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March 24

How to display surds on the TI-84 Plus

Does anyone know? Is there a program which I can download which does it?--124.171.116.21 (talk) 02:32, 24 March 2010 (UTC)[reply]

the volume of a solid bounded by the lines x=0, y=0, z=0 and the plane x+y+z=9

Just checking webwork ... isn't the volume 729/2? I'm attacking this from several angles -- e.g. double integration just plain ole geometric sense (it should be half the cube 9x9x9). But the system says I'm wrong ... John Riemann Soong (talk) 09:01, 24 March 2010 (UTC)[reply]

If you had

x+y\leq 9,\ 0\leq z\leq 9

you'd get a prism which is half the cube. Since you have

x+y+z\leq 9

you get a pyramid which is one third of the prism, or only 1/6 of the cube. Integration also gives 121.5. -- Meni Rosenfeld (talk) 09:21, 24 March 2010 (UTC)[reply]

You've got a pyramid one of whose faces is half of the square. The volume of a pyramid is 1/3 × base × height. So it's 1/3 × (half the square) × (height) = (1/3) × (1/2) × (volume of the cube). Michael Hardy (talk) 14:25, 24 March 2010 (UTC)[reply]

Poincare Conjecture and Ricci flow

Perelman used Ricci flow in proposing a proof of Poincare's conjecture.

Eddington in about 1934 said he proved, or seemed to prove, (as Kumar seemed to earlier) that if Pythagoras' theorem is extended to four or more dimensions then the sign of the fourth or further term is negative. For example, time in Einstein's metric contributes with the opposite sign to the space terms, ds squared = (say) dx1 squared + dx2 squared + dx3 squared - dt squared. Eddington said he did this to say something about Dirac's spin. Eddington's proof (?) involved group theory, but I'm not sure of its elimination of alternatives, for I only saw a semi-popular account of it.

Does this modify or limit the applicability of the Ricci flow method to Perelman's proof of Poincare's conjecture? Is Poincare's conjecture itself affected by the different sign on the fourth dimension, in a metric?

As a University of Sydney student, I had not much spare time to study or ask this.

122.152.132.156 (talk) 10:25, 24 March 2010 (UTC)[reply]

I should preface this by saying I don't know a whole lot about this topic. But the Poincare conjecture is purely a topological statement. It doesn't depend on any metric, even if a particular metric was used to prove it.

Also, the standard definition of a metric requires it be positive definite. Rckrone (talk) 17:47, 24 March 2010 (UTC)[reply]

Agree with above (I'm also not an expert here). Your question sounds similar to Marilyn vos Savant's infamous objection (later retracted) to Wiles' proof of Fermat's last theorem. She objected that the proof uses hyperbolic geometry which isn't allowed because the "real word" uses Euclidean geometry (which is only mostly true). The fact is that mathematicians use whatever abstract notions are appropriate for the setting, regardless of any considerations of the "real world", which is irrelevant in the context of an abstract proof. Staecker (talk) 22:41, 24 March 2010 (UTC)[reply]

The "metric" with -dt^2 is called the Minkowski metric which as Rckrone says is not a true metric in the usual mathematical sense. The geometric symmetry group of Minkowski space is the Poincaré group, if that's what you're thinking of. AFAIK it doesn't have anything to do with the Poincaré conjecture in topology. 66.127.52.47 (talk) 07:04, 25 March 2010 (UTC)[reply]

Beads on a necklace

If you had n beads, all different and unique, then would the number of unique orderings on the necklace, after allowing for rotations and reflection, be n!/2n ? 78.149.167.173 (talk) 22:09, 24 March 2010 (UTC)[reply]

No. If we plug in 2 beads, we get 2!/2(2) = 2/4 = 1/2. Obviously there isn't half a combination. Can you figure out what's wrong with your formula ? StuRat (talk) 00:49, 25 March 2010 (UTC)[reply]

The formula gives the correct answer for n=3 at least. 84.13.22.69 (talk) 14:09, 25 March 2010 (UTC)[reply]

You can cheat using the search function on wikipedia 9or google) and you'll find Necklace (combinatorics) Dmcq (talk) 01:24, 25 March 2010 (UTC)[reply]

The problem is fairly easy, and this article doesn't make it any easier at all. -- Meni Rosenfeld (talk) 09:58, 25 March 2010 (UTC)[reply]

You seem to be quite right, you'd need to know how to solve the problem above long before you ever turned to that article! Dmcq (talk) 14:30, 25 March 2010 (UTC)[reply]

The way I got the n!/2n was, if you had a line of n unique and different objects, then the number of different ways of ordering them would be n!. Join the ends together, then the number of different ways would be reduced by n as you could rotate the string of n beads into n positions, and by a further 2 because you could flip the necklace over. What is wrong with that reasoning? 84.13.22.69 (talk) 14:06, 25 March 2010 (UTC)[reply]

In fact that's quite reasonable but it still goes wrong for 0, 1 or 2 beads! :) For more than 2 you're quite right, you don't get exactly the same pattern flipping the necklace over. Dmcq (talk) 14:30, 25 March 2010 (UTC)[reply]

And the reason is that flipping the necklace over doesn't change the order, with 2 beads or less. StuRat (talk) 17:58, 25 March 2010 (UTC)[reply]

So is the formula correct provided that n>2 ? 84.13.22.69 (talk) 14:45, 25 March 2010 (UTC)[reply]

Indeed. -- Meni Rosenfeld (talk) 15:00, 25 March 2010 (UTC)[reply]

Would it be worth a mathematician noting that on the Bracelet (combinatorics) page, since itys not obvious to the casual reader that the complicatated formula given does reduce down to the above with the limitations given? 84.13.34.56 (talk) 13:14, 26 March 2010 (UTC)[reply]

I gave it a shot, using the material from this Q. Note that I'm not a mathematician. However, I think the original version having been written by a mathematicians was the problem, as this results in articles only readable by other mathematicians. StuRat (talk) 13:43, 26 March 2010 (UTC)[reply]

Unfortunately, you added it to the wrong article. Your example is about (what our articles refer to as) bracelets, not necklaces. Algebraist 13:46, 26 March 2010 (UTC)[reply]

Thanks, I moved it here: Bracelet_(combinatorics)#Example. Although, I don't care for the terms "necklace" and "bracelet", I much prefer "reversible necklace" and "nonreversible necklace", since those terms actually describe the difference. But, if the mathematicians have chosen the vague terms, we may be stuck with them. StuRat (talk) 13:56, 26 March 2010 (UTC)[reply]

So, if we allow rotation, but not flipping it over, what they call a "necklace", is the formula then n!/n (which is the same as (n-1)!) ? What are the restrictions on the range of n (obviously it can't be zero) ? StuRat (talk) 14:00, 26 March 2010 (UTC)[reply]

Except that the beads on a necklace are not assumed to have pairwise different colours, so this formula has nothing to do with the

N_{k}

and

M_{k}

counting functions in the article. The same problem is with bracelet counting by the way, so the example you added there is quite misleading, n!/2n has nothing to do with

B_{n}(n)

.—Emil J. 14:13, 26 March 2010 (UTC)[reply]

Please explain what "pairwise different colours" means, so we can bridge the gap. StuRat (talk) 14:20, 26 March 2010 (UTC)[reply]

Let n = 3, for example. The 3!/3 = 2 things you counted are 012 and 210 (as cyclic orders). The

N_{3}(3)=11

necklaces are 000, 111, 222, 001, 002, 110, 112, 220, 221, 012, and 210. Is it clear?—Emil J. 14:26, 26 March 2010 (UTC)[reply]

It looks like my example assumes each bead is unique (which is stated in the first sentence), while, in general, this assumption is not made. That seems OK to me, as long as the assumption is stated clearly. I can clarify it a bit more, if you like. StuRat (talk) 14:35, 26 March 2010 (UTC)[reply]

I added an example to the necklace article, too: Necklace_(combinatorics)#Example. Is it OK ? Specifically, is calling it a "2D" necklace right, since it closes back on itself, or should it still be considered 1D ? StuRat (talk) 15:34, 27 March 2010 (UTC)[reply]

