Xpath to select elements that has specific child
XPath - Get node with no child of specific type
How to change boot priority? How do you find time period as a function of time for a simple pendulum that is in the form of a hollow sphere that is filled with mercury and there is a hole in the bottom through which the mercury is constantly falling at a fixed rate? I tried creating a function for the time period to check how it varies with the mass of mercury, but I had too many variables.   EDIT: This question was a concept based multiple choice question in my homework book. The option were time period becomes erratic, time period increases, time period decreases, time period first increases then decreases. How to add tags to files in Windows 7 so that they appear in search results In Windows 7, I want to add tags to my files (PDF, Word, Images etc) so that they appear directly in the search results, regardless of their content, title etc.    shows how to search by tag, but it does not tell how to add a tag in any file.  In the long term, I'm planning to add such tags programmatically as well. Children's book with "square triangles" in parallel universe? Properties of p-adic valuations Let $n$ be a integer not equal to zero, $p$ a prime and $$|n|_p = p^{-\operatorname{ord}_p(n)}; |0|_p=0$$  Proposition: $|xy|_p=|x|_p|y|_p$  We want to show: $\operatorname{ord}_p(ab)=\operatorname{ord}_{p}(a)+\operatorname{ord}_p(b)$ So if we suppose there is a largest integral $v$ such that $p^{v}|(ab)$, then we can split up v such that $p^{v_1}|a,p^{v_2}|b$. If we suppose that there exist $v_1,v_2$ with them both being the largest integers such that $p^{v_1}|a, p^{v_2}|b$,  with $v_1+v_2= : v $ then $p^v|ab$. (I don't believe this to be a formal proof but rather my best own "idea" I can come up with...)  The script also mentions that the padic valuations' sums are bounded: $|x+y|_p\le \sup\{|x|_p,|y|_p\}$, how to show this?   Edit:   According to Robert Israel and Bill Dubuques answers, the first proposition is proven, if I understood correctly:   Given $ord(a)= v$ and $ord(b) = u$ which is the same as $a:= p^{v}m, b:= p^{u}n$ with $gcd(m,p)=gcd(n,p)=1$, because $ab=p^{u+v}mn$ with $gcd(mn,p)=1$ it follows that $ord(ab)=ord(a)+ord(b)$ and so $|xy|_{p}=p^{-ord(xy)}=p^{-(ord(x)+ord(y))}=|x|_{p}|y|_{p}$  Now to show: $|x+y|_{p}\le sup\{|x|_{p},|y|_{p} \}$  and the second can be tackled like this: set $a:=p^{v}m. b:=p^{u}n$, then checking two cases $1:(u=v)$ and $2:(v&lt;u)$: $(a+b)=p^{v}m+p^{u}n= p^{v}(m+p^{u-v}n)$ so in case of (u=v) it follows that  $ord(a+b)= u = sup\{ord(a),ord(b)\}$  and in case of $(v&lt;u)$ because the biggest v is $v:= u-1$: $ord(a+b)\le p^{u-1}(m+pn) \le p^{u}(m+n)= sup\{ord(a),ord(b)\}$  It seems that I did something wrong in both cases... Find index of nested item in python Capitalization of explanation of abbreviations How can I vertically align the numerators of two fractions? I am writing the following formula:  % arara: lualatex  \documentclass{article} \usepackage{mathtools} \usepackage{unicode-math}  \begin{document} \[y_\mathup{COG}=\frac{h}{3}\frac{l_\mathup{c,a}+2l_\mathup{c,i}}{l_\mathup{c,a}+l_\mathup{c,i}}\] \end{document}   which yields:    I don't like the way, the h and l_\mathup{index} is vertically aligned. Is that typical behavior? Good style? Could I avoid that? What novel had an isolationist Earth and the rest of humanity in a war of attrition against aliens? Importing google-play-services lib into Intellij IDEA 12 (and 13) If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator Let $V$ be a finite-dimensional inner product space over $\mathbb{C}$ and $T: V \to V$ a linear transformation.   Show that if every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal.   We need to show that $\forall v \in V,\ TT^*v=T^*Tv$.  