Skyrim mannequin duplication fix not working?
Mannequin Duplication glitch
L1 & L2 Regularization in Light GBM This question pertains to L1 &amp; L2 regularization parameters in Light GBM. As per official documentation:  reg_alpha (float, optional (default=0.)) – L1 regularization term on weights.  reg_lambda (float, optional (default=0.)) – L2 regularization term on weights  I have seen data scientists using both of these parameters at the same time, ideally either you use L1 or L2 not both together.   While reading about tuning LGBM parameters I cam across one such case: Kaggle official GBDT Specification and Optimization Workshop in Paris where Instructors are ML experts. And these experts have used positive values of both L1 &amp; L2 params in LGBM model. Link below (Ctrl+F 'search_spaces' to directly reach parameter grid in this long kernel)      I have seen same in XGBoost implementations.  My question is why use both at the same time in LGBM/XGBoost.  Thanks. How can I use sed to replace a multi-line string? I've noticed that, if I add \n to a pattern for substituting using sed, it does not match. Example:  $ cat &gt; alpha.txt This is a test Please do not be alarmed  $ sed -i'.original' 's/a test\nPlease do not/not a test\nBe/' alpha.txt  $ diff alpha.txt{,.original}  $ # No differences printed out   How can I get this to work? Time dependent Schrödinger equation with time independent potential and separation of variables suppose we have a potential that's independent of time $V(x,t) = V(x)$ so in Schrödinger equation we get: $$i\hbar \frac{\partial \Psi (x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi (x,t)}{\partial x^2}+V(x)\Psi(x,t)$$ since the LHS involves a variation of $\Psi$ with $x$ and the LHS involves a variation of $\Psi$ with $t$ can we say that a basis for the solutions will be with the form $\Psi(x,t)=\psi(x)T(t)$? What is the difference in meaning of "frustrated at" and "frustrated with" used in sentences like     He is frustrated with me   or      He is frustrated at me   When is each of these sentences above are used? I wonder  whether there is a function $f\colon\Bbb R\to\Bbb R$ with the folowing characteristic?   for every two real numbers $\alpha,\beta,\alpha\lt\beta$,  $$\{f(x):x\in(\alpha,\beta)\}=\Bbb R$$   I can't  say such a function does not exist, neither can I construct a example  Thanks a lot! Is there a CSS selector for elements containing certain text? Text processing - join every two lines with commas I have more than 1000 lines in a file. The file starts as follows (line numbers added):  Station Name Station Code A N DEV NAGAR ACND ABHAIPUR AHA ABOHAR ABS ABU ROAD ABR   I need to convert this to a file, with comma separated entries by joining every two lines. The final data should look like  Station Name,Station Code A N DEV NAGAR,ACND ABHAIPUR,AHA ABOHAR,ABS ABU ROAD,ABR ...   What I was trying was - trying to write a shell script and then echo them with comma in between. But I guess a simpler effective one-liner would do the job here may be in sed/awk.  Any ideas? How can I obtain a Cauchy distribution from two standard normal distributions? I am interested in      Let $X\sim N(0,1), Y \sim N(0,1)$ independently. Show $\frac{X}{X+Y}$   is a Cauchy random variable.   My work:  $f_{X,Y}(x,y)=\frac{1}{2\pi} e^{\frac{-1}{2}(x^2+y^2)}, -\infty&lt;x,y&lt;\infty$ by independence  Let $U=\frac{X}{X+Y},V=X+Y$. (Is there a better $V$ to choose for this bivariate transformation?)  Then, $X=UV, Y=V - UV$. So, $|J|=V$.  $f_{U,V}(u,v)=f_{X,Y}(uv,v-uv)|J|=\frac{v}{2\pi}e^{\frac{-v^2}{2}(2u^2-2u+1)},-\infty&lt;u&lt;\infty$  $f_U(u)=\frac{1}{2\pi}\int_{-\infty}^{\infty}ve^{\frac{-v^2}{2}(2u^2-2u+1)}dv$. Let $y=\frac{v^2}{2}(2u^2-2u+1)$, so $dy=v(1+2u^2-2u)dv$. Then,  $f_U(u)=\frac{1}{2\pi(2u^2-2u+1)}\int_0^{\infty}e^{-y}dy=(\pi[(\frac{u-1/2}{2})^2+1])^{-1}, -\infty&lt;u&lt;\infty$,   which is not exactly a Cauchy distribution. Where did I mess up? More importantly, how would you proceed in solving this problem? How can I hide username and computer in terminal command prompt?   In Terminal it says  Last login: Mon Jan 13 00:00:14 on ttys000 Whatever:~ UserName$    Is it possible to show just the current folder and $ sign? every permutation is either even or odd,but not both How we can show every permutation is either even or odd,but not both......I can't arrive at a proof for this ..... Can anybody give me the proof...  Thanks in advance... SQL Server 2012 and 2016 Standard:  If I put if-else logic in a stored procedure to execute one of two branches of code, depending on the value of a parameter, does the engine cache the latest version?  And if on the following execution, the value of the parameter changes, will it re-compile and re-cache the stored procedure, since a different branch of the code must be executed? (This query is quite expensive to compile.) Spider Identification Illinois I noticed a really huge, beautiful web on my porch last week, and tonight, I discovered its resident. I'm thinking it's an Araneus diadematus (Cross Orbweaver), but I was hoping for some confirmation.   