Recent content by Square

You mean how one would prove that AB = I provided that BA = I? If you assume that BA = I and thus that A^{-1} = B and A = B^{-1} you could try multiplying from the left with B^{-1} and from the right with A^{-1}, which gives you B^{-1}BAA^{-1} = B^{-1}IA^{-1} \ .

I sometimes use the notation A_{\bullet i} to denote the i-th column and A_{j\bullet} to denote the j-th row.

There is some identity which tells you that \mathrm{Tr}(AB) = \mathrm{Tr}(BA) (more generally one could state that the trace is invariant under cyclic permutations). Use it and your problem should be as good as solved.

Just to pick one example, in classical electromagnetism (the kind that's taught in the first electromagnetism courses on universities) relies heavily on vector calculus . You calculate all sorts of line- and flux integrals over magnetic and electric fields particular while using Maxwells equations.

http://www.iancgbell.clara.net/maths/vectors.htm About midway down this page you can see that the dot product in polar coordinates is \small (r_1,\theta_1) \tiny \bullet \small (r_2,\theta_2) = r_1r_2 \cos(\theta_1-\theta_2). One solution is to use this formula. The other one is just to...

The short answer is that the exponential function a^x increases faster than any power of x (x^{\alpha}, \ \alpha \in \mathbb{R}). The long answer is that you could prove that the limit \displaystyle \lim_{x\to\infty} \frac{a^x}{x^{\alpha}} (and thus that your given limit tends towards inf)...

Yes, [A,B] is in this case the Lie brackets since it got defined by the writer in equation (1.1) earlier on the same page. The ad stands for adjoint I believe. You can read more about the adjoint endomorphism on wikipedia: http://en.wikipedia.org/wiki/Adjoint_representation_of_a_Lie_algebra

Yep. Factor the numerator and you'll get \lim_{n\to \infty} \left| \frac{2(n+1)(2n+1)}{(n+1)^2} \right| = \lim_{n\to \infty} \left| \frac{2(2n+1)}{n+1} \right| which evaluates to 4 (divide numerator and denominator with n).

First: \lim_{n\to \infty} \left| \frac{\left(n \cdot (n-1) \cdot (n-2) \ldots \right) ^{2} \cdot (2n+2)(2n+1)}{\left( (n+1) \cdot n \cdot (n-1) \cdot (n-2) \ldots \right) ^{2}} \right| = \lim_{n\to \infty} \left| \frac{2(n+1)(2n+1)}{(n+1)^{2}} \right| = \lim_{n\to \infty}\left|...

I suspect it's angular velocity, which indeed has the unit rad/s.

I think you might be right, and the book wrong, since it says "saw the advertisement in at least one of the two media." Then it should be 1-(60/200) = 7/10 The complement of the probability that the customer did not see the advertisement at all. Which is the same as your result.

It should be P(A or B) = (60+50-30)/200 = 80/200 = 2/5

2. cosine and sine are the x and y-coordinates in the unit circle. At an angle v you can form a triangle with hypothenuse 1 and catheti x and y.