I have a valid multiple entry schengen business visa issued by the norwegian embassy for a duration of 1 year.  I was supposed to travel at the beginning of February to Norway and use this visa but unfortunately my business trip got cancelled. Therefore, I have not used this visa so far.  I wanted to know if I can use this visa for tourism and now can I enter Germany through this visa as it is valid for a year and is multiple entry visa for all schengen states ?
I am a non-EU national, permanent UK Resident. Recently, my work issued me with a multiple Schengen visa with Norway as the first (main) point of entry. Now I want to go on holidays to Greece - do I need to apply to Schengen visa again, as Greece is not Norway and I am not going there for work?
Aut $\mathbb Z_p\simeq \mathbb Z_{p-1}$ In the special relativistic action for a massive point particle, $$S=\int_{t_i}^{t_f}\mathcal {L}dt,$$ why is the Lagrangian $$\mathcal {L}=-E_o\gamma^{-1}$$ a negative number? example of compact operator Let $E=l^p=\left\{x=\left(x_n\right): x_n\in\mathbb{R}, \sum_{1}^{\infty}\left|x_n\right|^p&lt;\infty\right\}$ with $1\le p&lt;\infty$, $\left\|x\right\|_E=\left(\sum_{1}^{\infty}\left|x_n\right|^p\right)^{1/p}.$ Let $\left(\lambda_n\right)$ is a bounded sequence in $\mathbb{R}$ and consider the operator $T$ defined by $$T(x)=\left(\lambda_1x_x,\ldots,\lambda_nx_n,\ldots\right)$$ where $x\in l^p$. Prove that $T$ is a compact operator from $E$ to $E$ iff $\lambda_n\rightarrow 0$.  Can anyone give me some hints to solve this problem? Thank you! Determining best fitting curve fitting function out of linear, exponential, and logarithmic functions Context:  From a question on Mathematics Stack Exchange , someone has a set of $x-y$ points, and wants to fit a curve to it, linear, exponential or logarithmic.  The usual method is to start by choosing one of these (which specifies the model), and then do the statistical calculations.  But what is really wanted is to find the 'best' curve out of linear, exponential or logarithmic.  Ostensibly, one could try all three, and choose the best fitted curve of the three according to the best correlation coefficient.  But somehow I'm feeling this is not quite kosher. The generally accepted method is to pick your model first, one of those three (or some other link function), then from the data calculate the coefficients. And post facto picking the best of all is cherry picking. But to me whether you're determining a function or coefficients from the data it is still the same thing, your procedure is discovering the best...thing (let's say that which function  is -also- another coefficient o be discovered).  Questions:   Is it appropriate to choose the best fitting model out of linear, exponential, and logarithmic models, based on a comparison of fit statistics? If so, what is the most appropriate way to do this? If regression helps find parameters (coefficients) in a function, why can't there be a discrete parameter to choose which of three curve families the best would come from? Was mathematics invented or discovered? What would it mean to say that mathematics was invented and how would this be different from saying mathematics was discovered?   Is this even a serious philosophical question or just a meaningless/tautological linguistic ambiguity? Can't find Corruption I have this function:  $$ f(x,y) = \frac {xy}{|x|+|y|} $$  And I want to evaluate it's limit when $$ (x,y) \to (0,0)$$ My guess is that it tends to zero. So, by definition, if:  $$ \forall \varepsilon \gt 0, \exists \delta \gt 0 \diagup \\ 0\lt||(x,y)||\lt \delta , \left|\frac{xy}{|x|+|y|}\right| \lt \varepsilon $$ Then $$ \lim_{(x,y)\to(0,0)}\frac {xy}{|x|+|y|} = 0 $$ So:  $$ \left|\frac{xy}{|x|+|y|}\right| = \frac{|xy|}{|x|+|y|} = \frac{|x||y|}{|x|+|y|} \le 1 |y| \lt \delta $$  So for any $$\delta \lt \varepsilon$$ the inequality is true. Hence, the limit exists and is equal to zero.  . Am I wrong or is Wolfram|Alpha wrong? We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not.  My question is why.  As background there are plenty of mathematical schemes that can be used to define these operators. One of these is   where $H_0(a,b) = b+1$   (successor op) $H_1(a,b) = a+b$   (addition op) $H_2(a,b) = ab $   (multiplication op) $H_3(a,b) = a^b$   (exponentiation op) $H_4(a,b) = a\uparrow \uparrow b$   (tetration op: $a^{(a^{(...a)})}$ nested $b$ times ) etc.    Here it is not obvious to me why $H_1(a,b)=H_1(b,a)$ and $H_2(a,b)=H_2(b,a)$ but not $H_3(a,b)=H_3(b,a)$  Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion?  