Being on Ubuntu Server I am trying to write an automation script.  Here is what I have:  #!/bin/bash  while [ "$1" != "" ]; do     case $1 in         -p | --port )           shift                                 PORT=$1                                 ;;         -d | --domain )         shift                                 DOMAIN+=("$1")                                 ;;         -l | --location )       shift                                 PATH=$1                                 ;;         * )                     exit 1     esac     shift done  echo "Parsing finished. List of domain is: " for val in "${DOMAIN[@]}"; do     echo " - $val" done  echo "Generating -d domain list..." for val in "${DOMAIN[@]}"; do     domains="$domains -d $val" done  echo "-d list is: $domains"  echo "Port is: $PORT."  echo "Location to save is: $PATH"  certbot certonly --standalone --standalone-supported-challenges http-01 --http-01-port $PORT $domains --cert-path $PATH   When ran with just two parameters, it works correctly. When I specify the third parameter -l or --location, it gives:  cert-ggm.sh: line 35: certbot: command not found   I tried to rename the -l parameter to capital P, got the same result. The test echoes return correct value parse. What's the deal with the location parameter casting certbot command uncallable? Thank you in advance.
I've installed a version of Ubuntu Server and although I'm fairly still fresh with Ubuntu, I'm trying to create a bash file to run a program in the background. More importantly, I'm trying to get a file nameto display correct and create it before I start running the program (I like my debug stuff).  However, here is my bash script:    #!/bin/bash  PATH="/var/log/servers/Game/"  FILENAME="startup"  EXT=".log"  DATE=$(date +%d-%m-%Y)  NEWFILE=${PATH}${FILENAME}_${DATE}${EXT}   I have verified that the file does exist in the PATH directory. However, the issue is with the date command. When I run the script bash -x game.sh this is what is displays.   + PATH=/var/log/servers/Game/  + FILENAME=startup  + EXT=.log  ++ date +%d-%m-%Y  game.sh: line 5: date: command not found  + DATE=  + NEWFILE=/var/log/servers/Game/startup_.log  + echo /var/log/servers/Game/startup_.log   I don't understand why it would say the command doesn't work, when I can go to the main console and run date and get a valid response.  Any help is greatly appreciated!
Is the collection of all functions between two sets a set? Can't modify list elements in a loop Generating data with a given sample covariance matrix Possible Duplicate:          We have some "theme colors" that are reused in our CSS sheet.  Is there a way to set a variable and then reuse it?  E.g.  .css OurColor: Blue  H1 {   color:OurColor; } Sequence of a product of functions in $L^p. L^q$ with $p,q$ conjugate The question is: if $f_i$ is a sequence of functions in $L^p$ converging to $f$ and $g_i$ a sequence in $L^q$ converging to $g$ show that $f_ig_i$ converges to $fg$ in $L^1$ for $p,q$ finite and $\frac{1}{p}+\frac{1}{q}=1$. Does this result hold if $p=1, q=\infty$?  So I think I showed the first part. $||f_ig_i-fg||_1=||f_ig_i-f_ig+f_ig-fg||_1\leq||f_ig_i-f_ig||_1+||f_ig-fg||_1 = ||f_i(g_i-g)||_1+||g(f_i-f)||_1\leq||f_i||_p||g_i-g||_q+||g||_q||f_i-f||_p$   (by Holder's inequality) which goes to $0$ as $i\rightarrow\infty$.  However the second part of the question throws me off because I don't see why this proof doesn't work just as well for $p=1,q=\infty$.  Am I missing something? Does the second part also hold? If $\alpha_{1},\ldots,\alpha_{m}$ are vectors different from zero vector, then there is a linear functional $f$ such that $f(\alpha_{i})\neq 0$ Roots of $\,ax^2+bx+c\,$ are irrational if $a,b,c$ are odd integers How would I show the following.  For all odd integers a,b, and c if z is a solution of $ax^2+bx+c=0$ the z is not rational.  How would I show this? This is what I did  Let a be an odd integer then for a integer k it has the form $a=2k+1$  then $b=2t+1$ for if c integer.  then $c=2r+1$ if s is an integer then I do  $(2k+1)(x)^2+2t+1(x)+(2r+1)=0$  But I find my self stuck on what to do any hint are appreciated. See YouTube videos in Anaglyph mode How to see a YouTube video in  from a non-3D Android device? I am not able to enable Anaglyph mode from the YouTube application. Ubuntu 19.04 show applications box Showing $\gcd(2^m-1,2^n+1)=1$ A student of mine has been self-studying some elementary number theory. She came by my office today and asked if I had any hints on how to prove the statement      If $m$ is odd then $\gcd(2^m-1,2^n+1)=1$.    It's been a while since I took number theory and I'm not sure what to do. She said she is learning about congruences, primitive roots, and power residues.  