I want to make a clay material like in this photo. Not a clay render, a clay material; something for my sculptures I model in Blender. How could I make something like the material in this image?    I have been trying to make this material for hours and it's driving me crazy. please help!  this is what i have gotten after playing around with nodes:
Sometimes, I find that it is useful to do a Clay Render for my model, where all of the materials render with a single common "clay"-like material (see example image below). In the past, I have done this by:   Selecting all of my objects by using the A hotkey Manually deselecting the World object, camera, and lights in the Outliner Linking materials using Ctrl+L Manually setting to the same material in the Properties>Materials menu Render like normal   When I was done with my Clay render, I would revert to the old version by pressing Ctrl+Z a few time. However, this is a rather clunky solution, since I can't make any edits in the clay version without losing my old materials/textures data.  Is there a better way to do Clay renders, either with an addon or through some other technique such that I can make edits to my model and not lose my materials? If at all possible, I would like a single setting that allows me to quickly toggle clay-mode on and off in the same file.  Example "Clay Render":
Antonym of "deepen" Since there is no word "shallowen", is there a good antonym of "deepen"? The meaning of "deepen" that I am referring to here is "to cause to become deeper." It doesn't necessarily refer to digging a hole. For example, a temporary deformation in a soft surface:     Putting the anvil on the mattress deepens the depression in it, but removing the anvil ______ it.   I would also be happy with an antonym of the intransitive version "to become deeper." If we know the GCD and LCM of two integers, can we determine the possible values of these two integers? I know that $\gcd(a,b) \cdot \operatorname{lcm}(a,b) = ab$, so if we know $\gcd(a,b)$ and $\operatorname{lcm}(a,b)$ and we want to find out $a$ and $b$, besides factoring $ab$ and find possible values, can we find these two integers faster? Split-like environment inside cases environment I'd like to have a cases environment with some cases that are too long to fit on one line. I tried this:  \begin{cases}   \begin{split}     long expression \\ second line of long expression   \end{split} &amp; condition \\   ...   but I get the warning     Package amsmath Warning: Cannot use 'split' here;   trying to recover with 'aligned' on input line 201.   It automatically replaces the split with an aligned environment, which looks like this:    Is there any way to get these long expressions to behave like they were in split or multline environments, with the top line flush left and the next line indented a little? Canonical replacing Unity with gnome-shell in Ubuntu 18.04 what now? The default desktop environment in Ubuntu 18.04 is going to be gnome-shell instead of Unity 8.  Does this mean development of Unity is stopped? or just that we can expect it to be beta a bit longer?  source on  post by Mark Shuttleworth. phoronix also posted about it  and here about the I don't think I've seen any answers to this question, but I'm curious as to why, when you have a 64-bit OS installed (ex: Vista x64), IE defaults to the 32-bit version? Are there drawbacks to using the x64 version of IE (IE8, specifically)? Evaluating $\int \sqrt{1 + t^2} dt$? How do we solve (i.e. get the closed form of) $\int \sqrt{1 + t^2} dt$ ?   shows the closed form of it but not the steps in solving it.  I think I need some algebraic trick..   I thought of some hyperbolic functions and their derivatives and equalities:  If I make $ t = \sinh x$, then I can reduce the integral to  $$ \int \cosh ^2 x dx $$ though I'm not sure if this is valid. But from here, I cannot go on evaluating the integral... False proof that $ρe^{iθ} = ρ$ and so complex numbers do not exist? My professor showed the following false proof, which showed that complex numbers do not exist. We were told to find the point where an incorrect step was taken, but I could not find it. Here is the proof: (Complex numbers are of the form $\rho e^{i\theta}$, so the proof begins there) $$\large\rho e^{i\theta} = \rho e^{\frac{2 \pi i \theta}{2\pi}} = \rho (e^{2\pi i})^{\frac{\theta}{2\pi}} = \rho (1)^{\frac{\theta}{2\pi}} = \rho$$ $$Note: e^{i\pi} = -1, e^{2\pi i} = (-1)^2 = 1$$ Since we started with the general form of a complex number and simplified it to a real number (namely, $\rho$), the proof can claim that only real numbers exist and complex numbers do not. My suspicion is that the error occurs in step $4$ to $5$ , but I am not sure if that really is the case. I am a little bit confused by early stopping and in particular by how it can be inserted inside a CV framework. As far as I understand, I can fix the optimal number of epochs (for NN, or number of trees for XGB) by early stopping, that is:   pick a validation set, train with increasing