March 25

Circle

Hey all. I was surfing the web, looking for nothing in particular one night, and I came upon a short math quiz on some website. Most of the questions were pretty easy, but there's one that had me stuck. It goes something like, "I have three semicircles, with radii of 3, 2, and 1 unit respectively. The semicircle with radius 2 is externally tangent to the semicircle of radius 1, and both the semicircles of radii 1 and 2 are internally tangent to the one with radius 3. There is a circle which is externally tangent to the semicircles of radii 1 and 2, and internally tangent to that with radius 3. What is its radius?" At first I thought it had something to do with derivatives, but that just led me to a dead end. Thougths? —Preceding unsigned comment added by 99.13.219.136 (talk) 03:15, 25 March 2010 (UTC)[reply]

Just eyeballing it, it looks to be a bit smaller than the radii 1 circle, maybe (3^½)/2 or 0.866 ? StuRat (talk) 03:32, 25 March 2010 (UTC)[reply]

There is no diagram? Hope I'm getting this pictured correctly, the semi-circle (SC) of radius 3 contains SC 1, SC 2, and circle of unknown radius inside. Very interesting, I'll get back to you on that. --Kvasir (talk) 03:49, 25 March 2010 (UTC)[reply]

I'm getting hung up on why they're explicitly semicircles - the way I'm imagining things laid out (r1 & r2 side-by-side hills, r3 joining over them, the circle of unknown radius nestled in the valley between r1 & r2), the problem would be exactly the same if it was circles of radius 1, 2 & 3. The fact that they explicitly mention "semicircles" indicates that this may be a "gotcha" problem, where the "real" way to solve it is not the obvious one. -- 174.21.224.236 (talk) 05:08, 25 March 2010 (UTC)[reply]

I also drew a full R3, R1 and R2 circles thinking it wouldn't change the problem, except there would be 2 of the same unknown circles fitting on either sides of the R1 and R2 circles. I think one of the keys of solving this is to think outside the box and draw the full circles. I think it's still a geometry problem, not a system of equation type question. --Kvasir (talk) 06:05, 25 March 2010 (UTC)[reply]

The arrangement of semicircles you describe is called an arbelos. But I agree that in this case the fact that they're semicircles seems to be a bit of a distraction and it would be simpler to describe with just circles. Descartes' theorem is probably what you're looking for. —David Eppstein (talk) 07:16, 25 March 2010 (UTC)[reply]

I placed the circles in a Cartesian coordinate system so that C3 centre is (0,0) and C1 centre is (2,0), built a three equations system, solved it and got r and Cr centre... Err, is it OK to put the answer here...? --CiaPan (talk) 07:45, 25 March 2010 (UTC)[reply]

I expect Circles of Apollonius helps - but I haven't read it through yet.And Problem of Apollonius is definitely relevant. -- SGBailey (talk) 11:32, 25 March 2010 (UTC)[reply]

I sort of cheated with the help of Autocad, the radius of the circle is approximately 0.8571 unit. (AutoCad can draw a circle given 3 tangent points, so there must be some sort of logarithm.) I'm still trying to look at the geometric relationships there. Stay tuned. --Kvasir (talk) 14:43, 25 March 2010 (UTC)[reply]

That is accurate to 4 decimal places, but there are no logarithms involved, only rational numbers. Gandalf61 (talk) 15:19, 25 March 2010 (UTC)[reply]

oh yeah... I meant to say algorithm, hahah. --Kvasir (talk) 15:33, 25 March 2010 (UTC)[reply]

Hmm check out Bankoff circle. It's nice not needing to re-invent the wheel, so to speak. --Kvasir (talk) 15:42, 25 March 2010 (UTC)[reply]

That article is little more than a stub, and seems to be poorly written, in that the labels in the discussion don't appear to match those in the diagram (C₁, C₂, C₃, vs. C"₆ ?). I also can't follow what they mean by "r = AB/AC". From the diagram, "r = AB" would be correct. Finally, "R", I assume, is the radius of the red circle in the diagram, which is not the one we want. Would anyone be willing to fix this article and diagram ? StuRat (talk) 17:17, 25 March 2010 (UTC)[reply]

Yeah the Bankoff circle is not the circle referred to by the OP. The only thing that is helpful from the article is the diagram illustrating this particular problem here. --Kvasir (talk) 17:32, 25 March 2010 (UTC)[reply]

In case anyone has problem solving it, the exact value confirmed the previous estimation above using Descartes' Theorem suggested above. The answer is 6/7 ~ 0.8571. This was fun. --Kvasir (talk) 17:59, 25 March 2010 (UTC)[reply]

So my hand drawn sketch and eyeballing the answer were off a bit, but 0.866 is pretty good for that kind of estimate (just over 1% off). StuRat (talk) 18:34, 25 March 2010 (UTC)[reply]

Cute problem. I now think I can express the answer in what some might consider "closed form", although it's a bit hairy. Michael Hardy (talk) 21:41, 27 March 2010 (UTC)[reply]

A matrix M raised to the i-th power (i being imaginary), or the natural base raised to the M-th power (M being a matrix).

What is that? Eliko (talk) 10:36, 25 March 2010 (UTC)[reply]

The second one presumably refers to matrix exponential.—Emil J. 11:11, 25 March 2010 (UTC)[reply]

... and for the first one, if M has a matrix logarithm then the natural way to define Mⁱ would be Mⁱ = exp(i log(M)). Gandalf61 (talk) 11:15, 25 March 2010 (UTC)[reply]

...but that's not unique (even if it exists), is it?—Emil J. 11:21, 25 March 2010 (UTC)[reply]

No. Relatedly, e^M could be interpreted to mean exp(M log(e))=exp(M (1+2niπ)), which is also multivalued, with the standard matrix exponential as one of its values. Algebraist 11:42, 25 March 2010 (UTC)[reply]

For a sufficiently well-behaved matrix, you can use the principal logarithm. -- Meni Rosenfeld (talk) 11:46, 25 March 2010 (UTC)[reply]

So ? The i-th power of a real or complex number isn't unique either, for the same reason. Gandalf61 (talk) 12:16, 25 March 2010 (UTC)[reply]

Eliko, you say on your user page that you're a native speaker of English. Please get clear on this: it's a matrix, not a "matrice". The plural of "matrix" is "matrices". Michael Hardy (talk) 02:44, 27 March 2010 (UTC)[reply]

Yes, you are right. I've fixed it. Eliko (talk) 21:34, 27 March 2010 (UTC)[reply]

Dissections in Riemann Integration

Hi. I'm currently going through my notes on an Analysis course and am a bit confused on a Lemma regarding dissections. It says

"If D₁ and D₂ are any two dissections then $S(f,D_{1})\geq S(f,{D_{1}}\cup {D_{2}})\geq s(f,{D_{1}}\cup {D_{2}})\geq s(f,D_{2})$ , where S(f, D) denotes the upper sum of f wrt D and s(f, D) the lower sum."

My main problem comes from the two extremes, ie that $S(f,D_{1})\geq s(f,D_{2})$ . If we know nothing about these dissections then how can we make such a statement? Thanks. 92.11.43.155 (talk) 15:38, 25 March 2010 (UTC)[reply]

Any upper sum is greater than the integral, and any lower sum is less than the integral, so any upper sum is greater than any lower sum.

This of course is an intuitive explanation for cases when the integral exists - for a proof in the general case, that's exactly what the lemma gives. -- Meni Rosenfeld (talk) 16:48, 25 March 2010 (UTC)[reply]

Thank you Meni! 92.11.43.155 (talk) 17:42, 25 March 2010 (UTC)[reply]

Two dimensional representation of an arbitrary distance matrix

Given a set of items and a complete set of arbitrary pairwise "distances" between them (with the property that d(a,b) = d(b,a)), is there a technique for assigning 2D positions to them, such that the set of pairwise 2D euclidean distances "best" (e.g. in a least squares/expectation maximization sense) matches the set of starting distances? -- 174.21.224.236 (talk) 17:03, 25 March 2010 (UTC)[reply]

Would you want to keep all the distances, or toss out those that don't fit with the rest ? StuRat (talk) 18:30, 25 March 2010 (UTC)[reply]

Yes. From the distance matrix compute a kernel matrix, apply kernel PCA and take the first two components.

To find the kernel matrix, enumerate the items as

\{v_{1},\ldots ,v_{n}\}

. Let

S_{i}=\sum _{j}d(v_{i},v_{j})^{2}\;\!,

S={\frac {\sum _{i}S_{i}}{2n}}\;\!,

\|v_{i}\|^{2}={\frac {S_{i}-S}{n}}\;\!.

Then

K_{i,j}=(\|v_{i}\|^{2}+\|v_{j}\|^{2}-d(v_{i},v_{j})^{2})/2

.

