update the badge count with remote notifications
Push Notification Badge Count Not Updating
Proof : $2^{n-1}\mid n!$ if and only if $n$ is a power of $2$. I want to prove that:  $2^{n-1}\mid n!$ if and only if $n$ is a power of $2$. I bought a new SSD with 750 GB and made two partitions: 600 GB and 150 GB. The smaller one was thought to be a Bootcamp partition one day, but to date it was just another unused partition. When I needed more space, I decided to merge these two partitions. The partition was created with Mavericks, the merge was done with Yosemite.  I opened Disk Utility and the "minus" button. Disk Utility told me both partitions would be merged then. However it miserably failed. My main hard disc suddenly had errors which couldn't be repaired and even two of my other external hard discs could not be mounted any longer.  I eventually restored the hard disc from Time Machine, which worked well.  My second partition is now listed as free:    Unfortunately I can't resize the 600 GB partition nor can I create a new partition using the free space.  Diskutil reveils the following:  ~  diskutil list /dev/disk0    #:                       TYPE NAME                    SIZE       IDENTIFIER    0:      GUID_partition_scheme                        *750.2 GB   disk0    1:                        EFI EFI                     209.7 MB   disk0s1    2:          Apple_CoreStorage                         599.6 GB   disk0s2    3:                 Apple_Boot Recovery HD             650.0 MB   disk0s3 /dev/disk1    #:                       TYPE NAME                    SIZE       IDENTIFIER    0:                  Apple_HFS Macintosh HD           *599.3 GB   disk1                                  Logical Volume on disk0s2                                  XXXXXX-XXXX-XXXX-XXXX-XXXXXXXXXXXX                                  Unencrypted   Also diskutil brought up this:  ~  diskutil cs list CoreStorage logical volume groups (1 found) | +-- Logical Volume Group 9D2BA4BA-xxxx-xxxx-xxxx-xxxxxxxxxxxx     =========================================================     Name:         Macintosh HD     Status:       Online     Size:         599640592384 B (599.6 GB)     Free Space:   6111232 B (6.1 MB)     |     +-&lt; Physical Volume 29A9A02B-xxxx-xxxx-xxxx-xxxxxxxxxxxx     |   ----------------------------------------------------     |   Index:    0     |   Disk:     disk0s2     |   Status:   Online     |   Size:     599640592384 B (599.6 GB)     |     +-&gt; Logical Volume Family 2FA7892F-xxxx-xxxx-xxxx-xxxxxxxxxxxx         ----------------------------------------------------------         Encryption Status:       Unlocked         Encryption Type:         None         Conversion Status:       NoConversion         Conversion Direction:    -none-         Has Encrypted Extents:   No         Fully Secure:            No         Passphrase Required:     No         |         +-&gt; Logical Volume 7BF42B7B-xxxx-xxxx-xxxx-xxxxxxxxxxxx             ---------------------------------------------------             Disk:                  disk1             Status:                Online             Size (Total):          599282155520 B (599.3 GB)             Conversion Progress:   -none-             Revertible:            No             LV Name:               Macintosh HD             Volume Name:           Macintosh HD             Content Hint:          Apple_HFS   I have tried this:    fsck -fy (The volume Macintosh HD appears to be OK.)   but it wouldn't help. It said I couldn't recover because it wasn't created in a way it would support (sorry for missing the exact message).  I then have learned that Apple_CoreStorage might indicate my disk is encrypted, which might explain why I can't resize it. It doesn't explain why I can't create a new partition to me.  What are my options to make my hard disk available in a single partition with size 750 GB? Mail on Big Sur does not reliably find dollar values Show that a connected metric space is $\epsilon$-chainable for $\epsilon>0$ Show a connected metric space (X,d) is $\epsilon$-chainable for $\epsilon &gt;0$       $\epsilon$-chainable Definition      (X,d) is $\epsilon$-chainable if given any two points $a,b \in X$, $\exists$ a finite set $\{ x_0,x_1,...,x_n\}$ of points in X with $x_0 = a$, $x_n=b$ and $d(x_{n-1},x_n)&lt;\epsilon$, $\forall$ n      Clopen Definition      A set is clopen $iff$ its boundary is empty      Connected Metric Definition      Let $(X,d)$ be a metric space. It is connected $iff$ it has no proper subsets which are clopen      The following are equivalent:         $X$ is connected   The only subsets of $X$ that are clopen are $\emptyset$ and $X$   $\nexists$ non-empty, disjoint, open sets $A$ and $B$ such that $X=A\bigcup B$   $\nexists$ non-empty, disjoint, closed sets $A$ and $B$ such that $X=A\bigcup B$         Theorem      Let $(X,d)$ be a metric space, and let $S$ be any subset of $X$. Then $S$ is closed $iff$ $S^c$ is open.     So how would someone show this exactly?  