Thanks. There is a socket related function call in my code, that function is from another module thus out of my control, the problem is that it blocks for hours occasionally, which is totally unacceptable, How can I limit the function execution time from my code? I guess the solution must utilize another thread. Speed of light in a gravitational field? How to determine a matching color code from a physical sample? I need to make a restaurant menu design that matches restaurant wallpapers. For some reason my client cannot get color codes used in wallpaper, but he provided me with a sample. (see below)    How can I get the correct color codes from the wallpaper? (someone mentioned that there are scanners for this purpose, but I could not find any information regarding that)  Let me know if this question should be in another site. Investment options for f1 visa students in USA I am foreign graduate student in USA. I recently completed my undergrad (from USA) and have a bit of money saved up. I know that F1 students cannot work off campus and other business activities such stock trading etc are prohibited as well. I wanted to know if opening an account in an investment management firm like Charles Schwabb or Betterment is allowed. I would not be working but I will still be earning a small interest on my deposits. If that is not allowed, are there any good and legal ways for foreign students to earn a little interest on savings? $(this) inside of AJAX success not working For a given inner product space $V$, I would like to prove the inequality $$\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = 2(\|\mathbf{x}\|^2 + \|\mathbf{y}\|^2).$$  We see that $$\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = \langle\mathbf{x}+\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x}-\mathbf{y},\mathbf{x}-\mathbf{y}\rangle,$$  which by linearity $$=\langle\mathbf{x},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x},\mathbf{x}-\mathbf{y}\rangle - \langle\mathbf{y},\mathbf{x}-\mathbf{y}\rangle.$$  I've been trying to figure out how to go about this proof using linearity in the second argument of an inner product, but my textbook does not say that linearity necessarily holds in the second component. If it does, how would I go about the rest of this proof? If not, should I be potentially using some aspect of conjugate symmetry to prove this statement? Let's say I am building a logistic regression model where the dependent variable is binary and can take the values $0$ or $1$. Let the independent variables be $x_1, x_2, ..., x_m$ - there are $m$ independent variables. Let's say for the $k$th independent variable, the bivariate analysis shows a U-shaped trend - i.e., if I group $x_k$ into $20$ bins each containing roughly equal number of observations and calculate the 'bad rate' for each bin - # observations where y = 0 / total observations in each bin - then I get a U shaped curve.  My questions are:   Can I directly use $x_k$ as input while estimating the beta parameters? Are any statistical assumptions violated which might cause significant error in estimating the parameters? Is it necessary to 'linearize' this variable through a transformation (log, square, product with itself, etc.)? Suppose I attach an blur function to an HTML input box like this:  &lt;input id="myInput" onblur="function() { ... }"&gt;&lt;/input&gt;   Is there a way to get the ID of the element which caused the blur event to fire (the element which was clicked) inside the function? How?  For example, suppose I have a span like this:  &lt;span id="mySpan"&gt;Hello World&lt;/span&gt;   If I click the span right after the input element has focus, the input element will lose its focus. How does the function know that it was mySpan that was clicked?  PS: If the onclick event of the span would occur before the onblur event of the input element my problem would be solved, because I could set some status value indicating a specific element had been clicked.  PPS: The background of this problem is that I want to trigger an AJAX autocompleter control externally (from a clickable element) to show its suggestions, without the suggestions disappearing immediately because of the blur event on the input element. So I want to check in the blur function if one specific element has been clicked, and if so, ignore the blur event. Please anyone tell me when we have to use "has been, have been , had been" and something  started in past and still continuing is it possible to use has been ? Is there anything that is totally random? When I say totally random, I mean absolutely random, not pseudorandom. If I want to say "totally random" numbers such as 1,26,17,4,1 and 27, although I see them to be totally random, they aren't. These are numbers that I think are influenced by my childhood, ideology and everything that I've seen. So, what do you think? Is there anything completely random? Is there evidence of "think tank" meaning "the brain"? The  Random House Dictionary defines the expression  as:         a research institute or organization employed to solve complex problems or predict or plan future developments, as in military, political, or social areas.      According to  the above connotation  is an extention of an earliar sense  which referred to "the brain":        also think-tank, 1959 as "research institute" (first reference is to Center for Behavioral Sciences, Palo Alto, Calif.); it had been colloquial for "the brain" since 1905.      : shows usage of "think tank" only from the late '50s.  Is there evidence  of earlier usages of "think tank" meaning "the brain"? Despite  its colloquial   usage I assume  the expression was quite widespread before it was adopted by the Stanford University. DES Initialization Vector in C#