Make head if my figure smoother
How to avoid the wrinkle at the poles of the UV sphere?
Similar to , what's the best way to stay up to date with the latest stable releases of Thunderbird? I want to know how can I achieve this effect in LaTeX:    Basically it looks like this: {F}IRST {L}ETTER, where the letters in the braces are a little bigger than the other ones but they all the letters are uppercase.  Any suggestions? Evaluating $\int_0^{\tfrac{\pi}{4}}\ln(\cos x-\sin x)\ln(\cos x) dx - \int_0^{\tfrac{\pi}{4}}\ln(\cos x+\sin x)\ln(\sin x)dx=\frac{G\ln 2}{2}$ In order to compute, in an elementary way,  $\displaystyle \int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$  (see  )  i need to show, in a simple way, that:  $\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\cos x-\sin x)\ln(\cos x) dx - \int_0^{\tfrac{\pi}{4}}\ln(\cos x+\sin x)\ln(\sin x)dx=\dfrac{G\ln 2}{2}$  $G$ being the Catalan constant.  The reason of my interest for this question is, if i am right, that this formula permits to find out a relation between integrals:  $\displaystyle \int_0^1 \dfrac{\ln(1+x)\ln(1+x^2)}{1+x^2}dx$  $\displaystyle \int_0^1 \dfrac{\ln(1+x)^2}{1+x^2}dx$  $\displaystyle \int_0^{\tfrac{\pi}{4}}\big(\ln(\cos x)\big)^2 \dfrac{}{}dx$  and some constants. $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ (Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12)  For $c \gt 0$, consider the quadratic equation $x^2 - x - c = 0, x &gt; 0$.  Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ and then, if $n$ is an index for which $x_n$ has been defined, defining  $$x_{n+1} = \sqrt{c+x_n}$$  Prove that the sequence $\{x_n\}$ converges monotonically to the solution of the above equation.  Note: The answers below might assume $x_1 \gt 0$, but they still work, as we have $x_3 \gt 0$.    This is being repurposed in an effort to cut down on duplicates, see here: .  and here: . How to catch creation of DOM elements and manipulate them with jQuery $a+b+c = 3$, prove that :$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6$ $a, b,c $ are positive real numbers such that $a+b+c = 3$, prove that :$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6$  Any ideas ? What is the proper use of [square brackets] in quotes? I sometimes see square brackets used while quoting.  My assumption is that they are replacing a pronoun with what the object of the pronoun, but I never know for sure because I don't usually get to see what the original quote looks like before the modification.  What are these called and what are the rules of use? I want to make an animation in time to a song so when I hit play on the timeline I need to be able to hear the song so I can finely tune the animations to start and end at the right time.  Is it possible to have sound play on an animation while it is playing in the 3d viewport before it has been rendered. I have a skyscraper building but I only want the glass in the texture to be glossy.  How would I accomplish this?    Ok, I have done a test and here are the results.  Here is the original:   here is an adjusted image So I wanted to make some minecoins and I also heard that anyone become a marketplace creator could but I don't know where to apply. I'm in Minecraft Pocket Edition 1.14.30. Can you tell me where I can apply? Objective C const NSString * vs NSString * const How to create an Atlas based on point attributes? Essentially I am wanting to produce an atlas based on a categorical field in a point layer.   i.e. I have a point layer of childcare providers with the categorical field "Provision". I've categorized each feature in this field with "After School Club" , "Breakfast Club" etc, and I now want to produce a set of maps that iterate through each category and show only the points for each. One map of after school clubs, one map of breakfast clubs etc. The extents might be subtly different.  I could do it one by one but it seems like there should be a way to produce an atlas based on the extents of each category? (I do feel i'm missing something obvious :) )   Or alternatively is there a way of automating the creation of a polygon layer and using that as a hidden coverage for the atlas?  EDIT: I've made a little progress with this  - you can use rule based styling to switch on and off features relevant to the current atlas coverage feature.  it actually works fine if all you want to do is show a different set of points. I'm now looking at tying that back to a color scheme and reactive legend. In Dummit and Foote, Exercise 15.18 (p. 727 3rd edition) one is asked to prove that for a commutative ring with unity $R$, and $f$ a non-nilpotent element of $R$, that the localization $R_f$ is isomorphic to $R[x]/(xf - 1)$.  