Note that your distance matrix should satisfy the triangle inequality - otherwise the results may be unpredictable.

This is closely related to Nonlinear dimensionality reduction, you may find some relevant information there. -- Meni Rosenfeld (talk) 19:32, 25 March 2010 (UTC)[reply]

Thanks! Nonlinear dimensionality reduction was what I was looking for. A few followup questions: In the proceedure you suggest, am I correct in stating that you transform the distance matrix with the kernel K, and then do PCA on the transformed matrix? If so, I'm a little confused on how to use the first two principle components to compute the (x, y) positions of the original points. For principle component vectors C₁ and C₂, is the coordinate for item v_i simply ( C_1,i, C_2,i), or do I have to "backtransform" the components through the kernel? Additionally, is there a particular reason you selected kernel PCA from the list of techniques at nonlinear dimensionality reduction, and how did you select the kernel function? Finally, the matrix is only "distances" in that it's somewhat analogous to a separation between the two objects. I'm uncertain whether they will always satisfy the triangle inequality. Is there a (simple) way around that, or what do you mean by "unpredicable"? -- 174.31.194.126 (talk) 16:32, 26 March 2010 (UTC)[reply]

The orthogonal diagonalization of K will give you

K=UDU^{-1}

where

U^{T}U=I

and the diagonal entries of D are with decreasing magnitude. Letting

C_{1},\ C_{2}

be the first two columns of U and

\lambda _{1},\ \lambda _{2}

be the corresponding eigenvalues, the coordinates of point i are

\left((C_{1})_{i}{\sqrt {\lambda _{1}}},(C_{2})_{i}{\sqrt {\lambda _{2}}}\right)

.

The techniques of dimensionality reduction typically deal with data that is explicitly embedded in some high-dimensional Euclidean space, and they construct a distance matrix as part of the solution. If this is your "real" problem they may all be applicable, but you have given the distance matrix as part of the problem.

The kernel I've described is the matrix of inner products in the hi-dim Euclidean space in which the data is embedded (if there is one, and assuming wlog that the mean of all points is 0). So the output of the kernel PCA will be equivalent to the projection to a 2-dimensional subspace with the minimal total distance of the points from their projection.

If there is no triangle inequality, the points are not embedded in an Euclidean space and the theory behind PCA is void. The kernel matrix will have negative eigenvalues. It might still work and you may get good results, but I wouldn't count on it.

Multidimensional scaling is based on a different optimization criterion and it might be more suitable for arbitrary dissimilarity matrices (maybe the end result is the same for both, I'm not sure).

If the only distances that are meaningful are local, between nearby points, then Isomap and MVU are worth a look. Others might be good too but I have no familiarity with them (as opposed to these two with which I have little). -- Meni Rosenfeld (talk) 20:32, 27 March 2010 (UTC)[reply]

Perhaps I'll say a few more words on how this works.

PCA does the following:

Start with an $m\times n$ matrix X, representing a set of n points in $\mathbb {R} ^{m}$ whose total is 0.
Calculate the kernel matrix $K=X^{T}X$ .
Let $K=UDU^{-1}$ , $Y=D^{1/2}U^{T}$ .
It can be shown that Y (as a matrix whose columns represent vectors) is just a rotation of X, so they are geometrically equivalent. However, in Y the entries are uncorrelated and are in decreasing significance, so you can take $k$ entries of each vector and obtain a low-dimensional approximation for the data.

Kernel PCA is the same, but there is no explicit embedding in hi-dim space; rather, you obtain the kernel matrix some other way, and it represents what

X^{T}X

would be if you did embed the data explicitly. Taking all components will give you an n-dimensional embedding of just those n-points, equivalent to the subspace spanned by those points in some even higher-dim space. Taking a few components will give you an approximation, as usual.

Any positive semidefinite matrix K is valid. However, you need every row of K to sum to 0 (corresponding to the vectors totaling 0), if it's not this way to begin with you need to use the trick described in Kernel PCA to make it so.

In my response I have used the matrix that follows necessarily if we assume that all distances given should be observed. Many dim-reduction methods assume that only local distances need to be observed, and the key step is constructing a kernel which observes the local distances and "unfolds" the data to describe the global structure with as few dimensions as possible. -- Meni Rosenfeld (talk) 09:58, 28 March 2010 (UTC)[reply]

Normed field

I am looking at a problem in a book, which I know how to do, except that I wonder if the question is slightly off. The full question is:

Prove that in a normed field the following assertion holds: Let <a_n> be a Cauchy sequence, but not a null sequence. Prove there exist a number c > 0 and a positive integer N such that for all n > N either a_n > c or a_n < -c.

My question is, does the very last part make sense in an arbitrary normed field, a_n > c or a_n < -c? It does not say c is a real number. It just says it is a number, which could mean a field element. I mean, I think I can think of a counterexample, which would be the field of 2 elements with trivial norm and the sequence 1, 1, 1, 1, ... which is Cauchy but not null. However, there is no such c. I guess in such a field c > 0 may not even make sense? I don't know. What's going on here? Thanks! StatisticsMan (talk) 17:52, 25 March 2010 (UTC)[reply]

What, in the opinion of your book, is a normed field? Algebraist 17:59, 25 March 2010 (UTC)[reply]

(e/c) The statement as given only makes sense in an ordered field (interpreting c as a field element rather than a real number), and for it to hold you'd also need some compatibility conditions linking the order to the norm (which, I suspect, would imply that the field is Archimedean, and therefore a subfield of the reals, so you are back to square zero). Presumably the book wanted to say that the norm of a_n is at least c.—Emil J. 18:11, 25 March 2010 (UTC)[reply]

...where, with regard to Algebraist's question, I assumed that a normed field means a field equipped with an absolute value (algebra).—Emil J. 18:23, 25 March 2010 (UTC)[reply]

The book definition of a normed field is a field with a norm on it. The definition of field and norm are standard. Now that I've thought about it, I believe it's just that the author took a theorem about real numbers and didn't think to change it to work for a general normed field. So, I assume it should just be changed to "there exists a REAL number c > 0" and "such that for all n > N we have ||a_n|| > c". Thanks for your help. StatisticsMan (talk) 23:59, 25 March 2010 (UTC)[reply]

Periodic formula of an EKG shape?

Are there any formulas that would give me a repeating EKG type shape? --70.167.58.6 (talk) 19:07, 25 March 2010 (UTC)[reply]

A function of the form

f(t)=a_{0}+\sum _{k=1}^{n}(a_{k}\cos(k\omega t)+b_{k}\sin(k\omega t))

should give a good fit. The parameters

\omega ,a_{k},b_{k}\;\!

can be computed from data. See also Fourier series and Fourier transform. -- Meni Rosenfeld (talk) 19:40, 25 March 2010 (UTC)[reply]

In reality though, the heartbeat is never perfectly periodic; although it is often very close. We are all human and we all have quirks in our workings. A Fourier Series expansion is only any good for a short term model, under constant conditions. For example, the subject must not move, must not sleep, must not get excited, etc. •• Fly by Night (talk) 23:18, 25 March 2010 (UTC)[reply]

Reverse engineering a graph

This is a broader follow up to my question above. Are there any websites or apps that can reverse engineer a graph? For example, I draw my EKG shape in vector with Adobe Illustrator. Could I upload that image to a site that would spit out a trig formula that is similar to that curve? If I do it with a vector app, would there be some sort of PostScript data I could extract to make that curve as a trig function? --70.167.58.6 (talk) 19:12, 25 March 2010 (UTC)[reply]

Curve fitting has links, and says that it's a common feature for statistical programs. -- Finlay McWalter • Talk 19:16, 25 March 2010 (UTC)[reply]

Since the function is known to be roughly periodic, the generic techniques in that article are not the right tool for the job. You should attempt a Fourier transform. -- Meni Rosenfeld (talk) 19:44, 25 March 2010 (UTC)[reply]

Number of non-isomorphic graphs in n vertices?

