk) 03:19, 23 February 2008 (UTC)[reply]

\lim _{x\to 9}{\frac {x-9}{{\sqrt {x}}-3}}

$=\lim _{x\to 9}{\frac {({\sqrt {x}}-3)({\sqrt {x}}+3)}{{\sqrt {x}}-3}}$
$=\lim _{x\to 9}{\sqrt {x}}+3$
$={\sqrt {9}}+3$
$=6$
Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 15:15, 23 February 2008 (UTC)[reply]

Ftbhrygvn, your desire to help is appreciated, but we prefer not to give full solutions to what may be homework problems. -- Meni Rosenfeld (talk) 15:24, 23 February 2008 (UTC)[reply]

An alternative and lazier way to solve this problem is to apply l'Hôpital's rule. Differentiate the numerator and the denominator, so that you get

\lim _{x\to 9}2{\sqrt {x}}

, substitute x for 9 and you get 6. --Taraborn (talk) 18:17, 23 February 2008 (UTC)[reply]

question about functions

can we say that any known function,f(x)=y,satisfies the condition,limit[f(x)\x]→1,when x,either approches to infinity or to afixed value,x0?Husseinshimaljasimdini (talk) 11:44, 23 February 2008 (UTC)[reply]

I'm not sure I understand the question. Are you asking if there exists such a function? If so, then yes, f(x)=x has that property. If you're asking if all functions have that property, then no, take f(x)=0, for example. --Tango (talk) 13:58, 23 February 2008 (UTC)[reply]

alittle bit of help

i read here about that figure which has an infinite surface but afinite volume though.can i get some help to locate the article?Husseinshimaljasimdini (talk) 11:49, 23 February 2008 (UTC)[reply]

An example of such a surface is Gabriel's Horn. -- Xedi (talk) 13:50, 23 February 2008 (UTC)[reply]

A standard example of the lower dimensional case of a finite area and infinite perimeter is any landmass - assuming the coastline is effectively random at every scale, if you measure it with a smaller and smaller ruler you'll get a larger and larger answer (because you take into account smaller and smaller deviations from a straight line) and in the limit, the answer is infinite. --Tango (talk) 13:56, 23 February 2008 (UTC)[reply]

Help needed in understanding 1925 Biometrika paper

I'm reading the paper Tippet, LHC. On the Extreme Individuals and the Range of Samples Taken From a Normal Population. Biometrika vol 17, 364-387, 1925., temporarily available here.

The paper defines the function

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

,

where $\,\alpha _{p}$ initially was defined as $\int _{-\infty }^{x_{p}}\phi (x)dx$ , where $\int _{-\infty }^{+\infty }\phi (x)dx=1$ , i.e. it is a cumulative distribution function, and n is an integer greater than one. The function returns the expected value of the range of a sample of size n, taken from the probability distribution defined by $\,\phi (x)$ . If $\,\phi (x)=\Phi (x)$ is the standard normal distribution, w is known as the control chart constant d2 in the field of statistical process control.

I posted a related question recently about this function. I have found that the function

d_{2}(n)=\int _{-\infty }^{+\infty }(1-(1-\Phi (x))^{n}-\Phi (x)^{n})\,dx

is easily evaluated numerically, using a naive brute force approach such as the following C++ implementation:

double d2(int n)
{
    double x, dx;
    dx = 0.01;
    double sum = 0.0;
    for (x = -12; x <= 12; x = x + dx)
    {
        double alpha = cumulative_normal_distribution(x);
        sum = sum + (1 - pow(1-alpha,n) - pow(alpha,n))*dx;
    }
    return sum;
}

I know that there are far more efficient and elegant ways of evaluating integrals numerically. Nevertheless, this implementation works, and mirrors my simple-minded mental image of what is going on, so I'll stick with it for now. Tippett then defines as equation (10) the following:

\,\mu _{2}=\sigma ^{2}=2\int _{-\infty }^{+\infty }\int _{-\infty }^{x_{1}}(1-\alpha _{1}^{n}-(1-\alpha _{n})^{n}-(\alpha _{1}-\alpha _{n})^{n})dx_{1}dx_{n}-{\overline {w}}^{2}

and states (page 372) that

"The second moment, and hence the standard deviation of the distribution, was determined in several cases by equation (10). The work is very laborious, as it involves cubature, and even so, the result can only be given to a few figures. It is believed that the values given in Table IV are correct to the last figure."

