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How to find out who called a method? How to recover from a Nvidia fail on Ubuntu 16.04? I have an older Nvidia video card and my PC crashes when installing the driver for the video card in Ubuntu 16.04. When I get to the login screen but no further: after entering my password, the system went into a loop and returned always back to the login screen [with funny stripes present].   How do I fix this problem? How to Achieve this Requirement ???Can i use Jave Script , CSS and Visual force page? How do candles and wicks work? Solve recurrence relation $a_n = 3 a_{n - 3} + a_{n-1} + 2 n$ I am not able to solve the above recurrence relation which involves $n$. The initial conditions $a_0 = 1, a_1 = 2, a_2 = 3$, for $n \geq 0$. Please try to give closed form in terms of arbitrary $a_i \geq 1 $ for $i \geq 0$ After cruising through a lot of material online, and answers over here, my understanding of contravariant and covariant vectors are, in a finite-dimensional vector space, suppose we have a vector, which is contravariant to its basis. However, using the metric tensor, we can map this vector to a one form in the dual space, which acts as a vector space for this example. This one form, or covactor, varies similarly to the basis in our original vector space. However, the covectors have a different basis, which is the dual basis, and are expanded in terms of those. Here's where my confusion arises : Many diagrams over the internet give a geometric picture of the scenario, by claiming that contravariant components of an 'arrow' are found by drawing parallel lines along the axes, and checking where they intersect. Co-variant components could be found by sketching the perpendicular lines on these axes. So, these materials are treating covariant and contravariant as different representations of the same arrow, while I'm inclined to believe they are completely different. However, if we let go of rigour, in exchange for a 'more' geometric understanding, I think we are allowed to do this. The diagrams are usually of the form :  However, even if we assume that contravariant and covariant are different representations of the same object, this particular diagram still seems wrong to me. Particularly the locations of $x_1,x_2$. In this diagram, they are located on the span of the original basis. Shouldn't covector components be located on the span of the dual basis ? This diagram suggests to me, that co-variant components have the same basis, as the contravariant ones. Shouldn't the diagram look more like this ?  This second diagram seems to fit the concept better according to me. This is because the dual vector must be 'contravariant' with respect to the 'dual basis'. This means, their components must be found out by sketching parallel lines along the span of the dual basis. These lines intersect the original axis at a right angle, which is to be expected, as dual bases are orthogonal to the original bases. Moreover, this second picture can also show in a much better way, how scaling up of the original basis, scales down the 'contravariant components' and the 'dual basis', which in turn scales up the 'covariant' components.This shows that the components of the dual vector are covariant to the original basis. This is something that is not readily visible in the first diagram. So, am I correct in assuming this second 'geometrical' representation is correct ? I know this doesn't make much sense, because as the mathematicians said, that vectors are completely different from one forms, and one forms should be represented using the number of hyperplanes intercepted by our arrow. However, I've seen most course material refer to it in this way, and frankly it is easier to visualize. However, most of this material used the first picture. Can anyone point out my mistakes, if any, and tell me if the first picture is correct, or the second one ? Or in this case 'less-flawed'. Let $F$ be a field and let $a \neq 0$ be a zero of the polynomial $a_0 + a_1x + . . . +a_nx^n$ in $F[x]$.  I want to show that $\frac{1}{a}$ is a zero of the polynomial $a_n + a_{n-1}x + . . . + a_0x^n$  How can I do this? Fighting Spam - What can I do as an: Email Administrator, Domain Owner, or User? It is often stated that the Lagrangian formalism and the Hamiltonian formalism are equivalent.   We often hear people talk about eigenvalues of Hamiltonians but I have never ever heard a word about eigenvalues of Lagrangians.   Why is this so? Is it not useful? is it not possible to do it? Pattern in Isomorphisms between $\mathbb Z_m \times \mathbb Z_n$ and $\mathbb Z_{mn}$ How is Starkiller Base's beam visible from Takodana in The Force Awakens? On Takodana,      the main characters and various bar patrons can see the planet Hosnian Prime being destroyed by Starkiller Base with their (respective) naked eyes. How did they see this? Is the Hosnian system close enough to see? If it's not very close and they could just see the explosion, how did the light from it reach them so quickly? When I use any variation of English, US international (with dead keys,altGr dead keys or alternative) on my Linux Mint machine I always encounter this behaviour. When I press one of these keys: ' " and then follow them with a 'non-accentable' character like a [ or a b no output comes out at all.    