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How to only perform LTS upgrades? Can I create a Controller that simply returns an image asset?  I would like to route this logic through a controller, whenever a URL such as the following is requested:  www.mywebsite.com/resource/image/topbanner   The controller will look up topbanner.png and send that image directly back to the client.  I've seen examples of this where you have to create a View - I don't want to use a View. I want to do it all with just the Controller.  Is this possible? Interactive pixel information of an image in Python? Been doing tests in sandbox in which I've created about a million accountshares that I need to get rid of for testing.   I'm trying data loader in bulk api mode to export than delete but I keep getting:     InvalidBatch : Failed to process query: QUERY_TIMEOUT: Your query request was running for too long. Trying again later.;   Not sure if I should be waiting longer? Any other options or dataloader configurations I should be looking at?  EDIT  I've been trying with batch apex jobs yet still having issues.   Script  string str='select id from AccountShare where lastmodifiedbyid =\'0050P000007Kfy2\' limit 10000'; sharedeletebatch bdt=new sharedeletebatch(str); Database.executeBatch(bdt); system.debug(bdt);   Batch  global class sharedeletebatch Implements Database.batchable&lt;sobject&gt; {       global final string query;      global sharedeletebatch(string q){            query=q;      }       global Database.QueryLocator start(Database.BatchableContext BC){        return Database.getQueryLocator(query);      }      global  void execute(Database.BatchableContext BC,List&lt;SObject&gt; scope){          delete scope;      }     global void finish(Database.BatchableContext BC){     }   }      First error: [REQUEST_RUNNING_TOO_LONG] Your request was running for too long, and has been stopped. Can you explain the sentence structure 'In a hole in the ground there lived a Hobbit'? Why put the verb before the subject? The opening sentence to The Hobbit by J.R.R. Tolkien reads,     In a hole in the ground there lived [verb] a hobbit [subject].   I wonder if there are accepted stylistic purposes for such a structure.  When is it natural, and when is it unnatural? How can I create a /dev/null-like "blackhole" directory? I would like to create a "/dev/null" directory (or a "blackhole" directory) such that any files written to it are not really written, but just disappear.   I have an application that writes out large temporary files to a directory. I have no control over the name of the files and I don't really care about the content of these files. I could write a script that periodically clobbers these files, but the files are written out very quickly and fill my disk. I'm looking for something cleverer. I want the application to "think" that it is writing out these files, when in fact, the writes are just being discarded at the other end.  Also see  old related thread. When should I use "a" vs "an"? What is a good introductory book on quantum mechanics? I want to advance in the ranks of Senator Lavigny-Duval. How can I get merit points? I know I can participate in expansion goals, but is there a better way than that? Prove by induction $\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}$ for $n\ge1$ Prove the following statement $S(n)$ for $n\ge1$:  $$\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}$$  To prove the basis, I substitute $1$ for $n$ in $S(n)$:  $$\sum_{i=1}^11^3=1=\frac{1^2(2)^2}{4}$$  Great. For the inductive step, I assume $S(n)$ to be true and prove $S(n+1)$:  $$\sum_{i=1}^{n+1}i^3=\frac{(n+1)^2(n+2)^2}{4}$$  Considering the sum on the left side:  $$\sum_{i=1}^{n+1}i^3=\sum_{i=1}^ni^3+(n+1)^3$$  I make use of $S(n)$ by substituting its right side for $\sum_{i=1}^ni^3$:  $$\sum_{i=1}^{n+1}i^3=\frac{n^2(n+1)^2}{4}+(n+1)^3$$  This is where I get a little lost. I think I expand the equation to be  $$=\frac{(n^4+2n^3+n^2)}{4}+(n+1)^3$$  but I'm not totally confident about that. Can anyone provide some guidance? What are point releases in LTS versions? Etymology of "history" and why the "hi-" prefix? According to , history comes from the same root as story. If they are from the same word, where does hi- come from? Is it just because of the English habit of taking names from other languages verbatim or is there something more to it? I have been working on this exercise for a while now. It's in B.L. van der Waerden's Algebra (Volume I), page $19$. The exercise is as follows:     The order of the symmetric group $S_n$ is $n!