The ~~return values~~ square root of $\,\mu _{2}$ should correspond to the control chart constant d3, and Table IV confirms that this is indeed the case. My problem is that I haven't been able to evaluate $\,\mu _{2}$ , and I suspect I've misunderstood something, possibly something pretty elementary about double integrals (I have no mathematical training beyond this, and it's a long time ago). Here's my implementation, again using a similar brute force approach. I've highlighted in red the part that I was most unsure of:

double d3(int n)
{ 
    double x_1, x_n, dx_1, dx_n;
    dx_1 = dx_n = 0.01;
    double sum = 0.0;
    for (x_n = -12; x_n <= 12; x_n = x_n + dx_n)
    {
        double alpha_n = cumulative_normal_distribution(x_n);
        for (x_1 = -12; x_1 <= x_n; x_1 = x_1 + dx_1)
        {
            double alpha_1 = cumulative_normal_distribution(x_1);
            sum = sum + (1 - pow(alpha_1,n) - pow(1 - alpha_n,n) - pow(alpha_1 - alpha_n, n))*dx_1*dx_n;
        }
    }
    double variance = 2*sum - pow(d2(n),2);
    return sqrt(variance);
}

The return values of this function are way off the correct values. My question is: can someone spot the misunderstanding(s) in my interpretation of the paper, the maths, and/or the error(s) in my implementation? Your help would be much appreciated. Thanks in advance, --NorwegianBlue^talk 16:13, 23 February 2008 (UTC)[reply]

Mathematics(Algebra)

Railways authorities increased the fare by 6.25%,hence the sale of tickets was decreased by 5%. Even then the income by selling tickets was increased by Rs 15000.Find the original income by selling of tickets?18:07, 23 February 2008 (UTC)117.195.16.73 (talk)

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misevaluation, but it is our policy here to not do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn how to solve such problems. Please attempt to solve the problem yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. Thank you. -- Meni Rosenfeld (talk) 18:22, 23 February 2008 (UTC)[reply]

Your first step is to formulate the problem in terms of equations. Give each unknown value a symbol (eg the original fare could be F_o and the new fare F_n, etc.), then rewrite all the information in the question as equations using the symbols. Once you've done that, it's just a matter of solving the equations. --Tango (talk) 19:16, 23 February 2008 (UTC)[reply]

February 24

derivatives and second derivatives of rotation functions

hi, how do i find the derivative and second derivative of a function that rotates counter clockwise an angle theta? do i just take the derivative and second derivative of (-sin+cos, cos+sin)?

also, how might one prove that a harmonic function composed with a function that preserves dot product is still harmonic? thanks —Preceding unsigned comment added by 199.74.71.147 (talk) 03:33, 24 February 2008 (UTC)[reply]

For the first question I assume you mean rotation around the origin in the Euclidean plane, using the mapping

(x,y)\mapsto (x\cos \theta -y\sin \theta ,x\sin \theta +y\cos \theta )\,.

To get the n-th derivative with respect to the rotation angle θ, just work out

(x,y)\mapsto {\frac {d^{n}}{{d\theta }^{n}}}(x\cos \theta -y\sin \theta ,x\sin \theta +y\cos \theta )\,,

where a pair is differentiated coordinate-wise:

{\frac {d}{d\theta }}(X(\theta ),Y(\theta ))=({\frac {d}{d\theta }}X(\theta ),{\frac {d}{d\theta }}Y(\theta ))\,.

