Wikipedia:Reference desk/Mathematics

June 20

Determine a Period for irregular time series data

I have time series data (taken at irregular intervals) - Julian Date and Object Brightness. The objects brightness varies over time.

I am now trying to get a Power Spectrum that, for a series of test Periods, provides me with the 'strength of fit' of each of those periods to the data.

All the papers I have read refer to the use of a Fourier Transform defined by:

m(t) = C0 + Σ (aj*Cos(2*Pi*j*(t-E)/P) + bj*Cos(2*Pi*j*(t-E)/P)) {j=1 to J}

My math skills are basic. ie I can find A and B in X = A + B*Y where X and Y are supplied. I fear I may be out of my depth. Note that I have Mathematica 6 (if that helps).

Do you have reason to assume that the fluctuations of the brightness are periodic in time with a mixture of constant periods? Also, how dense are the data? If you plot the data you have, does it look like you could draw a smoothly sinuating curve through the points that does not requite too much fantasy to "connect the dots"? In other words, if you left out some of the points and drew the curve, would it pass near the points left out?  --Lambiam^Talk 07:37, 20 June 2007 (UTC)

Just curious

Why is the Fibonacci sequence always written (except on Wikipedia) as starting with {1,1,2,3,...} instead of {0,1,1,2,...} ? After all, zero plus one is one. -Haikon 05:06, 20 June 2007 (UTC)

Because it's hard to draw a picture of zero of something. Because zero (as distinct from "not") is a rather advanced concept. Because if we started with zero the layman would think "So I start by adding zero – where does that get me?". Pick one ... —Tamfang 05:24, 20 June 2007 (UTC)

I've seen it presented both ways, and have always just taken it as a convention. If starting with 1 is more common, it could be caused in part by the fact that all of the examples under "Origins" in the Wikipedia article naturally start with 1. 209.189.245.114 06:19, 20 June 2007 (UTC)

The Fibonacci numbers have many remarkable properties, such as that F_5n is a multiple of 5. Mathematicians have a strong tendency to start numbering the elements of a sequence at 1, as in a₁, a₂, ... . If you do that with the Fibonacci sequence starting at 0 (0, 1, ...), the expression of many of these properties becomes less pretty. Of course, the issue is resolved by counting like a₀, a₁, ..., as some would argue should be done all the time. For several sequences, including the Fibonacci sequence, you can actually extend the indexing to Z: F_−n = (−)ⁿF_n, giving a two-sided infinite sequence ..., 2, −1, 1, 0, 1, 1, 2, ... .  --Lambiam^Talk 07:18, 20 June 2007 (UTC)

It is not; see, for example, Graham, Knuth, & Patashnik, Concrete Mathematics (ISBN 978-0-201-14236-5). But if we are counting rabbit mates as they reproduce (per Fibonacci himself), it makes sense to start with one pair. --KSmrq^T 08:16, 20 June 2007 (UTC)

That is only the case when you assume that rabbits reproduce only starting from the second month. But if you take easter bunnies, which lay an egg every month starting from the first month and the eggs hatch after a month, then it makes sense to start with zero bunnies (and an egg).  :) – b_jonas 13:53, 21 June 2007 (UTC)

Our article defines F(0)=0, which is consistent with MathWorld and OEIS (to pick just two sources). Whether you start a list of Fibonacci numbers with F(0) or with F(1) is a minor point. Gandalf61 13:13, 20 June 2007 (UTC)

And a Google search [1] gives lots of other sources including 0. PrimeHunter 13:26, 20 June 2007 (UTC)

I see. Thank you reference desk people. -Haikon 17:09, 20 June 2007 (UTC)

choosing a projection hyperplane

I have a set of 306 points in 11 dimensions. I want to display them in 2 or 3 dimensions without collisions. One way to do that, it seems to me, is to enumerate the 46665 lines joining the given points, and find the 8- or 9-dimensional hyperplane that maximizes its minimum angle with any of those lines; the projection space, then, is the 3- or 2-dimensional hyperplane perpendicular to that one. How would I go about finding the desired hyperplane? —Tamfang 05:36, 20 June 2007 (UTC)

Maybe there will be better suggestions, but unless you have excessive computing power available, finding the optimal solution appears to me not quite feasible. What I'd try is finding an acceptably good solution, as follows:

1. Pick many random directions in 11-D space (many = 10000 or 100000 or more, depending on factors like available time).
   • For each direction picked, compute the projections of the points to the lower-dimensional space.
   • Find the pair of nearest neighbours – using methods from computational geometry that are much faster than O(n²), such as quad trees.
   • Keep the B best (largest minimum separation), where B is some number like 100 or 200.
2. Finally, use gradient ascent, simulated annealing, or other local optimization methods to optimize each of the B retained directions, and take the best result. Possibly you can also take a linear local approximation of the objective function and use linear programming, but I haven't given that idea any thought, and even if it works it may be overkill.

The "coarse" cycle 1 (many random directions) and the "refinement" phase 2 (local optimization) are separated here for clarity, but since the probability of a new random direction being better than any of the currently B best diminishes fast enough, you can actually immediately do the refinement for each addition to or replacement of the B best without too much additional cost, thereby having a best result available at all times, which may make it easier to decide when to stop.  --Lambiam^Talk 08:07, 20 June 2007 (UTC)

Thanks. First question: how do I pick a multidimensional "direction"? I know how to get three mutually perpendicular vectors in 3-space, but not in n-space! —Tamfang 20:23, 20 June 2007 (UTC)

Take the values of n independent normally distributed random variables with mean 0 and variance 1. The resulting n-vector has an isotropically random direction.  --Lambiam^Talk 22:49, 20 June 2007 (UTC)

Independently random vectors won't do: I need a three-dimensional hyperplane, generated by three mutually perpendicular vectors. Oh, I guess I see now, I can choose a vector in 11-space and then choose another from the orthogonal 10-space ... —Tamfang 01:29, 21 June 2007 (UTC)

(A seedy-looking character mumbles something incomprehensible about the possible utility of principal components analysis, then shuffles off. You are in a maze of twisty little dimensions, all different.) --KSmrq^T 16:38, 20 June 2007 (UTC)

Principal components analysis is what I tried first, and got 11 eigenvectors of roughly the same magnitude – all 11 dimensions are necessary. Now, accepting that any projection is going to lose most of the shape information, I'm looking instead for one that keeps it neat, i.e. a projection in which the images of the vertices don't collide. —Tamfang 20:23, 20 June 2007 (UTC)

On yet another hand, I might simply place my objects in a 2D array and "anneal" (once I learn in more detail what that means) with the original proximity measure as the function to be optimized. —Tamfang 20:23, 20 June 2007 (UTC)

It might help us in giving advice if we understood what purpose this should serve. Geometric properties that are preserved by projection are that points that are very close remain close, that angles that are small remain small, and that collinear points remain collinear. However, that can hardly be why you want a projection: points and lines that were far apart or not collinear may now become so. If tidyness is the goal, can't you take just any projection and then move points that come too close a bit apart?  --Lambiam^Talk 22:49, 20 June 2007 (UTC)

Well .. you've seen drawings of hypercubes, right? A well-chosen projection has the properties I'm after. I'd like it to be tidy and have the metric properties of a faithful projection. Thanks for asking, though; the question does bring me down to reality a bit. —Tamfang 01:29, 21 June 2007 (UTC)

In drawing a hypercube – or a 2D projection of a normal cube, for that matter – you want more than separation of the projected vertices. You also want vertices to remain separated from edges they are not incident with in the original.  --Lambiam^Talk 15:08, 21 June 2007 (UTC)

Polyhedral angles

If 3 plane faces meet at the vertex of a polyhedron, their angles at the vertex being alpha, beta and gamma, is there a standard relation between these angles? I'm thinking of something like their cosines having a constant product. What happens with 4 or more angles at a vertex?