Is there a known result to the following question; How many graphs are there in n vertices such that all of the graphs are non-isomorphic to each other graph? A math-wiki (talk) 23:49, 25 March 2010 (UTC)[reply]

OEIS: A000088. —David Eppstein (talk) 00:18, 26 March 2010 (UTC)[reply]

I figured as much, most of what that linked to was beyond my knowledge but em I correct in thinking the formula for the sequence they arrived at was $f(n)={\frac {2^{\frac {n(n-1)}{2}}}{n!}}$ A math-wiki (talk) 00:51, 26 March 2010 (UTC)[reply]

No. Stopping reading a formula at the first * sign is not a good route to understanding it. Algebraist 01:06, 26 March 2010 (UTC)[reply]

Then I must inquire as to what my result is missing. I arrived at that formula via a counting argument. I first asked how many possible edges (max) does a graph of n vertices have? This can by shown to be the sum of the first n-1 integers greater than 0 or ${\frac {n(n-1)}{2}}$ from the formula for sum of the first n integers. Then when thinking about the 'edge list' for any graph of n vertices any of those edges is either present or absent, and this there are 2 choices per edge in the 'list' and furthermore these choices are independent of each other. This gives $2^{\frac {n(n-1)}{2}}$ possible lists. This however counts the same graph numerous times as their are multiple ways to name some graphs (e.g. symmetric graphs) such that their 'edge list' remains unchanged, and furthermore there are different 'edge lists' that are, in fact, isomorphic to one another. The number of times each graph counted is precisely the number of ways its vertices can be named, since for any graph to be isomorphic to another, their must exist a bijection between their vertices that preserves their 'edge list.' Naming them from an ordered list (e.g. 1,2,3...) we can see their are n choices for the first name, n-1 for the second and so forth, so their are n! was to name each graph, so I concluded that $f(n)={\frac {2^{\frac {n(n-1)}{2}}}{n!}}$ gave the correct count for n vertices (not at most n, exactly n) It doesn't consider graphs with loops or with multiple edges between the same pair of vertices, but it does allow for non connected graphs and lone vertices with no edges connected with them. A math-wiki (talk) 02:00, 26 March 2010 (UTC)[reply]

If every graph were asymmetric (like the Frucht graph, having no nontrivial graph automorphisms), then

{\frac {2^{\frac {n(n-1)}{2}}}{n!}}

would be the correct formula: there are

2^{\frac {n(n-1)}{2}}

ways of writing down an adjacency matrix for a graph, and n! different adjacency matrices that result from the same graph. But when a graph has some symmetries, then there are fewer adjacency matrices that it can generate. Most graphs (in a technical sense, almost all of them) are asymmetric, so the formula is close to accurate, but it is not exact. The precise correction terms needed for situations like this, where you're trying to count things that may have symmetries, are given by the Pólya enumeration theorem.

There's also a very simple explanation for why your formula can't possibly be the correct answer: it's not an integer. The numerator has only factors of two in its prime factorization, while the denominator has other prime factors that can't be cancelled by anything in the numerator. As a non-integer it can't be the answer to a counting problem such as this one. —David Eppstein (talk) 02:39, 26 March 2010 (UTC)[reply]

Ok, that makes sense now, I should've noticed that it'd give non-integer answers most the time. A math-wiki (talk) 15:55, 28 March 2010 (UTC)[reply]

March 26

unsolved problems

are these problems still unsolved?

Equichordal Points: Can a closed curve in the plane have more than one equichordal point?
Is pi+e irrational?
(Tiling the Unit Square)
Will all the 1/k by 1/(k+1) rectangles, for k>0, fit together inside a 1 X 1 square?

—Preceding unsigned comment added by 208.79.15.130 (talk) 01:01, 26 March 2010 (UTC)[reply]

no(no) yes yes. I'm being minimalist today :) Dmcq (talk) 11:28, 26 March 2010 (UTC)[reply]

Dcmq, are you saying that the problem of whether pi plus e is irrational has been solved? Can you be specific? Who and when? Michael Hardy (talk) 02:42, 27 March 2010 (UTC)[reply]

I think he says yes, it is still unsolved. 66.127.52.47 (talk) 07:49, 27 March 2010 (UTC)[reply]

Ok,if pi+e is still unsolved problems,does this applied for all irrational numbers?and where can one look in wikipedia for this subject?respectfully. —Preceding unsigned comment added by 208.79.15.130 (talk) 19:47, 27 March 2010 (UTC)[reply]

See Irrational number. -- Meni Rosenfeld (talk) 21:05, 27 March 2010 (UTC)[reply]

Navier-Stokes equation

A University of Sydney student with not much spare time asks another question. If a metric is defined on solutions to the Navier-Stokes equation, then do 'distances' that diverge allow classification as turbulence? Then is fractal geometry or nonlinear dynamics the way to classify turbulence of Navier-Stokes solutions, a 'diverging' metric allowing turbulence to be inferred? Has this approach already been tried long ago? 122.152.132.156 (talk) 05:53, 26 March 2010 (UTC)[reply]

Just about everything has been tried. I don't understand much of this article but you might find it interesting. 66.127.52.47 (talk) 08:44, 28 March 2010 (UTC)[reply]

Wiener process properties

Hi, do you know how to calculate the expected value of the absolute value of the correlation between the time and the value of a Wiener process of some length t? I.e. for a sample function f what I mean would be

$\left\vert {\frac {Cov(time,value)}{Var(time)\cdot Var(value)}}\right\vert =\left\vert {\frac {{\frac {1}{t}}\cdot \int _{0}^{t}{ds\ s\cdot f(s)}-{\frac {t}{2}}\cdot {\frac {1}{t}}\cdot \int _{0}^{t}{ds\ f(s)})}{({\frac {t^{2}}{3}}-({\frac {t}{2}})^{2})\cdot ({\frac {1}{t}}\cdot \int _{0}^{t}{ds\ f(s)^{2}}-({\frac {1}{t}}\cdot \int _{0}^{t}{ds\ f(s)})^{2})}}\right\vert$

Thank you very much in advance. Hurugu (talk) 07:38, 26 March 2010 (UTC)[reply]

Two remarks that don't answer the question: (1) The correlation between X and Y is

{\frac {{\text{cov}}(X,Y)}{\sqrt {{\text{var}}(X){\text{var}}(Y)}}},

with the radical in the denominator. That makes correlation dimensionless. (2) In order to speak of the correlation between time and something else, you'd need to think of time as a random variable, with a probability distribution. Without that, it's unclear at best what the question means. Michael Hardy (talk) 19:15, 26 March 2010 (UTC)[reply]

.... OK, from your formula is looks as if you're thinking of time as uniformly distributed between 0 and t. Michael Hardy (talk) 19:18, 26 March 2010 (UTC)[reply]

Your proposed way of computing the variance of the value of the Wiener process looks weird. Squaring the density function (if that's what you intend ƒ to be) is not done. And why integrate from 0 to the time. The value of a Wiener process is not constrained to lie in that interval. And do you mean the value at the time that's uniformly distributed in that interval? Or the value at time t? Or what? The question is unclear. Michael Hardy (talk) 19:21, 26 March 2010 (UTC)[reply]

Thank you for trying to answer, I am sorry for the confusing phrasing (and the missing square root in the denominator). $f$ is not meant to be interpreted as a density function. Maybe I can explain the question better in terms of a discrete problem: Take a sample function of the Wiener process, and then pick the values at the times ${\frac {1}{2}}{\frac {t}{n}}$ , $(1+{\frac {1}{2}}){\frac {t}{n}}$ , $(2+{\frac {1}{2}}){\frac {t}{n}}$ , ..., $(n-{\frac {1}{2}}){\frac {t}{n}}$ for some natural number $n$ . Together with the corresponding times they can be thought of as points in a plane, like the patterns in the top right image in the article Correlation. I am looking for the expected value of the absolute value of this correlation, for the limit $n\rightarrow \infty$ . Hurugu (talk) 22:19, 26 March 2010 (UTC)[reply]

OK, I think the question is clear now. In effect the time is uniformly distributed between 0 and t and you mean the value of the Wiener process at that random time. Maybe more later.... Michael Hardy (talk) 00:17, 27 March 2010 (UTC)[reply]

One of the complications is that you said "absolute value".