number of epochs until a predefined metric evaluated on the validation set starts worsening that is the optimal number of epochs to prevent overfitting   and that's fine. But then I would like to insert this early stopping framework inside a CV framework: Suppose I have a model with 10 hyperparameters I want to fix via CV. And suppose that there is an eleventh hyperparameter, the number of epochs. My feeling is that one can do like this:   create the K resampled folds, for each of which you have a training and validation set choose a suitable grid for your 10 hyperparameters for each point on the grid train your model in each fold with early stopping, that is use the validation set of the fold to keep track of the preferred metric and stop when it gets worse take the mean of the K validation metric choose the point of the grid (i.e. the set of hyperparameters) that gives the best metric   Questions:   What number of epochs should I choose as optimal? in each of the K folds I have, in general, a different number of stopping epochs.  seems to suggest they take the mean of the K stopping epochs. Is this right? Is it actually a "fair" practice to use the validation metric coming from the early stopping as a proxy of the out-of-sample metric? As  in section 3.4.5, maybe the best thing to do would be:     ...if you want to do early stopping, then in each fold you should take your training set and split it again, holding out a small early-stopping-set to guide the early stopping, and then evaluate the model on the validation set of that fold.   But this seems to me a really intricate process...   Unfortunately, I wasn't able to find references where this issue is presented in a clear and transparent form. Electron Flow Confusion Using conventional notation in the diagram below, I understand that the circuit is not complete because the diode is in reverse bias. But knowing that the electrons actually flow from the negative to the positive terminal(direction of the arrows), wouldn't the electrons take a path from the negative terminal of the battery, go through the lamp and stop at the diode(the red path in the picture). If they would, then why doesn't the lamp light up with the first electrons that are able to flow through it before the voltage in the wire is built up and the current stops. Why does space expansion not expand matter? I am trying to make a histogram with the fitted curve plotted on it. Here's the mwe:   \documentclass[a4paper]{scrartcl}     \usepackage{tikz}     \usepackage{pgfplots,pgfplotstable}     \usepackage{filecontents}     \pgfplotsset{compat=1.10}     \usepackage[active,tightpage]{preview}     \setlength\PreviewBorder{2pt}     \newcommand{\plots}{0.611201}     \newcommand{\plotm}{2.19882}     \begin{filecontents}{data.dat}     2 0.0629921259843     3 0.0236220472441     4 0.0314960629921     5 0.125984251969     6 0.0629921259843     7 0.102362204724     8 0.110236220472     9 0.0551181102362     10 0.0629921259843     11 0.0314960629921     12 0.0236220472441     13 0.0314960629921     14 0.0629921259843     15 0.0551181102362     16 0.0393700787402     17 0.0472440944882     18 0.00787401574803     19 0.0393700787402     24 0.00787401574803     27 0.00787401574803     33 0.00787401574803     \end{filecontents}     \begin{document}     \begin{preview}      \begin{tikzpicture}     \begin{axis}[     yticklabel style={/pgf/number format/fixed},     scaled y ticks = false,     minor y tick num={1},     xtick pos=left,     legend cell align = left,     legend style={draw=none},     xlabel = {group size},     ylabel = {ratio}     ]     \addplot[blue,ybar,fill, fill opacity=0.3, bar width = 0.8,] table {data.dat};     \addplot[red, line width = 1,domain=1:40,samples=100] {1/(x*sqrt(2*pi)*\plots)*exp(-(ln(x)-\plotm)^2/(2*\plots^2))};     \legend{empirical,lognormal fit}     \end{axis}     \end{tikzpicture}     \end{preview}     \end{document}     As you can see, the legend is not really good. How can I change the legend so it looks more like this? Review of a proof for De Moivre's theorem using mathematical induction The following is a rigorous proof of De Moivre's theorem by means of mathematical induction. The theorem put simply is that:  Any complex number, $z = a+bi$, on a cartesian plane can be expressed in polar form, where $a=r\cos\theta$ and $b=r\sin\theta$ and $r$ is the absolute distance from the origin to the point $z$. In light of this, the expression for the complex number $z$ in polar form is $z=r(\cos\theta+i\sin\theta)$ In order to find the $n^{th}$ power of $z$, the following rule applies: $$\displaystyle z^n=(r(\cos\theta+i\sin\theta))^n=r^n(\cos n\theta+i\sin n\theta)$$ where $n\in\mathbb N$ (I have confined $n$ to the natural numbers given that I intend on using mathematical induction to prove the theorem.  