Let the set $A$ be the set of all elements in $(X,d)$ that can be reached from $a\in X$ in a finite number of steps with length less than $\epsilon$.  Does this mean that there should be no isolated points? (an isolated point would be at distance infinity from any other point, or distance zero from itself, since $\epsilon &gt;0$ we don't want this?)  Knowing that $X$ is connected $\implies$the only subsets of X that are clopen are $\emptyset$ and $X$  Also, by construction of A;   A is the set of all $\epsilon$-chains between any element and $a$  And, so $A^c = \emptyset$ if $A$ was $\epsilon$-chainable (because A has to have a path from all elements to a, for this to be true, the complement wouldn't have anything in it)  Leaving only $A$ as the remaining clopen set  Does this make any sense? Or should an alternative be used, and how to formalize it.  Thanks. How to keep Blender from adding frame numbers to output filename? Why is setting a variable before a command legal in bash? I've just encountered several answers such as to  that use the construct:  while IFS=, read xx yy zz;do     echo $xx $yy $zz done &lt; input_file   where the IFS variable is set before the read command.  I've been reading through the  but can't figure out why this is legal.  I tried  $ x="once upon" y="a time" echo $x $y   from the bash command prompt but got nothing echoed. Can someone point me to where that syntax is defined in the reference that allows the IFS variable to be set in that way? Is it a special case or can I do something similar w/ other variables? Textbook approach to modeling non-proportional hazards in the Cox model In Cox models with time varying coefficients, the effect of covariates on the hazard is allowed to change through time. In cases where a coefficient has a linear relationship with time, I am aware of at least two text books ( and an Appendix to ) that advocate for the following model, first in R code:  coxph(Surv(time, death) ~ disease + disease:time)   And now in mathematical notation:  $$ h(t|X) = h_0(t) \exp(\beta_1X + \beta_2X_T) $$ Where $h(t|X)$ is the instantaneous hazard rate at time $t$ given covariates $X$; $h_0(t)$ is the baseline hazard rate; and $T$ is the censoring or event times. $\beta_1$ represents the log hazard ratio for individuals with $T=0$ (which is sort of nonsensical since you can't have $T=0$ in the data), while $\beta_2$ is harder to interpret... $\beta_2$ shows how the log hazard ratio depends on censoring/event times, but that is circular because event times are also the response variable (this point is made in Terry Therneau’s vignette “”).  According to Therneau's vignette, to correctly specify a time varying coefficient, you need to use the following model, first in R code:  coxph(Surv(time, death) ~ disease + tt(disease), tt=function(x,t,...){x*t})   And now in mathematical notation: $$ h(t|X) = h_0(t) \exp(\beta_1X + \beta_2X_T) $$  The difference is that the time interaction is with $t$ in the model rather than with the event times $T$ in the data. The resulting parameters are easy to interpret: they are the intercept and slope of the linear relationship between the log hazard ratio of $X$ and time. This approach seems like the proper way to model a time varying effect.  Can anyone justify or advocate for the first approach to modeling non-proportional hazards in the Cox model? If it is really incorrect (as I suspect it is), what are the practical consequences of using this approach? How to make skill use on investigations "important" but failure not derail the plot? Well guys, following previous recommendations here I was able to build a Horror Themed Savage Worlds campaign that has made my players excited. However they're still getting used to the style, as well as I am. In order for them to hurt the major evils of the adventures they have to find their Form (physical appearance, name), Truth (How the creature came to be) and Reason (What the creature wants to), so they have to do research, kill lesser creatures, and run away from the big baddie of the adventure as they seek what it is.  Well, this is my problem:  How can I make them use skills significantly?  I'm used to railroading my players, even with mysteries, and I don't know how to put their skills for significant use. An example is, they usually require Streetwise or Investigation to gather clues and such, however, I wonder what to do if they fail the roll... If they fail they're not supposed to find anything, so chances they never finish the adventure are well, high, that and I have no idea how to give them additional, interesting ways for them to find the truth. So do I give it no matter what they rolled? How do I make use of their skills important and not just busy-work without having a failure on a roll completely stop their progress in the investigation? Quote unquote (or end-quote), unseparated by the actual quotation Crafty solutions to the following limit The following problem came up at dinner, I know some ways to solve it but they are quite ugly and as some wise man said: There is no place in the world for ugly mathematics.  