This arose when I was considering the possibility of 4 triangles, with suitably matching sides, hinging together to form a tetrahedron. Considering one vertex, if one of the triangles is taken to be the base, its angle there must be less than the sum of the angles of the other two (with the same being true no matter which is taken to be the base, i.e. there is an analogy with the side lengths of a feasible triangle). My particular problem: if the base triangle has angle gamma at the vertex and the others have angles alpha and beta, what is the angle to the base of the ridge line between the other triangles?86.132.236.13 10:36, 20 June 2007 (UTC)

Relevant links include [2] and Defect (geometry). nadav (talk) 10:54, 20 June 2007 (UTC)

I'm not sure I'm correctly understanding the question. Between the angles α, β and γ, the only relation that's necessary for an arbitrary convex polyhedron is that α+β+γ<360 degrees. You can see this by cutting out some triangles from construction paper and pasting them together. Donald Hosek 16:44, 20 June 2007 (UTC)

I'm a little confused. Are these angles the angles between pairs of planes, or the angles between intersections of two planes on the third plane? In other words, is alpha equal to the angle between planes 1 and 2, or if we look at the polygon on plane 3, is alpha equal to the angle of its vertex at that point? There's a difference. In the first case, the only restriction I can think of is that alpha + beta + gamma < 540 degress. In the second, as Donald Hosek says, it would be that alpha + beta + gamma < 360 degrees. Both of these say that the shape is too few degrees around at the vertex to lie in a plane, and must form a point. I'm afraid I don't understand the "base of the ridge line" question. Black Carrot 04:15, 21 June 2007 (UTC)

I'm the original poster, despite the changed number below. Regarding my first question, yes I now see that the only relation for the angles at a convex vertex is summing to less than 360 degrees. Regarding the second, let me try to be clearer. Three triangles meet at a polyhedral vertex, their angles there being α, β and γ (i.e. for each triangle, the angle between the two lines meeting at the vertex). At a convex vertex, it would seem to be necessary that each such angle is less than the sum of the others. Take the triangle with angle γ as the base, and define as the ridge the line common to the other triangles. What I want is the angle between the ridge and its perpendicular projection onto the base, in terms of α, β and γ.81.154.108.148 10:58, 21 June 2007 (UTC)

You're right, my mistake, each would indeed have to be less than the sum of the others, otherwise the figure would pull itself apart trying to wrap around, just like the triangle inequality. I think I have an answer now to your second question, but it's a bit involved. I'd prefer to tell you how I got it, as well as the result itself, because I think it's a safe bet you'll want more freedom of movement than angles-gives-angles. Here's the logic. Take the vertex in question to be facing you, the face with angle gamma to be the base, the face with ange alpha to be to the right, and the face with angle beta to be to the left. Then, I will call the unit vectors running along each edge A, B, and C. A is the vector along the base to the right, B is along the base to the left, and C is the ridge, with the vertex as the origin. Next, I work out the coordinates for each of these vectors. If you happen to know the coordinates of each vertex of the pyramid to begin with, you can skip this part, by just dividing each set of coordinates by the magnitude of the vector they describe. Make sure you've first set the vertex as <0,0,0>, with all the others shifting around to match. If you only have the angles, here's how you might work out the coordinates. Assume the X-axis is along A, making A = <1,0,0>. With both the base of the pyramid and the XY-plane flat against the table, we can find B. B, along with its components parallel to the X and Y axes, would form a right triangle, with gamma as one of the angles. Therefore, its coordinates would be <cos(gamma),sin(gamma),0>. Now, C is trickier. First, I draw a circle in the air where it might be, then I narrow it down to a point. The easier circle to draw is around A. If C was directly above A, it would have coordinates <cos(alpha),0,sin(alpha)>, much like B. However, it can be anywhere along a circle from there, because the only restrictions for now are that it be angle alpha away from A, and it be of unit length. So, spinning it and introducing a dummy variable k, the temporary coordinates for C are

$\left\langle \cos a,k,{\sqrt {\sin ^{2}a-k^{2}}}\right\rangle .$

If you take the norm of this, you'll see it's guaranteed to have length 1. As the X value never changes, it's confined to a plane parallel to the YZ plane, and within that the radius of the circle will be sqrt( k²+sqrt(sin²a-k²)² ) = sin alpha. You could put +/- before each square root, but it seems reasonable to assume you want the solution above the table, not below it, meaning we've actually got a half-circle. Now, using the formula for dot product, for any vectors a and b, and the angle between them theta, a dot b = abs(a)abs(b)cos(theta). The dot product of B and C, then, should be equal to (1)(1)cos(beta), or just cos(beta). The dot product is the sum of the products of each of their respective components, so cos(alpha)cos(gamma) + ksin(gamma) + 0 = cos(beta). Solving for k to narrow the semicircle to a point, we get

$k={\frac {\cos b-\cos a\cos g}{\sin g}}.$

Sorry, I'm not sure how to put greek letters in formulas. Now, we can substitute that into the coordinates for C.

$\left\langle \cos a,{\frac {\cos b-\cos a\cos g}{\sin g}},{\sqrt {\sin ^{2}a-\left({\frac {\cos b-\cos a\cos g}{\sin g}}\right)^{2}}}\right\rangle .$

Now that you can get A, B, and C, you can work out the angle between C and the plane containing A and B (the table) fairly easily. First, you take the cross product of A and B, to get their normal vector. Or, since we've set this particular one up along the axes, you could just take the vertical unit vector <0,0,1>. Then you use the formula for dot product again. Since a dot b = abs(a)abs(b)cos(theta), theta = arccos(a dot b / abs(a)abs(b)). To find the angle between C and the normal, take their dot product, divide by the product of their magnitudes, and take the arccosine of that. In this case, it would simplify to taking the arccosine of that really complicated term in C. Subtract the result from 90 degrees to get the angle between C and the table.

It's worth pointing out, by the way, that if I've made no mistakes, this should give you the shortest angle between the ridge and the plane containing gamma, regardless of whether that falls inside the pyramid. You may have to make adjustments if that's not what you want, for instance if the angles of the pyramid are obtuse and you want the internal angle. Black Carrot 16:58, 22 June 2007 (UTC)

Thanks for this - I'm just going away for a few days but will have a careful read through on returning.81.153.218.145 19:52, 22 June 2007 (UTC)

More complex analysis fun

Hi all, thanks for the help you've given me so far. I have another exercise I can't seem to figure out. (Again, this is not homework, this is preparation for a PhD qualifying examination.)

Let P(z) be a polynomial with real coefficients, say $P(z)=a_{n}z^{n}+\cdots a_{1}z+a_{0}$, satisfying $a_{0}>a_{1}>a_{2}>\cdots >a_{n}>0$. Prove that all the zeros $\zeta$ of P(z) satisfy $|\zeta |>1$.

At first, I thought it could be done by induction, since the base case n = 1 is so simple; but I can't even prove it for n = 2! Am I missing something easy here, or what? Any help would be appreciated. Thanks! –King Bee ^{(τ • γ)} 12:30, 20 June 2007 (UTC)

I haven't found a solution, but here are some ideas: The statement is clearly correct for the real roots; the imaginary roots come in conjugate pairs. This allows you to write $P(z)=(z-z_{0})(z-{\overline {z_{0}}})Q(z)$, which immediately solves the case $n=2$ and possibly simplifies other cases. Rouché's theorem might also be of use. -- Meni Rosenfeld (talk) 13:39, 20 June 2007 (UTC)

• Meni - Thanks for the idea! However, I think I have a nice, elegant, clever solution, but I can't get the strict inequality to quite work out. (Not my idea; fellow graduate student pointed me in the right direction.)

Note that $zp(z)=a_{n}z^{n+1}+\cdots a_{1}z^{2}+a_{0}z$, and look at what $zp(z)-p(z)$ is; we achive

$(z-1)p(z)=a_{n}z^{n+1}+(a_{n-1}-a_{n})z^{n}+(a_{n-2}-a_{n-1})z^{n-1}+\cdots +(a_{0}-a_{1})z-a_{0}$.

Suppose by way of contradiction that there exists a zero of p(z) inside the unit disk, say α. Then $|\alpha |\leq 1$. Substituting this in to the expression above, we get:

$0=a_{n}\alpha ^{n+1}+(a_{n-1}-a_{n})\alpha ^{n}+(a_{n-2}-a_{n-1})\alpha ^{n-1}+\cdots +(a_{0}-a_{1})\alpha -a_{0}$.

Adding $a_{0}$ to both sides and then taking modulus (and using the triangle inequality and the fact that $|\alpha |\leq 1$), we obtain:

$a_{0}\leq |a_{n}|+|a_{n-1}-a_{n}|+|a_{n-2}-a_{n-1}|+\cdots +|a_{0}-a_{1}|$,

and since all of the differences above are already real and positive, we can remove the absolute value signs and we get a telescoping sum which sums to $a_{0}$. However, this leads me to $a_{0}\leq a_{0}$, which is not a contradiction; should some inequality above be strict? I guess I can use this method to show that there aren't any zeros of p(z) inside the open unit disk, but then I would have to deal with the boundary case separately. Any help on that one? –King Bee ^{(τ • γ)} 16:12, 20 June 2007 (UTC)

This is quite simple, actually. My hint is: The triangle inequality is strict unless all the summands are on the same ray from the origin. -- Meni Rosenfeld (talk) 16:38, 20 June 2007 (UTC)

• slaps forehead* Excellent! I am a fool, thanks for that tidbit. =) –King Bee ^{(τ • γ)} 16:47, 20 June 2007 (UTC)

The theoretical definition of decimal expansion

Hello.

I can't prove the following statement from my book (Principles of Mathematical Analysis - Walter Rudin).

Consider the following definition of decimals. We are assuming that we have the Dedekind definition of the reals.

Let x>0 be real. Let n_0 be that largest integer such that n_0 <= x. Having chosen n_0, n_1, ... , n_(k-1) let n_k be the largest integer such that n_0 + (n_1/10) + ... + (n_k/10^k) <= x.

