Wikipedia:Reference desk/Archives/Mathematics/2010 January 17

= January 17 =

Archimidean Spiral Segment
What is the area of a segment of an Archimidean spiral r=a+bt between t0 and t1, where t0 and t1 are separated by no more than 180 degrees?

In fact, what I'm really looking for is the maximum distance between the chord and the piece of the spiral, but I'm guessing that the area is easier to calculate and will do as an approximation for my error calculation.

82.255.173.168 (talk) 17:15, 17 January 2010 (UTC)
 * $$ {1 \over 2} \int_{t_0}^{t_1} r^2 \, d\theta $$ will give you the area of a sector; subtract from that the area of a triangle and you get the segment. It shouldn't be difficult to write down the equations for the distance of an arbitrary point (parametrized by t) on the arc from the chord, and then differentiating and equating to 0 to obtain the equation for the maximizing argument. Maybe it even has a nice-looking solution, which you can plug in the the formula for the distance to obtain the maximum distance. -- Meni Rosenfeld (talk) 17:58, 17 January 2010 (UTC)


 * And what actually is that integral? I notice the article on Archimidean spirals has an expression for the arc length, but not the segment area, which makes me think that the integral is non-trivial to work out (it's certainly beyond me). 82.255.173.168 (talk) 19:55, 17 January 2010 (UTC)
 * Here the above is $$ {1 \over 2} \int_{t_0}^{t_1} (a+b\theta)^2 \, d\theta $$ can't you compute it?--84.220.119.148 (talk) 02:31, 18 January 2010 (UTC)
 * Unless it's $${{a^2 t + abt^2+b^2t^3+c} \over 2}$$, then no, I can't (I'm a software guy). Is that what it is? 82.255.173.168 (talk) 07:05, 18 January 2010 (UTC)
 * Almost. The integral is $${1 \over 2} \int_{t_0}^{t_1} (a+bt)^2 \, dt = \frac{(a+bt_1)^3-(a+bt_0)^3}{6b}$$, and the area of the triangle is $$\frac{(a+bt_0)(a+bt_1)\sin(t_1-t_0)}{2}$$. -- Meni Rosenfeld (talk) 08:19, 18 January 2010 (UTC)
 * Thanks, that's just what I needed. 66.240.20.113 (talk) 11:26, 18 January 2010 (UTC)
 * Ooops, it might not be just what I need. As a true empiricist, I tried a few extreme values to find out what happes. In the case of a=1, b=0, The segment evaluates to $${(1+0)^3 - (1+0)^3}\over{6 \times 0})$$. Would l'Hopital's rule have given me a more realistic value, or is there a blunder? 66.240.20.113 (talk) 13:06, 18 January 2010 (UTC)
 * Indeed, the formula was written for $$b\neq0$$, but this is just a notational issue. You certainly can use l'Hopital's rule to obtain the limit when $$b \to 0$$, which gives the expected result $$\frac{a^2(t_1-t_0)}{2}$$ for the integral. Or you can write the answer as $${1 \over 2} \int_{t_0}^{t_1} (a+bt)^2 \, dt = \frac{a^2(t_1-t_0)}{2} + \frac{ab(t_1^2-t_0^2)}{2} + \frac{b^2(t_1^3-t_0^3)}{6}$$ which works for any $$b \ge 0$$. -- Meni Rosenfeld (talk) 14:09, 18 January 2010 (UTC)

Identity Matrix Multiplication
I am a little confused on Identity Matrix Multiplication

$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & -2 & 5 \\ 0 & 2 & -3 \\ -1 & 4 & 0 \end{bmatrix} = \begin{bmatrix} 3 & -2 & 5 \\ 0 & 2 & -3 \\ -1 & 4 & 0 \end{bmatrix} $$

Whereas, Matrix Multiplaction requires us to go this way:

....
 * matrix 1 row 1 * matrix 2 column 1
 * matrix 1 row 1 * matrix 2 column 2
 * matrix 1 row 1 * matrix 2 column 3

Thus:


 * 1(3) + 0(0) + 0(-1) = 3
 * 1(-2) + 0(2) + 0(4) = -2
 * 1(5) + 0(-3) + 0(0) = 5

Thus:

$$ \begin{bmatrix} 3 & -2 & 5 \\ . & . & . \\ . & . & . \end{bmatrix} $$

However, Identity Matrix Multiplication seems to be:


 * matrix 1 row 1 * matrix 2 row 1 = solution matrix row 1?


