Wikipedia:Reference desk/Mathematics

May 20

Implicit equations and distance

Given a curve (or surface) defined implicitly by f(x,y)=0 (or f(x,y,z)=0) and a point P(x_P,y_P) (or P(x_P,y_P,z_P) ), what does f(x_P,y_P) (or f(x_P,y_P,z_P) ) correspond to ?

If the curve is a line, the equation is ax+by+c=0 and ax_P+by_P + c corresponds to a certain distance from the point to the line (as ${\frac {|ax_{P}+by_{P}+c|}{\sqrt {a^{2}+b^{2}}}}$ is the (perpendicular) distance from the point to the plane).

What generalisation is possible, if any ? --Xedi 19:45, 20 May 2007 (UTC)

f needn't mean anything very much, in general. For example, take your line ax+by+c=0. This can be rewritten as e^(ax+by+c)-1=0, but now f(x,y)=e^(ax+by+c)-1 means nothing in particular. The most useful thing I can think of is that (as long as f is well-behaved), the gradient of f is a vector orthogonal to the surface f=0. For example, taking f(x,y,z)=x²+y²+z²-1, we get f=0 is the unit sphere, and $\nabla f$ is (2x,2y,2z), which for (x,y,z) on the sphere is a vector pointing directly outwards. Algebraist 20:33, 20 May 2007 (UTC)

Well, yes, one just has to study the properties of f to grasp what it describes. Indeed taking the gradient does help, I'm sure many other things are also possible (if possible getting one variable in function of another, etc...). But what I precisely wanted to know was the signification of the number f(x_P,y_P). Does your answer actually mean there isn't any ? I understand, as you said, f can be about anything and many functions will describe the same surface (f and e^f-1 and ln(f+1) (under certain assumptions for f)). Obviously, even in the worst cases, f(x_P,y_P) has to represent some sort of distance, even if not well defined, as if f(x_P,y_P) is relatively small, the point P must be relatively near the surface (of course, the nearness varies according to f, 2x+3y-4 will not give the same than 200x+300x-400 but the distance stays the same). Thanks --Xedi 20:56, 20 May 2007 (UTC)

Under certain assumptions, primarily $f(\mathbf {x} _{p})$ being small enough, ${\frac {|f(\mathbf {x} _{p})|}{\|\nabla f(\mathbf {x} _{p})\|}}$ is roughly equal to the distance of $\mathbf {x} _{p}$ from the hypersurface. I doubt much more than that can be said (of course, you could try find better approximations for the distance using higher order derivatives of f, but this quickly becomes unwieldy). -- Meni Rosenfeld (talk) 21:23, 20 May 2007 (UTC)

1) Yes, I think I understand, that's under the assumption that the second derivatives are 0 ? So what would be the approximations using higher derivatives ?

Thanks. --Xedi 21:45, 20 May 2007 (UTC)

In fact, it is not enough for $f(\mathbf {x} _{p})$ to be small. We need $\mathbf {x} _{p}$ to actually be close to the hypersurface, which is not implied by the former. For example, take $f(x,y)={\frac {x^{2}+y^{2}-1}{x^{4}+y^{4}+1}}$. This gives a circle, but there are points arbitrarily distant from the circle with arbitrarily small f value. -- Meni Rosenfeld (talk) 21:30, 20 May 2007 (UTC)

2) But that only results of the fact that it's an approximation that may go wrong as P gets further away from the surface. No ? --Xedi 21:45, 20 May 2007 (UTC)

You can draw graphs of the level sets of f(x,y)=0. For instance take the circle f(x,y)=x^2+y^2-1. The the set f(x,y)=0 is the circle you first though of, f(x,y)=1 will be a bigger circle and f(x,y)=-0.5 is a smaller circle f(x,y)=-1 a single point and f(x,y)=-2 the empty set. A more interesting example is f(x,y)=x^2-y^2, f(x,y)=0 gives two lines crossing f(x,y)=1 and f(x,y)=-1 give two non interseting lines. Lots of fun can be had by examining other functions and this sort of analysis is the basis of a lot of Singularity theory.

One way to get a feel for the functions is to treat it as a graph let z=f(x,y) and plot the points (x,y,z). The implicit curve will be the intersection of this surface with the x-y plane. Its worth doing in the 2D case which might help you understand whats happening in general. The 3D case might prove tricky. --Salix alba (talk) 21:33, 20 May 2007 (UTC)

3) Yes, I'll try that, definately. --Xedi 21:45, 20 May 2007 (UTC)

(after double ec)Meni Rosenfeld: I was going to point out I have a polynomial counterexampling your claim in 1D. Indeed, mine has f/f' tiny a long way from the hypersurface (=point), as well as f. Algebraist 21:38, 20 May 2007 (UTC)

To Algebraist: Sorry for the edit conflicts :) Anyway, note that I had initially required a small f. A polynomial would not be a counterexample, since even if it did have tiny ${\tfrac {f}{\nabla f}}$ at distant points (which I don't see how is possible), f would still be large, a case I didn't aspire to deal with.

To Xedi: I have taken the liberty to enumerate your new questions.

1) Yes, sort of, if the second derivatives are 0 everywhere (not just at $\mathbf {x} _{p}$) then this will be exact, as you have pointed out above. The way to use higher order derivatives is to construct a power series expansion for f around $\mathbf {x} _{p}$, treat it as if it is exactly f, and solve the equation you get by equating it to 0. There will be in general (infinitely) many solutions - the one with the smallest distance would be a decent approximation under the right assumptions.

2) Yes, the approximation gets worse as P gets further away from the hypersurface - I tried to emphasize that we need P to be close for the approximation to work, but we can't deduce that P is close from the mere fact that $f(P)$ is small.

-- Meni Rosenfeld (talk) 22:00, 20 May 2007 (UTC)

2) Oh yes, all right. Seems quite obvious, then, because we have to divide by something that isn't everywhere equal. So obviously some points where f(x,y) is arbitrarily small might not be near because at that point the denominator is very small too.

Could you just give an example for what it would be with second derivatives if it's not too complicated and long to do ?

Thanks again. --Xedi 22:11, 20 May 2007 (UTC)

Apologies for wasting the OP's time with this, but one can find a poly f with the following properties (say): the only zeros of f in [-10,10] are near 0, f(5) is very small and positive, and there are points near 5 where f is very small but f' is very large. Proof: take a suitable continuous function, apply the Weierstrass approximation theorem. Algebraist 22:18, 20 May 2007 (UTC)

You're not wasting any time here ! Anyway, yes well as long as there is a suitable analytic function then there's a polynomial anyways.

But all this just depends on the fact that the approximation using only the first derivative isn't suitable for all functions. I also thought Meni Rosenfeld's principles for saying the approximation works quite well were a bit arbitrary, but I'm not really able to give any good principles anyway. --Xedi 22:24, 20 May 2007 (UTC)

Algebraist: Yes, you can find a polynomial with small f at distant points, but (for a fixed polynomial) not arbitrarily small f or aribtrarily distant points. For f a polynomial, it is still true that if f(P) is small enough, the point is close to the surface.

Xedi: Sorry, working out anything with second derivatives would be too much for me to write here. You're welcome to try, but I do believe it would be a waste of your time (much work for no obvious benefit). Don't get it wrong - this also will require the point to be close, but will just give a sharper estimate if the point is indeed close.

I'm not sure what you mean by "as long as there is a suitable analytic function then there's a polynomial" (is there a nonconstant periodic polynomial I'm not aware of, or a nonzero one which converges to 0 at infinity?), or by your last comment. -- Meni Rosenfeld (talk) 23:27, 20 May 2007 (UTC)

Ok, I see, so it's not really possible.

About the polynomial, I just really meant to say that a function as Algebraist described could well be a polynomial without even having to resort to Weierstrass's approximation theorem. Then I wasn't really talking about behaviour at infinity. I wasn't really meant to be anything of importance, just like saying that it appears quite logical that such a thing is possible. Then yes, Algebraist and I weren't mentioning arbitrarily small values for a fixed polynomial. --Xedi 17:06, 21 May 2007 (UTC)

A suprising amount can be done with interval analysis, especially if you know not only f at a point but some or all of its derivatives. This information can give bounds on the range of values the function can take. Techiques like Lipschitz continuity can used to give bounds on the posible values within a certain distance.

If your working with polynomials you can convert these to Bernstein polynomials which have a nice convexity property: if the coefficients are all positive, then the function will be strictly positive in a given range. I've exploited these to a large extent in an program I wrote for polygonizing implicit surfaces.

These techniques are often better at negative questions, where does the function not have zeros, than saying when it does have zeros.

