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January 10

Does this continued fraction converge?

I was playing around with [1;2,1,3,1,4,1,5....] at school today, and I'm fairly sure it converges to a value around 1.44. Is there a test that I can run to see if it does converge? Does anyone know if this particular series turns out to equal something interesting? Thanks. 24.18.51.208 (talk) 01:44, 10 January 2009 (UTC)[reply]

For a start you may find continued fraction interesting. Every continued fraction with positive integer coefficients will converge; yours converges to about 1.35804743869438. It is irrational (because the continued fraction is infinite), and furthermore is not the root of any quadratic equation (because the continued fraction is not periodic). The continued fraction looks similar to the continued fraction for

{\frac {\tan(1)-1}{2-\tan(1)}}=1.259415845

, which has continued fraction [1;3,1,5,1,7,1,9,...]. Perhaps if you can figure out how to calculate the continued fraction for tan (1) (which has continued fraction [1;1,1,3,1,5,1,7,1,9,...]) then that might give some ideas of how to find the value of [1;2,1,3,1,4,1,5,...]. Eric. 68.18.17.165 (talk) 04:23, 10 January 2009 (UTC)[reply]

You may also find this calculator interesting if you just want an evaluation. It gives 641/472 = 1.3580508474576272 = 1, 2, 1, 3, 1, 4, 1, 5 hydnjo talk 04:59, 10 January 2009 (UTC)[reply]

Numbers having such continued fractions are a special case of Hurwitz numbers [1]. Using the formula (4) from this article and manipulating it a bit to remove the unwanted preamble, we get

[1;2,1,3,1,4,1,5,\dots ]={\frac {\alpha -2}{3-\alpha }}\quad {\text{where }}\alpha ={\frac {J_{1}(2)}{J_{0}(2)}}{\text{,}}

using the bessel functions of the first kind.—Pt_(T) 02:28, 14 January 2009 (UTC)[reply]

Pentagons

With only a sheet of paper, a ruler and a pencil, how might I go about drawing a regular pentagon of which each side is 150mm long, without having any construction lines around it afterward? The easiest way seemingly might be to start with one line 150mm long and work out how far away horizontally and vertically from that the end of each other line would be, but how should I do that? 148.197.114.165 (talk) 11:52, 10 January 2009 (UTC)[reply]

Can you use a compass? If so, see Pentagon#Construction. You'll need the formula in the previous section to work out what radius to use. Zain Ebrahim (talk) 12:00, 10 January 2009 (UTC)[reply]

Actually, this is better because it gives links to details for each step. Zain Ebrahim (talk) 12:13, 10 January 2009 (UTC)[reply]

I think I've worked it out. Would this give the right shape?: To draw regular pentagon ABCDE, where s is the length of any side of the pentagon, create a rectangle of size (1+(5^1/2)/2)*s by (((1+(5^1/2)/2)*s)^2 - s/2^2)^1/2. Draw then the line AB, of length s along one of the longer sides of the rectangle, so both lines have their centres at the same point. Draw the line AE from A to the nearest of the shorter sides of the rectange such that its length equals s, then do the same for BC from B to the other side. Find the centre of the long side opposite line AB, mark this as point D and draw lines from this to both C and E.

For a pentagon with sides length 150mm, this rectangle would be 242.7mm by 230.82mm, with A and B 46.35mm from the corners.

Something like that? 148.197.114.165 (talk) 17:48, 10 January 2009 (UTC)[reply]

Try it and see. It's easy enough to tell if you have the right shape at the end. --Tango (talk) 23:32, 10 January 2009 (UTC)[reply]

max size of a k-component graph

Resolved

– --Shahab (talk) 14:24, 11 January 2009 (UTC)[reply]

This is related to a question I asked a few days back. What is the maximum number of edges possible in a n vertex graph having k connected components where each component has $n_{i}$ vertices. Obviously $\sum n_{i}=n$ and I need the maximum value of ${\frac {1}{2}}\sum n_{i}(n_{i}-1)$ . How should I proceed further? I want the final answer to be in terms of n and k. Thanks--Shahab (talk) 13:06, 10 January 2009 (UTC)[reply]

