How do you convert Audible AAX files to M4B audiobooks
I bought an audio book from audible and want to convert it to mp3 from aa because my phone doesn't support the format. Is there any way to do that?
How to execute multiple statements in a MATLAB anonymous function? I'd like to do something like this:  &gt;&gt; foo = @() functionCall1() functionCall2()   So that when I said:  &gt;&gt; foo()   It would execute functionCall1() and then execute functionCall2().  (I feel that I need something like the )  EDIT:  functionCall1 and functionCall2 are not necessarily functions that return values. (I'm going to link to this Stackexchange post concerning Nullary Operations: )  In general an  is a function $f:S^n \to S$, where $n$ denotes the arity of the operation, and a  is a function $f:S^0 \to S$.  It is clear that $S^n=S \times S \times \cdots \times S$ denotes the Cartesian product of a set with itself ($n$ times), however $S^0$ makes little sense to me.  One might define $S^0=\emptyset$. At first glance, that seems like a good choice. However then a nullary operation makes no sense, since it must be a function $f: A \to B$ where $A$ and $B$ are two sets representing the domain and codomain of the function. And note a function is defined in terms of relation, where a relation is a subset of the Cartesian product $A \times B$. So when we think of $A=S^0=\emptyset$ and $B=S$, so that we have a nullary operation $f:\emptyset \to S$, what is a a subset of $\emptyset \times S$? What even is $\emptyset \times S$? By definition of the Cartesian product, \begin{equation} \emptyset \times S = \{(a,b):a \in \emptyset\textrm{ and }b \in S\}. \end{equation} But there exists no $a \in \emptyset$! Therefore there exist no ordered pairs in our Cartesian product! Therefore $\emptyset \times S = \emptyset !$. Remember a function is a relation which is a subset of the Cartesian product. Therefore our relation is the empty set, and therefore also $f=\emptyset$.    Edit: We got the definition sorted out in the comments. Thanks especially to Hayden and Christoff for their great help. In summary, we figured out that $S^0=\{()\}$, the set containing only one element, the "empty $n$-tuple". Then if taking $S=\{1,2,...,k\}$, then a nullary operation $f:S^0 \to S$ is a function, i.e. some subset of $S^0 \times S = \{()\} \times S$. Looking at $\{()\} \times S$, this equals $\{((),1),((),2),...,((),k)\}$. Then a function is a special subset of this, so that each element of the domain (and there is only one element, namely $()$) is paired with exactly one element of the codomain. Therefore a function would have the form $f=\{((),a)\}$ where $a$ is one of $1,2,...,k$. Thank you for your help. Remove duplicate fields in a given column (after ID check) Renaming user name I made a mistake in the username at the time of installation. So I changed the name of the user after logging in by creating a guest account with SUDO rights. But when I checked the list of users using ls /home command I am not seeing the new name for the user instead Old name is appearing. But I use the new name to login.   Is it like I need to change the name inside Home directory separately? Also , if there is any other changes needs to be done .. ? If $A = \tan6^{\circ} \tan42^{\circ},~~B = \cot 66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$ My trigonometric problem is:     If $A = \tan6^{\circ} \tan42^{\circ}$ B = cot$66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$.    Working :   $$B = \cot 66^{\circ} \cot78^{\circ} = 1- \frac{\tan24^{\circ}+\tan18^{\circ}}{\tan42^{\circ}}$$  $$A= \tan6^{\circ} \tan42^{\circ} = 1- \frac{\tan6^{\circ} +\tan42^{\circ}}{\tan48^{\circ}}$$  but it seems  this is the wrong way of doing this...please suggest. Thanks! sstatic.net SSL certificate is for an incorrect domain Some images shown outside of preview rectange Some images shown outside of preview rectange before uploading, eg: Someone mentioned that "veiling glare" or "veiling flare" was to blame for reduced contrast in a photo. What does this mean exactly?  What causes veiling glare, and how can it be avoided? How does it relate to those floating polygonal lens flares one often sees in movies? The  states that:      The scores are created by comparing the word in the list in step 2* with all the 3-letter words. By using the scoring matrix (substitution matrix) to score the comparison of each residue pair, there are 203 possible match scores for a 3-letter word. For example, the score obtained by comparing PQG with PEG and PQA is 15 and 12, respectively with the BLOSUM62 weighting scheme.   I do not understand how they got a score of 15 for PQG v. PEG. What is a scoring matrix, how is it computed, and how is it used?      Step 2 on that page is Make a k-letter word list of the query sequence: “Take k=3 for example, we list the words of length 3 in the query protein sequence (k is usually 11 for a DNA sequence) sequentially, until the last letter of the query sequence is included”. Why can't I use switch statement on a String? Arithmetic operator overloading for a generic class in C# Function with same name but different signature in derived class I'm trying to find a discontinuous linear functional into $\mathbb{R}$ as a prep question for a test. I know that I need an infinite-dimensional Vector Space. Since $\ell_2$ is infinite-dimensional, there must exist a linear functional from $\ell_2$ into $\mathbb{R}$. However, I'm having trouble actually coming up with it.   I believe I'm supposed to find an unbounded function (although I'm not sure why an unbounded function is necessarily not continuous; some light in that regard would be appreciated too), so I thought of using the vectors $e^i$, which have all entries equal to zero, except for the $i$-th one. Then, you can define $f(e^i)=i$.  That'd be unbounded, but I'm not sure if it'd be linear, and even if it is, I'm not sure how to define it for all the other vectors in $\ell_2$.   A friend mentioned that at some point the question of whether the set $E=\{e^i:i\in\mathbb{Z}^+\}$ is a basis would come up, but I'm not sure what a basis has to do with continuity of $f$.   I'm just learning this topic for the first time, so bear with me please.   The space of sequences that are eventually zero (suggested by a few people) turned out to be exactly what I needed.  It also helped to cement the notions of Hamel basis, not continuous, etc. I've occasionally noticed an OP who will edit their own question and destroy it completely, always leaving some kind of graffiti-like scrawl in its place.  