In 1D Sturm's theorem can be used to tell you how many roots (zeros) a function has. This has been exploited by ray-tracing algorithms. --Salix alba (talk) 23:31, 20 May 2007 (UTC)

Meni Rosenfeld: apologies, I had misunderstood the order of your quantifiers (this is what comes of not communicating in first order predicate calculus at all times). Algebraist 23:54, 20 May 2007 (UTC)

Shall we try a visual metaphor? The real-valued function f(x,y) has a natural interpretation as a height at a point of the xy plane. The curve it implicitly defines is the shoreline at sea level. Physical terrain is, we expect, continuous; mathematical terrain need not be. Unless we place special constraints on the shape of the land, the height at a given point tells us nothing; it certainly need not predict the distance to the shoreline, nor the direction. Even if we are lucky enough to be able to measure the slope of the land (the gradient of f), we may still have difficulty finding the shoreline. This is a very practical challenge for algorithms that are meant to find zeros of functions, whether arbitrary functions or simply polynomials. We have much the same difficulty if we wish to find a mountaintop rather than a shoreline (that is, an optimization problem). --KSmrq^T 05:43, 21 May 2007 (UTC)

Nice. This can be considered another way, if you are on a mountain top and know a limit on the maximum slope, you can say that the shoreline cannot be within a certain distance. This can be useful for algorithms as eliminating the set of possible values reduces the search space. --Salix alba (talk) 11:35, 21 May 2007 (UTC)

Yes, well it now appears obvious that the function can really be anything - even if we assume continuity it still stays quite unbearable. Obviously, yes, f(x,y) won't really tell us much from the distance to the nearest zero, at best an indication. Same goes for the derivatives.

So this also means there can't be any reasonably simple way to know how far the function is from a zero only knowing f(x,y) and some derivatives. Then, yes, the only way to go would be the other way - for example, getting a formula giving the distance to the curve (f(x,y)=0) depending on x and y (minimizing the distance between the point and the curve).

So just for this last thing, how would this optimization problem be solved (analytically), other than doing it point per point ? Are conditions useful for this to be much simpler (without alienating the problem too much), like continuity or derivability ?

Thanks so much for all this insight. --Xedi 17:06, 21 May 2007 (UTC)

Differentiability is a big help. Working with polynomials, that is an algebraic varity does make things very much simpler, as d^nf/dx^n will be constant when n=deg. In 1D if you know the first derivative is strictly positive, then a simple sign check on the end points can tell you whether a zero exists or not. Similarly if you know second derivative is strictly positive, then the first derivative can only have one zero, and the function can have at most two zeros. This leeds to something along the lines of Descartes' rule of signs.

In 2 or more dimensions you can play with critical sets and singular points, that is the curves where df/dx or df/dy=0 and points where both vanish. These divide up the domain into sections where things behave nicely.

Again if you have polynomials there is some hefty computer algebra you can rely on. I've seen a technique, which uses CA to factor polynomials, this of course cannot be done over the reals but you can use other rings where it is possible.

There is a big litrature on this problem, particle systems are a nice method. This basically means take a set of points in your domain, let them move along the gradient vectors, eventually they will either leave the domain or arrive on the surface. This is sort of a Newton's method in higher dimensions. --Salix alba (talk) 18:14, 21 May 2007 (UTC)

Sorry, but I don't see how this is really relevant here. I'm not actually looking for the zeroes but more for a certain distance to the curve. Effectively the methods you give are an aid to find the zeroes of the function (following the gradient, like, as you said, in Newton's method). Still, thanks for the explanation, it's definitely very important to be able to find out the zeroes of functions nowadays. Maybe I'm just too used to functions of the type y=f(x) or z=f(x,y).

What I was more curious about was what it is possible to do with f(x,y) (for example, finding distance to a zero). So, yes, I realize not much can be done, because the function can behave in too many different ways to be able to draw conclusions from the value of f(x,y). And now I'm just asking, as f(x,y) won't, in general, give any kind of distance, even though Meni Rosenfeld's approximation works well in many cases (and does give an exact solution when it's a plane), is it possible to assign a "distance" value to each point easily ? For example (extremely simple case), with xy=0, the distance would be max(x,y). Because all I'm able to do is to calculate for each point (one at a time) the distance by an optimization problem (finding the minimum length between the point and the curve f(x,y)=0) (and in simple cases too).

I suppose the question is, as f(x,y) can't really define distance from the curve, what does, if anything ? Thank you for your time. --Xedi 19:47, 21 May 2007 (UTC)

Given a curve defined implicitly by f(x,y) = 0, the distance of a point P = (x_P,y_P) to the curve is the infimum of ||(x,y) − (x_P,y_P)|| for (x,y) ranging over all points satisfying f(x,y) = 0, and similarly for surfaces. Determining this in general is just as hard as general optimization; for specific functions f you might be lucky and find a simple answer, but not in general.  --Lambiam^Talk 21:34, 21 May 2007 (UTC)

Yes, that's what I meant, there's no proper way of finding a general expression for the distance in most cases as we would have to find the minimum for an infinity of points. Thanks for the precise answer. --Xedi 21:51, 21 May 2007 (UTC)

For a point on the curve to be a minimum of distance to P it requires that the normal at that point is in the direction of P, you can produce a formula g(x,y) which will be zero when the gradient is in direction of P. It then becomes a problem of solving the pair of equations f(x,y)=0, g(x,y)=0, which may be possible to do algebraically. Think of a circle centered P which is tangent to the curve. --Salix alba (talk) 22:52, 21 May 2007 (UTC)

May 21

What do I call this maths thing?

What do I call this maths thing, and where do I find it, what does it mean, and how does one get to the answer? Also, if there are any articles relating to this thing, can you please provide them? $5+{\ 14 \over {\ 5+{\ 14 \over {5\ +{\ 14 \over {5\ +...}}}}}}$ ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 19:05, 21 May 2007 (UTC)

I believe Continued fraction will have all the answers. -- Meni Rosenfeld (talk) 19:38, 21 May 2007 (UTC)

Thanks alot! I guess I should sharpen up my English skills lol :P ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 19:45, 21 May 2007 (UTC)

Although this thing is in fact a generalized continued fraction. The statement "... are precisely ..." about periodic continued fractions in the article Continued fraction also holds for the generalized ones if you make a small modification: omit the word "irrational" from that statement. In the generalized case you still get solutions of quadratic equations, but they may be rational. The articles do not actually say how to obtain the exact answer. It is rather simple, though. Call the whole thing x, so

$x=5+{\ 14 \over {\ 5+{\ 14 \over {5\ +...}}}}\,.$

Note that the (infinite) expression for x reappears as the denominator of the fraction whose numerator equals 14, so we can replace that denominator by x without change of meaning. This means that whatever the value of x is (assuming it has a value; but all continued fractions do), x satisfies

$x=5+{\ 14 \over x}\,.$

If you simplify this, you end up with a quadratic equation. It has two roots, but only one can be the right answer. It is not hard to figure out which one.  --Lambiam^Talk 21:54, 21 May 2007 (UTC)

Sorry, another question

Is it mathematically possible to determine k, if (x^3 - 7x^2 + kx + 13) / (x - 3) gives a remainder of 4? Isn't the value of x needed to find k? I've checked Polynomial remainder theorem and Factor theorem but I can't seem to find anything there. ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 20:51, 21 May 2007 (UTC)

What do you get when you apply the polynomial remainder theorem to find the remainder of x³ − 7x² + kx + 13 divided by x − 3?  --Lambiam^Talk 21:23, 21 May 2007 (UTC)

This is polynomial long division, not regular division. — Daniel 21:26, 21 May 2007 (UTC)

What a fun question. Working with integer remainders, recall that removing all multiples of divisor d from n leaves a number less than d. Similarly, when working with polynomials the remainder must have degree less than that of the divisor. Here the divisor has degree one, so the remainder must be a constant, independent of x. Of course, in this case the "constant" will be a symbolic expression involving k. If the abstractions are too confusing, try assigning k numeric values from 1 to 5 and calculate the remainder each time; chances are you will quickly understand. --KSmrq^T 21:43, 21 May 2007 (UTC)

[Edit conflict] You may be confusing two different things: The remainder of division of the integers $x^{3}-7x^{2}+kx+13$ and $x-3$, for some unspecified but fixed integer value of x, as opposed to the remainder of division of the polynomials $x^{3}-7x^{2}+kx+13$ and $x-3$. The latter is discussed in Polynomial remainder theorem, and Lambiam hints at how to solve the problem. In case you mean the former, I suppose the problem is indeed unsolvable. -- Meni Rosenfeld (talk) 21:29, 21 May 2007 (UTC)

May 22

Can this be done without using a computer

A company opens a bank account with a fixed interest of 10%. At the start of each 
year the company deposits X dollars, X[0] = $10 . The bank gives a simple annual 
interest of 10% . Also the amount the company deposits at the start of each year 
grows by 10% as well.

