I'm traveling to London, UK for tourism. I just received my passport from UK embassy with 6 months visa. My passport is about to expire and will renew it before travelling to London. So I'm travelling with 2 passports (one with the visa and one is new), I believe that's fine with the UK immigration. My concern is that the visa has the passport number and never travelled to uk before, is there any problem travelling with passport has new visa and holding another new passport?  (Note: travelling from Saudi Arabia to UK)
I have a valid (multiple entry) business visit visa to the UK on my old passport. I have recently obtained a new passport as I ran out of pages on my old passport. The purpose of my visit is still for business and there are no change in personal details on my new passport. There is however a change in address on the new passport. Incidentally it is the same address that I filled out for my UK visa application while answering the query "WHAT IS YOUR PERMANENT RESIDENTIAL ADDRESS AND CONTACT DETAILS?"  In light of the above, can I still travel to the UK by carrying both the old and new passports? Or, will I be required to either transfer the visa or make a fresh application for my new passport. Thanks.
find . -name *.rb   doesn't work on my ubuntu  I have to do  find . -name "*.rb"   to get it to work.  Why is that? Mutual Independence Definition Clarification I think this is a very recent design change, because I've not seen it before:    Where before a small box with the bounty value dislocated the title a couple of characters to the right, now the additional text saying how long until it expires dislocates the title by about a quarter of the width of the column.  I find that this makes it much harder to read the title, and is counterproductive to the goal of attracting attention. Is the vector cross product only defined for 3D? Here's a question that I share:  Show that if $D$ is a countable subset of $\mathbb R^2$ (provided with its usual topology) then $X=\mathbb R^2 \backslash D $ is arcwise connected. Subequation: Counter too large Regular cast vs. static_cast vs. dynamic_cast What ticket reservation do I need for a Schengen visa? I have a visa appointment this Friday 19 at the Italian Consulate in Philadelphia and I have a question about the flight reservation requirement: According to the Visa requirements "Confirmed flight reservation for a round trip showing complete itinerary."  Does this mean I have to reserve the tickets, pay and print the itinerary?  In this case if my visa gets denied then I will lose the money I paid or can I just show the planned itinerary (written not booked)? in this case the travel agency would not give me the Travel Insurance coverage until I book and pay for the reservation.   I am a US green card holder and applying from North Carolina. How do I complete the square from the point I have left off at, and is this correct so far?   I have a normal prior for $\beta$ of the form $p(\beta|\sigma^2)\sim \mathcal{N}(0,\sigma^2V)$, to get:  $p(\beta|\sigma^2)=(2\pi\sigma^2V)^\frac{p}{2}\exp[-\frac{1}{2\sigma^2}\beta^T\beta]$  where $\beta^T\beta$ is   $\sum\limits_{i=1}^p \beta_i^2$.  My likelihood has a normal distribtuion for the data points y of the form $p(y|\beta,\sigma^2)\sim\mathcal{N}(B\beta,\sigma^2I)  $  $p(y|\beta,\sigma^2)=(2\pi \sigma^2V)^\frac{n}{2}\exp[-\frac{1}{2\sigma^2}({\bf y}-{\bf B}{\bf \beta})^T({\bf y}-{\bf B}{\bf \beta})]$  (Note that $\beta$ is a matrix/vector too, \bf does not work.)  To get my posterior for $\beta$ I combined the above, took the exponential parts only, and then expanded to get:  $\exp[-\frac{1}{2\sigma^2}({\bf y}^T{\bf y}-{\bf y}^T{\bf B}\beta-\beta{\bf B}^T{\bf y}-\beta^T{\bf B}^T{\bf B}\beta)]\exp[-\frac{1}{2\sigma^2}({\bf \beta}^T{\bf B})]$.  I dropped the $({\bf y}^T{\bf y})$ term, as is not a function of $\beta$.  Putting into one expression without the exponential:  $-\frac{1}{2\sigma^2}(-{\bf y}^T{\bf B}\beta-\beta{\bf B}^T{\bf y}-\beta^T{\bf B}^T{\bf B}\beta+{\bf \beta}^T{\bf B})$.  I know I need to combine the similar terms and get into the form of the multivariate normal distribution, which is what I am aiming for, but I am unsure how to do this? I probably have to add an extra term to the expression to get it into the correct form?  Note: This is not homework, it's a project, but my Bayesian working knowledge is not good at all and so I need to understand the working out. I intend to integrate out the $\beta$ and then the $\sigma^2$ after getting into the multivariate form. Hi I can compute a Basic Loan Amortization Schedule, But now I'm trying to Calculate a Seasonal Loan Amortization Schedule . So for example A loan starts at January And Will go on for 3 years, with Monthly Payments. Interest is at 3% Annually and is compounded Annually. During the months of September, October, November, December It'll only accumulate Interest But no payments and principal. Principal is 100000. How Do I calculate Payments and interest. Both  (if you take a list of values, the distribution of the most significant digit is rougly proportional to the logarithm of the digit) and  (given a corpus of natural language utterances, the frequency of any word is roughly inversely proportional to its rank in the frequency table) are not theorems in a mathematical sense, but they work quite good in the real life.   Does anyone have an idea why this happens?  (see also ) How do I show that if $f : [0,1] \to [0,1]$ is continuous, then there is some $c$ such that $f(c) = c$? Theorem: Let $f:[0,1] \to [0,1]$. Assume $f$ is continuous. Then there exists   $c \in [0, 1]$ such that $f(c) = c$.         