Wikipedia:Reference desk/Archives/Mathematics/2007 September 14

= September 14 =

How does anyone feel about...
The set of boolean values being represented as $$\mathbb{B} = \lbrace 0, 1 \rbrace $$? Is $$\mathbb{B}$$ being used for anything else?

Then, $$\mathbb{B}^n$$ can be the set of all n-tuples, e.g. $$\mathbb{B}^3 = \lbrace (x_1, x_2, x_3) | x_i \in \mathbb{B} \rbrace$$


 * I think the standard notation is $$\mathbb{Z}/2\mathbb{Z}$$ or $$\mathbb{Z}_2$$. Gandalf61 14:14, 14 September 2007 (UTC)
 * But these notations imply that we are referring to the field, where 1+1=0, while I believe the OP is looking for a notation that will emphasize that we have 1+1=1. -- Meni Rosenfeld (talk) 15:17, 14 September 2007 (UTC)
 * Usually AND is implemented by the multiplication operation, so 1*1=1, whereas addition implements XOR, the "exclusive or", so 1+1=0. $$\mathbb{Z}_2$$ is the simplest example of a Boolean ring. Gandalf61 15:28, 14 September 2007 (UTC)


 * I've seen $$\mathbb{B}^n$$ used for the n-dimensional unit ball. The set {0,1} is sometimes written 2 or 2, which makes sense if you think of it as the ordinal 2. -- BenRG 16:56, 14 September 2007 (UTC)


 * I've seen the notation $$\mathbb{B} = \lbrace 0, 1 \rbrace $$ used in the context of computing science, as in this course text (see page 46) on Funmath. --Lambiam 23:25, 14 September 2007 (UTC)


 * I guess I wouldn't object, but the notation {0,1}n itself is quite widely used. Is there really that much gained by shortening {0,1} to $$\mathbb{B}$$? — PaulTanenbaum 02:22, 16 September 2007 (UTC)

sum
Sum (from n=0 to n=p) of (-1)n/(p-n)!n! (using 0!=1)

The sum = 0 except when p=0 in which case the sum=1. (from xp terms of polynomial expansion of:  exp(p) x exp (-p)  )

Can anyone give an alternative proof of this just out of curiousity (I've already got sum=0 when p=odd, but couldn't go further)83.100.255.59 15:04, 14 September 2007 (UTC)


 * Have you tried using the Binomial theorem? -- Meni Rosenfeld (talk) 15:16, 14 September 2007 (UTC)
 * erm, I used the Multinomial theorem (for a power of 2) to get the original result - not sure if there is an alternative method you meant? ignore - not the multinomial (a bit like it but not the same)83.100.255.59 15:40, 14 September 2007 (UTC)
 * Can't get that to work - would need to find a way to get the n! top term in the binomial coefficent to dissapear83.100.255.59 15:51, 14 September 2007 (UTC)
 * Assuming that I understand correctly that you are trying to prove that
 * $$\sum_{n=0}^p\frac{(-1)^n}{n!(p-n)!} = 0$$
 * The equality does not change if we multiply by $$p!$$, so we have to show that
 * $$0 = \sum_{n=0}^p(-1)^n\frac{p!}{n!(p-n)!} = \sum_{n=0}^p(-1)^n{p \choose n}$$
 * Now we can use the binomial theorem.
 * An alternative method is to prove it directly using induction. -- Meni Rosenfeld (talk) 15:52, 14 September 2007 (UTC)
 * Yes, thanks.83.100.255.59 16:32, 14 September 2007 (UTC)

question
A and B can run 200metres in 22 and 25 seconds respectively.


 * How far is B from the finishing line when A reaches it? —Preceding unsigned comment added by 122.164.236.110 (talk) 16:29, 14 September 2007 (UTC)


 * Do your own homework. Hint: Distance is speed times time, and the time when A crosses the finishing line to the time when B is behind A at this time is, well, the same.  x42bn6 Talk Mess  16:32, 14 September 2007 (UTC)


 * It might help to think that B will be 3 seconds behind when A crosses the line. So, B will have 3/25 of the 200 meter distance yet to go.  You take it from there. StuRat 04:45, 19 September 2007 (UTC)

Fourier analysis of resonance
In college I took a semester-long class in the phyics of resonance phenomena; the fundamental idea was that a resonant system has a preferred vibrational frequency, and application of a periodic driving force will drive the sistem to an amplitude that depends on how closely the frequency of the driving force corresponds to the preferred frequency of the system.

As far as I can recall, we considered only sinusoidal driving forces. I was thinking about this recently, and it seemed to me that for nonsinusoidal drivers, one could use Fourier analysis to decompose the driving force into a sum of sinusoids, and then deal with their effects on the system independently. Is that right?

My real question is, where can I read more about this?

Thanks. -- Dominus 16:38, 14 September 2007 (UTC)


 * This is indeed a standard technique, as long as the driving force is periodic and the differential equation representing the system is linear. As our article on the uses of trigonometry puts it: In almost any scientific context in which the words spectrum, harmonic, or resonance are encountered, Fourier transforms or Fourier series are nearby. You can find some theory and examples at Linear differential equation, although the link with resonance is not made explicit. An easy introduction is the treatment by example found here. --Lambiam 23:12, 14 September 2007 (UTC)


 * Seems like you're modeling the system as an LTI system, in which case you can get the response to arbitrary input by convolving with the impulse response in the time domain, or equivalently, multiplying by the transfer function in the frequency domain. —Keenan Pepper 03:11, 15 September 2007 (UTC)