Let E be the set of these numbers n_0 + (n_1/10) + ... + (n_k/10^k) (k=0,1,2,...)

Then x = sup E. The decimal expansion of x is n_0.n_1n_2n_3...

My problem is that I am not able to show that x = sup E. The difficulty seems to be the reason why a number of the form x - e (e>0) can't be a supremum.

Suppose x-e is the supremum. By way of contradiction we must now have an element of E strictly greater then x-e. But elements of E get progressively smaller and smaller increments in their values & so its not clear how to produce such an element. For example suppose that y = n_0 + (n_1/10) + ... + (n_k/10^k) for some k is some elemnent of E. By hypothesis since x-e is the supremum so y <= x-e. Taking the worst case y < x-e, we need to produce an element of E greater then x-e. For this we need to consider elements like y + (n_(k+1)/10^(k+1)), y + (n_(k+1)/10^(k+1)) + (n_(k+2)/10^(k+2)) etc. But as terms like (n_ (k+1)/10^(k+1)) are very small there is no guarantee that there addition to y will make it enough to surpass x-e. So one can't claim from the above procedure that x-e is not the supremem. I have a gut feeling that there are some more things required in the proof of which I am not aware. Can anyone shed some light on them. Thanks. --Shahab 14:18, 20 June 2007 (UTC)

Tidying the notation up a bit will help. We have:

• $S_{k}=\sum _{i=0}^{k}{\frac {n_{k}}{10^{k}}}$
• By definition of $n_{k}$,

$S_{k}\leq x$

$S_{k}+10^{-k}>x\;\!$

• $E=\{S_{k}|k\geq 0\}$

If $e>0\;\!$ and $x-e\;\!$ is an upper bound for $E\;\!$, then there is some $k\;\!$ such that $10^{-k}<e\;\!$. Can you see how to proceed from here? -- Meni Rosenfeld (talk) 14:56, 20 June 2007 (UTC)

A further remark: the last step above would tend to use the so-called Archimedean property of the real numbers. However, to prove that is a bit similar to what you're asked to prove here, so it is not clear you are allowed to use it. Your introduction of e is a bit of a red herring. Instead of x−e, consider any u less than x, and examine why u can't be the supremum. Since u < x, you know that x and u have different Dedekind cuts. It means that there exists some rational number by which you can tell them apart. What do you know about that number? You should from this point not use anything but properties of integers and rational numbers. Does that help?  --Lambiam^Talk 15:16, 20 June 2007 (UTC)

Complex Analysis

I can't figure out how to do this question:

If f is analytic and $|f(z)||f'(z)|\leq 1$ for each z then show that f must be constant. Thanks--Shahab 14:27, 20 June 2007 (UTC)

Hint: Consider $g(z)=f(z)^{2}\;\!$. -- Meni Rosenfeld (talk) 14:58, 20 June 2007 (UTC)

Thanks for the response. But I don't understand. As far as I can see then g'=2ff' and so g' is bounded and so constant. But what then? Can you be more explicit.--Shahab 17:40, 20 June 2007 (UTC)

Maybe this will help: Since $g'$ is constant, that means that $g$ is at best a polynomial of degree 1. But $g$ was defined above as the square of a different analytic function...–King Bee ^{(τ • γ)} 17:51, 20 June 2007 (UTC)

Still the fog persists in my mind. Is there any result that g' is constant implies that g is a poly of degree at best 1. Intuitively it is clear but how to prove it. Again if g(z)= az+b=f(z)^2, where is the contradiction? I am afraid that I must trouble you folks again.--Shahab 18:17, 20 June 2007 (UTC)

Yes, if $g'$ is constant, then $g$ is a polynomial of degree at most 1 (this you can cite freely; I doubt anyone would argue with you, but you can write up a proof if you wish). You have now that $f^{2}$ is either a linear function or a constant. Since $f$ is entire, which of these two is possible, and why? Just think about it for a while, and I think the light will come on eventually, and you'll be happy you did it yourself! –King Bee ^{(τ • γ)} 18:25, 20 June 2007 (UTC)

Thanks for the response. I think I have got it now. From the Taylor's expansion it follows that g' is constant implies g is a poly of degree at best 1. Also $f(z)={\sqrt {Az+B}}$ is not entire as the square root function isn't. So $f(z)={\sqrt {C}}$ i.e. f is a constant. Cheers.--Shahab 18:57, 20 June 2007 (UTC)

You don't need Taylor for that. Think: g is an antiderivative of g'. If g'(z) = A, how can you find the general antiderivative?  --Lambiam^Talk 22:57, 20 June 2007 (UTC)

Note that there is some subtlety in that last part: Since every nonzero complex number has two square roots, and it is not clear which one is "better", it is best to avoid the notation ${\sqrt {}}$ unless you are completely sure you know what you are doing. "$f(z)={\sqrt {Az+B}}$" is not a correct consequence of $f(z)^{2}=Az+B$. The correct fact you should cite is that there is no entire function (or even a continuous one) whose square is the identity function, from which it is easy to show that there is no entire function whose square is $Az+B$ with $A\neq 0$, thus $A=0$. So $f(z)^{2}$ is constant - but this, by itself, still doesn't mean that f is. However, showing now that f is constant is not hard, using the fact that f is continuous. -- Meni Rosenfeld (talk) 19:27, 20 June 2007 (UTC)

Write $f(z)=a+bz+z^{2}h(z)$ for some series expansion h. Then

$Az+B=g(z)=f^{2}(z)=a^{2}+2abz+b^{2}z^{2}+2a^{2}z^{2}h(z)+2bz^{3}h(z)+z^{4}h^{2}(z)$.

By uniqueness of series expansion, $z^{4}h^{2}(z)=0$, i.e. h=0. From $Az+B=g(z)=f^{2}(z)=a^{2}+2abz+b^{2}z^{2}$, we have $b^{2}z^{2}=0$, i.e. b=0. Therefore f=a is a constant. Is it acceptable? Twma 00:26, 21 June 2007 (UTC)

3 car purchases over a 3 year period - simple maths brain problem

My brain is not working very well maths wise today and i'm trying to figure out something a little 'odd'. Basically if I bought a car with a loan rate of 0%-APR (36 payments) for £7,000 owned it for a year, sold it for £6,500, used the money to repay the loan and get a new loan for £7,000 (same 0% APR/36 payments) owned it for a year, sold it for £6,500, use the money to repay the loan and got a new loan for £7k (same APR/payments and repeated. How much money would I have lost?

It is a 'scenario' so no running-costs/bank charges for paying early and i'm not bothered about loans being 'weighted' to paying more interest at the start of the loan. Sorry just my brain has packed in for the night due to poorliness/tiredness. What I want to know is how much money you would 'lose' over the ownership. My brain says £1,500 (£500 loss per car) is wrong but i'm not sure why at all- am I overcomplicating it? ny156uk 22:16, 20 June 2007 (UTC)

I think you only lost £1000 (£500 between the purchase of the 1st and 2nd cars and £500 between the 2nd and 3rd cars). Of course, you are then obligated to pay the £7000 loan on the third car, so have a total obligation of £8000. StuRat 04:53, 21 June 2007 (UTC)

June 21

Write naturally in math, automatically translate to TeX

Is there a product out there where I can write mathematical equations naturally on a surface, and it would automatically convert it to TeX for me? --HappyCamper 01:32, 21 June 2007 (UTC)

No. Learn to type TeX quickly; that should be more useful. (This unsigned comment added by 129.78.64.102.)

That sounds about right. State-of-the-art PDAs have enough trouble with simple letters and numbers, the mind boggles at what it'd take to get them to understand complex formluas. Black Carrot 04:06, 21 June 2007 (UTC)

Hmm...nevermind, I just found what I was looking for today. Thanks! --HappyCamper 11:41, 21 June 2007 (UTC)

Ah, wanna share? —Bromskloss 13:28, 21 June 2007 (UTC)

"Wanted: math grad student for typing job, $10/hr or ramen+shower." [edited] Tesseran 01:00, 22 June 2007 (UTC)

dodecagons and hexagons

How do I calculate the area of a dodecagon and the area of the hexagon if I only know the length of one side, let's say 7cm?Invisiblebug590 06:32, 21 June 2007 (UTC)

Dodecagon—12 sided figure

Hexgon—6 sided figure

You can't—unless you assume/guarantee something more about the shapes (e.g. that they are regular convex polygons). --CiaPan 07:18, 21 June 2007 (UTC)

Have you read our dodecagon and hexagon articles? they would seem to answer your question (for regular versions). 213.48.15.234 07:15, 21 June 2007 (UTC)

P.S. the dodecagon and hexagon are regular polygons.

Invisiblebug590 06:11, 23 June 2007 (UTC)

See Regular_polygon#Area 67.182.185.143 22:37, 23 June 2007 (UTC)

Maybe i've turned stupid

How would you solve $2^{x}+3^{x}=13$ for x?