If (capital) T is the time that is a random variable uniformly distributed between 0 and (lower-case) t, then cov(T, B_T) = 0. But for any particular sample path, the correlation is not 0; rather it is a random variable whose expected value is 0. But now we want the expected value of its absolute value. To be continued..... Michael Hardy (talk) 01:43, 27 March 2010 (UTC)[reply]

....and now I see that by ƒ you say you mean the sample function. Now that that's clear, the correlation you wrote above makes sense. Michael Hardy (talk) 02:36, 27 March 2010 (UTC)[reply]

This now seems like a harder problem than I initially thought it was. Possibly it can only be done numerically. More later maybe.... Michael Hardy (talk) 18:54, 27 March 2010 (UTC)[reply]

OK, maybe I'll have something shortly.... "Hurugu", you haven't enabled Wikipedia email. Are you still there? Michael Hardy (talk) 19:45, 30 March 2010 (UTC)[reply]

Still here, and enabled email. Hurugu (talk) 20:55, 30 March 2010 (UTC)[reply]

Here's something whose details I haven't worked out yet. Consider the line y = ms, where m is chosen so as to minimize the sum of squares of residuals:

\int _{0}^{t}(B_{s}-ms)^{2}\,ds,

where B_s is the value of the Brownian motion at time s. Then we can partition the total sum of squares

\int _{0}^{t}B_{s}^{2}\,ds

as the sum of an "explained" part

\int _{0}^{t}(ms)^{2}\,ds

and an "unexplained" part

\int _{0}^{t}(B_{s}-ms)^{2}\,ds.

Since m is so chosen as to minimize the unexplained part, some consequences follow, one of which is that

\int _{0}^{t}B_{s}-ms\,ds=0

(which explains why the total sum of squares really is the sum of the two other expressions I put above), and another consequence is that the square of the correlation should be the explained part of the sum of squares divided by the total sum of squares. Michael Hardy (talk) 00:37, 31 March 2010 (UTC)[reply]

.....OK, a bit more detail. The value of m that minimizes

\int _{0}^{t}(B_{s}-ms)^{2}\,ds

can be found by observing that

\int _{0}^{t}(B_{s}-ms)^{2}\,ds=\int _{0}^{t}B_{s}^{2}\,ds-2m\int _{0}^{t}sB_{s}\,ds+m^{2}\int _{0}^{t}s^{2}\,ds,

then differentiating with respect to m, setting that equal to 0, and solving for m, getting

m={\frac {\int _{0}^{t}sB_{s}\,ds}{\int _{0}^{t}s^{2}\,ds}}

Now substitute that for m in the expression

\int _{0}^{t}(B_{s}-ms)ms\,ds,

(But write it as

{\frac {\int _{0}^{t}rB_{r}\,dr}{\int _{0}^{t}r^{2}\,dr}}

so that the letter s won't be overworked), and you find that

\int _{0}^{t}(B_{s}-ms)ms\,ds=0\,

because of the way m was chosen. So now

{\begin{aligned}{\text{total sum of squares}}&=\int _{0}^{t}B_{s}^{2}\,ds\\&=\int _{0}^{t}((B_{s}-ms)+ms)^{2}\,ds\\&=\int _{0}^{t}(B_{s}-ms)^{2}\,ds+2m\int _{0}^{t}(B_{s}-ms)s\,ds+m^{2}\int _{0}^{t}s^{2}\,ds\\&=\int _{0}^{t}(B_{s}-ms)^{2}\,ds+0+m^{2}\int _{0}^{t}s^{2}\,ds\\&={\text{unexplained (or residual) sum of squares}}+0+{\text{explained sum of squares}}.\end{aligned}}

Now

{\begin{aligned}{\text{square of correlation}}=R^{2}&={\frac {\text{explained sum of squares}}{\text{total sum of squares}}}\\&={\frac {m^{2}\int _{0}^{t}s^{2}\,ds}{\int _{0}^{t}B_{s}^{2}\,ds}}=\left({\frac {\int _{0}^{t}sB_{s}\,ds}{\int _{0}^{t}s^{2}\,ds}}\right)^{2}\cdot {\frac {\int _{0}^{t}s^{2}\,ds}{\int _{0}^{t}B_{s}^{2}\,ds}}\\[6pt]&={\frac {\left(\int _{0}^{t}sB_{s}\,ds\right)^{2}}{\int _{0}^{t}s^{2}\,ds\int _{0}^{t}B_{s}^{2}\,ds}}.\end{aligned}}

So the thing that is random and Gaussian gets squared, and appears in both the numerator and the denominator, so we've got some non-linearity here. How then, do we find the probability distribution of R = the absolute value of the correlation? I haven't worked that out and I still don't know if it's going to require numerical methods. Michael Hardy (talk) 02:32, 1 April 2010 (UTC)[reply]

....and now I'm thinking, maybe the proposed idendity

{\text{square of correlation}}=R^{2}={\frac {\text{explained sum of squares}}{\text{total sum of squares}}}

isn't true in the usual sense of correlation unless you do the least-squares fit of both the slope and the intercept. But also, maybe the usual sense of correlation isn't the one that one should use in this context, where one knows that B₀ = 0 exactly. Michael Hardy (talk) 04:00, 1 April 2010 (UTC)[reply]

Wow, thanks for all the effort! Numerically, by running a kind of random walk, I get a value of about 0.595. Hurugu (talk) 17:05, 1 April 2010 (UTC)[reply]

Opposite Category of Sets

Let S be the category of sets. Consider the category with sets as objects and morphisms A -> B as preimages of functions f from B to A, is this the same as the opposite of S? Thanks:) 66.202.66.78 (talk) 10:26, 26 March 2010 (UTC)[reply]

What do you mean by preimage of f?—Emil J. 11:16, 26 March 2010 (UTC)[reply]

OK, here's a guess: If ƒ:B → A, what if we say

P(B) is the set of all subsets of B,

P(A) is the set of all subsets of A,

P(ƒ) is the function from P(A) to P(B) defined by

P(f)(C)=\{b\in B:f(b)\in C\}{\text{ for }}C\in P(A).\,

Does that give us an opposite category? Michael Hardy (talk) 02:27, 27 March 2010 (UTC)[reply]

PS: I don't know any standard definition of the concept of "preimage" of a function. Michael Hardy (talk) 02:30, 27 March 2010 (UTC)[reply]

Michael Hardy, I think your talking about the contravariant power set functor P from SET to SET. This is not the same as the opposite of SET, which has exactly the same objects and arrows as SET, with the domain of a function f:A->B equal to B instead of A. Composition is reversed, so fg (in the opposite category) is equal to gf (in SET). However P is covariant as a functor from the opposite of SET to SET. I don't know what the OP meant by preimage of a function. Money is tight (talk) 06:40, 27 March 2010 (UTC)[reply]

Do you generally take it to be part of the definition of opposite category that it has exactly the same objects? I had thought the category of Boolean algebras and Boolean homomorphisms is the opposite of the category of Stone spaces and continuous functions. Michael Hardy (talk) 18:56, 27 March 2010 (UTC)[reply]

The opposite of a category is just the same category with arrows reversed. Stone duality states that the category of Boolean algebras is equivalent to the opposite of the category of Stone spaces (and vice versa), not equal. Algebraist 19:04, 27 March 2010 (UTC)[reply]

Op here, sorry I was very sleepy when I posted this. What I meant was this,

let f:A -> B, define a function D(f):P(B) -> P(A) so D(f)(r) is the set of all x in A so that f(x) is in r. Then, D(fg) = D(g)D(f). Define a category, with Hom(B, A) the image of Hom(A, B) under D. Is this new category the opposite category of sets. *By preimage of functions, I meant taking sets in B to preimages of them in A for a given f; sorry this was so unclear. I apologize if this still lacks sense:) 66.202.66.78 (talk) 11:11, 31 March 2010 (UTC)[reply]

Finite differences in exponential powers

Hi. I took positive integers starting from 1 to the power of y, in the order of 1^y - 2^y, 3^y - 2^y and so on. I did this up to an exponent of 6, found the differences between the powers (first differences), then the differences between those (second differences), and so on, always subtracting the previous number from the next number. For example, the first differences for x₂² - x₁²... starting from 1² - 0 were 1, 3, 5, 7, 9, 11, and so on. Here are some of the things I've found by doing this:

x²

First differences are odd numbers.
Second differences are all 2.

x³

First differences are prime numbers, or multiples of prime numbers.
Second differences begin at 6 and are all multiples of 6.
Third differences are all 6.

x⁴

First differences have a final digit of 1, 5, or 9.
The remaining inverse pyramid of differences appear to be almost random; some examples include 14, 40, 306, 820, 154, 38, 112, 148, -74, 24, 4, -106, 324, -576, -1116, etc.
Fourth differences end with 4, 8 or 6.

x⁵

All first differences have a final digit of 1.
All second, third and fourth differences have a final digit of 0.
All fourth differences are multiples of 120, and start at 240.
All fifth differences are 120.

x⁶

All first differences are odd.
All second differences have a final digit of 2, and a second-final digit of 6, 0 or 1.
All remaining differences end with 0.
All sixth differences are >700 and <800.