As with the k=2 case, the maximal case is when all but one component has 1 vertex. Algebraist 17:00, 10 January 2009 (UTC)[reply]

I came to the same conclusion by the following procedure: As

\sum n_{i}^{2}=n^{2}-(k-1)(2n-k)-2\sum _{i<j}(n_{i}-1)(n_{j}-1)

so if all but component has 1 vertex then the number of edges is

{\frac {1}{2}}(n-k)(n-k+1)

. Since this acts as an upper-bound on the number of edges too (because of the negative sign in

-2\sum _{i<j}(n_{i}-1)(n_{j}-1)

) hence this is the maximal achievable value for the number of edges. Is this proof correct? Secondly, is there a way to maximize the function

{\frac {1}{2}}\sum n_{i}(n_{i}-1)

subject to the constraints

\sum n_{i}=n,n_{i}\in \mathbb {N}

. Cheers--Shahab (talk) 17:31, 10 January 2009 (UTC)[reply]

Shahab asks "is there a way to maximize the function $\textstyle {\frac {1}{2}}\sum n_{i}(n_{i}-1)$ subject to the constraints $\textstyle \sum n_{i}=n,n_{i}\in \mathbb {N}$ ?" The answer is to take $n_{1}=n$ . But perhaps Shahab meant to ask the related question: how to minimize the function $\textstyle \sum n_{i}$ subject to $\textstyle {\frac {1}{2}}\sum n_{i}(n_{i}-1)=N$ . Bob Robinson (UGA) and I solved this problem more than 20 years ago but never published it. Consider the greedy approach: choose $n_{1}$ as large as possible so that $\textstyle {\frac {1}{2}}n_{1}(n_{1}-1)\leq N$ , then continue in the same fashion with what is left. Our theorem was that this works (i.e. the greedy choice of $n_{1}$ is correct) for all but a finite set of values of N, which we determined. If I recall correctly, there were something like 20 exceptions and the largest was about 86,000. In the exceptional cases, $n_{1}$ should be chosen 1 or 2 less than the maximum possible. McKay (talk) 01:59, 11 January 2009 (UTC)[reply]

Thanks for the extra info. But my initial question was about maximizing

{\frac {1}{2}}\sum n_{i}(n_{i}-1)

subject to

\sum n_{i}=n,n_{i}\in \mathbb {N}

where I meant

\mathbb {N} :=\{1,2\cdots \}

. I cannot take

n_{1}=n

for then the graph is just

K_{n}

and I need k connected components in the graph. Cheers--Shahab (talk) 05:15, 11 January 2009 (UTC)[reply]

But that question is already answered above: take k-1 components equal to 1 and one component equal to n-k+1. The fact that it is best for k=2 proves that it is best for arbitrary k also. McKay (talk) 11:11, 11 January 2009 (UTC)[reply]

I am probably being dense here (& for this I ask you to indulge me) but I cannot understand your last statement. Isn't it possible for the max value to be different then

{\frac {1}{2}}(n-k)(n-k+1)

and yet be equal to

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

for k=2. Can you please clarify?--Shahab (talk) 12:46, 11 January 2009 (UTC)[reply]

McKay probably has in mind the following simple proof (also the proof I had in mind above, for what that's worth): suppose we have values n_i subject to

\sum n_{i}=n,n_{i}\in \mathbb {N}

, and suppose there are two values of i for which n_i is not 1. Then by the k=2 case, we can increase

{\frac {1}{2}}\sum n_{i}(n_{i}-1)

by changing one of these two values to 1 (and the other to their sum minus 1). Hence in the maximal case, all but one n_i must be 1. Algebraist 13:32, 11 January 2009 (UTC)[reply]

Exactly, thanks. McKay (talk) 14:28, 13 January 2009 (UTC)[reply]

OK. Thanks--Shahab (talk) 14:24, 11 January 2009 (UTC)[reply]

Multiplying integers by digit-swapping.

