I have this object:    {"wordFormId":"abandon",    "wordFormIdentity":1,    "ascii":97,    "wordId":"abandon",    "primary":true,    "posId":2}   How can I remove the ascii property? I know I could set it to null but I assume it would then still be there. What I would like to do is to completely remove it.
Say I create an object as follows: let myObject = {   &quot;ircEvent&quot;: &quot;PRIVMSG&quot;,   &quot;method&quot;: &quot;newURI&quot;,   &quot;regex&quot;: &quot;^http://.*&quot; };  How should I remove the property regex to end up with new myObject as follows? let myObject = {   &quot;ircEvent&quot;: &quot;PRIVMSG&quot;,   &quot;method&quot;: &quot;newURI&quot; };
Volume of bounded region between sphere and a plane A Hilbert space is, by definition, a complete inner product space. If $(V,|.|)$ is finite dimensional inner product space of dimension $n$ then it is (topologically) isomorphic to $\mathbb{R}^n$ which is of course complete. My instinct here is to say "and therefore, $V$ is also a Hilbert space". I'm not sure about this step however since completeness depends on the norm and the norm depends on the selected inner product (assuming the induced norm $||x|| = (x | x)^{1/2}$ is used). So, must we place some condition on the inner product or is the topological isomorphism between $\mathbb{R}^n$ enough to guarantee that V is Hilbert? How to handle dynamic changing ID's In XPath? Is the inverse of a linear transformation linear as well? A question from the field of Linear Algebra. If I have a linear transformation $T$ that is one-to-one and onto, would that mean that the $T^{-1}$ will also be linear? If so, is there any general proof for it?  Thanks "In a party people shake hands with one another" In a party people shake hands with one another (not necessarily everyone with everyone else).   (a) Show that 2 people shake hands the same no. of times.   (b) Show that the number of people who shake hands an odd no. of time is even.        I find this question to be very tough and I can't find a way to even start answering this. It would be great if I could get a small hint in order for me to at least start it. How to put small numbers outside the pie slices without making a mess I'm new to LaTeX and need some help with my figures. I would like to to put small numbers outside the pie slices without making a mess. Having found this thread , I tried it out, but didn't get all the way through with it. Right now, my pie looks like this:    I used the following code:  \usepackage{tikz,stackengine} \makeatletter  \tikzstyle{chart}= [legend label/.style={font={\scriptsize},anchor=west,align=left}, legend box/.style={rectangle, draw, minimum size=5pt}, axis/.style={black,thin,-&gt;}, axis label/.style={anchor=east,font={\tiny}}]  \tikzstyle{bar chart}=[ chart, bar width/.code={     \pgfmathparse{##1/2}     \global\let\bar@w\pgfmathresult}, bar/.style={thin, draw=black}, bar label/.style={font={\bf\small},anchor=north}, bar value/.style={font={\footnotesize}}, bar width=.75,]  \tikzstyle{pie chart}= [chart, slice/.style={line cap=round, line join=round, thin, draw=black}, pie title/.style={font={}}, slice type/.style 2 args={     ##1/.style={fill=##2},     values of ##1/.style={}}]  \pgfdeclarelayer{background} \pgfdeclarelayer{foreground} \pgfsetlayers{background,main,foreground}  \newcommand{\pye}[3][]{     \begin{scope}[#1]         \pgfmathsetmacro{\curA}{90}         \pgfmathsetmacro{\r}{1}         \def\c{(0,0)}         \node[pie title] at (270:1.3) {#2};         \foreach \v/\s in{#3}{             \pgfmathsetmacro{\deltaA}{\v/100*360}             \pgfmathsetmacro{\nextA}{\curA + \deltaA}             \pgfmathsetmacro{\midA}{(\curA+\nextA)/2}             \path[slice,\s] \c             -- +(\curA:\r)             arc (\curA:\nextA:\r)             -- cycle;             \pgfmathsetmacro{\d}{max((\deltaA * -(.5/50) + 1) , .5)} %%%              %\pgfmathprintnumberto{#3}{\fnum}  %\ifthenelse{\lengthtest{#3 pt &lt; 10pt}} %{\path (O) -- ++ (\midangle:\radius) %node[inner