So, my question is, what is the significance of this? Can it be applied to other areas of mathematics, and do we have an article excliciptly on this phenomenon? Thanks. ~A H 1^(TCU) 11:47, 26 March 2010 (UTC)[reply]

here's an idea...work out

(n+1)^{2}-n^{2}

. Use that to work out why the diffs are odd, and the second differences are 2. Then try the same thing for other powers. If you start out with n^k, and take differences, what is the highest power of n that appears? What does this mean if you take differences k times? Tinfoilcat (talk) 12:12, 26 March 2010 (UTC)[reply]

... and check your working. 4th differences of x⁴ sequence all have the same value; and, in general, nth differences of xⁿ sequence will be constant. Gandalf61 (talk) 13:04, 26 March 2010 (UTC)[reply]

Could this problem have any applications for dimensions above 3, or even manifolds? ~A H 1^(TCU) 00:13, 27 March 2010 (UTC)[reply]

"Multiples of prime numbers"?? What number is not a "multiple of a prime number"? Michael Hardy (talk) 01:44, 27 March 2010 (UTC)[reply]

Try Finite difference as a starting point. These patterns are well known but are rarely covered in standard math curricula, at least at an elementary level. This leads to them being rediscovered frequently. The nth differences of xⁿ is n!.--RDBury (talk) 03:03, 27 March 2010 (UTC)[reply]

This can also be understood with derivatives. In particular the fact that d(xⁿ)/dx = nx^n-1. Rckrone (talk) 06:59, 28 March 2010 (UTC)[reply]

I need your knowledge in all mathematical realms you are familiar with.

Let's assume that the concept of "even fraction" is defined as an irreducible fraction whose denominator is even. Unfortunately, no unique definiendum is received from this definition, even not from any of its sub-definitions referring to a given even denominator. Let's assume we would like to receive a unique definiendum. Fortunately, we know that there is a 'natural' surjection - from the class of pairs of an even denominator with an odd nominator - on the class of even fractions which are received by the original definition; Thanks to this surjection, we can receive a unique definiendum of "even fraction", by replacing the previous "indefinite" definition by a "definite" definition, which is received by deviding the original definition into sub-definitions, each of which defines the (unique) "even fraction" as the (unique) irreducible fraction whose given denominator is even and whose given nominator is odd, in such a way that this devision of the original definition into sub-definitions - succeeds to preserve the original class of "even fractions" received by the original "indefinite" definition.

Ignoring the very issue of "even fractions" (and fractions at all), do you know of any (well-known) similar process, in any mathematical realm you are familiar with? i.e., a process in which the classical definition of Y (whatever Y is) does not yield a unique definiendum; however, thanks to the existence of a 'natural' surjection (whatever this 'naturality' means) - from the class of X's - on the class of Y's received from the original definition, we can receive a unique Y, by replacing the previous "indefinite" definition by a "definite" definition, which is received by deviding the original definition into sub-definitions, each of which refers to a given X (by which Y can be defined uniquely), in such a way that this devision of the original definition into sub-definitions - succeeds to preserve the original class of Y's received by the original "indefinite" definition.

HOOTmag (talk) 12:19, 26 March 2010 (UTC)[reply]

Category theory is the closest, but overall it sounds to me more like All your base are belong to us :) Dmcq (talk) 18:07, 26 March 2010 (UTC)[reply]

More details? Does category theory deal with the uniqueness of definiendum? HOOTmag (talk) 18:42, 27 March 2010 (UTC)[reply]

I'd have guessed "even fraction" would mean one where the numerator is even, since then even numbers would be even fractions. Michael Hardy (talk) 19:11, 26 March 2010 (UTC)[reply]

The examples given here for "odd fractions" are f/3, f/5, f/7, etc. HOOTmag (talk) 18:42, 27 March 2010 (UTC)[reply]

I'm having trouble understanding what you're getting at, but maybe you're just talking about a "coding" or "construction" (as in constructive logic)? 66.127.52.47 (talk) 23:17, 26 March 2010 (UTC)[reply]

Any connection between my request and "coding" or "construction"? HOOTmag (talk) 18:42, 27 March 2010 (UTC)[reply]

In your example, you code the even fraction as a pair of integers, (even, odd). 66.127.52.47 (talk) 18:49, 27 March 2010 (UTC)[reply]

I code nothing. Look again at my first paragraph, where I explain how I explicitly define the even fraction (without coding anything), so that I receive a unique definiendum. HOOTmag (talk) 19:55, 27 March 2010 (UTC)[reply]

Hm. We could also say that you gave an axiomatic definition of an even fraction, then described a structure (the set of (even,odd) integer pairs) that interprets the definition. Does that help? There is a topic called constructive type theory that might also reach towards what you might be getting at. Unfortunately, our article about it is very technical. 66.127.52.47 (talk) 02:26, 28 March 2010 (UTC)[reply]

Curve Sketching

Hello. For some function f, f(a) = r, where r is a constant term; f'(x) is undifferentiable at x = a; $\lim _{x\rightarrow a^{-}}f'(x)=-\infty$ ; $\lim _{x\rightarrow a^{+}}f'(x)=\infty$ . While sketching a graph with the information above, is there a plausible function where $\lim _{x\rightarrow a^{-}}f(x)=-\infty$ and $\lim _{x\rightarrow a^{-}}f(x)=-\infty$ while f(a) = r? If so, name the function. Thanks in advance. --Mayfare (talk) 20:21, 26 March 2010 (UTC)[reply]

There are definitely functions that satisfy the conditions you mentioned: f'(x) is undifferentiable at x = a;

\lim _{x\rightarrow a^{1-}}f'(x)=-\infty

;

\lim _{x\rightarrow a^{1+}}f'(x)=\infty

. There is the classic $f(x)=1/(x-a)$ for example.

You can then define the function where it is defined at f(a)=r (a single point) such that you have a piecewise function. --Kvasir (talk) 22:01, 26 March 2010 (UTC)[reply]

Can there be a cusp at x = a with the information in the first sentence? --Mayfare (talk) 12:50, 27 March 2010 (UTC)[reply]

Yeah, that's possible too, and if you want the function to be continuous then it would have to be that way. For example let

f(x)={\sqrt {R^{2}-(x-a+R)^{2}}}+r

for a-R ≤ x ≤ a and

f(x)={\sqrt {R^{2}-(x-a-R)^{2}}}+r

for a ≤ x ≤ a+R, for some radius R > 0, which is two quarter circles up against each other. Rckrone (talk) 17:45, 27 March 2010 (UTC)[reply]

March 27

Lattice of quotient groups

Instead of using normal subgroups, we'll think of all partitions of a group corresponding to congruence relations. If I partially order them by Q1 is smaller than Q2 iff Q1 is finer than Q2, does the set form a complete lattice? Subgroups do... Money is tight (talk) 07:06, 27 March 2010 (UTC)[reply]

A congruence relation on a group is the same thing as a normal subgroup, isn't it? Algebraist 10:48, 27 March 2010 (UTC)[reply]

A partition is uniquely determined by the normal subgroup of elements congruent to the identify, so I think we can just look at those. A partition is finer than another one if the corresponding normal subgroups are contained in each other (one-line proof omitted). The converse is obviously true (the normal subgroup is one of the parts). Therefore, the set of partitions of a group will be a complete lattice iff the set of normal subgroups of a group is one, which it is. So the answer to your question is: yes. --Tango (talk) 20:29, 27 March 2010 (UTC)[reply]

maruschka doll and half-life - fractals?

Are these things fractals? What if you find two dolls within each maruschka doll, would it be a fractal? And the half-life of radioactive materials, if I plot it, is it a fractal?--Quest09 (talk) 18:32, 27 March 2010 (UTC)[reply]

I'd say the nesting dolls are fractal, especially if each is identical to the others, but they aren't always like that: [1]. I'm not quite following the half-life argument. Are you talking about the decay of radioactive elements ? There has to be some pattern which repeats at different scales, and I'm not sure if a radioactive decay curve is complex enough to be called a "pattern". After all, a straight line could be considered fractal if you wanted to include it. StuRat (talk) 18:43, 27 March 2010 (UTC)[reply]

I think a fractal is a set whose fractal dimension is not an integer. 66.127.52.47 (talk) 01:53, 28 March 2010 (UTC)[reply]

The first four iterations of the Koch snowflake

I don't agree with that. If we look at the illustration to the right, that shows the first 4 iterations of a fractal, with ~~dimensions~~ iterations 1-4 (or is it 0-3) ? I wouldn't normally call just an equilateral triangle a fractal, because there isn't yet any repetition of the pattern on a smaller scale. But the shapes after that are fractals, IMHO, and each has an integer fractal dimension. StuRat (talk) 14:16, 28 March 2010 (UTC)[reply]

All iterations have fractal dimension 1, and the limiting curve has dimension 1.26... . I think usually only the limiting curve is considered a fractal, not any finite iteration. Taking e.g. the 3rd iteration, it has no self similarity - it contains no smaller copies of itself, but rather of the 2nd iteration. -- Meni Rosenfeld (talk) 14:37, 28 March 2010 (UTC)[reply]

How is that 1.26 calculated ? StuRat (talk) 19:28, 28 March 2010 (UTC)[reply]

In each iteration, the scale is decreased x3 and the relative length is increased x4, so it's

\log _{3}4\;\!

