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May 19[edit]

Eigenvectors[edit]

I'm looking at this matrix:

\left[{\begin{matrix}1&1/2&1/8&1/48&1/384&\cdots \\[6pt]0&1/2&1/4&1/16&1/96&\cdots \\[6pt]0&0&1/8&1/16&1/64&\cdots \\[6pt]0&0&0&1/48&1/96&\cdots \\[6pt]0&0&0&0&1/384&\cdots \\[6pt]\vdots &\vdots &\vdots &\vdots &\vdots &\ddots \end{matrix}}\right].

It goes on forever. The pattern is this:

Below the main diagonal every entry is 0;
The entries in the first row are the reciprocals of the double factorials

2,\qquad 2\cdot 4,\qquad 2\cdot 4\cdot 6,\qquad 2\cdot 4\cdot 6\cdot 8,\quad \dots \dots \,

Each column is proportional to a row of Pascal's triangle; e.g. the last column shown above is proportional to 1, 4, 6, 4, 1.
Consequently the sequence of numbers on the main diagonal is the same as the first row.
Also consequently, if each column is summed, the sums are 1, 1, 1/2, 1/6, 1/24, the terms being the reciprocals of the factorials.
Each row is a shift of a scalar multiple of the first row.
This matrix is the matrix of coefficients in the "inversion formulas" section of this article.

Since it's an upper-triangular matrix, the eigenvalues are the numbers on the main diagonal. I've found the first three corresponding eigenvectors:

\left[{\begin{matrix}1\\0\\0\\0\\0\\\vdots \end{matrix}}\right],\quad \left[{\begin{matrix}1\\-1\\0\\0\\0\\\vdots \end{matrix}}\right],\quad \left[{\begin{matrix}5\\-14\\21\\0\\0\\\vdots \end{matrix}}\right]

What more can be said about the eigenvectors? Michael Hardy (talk) 23:44, 18 May 2010 (UTC) Michael Hardy (talk) 15:15, 20 May 2010 (UTC)[reply]

In case it helps, the following eigenvectors are {205, -987, 3243, -5405} and {111049, -761404, 3950262, -13752764, 24067337}. Also, for all eigenvectors up to at least the 40th, the entries of the eigenvector have alternating signs. -- Meni Rosenfeld (talk) 04:42, 20 May 2010 (UTC)[reply]

Thank you; I'll see if the Online Encyclopedia of Integer Sequences says anything about those. And otherwise see if I can spot any patterns that shed some light. Michael Hardy (talk) 15:03, 20 May 2010 (UTC)[reply]

Dart question[edit]

"A dart, thrown at random, hits a square target. Assuming that any two parts of the target of equal area are equally likely to be hit, find the probability that the point hit is nearer to the center than to any edge." I got that the probability would be 25%, but this seems too easy for a Putnam question. Can someone confirm my answer? 173.179.59.66 (talk) 02:51, 19 May 2010 (UTC)[reply]

I did a quick back of the enveloppe calculation and I find 2/3. By symmetry, you can look at the following area. If you put the four corners of the square at x = ±1/2 and y = ±1/2, then consider the area within the square between the positive x-axis and the line y = x. This area is 1/8. The distance to the closest edge will then always be the distance to the edge at x = 1/2 which is 1/2 - x. The distance to the center is

sqrt(x^2+y^2). You can then easily see that the region that is closer to the center is from x = 0 to x = 1/4 and y = 0 from y = sqrt(1/4-x). Count Iblis (talk) 03:18, 19 May 2010 (UTC)[reply]

I see where I messed up, thanks! 173.179.59.66 (talk) 04:01, 19 May 2010 (UTC)[reply]

I don't think it is that simple, unless these thoughts are unnecessarily over complicated... Looking at iblis's triangle with corners at (0,0) (0.5,0) and (0.5,0.5). To help think about it, Draw an arc radius 0.1 and a vertical line at x=0.9 - these points are all 0.1 from the centre / edge respectively. Again arc radius 0.2 and line at x=0.8. Fnally arc 0.25 and vertical line at x=0.25. These touch at (0.25,0) and this si an equidistant point. The space remaining is a "triangle" with the vertical line, the curved arc and the original diagonal edge as its three sizes. The locus of equidistant points will curve from (0.25,0) to (0.5*(1-sqrt(2),0.5*(1-sqrt(2)), but that locus isn't a straight line nor do I think it a circular arc. It should be possible to determine x as a function of y and then integrate... -- SGBailey (talk) 13:45, 19 May 2010 (UTC)[reply]

I made a mistake in my posting yesterday a bit after bedtime. When I was in bed I saw the mistake I made. You need to do as SGBailey says but I think you then end up integrating y = x from x = 0 to 0.5*(1-sqrt(2) and then from there to x = 1/4 the function

y = sqrt[(1/2-x)^2 - x^2] = sqrt(1/4 - x)