Year   X      1stJan    31stDec
 0    $10      $10       $11
 1    $11      $22       $24.20
 2    $12.10   $36.30    $39.93
 
 How much money is in the bank account after 5000 years (aka 31stDec year 4999)?

This question drove me nuts because I not allow to use a computer program to generate the answer (considered as cheating). Is there a way to get the answer without using an electronic computer or programable calculator?.

202.168.50.40 00:45, 22 May 2007 (UTC)

This problem is indeed solvable without a computer. Let T(n) be the total amount the company has in its account at the end of year n. Then $T(n)=1.1(T(n-1)+X(n))$. You should be able to figure out what X(n) equals. Once you plug that in, you will need to find a closed-form expression for this recurrence relation. The geometric series article should be useful for that. nadav (talk) 01:56, 22 May 2007 (UTC)

A good way to find such a closed-form expression involves 4 steps:

1. Calculate T(n) for n = 1, 2, 3, ... (how many values should you calculate? enough for you to do the next step).
2. Try and find a pattern in the T(n)s - try rearranging them so they look similar.
3. Guess what the formula might be for a general n.
4. Prove your guess via mathematical induction (depending on circumstances, you may not have to do the full, formal proof, just enough to justify to yourself that it's correct).

Having done that, you will have your expression for T(n) and can just pop in the appropriate value of n. Confusing Manifestation 02:55, 22 May 2007 (UTC)

$T(n)=1.1(T(n-1)+X(n))$. You should be able to figure out what X(n) equals.

$T(n)=1.1(T(n-1)+10*1.1^{n})$ where T(-1)=0

T(0) = 1.1 (T(-1) + 10 * 1.1 ^ 0) = 1.1 (0 + 10) = 11

T(1) = 1.1 (T(0) + 10 * 1.1 ^ 1) = 1.1 (11 + 11) = 24.20

This is not helping me because I don't want to do this 5000 times. 202.168.50.40 02:58, 22 May 2007 (UTC)

Damn it, someone else in class claims to have the answer already. He says

f(n) = 11 * n * 1.1 ^ (n - 1)

f(1) = 11.00

f(2) = 24.20

f(3) = 39.93

f(5000) = 11 * 5000 * 1.1 ^ 4999

How the hell did he get the answer so quick?202.168.50.40 03:02, 22 May 2007 (UTC)

Damn it! That smartarse is going to the front of the class to explain how he got the answer. What is he saying? He use the dollar, euro and pound method.

At the end of 1st year, the amount is 11 dollars.

He converted it to euro using a fake conversion rate of 1 euro = 1.1 dollar. So 11 dollars = 10 euro

At the second year, starts with 10 euros and deposit 10 euros for the end of year amount of 1.1 * (10 euro + 10 euro) = 22 euros (or 24.20 dollars)

He converted it to british pound using a fake conversion rate of 1 pound = 1.1 euro. so 22 euros = 20 pounds.

At the third year, starts with 20 pounds and deposit 10 pounds for the end of year amount of 1.1 * (20 pounds + 10 pounds) = 33 pounds (or 36.30 euros or 39.93 dollars)

In sumary he says

1st year 11 dollars

2nd year 22 euros

3rd year 33 pounds

so f(n) = 11 * n * 1.1 ^ (n - 1)

202.168.50.40 03:12, 22 May 2007 (UTC)

Lol, I really hope he is joking. Further suggestion: When you follow Confusing's method above, try to use variables to replace some of the numbers. For example, replace 1.1 with r, and 10 with X(0), say. This should help elucidate the pattern. My guess is that you will be able to find the pattern without going beyond T(4). nadav (talk) 03:52, 22 May 2007 (UTC)

Yeah, I would point out that you don't want to put it into a number, because then you almost certainly won't see the pattern. Like Nadav suggests, replace a few of the numbers with variables, or at least write it out in as much detail as possible - I'll give you the first couple of steps to start you off, just this time :)

T(0) = 11

T(1) = 11 * 1.1 + 10 * 1.1 = (11 + 10) * 1.1

T(2) = ((11 + 10) * 1.1) * 1.1 + (10 * 1.1 * 1.1) = (11 + 10 + 10) * 1.1 * 1.1

T(3) = ?

Does that give you an idea of what the pattern is? Confusing Manifestation 04:02, 22 May 2007 (UTC)

Why does he have to be joking? let's say unit(n) = 1.1^n dollars, so unit(0) = dollar, unit(1) = euro, unit(2) = pound etc. This way, if you have how many of unit(x) in year(x), you can calculate dollars with the inverse, which is dollars=unit(n)*(1.1^n). Get it? By the way, it would be esier to use 1st year = 10 euros, 2nd year = 20 pounds ect.— Daniel 19:17, 22 May 2007 (UTC)

So you are saying that

f(n) = 10 * n * 1.1 ^ n

f(1) = 11.00

f(2) = 24.20

f(3) = 39.93

f(5000) = 10 * 5000 * 1.1 ^ 5000

202.168.50.40 21:53, 22 May 2007 (UTC)

That's about right, and since it fits your first calculations, it's quite definitely right (for December 31st). — Daniel 23:44, 22 May 2007 (UTC)

Rotating a sine/cosine graph

Hey guys, how would I rotate a sine or cosine graph by, say, 45 degrees? I'm trying to make it tangent to y=x+1 and y=x-1. I don't want to use rotation matrices. Thanks, Fbv65edel / ☑t / ☛c || 03:11, 22 May 2007 (UTC)

$f(x)=\sin {x}+x$

Does that work? --0rrAvenger 03:33, 22 May 2007 (UTC)

That is a shear rather than a rotation, but it may be suitable for the desired tangents. - Rainwarrior 03:41, 22 May 2007 (UTC)

Whether or not you want to use rotation matrices, you will still have to do all of the math involved in the rotation (perhaps you should look at the Rotation (mathematics) article, which might be easier to understand?). If you are trying to rotate: $y=\sin(x)$, the usual way to rotate (x,y) looks like:

$x'=x\cos \theta -y\sin \theta \,$

$y'=x\sin \theta +y\cos \theta \,$

So from here, substitute these into your equation:

$x\sin \theta +y\cos \theta \ =\sin(x\cos \theta -y\sin \theta \ )$

Now the trick is to pull y back out and put it on the left side by itself. I'll leave that for you. (Careful! Depending on the angle there may be multiple values for y.) - Rainwarrior 03:38, 22 May 2007 (UTC)

Well, it turns out a shear what I was looking for, but the rotation using sin and cos of theta is something for me to explore as well. Thanks a lot! --Fbv65edel / ☑t / ☛c || 03:47, 22 May 2007 (UTC)

Actually, a much easier suggestion than trying to find y in terms of x is to define two orthogonal vectors, such as (1,1) and (1,-1), and then map the original function's x and y onto these, so: y = sin(x) gives you x and y, and then to get your coordinate, take: y*(1,1) + x*(1,-1). (Which is: y' = y-x, and x' = y+x.) - Rainwarrior 03:57, 22 May 2007 (UTC)

And if you use generic orthonormal vectors (cos θ, −sin θ) for x and (sin θ, cos θ) for y, you actually get rotation over an arbitrary angle – which you'd need if you want the rotated curve to be tangent to a given line.  --Lambiam^Talk 08:44, 22 May 2007 (UTC)

xkcd

This is based on [1]. (Hover your mouse above the comic for a moment to see it.) Is there such a thing as a "tightly closed nonorientable space," and if so, what would it be like? Would "nonorientable" in such a context include something like the Klein bottle? Black Carrot 06:52, 22 May 2007 (UTC)

See Projective space.  --Lambiam^Talk 08:38, 22 May 2007 (UTC)

(Note visibility of the popup comment is browser-dependent.) Since "tightly closed" suggests a metric, perhaps it is meant as a colloquial substitute for "compact". A Klein bottle is a compact nonorientable surface, but his suggestion would seem to require a three-dimensional space. Euclidean 3-space is not compact, and a 3-ball is orientable, so Lambiam has suggested the most obvious simple example. One way to model the projective space RP³ is to begin with a 4-sphere, such as {(w,x,y,z):w²+x²+y²+z²=R}, for some fixed radius R. (Make it a "tight" sphere, if desired.) Now identify (w,x,y,z) with (−w,−x,−y,−z). An equivalent, slightly more complicated approach, begins with a 3-ball. As for the rest of his description, I think it refers to a homomorphic operator on curvilinear forms, but this goes beyond my expertise. --KSmrq^T 09:54, 22 May 2007 (UTC)