 * and when the element (in matrix 2) is multiplied by 0 it remains the same; i.e, assume it is multiplied by 1?


 * and when the element is 0 (matrix 2 row 3 column 3) and it is multiplied by 1 it remains 0?

--33rogers (talk) 20:13, 17 January 2010 (UTC)
 * No, matrix 1 row 1 * matrix 2 col 1 is answer matrix element in row 1 and column 1. (Multiply a row by a column, by multiplying element by element and adding up the products.) Matrix 1 row 1 * matrix 2 col 2 is answer matrix element in row 1 and column 2. So the answer matrix has the number of rows of the first matrix and the number of columns of the second matrix. And your calculations for the other rows are:
 * Second row:
 * 0(3) + 1(0) + 0(-1) = 0
 * 0(-2) + 1(2) + 0(4) = 2
 * 0(5) + 1(-3) + 0(0) = 3


 * Third row:
 * 0(3) + 0(0) + 1(-1) = -1
 * 0(-2) + 0(2) + 1(4) = 4
 * 0(5) + 0(-3) + 1(0) = 0
 * Get it now? I hope this helped! Remember the paratroopers - they fly in horizontally from the left and parachute down vertically on the right! RupertMillard (Talk) 20:40, 17 January 2010 (UTC)
 * That is when it normal matrix multiplication. My question is regarding Identity Matrix Multiplication. --33rogers (talk) 20:43, 17 January 2010 (UTC)
 * Not sure if this is what is causing the confusion, but: 1(-2) + 0(2) + 0(4) = - 2 (not +2, as you have above). Abecedare (talk) 20:57, 17 January 2010 (UTC)
 * Thanks for that (now fixed).--33rogers (talk) 21:21, 17 January 2010 (UTC)
 * There is no separate such thing as 'Identity Matrix Multiplication'. There is an identity matrix for each size matrix and if you do matrix multiplication using an identity matrix then the result is the same as the other operand. Dmcq (talk) 21:02, 17 January 2010 (UTC)
 * Thanks. --33rogers (talk) 21:21, 17 January 2010 (UTC)

Weird summation notation
Hi all,

I have an article and there's a part I don't quite understand. Here it is, quoted:


 * $$v\displaystyle\sum b_{i_1}b_{i_2}\dots b_{i_{h-1}}b_{i_1+\dots+i_{h-1}}$$

where the sum is over $$i_1, \dots, i_{h-1}$$ and the index of the last term is to be interpreted modulo v.

I don't quite understand what this means: The summation part or the part saying "where the sum is over $$i_1, \dots, i_{h-1}$$". The prior parts of the article make no reference to a double subscripted variable - just $$b_i$$ and $$b_{ij}$$. Could anyone shed some light on this?

This is the paper of interest: Page 222.

Thanks in advance. x42bn6 Talk Mess 21:27, 17 January 2010 (UTC)


 * I can't read the paper, but the notation is standard and unambiguous, provided it is clear where each of the $$h-1$$ indices $$i_1,\dots, i_{h-1}$$ varies; e.g. from 1 to v (or whatever). So you have $$v^{h-1}$$ of these (h-1)-ples $$(i_1,\dots,i_{h-1})$$, namely $$(1,\dots,1)$$ up to $$(v,\dots,v)$$. Suppose for instance $$v=h=3,$$ then there are 9 pairs and the sum is
 * $$b_1b_1\,b_2+b_1b_2\,b_3+b_1b_3\,b_1+b_2b_1\,b_3+b_2b_2\,b_1+b_2b_3\,b_2+ b_3b_1\,b_1 +b_3b_2\,b_2+b_3b_3\,b_3

$$--pm a 23:08, 17 January 2010 (UTC)

