I followed the Pierre Hermé éclairs recipe:     60mL water    75mL whole milk    3g salt    3g caster sugar    56g butter    75g flour    A little over 2 eggs   I did the recipe and it called for 45 minutes of baking at 185°C.  I baked them for 40 minutes and left them in the oven for an extra 10 minutes before removing them.  The result was that the exterior was terrific, but the inside was too doughy and I couldn't fill it with crème pâtissière (custard).  Is skipping the 5 last minutes of baking the reason it failed?
Last time I tried to make Chocolate eclairs (similar to cream puffs), the choux pastry just sat in the oven and basically fried.  It didn't rise/grow and so I couldn't hollow out the shells.  Any suggestions?  Recipe was from The Australian Women's Weekly Original Cookbook (Golden Press Pty Ltd, 1977) p. 204  It is almost identical to the one published on their website , with the exception that the cookbook recipe used 1C water and 1C plain flour.
Any positive linear functional $\phi$ on $\ell^\infty$ is a bounded linear operator and has $\|\phi \| = \phi((1,1,...)) $ How can I check which songs or videos that I purchased from the iTunes Store have DRM restrictions?  All of the mobile OSes are very impressive to me at the moment, and I'm thinking about trying a non-iOS device again. I want to know now the impact it has on my Apple-sourced music collection though.  Is there a date cut-off for purchases when items went DRM free? I understand how Insight and Perception can be used passively: for the DM to determine if an enemy Deception or Stealth attempt is successful without alerting the players that someone is deceiving or sneaking.    But the Observant feat in the Player's Handbook states:      [...] You have a +5 bonus to your passive Wisdom (Perception) and passive Intelligence (Investigation) scores.     So in what situation would a DM be rolling something against the player's Investigation without letting them know? Would the Animate Objects spell end an ongoing spell affecting the targeted object(s)? Using Unicode Zero Width Spaces allows all space comments Proof of when is $A=X^TX$ invertible? Expected number of hamiltonian paths in a tournament The following theorem is from Alon&amp;Spencer's The probabilistic method, in the beginning of chapter 2:  Theorem 2.1.1: There is a tournament $T$ with $n$ vertices and at least $\frac{n!}{2^{n-1}}$ Hamiltonian paths.  Briefly the theorem is proved by looking at $X_\sigma$ the indicator random variable for a permutation $\sigma$ giving a Hamiltonian path, that is, satisfying $(\sigma(i),\sigma(i+1))\in T$ for $1\leq i \lt n$. Then   $\sum \mathbb{E}\left[X_\sigma\right] = \frac{n!}{2^{n-1}}$.  I don't understand where did this value ($\frac{n!}{2^{n-1}}$) came from! could someone explain it please? Each Stack Exchange site has a weekly email newsletter.  What is in the newsletter? Who composes them and selects what is included? How do I subscribe and unsubscribe to a newsletter? How does iCloud Photo Library's "Optimize iPhone Storage" setting work in relation to other photo apps on iOS? If I have Optimize iPhone Storage enabled on my iPhone and my phone storage is completely full, I understand that iOS will only store a low-res thumbnail of many of my photos on the device. (The full high-res version of the image will only exist in iCloud and on devices with Download and Keep Originals enabled.)  But when only thumbnails exist on the device, how does that work in relation to other photo apps on iOS? For example, do third-party cloud-based photo apps (like Google Photos, Dropbox's Carousel, or Amazon Prime Photos) have access to the full high-res images from the iCloud Photo Library? Or will they only see the low-res thumbnails in some cases? How to Map a subset of list elements to a function? Passive aggressive people will sometimes veil insulting, critical, derogatory or generally aggressive comments with humor. The patina of humor makes the comment seem like a joke, not to be taken seriously, all in good fun, and it safeguards the aggressive person. If someone gets offended, they are being too serious, can't take a joke, and so they cannot voice their upset without seeming to be the one escalating the situation, when in fact it was the aggressor who did that.   This behavior seems to lie on a spectrum. On one side are comments which are simultaneously aggressive and genuinely funny (though of course both, especially the latter, are fairly subjective). On the other side would be lame attempts at the behavior, something like "you are fat!... just joking!" in which nothing in the tone or delivery signals a register of humor. In the middle of the spectrum, I imagine, are aggressive statements genuinely delivered as jokes. Their register signals that they are jokes, but they are aggressive rather than funny. They are aggressive comments masquerading as jokes.      