I can see that the answer's x=2 but i have no idea how to go about getting the solution through manipulation of the equation. 213.48.15.234 07:40, 21 June 2007 (UTC)

I had enough trouble sleeping before you asked. :) The only thing I know is that you need logarithms to remove the x variables from the exponent. I just don't see how to do it because the "simple" rules of logarithms don't give a simplification for log of the left side of your equation. YechielMan 07:29, 21 June 2007 (UTC)

X=2 is certainly one answer, but I doubt it's the only correct answer. -- JackofOz 07:32, 21 June 2007 (UTC)

I've tried both logarithms and exponent manipulation, but the old not being able to do much with added exponents of different bases is throwing me off. Jack, I can't see another real answer, maybe there are some imaginary ones. 213.48.15.234 07:36, 21 June 2007 (UTC)

Since $2^{x}+3^{x}$ is a strictly increasing function of $x$, there can be only one solution. —Bromskloss 08:01, 21 June 2007 (UTC)

$\log(a+c)=\log a+\log(1+e^{\log c-\log a})$ Assuming Natural Log.

Put a = 2^x and put c=3^x

$\log(2^{x}+3^{x})=x\log 2+\log(1+e^{x\log 3-x\log 2})$

$\log(2^{x}+3^{x})=x\log 2+\log(1+e^{x(\log 3-\log 2)})$

Now solve

$x\log 2+\log(1+e^{x(\log 3-\log 2)})=\log 13$

$xF+\log(1+e^{xG})=H$

where F = Log 2

where G = Log 3 - Log 2

where H = Log 13

So we have

$x+{\frac {\log(1+e^{xG})}{F}}={\frac {H}{F}}$

$x+{\frac {\log(1+e^{xG})}{\log K}}=L$

where e^K=F

where L=H/F

$x+\log(1+e^{xG}-K)=L\,$

Raise to the e

$e^{x}\times (1+e^{xG}-K)=e^{L}\,$

$e^{x}\times (M+e^{xG})=N\,$

where M=1-K

where N=e^L

Now expand

$Me^{x}+e^{x}e^{xG}=N\,$

$e^{x+P}+e^{x(1+G)}=N\,$

$e^{x+P}+e^{xQ}=N\,$

211.28.131.152 07:47, 21 June 2007 (UTC)

Well, i could do all of that from your first formula, that's exactly the point I got stuck, though :p 213.48.15.234 07:53, 21 June 2007 (UTC)

shoot he's still going 213.48.15.234 08:11, 21 June 2007 (UTC)

This transformation seems incorrect to me:

$x+{\frac {\log(1+e^{xG})}{\log K}}=L$

$x+\log(1+e^{xG}-K)=L\,$

Logarithmic identity is ${\tfrac {\log A}{\log B}}=\log _{B}A\neq \log {(A-B)}.$

--CiaPan 08:34, 21 June 2007 (UTC)

I believe you're correct there CiaPan, i'm still only a little closer to a solution. :/ 213.48.15.234 09:00, 21 June 2007 (UTC)

If simple manipulations could solve these kinds of equations, I'm sure Fermat's last theorem would have been proven much earlier. Some related (but much tougher obviously) problems are described in Mihăilescu's theorem. nadav (talk) 09:39, 21 June 2007 (UTC)

Let me just emphasize these statements:

• X=2 is certainly one answer, but I doubt it's the only correct answer. -- JackofOz 07:32, 21 June 2007 (UTC)
• Since $2^{x}+3^{x}$ is a strictly increasing function of $x$, there can be only one solution. —Bromskloss 08:01, 21 June 2007 (UTC)

Now, for the complex solutions, $n^{r\cdot \operatorname {cis} (\varphi )}=n^{r\cdot \cos(\varphi )}\cdot \operatorname {cis} (r\cdot \ln(n)\cdot \sin(\varphi ))$, so we get (for "cis", see this section):

$2^{r\cdot \cos(\varphi )}\cdot \cos(r\cdot \ln(2)\cdot \sin(\varphi ))+3^{r\cdot \cos(\varphi )}\cdot \cos(r\cdot \ln(3)\cdot \sin(\varphi ))=13$

$2^{r\cdot \cos(\varphi )}\cdot \sin(r\cdot \ln(2)\cdot \sin(\varphi ))+3^{r\cdot \cos(\varphi )}\cdot \sin(r\cdot \ln(3)\cdot \sin(\varphi ))=0$

Try solving this system, and let me know what you get. -- Jokes Free4Me 09:44, 21 June 2007 (UTC)

Oh, are we looking for non-real solutions as well? —Bromskloss 09:51, 21 June 2007 (UTC)

Not really, I was just speculating, however I would like to see if it's possible to find x=2 from the equation. From the answers i've gotten so far, it doesnt look like it is easy to do. 213.48.15.234 09:55, 21 June 2007 (UTC)

One needs to use a root-finding algorithm.--Patrick 11:31, 21 June 2007 (UTC)

I haven't worked out the details, but I strongly suspect there are no complex solutions beyond x = 2. Putting x = a+bi, |2^x + 3^x|² is at least (3^a – 2^a)² and at most (3^a + 2^a)²; if 13² is not in this range, we can't have a solution. This implies that 2 ≤ a < 2.61. Examining this range in small increments, with 0 ≤ b < 2π, does not suggest any second solution lurking there.  --Lambiam^Talk 15:54, 21 June 2007 (UTC)

Your intuition is wrong. I did a quick graph of $2^{x+yi}+3^{x+yi}=13$ in the OS X grapher application and got what looked like a periodic curve (but probably isn't), which shouldn't be terribly surprising. Consider the simpler case of $e^{z}=c$ for some real $c>1$. Remember that $e^{z}$ is a transcendental function and periodic. The functions $2^{z}$ and $3^{z}$ are similarly shaped and adding them gives you something which is vaguely periodic (I think that the periods will interact in an irrational fashion so there probably does not exist non-zero $w\in \mathbb {C}$ such that $2^{z+w}+3^{z+w}=2^{z}+3^{z}$. There are in fact an infinite number of solutions to $2^{z}+3^{z}=13$ but only one real solution. Donald Hosek 16:57, 21 June 2007 (UTC)

My intuition here is mostly useless, but empirically looking at the imaginary part of the sum via a contour plot in Mathematica, with x between -3 and 4, and y between -3 and 3, and one contour line at 0 while sampling on a grid of 3000 points on each axis, shows one contour which strongly seems to lie on the real axis. Zooming the y axis in to between +/- 0.001 supports this hypothesis. This doesn't prove anything, but does suggest only the aforementioned root 2. Baccyak4H (Yak!) 17:28, 21 June 2007 (UTC)

Have mathematica give you a graph of solutions to $2^{z}+3^{z}=13$. I'm not sure what you're trying to prove with looking at the imaginary part of the graph. I'd spend some time also just looking at graphs of the exponential functions. Look at the real part, imaginary part and modulus (absolute value) to get some sense of what's going on, then you should be able to get some sense of what happens with the function $f(z)=2^{z}+3^{z}$ and what the set of solutions to $f(z)=13$ should look like. Mathematica should be able to graph those for you as well (you may need to use the $(x,y)$ form of the equation I cited above to get it to graph the solutions in a real plane... I'm too old and cheap to know mathematica well). Donald Hosek 17:57, 21 June 2007 (UTC)

Sorry if I wasn't clear. I did just what you suggested: made a contour plot of Im( 2^(x+i*y) + 3^(x+i*y) ) with x and y being the two axes of the plot. The idea is for the sum to be 13, the imaginary part must be 0. Allowing one contour value at 0 shows where this happens (indeed, the intersection of those contours with those of the real part at value 13 are in principle all of the roots). The only contour which took value zero was apparently coincident with the x axis, meaning y=0, and thus implying all roots indeed are real. Baccyak4H (Yak!) 18:17, 21 June 2007 (UTC)

Update: This suggests a way to prove it. Ignore the 13 for a moment and try to show that 2^x + 3^x is real implies x is real. Baccyak4H (Yak!) 18:19, 21 June 2007 (UTC)

Not sure how accurate it is, but Maple turned up no complex solutions. Dlong 18:57, 21 June 2007 (UTC)

There are 2 complex solutions...somewhere around r = 6.524959 and theta = +- 1.1651783. --HappyCamper 19:51, 21 June 2007 (UTC) <--- update -- actually, I take that back. I think there are an infinite number of solutions. It's easier to see them by solving with r and theta. --HappyCamper 19:55, 21 June 2007 (UTC)

Good stuff... my efforts didn't look far out enough. Baccyak4H (Yak!) 20:06, 21 June 2007 (UTC)

If someone could double check these, it would be great. I don't have Mathematica at my disposal. Looks like the next furthest solution is r = 11.3275949 and theta = 1.362440594. --HappyCamper 20:37, 21 June 2007 (UTC)

Check. Also r = 17.4819... and θ = 1.4525....Baccyak4H (Yak!) 20:48, 21 June 2007 (UTC)

• Check. I get r = 17.48192845163..., θ = 1.452539552949... --HappyCamper 21:24, 21 June 2007 (UTC)

Come on guys, take a look at the graph of the exponential function. Note that the real values for $e^{z}=c$ for $c\in \mathbb {R} ^{+}$ are not isolated points, but are actually continuous sets. It shouldn't take a genius to realize that $a^{z}=c$ for $a$ positive and not 1 will be similar other than scaling (think back to high school algebra for change of base). Now with a bit more effort, you should be able to come up with the idea that the solutions for $2^{z}+3^{z}=c$ will also be continuous sets. Donald Hosek 20:55, 21 June 2007 (UTC)

Whether you meant it to or not, your tone seems like it could offend some people, especially since they're trying their best to help with a problem that resists immediate solution. That aside, though, what you've written is false. If c > 0 is real, for example, the solution set of z such that e^z = c consists of exactly those points of the form Log c + 2kπi for k some integer. This set, as you probably know, is discrete (made up of isolated points), rather than continuous. In fact, the solution set is discrete for any complex c; this is captured by the observation that the exponential map is a covering map (onto the plane minus the origin).