You could be right, he might mean compactness. He's usually pretty careful about that kind of thing, though. What would tight closure do to it? If I'm understanding the beginning of the tight closure article correctly, the space would have to be a specialized type of ring. Is it possible to define addition and multiplication such that a real projective plane or 3-space is tightly closed? Since it seems like that would effect measure and movement, would that change what it was like to live inside it? Black Carrot 00:00, 23 May 2007 (UTC)

Problem on Pearson Correlation

I encountered a problem on correlation that is rather strange to me. I have got two sets of time-series data (N = 115) which have correlation coefficient of -0.0636 (Significance 0.01 - two tailed). However if I divide the data into two halves (N1 = 58 and N2 = 57) the correlation coefficient of first half becomes +0.6134 (Significance 0.01 - two tailed)and the second half +0.2317 (Significance 0.08 - two tailed). Previous studies say that the correlation coefficient between the data should be positive and significant. I am not a very statistics person. Can anyone help me what might be the reason for these results, in non-technical terms? How to interpret such results and relate it to the previous studies?

The data sets are stock market trading volume and volatility, if it helps you to answer more effectively.

Thanks in advance.--202.52.234.141 08:45, 22 May 2007 (UTC)

Plotting the data with one set on the x-axis and the other on the y-axis is always a good first step. You may find that you have data which looks like

                  x
                 x                     x
                x                     x
               x                     x
                                    x

Taking as a whole you could fit a slightly downward pointing line. But each group individually would fit an upwards pointing lines. Is there a reason for the two halves, say if the data comes from two seperate days there might be an explination. --Salix alba (talk) 09:05, 22 May 2007 (UTC)

Square root of pi

What is the square root of pi? —The preceding unsigned comment was added by Amleth (talk • contribs).

Approximately 1.77245385? Something tells me you want something a little bit more than just a number. x42bn6 Talk Mess 20:18, 22 May 2007 (UTC)

(after edit conflict) It is the positive number whose square equals the number π. Just like π, it is a transcendental number, so any numeric values we give are only approximations. Then

${\begin{array}{rcl}\pi &=&3.14159265358979323846264338327950288419716939937510...\\{\sqrt {\pi }}&=&1.77245385090551602729816748334114518279754945612238...\end{array}}$

Does that answer your question?  --Lambiam^Talk 20:28, 22 May 2007 (UTC)

positive *real* number ^^ —The preceding unsigned comment was added by 84.187.55.83 (talk • contribs) 20:41, May 22, 2007 (UTC) - Please sign your posts!

? Is there a positive *not-real* number whose square equals π? If not, what is the point of this addition?  --Lambiam^Talk 21:54, 22 May 2007 (UTC)

If I'm remembering my complex algebra correctly, there are only two numbers that, when squared, equal π, and neither one has an imaginary component. Thus, the qualifier "real" is unneccessary. --Carnildo 21:58, 22 May 2007 (UTC)

Yes, it was a fun addition, but you should at least think a moment about the possibility of additional complex roots. You can prove that there is no additional root by considering the polar form. Sqaring is doubling the angle and it becomes obvious, that there can be no additional square root. There are however additional fourth roots.

I just want to point out that there are two real numbers whose square is pi. I'm quite positive about one of them but I am negative about the other one. 202.168.50.40 22:02, 22 May 2007 (UTC)

Yes, if you are talking about solutions to the equation x² = π. However, that was not what was asked, and it is convention to consider sqrt, ie., the square root, as positive and single-valued.

Can you clarify this for me? My memory (such as it is) tells me that the words "the square root of <whatever>" means both positive and negative roots, and in an exam you'd be wrong to provide only the positive one. But if the question is "What is ${\begin{array}{rcl}{\sqrt {\pi }}\end{array}}$", then the use of the symbol means that all you're being asked about is the positive root, and in an exam you'd be wrong to provide the negative one as well. I know this is not an exam, but isn't this the technical difference between the words "square root" and the square root symbol? JackofOz 03:14, 23 May 2007 (UTC)

I use "a square root" if I mean any number whose square is equal to the argument, and "the square root" for the principal square root. I don't recollect ever seeing or hearing a mathematician use the locution "the square root of" in another meaning.  --Lambiam^Talk 10:13, 23 May 2007 (UTC)

If that's true, what use is letting $x={\sqrt {2}}$ when, using the definition you gave, leads to ambiguity? This notation almost always implies the positive root for people who are familiar with the notation, and reads as "let x equal the square root of 2". There is a reason we have "$\pm$". However, $x=\pm {\sqrt {2}}$ can be rather confusing since x actually stands for two values. I suspect this is where most peoples confusion lies. However, I cannot say that in some context (possibly where people are ignorant of convention or are abusing notation) that ${\sqrt {x}}$ may mean something else other than the principle square root of x. Root⁴(one) 03:46, 23 May 2007 (UTC)

(edit conflict) Nod, the fourth root of one (yes, I'm spamming myself!) is generally only considered to be one, although there are 4 roots to $x^{4}=1$. To generalize the convention to the complex plane, if a number x be expressed as $(r,\theta )$, you calculate the nth root of x to be $({\sqrt[{n}]{r}},\theta /n)$. I believe this concept is called the "principle root". One reason for this name may be that this is the only root from which you can calculate the rest of the n-1 roots by incrementally adding $2\pi /n$ to the angle and not have to calculate the angle additions modulo $2\pi$. That is, you don't have to subtract $2\pi$ from any resulting angle so that $0\leq \theta <2\pi$. Root⁴(one) 03:30, 23 May 2007 (UTC)

To JackofOz: I think using what's likely to be marked correct on an exam is a silly metric to use for almost anything. On an exam, you should put down whatever the convention established in your class by your professor, whether right or wrong -- that doesn't make it mathematically correct. To Root4: I don't know any context in which the fourth root of 1 would be only considered to be 1 (of course there is no argument that 1 is a fourth root of 1). In fact, the only time I can think of for preferring one nth root of unity over another is when you want to focus on primitive roots, and of course 1 is never an primitive nth root of unity. Tesseran 04:55, 23 May 2007 (UTC)

I realise that words and symbols are mathematically "correct" only to the extent that there is agreement as to what they mean in mathematical contexts. I was taught the above difference by a number of teachers, both in school and university, as an established convention that I understood was universally agreed on. But maybe this isn't the case. I don't understand what you mean by "a silly metric". Surely professsors don't just make up their own meanings for mathematical symbols and expect their students to comply. Do they? JackofOz 05:10, 23 May 2007 (UTC)

This is not a very deep topic. In formulas, there is never any confusion since the symbol always denotes the positive root as you said. When words are used, then it depends on the context. nadav (talk) 05:32, 23 May 2007 (UTC)

Thank you, that's all I wanted to know. (Sorry for dragging you into shallow waters, but they tend to be warmer and permit more illumination). JackofOz 05:39, 23 May 2007 (UTC)

It should be emphasized that the word "the" in mathematics can only be used when the object described is unique. For example, the expression "the real number whose square is 4" is meaningless, as there are two such numbers. Similarly, using "the square root of 4" to describe either the positive or the negative root is incorrect. -2 is a square root of 4, not the square root of 4. By convention, the phrase "the square root of x" is shorthand for "the principal square root of x", which is the nonnegative one, and denoted ${\sqrt {x}}$. I can understand high school teachers being confused about this point, but not university teachers, so you may want to discuss the matter with them (making sure you understood their convention correctly). -- Meni Rosenfeld (talk) 11:53, 23 May 2007 (UTC)

Excellent explanation, Meni. I parted company with my university mathematics teachers over 25 years ago, so I'm not inclined to hunt them down at this remove. Obviously I had the wrong impression. (Quite strange - this is only the second time I've ever been wrong. The first time was many years ago, when I thought I'd given a wrong answer to a question. As it turned out, I was right and my examiner was wrong; but I was wrong to have thought I was wrong.)  :) JackofOz 00:17, 24 May 2007 (UTC)

Can anyone think of any interesting properties of the square root of π? All I can think of is that it's the area under the Gaussian function. nadav (talk) 02:50, 23 May 2007 (UTC)

Silly me, I forgot about the gamma function. nadav (talk) 05:24, 23 May 2007 (UTC)

This has already been asked. — Daniel 18:43, 23 May 2007 (UTC)

Thanks. You have a good memory. But I'm somewhat disappointed because I expected a lot of interesting-looking Ramanujan type identities besides the two properties above and Stirling's approximation, nadav (talk) 19:13, 23 May 2007 (UTC)

${\sqrt {\pi }}=\pm {\sqrt {\pi }}$

but...

