I'm just doing a basic topograph and the interpolate line tool does not work. The DEM is selected and I even made a hillshade to see if that would solve the problem. After I finish the line and double click the line disappears. There is no error message, just a sound as if I was clicking on the wrong window but there are no other windows open. I've restarted GIS, I've made a new map, I've changed where the topo data goes, I've deactivated and reactivated 3D analyst and I've changed the coordinate system to a new one and back to the original and it still does not work.
I'm attempting to make an elevation profile from a DEM in ArcMap using the Interpolate Line tool.  I've done this before, and it has worked great--basically I just draw a line with Interpolate Line, then click Profile Graph to get the elevation profile.  However, for some reason, it's not working now: when I draw out my line, nothing shows up on the map.  I don't get a line, and I can't figure out what I'm doing wrong.  I've searched online for an answer, but to no avail.    Any tips?
How do I clear keyboard/autocorrect history on the Galaxy s4? I let my younger cousins play on my S4 from time to time, but now I have found that all the words they have used are the first to come up when I use my phone to text. It has gotten quite irritating. Samsung seems to have very hard to find settings compared to my old Sony Ericsson so I cannot find where to get rid of the word history typed onto my phone's keyboard. Can somebody give me an example of a graph on eight vertices that has a pair of pseudosimilar vertices? Or, is it not possible? If it is not possible, please do not give me a proof, as I would like to figure out how to prove it myself.     Definition: vertices $u$ and $v$ in a graph $X$ are similar if there is an automorphism of $X$ that maps $u$ to $v$. If $u$ and $v$ are similar, then the vertex-deleted subgraphs $X-u$ and $X-v$ are isomorphic. If $X-u$ and $X-v$ are isomorphic but $u$ and $v$ are not similar, we say that they are pseudosimilar. Subscribe to tags I am interested in? How do I subscribe to the tags that I am interested in?  I like to get notified via email or so, when someone asks new questions or if someone responds to my questions. Proving that $\lim_{x \to 0} \frac{f(x)}{|x|}=1$ implies $f$ is not differentiable at $0$. Proving $\sin(x^2)$ Continuous but Not Uniformly Continuous with non-standard Definition Show that the bounded function $f(x) = \sin(x^2)$ is continuous, but not uniformly continuous, on the interval $[0,\infty)$.  The problem should be solved using the following definitions: Definition of Continuous Function Let $X$ and $Y$ be metric spaces. We sat that a function $f:X\rightarrow Y$ is continuous if given any open set $V \subseteq Y$, its inverse image $f^{-1}(V)$ is an open subset of $X$. Definition of Uniformly Continuous Function Let $(X,d_x)$ and $(Y,d_y)$ be metric spaces. If $E \subseteq X$, then we say that a function $f: X\rightarrow Y$ is uniformly continuous on $E$ if for every $\epsilon &gt; 0$ there exists a $\delta &gt; 0$ such that for all $x,y \in E$ we have $$d_x(x,y) &lt; \delta \longrightarrow d_y(f(x),f(y)) &lt; \epsilon$$ For the continuity part it is clear that $f^{-1}(a,b), (a,b) \subseteq (-1,1) = \{x:f(x) \in (a,b)\}$ which appears to be an open set however I doubt this is sufficient as a proof since it holds for any function $f$. Not sure how to go about proving $f$ not uniformly continuous. Seems to be difficult to prove from the definition alone. Shapefile layer no longer editable (toggle edit option greyed out) in QGIS What would be the symbol for 'multiplication on division'? The symbol I would like to obtain ought to be like the symbol '\pm'. When one wants to list various cases/classes/categories/types of objects in a string of conjunctions, is it preferable (or even mandatory) to keep on using (the same) preposition in front of each one them?  Moreover, does singular or plural form of the word "case", "class", "category" make any difference in the previous question?  Examples:   In the present article, we study the case of X, of Y and of Z objects. (in this example, as in most of the ones I have in mind, it is the preposition "of" which is relevant to my question). We study the cases/classes of X and of Y objects. vs. We study the cases/classes of X and Y objects.   Although it is rather clear that, in the singular case, the absence of "of" in later conjuncts may lead to a logical confusion (e.g., mistakenly taken as the one class/case of objects which are both X and Y), when it comes to plural of the words "case", "class", etc., this might not be an issue. Still, there might anyway be a rule in the plural setting as well. "Didn't use to get" or "Didn't used to get"? I just made a minor , and  the grammar correction was unnecessary.  The original text was     Another factor--sometimes life happens and you can't fly. In the old days you could simply sell your ticket to someone else, now you either have to eat a hefty change fee or lose it outright. That's money in their pockets that they didn't used to get.   It was corrected to:     Another factor--sometimes life happens and you can't fly. In the old days you could simply sell your ticket to someone else, now you either have to eat a hefty change fee or lose it outright. That's money in their pockets that they didn't use to get.    Was this grammatical correction an improvement? Was the original text grammatically correct? Is the edited text grammatically correct? I'm from the State of Kuwait and hold an Eritrean passport. I'm planning to visit UK in October but my visa will expire in November.    