I started by picking $v$ to be an eigenvector of $T$, hence $Tv=\lambda_1 v$, and $T^*v=\lambda_2 v$. Therefore: $$TT^*v=T(\lambda_2 v)=\lambda_2 Tv=\lambda_1 \lambda_2 v=...=T^*Tv$$  But I'm stuck when $v$ is not an eigenvector.  (Also, this question is taken from a chapter about Jordan forms, but I can't see how it relates) Finding the radius of a third tangent circle Sorry if this is a foolish question, but I'm having difficulty understanding how to solve for $r_3$ in the following diagram...    According to WolframAlpha's page on tangent circles, the radius of $c_3$ can be calculated using the following formula $r_3=\frac{r_1 \times r_2}{(\sqrt{r_1}+\sqrt{r_2})^2}$, which can be simplified to $\frac{1}{\sqrt{r_3}}=\frac{1}{\sqrt{r_1}} + \frac{1}{\sqrt{r_2}}$.  To be completely honest, I'm having a hard time understanding understanding how this formula works (as I'm a very visual person). What exactly is happening here? Is there a function whose inverse is exactly the reciprocal of the function, that is $f^{-1} = \frac{1}{f}$? Is there a function whose inverse is exactly the reciprocal of the function? That is $f^{-1} = \frac{1}{f}$.  We know that the inverse of a function is not necessarily equal to its reciprocal in general. Executing .bat file Children's book with "square triangles" in parallel universe? In the 1960s I read book--an older children's book, I think--which I loved, and still remember, roughly.  I have no idea what the title is or who the author is, and I'm wondering if anyone knows this one:  Scientists had invented a device which they didn't fully understand, and which had puzzling effects.  For example, when a pencil was put into the machine, it disappeared, and then reappeared inside out: The lead was on the outside.  When people went into the machine, they came back dead, having died from fright.  Finally they sent a very open-minded person into the machine.  She was an artist, I think.  She was able to return alive, and described an alternate universe in which geometry was very different.  I distinctly remember that there were supposedly square triangles or four-sided triangles--or was it three-sided squares?  (This is, of course, a literally contradictory claim, but I didn't see the matter as clearly then.)  Thanks! Is there any physical or mental cost to casting spells? How to use 'hover' in CSS An equality involving the mean value property I am trying to establish that  $$\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\cos^2\left(\frac{\theta}{2}\right)\;\mathrm d\theta=2f(0)+f^\prime(0)$$  for $f$ analytic on $|z| &lt; R,\; R &gt; 1$  I think I am making a mistake somewhere, please point it out to me.Here's what I've done unsing the mean value property and Cauchy's integral formula:  $$2f(0)+f&#39;(0)=2(\frac1{2\pi}\int_0^{2\pi} f(e^{i\theta})d\theta + \frac1{2\pi i}\int_0^{2\pi}\frac{f(e^{i\theta})ie^{i\theta}}{e^{i2\theta}}=\frac{1}{\pi}\int_0^{2\pi} f(e^{i\theta})d\theta\frac1{2\pi}\int_0^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta}}\quad\text{(1)}$$  But when I develop the expression $\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\cos^2(\frac{\theta}{2}) d\theta$ I get:  $$\frac1{\pi}\int_0^{2\pi} f(e^{i\theta})d\theta+\frac1{2\pi}\int_0^{2\pi}f(e^{i\theta})[{e^{i\theta}}+e^{i-\theta}]d\theta \quad\text{(2)}$$  The second integral in (1) differs from the second integral in (2). Thanks. In the beginning of the series, it's mentioned that Muggle-born wizards/witches are identified due to inadvertent spell-casting. However, it seems like later in the book, disarmed wizards/witches are completely helpless.   What is the explanation for why some spells can be cast without a wand and is there an in-universe explanation for why this isn't done? Also, are there spells that can be cast non-verbally (with/without a wand)? PHP Error - Uploading a file