For any function $f$, let $g(x) = f(x) − x$. If the theorem is true, what does that say about $g(c)$?   Write a proof of the theorem. Do it by applying the Intermediate Value Theorem   in the appropriate way to the function $g(x) = f(x) − x$ with $a = 0$ and $b = 1$. Clearly state why the hypotheses of the IVT are satisfied here.        Well if the theorem is true, then there exists $c$ such that $f(c) = c$. Then $f(c) - c = 0$. Then $g(c) = 0$. Intermediate Value Theorem: Let $I$ be an interval and $f$ a function whose domain contains $I$. If $f$ is continuous, then for all $a, b \in I$ with $a &lt; b$ and all real numbers $k$, if $k$ is strictly between $f (a)$ and $f (b)$, then there exists $c$ such that $a &lt; c &lt; b$ and $f (c) = k$.   Proof Attempt: Take $a = 0$ and $b = 1$. Then $a,b \in[0,1]$ and $a &lt; b$. Since $f:[0,1] \to [0,1]$, it follows that for all real numbers $k$ such that $f(a) &lt; k &lt; f(b)$, then there exists $c$ such that $a &lt; c &lt; b$ so $0 &lt; c &lt; 1$ and $f(c) = k$. Then $f(c) - k = 0$. Thus, since $g(x)=f(x)-x$, it follows that $g(c) = 0$.   Looking to see if I did this correctly and/or if there is a more elegant way to prove problem 2. I am a little confused why the first theorem is true, but I went with it. How can I make an object follow a path exactly Working in Cycles.  Say I have a "disco ball" with a glossy shader. I'm pointing a spot light to it, and I'm trying to have the reflections of that light illuminate the diffuse material on the walls with the patterns from the ball.    I get this:    but I'd like to get something like:    For the illustration I placed a point light inside the ball an removed faces. But what I'd like to do is move the light around the ball so that the light reflected from it changes like it would do in real life.   Is that possible at this point?  EDIT:  Just for the record here's the image rendered in LuxRender (yet another program to learn...) How to prove that the set $A = \left\{ {p:{p^2} < 2,p \in {\Bbb Q^+}} \right\}$ has no greatest element? More specifically, I have to find a $q$ in $A$ such that $q$ is larger than any $p$ in $A$.  The only thing I can think of is using (${2 - {p^2}}$) somehow, or ${(p+x)^2}&lt;2$, but other than that I don't know how to proceed. This exercise is taken from the very first chapter of Walter Rudin's Principles of Mathematical Analysis. Quoting the exercise:     To do this, we associate with each rational $p &gt; 0$ the number:     $$q = p - \frac{{{p^2} - 2}}{{p + 2}} = \frac{{2p + 2}}{{p + 2}}$$   But he didn't explain how he came up with that number.   The only thing I've thought of is the following procedure:  $$\eqalign{   &amp; {p^2} + (2 - {p^2}) = 2 \Rightarrow {p^2} + (\frac{{2 - {p^2}}}{2}) &lt; 2,{\text{ and since }}2 - {p^2}{\text{ is always positive,}}  \cr    &amp; {p^2} &lt; \left[ {{p^2} + (\frac{{2 - {p^2}}}{2})} \right] &lt; 2  \cr    &amp; {\text{Therefore, let }}q = \sqrt {{p^2} + (\frac{{2 - {p^2}}}{2})} ,{\text{ then:}}  \cr    &amp; p &lt; q &lt; \sqrt 2  \cr} $$  However, my result doesn't necessarily belong to ${\Bbb Q}$. I wonder how did he obtain the value of $q$ in the book.  EDIT: Feel free to point out if there are more appropriate tags for this question. I've imported several shapefiles which where exported from a Mapinfo .tab. Several  .tab-files are imported resulting in 20+ layers. Afterwards I want to export it to GeoJSON; but I'm reluctant to select each layer and export it manually. Is there a way to merge all the layers into one using QGIS? 301 redirect instead of 404 when URL is a prefix of a post or page name? hey all! i have an odd problem with URL routing. when i request a URL that doesn't exist, e.g. , wordpress correctly returns a 404. however, if that URL is a prefix of a post or page name, it instead returns a 301 redirect to the post or page.  for example, if i have a post on 10/1/2010 named Food Post, it will return a 301 with Location:  (my permalink structure). likewise, if i have a page named Food Page, it will return a 301 with Location: .  you can see this in action on my live site,  . e.g.  redirects to  .  turning off permalinks (ie switching to "default") fixes it, for both posts and pages, but naturally i don't really want to do that.  i see this on three different installations, all wordpress 3.0.1 and apache 2.2, two ubuntu lucid/mysql 5.1 and one freebsd 7.3/mysql 5.0. i've deactivated all plugins and removed everything from my .htaccess except the lines for wordpress below, verbatim, but no luck.  RewriteEngine On RewriteBase / RewriteRule ^wordpress/index\.php$ - [L] RewriteCond %{REQUEST_FILENAME} !-f RewriteCond %{REQUEST_FILENAME} !-d RewriteRule . /wordpress/index.php [L]   thanks in advance... How to make a disk image and restore from it later? There are many visa applicants who are working as an unregistered freelancer (photographer, construction contractor, fashion designer, journalist etc ). Some of these freelancers make really handsome monthly income and don't really need to work as a full time employee. Most of them receive variable amount of payments (cash, cheques, demand drafts etc) When it comes to filling a visa application form (USA, UK, Schengen etc), there is always an employment section in all visa form, eg for Schengen visa employment form :    This is a very important factor to make an application really strong.  Questions:   1 Unregistered freelancers can just simply write their profession? what documents to produce along with the application?  2 Bank statements with variable methods of payments could convince the visa officer that these are all legitimate payments? How to add a browser tab icon (favicon) for a website?