I am trying to change a blockdata as an entity (ArmorStand, but at his x y and z) it doesn't work! Any solutions?  /execute @e ~ ~ ~ detect ~ ~ ~ minecraft:dropper 0 {Items:[0:{Slot:4b,id:"minecraft:stone",Count:1b,Damage:0s}],id:"Dropper"} /blockdata ~ ~ ~ {......}
I was using this command to use a custom crafting table:  /execute @a ~ ~ ~ detect ~ ~-1 ~ dropper 1 {Items:[{id:minecraft:blaze_rod,Slot:1b,Count:1b,Damage:0s},{id:minecraft:blaze_rod,Slot:4b,Count:1b},{id:minecraft:stick,Slot:7b,Count:1b}]} /scoreboard players set @p LSC 1   But I get this error message when I stand on a dropper with these items:     Failed to execute '{Items:[{id:minecraft:blaze_rod,Slot:1b,Count:1b,Damage:0s},{id:minecraft:blaze_rod,Slot:4b,Count:1b},{id:minecraft:stick,Slot:7b,Count:1b}]} /scoreboard players set @p LSC 1' as TrevMCC38   By the way, Im in 16w06a
C++ templates that accept only certain types The cartesian product $\mathbb{N} \times \mathbb{N}$ is countable I'm examining a proof I have read that claims to show that the Cartesian product $\mathbb{N} \times \mathbb{N}$ is countable, and as part of this proof, I am looking to show that the given map is surjective (indeed bijective), but I'm afraid that I can't see why this is the case. I wonder whether you might be able to point me in the right direction?  Indeed, the proof begins like this:  "For each $n \in \mathbb{N}$, let $k_n, l_n$ be such that $n = 2^{k_n - 1} \left(2l_n - 1 \right)$; that is, $k_n - 1$ is the power of $2$ in the prime factorisation of $n$, and $2 l_n - 1$ is the (necessarily odd) number $\frac{n}{2^{k_n - 1}}$."  It then states that $n \mapsto \left(k_n , l_n \right)$ is a surjection from $\mathbb{N}$ to $\mathbb{N} \times \mathbb{N}$, and so ends the proof.  I can intuitively see why this should be a bijection, I think, but I'm not sure how to make these feelings rigorous?   I suppose I'd say that the map is surjective since given any $\left(k_n , l_n \right) \in \mathbb{N} \times \mathbb{N}$ we can simply take $n$ indeed to be equal to $2^{k_n - 1} \left(2l_n - 1 \right)$ and note that $k_n - 1 \geq 0$ and thus $2^{k_n - 1}$  is both greater or equal to one so is a natural number (making the obvious inductive argument, noting that multiplication on $\mathbb{N}$ is closed), and similarly that $2 l_n - 1 \geq 2\cdot 1 - 1 = 1$ and is also a natural number, and thus the product of these two, $n$ must also be a natural number. Is it just as simple as this?  I suppose my gut feeling in the proving that the map is injective would be to assume that $2^{k_n - 1} \left(2 l_n - 1 \right) = 2^{k_m - 1} \left(2 l_m - 1 \right)$ and then use the Fundamental Theorem of Arithmetic to conclude that $n = m$. Is this going along the right lines? The 'implicit' definition of the mapping has me a little confused about the approach.    On a related, but separate note, I am indeed aware that if $K$ and $L$ are any countable sets, then so is $K \times L$, so trivially, taking the identity mapping we see trivially that this map is bijective and therefore that $\mathbb{N}$ is certainly countable (!), and thus so is $\mathbb{N} \times \mathbb{N}$. Hence, it's not really the statement that I'm interested in, but rather the exciting excursion into number theory that the above alternative proof provides. Evaluating a PDF to find an unkown constant Evaluate the PDF to find K  $1 = \int_0^{\infty} \frac{k}{(1+x)^4}$  using u substitution I got:  $u=1+x$ and $du= dx$  $\int_0^{\infty} \frac{k}{u^4} \to \frac{k}{\frac{-1}{3} u^3}$   Since $u^4 du = \frac{u^{-4+1}}{-4+1}=$ $\frac{u^{-3}}{-3}$  which evaluating gives:  $1=\frac{k}{\frac{-1}{3(1+x)^3}} \vert_0^{\infty} \to \frac{-3k}{(1+x)^3} \vert_0^{\infty} = 0 -3k$   and $k= \frac{-1}{3}$  somehow though they came up with $\frac{-k}{3} \frac{1}{(1+x)^3}$ after integrating $\int_0^{\infty} \frac{k}{(1+x)^3}$   Was my approach correct to use substitution? Where does my work go wrong? Folland, "Real Analysis", Chapter 5.3, Exercise 36. Folland, "Real Analysis", Chapter 5.3, Exercise 36:     Let $\mathcal{X}$ be a separable Banach space and let $\mu$ be   counting measure on $\mathbf{N}$.  Suppose that   $\left\{x_n\right\}_1^\infty$ is a countable dense subset of the unit   ball of $\mathcal{X}$, and define $T \colon L^1(\mu) \to \mathcal{X}$   by $Tf = \sum_1^\infty f(n) x_n$.  (a)  $T$ is bounded.  (b)  $T$ is   surjective.   I have proved (a).  I would like help on (b).  Here are my ideas so far.    Say we want to show $y \in \mathcal{X}$ is in the image of $T$.  Reduce to the case $||y|| = 1$ and let $x_{n_m} \to y$.  My thought is to produce for each $N$ a function $f_N \in L^1(\mu)$ such that $||y - \sum_n^\infty f(n)x_n|| &lt; 1/N$ and also such that the $f_N$ converge in $L^1(\mu)$ to some $f \in L^1(\mu)$.  Then use the continuity of $T$ to conclude $T(f) = y$.    The way I have been setting up my inequalities is as follows. Pick $M_N$ such that for all $m &gt; M_N$, $||y - x_{n_m}|| &lt; 1/N$.  Then I want to define $f_N$ so that $$\sum_n^\infty f_N(n) = \sum_{m &gt; M_N}^\infty f_N(m) = 1, $$ hence $$||y - T(f_N)|| = ||\sum_{m &gt; M_N}^\infty f_N(m) y + \sum_n^\infty f_N(n)x_n|| \leq \sum_{m &gt; M_N}^\infty f_N(m) || y - x_{n_m} || &lt; \frac{1}{N}. $$  Presumably, if the $f_N$ are appropriately chosen, then I can find a dominating function and apply the Dominated Convergence Theorem to finish up.    However, getting all these hypotheses to hold simultaneously has been difficult.  In essence, the difficulty seems to be in requiring that the $f_N$ be "normalized" as well as possess a dominating function.  If this method is feasible, at least it seems to require a bit of analysis to show the desired series all converge.  Perhaps I am missing something cleaner.  If so, a simpler solution would be much appreciated.    -Thanks. Non-numerical citation mark like [Str91] [Samar] [Tro77] Taking back turns, and forgotten privileges What is the " := " operator in (VB).NET or what is it good for? How to make the term "page" show in the Glossary list? Black and white beads on a circle There are $n$ beads placed on a circle, $n\ge 3$. They are numbered in random order as viewed clockwise. Beads for which the number of the previous bead is less than the number of a next bead are painted in white color,and others - in black.  Two colourations that can be made equal by rotation are considered identical. How many different colourations can occur?  I've write a programm and for $n=3...11$ I've got answers $2 , 1 , 6 , 7 ,  18 , 25 , 58 , 93 , 186$ In Newtonian mechanics, if we throw an object in against direction of gravity with speed $v$ and it achieve max height of $h$. Now if we allow object to fall from that height $h$, it will eventually attain speed $v$ when it reach position where we launch it.   Now applying same idea to a black hole in general relativity. Speed require to escape black hole gravity is greater than $c$, so if we throw something into black hole with almost the speed of light, the object speed will exceed speed of light $c$ before hitting black hole surface! How relativity explain this? Can space-time curvature reduce speed of this freely falling object from attaining speed of light? is there a way to find suitable ratio for matches to controls (using "ratio" in Matchit package in R or other programs) I am using matchit package in R to find a control set for my treated units and I am testing the model with different ratio to get suitable control set. Is there a way to find suitable ratio and I will use that to find my control set. What's the derivative of: $ \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$? let $ y=\displaystyle \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$, i'm really interesting to know how do I find :  $\displaystyle \frac{dy}{dx}$ ?.  Note: I have used the definition of derivative of the square root function but i don't succed .  Thank you for any help I've got a Windows 10 Computer (home edition), which has 4GB of Ram and an i5 processor. I need to run an old 32 bit accounting programme. The vendor said it won't release a 64 bit version (although a 64bit version exists, it's missing key features from the older version).  The programme installs fine, but when I click on the programme icon, it says the programme won't work on a 64 bit system.   What options do I have?    I've done a fair bit of research, but it seems to run up dead ends.   1) WoW64     Apparently, this is supposed to let you run 32 bits on 64 bit Windows. However, I am not sure how to install it. I think it's for web servers only. However,  says it can work on Windows 7, but doesn't explain how to set it up (or if it will work on Windows 10).  