I am working on an adventure mode map on the Nintendo Switch Bedrock Edition, and so I am trying to give myself a diamond pickaxe which can only break mob spawners and is unbreakable. I have currently:  /give @s diamond_pickaxe 1 0 {CanDestroy:["minecraft:mob_spawner"],Unbreakable:1}   and whenever I enter the command I get:      Line 1, Column 2 missing '}' or object member name.   I have gone off of a couple tutorials and this is the closest I can get to having it work.  Can anyone please tell me where the error is?
This question was recently split into different questions for each workaround. See   meta posts for more information.   Can you use NBT tags in commands? I’ve tried commands that normally work in Java Edition like: give @s minecraft:planks 30 1 {display:{Lore:[&quot;Wooden planks&quot;]}} give @s stone_pickaxe{Unbreakable:1b} give @s item_frame{EntityTag:{Invisible:1b}} summon villager ~ ~ ~ {Offers:{Recipes:[{buy:{id:&quot;emerald&quot;,Count:5b},sell:{id:&quot;diamond_sword&quot;}}]}} summon tnt ~ ~ ~ {Fuse:20s}
Let $y$ be a differentiable function on $\mathbb{R}$. If  $$\lim_{x\rightarrow \infty}(y(x)+y'(x))=0$$ Then how does one show that $$\lim_{x\rightarrow \infty}y(x)=0$$ I'd appreciate some help on this problem. Thanks a lot. Error sending edittext in java server. See code Companion affinity change notification What's a concise way to check that environment variables are set in a Unix shell script? Prediction interval for binomial random variable What is the formula (approximate or exact) for a prediction interval for a Binomial random variable?  Assume $Y \sim \mathsf{Binom}(n, p)$, and we observe $y$ (drawn from $Y$). The $n$ is known.  Our goal is to obtain a 95% prediction interval for a new draw from $Y$.   The point estimate is $n\hat{p}$, where $\hat{p}=\frac{y}{n}$. A confidence interval for $\hat{p}$ is straightforward, but I cannot find a formula for a prediction interval for $Y$. If we knew $p$ (rather than $\hat{p}$), then a 95% prediction interval just involves finding the quantiles of a binomial. Is there something obvious I'm overlooking? Combinatorial Proof Question: Vandermonde convolution identity I'm really iffy on combinatorial proofs in general and now that there is a sum, it's just confused me even more. Can someone try and walk me through this proof?  $$ \binom{m + n}{r} = \sum_{k=0}^r \binom{m}{k}\binom{n}{r - k} $$ Another meta question identified a bug that was marked as designed. To summarize the bug, the little number that appears next the Review link in the header shows an incorrect value most of the time.         I would like to propose that the feature be dropped. Honestly, I would prefer it be fixed, but the response to the question above gives me little faith that it will be. Half of the time, I click that little brown box and find out that I was lied to. There's nothing to review.  The little number no longer does the job it was created to do.        Oh sweet! Three things I can review!        Just kidding... ASCII to Binary and Binary to ASCII conversion tools? Which is a good tool to convert ASCII to binary, and binary to ASCII?  I was hoping for something like:  $ echo --binary "This is a binary message" 01010100 01101000 01101001 01110011 00100000 01101001 01110011 00100000 01100001 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00100000 01101101 01100101 01110011 01110011 01100001 01100111 01100101   Or, more realistic:  $ echo "This is a binary message" | ascii2bin 01010100 01101000 01101001 01110011 00100000 01101001 01110011 00100000 01100001 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00100000 01101101 01100101 01110011 01110011 01100001 01100111 01100101   And also the reverse:  $ echo "01010100 01101000 01101001 01110011 00100000 01101001 01110011 00100000 01100001 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00100000 01101101 01100101 01110011 01110011 01100001 01100111 01100101" | bin2ascii This is a binary message   PS: I'm using bash  PS2: I hope I didn't get the wrong binary In a question related to introductory Galois theory, I was asked to given an example of a tower of fields $F \subset K \subset E$ such that $E$ is a splitting field for some polynomial $f(x) \in F[x]$, but that $K$ need not be a splitting field. The example that comes to mind for me is $\mathbb Q \subset \mathbb Q(\sqrt[3]{2}) \subset \mathbb Q(\omega_3, \sqrt[3]{2})$.  So I know that the minimal polynomial for $\mathbb Q(\omega_3, \sqrt[3]{2})$ over $\mathbb Q$ is $x^3-2$. This polynomial does not split in the intermediate field $\mathbb Q(\sqrt[3]{2})$.  I wanted to make sure that I'm justifying that this intermediate field is not a splitting field correctly. Now I may be confused about the books terminology. When we say something is not a splitting field, we always have to refer to a specific polynomial correct? I've read a term called a normal extension which I think refers to one in which every polynomial splits, but my book doesn't mention these.   If splitting field means with respect to the polynomial $x^3-2$, then I've already demonstrated that $\mathbb Q(\sqrt[3]{2})$ is not a splitting field. Also, it couldn't be a normal extension either because that polynomial doesn't split.  