Is it OK to make a journey in the last month of my visa, provided that I make sure that I book my plane back before expiring date? Creating symbol with size and color based on two variables in ArcMap? Convert IEnumerable to DataTable How to copy files via terminal? I know how to solve "simple" recurrence relations. For instance, say you have:  $$c_0 = 20$$ $$c_1 = 30$$ $$c_n = 3 c_{n-1} - 2 c_{n-2}$$  We can write the characteristic equation as:  $$3x^{n-1} - 2x^{n-2} = x^n$$  Solving this with $n=2$, we get $x = 1$ or $x = 2$. This lets us write the relation $c_n = \alpha_1 1^n + \alpha_2 2^n$, and we can solve for $\alpha_1$ and $\alpha_2$ with the initial states $c_0$ and $c_1$.  However, this depends on the fact that $3x^{n-1} - 2x^{n-2} = x^n$ has two roots.  Now, I'm stuck on another problem where the characteristic equation has fewer roots than terms.  Say I have this recurrence relation instead:  $$a_0 = 0$$ $$a_1 = 2$$ $$a_2 = −1$$ $$a_n = 9a_{n-1} - 15a_{n-2} - 25a_{n-3}$$  The characteristic equation would be:  $$9x^{n-1} - 15x^{n-2} - 25x^{n-3} = x^n$$  However, solving with $n=3$, we only get two roots: $x=-1$ or $x=5$. There are not enough roots to write a relation in the form of $a_n = \alpha_1 r_1^n + \alpha_2r_2^n + \alpha_3r_3^n$. How do I proceed? Can I list-initialize a vector of move-only type? I have just booted my clean Ubuntu 18.04 LTS VM and tried to install some application with sudo apt install.  But got error message:     E: Could not get lock /var/lib/dpkg/lock-frontend - open (11: Resource temporarily unavailable)       E: Unable to acquire the dpkg frontend lock (/var/lib/dpkg/lock-frontend), is another process using it?   What process created the /var/lib/dpkg/lock-frontend file in Ubuntu? How to gently stop it and take manual control on APT? Counting overlapping polygons in GeoJSON to result in new set of polygons? Question on groups of order $pq$ I have a simple question regarding "conditional probability" and "Likelihood". (I have already surveyed this question  but to no avail.)  It starts from the Wikipedia . They say this:     The likelihood of a set of parameter values, $\theta$, given   outcomes $x$, is equal to the probability of those observed outcomes   given those parameter values, that is      $$\mathcal{L}(\theta \mid x) = P(x \mid \theta)$$   Great! So in English, I read this as: "The likelihood of parameters equaling theta, given data X = x, (the left-hand-side), is equal to the probability of the data X being equal to x, given that the parameters are equal to theta". (Bold is mine for emphasis).  However, no less than 3 lines later on the same page, the Wikipedia entry then goes on to say:     Let $X$ be a random variable with a discrete probability distribution   $p$ depending on a parameter $\theta$. Then the function      $$\mathcal{L}(\theta \mid x) = p_\theta (x) = P_\theta (X=x), \, $$      considered as a function of $\theta$, is called the likelihood   function (of $\theta$, given the outcome $x$ of the random variable   $X$). Sometimes the probability of the value $x$ of $X$ for the   parameter value $\theta$ is written as $P(X=x\mid\theta)$; often   written as $P(X=x;\theta)$ to emphasize that this differs from   $\mathcal{L}(\theta \mid x) $ which is not a conditional probability,   because $\theta$ is a parameter and not a random variable.   (Bold is mine for emphasis). So, in the first quote, we are literally told about a conditional probability of $P(x\mid\theta)$, but immediately afterwards, we are told that this is actually NOT a conditional probability, and should be in fact written as $P(X = x; \theta)$?   So, which one is is? Does the likelihood actually connote a conditional probability ala the first quote? Or does it connote a simple probability ala the second quote?  EDIT:  Based on all the helpful and insightful answers I have received thus far, I have summarized my question - and my understanding thus far as so:    In English, we say that: "The likelihood is a function of parameters, GIVEN the observed data." In math, we write it as: $L(\mathbf{\Theta}= \theta \mid \mathbf{X}=x)$. The likelihood is not a probability.  The likelihood is not a probability distribution.  The likelihood is not a probability mass. The likelihood is however, in English: "A product of probability distributions, (continuous case), or a product of probability masses, (discrete case), at where $\mathbf{X} = x$, and parameterized by $\mathbf{\Theta}= \theta$." In math, we then write it as such: $L(\mathbf{\Theta}= \theta \mid \mathbf{X}=x) = f(\mathbf{X}=x ; \mathbf{\Theta}= \theta) $ (continuous case, where $f$ is a PDF), and as $L(\mathbf{\Theta}= \theta \mid \mathbf{X}=x) = P(\mathbf{X}=x ; \mathbf{\Theta}= \theta) $ (discrete case, where $P$ is a probability mass). The takeaway here is that at no point here whatsoever is a conditional probability coming into play at all.  In Bayes theorem, we have: $P(\mathbf{\Theta}= \theta \mid \mathbf{X}=x) = \frac{P(\mathbf{X}=x \mid \mathbf{\Theta}= \theta) \ P(\mathbf{\Theta}= \theta)}{P(\mathbf{X}=x)}$. Colloquially, we are told that "$P(\mathbf{X}=x \mid  \mathbf{\Theta}= \theta)$ is a likelihood", however, this is not true, since $\mathbf{\Theta}$ might be an actual random variable. Therefore, what we can correctly say however, is that this term $P(\mathbf{X}=x \mid \mathbf{\Theta}= \theta)$ is simply "similar" to a likelihood. (?) [On this I am not sure.]   EDIT II:  Based on @amoebas answer, I have drawn his last comment. I think it's quite elucidating, and I think it clears up the main contention I was having. (Comments on the image).    EDIT III:   I extended @amoebas comments to the Bayesian case just now as well: Let $(a_n)_{n}\in\ell^2:=\ell^2(\mathbb{R})$ be a fixed sequence. Consider the subspace $$C=\{(x_n)_{n}\in\ell^2 : |x_n|\le a_n\text{ for all }n\in\mathbb{N}\}.$$  According to the book [Dunford and Schwartz, Linear operators part I, page 453] $C$ is compact in the $\ell^2$-norm, but there is no proof. How can I show that $C$ is indeed compact in $\ell^2$ ?