A friend once called this "joking on the square", but I have never heard or seen this phrase. This type of comment is both pervasive and psychologically subtle and in my experience there is often an exact word describing such things. If not a word, is there an astute phrase which cuts to the heart of this? Is the tag info page not available on mobile theme? I was surfing SO on my iPhone, and wanted to see the info (tag-wiki) page for a tag, but didn't find it. Clicking on the tag always leads to the list of question, and there is no other link to the tag wiki. Is it not available on the mobile site? I can't execute newest Blender 2.80, it says that I need a graphics card that supports OpenGL 3.3 or higher ERROR: A graphics card and driver with support for OpenGL 3.3 or higher is required  Processor: AMD A6-3670 APU with Radeon(tm) HD Graphics, 2695MHz BIOS Version: HPQOEM - 1072009 Dedicated GPU:   This is a very old PC, but when I search online it seems that theoretically Blender 2.8 should run since my openGL is 4.4 find command with regex quantifier e.g. {1,2} I have been trying to create a find command string that will find all files that end with a number 1-99 but exclude all others.  e.g. I want to find myfile1 myfile99 but not myfile456 and not myfilebackup  The regex I'm thinking of is myfile[1-9]{1,2} but I can't get this to work with find.  find . -regex '.*myfile[0-9]{1,2}' OR  find . -iname 'myfile[0-9]{1,2}'  From what I can see it's the {1,2} part that is not working.  (by the way can you use -name -regex interchangably?)  Any help appreciated. Two different solutions to integral Find all primes $p$ such that  $ p^3-4p+9 $ is a perfect square.   I tried a few different values for $p$, namely $2,3,5,7,$ and $11$. The prime $p =2,7,11$ all worked but $p =13$ didn't so it makes me wonder. How can I find all primes such that it is a perfect square?  EDIT: it turns out this is problem P25 in post number #63 at How did Montague Apparate into the Hogwarts bathroom? In year 5, Fred and George jam Montague into the Vanishing Cabinet. He's found weeks later in a toilet.      ‘He never managed to get all the words out,’ said Fred, ‘due to   the fact that we forced him head-first into that Vanishing Cabinet   on the first floor.’      […]      ‘They’ve found Montague, sir, he’s turned up jammed inside a toilet   on the fourth floor.’   ‘How did he get in there?’ demanded Snape.   ‘I don’t know, sir, he’s a bit confused.’   In year 6 Malfoy expands on that.      ‘I had to mend that broken Vanishing Cabinet that no one’s used for years. The one Montague got lost in last year.’      […]      ‘The other’s in Borgin and Burkes,’ said Malfoy, ‘and they make a kind of passage between them. Montague told me that when he was stuck in the Hogwarts one, he was trapped in limbo but sometimes he could hear what was going on at school, and sometimes what was going on in the shop, as if the Cabinet was travelling between them, but he couldn't make anyone hear him ... in the end he managed to Apparate out, even though he’d never passed his test. He nearly died doing it.   What happened here? Did he Apparate out into Hogwarts, Apparate into Borgin and Burkes and somehow wander back to Hogwarts and get stuck into a toilet? I was playing around with some integrals and noticed that some integrals of the form:  $$I(a,b,c)=\int_a^\infty \frac{\arctan(x+b)}{x^2+c}dx$$ Does have a closed form. I am trying to find for what constant $c$ will this work. In case you wonder why only $c$ is problematic, I will try to show by an example.  $$I=I(1,3,16)=\int_1^\infty \frac{\arctan(x+3)}{x^2+16}dx$$ Let's start by letting $x-1=t\,$ thus: $$I=\int_0^\infty \frac{{\arctan(\color{blue}{t+4})}}{t^2+2t+17}dt$$ With $\displaystyle{t=\frac{17}{y}\rightarrow dt=-\frac{17}{y^2}dy}$ $$I=\int_0^\infty \frac{{\arctan\left(\color{red}{\frac{17}{y}+4}\right)}}{\left(\frac{17}{y}\right)^2 +\frac{34}{y}+17 }\frac{17}{y^2}dy\overset{y=t}=\int_0^\infty \frac{{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}{t^2+2t+17}dt$$ $$2I=\int_0^\infty \frac{{\arctan(\color{blue}{t+4})+{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}}{t^2+2t+17}dt$$ $${\arctan(\color{blue}{t+4})+{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}=\arctan\left(\frac{\color{blue}{t+4}+\color{red}{\frac{17}{t}+4}}{1-(\color{blue}{t+4})\left(\color{red}{\frac{17}{t}+4}\right)}\right)$$ $$=\arctan\left(\frac{x^2+8x+17}{x}\frac{x}{-4(x^2+8x+17}\right)=\pi-\arctan\left(\frac14\right)$$ Above follows since the original integral is positive so we take $\arctan(-x)$ as $\pi-\arctan x $ and therefore getting a negative answer will not be an issue. $$I=\frac12 \left(\pi -\arctan\left(\frac14\right)\right)\int_0^\infty \frac{1}{t^2+2t+17}dt$$ Well, now the inner integral is not hard to compute and the final answer happens to be: $$I=\frac12 \left(\pi -\arctan\left(\frac14\right)\right)\frac14\arctan\left(\frac{t+1}{4}\right)\bigg|_0^\infty =\frac{\pi^2}{16}-\frac{3\pi}{16}\arctan\left(\frac14\right)+\frac18\arctan^2\left(\frac14\right) $$ There are more examples that I found by checking and try such as:  $$I(1,2,9)=\int_1^\infty \frac{\arctan(x+2)}{x^2+9}dx$$ $$I(2,1,6)=\int_2^\infty \frac{\arctan(x+1)}{x^2+6}dx$$ $$I(2,2,13)=\int_2^\infty \frac{\arctan(x+2)}{x^2+13}dx$$ And so on... All those can be solved by the same method: First substitute $x-a=t$, then let $t=\frac{\alpha}{y}$, where $\alpha$ is the "free of x" coefficient from the denominator.   The problem is that I tried more than $100$ combinations to get those integrals which is not that nice. How can we "smartly" find $c$ so that $I(a,b,c)$ is evaluable by symmetry? Or put in other word what should be $c$ if one wants to compute by symmetry $I(7,13,c)$? How do I copy to the clipboard in JavaScript? Which is the correct use of these two words, and in which context should one be used rather than the other?