$\pi ^{\frac {1}{2}}=+{\sqrt {\pi }}$MHDIV ɪŋglɪʃnɜː(r)d(Suggestion?|wanna chat?) 22:25, 23 May 2007 (UTC)

Um, no, because if the square root sign implied both roots, then we wouldn't have to write the plus-or-minus sign. ${\sqrt {}}$, as a function over the positive reals, is defined as the positive square root. Even over the complex numbers, it's generally taken to be the principal root. Confusing Manifestation 22:50, 23 May 2007 (UTC)

OK, folks, drop the stick and slowly back away from the horse carcass.  --Lambiam^Talk 23:05, 23 May 2007 (UTC)

LOL! :-D --KSmrq^T 15:23, 24 May 2007 (UTC)

Me too. —Bromskloss 13:54, 25 May 2007 (UTC)

May 23

quadratic function

Please Help!! How do I write a bio-poem for the quadratic function? Thanks! arn4utoo

(in case anyone was wondering what a bio-poem is: [2], [3]). We're here to help with math, not to write poetry for you. All I can suggest is to look at Quadratic function and use the special properties described in the article in a way that's similar to the usual bio-poem format. nadav (talk) 03:45, 23 May 2007 (UTC)

what is criteria for mathmatical oprator

Respected sir, i am confused about the use of mathematical operators in certain formula, i want to know about criteria for a operator i.e. how do we choose one specific operator.

For example in given formula

W=m.g

here weight is function of mass, we can write this as

W=f(m)

where f is function of W.
Then my question is how do we choose only multiplication as a operator why not addition, subtraction, division.
Is there any specific criteria? Please, explain.

We'll need more information to understand your question. In the formula you gave, multiplication is used for physical—not mathematical—reasons (see Newton's second law). There is no one general reason for the choice of operations used in equations. nadav (talk) 07:35, 23 May 2007 (UTC)

W=f(m). Don't you mean f is a function of mass?Zain Ebrahim 08:51, 23 May 2007 (UTC)

A very strange question... Maybe you'll benefit from reading multiplication, division, addition and subtraction in order to understand how they differ and in which situations they are used. 213.48.15.234 09:55, 23 May 2007 (UTC)

I'm not sure if this is going to help but the operators are used based on the definition of the thing you're trying to quantify. So if you look at Newton's second law or Weight then you'd understand why we say W=mg and not (for example) W=m/g. Zain Ebrahim 10:58, 23 May 2007 (UTC)

(after edit conflict) I think the first response by Nadav1 is more to the point than the next two. The above equation, giving weight as a function of mass and gravitational acceleration, can be found in our article Weight (in the section "Conversion between weight (force) and mass"), and in Force (twice even; once in the section "Examples", and once on the section "Quantitative definition"). As the latter article duly points out, this is true by definition: weight is by definition the magnitude of the force of an object acted upon by gravity, so, using the definition of force F = ma from Newton's second law, replacing F by W and setting a = g, we have W = mg. (For nitpickers: I know this is not the "official" definition of force, but a simplified one for constant mass; for the sake of simplicity I've also omitted the vector aspect). Now this is not really a satisfactory answer to the rationale.

For one thing, mass and acceleration have different dimensions. Adding mass and acceleration is meaningless. Ask yourself: what would it mean to add 10.7 liters to 1.6 kilometers? OK, but why is the law not F = ma², for example? The answer is not mathematical. This is a law of physics, and the laws of physics are those formulas and equations that have proven to give a good quantitative description of the observed behaviour of things in reality. If you want to send a rocket to the Moon and you use Newton's laws to calculate its trajectory, you have a chance it may arrive. If you replace these laws by a randomly chosen other set of laws, no way it will get there.  --Lambiam^Talk 11:01, 23 May 2007 (UTC)

Actually, it can't be F = ma², because again you have a dimensional problem - this time, you're trying to say that a force (measured in Newtons, or kilograms metres per second per second) is equal to a mass times an acceleration squared, in units of kilograms metres metres per second per second per second per second (of course you'd actually say something like "kilograms metres squared per second to the fourth power" but this shows the difference more clearly). Then again, there could always be some kind of fundamental constant with appropriate dimensions to make it work out. See dimensional analysis to see how you can go from knowing which variables may influence each other to a valid (though not necessarily correct) formula relating them. Confusing Manifestation 22:45, 23 May 2007 (UTC)

In some other universe, with different laws of nature, the dimension of force could be ML²T⁻⁴. Before Newton formulated his law, he did not know force was supposed to have the same dimension as the newton. (The last sentence is also true in the alternative universe, in which 1 newton is 1 kg m² s^-4.)  --Lambiam^Talk 22:59, 23 May 2007 (UTC)

Huh ? Since force is defined as rate of change of momentum and momentum is defined as the product of mass and velocity, how can force have any other dimension than MLT^-2 - unless we are changing definitions, in which case everything is wibble. Gandalf61 21:48, 24 May 2007 (UTC)

Yeah, from what I hear "force" is an arbitrary concept that is not really used in more advanced physics (e.g. Lagrangian mechanics), which focuses more on energy and potential. But this question is better served by the science ref desk. nadav (talk) 06:52, 25 May 2007 (UTC)

A bit confused...

How can I find the height of a pyramid if I know the base area and volume?

This is what I have: "I would subtract the area of the base from the area of the triangular sides. Next I would... Here I stop."

This is what someone mentioned to me: "It includes division and multiplication by 3."

Colud someone tell me where all this ties in?? Thanks, IP 20:53, 23 May 2007 (UTC)

There's a formula relating the height, area of the base, and volume of a pyramid. It made quite an impression on me in Grade 8 that the volume of a flat-bottomed object that comes to a point on top is equal to a third of the area of the base multiplied by the height. iames 21:13, 23 May 2007 (UTC)

Let us call the volume $V$, the base area $A$ and the height $h$. The relation between these quantities is

$V={\frac {1}{3}}Ah$.

You can find the height from that equation. Do you know how to do that? —Bromskloss 21:16, 23 May 2007 (UTC)

This formula can be found in the article Pyramid (geometry).  --Lambiam^Talk 21:53, 23 May 2007 (UTC)

H = V · 3 ÷ B? I believe that is it. I ran it through many tests and it came out right. IP 16:47, 24 May 2007 (UTC)

Simple algebraic manipulation. $V={\frac {1}{3}}Ah\to 3V=Ah\to Ah=3V\to h=3{\frac {V}{A}}$. Yes you are correct. Obscurans 03:19, 25 May 2007 (UTC)

Square Roots mod n

How can I find all the square roots of a number x mod n. Is it easier if x=a² and n=ab for two numbers a and b? Thanks *Max* 23:25, 23 May 2007 (UTC)

If you can read PostScript files, you'll find an algorithm at http://zoo.cs.yale.edu/classes/cs460/Spring98/quadr.ps. (I haven't checked it, but this should be a reliable source.)  --Lambiam^Talk 23:37, 23 May 2007 (UTC)  On further inspection, this seems only to cover the case that n is prime. 23:41, 23 May 2007 (UTC)

IIRC, you want to look at Euler's criterion and Legendre symbol.

(e/c) For prime moduli, the problem is relatively easy and there a number of algorithms on offer. When n is composite, then it is equivalent to the integer factorization problem since we need the factorization of n. A friendly introduction to anything you ever wanted to know about square roots, including modular square roots, is Ezra Brown's "Square Roots From 1; 24, 51, 10 to Dan Shanks," published in The College Mathematics Journal but also available at [4]. It explains the Shanks-Tonelli algorithm, one of the algorithms for finding the root when n is prime. (also see quadratic residue) nadav (talk) 03:16, 24 May 2007 (UTC)

May 24

Anybody from Research?

Hi! I hope this is the suitable place to post this question. I have got time series data for daily stock returns. I want to find whether its volatility (as measured by its standard deviation)increased during a particular event. My initial idea was to calculate the S.D. of returns of a few weeks surrounding that event and compare it to the overall S.D. to find out whether there lies a difference. Is it workable? If not what can be a better way? Please suggest. Thanks.--202.52.234.140 07:21, 24 May 2007 (UTC)

You could replace the data (from during the event) with "dummy" data and compare the SD of that time series with the original one. "Dummy" data could be based on some sort of moving average that wouldn't significantly alter the SD so that any change in SD can be attributed entirely and directly to the event.Zain Ebrahim 08:35, 24 May 2007 (UTC)

One word of caution: if the hypothesis of increased volatility was occasioned by looking at the same data to be used for the test, then no test will serve to establish significance, due to irreparable observator bias.