Certain repeating digits
Prob'ly just an amusing bit of recreational math, but tell me if there's something deeper here. Just two examples:
 * $$ 2^{25} = 33554432. \, $$

And if one finds the probability of at least 10 successes in 12 independent trials with probability 0.6 of success on each trial one gets (the denominator has the 11th power of 5 rather than the 12th power because of a cancellation with a 5 in the numerator): So why do these odd repetitions happen? Are these both instances of a common phenomenon? (Of course, dividing by 5 is the same as multiplying by 2 and moving a decimal point.) Michael Hardy (talk) 22:12, 17 January 2010 (UTC)
 * $$ \frac{3^{10}\cdot 69}{5^{11}} = 0.08344332288. $$


 * Without knowing the sample space in which you're looking it's impossible to say much. E.g. in all the above you've powers of 2, 3, and 5 and a × 69. Looking at those up to powers of 25 (though your bound could be higher), with the other number going up to say 100, and its over a million possibilities, and that's not counting the number of ways you can arrange them. So if you look at all those sure you're going to see a few odd patterns of numbers, rather like if you look at the first million digits of π. Not sure I'd read anything else into it. -- JohnBlackburne wordsdeeds 00:03, 18 January 2010 (UTC)

A "sample space", as I usually understand that term, is a probability space. No "sample space" is involved in my first observation; it's just a fact of arithmetic, involving no probability. For the second observation, I was completely explicit in identifying the sample space. If you prefer, I could say we're talking about a binomial distribution, and anyone familiar with probability will know that when they read what I wrote. Michael Hardy (talk) 03:37, 18 January 2010 (UTC)


 * As I read his comment, John does not refer to the probabilities in the problem you described. Instead, he assumes (or models the situation so) that you have drawn these two formulas from some probability distribution over formulas, and surmises that the set of possible formulas (the sample space) is large enough that the oddity could be adequately explained by mere chance (i.e. given your rate of seeing "random" formulas, you'd expect to encounter a few odd ones once in a while). The probability distribution he means might be something like $$\operatorname{Pr}(\mathcal{F} = cb^n) \propto 1$$, with $$(c, b, n)\in \{1, \ldots, 100\}\times\{2, 6\}\times\{1, \ldots, 25\}$$, and $$\mathcal{F}$$ denotes a formula.
 * I'd write your latter example as 8344332288 = 611×23 to keep it as an integer. This way in both of your examples there is a relatively high power of a small number multiplied by a small number. Some more numbers of this type would be e.g. 1678822119 = 317×13, 1451188224 = 611×4 and 9886633715 = 711×5. No insight to offer though, sorry. -- Coffee2theorems (talk) 05:07, 18 January 2010 (UTC)
 * Well, my example of 8344332288 = 611×23 has four of those pairs and only two digits not participating. So I'm winning.  (But don't read this comment.) Michael Hardy (talk) 05:33, 18 January 2010 (UTC)


 * Heh. I was just a little curious what the other examples look like in case there's an obvious pattern, so I sicced the computer at it. If this were a competition, I guess using a computer would be grounds for disqualification :) -- Coffee2theorems (talk) 14:07, 18 January 2010 (UTC)

Well, the repeated digits are solutions to N = a×11 + b×11×100 + c×11×104 + ... for integers 0<a,b,c<10. I have no particular insight, other than the general experience that patterns are endemic when fiddling with integers in this way. Instead of base-10, how about other bases? Instead of 11, how about other values? Is there a base B and multiplier M (instead of 11) for which this phenomenon does not occur? linas (talk) 01:10, 20 January 2010 (UTC)