Does anyone know if is there an effect to make something like the following images with photoshop? I thought it was halftone lines but this lines aren't straight.
I want to turn a photo in a engraving style picture, like  using GIMP and, if possible, then make an SVG out of it with Inkscape.  Can you recommend any filters/tools/plugins/tutorials for this?
"This question is not available, it may have been deleted" when submitting a question on the Android app after losing internet connection Every closed subset $E\subseteq \mathbb{R}^n$ is the zero point set of a smooth function In Walter Rudin's Principles of mathematical analysis Exercise 5.21, it is proved that for any closed subset $E\subseteq \mathbb{R}$, there exists a smooth function $f$ on $\mathbb{R}$ such that $E=\{x\in \mathbb{R}\mid f(x)=0\}$.   Since $E^c$ is open in $\mathbb{R}$, $E^c$ is a countable disjoint union of open intervals $(a_i,b_i)$. On an interval $(a,b)$, if we let $f(x)=\exp(\frac{1}{(x-a)(x-b)})$, $x\in(a,b)$; $f(x)=0$, $x\notin(a,b)$, ($f(x)=\exp(\frac{1}{(x-a)})$ or $f(x)=\exp(\frac{1}{(x-b)})$ on $(a,\infty)$ and $(-\infty,b)$ resp.), then $f$ is smooth. Hence we can get a smooth function on $\mathbb{R}$ such that $E$ is the zero point sets of $f$.  (1). How about a closed subset $E$ of $\mathbb{R}^n$? is it true that for any closed subset $E\subseteq \mathbb{R}^n$, there exists a smooth function $f$ on $\mathbb{R}^n$ such that $E=\{x\in \mathbb{R}^n\mid f(x)=0\}$?  An open set in $\mathbb{R}^n$ cannot be written as a disjoint union of countable open balls hence the proof above is not valid. The partition of unity only claims that for an  open set $U$, there exists open $V$ in $U$ such that $\bar V\subseteq U$ and a smooth function $f$ on $\mathbb{R}^n$ such that $supp f\subseteq U$, $f|_V=1$.   (2). The taylor series of $f(x)=\exp(\frac{1}{(x-a)(x-b)})$ does not converge around points $a,b$ hence $f$ is not analytic at $a,b$. Is it true that for any closed subset $E\subseteq \mathbb{R}$, there exists an analytic function $f$ on $\mathbb{R}$ such that $E=\{x\in \mathbb{R}\mid f(x)=0\}$? A  not only grants me a Sentinel Raven, but also the handy ability to see through its eyes.  Unlike the telepathic connection with a regular familiar, I don't have to use an action to do so.    If I am blinded, but my Sentinel Raven is not (and within telepathic range), can I use my raven's sight to cast spells that require me to see my target? Would this negate any disadvantage resulting from me being blinded? Or would I still suffer disadvantage, for example, because I have to cast from my raven's perspective and not my own point of view? Ellipsis in array initialization in C kernel module KDB: apply dyadic function across two lists How do I check which shell I am using? I read that terminal is nothing but shell, and Unix provides different flavors of shells:   Bourne shell (sh) C shell (csh) TC shell (tcsh) Korn shell (ksh) Bourne Again shell (bash)   Questions:   When I open a terminal window, which shell is opened by default? How do I check how many shells are installed? How do I change the shell used from my account? How to completely disable auto-save and versions in Mac OS X Lion? I hate the automatic saving and versioning in Mac OS X Lion.  Is there a way to completely turn it off?  (This has been asked before , but the answer and the linked article are completely wrong, so I thought I'd try again.) In  (scrolled to 5:55) there's a claim that replacing an AC high voltage transmission line with a high voltage DC transmission line allows for having seven times fewer powerlines (the video shows how seven lines evaporate, then one line appears and the cutting in the forest gets narrower).  How is that possible? I've read the  and it claims that     The power delivered in an AC system is defined by the root mean square (RMS) of an AC voltage, but RMS is only about 71% of the peak voltage. The peak voltage of AC determines the actual insulation thickness and conductor spacing.   but that doesn't explain seven times less power lines.  How is such massive (seven times) saving possible when switching from AC to DC transmission? could you help me on this integral ? $$\int \frac{1}{\cos(x)}\,\mathrm dx$$ Here's what I've started : $$\int \frac{1}{\cos(x)}\,\mathrm dx = \int \frac{\cos(x)}{\cos(x)^2}\,\mathrm dx = \int \frac{\cos(x)}{1-\sin(x)^2}\,\mathrm dx$$ Now, I did : $u = \sin(x)$, so $\mathrm du = 1$. Now I have : $$\int \frac{\text{???