sep=0,align=left,anchor=west,rotate=\midangle](nums#3)    {~~\fnum};} %{\path (O) -- ++(\midangle:\innerpos)  %node {\scalefont{#3}\beforenumber\fnum\afternumber};}           %%%                         \begin{pgfonlayer}{foreground}                 \path \c -- node[pos=\d,pie values,values of \s]{$\v\%$} +    (\midA:\r);             \end{pgfonlayer}             \global\let\curA\nextA}     \end{scope}}  \newcommand{\legend}[2][]{     \begin{scope}[#1]         \path         \foreach \n/\s in {#2}         {++(0,-5pt) node[\s,legend box] {} +(2pt,0) node[legend label]     {\n}};     \end{scope}}      \begin{document}         \begin{center}      \begin{figure}[H]     \centering \begin{tikzpicture} [pie chart, slice type={A}{black!0}, slice type={B}{black!10}, slice type={C}{black!20}, slice type={D}{black!30}, slice type={E}{black!40}, slice type={F}{black!50}, slice type={G}{black!60}, slice type={H}{black!70}, pie values/.style={font={\small}}, scale=2]  \pye{1994}{6.1/A, 2.9/B, 0.2/C, 54.3/D, 25.2/E, 1.4/F, 6.7/G, 3.2/H} \pye[xshift=2.2cm,values of coltello/.style={pos=1.1}]% {2005}{6.8/A, 3.0/B, 0.5/C, 52.3/D, 25.6/E, 1.8/F, 7.1/G, 2.8/H} \pye[xshift=4.4cm,values of caffe/.style={pos=1.1}]% {2015}{7.2/A, 3.1/B, 0.6/C, 50.6/D, 26.5/E, 2.3/F, 7.5/G, 2.2/H}  \legend[shift={(0cm,-1.5cm)}]{{A}/A, {B}/B}  \legend[shift={(1.4cm,-1.5cm)}]{{C}/C, {D}/D}  \legend[shift={(2.8cm,-1.5cm)}]{{E}/E, {F}/F} \legend[shift={(4.2cm,-1.5cm)}]{{G}/G, {H}/H} \end{tikzpicture} \end{figure}  \end{document}   This looks quite the way I want it to, only the small numbers should be placed outside the pies.  Thanks in advance :) Question 5.33 from "Topology and its Applications"  by Baesner is to compute the fundamental group of the torus ($T^2$) with $n$ points removed. I can "see" in my mind that if we remove one point we get a bouquet of two circles. Less clear is what happens when we remove two (or more) points. Any hints? My keyboard has no "media" keys; can I control media without them? My USB keyboard does not have "media" keys -- that is, dedicated keys for play, stop, next, prev, volume up/down, etcetera.    For the sake of this question, I would prefer not to install additional software if I can avoid it.  Is it possible to issue some standard key sequence on a generic USB keyboard that emulates these play, stop, etc. multimedia keys?  Like Ctrl+Alt+Shift+F12 or something obscure like that? Variable Sized Arrays vs calloc in C On the discussion of dynamic memory here:   The author states:     A memory block like this can effectively be used as a more flexible array. This approach is actually much more common in real-world C programs. It's also more predictable and flexible than a "variable size array"   The type of memory block he is talking about is as such:  const int size = 5; int * array = calloc(size, sizeof(int));   and then using another pointer to iterate over the array:  int * index = array; for (i = 0; i &lt; size; i++) {     *index = 1; // or whatever value     index++; }   My question is how is this method better than a standard variable sized array like this?:  int array[variable];   or dynamic:  char name[] = "Nick";   The author doesn't really shed much light as to why I should prefer the former method to the latter. Or more specifically: How is it more "predictable and flexible"? Is the coefficient of friction always less than unity? What are the cases in which coefficient of friction is greater than unity? Built-in unit strings recognized by Quantity? Want \section* to change the page header In a book, I have several numbered sections followed by an unnumbered section. I have placed an \addcontentsline following the \section* that introduces the unnumbered section, so that it will appear (unnumbered) in the ToC; that works fine. However, the page headers for the unnumbered section are continued from the last numbered section.  