Your second observation is correct: the base e here is not special. Your last claim is also false; the best way to see this in general is a dimension argument. The function that sends z to 2^z + 3^z is an entire function, and in particular a smooth map from the complex plane (a 2-dimensional real manifold) to itself. Then any regular value (here, any regular value of c) has discrete, or 0-dimensional, preimage (solution set). By Sard's theorem (PM), the set of regular values has full measure, which means that if there are any values of c such that the solution set is continuous, those values form a set of measure 0. Thus your assertion holds extremely rarely, if at all. Tesseran 00:59, 22 June 2007 (UTC)

The dimension argument needs a little more care -- zero-dimensional is not in general synonymous with "discrete". For example the Cantor space is zero-dimensional. But I seem to recall that, if the zeroes of an entire function accumulate, then the function is constantly zero. I think you just figure out what the Taylor series has to be, centered at the accumulation point. That should show that the solution set is discrete, without invoking Sard's theorem. --Trovatore 19:39, 23 June 2007 (UTC)

In this case, we know that the solution set is a manifold, and a 0-dimensional manifold is discrete (has the discrete topology). As you say, any entire function which vanishes on a set containing a limit point is identically 0, and this implies discreteness immediately. (A much nicer argument.) Tesseran 19:40, 24 June 2007 (UTC)

Thanks! --Trovatore 22:49, 24 June 2007 (UTC)

It looks like the solutions a+bi with 2 ≤ a < 2.61 occur with an average frequency for b of once in every interval of length $\pi /((1/ln(2))-(1/ln(3)))=5.9001941$, that is, twice for every time that the argument of $3^{z}$ overtakes that of $2^{z}$.--Patrick 08:28, 22 June 2007 (UTC)

Actually, the 2 ≤ a < 2.61 I quoted from above should be 2 ≤ a < 2.7047.--Patrick 08:39, 22 June 2007 (UTC)

$2^{x}+3^{x}=2^{2}+3^{2}$

That's all there is to it. The above demonstrates quite clearly how 2 can be the only real solution.

As for imaginary solutions, put x = a + bi and you equate real and imaginary parts and get our original equation. I don't think there are any imaginary solutions.

Fan pressure

I have a room which in the room a person doing painting and sanding. There are a lot of dust. I want to suct all the dirt particles. Is there any formula to calculate the static pressure that we need to able the fan suct all the dirt particles?

I don't think we can help you practically, other than to advise that you use a vacuum cleaner, but try to clarify the situation. Are the particles floating in the air or sitting on the floor? And how do you know when they have been removed? YechielMan 15:12, 21 June 2007 (UTC)

Any changes in static pressure wont help you, as they will cause no net force on your dust particles. The only way your fan will help is if it causes an air flow strong enough to remove your particles. I agree that a hoover is the best option, though a fan may help ventilate your room to get rid of the smell of paint, as well as the smallest floating dust particles. Good luck. 195.137.96.79 06:27, 22 June 2007 (UTC)

Be sure to use one fan blowing in and another blowing out on the opposite side of the room, for maximum airflow. Close any other windows, as they would interfere with proper airflow. Also, beware that air saturated with saw dust can explode if exposed to an ignition source. So, avoid smoking or any other activity with an open flame in that room. StuRat 06:45, 22 June 2007 (UTC)

You may find health and safety regulations could be of help. If you are running a real buisness then there is surly regulations as to the amount of particles in the air, and there may well be standards for type of ventilation you need. Ignore advice about domestic vacuum cleaners which will not be upto the job. --Salix alba (talk) 19:08, 22 June 2007 (UTC)

Unless it's a domestic setting, and you're just doing a bit of DIY. A vacuum cleaner would be just fine for this. 195.137.96.79 03:40, 24 June 2007 (UTC)

June 22

Coupled differential equation

Just for kicks, I'd like to solve this coupled set of differential equations

${\frac {df}{dx}}=\alpha g(x)-xf(x)$

${\frac {dg}{dx}}=\alpha f(x)+xg(x)$

Is there a closed form power series for the answer? --HappyCamper 02:08, 22 June 2007 (UTC)

I think so, and it doesn't look too bad. If we let $f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\ldots$ and $g(x)=b_{0}+b_{1}x+b_{2}x^{2}+\ldots$, then you can substitute that all into the DE to get recurrence relations for $a_{n}$ and $b_{n}$, which are coupled in two groups - the $a_{2m}$ and $b_{2m+1}$, and the $a_{2m+1}$ and $b_{2m}$. Working them out for the first few terms in each series suggests that you could probably find a closed formula, since each one comes in a form of either $p(\alpha )a_{0}$ or $p(\alpha )b_{0}$, where p is a polynomial of degree n (when looking at the nth coefficient). Confusing Manifestation 03:34, 22 June 2007 (UTC)

Mathematica produces a direct closed form solution:

${\begin{cases}f(x)=C_{2}\,e^{{\frac {x^{2}}{2}}-ax+\ln(e^{2ax}-e^{C_{1}})}\\g(x)=f(x){\frac {e^{2ax}+e^{C_{1}}}{e^{2ax}-e^{C_{1}}}}\end{cases}}$

which you could calculate the power series for if you want. --mglg_(talk) 21:43, 22 June 2007 (UTC)

Wow - even better! Thanks! --HappyCamper 23:19, 22 June 2007 (UTC)

Obviously, I could have re-written that answer more simply as:

${\begin{cases}f(x)=C_{2}\,e^{{\frac {x^{2}}{2}}-ax}(e^{2ax}-C_{3})\\g(x)=C_{2}\,e^{{\frac {x^{2}}{2}}-ax}(e^{2ax}+C_{3})\end{cases}}$

--mglg_(talk) 00:13, 23 June 2007 (UTC)

Both were very useful to me. Thanks again!! --HappyCamper 15:05, 23 June 2007 (UTC)

Oops, I was full of it (as usual?): my expressions above are not solutions to your problem but to the much simpler problem without the minus sign. With the correct sign, I fail to find anything very useful (Mathematica spits out integrals of complicated sums of hypergeometric functions). --mglg_(talk) 02:08, 24 June 2007 (UTC)

Why does the extra minus sign complicate things so much? Is it because the symmetry is sort of removed from the system? --HappyCamper 17:59, 24 June 2007 (UTC)

Do maximal ideals of the homogeneous coordinate ring correspond to the points on it?

Hello,

let $Y$ be a projective variety, in $\mathbb {P} ^{n}$ . Let$S$ be the ring of polynomials in the variables $x_{0},\ldots ,x_{n}$. Let $I(Y)$ be the homogeneous ideal, generated by the homogeneous polynomials vanishing on $Y$. Let $S(Y)=S/I(Y)$ be the homogeneous coordinate ring.

Now, how much is this "like the affine case"? I mean, does every (homogeneous?) maximal ideal of $S(Y)$ correspond to a specific point on the projective variety $Y$ ?

Note that I am cautious when I write homogeneous, since I'm not even sure whether or not$S(Y)$ is a graded ring.

