secondary or not? Another noob question. I have been going back and forth on "im going to secondary my double IPA" and "Im just gonna leave it primary til completion" I keep reading contradicting theories and It's driving me insane. I dont want off flavors from the yeast, but i don't want worse off flavors from oxidation or contamination. Weigh in here guys, what is the common consensus. I am dry hopping the double IPA if that adds anything. My original schedule was going to be: Primary: 10 days Rack to secondary for 7 days, dry hop, then leave for 7 more before bottling. But now i dont know if I should leave out secondary and just dry hop a week from bottl
If/When to move to secondary fermentation Digression from   What are the good rules of thumb for determining a. whether to use secondary fermentation at all and b. when to move from primary to secondary.  Preferably any rule of thumb should have both chemical and anecdotal rationale supporting it.
If $A|B$ and $B|A$ then prove $A=\pm B$ If $A|B$ and $B|A$ then prove $A=\pm B$   So far I have $A|B \iff AX=B$ and $B|A \iff BY=A$ with $X,Y \in \mathbb{Z}$  Not sure how to finish, any help. Best resources to learn bash scripting? I would like to use Zapf's Palatino font for the text in my document, which also includes a lot of maths. And I would like a good font for the maths. This is a snapshot of what I am doing.  \documentclass[10pt,letterpaper]{memoir} \usepackage{amsmath} \usepackage{amssymb} \usepackage{mathpple} \usepackage{upgreek} \usepackage{mathpazo} \usepackage{lipsum}  \begin{document}  \chapter{Demo}  Some regular text. \lipsum[1]  Now for some maths.  \begin{displaymath} P_{N-1 + m} = \frac{C}{N-{1} + m} \binom{N_{2} - N_{1}}{m}         \alpha^{m}\beta^{(N_{2} - N_{1}) -m} \end{displaymath}  The number 27 is the sum of the first 5 odd prime numbers: $27 = 1+3+5+7+11$  \end{document}   To me the text and math fonts seem a good fit but the text does not look like Palatino. Let $(a_n)_{n \in \mathbb{N}}$ be a convergent sequence with limit $a \in \mathbb{R}$. Show that the arithmetic mean given by: $$s_n:= \frac{1}{n}\sum_{i=1}^n a_i \tag{A.M.} $$ also converges to $a$.  I have read:  but unfortunately the answers there don't help me much because I don't understand their substitutions and most of it all, why their substitutions seem to work.   What I know from the problem is that since $a_n$ is convergent and $\epsilon &gt;0$ is given, I can say that:   $$\exists N_1 \in \mathbb{N}, \forall n \geq N_1: |a_n-a|&lt;\epsilon_1 $$ I also know that since $a_n$ is convergent, it is bound, so $(a_n) &lt; M, \ \forall n \in \mathbb{N}$   I need to show that: $$\exists N_2 \in \mathbb{N}, \forall n \geq N_2: |s_n-a|&lt; \epsilon_2$$ I started as follows: $$ \left|s_n-a \right| = \left|\frac{1}{n}\sum_{i=1}^na_i -a\right|= \left|\frac{1}{n}\sum_{i=1}^m(a_i-a)+\frac{1}{n}\sum_{i=m+1}^n(a_i-a) \right| \\ \leq \frac{1}{n} \sum_{i=1}^m|(a_i-a)|+\frac{1}{n}\sum_{i=m+1}^n|a_i-a|$$ I believe to understand that the left sum after the $\leq$ is finite, bound and doesn't depend on the values that $n$ takes on. However, I don't understand where all the substitutions come from and make this proof so seemingly easy to complete.   Is there a general idea I can follow to complete such proofs? Because I know that the last step is to show that the given sum is smaller than $\epsilon$. I also know that I should bring the condition $a_n &lt; M$ into place somewhere, but I don't know where.  If I choose an $n \geq N: |a_n-a|&lt; \epsilon'$ what does that tell me about $|a_i-a|$? I know that they are seemingly the same just with a different index. How does the Commander interact with Banishing Light or Oblivion Ring? What is the minimum $ \sigma$-algebra that contains open intervals with rational endpoints What are the minimum $\sigma$-ring and $\sigma$-algebra on $\mathbb R$ which contain the open intervals with rational endpoints?  