}}{1-u^2}\,\mathrm du$$ But at this point, I think I did the most of the job but I'm stuck. Could you help me to solve this integral please (to the integration by substitution at the end) ? Thanks EDIT : Now I follow the steps and I got : $$\int \frac{1}{1-u^2}\,\mathrm du$$ Doing the partial fraction I got $A = 1/2$ and $B = 1/2$. So basically I have : \begin{align} &amp; \int \frac{1}{1-u^2}\,\mathrm du = \int \frac{1/2}{1+u}\,\mathrm du + \int \frac{1/2}{1-u}\,\mathrm du \\[8pt] = {} &amp; \frac 1 2 \left(\int \frac{1}{1+u} \, du - \int \frac{1}{1-u} \, du\right) \\[8pt] = {} &amp; \frac 1 2 \ln\left(\frac{1+u}{1-u}\right) = \ln\left(\left(\frac{1+\sin(x)}{1-\sin(x)}\right)^{1/2}\right) \\[8pt] = {} &amp; \ln \left(\frac{\sqrt{1+\sin(x)}}{\sqrt{1-\sin(x)}}\right) = \ln\left(\frac{\sqrt{1+\sin(x)}}{\sqrt{\cos(x)^2}}\right) \\[8pt] = {} &amp; \ln\left(\frac{\sqrt{1+\sin(x)}}{\cos(x)}\right) \end{align} Here's what my teacher got :  What's wrong with what I did ? Did I miss something ? WMI object to get current sessions with client name? Odd behavior in a QGIS plugin: my function is triggered twice I have a plugin with a function (triggered by a button) that selects a specific layer by name, then shows its attribute table. When I close the plugin and open it again, the attribute table pops up twice, and repeating this keeps adding another table popup. What's wrong?  Code:  def showTable(self):     for lyrs in QgsMapLayerRegistry.instance().mapLayers().values():         if lyrs.name() == "TestFeatures":             lyrTest = lyrs     self.iface.showAttributeTable(lyrTest) How to mesh codeigniter with ajax Show $\text{Hom}_{Ab}(A,\mathbb{Q}) = \{0\}$ for finite $A$ I have a finite abelian group $A$ and I have shown that $A\otimes\mathbb{Q} \cong \{0\}$, the trivial group.  Now I want to show that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$, i.e we only have the trivial zero group homomorphism from $A$ to $\mathbb{Q}$.  The question I'm trying to solve is phrased in such a way that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$ should fall out almost automagically once you have that the tensor product collapses but I'm really stuck on this even though some people I talked to said it was really easy.  Some thoughts:   I've tried to assume the existence of a nontrivial homorphism $A \to    \mathbb{Q}$ to generate a contradiction but with no success. We have the Tensor-hom adjunction theorem stating that for abelian groups $A,B,C$ we have a natural bijection  $$\text{Hom}_{\textbf{Ab}}(A\otimes B,C) \cong \text{Hom}_{\textbf{Ab}}(B ,\text{Hom}_{Ab}(A,C))$$ and by plugging in our particular groups we get that:   $$\text{Hom}_{\textbf{Ab}}(A\otimes \mathbb{Q},\mathbb{Q})\cong\text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q}))$$  Since $A\otimes\mathbb{Q} \cong \{0\} \implies \text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q})) = \{0\}$ so we only have the trivial morphism from $\mathbb{Q} \to \text{Hom}_{Ab}(A,\mathbb{Q})$ but I don't think this implies that $\text{Hom}_{Ab}(A,\mathbb{Q}) = \{0\}$ and I'm not sure if this reasoning helps me at all. What *is* an Artificial Neural Network? As we delve into  literature, we get to identify other methods with neuromorphic topologies ("Neural-Network"-like architectures). And I'm not talking about the . Examples are given below.  Then, it makes me wonder: what is the definition of an artificial Neural Network? Its topology appears to cover everything.    Examples:  One of the first identifications we make is between PCA and a linear Autoencoder with tied-weights in the encoder and decoder and thresholded activations in the bottleneck layer.  Also, a common identification is done between linear models (logistic regression in special) and a Neural Network with no hidden layer and a single output layer. This identification opens several doors.  . ? ANN. Gaussian Process? ANN (with single hidden layer with infinite hidden units).  And so, just as easily, we can incorporate arbitrary regularized versions with specialized loss functions of these algorithms into a Neural Network framework.  