If you use the overall s.d. for comparison, you may be diluting the effect (if real) too much. It is probably better to take a smaller segment, for about as long as a notable effect of the event would be expected to persist (but not too short because of the customary √(n/(n−1)) correction counteracting the s.d. bias in small samples).  --Lambiam^Talk 11:24, 24 May 2007 (UTC)

Trying to detect an occasional change in some statistic of time-series data is intrinsically difficult, as the magnitude (and genuineness) of the change is wanted and when it occurs. That's why statistical control charts were devised, and I suggest you consider something of the sort. Assuming that "current" SD can be estimated (from the most recent n readings, where the number is a compromise between over- and under-sensitivity), an ongoing plot against time could show a change. The obvious plot would be SD v time, but a small change of slope of a straight line is easier to see than a small change of level, so a cusum (no Wiki article, but just Google the term) plot would be better. In essence, decide a reasonable value of the statistic in question, then plot accumulated deviations from this. --86.132.239.228 13:28, 24 May 2007 (UTC)

Just like there is no such thing as "instantaneous frequency" per Uncertainty Principle, I would seriously doubt there is such a thing as "instantaneous volatility". You can measure accurate time, or accurate volatility, but not both simultaneously. Oh, you can calculate something you might consider calling an "instantaneous frequency", but really that frequency is over some specific window, and not for that particular time-point. You need to make sure the window size you choose for the volatility measurement is decent for your application. How to choose this? I'm not sure. Sometimes, if there is a standard window others have been using for some particular application, it is good to use the same sized window so you can compare your results with theirs. Also, the wider the window, the less times you need to actually calculate the volatility. If you calculate the volatility around two successive points $x_{1}$,$x_{2}$, the new "data" you get by calculating the volatility around $x_{2}$ is almost nil, as $x_{2}$ is highly correlated to $x_{1}$. If you are calculating volatility for a set of 50 points, you only need to calculate volatility every 50 points, although sometimes if you want to view smoother results, you could calculate it more often (however, the more redundant points you calculate then there's a chance you may "see" something that isn't actually there.)

Note: This posting is a result of my original research. I AM NOT A PROFESSIONAL STATISTICIAN! Which means I may be a dumbass for even posting. My main point is that sometimes it makes sense to analyze data sequences in something analogous to to the methods used for analyzing time/frequency distributions, even if you are not at all interested in the sinusoidal frequencies. Take with a huge grain of salt.

(Mutters to himself, Did he really help the poster?) Root⁴(one) 16:13, 24 May 2007 (UTC)

Mann-Whitney U-test

In the MW U-test, how do you interpret p values? The article does not explain this. Thanks very much for your help in advance. Aaadddaaammm 08:37, 24 May 2007 (UTC)

Our Mann-Whitney U test article doesn't even mention P-values, but our article on them has a section on their interpretation. Or did you perhaps mean the ρ (rho) statistic described at the end of Mann-Whitney_U#Relation_to_other_tests? -- Avenue 10:08, 24 May 2007 (UTC)

Yea I know on a basic level what p values mean, but am not sure what exactly they mean in this case. Is it the probability that the two groups of observations have arisen by chance? Aaadddaaammm 10:46, 24 May 2007 (UTC)

Do you understand the principles of statistical hypothesis testing in general? The Mann-Whitney U test gives you a statistic that you would use just like any other statistic to test your hypothesis. For that test, you need to know the p-value of the statistic (or, alternatively and more conveniently, the value of the statistic for which the p-value equals a predetermined significance level, giving you the critical region).  --Lambiam^Talk 11:04, 24 May 2007 (UTC)

Kurtosis

My professor says the effect of excess kurtosis is to increase the probability of very large values and very small values. As per my understanding a distribution has excess kurtosis if too many data points revolve around the mean (the frequency of observations around the mean is high) and make the curve more picked than the normal distribution. For example, there are 100 observations in a sample. Ninety six of the observations are 0, two are +1 and remaining two are -1. This is a distribution with excess kurtosis. Here we can clearly see that the probability of the value 0(intermediate value, the mean) is much higher than the probabilty of -1 and +1 (the extreme values). Then how come the probability of extreme values is high in a leptokurtic distribution? Rather the probability of intermediate values should be high, isn't it?--202.52.234.140 09:41, 24 May 2007 (UTC)

As an illustrative example generalizing yours, take the discrete distribution D_V, having a parameter V, where with 96% probability you observe 0, 1% gets you +1, 1% is for −1, 1% for +V, and the last 1% is for −V. Then the excess kurtosis Kurt(D_V) = 50(V⁴+1)/(V²+1)² − 3. The first few values:

Kurt(D₀) = 47

Kurt(D₁) = 22

Kurt(D₂) = 31

Kurt(D₃) = 38

Observe how the value first goes down, and then increases with the value of |V|. The decrease can be "explained", if that is the right term, by the peak going down from 98% to 96%. The further increase can be "explained" by the likelihood of large deviations increasing. For example, D₃ has a dramatically larger probability for observations whose magnitude is at least 3 than D₂. Also when viewed as a continuous function of real-valued V, Kurt(D_V) attains a minimum for |V| = 1. It's all in the formula; kurtosis is not an easy thing to get a "feel" for.  --Lambiam^Talk 10:38, 24 May 2007 (UTC)

Kurtosis is measured with respect to standard deviation; if we "peak" a distribution but simultaneously keep its s.d. constant, we must be extending its tails to make up for the concentration of probability at the mean. The result will, of course, have a very high probability of giving values near its mean, but will also have a higher probability than the original distribution of giving values very far (in terms of s.d.) from the mean. --Tardis 14:50, 24 May 2007 (UTC)

Here's a link to kurtosis, for those who are confused. StuRat 19:41, 25 May 2007 (UTC)

Quantities without symbols

Is there, in mathematical notation, a preferred way to write quantities that are represented by words rather than by an ordinary mathematical symbol? E.g., I might want to say that antenna efficiency is defined as radiated power divided by input power without introducing symbols for each of them. Rather than just writing

$\mathrm {antenna\;efficiency} ={\frac {\mathrm {radiated\;power} }{\mathrm {input\;power} }}$,

how should I distinguish them from ordinary mathematical symbols? —Bromskloss 09:46, 24 May 2007 (UTC)

I don't know about "preferred", but the way you used, with the words in roman rather than the italics conventionally used for mathematical variables, is fairly common. For obvious reasons, this is not recommended for any but the simplest formulas; just imagine writing out the symbols in

$V_{a}={{\sqrt {R_{a}G_{a}}}\,\lambda \cos \psi \over {\sqrt {\pi Z_{\circ }}}}E_{b}$.

 --Lambiam^Talk 10:47, 24 May 2007 (UTC)

If phrases are used in a formula, some authors parenthesize them.

$({\text{antenna efficiency}})={\frac {({\text{radiated power}})}{({\text{input power}})}}$

More commonly, symbols are defined, either temporary or persistent.

${\text{Let }}E={\text{antenna efficiency}},P_{\text{out}}={\text{radiated power}},P_{\text{in}}={\text{input power}};{\text{then}}\,\!$

$E={\frac {P_{\text{out}}}{P_{\text{in}}}}.$

There is often little benefit in defining symbols to be used exactly once, so authors tend to choose accordingly. --KSmrq^T 15:56, 24 May 2007 (UTC)

Area of a quadrilateral

A surveyor's map of a plot of land shows it to be a convex quadrilateral. The length of each side is given along with its compass bearing from which I can calculate each internal angle. Wanting to find the area I searched for an on-line (area) calculator without success. I do recall (from long ago) being able to find a solution knowing the four sides and two opposite internal angles but I cant even find that! So, I'm asking for a link to a calculator or the formula. Thanks, hydnjo talk 19:27, 24 May 2007 (UTC)

My wikimarkup blows but

K = pq sin(theta)/2 where p and q are diagonals, and theta is the internal angle between them. Further derrivation at the provided source. Source: [5]

Hipocrite - «Talk» 19:34, 24 May 2007 (UTC)

Hmm, I don't know the diagonals or how to calculate them but, from your source link, this:

K = sqrt[(s-a)(s-b)(s-c)(s-d)-abcd cos2([A+C]/2)] (Bretschneider's Formula)

seems to do it. Thanks, hydnjo talk 20:15, 24 May 2007 (UTC)