Is there a relation between this $\sigma$-algebra and Borels? Bluetooth Broadcom 43142 isn't working If I create just a basic hello.tex file an compile it using   pdflatex hello.tex   The pdflatex program is creating a plethora of file which is a little annoying. Currently it writes all those files about which I really don't care:  hello.aux, hello.log, hello.nav, hello.out, hello.snm, hello.toc   Is there a way to tell pdflatex to only write hello.pdf if the build is successful ? If not, can I specify the output directory of those files ? Maybe I could just write them in /tmp. Expansion of $(1+\sqrt{2})^n$ I'm using leaflet to display a map.  I have a control section that I'd like to be open by default, but I can't find a way to do it through the . Is there a way to do this? Undefined Control Sequence in \multido or \pspicture? Let $V$ be an n-dimensional vector space over $\mathbb{R}$ and $T:V \rightarrow V$ be linear. Call a vector $v \in V$ cyclic if $V$ is spanned by $\{v, \ Tv, \ T^2v,..\}$.  Question:  Show that $v$ is cyclic for $T$ then there is some $k \geq 1$ such that   $$B=\{v, \ Tv,..T^{k-1}v \}$$ is linearly independent in $V$ and that $T^kv$ lies in  the span of $B$.  Thoughts:  If we suppose that no such $k$ exists, then every subset of the form $\{v, \ Tv,..T^{l-1}v \}$ is linearly dependent, or it is linearly dependent with $T^lv$ not lying in it span.  There are in effect then two cases to consider.  Suppose every subset of the form $\{v, \ Tv,..T^{l-1}v \}$ is linearly dependent. Then $T^{l-1}$ can be expressed as a linear combination of $v,...,T^{l-2}v$. Now, $\{v, \ Tv,..T^{l-2}v \}$ is also linearly dependent by assumption and so we can express $T^{l-2}$ as a linear combination of  $v,...,T^{l-3}v$ etc... Continuing in this way, we show that $T^{l-1}v= \lambda v$ for some $\lambda \in \mathbb{R}$. Thus $$B=\{ \mu v \ | \ \mu \ \text{is an eigenvalue of} \ T^mv \ \text{for some} \ m \}$$ This is 'at best' countably infinite, and therefore could not span $V$ even if it were 1-dimensional, wince our base field is $\mathbb{R}$, which is uncountable.  Suppose then that $B$ is linearly independent for some $k$ but that $T^kv$ does not lie in its span. I can't see how to progress this argument.  Note: This question is for second year Undergraduates, and I think there must be some more simple approach; even the first part surprises me if we need to argue using the uncountability of $\mathbb{R}$, so any alternative approaches are also appreciated. Subset of a countable set is itself countable How is it proved that if $$A \subset B\ \text{with}\ B\ \text{countable} $$ then $A$ is either countable, finite, or empty? I think the proof involves a $1-1$ correspondence between $\mathbb{N}$ and $A$ but other than that I do not know how to proceed.  EDIT: I have checked the solution and it advises me to proceed as follows. " As a start to a definition of $$g: \mathbb{N} \rightarrow A$$ set $$g(1)=f(n_1) $$ Then show how to inductively continue this process to produce a $1-1$ function $g$ from $\mathbb{N}$ onto $A$."  So the proof according to my book involves induction. Reading a 5 Band resistor with Gold as a middle band? Can anyone read the resistor that the arrow is pointing to?  From what I know you can not have a gold band as a number and if I am reading it correctly it has a tolerance of 0 which does not seem right. Would it be considered cheating to ask for homework help on the internet, if I openly acknowledge the help received? How to remove the Mail icon indicator applet? How to sync bookmarks across Google Chrome and Mozilla Firefox bookmarks? A set $A \subset l_1$ is compact if and only if closed, bounded, and one other condition A set $A \subset \ell_1$ is compact if and only if $A$ is closed and bounded and given any $\epsilon &gt;0$, there exists $n_0$ such that $\sum_{k=n}^{\infty} |x_k| &lt; \epsilon$ for all $n&gt; n_0$ and $x \in A$.   It's easy to prove the part assuming that $A \subset l_1$ is compact, but I am finding difficulty to prove the other part. Why use apparently meaningless do-while and if-else statements in macros? Scroll to bottom of div? I am creating a chat using Ajax requests and I'm trying to get messages div to scroll to the bottom without much luck. I am wrapping everything in this div: #scroll {     height:400px;     overflow:scroll; }  Is there a way to keep it scrolled to the bottom by default using JS? Is there a way to keep it scrolled to the bottom after an ajax request?