Does it sound like I'm understanding this correctly or are there some subtleties I might be missing? Is there a way to detect if a browser window is not currently active? How to change a default application in Ice Cream Sandwich? I used the ADWLauncher EX by default while having Gingerbread 2.3.6.  Today I have manually updated my Nexus S to Ice Cream Sandwich 4.0.3, and I wanted to give the new default launcher a shot, but I cannot find anymore where to select / forget the default application for an action.  I've turned the settings of my phone inside out, googled and crawled the forums, but without success. If anybody can tell me how to change a default application in ICS, I'd be grateful!      Yeah, I never thought I'd want to see that Complete action using... window again! :D Best way for a 'forgot password' implementation? I've recently installed Ubuntu on Oracle's Virtualbox.  Now I want to install the guest additions.  I started up my Ubuntu guest OS, clicked "Devices" and chose "Install Guest Additions". Nothing happened, and I don't see the CD icon on my desktop.  How can I make this work? Is a constant condition more costly than switching shaders? In general, branching in shaders is not a good idea. But now I have a shader with a condition that is constant with respect to the entire draw call. So the branch that is executed is always the same for one draw call.  Is such kind of branching still more costly than having multiple shaders without these branches and switch between them? Referencing bib files with spaces in the filename When I'm editing a stack snippet, I often want to copy code from the original question into the snippet. But while the snippet editor is on display, there's no way to access the question, other than opening another tab/window. What I often do instead is copy all the code in the question to the clipboard, open the snippet editor, then paste it all in and revise it to what I want. But when the original code is poorly designed, that ends up being more work than just copying the bits and pieces that I want.  There could be a "View Question" button that pops up the question over the snippet editor and allows copying. Or maybe the Snippet Editor could be draggable and resizable, so you can reveal the question and copy from it; but this would require that it not be purely modal. Sum of infinite series $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}$ Lets define the following series $$S_n =\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}.$$ I know that $S_0$ does not converge so let's suppose $n \in N$ and define $S_n$ for some $n$. We have $S_1=1$ , $S_2=\frac1 4$ , $S_3=\frac 1 {18}$ , $S_4=\frac 1{96}$ , $S_5=\frac 1{600}$ etc.. the numerator in all results is 1the pattern in denominator is $[1,4,1,96,600...]$ and   that is equal to $n*n!$ Finally I want to prove the general equality: $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}=\frac 1 {n*n!}$ Relevance of cycle detection in Pollard's rho integer factorization algorithm I've recently learned about Pollard's rho factorization algorithm, and the way I understand it, it works by using a pseudorandom sequence of integers $x_{k}$ ranging from $0$ to $n-1$, where $n$ is the number to be factored, to find repetitions in an unknown sequence $y_{k} = x_{k} \bmod p$, where $p$ is an unknown, non-trivial factor of $n$. Such a repetition $y_{i}=y_{j}$ occurs when $x_{i}\equiv x_{j} \bmod p$, which can be detected by checking whether $\gcd(x_{i}-x_{j}, n)\gt 1$.  From what I could find on the Internet, Floyd's cycle detection algorithm or a variant thereof is used to find such repetitions. And the reason for this is what I don't understand.   First, Floyd's algorithm requires that each element in the sequence depends solely on the element before it, which, even if it is the case for $x_{k}$, is not necessarily the case for $y_{k}=x_{k}\bmod p$. For example, assuming we want to factorize $6$, let's take the sequence $5, 3, 2, 4, 1, 0, 5, 3, 2, 4, 1, 0,\ldots$. The cycle period is $5, 3, 2, 4, 1, 0$, where every element is contained only once. However, if we take this sequence $\bmod 2$, we get $1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0,\ldots$. Floyd's algorithm cannot be applied here to find the cycle period $1, 1, 0, 0, 1, 0$. Besides, we are not even interested in a cycle, only in repetitions. Whether we find the congruence $\bmod 2$ between $5$ and $3$, $5$ and $1$, or $4$ and $2$ in the first sequence is irrelevant.   Please correct me if my understanding of Pollard's rho algorithm as described above is wrong. But if it is not, it seems to me that one could just as well try out some random numbers between $2$ and $n-1$ and see if they share a divisor with $n$ other than $1$. So what is the benefit of using a cycle detection method like Floyd's algorithm?  I also found , but it is not answered, and the comment on the question does not really answer the question either. Are there names for consonant-shifts when suffixes are added? Prove if $|a_{n+1}|/|a_n| \leq |b_{n+1}|/|{b_n}|$ and $\sum |b_n|$ is convergent, then $\sum |a_n|$ is convergent