My oops, this shoud have been asked at /Math (not /M) so I moved it here. Sorry, hydnjo talk 20:15, 24 May 2007 (UTC)

I usually chop quads into two triangles and use something like the old (base x height)/2 trick for the area of each triangle then add the two together - but there are a bazillion ways to find the area of a triangle - and they are a lot easier to remember and find online than the quadrilateral case. In general, memorize the triangle equations and you can use them for any polygon providing you have the patience to chop it up! I like that you can chop a circle up into a gazillion tiny triangles - the height of each one is the radius of the circle and the bases of all of them add up to the circumpherence - hence the area of a circle is half the circumpherence times the radius. SteveBaker 20:31, 24 May 2007 (UTC)

I agree with SteveBaker, but since finding the height of a triangle is slightly harder, I would prefer Heron's formula. For this you will need the diagonals, which can of course be found using the Law of cosines. -- Meni Rosenfeld (talk) 21:03, 24 May 2007 (UTC)

I think Heron's formula is good if you only know all sides, but damned if that's a lot of calculating! (more calculating can imply more round-off error) Besides, assuming you know the interior angle θ, ${\frac {1}{2}}s_{1}s_{2}\ sin(\theta )$ is good (you don't need to think in terms of bases and heights). I think the best one, which requires the least technical math (I.E. no trig or square root) for triangles oriented in two dimensional euclidean space is the determinant trick listed on the triangle page. As his data is from a map plot, that may be the simplest (computational) answer. Root⁴(one) 21:55, 24 May 2007 (UTC)

Thanks to all. I think that Bretschneider's formula best suits this problem (for me) because of its plug-inability. hydnjo talk 22:16, 24 May 2007 (UTC)

${\frac {1}{2}}s_{1}s_{2}\ sin(\theta _{1})+{\frac {1}{2}}s_{3}s_{4}\ sin(\theta _{2})$ is not plug-in able? Root⁴(one) 22:22, 24 May 2007 (UTC)

Oops - I wrote out my previous response off-line before actually reading your post :-( and your approach certainly is much easier to "plug-in" than Bretschneider's formula. Thank you, hydnjo talk 22:48, 24 May 2007 (UTC)

I'm wondering if I'm truly the one who should have said "OOPS". Geez. One think one knows some simple trig. Reply coming shortly. Root⁴(one)

Area of triangle 1/2bh. Any triangle has three heights and three bases (sides). Right angled triangles are a special case --two of the heights ARE the sides. In any case, assume you have side-angle-side measurements, that is sides AB and BC and angle ABC -- the angle is formed by sides AB and BC. Let AB be your base. Then the height of your triangle for this particular base is the shortest distance between C and the line collinear with AB. We can find this by assuming BC to be a hypotenuse for a right triangle defined by BC being the hypotenuse and angle ABC being the angle opposite the height. sin(ABC) is the ratio of the length of side opposite to the angle over the length of the hypotenuse, thus to find length h, we find length(BC)*sin(ABC). Since the area of a triangle is 1/2 * base length * height length, we have 1/2 length(AB)*sin(ABC)*length(BC). There's nothing in our proof that would disallow us to have assumed BC to the base and then come up with sin(ABC)*length(AB) as the height for that base.

I suppose there was no reason to doubt myself... I just rushed to judgment after a later second reading before I was absolutely convinced. That and my mind was a bit flustered after my last martial arts practice, leading me to have some significant doubts about things and my reasoning abilities in general. That's actually not a good excuse. I was right the first time and then I had to go correct myself :P Root⁴(one) 02:48, 25 May 2007 (UTC)

Your reasoning abilities are excellent. You don't need to worry. nadav (talk) 03:24, 25 May 2007 (UTC)

Your TeX could use a little refinement, however. Instead of multiplying "s", "i", and "n":

${\frac {1}{2}}s_{1}s_{2}\ sin(\theta _{1})+{\frac {1}{2}}s_{3}s_{4}\ sin(\theta _{2})$

you should use the operator name "\sin".

${\tfrac {1}{2}}s_{1}s_{2}\ \sin(\theta _{1})+{\tfrac {1}{2}}s_{3}s_{4}\ \sin(\theta _{2})$

It's a common mistake, so I'm saying this as much for the lurking masses as for you personally. --KSmrq^T 04:01, 25 May 2007 (UTC)

No offense taken. I appreciate constructive criticism, as well as the occasional complement. Thank you both. Root⁴(one) 12:04, 25 May 2007 (UTC)

Interestingly, the proof of Bretschneider's formula in our article begins with Root4(one)'s simply stated and intuitive area formula and goes on to prove an area formula which appears to be more complex. No? hydnjo talk 19:00, 25 May 2007 (UTC)

May 25

Math Problem

My friend and I have some trouble while arguing who is right and who is wrong. The math problem, a very simple problem, is (+17)-(-3). What is the answer? 74.113.188.65

If that's arithmetic, 17-(-3) = 17+3 = 20. Black Carrot 00:42, 26 May 2007 (UTC)

Negative_numbers#Addition_and_subtraction Black Carrot 00:43, 26 May 2007 (UTC)

You and your friend have encountered the famous "double negative equals positive" conundrum. Knowing the correct answer to this specific example is less important than understanding the reasoning. What are the proposed answers, and what is the thinking behind them, for each of you?

Meanwhile consider the following line of thought. Suppose the correct answer is A, so that

$(+17)-(-3)=A.\,\!$

If we add equal quantities to both sides, we will preserve the equality. In particular, we can add (−3) to both sides:

$(+17)-(-3)+(-3)=A+(-3).\,\!$

On the left, we have both added and subtracted (−3), so the two should cancel. On the right, we have added (−3), which is the same as subtracting (+3). Thus

$17=A-3.\,\!$

In other words, if we subtract 3 from the correct answer we should get 17. Finally, we can add (+3) to both sides. Thus we arrive at a value for A without ever explicitly resolving a double negative. Whatever the resolution, it must be consistent with this answer if it is to be consistent with the other ordinary laws of arithmetic. --KSmrq^T 08:43, 26 May 2007 (UTC)

Fixed point theorem

Gah, this unexpectedly popped up in my Analysis exam today, and I had no idea what to do. Anyway, I tracked down the proof in Fixed point property, and I understand what's going on - but I'm not sure why the multiple counter-examples I've thought of don't prove it wrong. The [0,1] x [0,1] "box" surely has multiple functions that do not intersect with f(x) = x, and so there are multiple functions that do not have fixed points, surely? Icthyos 22:41, 25 May 2007 (UTC)

Well, I myself am rather confident in the Brouwer fixed point theorem. Perhaps if you gave one of the counter examples we could figure out the problem you have with the theorem. nadav (talk) 22:59, 25 May 2007 (UTC)

Perhaps I'm misunderstanding, since I've not done much deep Analysis, but what about, say, f(x) = x + (1/2)? It can be restricted to a function from [0,1] -> [0,1] (though obviously not bijective, but that's neither here nor there as far as I can tell), and it has no fixed point...that I can see. Icthyos 23:14, 25 May 2007 (UTC)

How is it a function from $[0,1]\!\,$ to $[0,1]\!\,$? Is $f(0.75)\in [0,1]$? -- Meni Rosenfeld (talk) 23:22, 25 May 2007 (UTC)

In case you meant

$f(x)={\begin{cases}x+{\tfrac {1}{2}}&0\leq x\leq {\tfrac {1}{2}}\\1&{\tfrac {1}{2}}\leq x\leq 1\end{cases}}$

Then $f(1)=1$. -- Meni Rosenfeld (talk) 23:37, 25 May 2007 (UTC)

I think you are confronting the conditions of the theorem, which is good. In the geometry of the Euclidean plane, a translation is an example of a continuous map of the plane to itself that leaves no point fixed; so is a glide reflection. If we then convert point w to w/(1+||w||), which is a bijection, we can create a continuous map of the open unit disk to itself without fixed point. Restricting attention to the closed unit disk, consider the map taking every point to the origin except the origin itself, which maps elsewhere; clearly this example has no fixed point, but neither is it continuous.

