I would like to produce an equation over two lines with a bracket opening in the first line and closing in the second line. Unfortunately, however, using the "\left" and "\right" commands (to have the size of the brackets automatically adjusted) leads to an error.  \begin{align*} L &amp;= \left[ \frac{a}{b} \\ &amp;+ a \right] \end{align*}   I would be very grateful for any advice on how to solve this problem! :)
I have a multi-line equation that I want to have matching sized parentheses. The advice I read online was to use \left( equation \right. \\ left. rest-of-equation \right) . However, this syntax leaves first half of the equation unaware of what is in the second half. For example, if you include a square root term after the break, the second parenthesis will be larger than the opening parenthesis. When you are using the aligned environment to split a long equation this can look somewhat offputting.    Compare the top equation to the bottom one. The parentheses match in size for the bottom equation, but the left hand parentheses for the top one is  noticeably smaller.  Top Code:   \begin{aligned}[t]   \gamma_{\text{u}} + \frac{1}{108}   \left(57\gamma_{\text{er}} \right. &amp;+  38\gamma_{\text{se}}\\ &amp; {}\pm  \left.\sqrt{2601 \gamma_{\text{er}}^2 + 3648  \gamma_{\text{er}}\gamma_{\text{se}} + 1444 \gamma_{\text{se}}^2}  \right)   \end{aligned}   Bottom Code:  \begin{aligned}[t] \gamma_{\text{u}} + \frac{1}{16} \left(9\gamma_{\text{er}}  +  6\gamma_{\text{se}} \pm  \sqrt{49 \gamma_{\text{er}}^2 + 76 \gamma_{\text{er}}\gamma_{\text{se}} + 36 \gamma_{\text{se}}^2} \right) \end{aligned}
Getting a collection of index values using a LINQ query Is there a better way to do this?  string[] s = {"zero", "one", "two", "three", "four", "five"};  var x =  s .Select((a,i) =&gt; new {Value = a, Index = i}) .Where(b =&gt; b.Value.StartsWith("t")) .Select(c =&gt; c.Index);   i.e. I'm looking for a more efficient or more elegant way to get the positions of the items matching the criteria. Multiple chrome.exe processes in Task Manager I know that each new tab on Google Chrome have its own process, but I have 14 chrome.exe processes running actually as I can see it on the Task Manager, and only two tabs are opened on Google Chrome.    Could it be due to Chrome extensions? What other reason can open more chrome.exe process and how to avoid it? Error Code: 1215. Cannot add foreign key constraint (foreign keys) setting up subdomain on ubuntu server $\zeta_{15}$ is $15$th primitive $n$th root of unity.   Question: Find the structure of the group $Gal(\mathbb{Q}(\zeta_{15})/\mathbb{Q})$  I know that if $p$ is prime then $G=Gal(\mathbb{Q}(\zeta_{p})/\mathbb{Q})=\mathbb{Z_{p}}^*$ but when $p$ is not prime, I am not show how to solve this if not prime.  Does my required group have any relation to some cyclic group $\mathbb{Z_n}$, i.e. abelian  Would really appreciate your guidance as I have not got my head around these concepts of cyclotomic fields and Galois structure. Thanks How to Understand the Definition of Cardinal Exponentiation I'm having trouble understanding the definition of cardinal exponentiation.  Let's start with the definitions / claims I've been given:  For any finite sets $A,B$, such that if $|A|=a$ and $|B|=b\neq 0$ then the number of functions from $A$ to $B$ is $b^a$. ( I'm guessing this is only used when both sets are finite ).  For any sets $A,B$, $B^A$ is the set of functions from $A$ to $B$.  For every set $A$, $|P(A)|=|\left \{ 0,1 \right \}^A|$  Now for the definition and why I don't understand it:  Let $a,b$ be cardinal numbers.  Let $A,B$ be sets such that $|A|=a, |B|=b$.  Then $|B^A|=b^a$.  How analagous to normal exponentiation is this and where does in differ when we introduce cardinals like $\aleph_0$ or $C$?  Is $2^2=4$ and is $\frak c^2=c\cdot c=c\times c$? For macro photography, how do I choose between a compact camera or a mirrorless interchangeable-lens camera (without considering a lens)?   What indicator(s) should I look for when making the choice? The easy way: Install Nvidia drivers, CUDA, CUDNN and Tensorflow GPU on Ubuntu 18.04 What's the easiest way to run GUI apps on Windows Subsystem for Linux as of 2018? Transit visa at Istanbul Airport, Turkey, during a Turkish Airlines arranged stopover? Do I need a transit visa at Istanbul Airport to enter Istanbul for city sightseeing or hotel accommodation?. All services provided free by the Turkish Airline. I am Nigerian citizen and booked Turkish Airline flight to Nigeria with 12hours layover at Istanbul Airport. unbalaced data set in classification tasks How do you make a deep copy of an object? The Fibonacci sequence is defined recursively by $F_1 = 1, F_2 = 1, \; \&amp; \; F_n = F_{n−1} + F_{n−2} \; \text{ for } n ≥ 3.$ Prove that $2 \mid F_n \iff 3 \mid n.$   Proof by Strong Induction : $\bbox[5px,border:1px solid green]{\color{green}{n = 1 }}$ $2 \mid F_1$ is false. Also, $3 \mid 1$ is false. The implication [False $\iff$ False] is vacuously true.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Induction Hypothesis}}}$ Assume that $2 \mid F_i \iff 3 \mid n$ for every integer $i$ with $1 ≤ i ≤ k$.  $\bbox[5px,border:1px solid green]{\color{green}{k + 1 \text{th Case}}} \;$  To prove:  $\quad 2 \mid F_{k + 1} \iff 3 \mid k + 1.$   $\bbox[5px,border:1px solid green]{\color{green}{n = k + 1 = 2}} \;$ $2 \mid F_2$ is false. Also, $3 \mid 2$ is false. So [False $\iff$ False] is vacuously true.  Hence assume that $k + 1 ≥ 3.$ We now consider three cases:  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 1: } k + 1 = 3q}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 \require{cancel}\cancel{\mid} (k − 1)$. By the ind hyp, $3 \require{cancel}\cancel{\mid} k    \iff F_k$ odd &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Since $F_{k+1} = F_k + F_{k−1}$, thus $F_{k+1}$ = odd + odd = even.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 2: } k + 1 = 3q + 1}}$  Thus $3 | k$ and $3 \require{cancel}\cancel{\mid} (k − 1).$ By the ind hyp, $3 | k \iff F_k$ even &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Thus $F_{k+1}$ odd.  $\bbox[5px,border:1px solid green]{\color{green}{{\text{Case 3: }} k + 1 = 3q + 2}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 | (k −1).$ By the ind hyp, $3 \require{cancel}\cancel{\mid} k \iff F_k$ odd and $3 \mid (k − 1) \iff F_{k - 1}$ even. Thus $F_{k+1}$ odd. $\blacksquare$  $\Large{1.}$ Does the proof clinch the $(\Leftarrow)$ of the $(k + 1)$th case?   $\Large{2.}$ Since the recursion contains $n, n - 1, n - 2$, thus the recursion "time lag" is $3$ here. So shouldn't $3$ base cases be checked?   $\Large{3.}$ Further to #2, shouldn't "assume $k + 1 \geq \cancel{3} 4$" instead?   $\Large{4.}$ Shouldn't the $n = k + 1 = 2$ case precede the induction hypothesis?  I referenced . Source: Exercise 6.35, P152 of Mathematical Proofs, 2nd ed. by Chartrand et al    Supplement to peterwhy's Answer:  $\Large{1.1.}$ I wrongly believed that all 3 Cases proved the $\Leftarrow$. I now see that Case 1 is $\Leftarrow$ via a Direct Proof. Cases 2 and 3 are $\Rightarrow$ via a Proof by Contraposition. Nonetheless, how would one foreknow/prevision to start from $3 \mid n$  for both directions of the proof? What potions actually grant fame or glory, or put a stopper in death? Are there any? ‘You are here to learn the subtle science and exact art of potionmaking,’ [Snape] began. He spoke in barely more than a whisper, but they caught every word – like Professor McGonagall, Snape had the gift of keeping a class silent without effort. ‘As there is little foolish wand-waving here, many of you will hardly believe this is magic. I don’t expect you will really understand the beauty of the softly simmering cauldron with its shimmering fumes, the delicate power of liquids that creep through human veins, bewitching the mind, ensnaring the senses ... I can teach you how to bottle fame, brew glory, even stopper death – if you aren’t as big a bunch of dunderheads as I usually have to teach.’&#185;   Philosopher's Stone - page 102 - Bloomsbury - chapter 8, The Potions Master   Are there any canon potions that actually grant fame or glory, or stopper death? I'm looking for specific canon-based answers from sources such as the Harry Potter books, interviews with J.K. Rowling, or information from Pottermore. Maybe because it's early I'm overlooking the obvious, but offhand I can't think of any canon potions that fit this criteria exactly.   &#185;I edited in the entire paragraph that I can teach you how to bottle fame, brew glory, even stopper death comes from for clarity. Hopefully this will be helpful and provide a better context for the comment in question. Recursive time series forecasting model In a recursive forecasting model, let's say you are trying to predict sales of Target for the next month and you will append that prediction to your input and predict the month after. Basically, your target is Sales quantity, which you lagged. But let's say you have other continuous numerical features like Price which you will lag as well. As your model is making predictions on Sales_QTY, it will feed it back to sales_qty as input so you will never ran out of values for that one. So how do you deal with your other features (other than the target) in order to generate them as next months inputs in a recursive forecasting model (because eventually you will run out of them)? DO you create a sub-model and try to predict them as well? If $x,y,z\gt 0$ and $xyz=1$ Then minimum value of $\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$ If $x,y,z\gt 0$ and $xyz=1$ Then find the minimum value of $\displaystyle \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$   $\bf{My\; Try::}$Using Titu's Lemma $$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge \frac{(x+y+z)^2}{2(x+y+z)} = \frac{x+y+z}{2}\ge 3\frac{\sqrt[3]{xyz}}{2} = \frac{3}{2}$$  and equality holds when $$x=y=z=1$$  My question is how can we solve it without the above lemma, like using Jensen's Inequality or other inequality.  Please explain me.  Thanks What is a glue record? This is a  about DNS glue records.   What exactly (but briefly) is a DNS glue record?  Why are they needed and how do they work? 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