. Some more information is available in fractal dimension and Koch snowflake. -- Meni Rosenfeld (talk) 19:44, 28 March 2010 (UTC)[reply]

More specifically, the Koch snowflake has a Hausdorff dimension of log(4)/log(3), which is approximately 1.26186. As pointed out below, having non-integer Hausdorff dimension is not sufficiently inclusive to be a defining characteristic of fractals, as there are fractals with an integer Hausdorff dimension - see list of fractals by Hausdorff dimension for examples. But self-similarity is too inclusive, as a straight line and an exponential curve are self-similar but would not usually be considered to be fractals. Gandalf61 (talk) 09:37, 29 March 2010 (UTC)[reply]

I thought that the Peano curve was considered by most to be a fractal, though it has an integer dimension in the limit. -- 174.31.194.126 (talk) 21:08, 28 March 2010 (UTC)[reply]

Rather than fractional Hausdorff dimension, a better working definition of a fractal is that its Hausdorff dimension is strictly larger than its topological dimension. This includes all sets with fractional Hausdorff dimension because the topological dimension is always an integer less than or equal to the Hausdorff dimension. It still excludes some of the examples (plane-filling curves, because the property does not distinguish between a curve and its image, or the Mandelbrot set, because it has nonempty interior), but it does include several fractals with integer dimension (such as the Smith–Volterra–Cantor set or the boundary of the Mandelbrot set).—Emil J. 15:59, 29 March 2010 (UTC)[reply]

March 28

How much should my game cost?

Say I developed a games for the PC which I wishes to sell by mailorder.

Here are my values

FixedCost = $100000 (I send 100 000 dollars in a year to develop the game)

MarginalCost = $3 (it cost me $3 to stamp out a DVD-ROM and mail it )

DemandCurve or SalesVolume(price) =1000000*(1/(1+price)) (The amount of game units sold for each price in dollars)

When I tried to calculate the sales price for my game (to maximize my profits). My calculations kept on crashing saying I should sell my game at $999 999 dollars per game which is a ridiculous amount for sell a computer game to teenagers. What is wrong with my calculations?

122.107.207.98 (talk) 03:10, 28 March 2010 (UTC)[reply]

I get that there's no right answer, which indicates that your demand curve isn't realistic. We realize that (for x=price) total profit = income-expense = (x-3)*1000000*(1/(1+x))-100000. We want to maximize this number, so we take the derivative to find out how profit depends on price. We get dprofit/dx = 4000000/(1+x)^2 (sorry, I can't use the math type proficiently), which we see is always positive. That means that if you increase the price, you'll make more money, for all prices... not realistic. You can see that your demand curve isn't realistic by the limiting case: notice that as price increases to infinity, demand drops, not to zero (like in real life), but to one. Buddy431 (talk) 03:44, 28 March 2010 (UTC)[reply]

The demand does tend to 0 as

p\to \infty

. But yes, the demand curve is not realistic.

My bad. Buddy431 (talk) 21:27, 28 March 2010 (UTC)[reply]

Maybe there are consultants with experience on how the demand curves for this kind of game look like. Otherwise I suggest picking a reasonable number like $10 rather than making accurate calculations on made up data - Garbage In, Garbage Out.

BTW, have you considered some form of digital distribution? -- Meni Rosenfeld (talk) 08:14, 28 March 2010 (UTC)[reply]

To expand on "picking a reasonable number like $10", you can look at what competitors charge for comparable products and assume that your cost structure will be similar to theirs. That may sound like a lot of assumptions, but there are also a lot of assumptions which go into the demand curve and even the fixed and variable cost curves. So, the formal method really isn't any more accurate than "just guessing". StuRat (talk) 14:02, 28 March 2010 (UTC)[reply]

Tricky trig integral

Is there a way to evaulate $\int \sin ^{4}x\cos ^{6}x\;dx$ without the use of general $\int \sin ^{m}x\cos ^{n}x\;dx$ reduction formulae? I've been able to come up with concrete methods of how to evaluate all integrals of the form $\int \sin ^{m}x\cos ^{n}x\;dx$ by using the Pythagorean and double angle identities alone except for the ones where $m$ and $n$ are nonequal even powers both greater than or equal to four. Could someone help me evaluate the integral above and ones like it using only trig identities and u-substitution? Thank you! Yakeyglee (talk) 04:35, 28 March 2010 (UTC)[reply]

\sin ^{4}x\cos ^{6}x=\left({\frac {\sin(2x)}{2}}\right)^{4}\left({\frac {1+\cos(2x)}{2}}\right)^{2}

and then use one of the other methods which you have got. (Igny (talk) 04:59, 28 March 2010 (UTC))[reply]

I don't see how that helps to evaluate it if it's an integral. :-/ Yakeyglee (talk) 05:11, 28 March 2010 (UTC)[reply]

\int \sin ^{4}x\cos ^{6}x\,dx=\int \left({\frac {\sin(2x)}{2}}\right)^{4}\left({\frac {1+\cos(2x)}{2}}\right)^{2}\,dx

Multiply out, use substitution $u=2x$ and voila, you reduced the power, and you can use some of the methods which you said you already have. (Igny (talk) 05:15, 28 March 2010 (UTC))[reply]

How to tell if a time series is nonlinear or not?

I have some time-series I am interested in. How can I tell if they are non-linear or not? If they are not non-linear, then it means that I will not have to buy and read a thick expensive text-book about non-linear time series. Thanks 78.144.250.185 (talk) 15:47, 28 March 2010 (UTC)[reply]

To be honest, this is the very first time I heard about this topic. The third section of this paper may be a good place to start reading about NL testing. Pallida Mors 17:10, 29 March 2010 (UTC)[reply]

Depends on what kind of non-linearity. If the model is non-linear in parameters, you'll need to use more complicated techniques to analyze the data. If the model is non-linear in regressors, you can simply try some non-linear forms of the regressors alongside the linear forms and see which show up as significant. Also, check the Box-Cox power transform. Wikiant (talk) 18:01, 29 March 2010 (UTC)[reply]

If I used nonlinear methods on a linear time series, would I still get valid results? Is it true that everything must be either linear OR nonlinear, or can common time series have elements of both? Thanks 78.146.84.14 (talk) 21:39, 29 March 2010 (UTC)[reply]

Probability

OK so this is homework but I have done my best to figure it out and still can't solve this.

A and B are independent. P(A union B) = 5/8. P(A intersect B') = 7/24.

First question was to find P(B) which I found by P(A union B) - P(A intersect B') = 1/3.

Second question is to find P(A intersect B). That's the one I'm struggling with.

Third question is to find P(A) which I know I can solve once I solve the second question. Since they are independent then P(A intersect B) = P(A) X P(B) so to find P(B), divide answer of second question by 1/3. I also could use the formula P(A) = P(A intersect B) + P(A union B) - P(B).