Many thanks Evilbu 12:02, 22 June 2007 (UTC)

There is an analog of the Nullstellensatz for projective varieties [3]. Also, for a graded ring A and homogeneous ideal I, $A/I$ will always be graded. See Graded algebra. nadav (talk) 06:26, 24 June 2007 (UTC)

Thank you very much. That's a very clear explanation, but unfortunately it doesn't really goes into depth about the maximal ideals. Do you know something about it. My instructor tried to convince me that points correspond to maximal ideals, but I don't see it?Evilbu 10:38, 24 June 2007 (UTC)

Applications of Pure Math

I've heard several times that lots of things that were once thought the silly, idealistic, nigh-childish fancy of pure mathematicians have found actual uses. I fact, I've heard there's some famous quote along the lines of "All results in mathematics, no matter how abstract, are in danger of being applied." The only examples I can think of are certain parts of number theory in the form of RSA (and related cryptographic systems), and nonEuclidean geometry in the form of relativity. Does anyone know of others? Black Carrot 17:16, 22 June 2007 (UTC)

Maybe non-Archimedean metrics have a place in quantum physics? (Google p-adic analysis physics or see p-adic and adelic physics) iames 19:34, 22 June 2007 (UTC)

Our article on group theory lists, next to cryptography, applications in the sciences. There is also the use of category theory in physics; see for example http://math.ucr.edu/home/baez/categories.html and http://math.ucr.edu/home/baez/quantum/. The concept of Von Neumann algebras can be considered to belong to abstract algebra, but was of course motivated by the possible application to quantum mechanics. See also Wikipedia:Reference desk/Archives/Mathematics/2007 June 18#Useless mathematics for an almost perfectly complementary question.  --Lambiam^Talk 06:22, 23 June 2007 (UTC)

If uses of category theory in theoretical physics (also computer science) are considered applications, then perhaps uses of foundations of mathematics topics in philosophy also qualify. Boolean logic was famously applied as a model for digital electronics. nadav (talk) 06:44, 23 June 2007 (UTC)

Thanks, that's a good start. Lot of good suggestions. Yeah, I remember that discussion. It looks like the only example he got was hyper-inaccessible cardinals, which someone else said actually do have applications. Real shame. Let me know if you think of any more. Black Carrot 04:23, 24 June 2007 (UTC)

June 23

Euclidean spaces

Hello. Suppose $k\geq 3$, x,y are in R^k, |x-y| = d > 0 and r > 0. Prove that if 2r > d, there are infinitely many z in R^k such that |z-x| = |z-y| =r. If 2r = d, there is exactly one such z and if 2r < d there is no such z.

All I can think of is to take a sphere centered at x with radius r and then somehow taking a cone with vertex y and base containing that circle on the sphere for which |z-y| = r. But I have no idea how to prove anything analytically. Any help will be appreciated. --Shahab 10:18, 23 June 2007 (UTC)

For the "no such" part, think triangle inequality, which for R^k is a consequence of the Cauchy–Schwarz inequality. The case of strict equality can be used to handle "exactly one such", although it is not hard to do that from first principles. For 2r > d, visualize two equally large spheres with radius r centred at, respectively, x and y. One sphere is the locus of ||z−x|| = r, the other of ||z−y|| = r. Where do you have that ||z−x|| = ||z−y|| = r? It should not be hard to come up with a formula or construction for some vector z_c parametrized by a value c such that ||z_c-x|| = ||z_c-y|| = r for all c, where different choices of c give you infinitely many different vectors z_c.  --Lambiam^Talk 10:53, 23 June 2007 (UTC)

Thankx for the reply. However I still have trouble in constructing z_c. In the case k = 3 actual spheres intersect and so I can use the result that their intersection is a circle. But I don't know how to do that in general. I think I can handle the other cases. Can you explain the construction of z_c?--Shahab 16:47, 23 June 2007 (UTC)

A fact that may be helpful is that the set of points equidistant from two given points, the perpendicular bisector, is flat. In 2D it's a line; in 3D, a plane; and so on. The set of points at a fixed distance from one given point is a "sphere", in a generalized sense. In 2D it's a circle; in 3D, a spherical surface; and so on. Now intersect. --KSmrq^T 19:23, 23 June 2007 (UTC)

WLOG we can assume that $\mathbf {x}$ lies at the origin and $\mathbf {y} =d\mathbf {e} _{1}+0\mathbf {e} _{2}+0\mathbf {e} _{3}+\cdots +0\mathbf {e} _{k}$. Let $\mathbf {z} (\varphi )=(d/2)\mathbf {e} _{1}+{\sqrt {r^{2}-(d/2)^{2}}}(\mathbf {e_{2}} \sin \varphi +\mathbf {e_{3}} \cos \varphi )$, $0\leq \varphi <2\pi$. These form a circle around $(d/2)\mathbf {e} _{1}$, with the axis along $\mathbf {e} _{1}$ and with radius ${\sqrt {r^{2}-(d/2)^{2}}}$. We verify

${\begin{aligned}||\mathbf {z} (\varphi )-\mathbf {x} ||&={\sqrt {(d/2)^{2}+({\sqrt {r^{2}-(d/2)^{2}}}\sin \varphi )^{2}+({\sqrt {r^{2}-(d/2)^{2}}}\cos \varphi )^{2}}}\\&={\sqrt {(d/2)^{2}+r^{2}-(d/2)^{2}}}=r\end{aligned}}$

and

${\begin{aligned}||\mathbf {z} (\varphi )-\mathbf {y} ||&={\sqrt {(d-d/2)^{2}+({\sqrt {r^{2}-(d/2)^{2}}}\sin \varphi )^{2}+({\sqrt {r^{2}-(d/2)^{2}}}\cos \varphi )^{2}}}\\&={\sqrt {(d/2)^{2}+r^{2}-(d/2)^{2}}}=r\end{aligned}}$.

Did I just do someone's homework? ;-) You could complete it by proving that the number of $\mathbf {z} (\varphi )$, $0\leq \varphi <2\pi$, are infinite. You could also do better than me and find all $\mathbf {z}$ that satisfies the criterion. I only give one circle of solutions, no matter how high the dimension $k$ is. (And find my mistakes. No intentional ones, I promise!) Bonne chance. —Bromskloss 20:31, 23 June 2007 (UTC)

June 24

Shuffling

I heard something about odds that I don't think is right, but I'm not sure I've worked it out correctly. Here's how you set it up: Get a deck of cards, and have someone shuffle it thoroughly. Ask them to name two numbers Ace through King (like Jack and 5, say). Bet them that a Jack and a 5, somewhere in the deck, have at most one card between them. The claim is that you have more than a 99% chance of winning that bet. I think the odds of winning would be less than 82%, probably a fair bit less. What do you think? Black Carrot 00:06, 24 June 2007 (UTC)

Well, you have 16 different jack+five pairs. The probability of a given five being separated by zero or one cards from a given jack is approximately 4/52. The probability of none of the 16 pairs being that close is approximately (1-4/52)¹⁶. So I would say the probability you are looking for should be approximately 1-(1-4/52)¹⁶~72.2%. Taking into account end-of-deck effects, but still approximating the pairs as independent, I end up with 1-(1-202/52²)¹⁶~71.1%. --mglg_(talk) 01:21, 24 June 2007 (UTC)

While the magnitude there sounds about right, I would point out that those 16 pairs are strongly non-independent, most notably that other Js and 5s can be between any one particular pair of them!. This suggests looking at the pairs in a consecutive fashoin, where there are only seven pairs, and the only non independence is that a card between one such pair cannot be between another.

The 8 Js and 5s are in a certain order, and in between them are 7 gaps, with two gaps on either end. To lose the bet for any ordering of those 8, each of the inner 7 gaps need to have at least 2 other cards in it. One can combinatorially compute how many all the gaps are filled up with two cards. For any assignment of pairs of cards to the gaps, one can combinatorially compute how many ways the rest of the cards can be inserted anywhere (including at the ends). Thus the number of shuffles which lose is a product of three terms: # orderings of the 8 cards, # ways separating all 7 adjacent pairs, # ways to insert the rest. Then divide into the total number of ways to arrange 52 cards and subtract from one. There may be other ways of breaking the problem down into independent nested binnings which are easier to calculate, but I strongly suspect that considering the 8 cards sequentially like this will makes things loads easier. Baccyak4H (Yak!) 02:41, 24 June 2007 (UTC)

See, the thing about that is, getting an accurate answer would take a lot of work (even more than Baccyak suggests, since in many orderings of the 8 cards two of the same could land right next to each other) and most answers I could get including the word "about" seem a bit vague. It's hard to tell how close they'd land. So, I tried to get a strict upper bound on it. The way I saw it, we could assume the Jacks had been placed first, then place the 5s. There are four Jacks, and at most 16 cards within two of a Jack, meaning at least 32 remaining. Placing the first 5, the odds are 32/(16+32)=2/3 that it would land away from the Jacks. Assuming all of them landed away from Jacks, the odds for each would be the same but with the 32 (the count of cards still unclaimed that aren't near Jacks) going down once each time. Their product, the lower bound on the odds of losing the bet, would be 1798/9729 = 0.1848, putting an easy upper bound on the odds of winning at 82%. Does that sound right? Black Carrot 03:17, 24 June 2007 (UTC)

I agree with your upper bound, Black Carrot – I just came back here to enter the same thing! I was going to express it as $1-{\binom {32}{4}}/{\binom {48}{4}}={\frac {7931}{9729}}\approx 81.52\%$. As you say, the exact answer is likely to be considerably lower than this, because there are large probabilities of having less than 16 "winning slots" free after placing the jacks. But the upper bound suffices to prove that the "99%" claim is false. --mglg_(talk) 04:12, 24 June 2007 (UTC)