But you need to carefully check all this. Count Iblis (talk) 14:09, 19 May 2010 (UTC)[reply]

To SGBailey: the curve you described, i.e. the locus of points equidistant from the given point and a given line is a parabola, the point and the line are called the parabola's focus and directrix (see Conic section#Eccentricity and Conic section#Parameters). --CiaPan (talk) 05:54, 21 May 2010 (UTC)[reply]

To be equally close to the center and to the top edge, is to be on a parabola whose focus is the center and whose directrix is the top edge. Similarly for the other edges. Thus the target region is bounded by arcs of four parabolas. Michael Hardy (talk) 16:30, 19 May 2010 (UTC)[reply]

Yeah this is what you want. If you let the bottom edge be y=0 and the center be (0,1) the parabola for that edge is y = x²/4 + 1/2 running from 2-√6 to √6-2. Rckrone (talk) 19:06, 19 May 2010 (UTC)[reply]

I did a polar integral of the area within which the dart would be closer to the center. I assumed that $\theta$ is an angle in standard position from the horizontal going through the center (i.e. - counterclockwise from above the horizontal on the right). I also said that the sides of the square each are $2r$ , just making calculation easier. Thus, the distance along the horizontal from the center to the right edge is $r$ . As $\theta$ goes from $0$ to ${\frac {\pi }{4}}$ , (i.e. - top right corner), the distance to the edge along that line can be found by ${\frac {r}{\cos \theta }}$ . Since we are integrating the area where it'd be closer to the center, we simply need to halve this expression. I called this distance $d$ . Thus, $d(\theta )={\frac {r}{2\cos \theta }}$ . Now, because of symmetry, I did a polar integral from $0$ to ${\frac {\pi }{4}}$ and just multiplied the result by eight. Thus, $8\cdot {\frac {1}{2}}\int _{0}^{\frac {\pi }{4}}\left({\frac {r}{2\cos \theta }}\right)^{2}d\theta$ . Evaluating this integral gives $r^{2}$ (it's an easier one than it looks). The area of the whole thing is the square of $2r$ , which is simply $4r^{2}$ . Dividing the smaller area by the bigger gives ${\frac {1}{4}}$ . I contend that the original answer was correct. :-) — Trevor K. — 02:17, 26 May 2010 (UTC)

Sorry! I realized the flaw in my solution. I did it again and I'll post how I did it here soon! I think I remember my answer to be something near 21%. It was an irrational number. — Trevor K. — 14:22, 27 May 2010 (UTC)

Constant continuous maps between a topology on the natural numbers and the reals[edit]

Hi all,

I'm wondering how to show that if $\tau$ is a topology on the natural numbers 1, 2,... such that the open sets are $\mathbb {N} ,\,\phi$ , and all sets of the form $\{1,\,2,\,...,\,n\},\,n\in \mathbb {N}$ , then any continuous map from $(\mathbb {N} ,\,\tau )$ to the Euclidean topology on $\mathbb {R}$ is constant.

If I can, I'd like to show this directly without relying on much more than the definitions (e.g. without quoting results, 'a homeomorphism preserves connectedness' etc). I'm always very happy visualizing these sorts of problems in a metric space with a concept of distance, but my intuition falls short when we fix our open sets by a specified topology instead.

Could anyone suggest the nicest way to go about a problem like this?

Thankyou all very much! Typeships17 (talk) 18:53, 19 May 2010 (UTC)[reply]

Any two nonempty open sets in your topology have nonempty intersection. This gives a quick proof. Algebraist 18:59, 19 May 2010 (UTC)[reply]

Argh, what I wouldn't give to be as good as you are at Mathematics! I do envy your grasp of what seems to be just about everything under the mathematical sun. Thankyou, that's been a great help :) Typeships17 (talk) 23:01, 19 May 2010 (UTC)[reply]

More generally, any continuous function from a hyperconnected space to a Hausdorff space must be continuous (the proof uses the same idea as Algebraist notes). PS T 01:53, 20 May 2010 (UTC)[reply]

In fact, you may find it interesting to know that the topological space you mention is what is known as a Toronto space. A Toronto space is a topological space that is homeomorphic to every proper subspace of the same cardinality (it may be a good exercise to check that the topological space you mention is indeed a Toronto space). It is also true (though perhaps not immediately obvious) that the only countable Hausdorff Toronto space is the discrete space (up to homeomorphism, of course). The Toronto problem asks whether every uncountable, Hausdorff Toronto space must be discrete (and, as far as I know, this problem still remains open). PS T 02:13, 20 May 2010 (UTC)[reply]

I must confess it is exam season for me at the moment so I probably won't have time to read up on that too much right now, but I'll be sure to check it out in a couple weeks when I'm free again - thanks! Typeships17 (talk) 03:21, 20 May 2010 (UTC)[reply]