Base step: We must prove that what is stated is true for $n=1$, hence $$z^1=r^1(\cos 1\cdot\theta+i\sin1\cdot \theta)$$ $$=r(\cos\theta+i\sin\theta)$$ which is true. Next we assume it is true for $n=k$, and therefore proceed to the inductive step. Inductive step: When calculating $z^{n+1}$, it's the same as calculating $z^nz$, hence $$z^{n+1}=(r^n(\cos n\theta + i\sin n\theta))(r(\cos \theta+i\sin \theta))$$ $$=r^nr(\cos n\theta+i\sin n\theta)(\cos\theta +i\sin \theta)$$ $$=r^{n+1}(\cos n\theta\cdot\cos\theta+\cos n\theta \cdot i\sin\theta+i\sin n\theta\cdot\cos\theta+i^2\sin n\theta\cdot\sin\theta)$$ from here, $i^2=-1$, and therefore $+i^2\sin n\theta\cdot\sin\theta$ becomes $-\sin n\theta\cdot\sin\theta$. In addition to this, the following trigonometric identities will be used: $$\sin(n\theta + \theta) = \sin n\theta\cdot\cos\theta+\cos n\theta\cdot\sin\theta$$ $$or$$ $$k\sin(n \theta+\theta)=k\sin n \theta\cdot\cos \theta + \cos n\theta\cdot k\sin\theta$$ $$and$$ $$\cos(n \theta + \theta)=\cos n\theta\cdot\cos\theta-\sin n\theta\cdot\sin\theta$$ by using these, the original equation has now become $$=r^{n+1}(\cos(n \theta + \theta) + i\sin(n \theta + \theta))$$ then, after factoring out $\theta$, the theorem becomes $$=r^{n+1}(\cos((n+1)\theta)+i\sin((n+1)\theta))$$ which completes the proof. Question: is this proof correct all the way through? Or have I missed any conditions? If all is well in the case of this proof, could somebody aid me in proving it for $\{n|n\in\mathbb R,n\geq1\}$ How should one interpret the comparison of means from different sample sizes? Take the case of book ratings on a website. Book A is rated by 10,000 people with an average rating of 4.25 and the variance $\sigma = 0.5$. Similarly Book B is rated by 100 people and has a rating of 4.5 with $\sigma = 0.25$.   Now because of the large sample size of Book A the 'mean stabilized' to 4.25. Now for 100 people, it may be that if more people read Book B the mean rating may fall to 4 or 4.25.   how should one interpret the comparison of means from different samples and what are the best conclusions one can/should draw?    For example - can we really say Book B is better than Book A. Must I enter Europe with Schengen visa from the country where I applied? If I applied for a Schengen visa from "country A" and I got the visa but I had some commitments in "country B" and I have to finish it before going to Europe should I get back to "country A" to travel to Europe or can I use the visa from any where? Java code needs a system.out.println statement to run Presheaf which is not a sheaf -- holomorphic functions which admit a holomorphic square root I'm thinking about a problem in Ravi Vakil's algebraic geometry notes  (Exercise 3.2B) and I'm having trouble with the second part of the exercise as my understanding of complex analysis is quite rudimentary.   If to every open set $U\subset \mathbb{C}$ we associate the ring of holomorphic functions which admit a holomorphic square root, this defines a presheaf on $\mathbb{C}$ (I can see this).   Apparently however this is not a sheaf -- supposedly it fails to satisfy the gluing property.   To prove this, we need to take an open cover $\{U_i\}$ of some open $U\subseteq \mathbb{C}$, a holomorphic function $f_i:U_i\to \mathbb{C}$ for each $i$ which has a holomorphic square root (i.e. there exists a holomorphic function $g_i$ for each $i$ such that $f_i(z)=g_i(z)^2$ for all $z\in U_i$), which all agree nicely on the intersections of the various open sets, but such that it is impossible to find a holomorphic function $f:U\to \mathbb{C}$ with a holomorphic function which restricts to $f_i$ on each $U_i$.   Any help would be great -- my problem is in showing that such a function cannot exist. Use of "though" versus "however" Consider the sentence      E-books are on the rise, but they haven't suppressed paper books though.    This usage seems to be quite common, but when I learned English  I was taught to use "however" where I now read "though". My questions on this use of "though" are:     Is this a new trend? Is it restricted to American English? Is "though"  considered more colloquial than "however" ?    And on a syntactic level: Do you put a comma before "though"? If $\cos x=\cos y$ and $\sin x=\sin y$ is $x=y$ or $x=2\pi,y=0$ or $x=0,y=2\pi$? $x,y\in[0,2\pi]$ I know that if $\sin x=\sin y$ then $x=n\pi+(-1)^ny$  and $\cos x=\cos y$ implies that $x=y+2k\pi$ or $x=-y+2l\pi$. Since the latter does not fit the form above $x=y+2k\pi$ which implies that $k=0,1,-1$. Hence, we have x=y or $x=2\pi,y=0$ or $x=0,y=2\pi$. Why is there no absolute maximum temperature? If temperature makes particles vibrate faster, and movement is limited by the speed of light, then I would assume that temperature must be limited as well. Why is there no limit? Searching a virtual printer, that sends emails I need a software, that emulates a printer, converts the printed documents to PDF/TIFF and send it via eMail to a specified address (for windows).  The virtual printer / PDF conversion is possible with a lot of PDF-converters. But I need a software, which is also able to directly mail me these PDFs. It's always the same mailaddress, and I have an SMTP server.  Any suggestions?