Thanks. Great minds think alike, I guess. :) Black Carrot 04:25, 24 June 2007 (UTC)

That, of course, being equally true about mediocre minds... --mglg_(talk) 04:38, 24 June 2007 (UTC)

I've run a simulation and, modulo mistakes in my implementation, the result is around 73.5%. -- Meni Rosenfeld (talk) 11:06, 24 June 2007 (UTC)

Wow, he hit it about on the head. I've been trying to work it out Baccyak's way, but the result I've gotten is a bit hard to believe. I should have gotten a much tighter upper bound, but instead I got that the odds of winning are strictly less than 94%. I'd like someone to check my work. I get the number of ways of placing eight cards in eight spots as 8!, the number of ways of choosing 14 cards from what remains as (44 C 14), the number of ways of arranging those between the original 8 as 14!, the number of ways of placing the remaining 30 cards as f(30) to be defined in a moment, and the number of ways of ordering those 30 as 30!. The total number of orderings that would lose is greater than that, because it counts nothing that does not lose but leaves out situations where cards of the same type (5 and 5, say) can be closer than two from each other. However, it seems like it should provide a tighter bound than the previous one, because it counts all the same cases and more. f(30) is the number of ways of dropping 30 identical pegs into 9 baskets, which would be a Figurate number of 8 dimensions and length 31. That makes it equal to (38!/30!)/8!. The total product, divided by 52!, simplifies to 44!38!/(52!30!), about 6.5%. Black Carrot 13:29, 24 June 2007 (UTC)

The way I looked at it (to find the upper bound) is to first assume the jacks are evenly spaced, so there are 4 different cards near each jack. This gives us a total of 16 cards of the remaining 48 non-jacks which we're concerned with. The chances that each five will be in those 16 cards is thus 16/48 = 1/3. The chance that each five is not one of those 16 cards is therefore 2/3. This gives us the probability that none of the 4 fives is near a jack as (2/3)⁴ or 16/81 = 19.75%, and the P that one or more five is near a jack is 80.25%. However, if we allow for the possibility that the jacks are near each other, or near the top or bottom of the deck, both would lower the probability of a nearby five to below 80%. StuRat 01:27, 25 June 2007 (UTC)

StuRat, your result is an approximation of the upper bound that has already been reported twice above. Your approximation is to ignore the fact that no more than one five can occupy the same position. This is why you get 65/81 instead of the exact answer 7931/9729 for the upper bound. --mglg_(talk) 01:50, 25 June 2007 (UTC)

Since the question was "is 99% correct" or possibly "is less than 82% correct", an exact answer is not needed, we only need to prove that 99% (or 82%) is above the upper bound to answer the question, which I have done. I believe my method was the simplest approximation which gave an answer accurate enough to answer this question (so simple, in fact, that you can do the math in your head). It is also helpful to confirm earlier answers, especially when using a slightly different method, as a check. StuRat 15:44, 25 June 2007 (UTC)

I can report an exact value for the probability of 3230755238/4389805875 = 0.735967678297..., which is consistent with Meni Rosenfeld's reported experimentally determined 73.5%. What I've done is count, for various lengths, the number of sequences (shuffle prefixes) for each of the patterns *···*@@, *···**J, *···**5, *···*J@, and *···*5@, in which * is any card, J is a Jack, 5 is a 5, and @ is any other card than Jack or 5; additionally, in the sequence no J and 5 must occur as neighbours or almost neighbours. Given all counts for length n for each of these patterns of the sequences with exactly u Js and v 5s, the same counts can be cheaply computed for length n+1. From the resulting data, with a little effort the probability can be determined.  --Lambiam^Talk 16:27, 25 June 2007 (UTC)

Thanks Lambiam! I knew you would come through and enlighten us in the end. --mglg_(talk) 17:55, 25 June 2007 (UTC)

I don't understand. Could you explain that a little more? Black Carrot 04:06, 26 June 2007 (UTC)

A rethink

Carrot, up above you came up with an approximation of about 6.5%, trying my line of reasoning. In hindsight, I realize the exact formulation I suggested was needlessly complicated, but nonetheless (IMO) still on the right track. I tried a rethink and want to share it here.

Forget which cards are which, except for a) eight target cards; b) the rest. This can be seen by noting your problem is isomorphic to selecting eight integers from 1 to 52 without replacement and seeing whether the smallest difference between any two integers is three or greater (three because there are two integers between integers that are three apart). I will tackle this formulation.

How many total ways can this be done, win or lose? This is 52 choose 8. Only the positions of the key cards matter here. Which particular key card is, e.g., J♥ or which non-key card is 2♣ is immaterial.

How can we enumerate the winning shuffles? Here, piggybacking on Lambiam's notation, let "*" be a key card (J or 5, in your example), let "x" be any single non-key card, and let "@" be any sequence (perhaps of length zero) of non-key cards. These are placeholders for the cards' positions, i.e., 1 to 52. Then a winning hand (does not have an interval of at most one card) can be enumerated as such (spacing for clarity):

@ *xx @ *xx @ *xx @ *xx @ *xx @ *xx @ *xx @ * @

Note the last "*" is not followed by "xx". (We could have written "@ xx* @ xx* ...", but the results will be the same.) The "*"s are exchangeable, as are the "x"s, so we can ignore their exact identity. Now each such winning shuffle (or sample of 8 numbers) is completely determined by the "@"s. These form a composition of (52 - 3*(8-1) - 1) = 30 into 9 parts, including the possibility of zeros (if two "*xx" are adjacent). The formula for the number of such compositions of n into r parts is (n + r - 1) choose (r - 1). In this case, it is 38 choose 8.

Thus the number of winning hands is

${\frac {\dbinom {38}{8}}{\dbinom {52}{8}}}$

which reduces to 479446 / 7377825 or about 6.49847%. The losing chance is thus about 93.50153%. I have done a simulation in R, and with 4000000 runs, am getting 6.5090% with SEM of 0.0123%. This is very close to your estimate above. (Code available on request—I am working to generalize this to any number of key cards from any size desk with any size gaps.)

An aside. The change of losing the bet if asked that there would be no cards adjacent to each other (that is, there was at least one card between each of the Js and 5s), then I get ~71.2% I wonder if this is what was really calculated elsewhere in several places. Baccyak4H (Yak!) 19:45, 26 June 2007 (UTC)

"This can be seen by noting your problem is isomorphic to selecting eight integers from 1 to 52 without replacement and seeing whether the smallest difference between any two integers is three or greater (three because there are two integers between integers that are three apart). " If I understand you correctly, this isn't correct, since you're counting, e.g., Jack-something-Jack as a successful hit. Donald Hosek 20:55, 26 June 2007 (UTC)

I think there's been a misunderstanding. If I challenged you to this, I would win if a Jack and a 5 were in close proximity (with about 73.5% likelyhood), and lose if no Jack was near a 5 (with about 26.5% likelyhood.) Whether a Jack is near a Jack, or a 5 near a 5, has no influence on the bet. That's why all the other analyses, except for my use of yours (which arrived at the same fraction you did :) ), specifically mentioned Jacks and 5s, in two groups of four. Black Carrot 02:46, 27 June 2007 (UTC)

J-something-J is a win for the poser (you lose!). There are 8 Js and 5s (in this case) total. Do not be confused between Js and 5s. Here, the important distinction is between 8 key cards (both Js and 5s in your example) and the rest. If (say) 2 Js were only one card apart, you would lose. But you would as well if 2 5s were only one card apart, or any J and 5. Thus you can look at the 4 Js and 4 5s as being from a group of 8 "key" cards. I am reading Carrot's reply to be questioning of Donald's conclusions; if so, I agree.

Can anyone point out a flaw in my reasoning? If not, I can provide a general solution (as alluded to above) to the probabilities of winning such a generalized bet. Baccyak4H (Yak!) 03:24, 27 June 2007 (UTC)

Um, no. Black Carrot's reply (as well as his original problem statement) confirms Donald's observations, and questions yours. You win if at any place there are a J and a 5 (not J and J or 5 and 5) close to each other (separated by at most one card). So the Js and the 5s are different, you can't treat them as just "8 key cards". -- Meni Rosenfeld (talk) 07:33, 27 June 2007 (UTC)

Mystery Series

Something came up on a Facebook forum, and I can't remember the answer to it. Well, I can, but it seems I'm remembering it wrong. Imagine a tower of identical bricks, each shifted to the right as far as it will go without falling off. In other words, the center of balace of each is right on the edge of whatever it's sitting on. How far over is each one, and how far does the tower extend to the side? I remember it being an example of the harmonic series in action, and I remember the result being that, given an impractically tall tower, you can extend the right edge arbitrarily far. I'm working it out now, though, and I'm getting the displacement of each to be 1/2^n from the one below it, with the result that you can never get even a block's length away from the base. As this is patently false, experiment being my guide, I have to assume I made a mistake somewhere. Any ideas? Black Carrot 23:58, 24 June 2007 (UTC)

For the tower to have the maximum stable extension, the edge of each brick should coincide with the combined center of mass of all the bricks above it. If you work this out, brick n (from the top) will extend by 1/(2 n) beyond the brick below it, for unit length bricks. This means that the total overhang of an n-brick tower will be ${\frac {L}{2}}\sum _{k=1}^{n-1}{\frac {1}{k}}$, which does go to infinity with n, albeit very slowly. Not all sequences that start with 1/2 and 1/4 are 1/2ⁿ... --mglg_(talk) 00:36, 25 June 2007 (UTC)

The first Google hit on "harmonic series" bricks is http://www.antiquark.com/2005/04/harmonic-series-and-bricks.html which includes 2 nice pictures. I have seen the problem before and the Google search has many relevant hits. Maybe we should mention this nice problem somewhere (I searched shortly and couldn't find it in Wikipedia). PrimeHunter 00:45, 25 June 2007 (UTC)

On page 3 of the Google search I get:

"In response to a legal request submitted to Google, we have removed 1 result(s) from this page. If you wish, you may read more about the request at ChillingEffects.org."