I am using fancyhdr with the following setup:  \pagestyle{fancy} \fancyheadoffset{0in} \renewcommand{\chaptermark}[1]{\markboth{#1}{}} \renewcommand{\sectionmark}[1]{\markright{\thesection\ #1}}  \fancyhf{} \fancyhead[LE,RO]{\thepage} \fancyhead[C]{\bfseries \draftinfo} \fancyhead[LO]{\sf\rightmark} \fancyhead[RE]{\sf\leftmark} \fancyfoot[C]{\bfseries \draftinfo} \fancypagestyle{plain}{%   \fancyhead{} % get rid of headers   \renewcommand{\headrulewidth}{0pt} % and the line } Similar to my old question about  are there any limits to the animal forms an Animagus can be?     We know that the animal form can be much smaller than human (cat, rat, beetle) so I don't think size is a factor.   I'm not aware of any Animagus who transforms into a magical creature (e.g. unicorn or phoenix) so maybe those aren't possible.   Ultimately, what I'd like to know is whether there could be a T-Rex Animagus... (prompted by ) Can a particle be physically observed inside a quantum barrier? I understand that if a particle approaches a finite potential barrier of height $V_0$ with energy $E &lt; V_0$, there is still a finite probability of finding the particle on the other side of the barrier due to quantum tunneling.  My question is, since the wavefunction is nonzero inside the barrier region, is it possible to actually make a position measurement and locate a particle inside the barrier?  I mean, if we can say that "there is a nonzero probability that the particle is inside the barrier", surely this would suggest we can do so?  If not, why not? Am I understanding the whole wavefunction/probability distribution thing right? Diagonalizable matrix with parameters $$     \begin{bmatrix}     -2 &amp; -2 &amp; 0 &amp; 0 \\     -b &amp; -b &amp; 0 &amp; 0 \\      0 &amp; 0 &amp; -4 &amp; 4 \\      0 &amp; 0 &amp; b &amp; -b \\     \end{bmatrix} $$  I have this matrix and I need to find the values for $b$ that this matrix won't  be diagonalizable.  For what I got this matrix is diagonalizable for every $b$ value.  Please write me also the final answer, thank you very much. Spring Data JPA Update @Query not updating? I keep going around in circles with this problem:  Given the functions $f:X\to Y$ and $g:Y\to Z$, if $g\circ f:X\to Z$ is bijective, then prove $f$ must be one-to-one and $g$ must be onto. I was able to prove this without difficulty. However the second part of the problem has me stumped:  a) Find functions $f,g$ where $g\circ f$ is bijective but $f$ fails to be onto.  b) Find functions $f,g$ where $g\circ f$ is bijective but $g$ fails to be one-to-one.  My attempt for (a): Let $f:\mathbb{R}\to \mathbb{R}^+$ be defined by $f(x)=e^x$ and $g:\mathbb{R}^+\to \mathbb{R}$ be defined by $g(x)=\ln(x)$. Then $g\circ f(x)=g(f(x))=g(e^x)=\ln(e^x)=x$ which is bijective, $g$ is onto and $f$ fails to be onto... or so I thought. Upon closer inspection $f$ $\underline{is}$ onto for the given codomain $\mathbb{R}^+$.  And now I'm stuck in a rut. Anyone out there have any hints? I am looking for a preferably introductory text on Yang-Mills theory with mathematical basis and language. (Differential forms) I have a background in QCD and QED, however, I want to go more in details in geometrical interpretations of Yang-Mills theory. I am sorry if this question has asked before. Thanks in advance. Can I reduce the MAC size in TLS? The MAC size is reduced to 80 bits in  , which says "forging of MAC values cannot be done off-line: in TLS, a single failed MAC guess will cause the immediate termination of the TLS session."  Question 1 (main question): So can I reduce MAC size to something even smaller, say 32 bits?  Question 2: The truncated_hmac extension does not effect AEAD. Can I reduce the tag's size in AEAD, e.g. GCM or CHACHA-POLY? The vast majority of intelligent species in the Star Trek universe are bipedal, humanoid creatures.  From the lot I've seen of TOS, TNG, DS9, Voyager, Enterprise, and the movies, it seems that this coincidence goes largely unquestioned.  Metaknowledge grants that it is much easier to have human actors portray human-like beings, and so this was probably done due to budgetary/technological limitations.  However, I'm wondering if an in-universe reason for this was ever given or explored?