Recent content by SP90

http://arxiv.org/pdf/0807.1310v5.pdf Equation 4.6 strikes me as wrong. The integral is \int_{0}^{\infty} B \mathrm{d}B The limits of the integral seem to be values of z, but the integral seems to be wrt B. If the limits were values of B, the answer would be ∞. It's clearly not, but I don't...

This is now solved. It can be solved by extending the left triangle to a right-angled triangle, with a base of V+k (hence where the single V comes from). Then \cos{\delta}=V+k k is given by x \cos{(\pi-(\theta + \beta))}=-x \cos({\theta + \beta}). Using the cosine sum rule I stated above...

Homework Statement Given the following two triangles: Show that v \cos{\delta} = V(1-\cos{\beta})+u\cos(\alpha - \beta) The Attempt at a Solution Using the cosine law I've got: v^{2}=x^{2}+V^{2}-2xV\cos{(\theta + \beta)} and u^{2}=x^{2}+V^{2}-2xV\cos{(\theta)} I figured maybe using the...

But because the joint PMF is a multiplication of the individual PMFs, the term becomes \lambda_{1}^{n_{1}}\lambda_{2}^{n_{2}}\lambda_{3}^{n_{3}}... which means the term \sum_{i=1}^k N_i = n doesn't appear in the joint PMF (although it would if the \lambda_{i} were equal), so I don't understand...

I don't understand how the \sum_{i=1}^k N_i = n term constrains the joint distribution, given that each \lambda_{i} can be distinct. Also, to evaluate P(\sum_{i=1}^k N_i = n), wouldn't this be the joint distribution, evaluated at N_{1}=n-\sum_{i=2}^{k}N_{i} and then...

Homework Statement The Attempt at a Solution I have that the joint probability mass function would be \Pi_{i=1}^{k} \frac{\lambda_{i}^{n_{i}}}{n_{i}!} e^{-\lambda_{i}} How would I go about applying the conditional to get the conditional distribution?

Isn't \nabla \cdot v \otimes n = (n_{1}(\frac{dv_{1}}{dx_{1}}+\frac{dv_{2}}{dx_{2}}+\frac{dv_{3}}{dx_{3}}), n_{2}(\frac{dv_{1}}{dx_{1}}+\frac{dv_{2}}{dx_{2}}+\frac{dv_{3}}{dx_{3}}), n_{3}(\frac{dv_{1}}{dx_{1}}+\frac{dv_{2}}{dx_{2}}+\frac{dv_{3}}{dx_{3}})) Which is n \cdot \nabla v? This would...

Homework Statement Homework Equations So I have that v \otimes n = \left( \begin{array}{ccc} v_{1}n_{1} & v_{1}n_{2} & v_{1}n_{3} \\ v_{2}n_{1} & v_{2}n_{2} & v_{2}n_{3} \\ v_{3}n_{1} & v_{3}n_{2} & v_{3}n_{3} \end{array} \right) The Attempt at a Solution I've tried applying the...

Homework Statement Suppose v is a vector satisfying: \alpha v + ( a \times v ) = b For \alpha a scalar and a, b fixed vectors. Use dot and cross product operations to solve the above for v. Homework Equations The unique solutions should be: v=\frac{\alpha^{2}b- \alpha (b \times a) + (b...

Ah, I see. So instead I use \hat{n}=\frac{1}{\sqrt{6}}(1i-2j+1k) and that gives Pv=\frac{1}{6}(25i+10j-5k). And Tv=\frac{1}{6}(32i-4j+2k). Is that right now?

Makes sense, thanks for the quick response. I get 10i-10k+5k which, when dotted with n obviously gives 0. Correcting the sign error for the second yields 17i-24j+12k. Thanks for your help

Homework Statement Given a plane \Pi with normal n=i-2j+k and a vector v=3i+4j-2k calculate the projection of v onto \Pi and the reflection of v with respect to \Pi. The Attempt at a Solution I need to check that I'm doing this is right. I think I need v - (v \cdot n)n =...

I've solved this now, no need for a response, but if anyone ever comes across a similar problem here is the solution: Take x_{n}=(1, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}}, ..., \frac{1}{\sqrt{n}}, 0, 0, 0 ,0,...) Obviously, as a series with a finite number of non-zero terms, it is square...

I thought if I took x_{n}=(\frac{n}{1},\frac{n}{2^{2}},\frac{n}{3^{2}}...), that would work, since x_{n} \rightarrow \frac{n\pi^{2}}{6}. It's clear that T(x_{n}) exists for all n, but x_{n} itself diverges. But then obviously the definition of the range being closed is that when T(x_{n}) has a...

Homework Statement Let T: \ell^{2} \rightarrow \ell be defined by T(x)=x_{1},\frac{1}{2}x_{2},\frac{1}{3}x_{3},\frac{1}{4}x_{4},...} Show that the range of T is not closed The Attempt at a Solution I figure that I need to find some sequence of x_{n} \rightarrow x such that...