I have done this, using the strategy outlined in the problem, using the universal property of $R_f$ proved in Theorem 36. The problem I have is this:  If $R_f \cong R[x]/(xf - 1)$ and $\phi:R[x] \to R_f$ is a surjective homomorphism, it stands to reason that $\text{ker}(\phi)$ should be $(xf - 1)$. I cannot show this directly. The surjective homomorphism I have in mind is, for:  $p(x) = a_0 + a_1x + \cdots + a_mx^m$  $\phi(p) = \dfrac{a_0}{1} + \dfrac{a_1}{f} + \cdots + \dfrac{a_m}{f^m}$  It is clear to me that $\phi$ annihilates $(xf - 1)$, so that ideal is contained in the kernel. What I cannot show is the other containment, or equivalently:  If $\phi(g) = 0_{R_f}$ then $g(x) \in (xf - 1)$. The trouble I am running into is that $R$ might be a "bad" ring, containing a lot of zero-divisors, $f$ might even be a zero-divisor (being non-nilpotent is a considerably weaker property), so the usual polynomial tricks of the division algorithm or using the irreducibility of $xf - 1$ aren't much use, here.  I feel I am missing something obvious....what is it? $$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$  Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable.  The answer is $Θ (n \log \log n)$.  Can anyone arrive at the solution. Automatically switch gpu graphics driver when gaming I have installed Ubuntu GNOME 15.10. I installed the nvidia tested proprietary driver and then I restarted. Afterr that I selected as primary GPU the Intel driver for normal use. However, when I enter a steam game(Portal) the graphics is bad so In order to play normal I have to switch manually to the nvidia driver again and reboot again. After the game I have to manually select as primary driver the Intel one and reboot again. It is so boring. QUESTION: How to automatically switch gpu graphics driver when I enter a game or something that needs more GPU? With English not being my first language, I usually need support for at least two languages when using features like auto-correct and dictation.  Is it possible to toggle the language which is used for dictation via a shortcut?  As of now, I always have to manually navigate through the menus in the System Preferences which if far from convenient. I've tried to see if there are any pre-defined actions in Automator, but found none. How to find $\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r}$? Problem was to find $$\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r}.$$ My partial progress i tried to motivate such that upper term in binomial terms gets constant rather than variable , so i thought of changing it to form $\binom{n}{n-r}\binom{n-r}{n-2r}$. Now it's of the form $\binom{n}{m}\binom{m}{k}$. We know it is equivalent to $\binom{n}{k}\binom{n-k}{m-k}$. Doing this we get $\binom{n}{2r}\binom{2r}{r}2^{n-2r}$. Now what I need to do as next step as such still nothing the upper or lower terms r any constant values(sum to ) , all are variables still. Now by one answer by Robpratt i got by snake oil , is there any other way to solve this problem ? Let $\mu(\Omega)$ be a probability measure (i.e. $\mu(\Omega) = 1$), and let $f,g$ be non-negative measurable functions on $\Omega$ such that $fg \geq 1$. Show that $1 \leq (\int f^p )(\int g^p)$ for all $0 &lt; p &lt; \infty$   I'm a bit at a loss of what to do. I know that since $fg \geq 1$, we may obtain:  $$\int_{\Omega} d\mu = \mu(\Omega) = 1 \leq \int_{\Omega} fg = \int_{\Omega} |fg|$$ The right-most integral is true since $fg$ is positive and it is equal to $||fg||_1$.  By Hölder's inequality, we obtain:  $$1 \leq \int |fg| \leq ||f||_p ||g||_q = (\int f^p)^\frac{1}{p}(\int g^q)^\frac{1}{q}$$  where $p$ and $q$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$. Raising the left-most and right-most sides of this inequality to the power $p$, I may obtain:  $$1 \leq (\int f^p)(\int g^q)^\frac{p}{q}$$  I now wish to show that $(\int g^q)^\frac{p}{q} \leq \int g^p$, but I've been at a loss of how to do it. Any tips or hints would be greatly appreciated. I predicted the formula by finding total number of non-isomorphic labelled trees of order 1 is 1 , order 2 is 1,order 3 is 3,order 4 is 16,order 5 is 125.But how do i prove it ? I am beginner in graph theory so i will be very thankful if i get some simple approach to solve the problem(Not by using some difficult theorems). We all know , but that guidance is, naturally, geared toward main sites, where we want to keep things clean and focused.  Meta, on the other hand, is a site's "second place", with some different rules and conventions.  For example, questions on main sites can be closed as primarily-opinion-based, but meta is all about opinions.  There's a "discussion" tag and such questions collect many opinions.  Meta is different.  Occasionally I see comment flags on metas and I'm never sure what to do with them.  Setting aside things that are blatantly offensive or spam (that's a no-brainer), is there such a thing as "too chatty" or a comment discussion that should be moved to chat on meta?  I am very conservative in deleting things on per-site metas (except on  posts, which like main should stay focused).  Are there any guidelines specific to per-site metas?  I'm specifying per-site metas and not MSE because MSE seems to have conventions that fall somewhere between "main site" and "per-site meta".  If the answer is the same for all metas then great, but I don't want to assume it'll be the same.