Whereas in Windows US-International it would print [ or b. If I wanted to type this I would have to escape each dead key with a space instead of with any 'non-accentable' character. This is annoying when programming (not really, but I trained with the Windows 'Qwerty International' on typing.io and switching back and forth between the systems is irritating).    Is there any way to change that so it works like in Windows? How to draw arrows between parts of an equation to show the Math Distributive Property (Multiplication)? There is a part of my paper where I use the distributive property and it might not be clear to the reader to see the operation result (I'm using in proposional logic for a proof). The logic equation (part of) is: $$ (l_{1}) \vee \big( (p \vee q) \wedge (\neg p \vee q) \wedge (p \vee \neg q) \wedge (\neg p \vee \neg q)\big) $$ $$ (l_{1} \vee p \vee q) \wedge (l_{1} \vee \neg p \vee q) \wedge (l_{1} \vee p \vee \neg q) \wedge (l_{1} \vee \neg p \vee \neg q) $$  So I was wondering if there is a way to represent this operation like a 'shower' (see another example in pic ). Maybe there is some hidden magic mathmode command or pkg for this. At last we can also rely on tikz. What is the analytic continuation of the Riemann Zeta Function Binary search (bisection) in Python What is the adjective to describe someone who views or represents someone/something as a stereotype? What I want to know is whether there is an adjective within the same word family as "stereotype" for that.  So, if it exists, it would be something like "stereotypist" or "stereotyping".  I know that "stereotypist" exists as a noun and "stereotyping" exists as a present participle but I wonder whether either (or both) of them has/have also been used in literature as an adjective. I want to be able to calculate the confidence interval from the estimated coefficient and respective standard errors.  I have a linear regression model which can be summarized (in R):  summary(fit1)  Call: lm(formula = bwt ~ height + weight + parity, data = data)  Residuals:     Min      1Q  Median      3Q     Max  -66.913 -10.624   0.991  10.979  55.621   Coefficients:             Estimate Std. Error t value Pr(&gt;|t|)     (Intercept) 31.19217   13.56879   2.299   0.0217 *   height       1.24964    0.23083   5.414 7.48e-08 *** weight       0.06781    0.02823   2.402   0.0164 *   parity1     -1.83309    1.19838  -1.530   0.1264     --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1  Residual standard error: 17.9 on 1170 degrees of freedom Multiple R-squared:  0.04898,   Adjusted R-squared:  0.04654  F-statistic: 20.08 on 3 and 1170 DF,  p-value: 1.071e-12   With this model I can calculate the respective confidence intervals:  &gt; confint(fit1)                   2.5 %     97.5 % (Intercept)  4.57029503 57.8140351 height       0.79676227  1.7025207 weight       0.01243198  0.1231932 parity1     -4.18429933  0.5181151   I would expect the intervals of the predictor height to be given by  $$ 1.24964 \pm (1.96*0.23083) = [0.7972132,1.702067] $$  where 1.24964 is the estimated value for the coefficient and 0.23083 is the standard error for this coefficient. The numbers are close but not quite the same.  What am I doing wrong? I am a graduate student in economics who recently converted to R from other very well-known statistical packages (I was using SPSS mainly). My little problem at the moment is that I am the only R user in my class. My classmates use Stata and Gauss and one of my professors even said that R is perfect for engineering, but not for economics. He said that many packages are built by people who know a lot about programming, but not much about economics and therefore are not reliable. He also mentioned the fact that since no money is actually involved in building an R package, there is therefore no incentive to do it correctly (unlike in Stata for example) and that he used R for a time and got some "ridiculous" results in his attempts to estimate some stuff. Moreover, he complained about he random number generator in R which he said was "messy".  I've been using R for just a little more than a month and I must say I have fallen in love with it. All this stuff I am hearing from my professor is just discouraging me.  So my question is: "Is R reliable for the field of economics?". What is the difference between the memory usage report in System Monitor and the one by free? I am using System Monitor 2.28.0.  When I look in the Resources tab, I see a nice graph with for memory and swap.  Memory is about 60% 2.3 GiB of 3.8.  When I type the command free, I got :               total       used       free     shared    buffers     cached Mem:       3994908    3962396      32512          0     100852    1477032 -/+ buffers/cache:    2384512    1610396 Swap:      8000328      28468    7971860   cat /proc/meminfo | grep MemFree give  MemFree:           34536 kB   The situation has remained the save for several minute.  I started a lot of process with a script and the script is waiting for the free memory to get lower.  According to what I am seeing in the Process tab (or with top), the number in System Monitor seem a lot closed to the total of the memory of the process that the one reported by free.   Thanks While proving a universal quantification, we begin by "Let ... in ..." or "Arbitrary choose ..."?