I haven't seen such a message before. The link doesn't work for me. PrimeHunter 00:49, 25 June 2007 (UTC)

I found a name. Does somebody (not me) want to write Leaning tower of Lire? [4] Or maybe it should just be a section somewhere. PrimeHunter 01:02, 25 June 2007 (UTC)

In reality, I assume the weight of the bricks would creat an enormous amout of stress that would eventually break any material. nadav (talk) 01:31, 25 June 2007 (UTC)

That's what it was! Thanks. Black Carrot 03:35, 26 June 2007 (UTC)

June 25

Additional Mathematics Project Work 2007 (Form 5)

Question: Ang, Bakar and Chandran are friends and they have just graduated from a local university. Ang works in a company with a starting pay of RM2000 per month. Bakar is a sales executive whose income depends solely on the commission he receives. He earns a commission of RM1000 for the 1st month increases by RM100 for each subsequent month. On the other hand, Chandran decides to go into business. He opens a cafe and makes a profit of RM100 in the first month. For the first year, his profit in each subsequent month is 50% more than that of the previous month.

In the second year, Ang receives a 10% increment in his monthly pay. On the other hand, the commission received by Bakar is reduced by RM50 for each subsequent month. In addition, the profit made by Chandran is reduced by 10% for each subsequent month.

1. a) How much does each of them receive at the end of the 1st year? (2 or more method are required for this ques.) [max: besides progressions method, what can i use?? show me..]

b) What is the percentage change in their total income for the 2nd year compared to the 1st year? Comment on the answers.

c) Ang, Bakar and Chandran, each decided to open a fixed deposit account of RM10000 for 3years without any withdrawal. - Ang keeps the amount at an interest rate of 2.5% per annum for a duration of 1month renewable at the end of each month. - Bakar keeps the amount at an interest rate of 3% per annum for a duration of 3months renewable at the end of every 3months. - Chandran keeps the amount at an interest rate of 3.5% per annum for a duration of 6months renewable at the end of every 6months.

(i) Find the total amount each of them will receive after 3years. (ii) Compare & comment on the difference in the interests received. If you were to invest RM10000 for the same period of time, which fixed deposit account would you prefer? Give your reasons.

Further Exploration

2. a) When Chandran’s 1st child, Johan is born, Chandran invested RM300 for him at 8% compound interest per annum. He continues to invest RM300 on each of Johan’s birthday, up to and including his 18th birthday. What will be the total value of the investment on Johan’s 18th birthday?

b) If Chandran starts his investment with RM500 instead of RM300 at the same interest rate, calculate on which birthday will the total investment be more than RM25000 for the 1st time.

PLEASE GIVE ME THE FULL ANSWERS A.S.A.P..THANK YOU..

Go away and do your homework. Come back when you at least have started some of the problems.— Preceding unsigned comment added by 203.49.208.227 (talk • contribs)

Non zero y so that x.y = 0

Hello.

My question is as follows:

If $k\geq 2$ and x is in R^k, prove that there exists y in R^k such that y is non zero but x.y = 0. Is this true if k = 1?

The way I have thought is as follows: R is an integral domain so the result is not true for non zero x if k is 1. For $k\geq 2$ we want that for non zero x,

$\sum _{i=1}^{k}x_{i}y_{i}=0$ there exists a non zero $(y_{1},\ y_{2}\cdots y_{k})$. But this is obvious as the set $\{x_{1},\ \cdots x_{k}\}$ is linearly dependent in R. Hence the result. What I want to know is there some other simpler way of constructing y. Thanks.--Shahab 09:41, 25 June 2007 (UTC)

Without giving it away, consider what it means for a sum to be zero. Some of the summands must be negative. This tells you the sign of some of the components of the vector y. Make another manipulation to easily force the sum to be zero. That will handle the case of k even. The case where k is odd is handled similarly, but with a minor twist.

Seems to me the simplest and clearest approach for k>1 is to construct a specific example. Set $y_{1}=y_{2}=...=y_{k-1}=1$. What value does $y_{k}$ have to take to make x.y equal 0 ? Is the vector y in R^k? If so, you are done. Gandalf61 11:30, 25 June 2007 (UTC)

This approach is pretty much equivalent also.

Gandalf's solution, as presented, fails in the case $x_{k}=0$. Of course, for a nonzero x, at least one coordinate is nonzero, and you can adapt the argument accordingly. -- Meni Rosenfeld (talk) 13:36, 25 June 2007 (UTC)

Ah, yes. Let's just say "details are left to the reader" ! Gandalf61 14:22, 25 June 2007 (UTC)

help me in my add maths project work

how to solve the questions.pls help me.i am lack of time.

Wikipedia's reference desk is not a service to do people's homework, so you should not waste what little time you have asking us to do so. However, if there is anything you do not understand and wish us to clarify, we will be more than happy to help. -- Meni Rosenfeld (talk) 15:08, 25 June 2007 (UTC)

Ellipse

I'm trying to create a drawing program much like paint and the circle drawing tool can obviously draw ellipses from knowing where the mouse's last click position was and where it is now. So how to find out an ellipe's two focus points from it's height and width? I saw nothing about it in the ellipse article, or it was too obscure to decipher. 62.20.156.137 16:51, 25 June 2007 (UTC)

Why do you want to find the foci? The easiest way to draw an axis-parallel ellipse is with the parametric equations $x=a\cos {t}$ and $y=b\sin {t}$, where a is the horizontal radius (half the width), b is the vertical radius (half the height), and t varies from 0 to 2π. However, if you insist on finding the foci, the relevant information is in the first section ("Eccentricity") of the ellipse article - the distance of each from the center, assuming $a>b$, is $a\epsilon$, where $\epsilon ={\sqrt {1-{\frac {b^{2}}{a^{2}}}}}$ is the eccentricity. If $a<b$ you need to reverse their roles. -- Meni Rosenfeld (talk) 17:21, 25 June 2007 (UTC)

counter intuitive probability

A friend of mine blew my mind with a bit of probability the other day, but she did a lousey job explaining it. The scenario is of a game show where one is offered three curtains, one of which is a prize and the others are something else. After one is picked, one of the false curtains is revealed and the contestant is offered the chance to change thier curtain of choice. She said that one should always change curtains, but couldn't articulate why. Is there a name or an article for this scenario?

142.33.70.60 17:45, 25 June 2007 (UTC)

Yes, the article is called Monty Hall problem. --mglg_(talk) 17:52, 25 June 2007 (UTC)

And you are welcome back if you want more persuasion that it's actually true. :-) —Bromskloss 18:48, 25 June 2007 (UTC)

Thanks :) Actually, it turns out that the arguement we were having revolved around a glossed over detail, covered in the article--that it all depends on Monty's rules(If he -always- has to show a goat curtain or not). I was positive that it was 50/50, but now that I know Monty always has to show the curtain, I understand what's going on.142.33.70.60 20:12, 25 June 2007 (UTC)

solve the equation

find the value of x and y where x^y+y^x=17 and x^x+y^y=31

Well I notice that in both equations, we're looking at examples of symmetric functions, meaning that we can exchange $x$ and $y$ without changing the equations. So if there are any solutions (and there are), they'll come in pairs. I'm not sure if that helps at all, though. Donald Hosek 17:59, 25 June 2007 (UTC)

Guess some simple values of x and y and try them. Then keep in mind Donald's tip. --mglg_(talk) 19:01, 25 June 2007 (UTC)

Finding a solution to this problem by guessing is pretty straightforward (even easier if you have a grapher which can graph the equations, the domains of the functions are kind of interesting, actually). I'm starting to wonder if there's any way of leveraging the symmetry to be able to generalize this to solving the systems $x^{y}+y^{x}=k$, $x^{x}+y^{y}=m$. Is there anything interesting to the values of $(k,m)$ which have integral solutions in $(x,y)$? Donald Hosek 19:50, 25 June 2007 (UTC)