These methods are using l'Hôpital, but that becomes quite hideous very quickly or by using series expansions.  So I'm looking for slick solutions to the following problem:     Compute $\displaystyle \lim_{x \to 0} \frac{\sin(\tan x) - \tan(\sin x)}{\arcsin(\arctan x) - \arctan(\arcsin x)}$.   I'm curious what you guys will make of this. What is the difference between conditioning on regressors vs. treating them as fixed? Is it necessary to use commas in lists before the conjunction? I have seen many sentences like this:     "Capitalism, Socialism, and Anarchism"   and this:     "She was happy, and cheerful"   Do we have to use a comma before "and", or is it incorrect?  If it is not, then why? Software gvSIG OA Digital Edition 2010 have tools topology for finding pseudo nodes in linear geometry. I set cluster tolerance 0.00002 and maximum number of errors -10000 for 20000 link count linear geometry. But unsuccessful result.   Are there any solutions that find pseudo nodes in free GIS software?   I need to layer pseudo nodes (one solution to this problem - to use tools topology of ArcInfo, but priority for me is to use free software). Linear geometry created several users in QGIS 1.8.0 in PostGIS (v. 2.0.1) database.  Add new image: 12 linear features with three pseudo nodes in A (line 4/5), B (line 6/7), C (line 9/10). Pseudo nodes should be points instead - two linear features with intersection in one point (node) should be one linear feature (line 4/5 - line 4, ...).  Is it possible to make a request in PostGIS, which will result in a layer of pseudo nodes?  Add new image of examples pseudo nodes: if I receive for linear layer point layer pseudo nodes (blue rects) I corrected following errors in linear layer: A - add missing geometry, B - snapped line in intersection, C - remove pseudo node. When & How often do 4e Monster powers recharge? How often should I roll power recharge rolls and at which point in the monster's turn do I re-roll them?  I'm trying to find the official definition of the recharge mechanic in the online compendium and I'm not seeing it there. Is it legal to study while on a tourist Schengen visa? If, say, you acquired a 90-day tourist Schengen visa, would it be legal to use it for the purpose of visiting a language course or a university class? Or do you have to apply for a special education visa if you have such intents?  Answers linking to official EU pages on that matter would be highly appreciated :) How to write uncertainties with the \pm symbol "10 ± 1 K" by using siunitx? Using the siunitx package, if I specify an error with the \pm symbol then it automatically converts it to a form using parenthesis. I would like it to not do this and keep the \pm symbol, but still space the elements of the quantity and symbol in a good way.  e.g. \SI{10 \pm 2}{\kelvin} would produce 10(1) K  I would like it to produce: 10\pm1 K or something similar with spacing taken care of. Let $F$ be a field, let $\omega$ be a primitive $n$th root of unity in an algebraic closure of $F$. If $a$ in $F$ is not an $m$th power in $F(\omega)$ for any $m\gt 1$ that divides $n$, how to show that $x^n -a$ is irreducible over $F$? Cannot get onto Apple Developer Forums Every time I attempt to log on to the apple developer forums it tells me I need to create a new user account (even though I already have one), and when I try to sign up it gives me an error:   Anyone else have this issue and/or know how to fix it? Thanks! Finding the n-th lexicographic permutation of a string I have an ordered set of symbols S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }. I want to find the 1,000,000-th permutation in lexicographic order of S. It is a programming puzzle, but I wanted to figure out a way without brute-forcing the task.  So my thinking was like this:  For 10 variable symbol positions, we have 10! permutations, obviously. Now we want to find the first symbol.  If we fix the first symbol, then the remaining 9 symbols can have 9! combinations.  That means that 0 or 1 cannot be the first symbol, because the highest possible position is 2*9! = 725,760, which is lower than 1,000,000.  The lowest position for a leading 3 is 3*9! + 1 = 1,088,641, so it can't be 3 or higher, either.  Therefore, the first number must be 2. 2*9! is the largest multiple of 9! no greater than 1,000,000, so I need the 2nd symbol (zero-based) from the current set.  So now the question becomes, of the remaining S := S \{2}, which permutation of these symbols is at lexicographic position (1,000,000 - 2*9!) = 274,240?  6*8! = 241,920 is the largest multiple of 8! which is smaller than 274,240, so I need the 6th-smallest symbol of the remaining set, which is 7. So the prefix should be 27 by now.  That way, I keep going and finally arrive at: 1,000,000 = 2*9! + 6*8! + 6*7! + 2*6! + 5*5! + 1*4! + 2*3! + 2*2! + 0*1! + 0*0!  which results in "2783905614" as my solution.  However, according to the , (free reg. req.) that is incorrect.  Where did I go wrong in thinking or applying? Evaluation of ratio of two binomial expression