But the more we dig, the more similarities appear. I just stumbled into , which makes the identification of a specific ANN architecture with decision trees, allowing these to be learned by ANN methods (such as Gradient Descent backpropagation). From this we can construct Random Forests and Gradient Boosted Decision Trees from solely Neural Network topologies.  If everything can be expressed as an Artificial Neural Network, what defines an Artificial Neural Network? I am trying to solve:  $$\int_{-\infty}^{\infty}\frac{x\sin{x}}{x^2 + a^2}dx$$  and show that it equals $\pi e^{-a}$ for all $a &gt; 0$.  I translate it to its complex equivalent:  $$lim_{R \rightarrow \infty} \int_{-R}^{R}\frac{z\sin{z}}{z^2 + a^2}dz = \int_{C}\frac{z\sin{z}}{z^2 + a^2}dz - \int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz$$  where C is a half circle with radius R, above the real axis, and $\gamma$ is only the curved (top part) of the half circle.  I've shown that $$\int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz$$ tends to 0 as $R \rightarrow \infty$.  So now:  $$lim_{R \rightarrow \infty} \int_{-R}^{R}\frac{z\sin{z}}{z^2 + a^2}dz = \int_{C}\frac{z\sin{z}}{z^2 + a^2}dz$$  By residue theorem,   $$\int_{C}\frac{z\sin{z}}{z^2 + a^2}dz = 2\pi i \cdot res_{ai}f$$  as $ai$ is the pole of $f$ that lies above the real axis.  $$2\pi i \cdot res_{ai}f = 2\pi i \cdot lim_{z \rightarrow ai} (z-ai)\frac{z sin{z}}{z^2 + a^2}$$  By L'Hospital's rule, this is equivalent to:  $$2\pi i \cdot lim_{z \rightarrow ai} (z-ai)\frac{z^2\cos{z} + 2z\sin{z} - zai\cos{z} - ai\sin{z}}{2z} = 2\pi i \cdot \frac{\sin{ai}}{2} = \pi i \sin{ai}$$  But how is $\pi i \sin{ai}$ equivalent to $\pi e^{-a}$? Did I make a mistake? Size in bytes is significantly less ( half the size) then size in MB in get info window In the get info window it says 387,037,139 bytes which is 387 MB whereas the size on disk is 716.9MB which is 716,900,000 bytes.   du -sh paramgen  result (in MiB) 684M    paramgen  Why is there such a large difference? Finding the dual basis for the space of polynomials of degree $\leq$ 2 Way to list "Contact Us" email address on web site, yet reduce likelihood of spam? What's a good way to list a "Contact Us" email address on a web site, while reducing the likelihood it will get spammed?  Is putting the email address in an image the best technique, or are there others? Learn QM algebraic formulations and interpretations I have a good undergrad knowledge of quantum mechanics, and I'm interesting in reading up more about interpretation and in particular things related to how QM emerges algebraically from some reasonable real world assumptions. However I want to avoid the meticulous maths style and rather read something more meant for physicists (where rigorous proofs aren't needed and things are well-behaved ;) ) I.e. I'd prefer more intuitive resources as opposed to the rigorous texts.       Can you recommend some reading to get started? For $G$ a group and $H\unlhd G$, then $G$ is solvable iff $H$ and $G/H$ are solvable? I recently read the well known theorem that for a group $G$ and $H$ a normal subgroup of $G$, then $G$ is solvable if and only if $H$ and $G/H$ are solvable. In my book, only the fact that $G$ is solvable implies $H$ is solvable was proven. I was able to show that if $H$ and $G/H$ are solvable, then so is $G$, but I can't quite show that $G$ is solvable implies $G/H$ is solvable.   My idea was this. Since $G$ is solvable, there exists a normal abelian tower $$ G=G_0\supset G_1\supset\cdots\supset G_r=\{e\}. $$ I let $K_i=G_i/(H\cap G_i)$, in hopes of getting a sequence $$ G/H\supset K_1\supset\cdots\supset K_r=\{e\}. $$ My hunch is that the above is also a normal abelian tower. However, I'm having trouble verifying that $K_{i+1}\unlhd K_i$ and that $K_i/K_{i+1}$ is abelian.  Writing $H_i=H\cap G_i$, I take $gH_{i+1}\in K_{i+1}$ for some $g\in G_{i+1}$. If $g&#39;H_i\in K_i$, then I want to show $g&#39;H_igH_{i+1}g&#39;^{-1}H_i$ is still in $K_{i+1}$, but manipulating the cosets threw me off. I also tried to use either the second or third isomorphism theorems to show that $K_i/K_{i+1}$ is abelian, but I'm not clear on how to apply it exactly. I'd be grateful to see how this result comes through. Thank you.