2) XP mode     A 32bit Windows XP emulator. However, it seems to be for Windows 7 only. Also according to this , you need Windows Professional.   3) Virtual Box Machine    Let's you emulate Windows. Seems overkill for my purposes. Plus I would need to buy a 32 bit Windows license. I think it would run slowly with 4GB of Ram.  4) Partition drive into 32bit and 64 bit versions Apparently, you can use the free tool  to split the drive into two and then install separate 32bit and 64 bit versions.   Again, seems overkill and it looks like I would need buy a 32 bit Windows 10 license.   Are these my only options?     Images  Installation    First run (From the error message I guess it's a 16-bit program)   Clicking on programme Is there an equivalent of 'which' on the Windows command line? As I sometimes have path problems, where one of my own cmd scripts is hidden (shadowed) by another program (earlier on the path), I would like to be able to find the full path to a program on the Windows command line, given just its name.  Is there an equivalent to the UNIX command 'which'?  On UNIX, which command prints the full path of the given command to easily find and repair these shadowing problems. When I gain control of a creature, do I also gain control of any auras or equipment attached to it? What would happen if I 'd a creature that had been enchanted by a bestow creature and cast a spell to give it protection from the enchantment's color on it to remove the bestowed creature?  For instance:  They have just bestowed a  on a  and passes the turn. I use Portent of Betrayal to take control of the Asphodel Wanderer, attack and in the second main I cast Gods Willing to give the Asphodel Wanderer protection from black. Nighthowler detaches and where does it go, my control, opponents control or graveyard?  As an additional question, in the same vein:  What if I destroyed the creature while it was on my side? So change the gods willing to a hero's downfall in the example.  Relevent (under state-based actions):     704.5n If an Aura is attached to an illegal object or player, or is not attached to an object or player, that Aura is put into its owner's graveyard. I have a Python regular expression that contains a group which can occur zero or many times - but when I retrieve the list of groups afterwards, only the last one is present. Example:  re.search("(\w)*", "abcdefg").groups()  this returns the list ('g',)  I need it to return ('a','b','c','d','e','f','g',)  Is that possible? How can I do it? Performing multiple layer intersection in QGIS? I have a set of 10 tables and each has an extra column that I don't want. Specifically, these tables are descriptions of database tables exported by phpMyAdmin, and they all have a "MIME" column that I would like to remove. Is there a way to do this other than using string manipulation / find and replace?   I have tried setting the column to p{0in}, but the column title still remains. PHP regex delimiter, what's the point? The Cartesian product of a family $(A_i\mid i\in I)$ is defined as $$\prod\limits_{i\in I}A_i=\{f:I\to\bigcup A_i\mid f(i)\in A_i\}$$ Let $I_n=\{i \in \mathbb N \mid i \le n\}$ for $n\in \Bbb N$ and $A_i$ be countable for all $i\in I_n$. Then $\prod\limits_{i\in I_n}A_i$ is countable.   My attempt: Lemma 1: $(\prod\limits_{i\in I_n}A_i) \times A_{n+1}$ and $\prod\limits_{i\in I_{n+1}}A_i$ are equinumerous for all $n \in \mathbb N$. (I presented a proof ) Lemma 2: If $A$ and $B$ are countable, then $A \times B$ is countable. (I presented a proof ) I will prove the theorem by induction on $n$. The theorem is trivially true for $n=0$. Assume that the theorem is true for $n=k$, then $\prod\limits_{i\in I_k}A_i$ is countable. Since $\prod\limits_{i\in I_k}A_i$ is countable (by inductive hypothesis) and $A_{k+1}$ is countable, then by Lemma 2 $(\prod\limits_{i\in I_n}A_i) \times A_{n+1}$ is countable. Furthermore, by Lemma 1 $(\prod\limits_{i\in I_k}A_i) \times A_{k+1}$ and $\prod\limits_{i\in I_{k+1}}A_i$ are equinumerous, then $\prod\limits_{i\in I_{k+1}}A_i$ is countable too. Thus the theorem is true for $n=k+1$. This completes the proof.   Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!