She has not taken any group theory. In An Introduction to Statistical Learning (James et al.), in section 3.7 exercise 5, it states that the formula for $\hat{\beta}_1$ assuming linear regression without an intercept is  $$\hat{\beta}_1 = \dfrac{\displaystyle\sum\limits_{i=1}^{n}x_iy_i}{\displaystyle \sum\limits_{i=1}^{n}x_i^2}\text{,}$$ where $\hat{\beta}_0 = \bar{y}-\hat{\beta}_1\bar{x}$ and $\hat{\beta}_1 = \dfrac{\displaystyle S_{xy}}{S_{xx}}$ are the usual estimates under OLS for simple linear regression ($S_{xy} = \displaystyle\sum\limits_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y})$).  This is not the actual exercise; I am merely wondering how to derive the equation. Without using matrix algebra, how do I derive it?   My attempt: with $\hat{\beta}_0 = 0$, we have $\hat{\beta}_1 = \dfrac{\bar{y}}{\bar{x}} = \dfrac{S_{xy}}{S_{xx}}$.  After some algebra, it can be shown that $\displaystyle S_{xy} = \sum\limits_{i=1}^{n}x_i^2 - n\bar{x}\bar{y}$ and $\displaystyle S_{xx} = \sum\limits_{i=1}^{n}x_i^2 - n\bar{x}^2$. From here, I'm stuck. It's not clear to me how to choose Spells Known when creating a new character at a higher level.  Let's take the Bard as a working example first:  Bard level 2:   Spells Known: 5 Slots per Spell level: 1st &times; 3   Bard level 3:   Spells Known: 6 Slots per Spell level: 1st &times; 4, 2nd &times; 2   I'm interpreting this to mean that, if an existing 2nd-level Bard gains a new level (level 3), he's gaining 1 (one) new spell known. He can then choose the new spell known to be either 1st or 2nd, correct?  What I don't understand then, is the case of creating a new character starting at a higher level; say, creating a new Bard at level 3.  In that case you're forced to choose 6 spells, which must be from the list of 1st- and 2nd-level spells, correct?  When making this 3rd-level Bard from scratch, though, can I select 2nd level spells for all 6 spells known?  As another example, say you're creating a very high level Bard from scratch, starting at level 15. The book says your Spells Known is 19, and your spell slots are 1st × 4, 2nd × 3, 3rd × 3, 4th × 3, 5th × 2, 6th × 1, 7th × 1, and 8th × 1.  So, for those 19 Spells Known, you can choose any spells from 1st through 8th spell level. Am I free then to select as many spells as I can at 8th spell level (5 in number), working backwards, and in the end ignoring the 1st, 2nd, 3rd level spells entirely?  As another example, the Ranger doesn't know any spells at 1st level, nor have any spell slots, but the book says “You know two 1st-level spells of your choice from the ranger spell list.” Does it mean at 2nd level and above?  Long story short: When choosing Spells Known for a spellcasting character created above 1st level, am I allowed to choose my Spells Known at whatever spell level I like, just so long as I have at least one spell slot of that level?  I'm working on a D&amp;D 5e character generator and need to clarify this. How do I turn off equation auto numbering I wrote an algorithm to randomly generate terrain and I am trying to load and edit my results with blender.  The problem is that when I load my terrain in blender, the individual triangles are not linked together.     You can see when I shrink/flatten a subset the the faces, the rest don't follow and deform as a single mesh.  I tried deleting the faces and edges and using F to create faces from the points, but it does not give usable results.    How can I go about solving this problem? "This box of matches is empty" There are several Japanese books teaching Japanese students how to write in English. I found this example in  (Correcting Errors in English Composition Manuals) by Michio Kawakami and J.D. Monkman.  The authors of this book claim that this sentence is incorrect:     This box of matches is empty.   They note that a “box of matches” is a box containing matches (and not a box made for keeping matches), hence the sentence should instead be:     This matchbox is empty.   Likewise, an “empty bottle of beer” should instead be an “empty beer bottle.”  Is this actually an error that I should avoid in writing English? If so, is it a grammatical error or something else? How do I delete all empty directories in a directory from the command line? Say I have a directory named foo/. This folder includes subdirectories. How can I delete all the empty directories in one command? Big floor symbols When I write  \lfloor\dfrac{1}{2}\rfloor   the floors come out too short to cover the fraction. How can I lengthen the floor symbols? Where is Jon going? Why are integer GCDs positive? [unit normalization of GCDs] The definition in my text reads, An integer $d$ is said to be the greatest common divisor of two non-zero integers $a$ and $b$ iff,  $d|a$ and $d|b$ and if $k$ is any other common divisor of $a$ and $b$ then $k|d$  Now here's the thing, if $d|a$ and $d|b$ then surely $-d|a$ and $-d|b$ as well, also $k|-d$ What I take from this? GCD is not unique! That is if $\mathrm{gcd}(12,8)= 4$ then by the definition, $\mathrm{gcd}(12,8) = -4$ as well. Yet I never ever seen a negative gcd. Someone please explain. Maybe, $4&gt;-4$, and we want the &quot;greatest common factor&quot; so...? But that still doesn't justify the definition. How to prove: if $a,b \in \mathbb N$, then $a^{1/b}$ is an integer or an irrational number?