This seems to happen on low-quality questions by low-rep users, and the moment of self destruction tends to occur after the question is closed or otherwise devolves into a communication breakdown between the poster and the community.  Here are some examples:        It seems like the poster usually does this in anger (1st and 3rd example), or else is just confused about how to delete their question (2nd example).  My question is about how this should be handled. In the past I wasn't sure, so I always flagged for mod attention. However I notice the solution on all of these was to just do a rollback (I did this myself on the 2nd example after nothing happened). Is this a good enough approach, or should a flag be used?  What about with particularly obnoxious users like in the third example, where the guy came back to do it again? What is the T-SQL To grant read and write access to tables in a database in SQL Server? What is the exact SQL to assign db_datareader and db_datawriter roles to a user in SQL Server?  The user name is MYUSER and the database is MYDB. What are your favorite English language tools? To prevent myself from asking an obvious, silly question multiple times: What are the English language tools you found most useful?  I found  extremely useful for looking up .   Please, one tool per answer. Starting scheduled task by detecting connection of USB device I know it has been discussed that it is not possible to start an application from an usb drive upon connection due to limitations of autorun (or autoplay??) in Win 7. But it is possible to create a scheduled task having an event type trigger. Surely there must be an event occurring when the drive -- or any USB device, for that matter -- is connected.    Does anybody have the slightest idea which Event ID should I use? Or at least what type of event? Where can I find the event in event viewer? How can I do a full shutdown of windows 10 without logging in? Proving $\sum\limits_{l=1}^n \sum\limits _{k=1}^{n-1}\tan \frac {lk\pi }{2n+1}\tan \frac {l(k+1)\pi }{2n+1}=0$ Prove that $$\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$$     It is very easy to prove this identity for each fixed $n$ . For example let $n = 6$; writing out all terms in a $5 \times 6$ matrix, we obtain:  $\begin{matrix}  \tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}  &amp;  \tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13} &amp; \tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13} &amp; \tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13} &amp; \tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}  \\  \tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}  &amp;  \tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13} &amp; \tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13} &amp; \tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13} &amp; \tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13} \\ \tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13} &amp; \tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13} &amp; \tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13} &amp; \tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13} \\ \tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13} &amp; \tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13} &amp; \tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13} &amp; \tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13} \\ \tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13} &amp; \tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13} &amp; \tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13} &amp; \tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13} &amp; \tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13} \\ \tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13} &amp; \tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13} &amp; \tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13} &amp; \tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}  \end{matrix}$  one can notice then, that the first column vanish the fourth one :  $\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}$  $\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}$  $\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}$  $\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}$  $\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}=-\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}$  $\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}$  and the third column vanish the fifth one :  $\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}$  $\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}$  $\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}$  $\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}$  $\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}$  $\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}=-\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}$  while the second column is self-vanishing:  $\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}$  $\tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}$  $\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}$ .  So the equality occurs. But how to generalize the proof? I'm using an old iBook G4 (PowerPC) running Tiger to serve iTunes to the house. For the past week or so, the mac boots fine, and then sometime during the day starts generating   An unexpected error occurred (error code -50).   errors when iTunes attempts to save its library file.  Checking reveals that the boot drive (main drive, in the laptop) behaves as if it is write-protected - I get the same error code when I attempt to create a folder on the desktop, for instance. If I attempt to create files via the terminal, e.g. via     ls > foo.txt I get an "invalid argument" error.   I've set the iBook to run in Target Disk Mode and run Disk Utility to repair the disk (from a Macbook running Snow Leopard) and get "The volume appears to be ok".   I don't have the option in Disk Utility to Repair Permissions in Target Disk Mode. When I ran Repair Permissions on the iBook itself, it came back with no errors, no permissions corrected, but did not in fact fix the problem.  Rebooting always solve the problem - for the next 5-10 hours or so.   Would Disk Warrior help, or some other software fix, or is it time to buy a new drive? (And, in any case, what's going wrong?)