=\prod_{1}^{n}\nu$. (Mathematical induction on $n$.)   I don't comprehend how we can logically use induction here. It seems that the first step would be proving $S_1$ has $1!=1$ elements. This is simply justified: There is only one permutation of $1$, the permutation of $1$ to itself.  The next step would be assuming that $S_n$ has order $n!$. Now here is where I get stuck. How do I use this to show that $S_{n+1}$ has order $(n+1)!$?  Here is my attempt: I am thinking this is because all $n!$ permutations of $S_n$ now have a new element to permutate. For example, if we take one single permutation $$ p(1,\dots,n) = \begin{pmatrix} 1 &amp; 2 &amp; 3 &amp; \dots &amp; n\\ 1 &amp; 2 &amp; 3 &amp; \dots &amp; n \end{pmatrix} $$ We now have $n$ modifications of this single permutation by adding the symbol $(n+1)$:  \begin{align} p(1,2,\dots,n,(n+1))&amp;= \begin{pmatrix} 1 &amp; 2 &amp; \dots &amp; n &amp; (n+1)\\ 1 &amp; 2 &amp; \dots &amp; n &amp; (n+1) \end{pmatrix}\\ p(2,1,\dots,n,(n+1))&amp;= \begin{pmatrix} 1 &amp; 2 &amp; \dots &amp; n &amp; (n+1)\\ 2 &amp; 1 &amp; \dots &amp; n &amp; (n+1) \end{pmatrix}\\ \vdots\\ p(n,2,\dots,1,(n+1))&amp;= \begin{pmatrix} 1 &amp; 2 &amp; \dots &amp; n &amp; (n+1)\\ n &amp; 2 &amp; \dots &amp; 1 &amp; (n+1) \end{pmatrix}\\ p((n+1),2,\dots,n,1)&amp;= \begin{pmatrix} 1 &amp; 2 &amp; \dots &amp; n &amp; (n+1)\\ (n+1) &amp; 2 &amp; \dots &amp; n &amp; 1 \end{pmatrix} \end{align}  There are actually $(n+1)$ permutations of that specific form, but we take $p(1,\dots,n)=p(1,\dots,n,(n+1))$ in order to illustrate and prove our original statement. We can make this general equality for all $n!$ permutations: $p(x_1,x_2,\dots,x_n)=p(x_1,x_2,\dots,x_n,x_{n+1})$ where $x_i$ is any symbol of our finite set of $n$ symbols and $x_{n+1}$ is strictly defined as the symbol $(n+1)$.  We can repeat this process for all $n!$ permutations in $S_n$. This gives us $n!n$ permutations. Then, adding in the original $n!$ permutations, we have $n!n+n!=(n+1)n!=(n+1)!$. Consequently, $S_n$ has order $n!$.  How is my reasoning here? Furthermore, is there a more elegant argument? I do not really see my argument here as incorrect, it just seems to lack elegance. My reasoning may well be very incorrect, however. If so, please point it out to me. Misplaced noalign after rowcolor in multicolumn How to change boot splash screen in 18.04 Measures: Atom Definitions Let $\Omega$ be a measure space with measure $\mu$. (Here, a measure is only meant to be countable additive!)  Consider a subset $A\in\Sigma$.  Then according to the  it is an atom if: $$(1)\quad\forall E\subseteq A:\quad\mu(E)&lt;\mu(A)\implies\mu(E)=0\quad(\mu(A)&gt;0)$$ and according to the  it is an atom if: $$(2)\quad\forall E\in\Sigma:\quad\mu(E\cap A)=0\lor\mu(E^c\cap A)=0\quad(\mu(A)&gt;0)$$  Now, these definitions agree for the really trivial cases: $$\mu\equiv0$$ $$\mu\equiv\infty$$ but they differ for the less trivial case: $$\quad\mu(E\neq\varnothing):=\infty,\,\mu(\varnothing):=0$$ namely the atoms are w.r.t. (1) all nonempty subsets whereas w.r.t. (2) only the singletons.  Excluding this pathological case, are the definitions equivalent? (More precisely, assume there exists a measurable subset $0&lt;\mu(F)&lt;\infty$.)  So far I checked that: $$(2)\implies(1):\quad\mu(E)&lt;\mu(A)\implies\mu(A\setminus E)&gt;0\implies\mu(E)=0$$ but what about the other direction? let $\gamma$ be a closed continuosly differentiable path in the upper half plane  not passing through $i$. What are the possible values of the integral $$\frac{1}{2\pi i}\int_{\gamma}\frac{2i}{z^2+1}dz$$ well the integral can be broken like  $$\frac{1}{2\pi i}\int_{\gamma}\frac{2i}{(z+i)(z-i)}dz=$$  $$\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z-i}dz-\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z+i}dz=$$ by Cauchy Integral Formuale $$f(i)-f(-i)$$,so the second integral is $0$ as it is analytic in the upper half plane, but the first integral iss just $n(\gamma,i)$, which iss the windding number of $\gamma$ around $i$ what more I can say? Thank you. Should 'fuck' be present in review audits (intentionally fake suggested edit reviews)? Why sometimes evidence is said to be too complex to compute and other times is negelected cause it is fixed? Change .jpg quality of terminal screenshots