The given answer is 7/48 but I need to understand how to work things out! Please help quickly, thanks! —Preceding unsigned comment added by 59.189.218.247 (talk) 16:22, 28 March 2010 (UTC)[reply]

have you made a typo, or are you having one of those (all too frequent with me) 'duh' moments? P(A intersect B) is a given in the problem, it equals 7/24. --Ludwigs2 16:36, 28 March 2010 (UTC)[reply]

No, what's given is P(A∩B′), where B′ means the complement of B. —Bkell (talk) 16:41, 28 March 2010 (UTC)[reply]

you're right. sorry, misread it. --Ludwigs2 16:57, 28 March 2010 (UTC)[reply]

(ec) You have all the pieces you need; you just need to put them together. One way to find P(A) is to use both of the two formulas you've given, P(A)=P(A∩B)+P(A∪B)−P(B) and P(A∩B)=P(A)×P(B), and put in everything you know. You don't know P(A∩B) yet, but you can write it as P(A)×P(B). This will give you an equation which you can solve for P(A). Once you have that, it's easy to get P(A∩B). Another way to go about it is to reason like this: Since P(B)=1/3, and A and B are independent, then P(A∩B) should be 1/3 of P(A). That means that P(A∩B′) should be 2/3 of P(A). You know P(A∩B′), so you're set. —Bkell (talk) 16:40, 28 March 2010 (UTC)[reply]

March 29

Rouché's theorem and real solutions

Hi all,

I'm trying to use Rouché's theorem to show that all the solutions to $z\sin(z)=1$ are real, and I've shown (fairly trivially) that in the interval $[-(n+1/2)\pi ,\,(n+1/2)\pi ]$ there are 2(n+1) real roots, and now want to use Rouché to show that in the disc $D(0,(n+1/2)\pi )$ there are also 2(n+1) roots, so they must -all- be real. I'm using the 'symmetric form' of Rouché's theorem as referred to in the article, and trying to show that $|(z\sin(z)-1)-z\sin(z)|=|g-f|=1<|z\sin(z)|=|f|$ on the boundary of the disc, $\partial {D}=\{z:|z|=(n+1/2)\pi \}$ , which would imply f ; but I'm not sure how to actually go about showing this (if indeed it is correct). It seems like it should be correct, because that would give me the correct number of zeroes in the disc (counted with multiplicity - i.e. 2 at z=0); 2(n+1) roots of zsin(z)=0 in the disc - but I can't see how strictly to prove it. I worked out that $|\sin(z)|={\sqrt {\sin ^{2}(x)+\sinh ^{2}(y)}}$ where z=x+yi, and I can see graphically the minimum for this is roughly around $y=0,\,x=\pm (n+1/2)\pi$ , but I can't see how to prove it, and of course I'd rather not get into lots of messy calculus of minima if it can be avoided. Could anyone suggest anything? :)

Thankyou very much, it's greatly appreciated! Spalton232 (talk) 15:37, 29 March 2010 (UTC)[reply]

Hints. Consider the strip {z : |Im(z)|<1}. Prove that there are no zeros of g(z):=zsin(z)-1 outside of it (use the expression for |sin(z)| you wrote, and note that |sinh(y)|>1 for any real y with |y|≥1). Take f(z):=z sin(z). Compute all zeros of f, with multeplicity. Prove that |f(z)-g(z)| = 1 < |f(z)| holds true either if |Im(z)|=1 or if |Re(z)|=(k+1/2)π (for any positive integer k), and use Rouché's theorem in consequence. Conclude as you were doing, counting with multeplicity the real zeros of f and g in the interval |x|≤(k+1/2)π. --pm a 08:04, 30 March 2010 (UTC)[reply]

Birthday Probabilities

I have some probability questions. One leads on from the other.

What is the probability that in a room of n people, exactly two people will share a birthday?
What is the probability that in a room of n people, at least two people will share a birthday?
What is the probability that in a room of n people, k people will have a birthday falling anywhere within a period of p consecutive days?

In regards to the last point, for example, what is the probability that in a room of ten people, six of them have birthdays falling in a given week? Although I would like a general solution; if at all possible. Thanks in advance. •• Fly by Night (talk) 19:40, 29 March 2010 (UTC)[reply]

Birthday problem may help 129.67.37.143 (talk) 20:21, 29 March 2010 (UTC)[reply]

It helps with the first question. I am primarily interested, as my example might show, in the third question. I hoped that a logical sequence of questions might help the thread. I know that by the strictest sense of the reference desk that I ought to be pointed to article so that I might help myself; but I was hoping for a more contemporary style reference desk answer, i.e. a solution with some links to help me understand the solution. Don't worry: it's not a homework question. I'm far too old for homework. •• Fly by Night (talk) 21:40, 29 March 2010 (UTC)[reply]

Actually it deals with all of your questions, but only the k=2 case of q3 - have a look at "near matches". There's a reference in that section that might be helpful with the arbitrary k case, and another to an article called "generalised birthday problem" that might help. I'm sorry I can't give you a better solution. 129.67.37.143 (talk) 23:19, 29 March 2010 (UTC)[reply]

Thanks for making an effort. But it doesn't actually deal with all of my questions. It deals with a very special case of my third question. Special cases are not very useful really. The (positive) number 2 is both even and prime. How many other (positive) prime numbers are even? If anyone can answer my questions as written, i.e. give a proper solution to the questions, then I would be very grateful. •• Fly by Night (talk) 23:32, 29 March 2010 (UTC)[reply]

If you assume that ever day of the year has the same probability of birth, which is an acceptable assumption, then there is a 1/365.25 chance that a person is born on some given day. If you are looking at a 2-day span, there is a 2/365.25 chance that a person is born on those two days. If you look at a 3-day span, there is a 3/265.25 chance. If you look at p days, there is a p/365.25 chance. If you have 1 person with a p-span of days, there is a p/365.25 chance that the person will be born on that day. If you have 2 people, there is a p/365.25 chance per person, for (p/365.25)². If you have 3 people, there is a (p/365.25)³ chance of the three being born on those p days. For k people, there is a (p/365.25)^k chance. Is that what you are asking? Obviously, this would be much easier without leap years. -- kainaw ™ 23:41, 29 March 2010 (UTC)[reply]

This is almost there. I want to start with n people and I want exactly k of them to have a birthday within a fixed period, of say p days. This is different to having n people and wanting them all to have a birthday within the same period of p days. The must be some kind of averaging or something like that... •• Fly by Night (talk) 00:10, 30 March 2010 (UTC)[reply]

Is this not given by the multinomial distribution? 66.127.52.47 (talk) 02:55, 30 March 2010 (UTC)[reply]

If you fix a particular week, then the probability that k people from n have birthdays in that week is multinomial. However this can't be used immediately to solve qn3 because the probability that k people have a birthday in a period w_1 and the probability that k people have a birthday in period w_2 are not independent. Inclusion-exclusion is needed. Let W_p be the set of all periods of length p consecutive days, and if w is such a period B_w be the event that at least k people have their birthday in w. Qn 3 asks for

P(\bigcup _{w\in W}B_{w})

. Inclusion-exclusion (and some kind of argument inductive on p) can be used, but it won't be straightforward. 129.67.37.143 (talk) 09:49, 30 March 2010 (UTC)[reply]

March 30

Guests at a dinner party

Having decided above that for unique beads on a necklace there are n!/2n ways of ordering them, n>2, allowing rotation and reflection.

How many ways are there of arranging guests around the table at a dinner party, such that every guest has for each ordering a) at least one different neighbour, b) two different neighbours? Thanks 89.242.246.24 (talk) 09:37, 30 March 2010 (UTC)[reply]

@@ Line 133: / Line 133: @@
 :::::::This is almost there. I want to start with ''n'' people and I want exactly ''k'' of them to have a birthday within a fixed period, of say ''p'' days. This is different to having ''n'' people and wanting them all to have a birthday within the same period of ''p'' days. The must be some kind of averaging or something like that... ••&nbsp;[[User:Fly by Night|'''''Fly&nbsp;by&nbsp;Night''''']]&nbsp;([[User talk:Fly by Night|<i>talk</i>]]) 00:10, 30 March 2010 (UTC)
 ::::::::Is this not given by the [[multinomial distribution]]? [[Special:Contributions/66.127.52.47|66.127.52.47]] ([[User talk:66.127.52.47|talk]]) 02:55, 30 March 2010 (UTC)
+::::::::: If you fix a particular week, then the probability that k people from n have birthdays in that week is multinomial.  However this can't be used immediately to solve qn3 because the probability that k people have a birthday in a period w_1 and the probability that k people have a birthday in period w_2 are not independent.  Inclusion-exclusion is needed.  Let W_p be the set of all periods of length p consecutive days, and if w is such a period B_w be the event that at least k people have their birthday in w.  Qn 3 asks for <math>P ( \bigcup _{w\in W} B_w ) </math>. Inclusion-exclusion (and some kind of argument inductive on p) can be used, but it won't be straightforward. [[Special:Contributions/129.67.37.143|129.67.37.143]] ([[User talk:129.67.37.143|talk]]) 09:49, 30 March 2010 (UTC)
 = March 30 =