These examples tell us that the theorem demands a continuous function of the closed disk to itself, or the equivalent. Too often when we learn a mathematical fact we ignore the fine print, the preconditions. We over-generalize, like children who pluralize every noun by adding "s", including "cats" and "dogs", but also "*foots" and "*childs". This is presumably a natural attempt to simplify; but like the children, we eventually must learn better. --KSmrq^T 09:38, 26 May 2007 (UTC)

Ahh, thank you all. I was confusing myself with my counter-example, by only treating a sub-interval of the function's domain, and ignoring what happened when x > (1/2) - obviously at these points, f(x) > 1. I was having trouble imagining it before, but I see now - if you've got a function that doesn't have a fixed point at 0, then it has to cross the line f(x) = x at some point, or have a fixed point at 1. Oh well, there go six exam marks, heh. I scribbled something down about the intermediate value theorem, though not with any degree of relevance - but at least I see it was used in the fixed point property proof.

(I swear, I'll learn TeX at Honours next year...) Icthyos 10:51, 26 May 2007 (UTC)

TeX isn't hard, at least not the part you need in order to write formulae in Wikipedia - Help:Formula is a good reference, which you can use together with copying other people's markup until you get accustomed with it. -- Meni Rosenfeld (talk) 12:48, 26 May 2007 (UTC)

May 26

Incredibly convulted integral

Hello, I need help solving this:

$\int _{\pi }^{\pi ^{2}}{\frac {\operatorname {erf} (x)}{{\rm {Si}}(x)}}(\cos x)^{x}(e^{\csc x})dx$

erf(x) is the error function, while Si(x) is the Sine integral.

How do I start? THANK YOU VERY MUCH.--0rrAvenger 00:59, 26 May 2007 (UTC)

Is a numerical approximation good enough? Cause I have no idea how to even begin to solve this integral analytically. nadav (talk) 02:36, 26 May 2007 (UTC)

Can you explain the origin of this integral? It looks somewhat implausible.  --Lambiam^Talk 09:02, 26 May 2007 (UTC)

High school math assignment... Just kidding. My friend made it up.--0rrAvenger 15:14, 26 May 2007 (UTC)

For π ≤ x < ³/₂π, and other parts of the range of integration, cos x is negative, and raising a negative number to an irrational value, as is required then by the factor (cos x)^x, is somewhat undefined. Our article Exponentiation ascribes the meaning exp (b·log a) to a^b, but that requires a somewhat arbitrary choice for log a when a is negative. For a = –1 and b irrational (corresponding to x = π), the possible values are dense on the unit circle in the complex plane.  --Lambiam^Talk 09:29, 26 May 2007 (UTC)

Wonky Room

Imagine a rectangular room. If you leave through one wall, you come back in on the opposite wall, rotated 180 degrees about the center of the wall - that is, walking on the ceiling. If you leave through the ceiling (or floor), you come back in through the floor (or ceiling) rotated 180 degrees about the center of each. Would this be an acceptable model of real projective 3-space? I base it on the idea of gluing together opposing boundary points on what's topologically a closed ball. Black Carrot 03:30, 26 May 2007 (UTC)

Yes, I believe the Wonky Room is an apt model of the projective space. You're basically identifying antipodal points on the boundary of a ball in a way that is analogous to the fundamental polygon for RP^2. nadav (talk) 08:23, 26 May 2007 (UTC)

Sounds a lot like the demo of Portal (computer game). -Wooty Woot? contribs 08:58, 26 May 2007 (UTC)

Yes the addition of "portals" to first-person shooter games so complicated the topology of the maps that companies were forced to redesign their game engines from the ground up. This was the problem 3D Realms ran into when they made Prey (video game), the first game to use portals. nadav (talk) 09:12, 26 May 2007 (UTC)

The Bot Has Failed!

More discussion at the reference desk discussion page. Root⁴(one) 04:25, 26 May 2007 (UTC)

Factor Theorem

Is the factor theorem true for expressions (polynomials) of multiple variables? For example to factorise a² + ab - ab² - b³, Can I treat it as a function of a - ie. f(a) = a² + ab - ab² - b³ and then say that since f(b²) = 0, then (a - b²) is a factor of this expression. I know it's true in this case but I want to know whether the proof is valid. --AMorris (talk)●(contribs) 09:16, 26 May 2007 (UTC)

The short answer is: yes. If the expression can be viewed as a polynomial P[V] in some variable V, and vanishes when some expression E is substituted for V; that is, P[E] can be simplified to 0 using the normal rules of algebra, then P[V] can be rewritten in the form (V−E) × Q[V], in which Q[V] is again a polynomial in V. The proof applies without change.  --Lambiam^Talk 09:58, 26 May 2007 (UTC)

Something I don't understand about integration

We'll be studying Integration in Additional Mathematics next term. We have already studied Differentation, and I understand Integration is the reverse process of differentiation. So if you differentiate a function and then integrate the result, you get the original function again. Right?

But there's something I don't understand about Integration. It's difficult to explain, so I'll use an example.

Let y = x^2 + 3x + 4

So dy/dx = 2x + 3. Right?

Now, let's say I am asked to integrate 2x + 3.

(My keyboard doesn't have an integration sign.)

When 2x is integrated, I get x^2 (I did this mentally, not using any formula).

When 3 is integrated, I get 3x.

So when I integrate 2x + 3, I get x^2 + 3x. Am I correct?

Now my question is, what happened to the 4? When I'm asked to integrate 2x + 3, how do I get back the 4? How do I even know that the last number's 4? Change the 4 in the original function to 5, and dy/dx is still 2x + 3.

--Kaypoh 11:16, 26 May 2007 (UTC)

For this reason, you always have $+C$ at the end of any integral without limits, where C is a constant (the 4 in your example). When the integral has limits, you don't need the $+C$.

(e/c) An excellent question. The answer is that when you perform the integration (that is, when you find the antiderivative), you must not forget to add the Arbitrary constant of integration. As you point out, you do lose the specific constant term when you differentiate and then antidifferentiate. nadav (talk) 11:37, 26 May 2007 (UTC)

Thanks for the quick response! Does that mean that there's no way to find out whether the nunmber is 4, 5 or something else, so I use c to represent the number? --Kaypoh 11:50, 26 May 2007 (UTC)

Yes, that's exactly right, unless you have additional information that tells you which value of C to choose. If all you know is that you are looking for a function F whose derivative is some function f that you are given, then you must include the + C. Sometimes, you are given additional information about this function F you are seeking, for example you might be told that F(2)=3. In those cases (called Initial value problems), you can solve for the exact value of C that's required for the equation to hold. nadav (talk) 12:04, 26 May 2007 (UTC)

I would say that you have misunderstood the fundamental theorem of calculus in a common way. Differentiation loses information; the derivative of a function tells us how it is turning at each point, but not an absolute position. Integration starts at a given position and accumulates all the twists and turns to give an absolute position everywhere. So we can integrate and then differentiate without loss, but if we differentiate then integrate we necessarily introduce uncertainty. A traditional way to denote that uncertainty is with a "constant of integration",

$\int f(x)\,dx=F(x)+C.\,\!$

It is easy to forget the constant, because it is irrelevant to definite integrals. Want a vivid mental picture? Suppose I tell you to go one block north and turn left. That's derivative-style information. But are we in New York or Hong Kong or Moscow? That's the constant of integration. And notice that no matter which city we are in, from the path we travel (the integral) we can recover the directions (the derivative). --KSmrq^T 12:44, 26 May 2007 (UTC)

This is a very useful way to think about the general problem, but I imagine it would be difficult to understand the concept of accumulating the twists and turns taken by the function without having studied definite integrals first. nadav (talk) 13:05, 26 May 2007 (UTC)

The easiest way to see that differentiation loses information is to consider what the derivative of 4 is, in your example. It's zero. It disappears. Now ask yourself what the integral of 0 is. Remember that the integral of a function f(x) that when differentiated gets you f back. Let f(x) = 0. But the derivative of any constant function is 0. We know that some constant must have been in the integral, so we have to put it back (hence the +C that you see).

x₀

How do you pronounce x₀? My professor says "x not", but I have no clue why he says it like this. Thanks in advance, --Abdull 15:06, 26 May 2007 (UTC)

x naught.--0rrAvenger 15:12, 26 May 2007 (UTC)

Definitions can be found at Wiktionary and Dictionary.com. nadav (talk) 15:19, 26 May 2007 (UTC)

Thanks to both of you! --Abdull 15:28, 26 May 2007 (UTC)

Some emphasize that the 0 is a subscript by saying "x sub zero" or "x sub naught". -- Meni Rosenfeld (talk) 17:09, 26 May 2007 (UTC)

Klein bottle

If and how can i graph a klien bottle on my ti-84 plus silver edition?

A Klein bottle is a three-dimensional figure which requires an extra dimension to be comprehended fully. What do you mean by "graph", since the ti-84 can only graph 2 dimensions.--0rrAvenger 17:30, 26 May 2007 (UTC)