Using matrix–vector notation, the mapping can be written as

{\begin{bmatrix}x\\y\end{bmatrix}}\mapsto M(\theta ){\begin{bmatrix}x\\y\end{bmatrix}}\,,

where the 2×2 rotation matrix M(θ) is given by:

M(\theta )={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\,\,\,\,\cos \theta \end{bmatrix}}\,.

Here you have to find ^d⁄_dθM(θ), which can be done entry-wise:

{\frac {d}{d\theta }}M(\theta )={\begin{bmatrix}{\tfrac {d}{d\theta }}\cos \theta &-{\tfrac {d}{d\theta }}\sin \theta \\{\tfrac {d}{d\theta }}\sin \theta &\,\,\,\,{\tfrac {d}{d\theta }}\cos \theta \end{bmatrix}}\,.

The second question is not clear. What does it mean for a function to preserve dot products? Normally you expect that to mean f(u·v) = f(u)·f(v), but the dot product turns two vectors into a scalar, so if u and v are vectors, on the left-hand side f is operating on a scalar and on the right-hand side on vectors. How can this be? Also, is it given in which order the harmonic function is composed with the dot-product preserving function? An example might help to clarify this. --Lambiam 06:26, 24 February 2008 (UTC)[reply]

Probably means orthgonal, u·v = f(u)·f(v). With some very minor condition (possibly none), an orthogonal function is an orthogonal matrix. Then it is just asking to show that the laplacian is invariant under orthogonal coordinate changes, which is definitely a very common exercise to give (so likely to be his question). JackSchmidt (talk) 06:44, 24 February 2008 (UTC)[reply]

that probably is my question, but i am unfamiliar with most of the terminology you used as my professor is rather scatterbrained in lecture. could you please give me some pointers on where to start if i were to prove this? thanks! —Preceding unsigned comment added by 199.74.71.147 (talk) 08:04, 24 February 2008 (UTC)[reply]

Basically you use chain rule. If g:R^2->R is harmonic and f:R^2->R^2 preserves dot products, then f is actually given by an orthogonal matrix (which looks almost exactly like a rotation matrix). Define h:R^2->R by h(x) = g(f(x)). Write out what that means fairly explicitly in terms of the matrix entries of f (just label them f11, f12, f21, and f22; don't use cosines in my humble opinion). Then use chain rule to take the first and second derivatives. The derivatives will involve dot products of rows of f. f is orthogonal, so the dot products will be 0 for the mixed derivatives of g, and 1 for the repeated derivatives of g. In other words, lap(h) evaluated at x is equal to lap(g) evaluated at f(x). JackSchmidt (talk) 08:17, 24 February 2008 (UTC)[reply]

Integral Powers of Trinomials

For expanding binomials, we have the binomial expansion which gives us all the info we need to know about expanding a binomial to (any) power. For example, if I want to find out what is the full expansion of $(x+y)^{25}$ , I can do so. My question is, is there a similar expansion formula for $(x+y+z)^{25}$ . Is there a shorthand way to write this expansion in terms of the coefficients of the expansion?A Real Kaiser (talk) 05:53, 24 February 2008 (UTC)[reply]

Memory of high school maths a bit hazy, but you can break

(x+y+z)^{25}

into

((x+y)+z)^{25}

, which gives

\sum _{k=0}^{n}{C_{k}^{n}}(x+y)^{n-k}z^{k}

, and then expand

(x+y)^{n-k}

binomially.

This gives something like

\sum _{k=0}^{n}\sum _{l=0}^{n-k}{C_{k}^{n}}{C_{l}^{n-k}}x^{l}y^{n-k-l}z^{k}

--PalaceGuard008 (Talk) 06:19, 24 February 2008 (UTC)[reply]

You are looking for the Multinomial theorem. You are basically counting how many ways to rearrange "mississippi", or other such combinatorial questions. JackSchmidt (talk) 06:26, 24 February 2008 (UTC)[reply]