Can anyone confirm this or offer a better suggestion? Prove that every Lebesgue measurable function is equal almost everywhere to a Borel measurable function Suppose $(\mathbb{R},\Sigma(m),m)$ is our measure space, where $m$ is Lebesgue measure.  Also, suppose $f : \mathbb{R} \to [-\infty, \infty]$ is a Lebesgue measurable function.  The problem:     Prove that $f$ is equal almost everywhere to a Borel measurable function.   My attempt:  We only have to prove this assertion for a non-negative Lebesgue measurable function $f$ because the result will follow for all Lebesgue measurable functions.  Furthermore, since any Lebesgue measurable function can be approximated by a sequence of non-negative, monotonically increasing, Lebesgue measurable simple functions $s_{n}$, we only need to prove the claim for an arbitrary Lebesgue measurable simple function.  So, let $s: \mathbb{R} \to [-\infty, \infty] $ be a Lebesgue measurable simple function.  We can write $s$ canonically as $$ s(x) = \sum \limits_{i = 1}^{n} \alpha_{i} \chi_{A_{i}}(x)$$ where $\alpha_{i} \in [-\infty, \infty]$, $\bigcup \limits_{i = 1}^{n} A_{i} = \mathbb{R}$, and $A_{i} \cap A_{j} = \emptyset$ if $i \neq j$.  For each $i$, since $A_{i}$ is Lebesgue measurable, we can find a Borel set $B_{i}$ such that $A_{i} \subseteq B_{i}$ and $m(B_{i} \setminus A_{i}) = 0$.  Clearly, this implies that $\bigcup \limits_{i = 1}^{n} B_{i} = \mathbb{R}$.  Now we just need the $B_{i}$'s to be pairwise disjoint, with each $B_{i}$ still retaining $A_{i}$.  To make them pairwise disjoint, I constructed the following sets:  $\tilde{B_{i}} = [B_{i} \setminus (\bigcup \limits_{j \neq i} B_{j})] \cup A_{i}$.  This construction gives us that the $\tilde{B_{i}}$'s are pairwise disjoint (I think....) and $A_{i} \subseteq \tilde{B_{i}}$.  But I don't know that $\tilde{B_{i}}$ is necessarily still a Borel set. :( :(  Am I approaching this problem all wrong? Cut part from video file from start position to end position with FFmpeg I have a video file of 30 minutes, but I want to extract a video from 00:09:23 to 00:25:33.   I can define the startposition with -ss, but I couldn't find one for the end position. Any help please? I have a small Problem that occurs from time to time.   Sometimes, when I copy things or when I put an object from the symbol library on my page, the whole file gets messed up. The screenshot shows it. I make the graphic on the right, Illustrator the ones on the left.   I have pixel snapping checked. Is there a way to tell Illustrator to align everything to pixels no matter what? If N and every subgroup of N is normal in G then G/N is abelian . Let $N$ be a normal subgroup of a group $G$ such that every subgroup of $N$ is normal in $G$ and $C_G(N)\subseteq N $. Prove that $G/N$ is abelian.   Here, as usual, $C_G\left(N\right)$ means the centralizer of $N$ in $G$ (i.e., those elements of $G$ that commute with everything in $N$).  I think we need to use that every subgroup of $N$ is normal in $G$ but i can't use .Please help me with Hints. Define the content of a polynomial (over an arbitrary commutative ring $A$) to be the ideal generated by its coefficients, denoted $c(f)$. I want to show that   $$c(fg) = c(f)c(g).$$      (I was told this is true.)    What I was able to show was that $c(fg) \subseteq c(f)c(g)$ (this is obvious), and that their radicals are the same. My reasoning for the latter was as follows: let $f = a_0 + \dotsb + a_nx^n,$ $g = b_0 + \cdots + a_mx^m,$ and consider the matrix  $$ \begin{pmatrix}  a_0 b_0 &amp; \cdots &amp; a_0b_m\\ \vdots &amp; \ddots &amp; \vdots\\ a_nb_0 &amp; \cdots &amp; a_nb_m \end{pmatrix} .$$  Then $c(f)c(g)$ is generated by the entries of this matrix, while $c(fg)$ is generated by the sums along the diagonals. Let $\mathfrak p$ be a prime ideal of $A$ not containing $c(f)$ or $c(g)$, and let $i,\,j$ be minimal such that $a_i,\, b_j \notin \mathfrak p$.  Then all the terms in the generator of $c(fg)$ corresponding to the coefficient of $x^{i+j}$ are in $\mathfrak p$ except for $a_ib_j$, so that $c(fg) \not\subset\mathfrak p$. It follows that $c(fg) \subseteq \mathfrak p \Longleftrightarrow c(f)c(g) \subseteq\mathfrak p$, and hence $\sqrt{c(fg)} = \sqrt{c(f)c(g)}$.     Can I go all the way and show that in fact $c(fg) = c(f) c(g)$? Putting a matrix inside of caption gives \caption has an extra } For-each over an array in JavaScript I'm trying to create a functioning sub theme of Omega 4.4. I've worked  with Drush before with it, but the site is on a host that I'm not administrator of. I've tried the  but not succeeded in executing the Omega Wizard. In short, the Omega Wizard may be extremely helpful, but I can't get it to work.   Next, I've tried to manually create a sub theme using the . However, the results of this is a theme catalogue in the Themes-folder, but the theme not showing up in Drupal. Perhaps I'm using the wrong instructions for the wrong version of Omega?  Anyhow, does anyone have instructions for how to install an Omega 4.4 sub theme manually for Drupal 7?