I've been playing around with integers which are multiplied by a factor (≤9) when the RH digit is moved to the LH end. For example, 102564 is quadrupled when the 4 is moved to the start. There is obviously an infinite number of these, seen by considering 102564102564, so I'm interested in the smallest integer for each multiplier. For multiplier k, my analysis gives that the integer comes from an expression with 10k-1 in the denominator, so 102564, for example, must have something to do with thirty-ninths. And 1/39 = 0.0256410256410..., so as a hypothesis the smallest integer for a given multiplier is found by looking at the decimal expansion of 1/(10k-1) and selecting the cycle starting 10. Which brings me to the question - acting on the basis outlined, I've found that the 58-digit integer 1,016,949,152,542,372,881,355,932,203,389,830,508,474,576,271,186,440,677,966 is multiplied by 6 when the final digit is moved to the front, but is there a smaller one?→81.151.247.41 (talk) 19:51, 10 January 2009 (UTC)[reply]

There is certainly a name for an integer whose digits get swapped (as you have described) when multiplying with a positive integer less than or equal to 9. Although I can't seem to remember... Can any number theorist help out here? —Preceding unsigned comment added by Point-set topologist (talk • contribs) 20:38, 10 January 2009 (UTC)[reply]

According to Sloane's, these are called k-parasitic numbers and the OP's number is indeed the least 6-parasitic number. Algebraist 20:42, 10 January 2009 (UTC)[reply]

3x3 Antisymmetric Matrices

Let A be a real 3×3 non-zero antisymmetric matrix. How does one show that there exist real vectors u and v and a real number k such that Au = kv and Av = −ku? What relation do these u,v and k bear to the exponential of the matrix A?

I know the exponential of the matrix A is a rotation matrix, but having not seen an appropriate approach to the first part of the question, I'm unsure as to how to progress with this - thanks.

86.9.125.104 (talk) 23:40, 10 January 2009 (UTC)Godless[reply]

If Au = kv and Av = -ku, then A²u = kAv = -k²u, and similarly A²v = -kAu = -k²v. So one approach would be to see what you can show about the eigenvalues and eigenvectors of A², given that A is 3x3 antisymmetric. Gandalf61 (talk) 09:31, 11 January 2009 (UTC)[reply]

Three '2's Problem

You have 3 twos, and you must use all of them in any expression of a number. Operations allowed are:

addition (e.g 2+2+2=6)

subtraction (e.g 2-(2-2)=2)

multiplication (e.g 2*(2+2)=8)

division (e.g 2*(2/2)=2)

raising to a power (e.g 2^(2^2)=16)

rooting - i.e. raising to the power of 1/n, where n is an integer (e.g 2^(1/(2+2))=2^(1/4))

logarithm to base 2 or base e - if to base 2, you have to indicate, and that would mean using one of your three twos: otherwise it would be to the base e. (e.g log2(2*2)=2, or log(2*2*2)=3ln(2), although the latter is not an integer)

concatenation - i.e combining say 3 and 4 to get 34 (Using "~" for this: e.g 2~(2/2)=21)

How many distinct positive integers can you form in this way? (note that for example 2+(2+2)=2+(2*2) so this only counts as 1 distinct positive integer).

Thanks,

Mathmos6 (talk) 23:58, 10 January 2009 (UTC)Mathmos6[reply]

This problem is fairly trivial: I suggest that you have a go yourself. We won't give you answers here; we can only give you hints at the most. Please first show us your progress (tell us what is the smallest and highest possible positive integer that you can obtain). —Preceding unsigned comment added by Point-set topologist (talk • contribs) 13:01, 11 January 2009 (UTC)[reply]

Without concatenation, it's pretty trivial. With concatenation, however, you can get significantly larger integers. --Tango (talk) 16:47, 11 January 2009 (UTC)[reply]

Also, RIES is a useful tool for these sorts of problem. **CRGreathouse** (t | c) 04:08, 16 January 2009 (UTC)[reply]