If $f: \mathbb{R}^2 \to \mathbb{R}$ is polynomial in each argument, $f$ is polynomial
Is a bivariate function that is a polynomial function with respect to each variable necessarily a bivariate polynomial?
Grammar-checking tool for use with LaTeX? I am looking for a grammar-checking tool, i.e., something that checks for mistakes such as "two dog" or "a books".  It has to work on Linux (Ubuntu).  Also, since I use macros to generate some of the prose, it has to somehow work with LaTeX.  I don't think that such checkers are LaTeX-aware, so it has to work with some compilation product of LaTeX: DVI or PDF.    If you suggest a grammar checker that checks plain text, please explain how to you extract the text.  catdvi does a nice job, but it leaves some garbage like page numbers. Prove that $\Vert f(b)-f(a)-f'(a)(b-a) \Vert\leq \sup_{x\in [a,b]} \Vert f''(x)\Vert\Vert b-a\Vert^2$ Let $\Omega$ be an open set in $\Bbb{R}^n$ and $f:\Omega\to \Bbb{R}^m $ be of class $\Bbb{C}^2.$ Let $a,b$ be in $\Omega$ such that $[a,b]\subset \Omega.$ Prove that  \begin{align}\Vert f(b)-f(a)-f'(a)(b-a) \Vert\leq \sup_{x\in [a,b]} \Vert f''(x)\Vert\Vert b-a\Vert^2.\end{align} where $$f'(a)(b-a)=\big&lt;\nabla f(x),(b-a)\big&gt;=\sum^{n}_{k=1}\frac{\partial f_i}{\partial x_k}(a)(b_k-a_k).$$  I  think the idea of Mean Value Theorem for $C'$ functions in $\Bbb{R}^n$ () will help me in showing it but I don't know how to! Please, can anyone show me a proof or reference? Multiply cited reference in footnotes Consider an isosceles triangle. Let r be the radius of the circumscribed circle and p be the radius of the inscribed circle. Prove that the distance d between the centers of these two circles is $$d = \sqrt{r(2p-r)}$$ Harmonic Motion I'm using 17.04 ( Clean Install ), and as far as I remember, every try to install any sofware- it fails with that error code.  I tried sudo apt-get update &amp;&amp; apt-get upgrade and problem persists.  any idea why?   edit: Using software Updater gave this result :   edit : output of update  guy@LG-LAPTOP:~$ sudo apt-get update &amp;&amp; apt-get upgrade [sudo] password for guy:  Ign:1 http://il.archive.ubuntu.com/ubuntu zesty InRelease Ign:2 http://il.archive.ubuntu.com/ubuntu zesty-updates InRelease               Ign:3 http://il.archive.ubuntu.com/ubuntu zesty-backports InRelease             Err:4 http://il.archive.ubuntu.com/ubuntu zesty Release                           503  Service Unavailable Get:5 http://security.ubuntu.com/ubuntu zesty-security InRelease [89.2 kB]      Err:6 http://il.archive.ubuntu.com/ubuntu zesty-updates Release                   503  Service Unavailable Get:7 http://archive.canonical.com/ubuntu zesty InRelease [10.2 kB]             Err:8 http://il.archive.ubuntu.com/ubuntu zesty-backports Release                 503  Service Unavailable Get:9 http://archive.canonical.com/ubuntu zesty/partner Sources [1,416 B]       Get:10 http://archive.canonical.com/ubuntu zesty/partner amd64 Packages [1,616 B] Get:11 http://archive.canonical.com/ubuntu zesty/partner i386 Packages [1,612 B] Get:12 http://archive.canonical.com/ubuntu zesty/partner Translation-en [708 B] Get:13 http://security.ubuntu.com/ubuntu zesty-security/main amd64 DEP-11 Metadata [11.7 kB] Get:14 http://security.ubuntu.com/ubuntu zesty-security/main DEP-11 64x64 Icons [9,219 B] Get:15 http://security.ubuntu.com/ubuntu zesty-security/universe amd64 DEP-11 Metadata [14.4 kB] Get:16 http://security.ubuntu.com/ubuntu zesty-security/universe DEP-11 64x64 Icons [31.0 kB] Reading package lists... Done                                          E: The repository 'http://il.archive.ubuntu.com/ubuntu zesty Release' does no longer have a Release file. N: Updating from such a repository can't be done securely, and is therefore disabled by default. N: See apt-secure(8) manpage for repository creation and user configuration details. E: The repository 'http://il.archive.ubuntu.com/ubuntu zesty-updates Release' does no longer have a Release file. N: Updating from such a repository can't be done securely, and is therefore disabled by default. N: See apt-secure(8) manpage for repository creation and user configuration details. E: The repository 'http://il.archive.ubuntu.com/ubuntu zesty-backports Release' does no longer have a Release file. N: Updating from such a repository can't be done securely, and is therefore disabled by default. N: See apt-secure(8) manpage for repository creation and user configuration details. guy@LG-LAPTOP:~$    output of apt get  guy@LG-LAPTOP:~$ sudo apt-get install putty Reading package lists... Done Building dependency tree        Reading state information... Done The following additional packages will be installed:   putty-tools Suggested packages:   putty-doc The following NEW packages will be installed:   putty putty-tools 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 661 kB of archives. After this operation, 2,713 kB of additional disk space will be used. Do you want to continue? [Y/n]  Err:1 http://il.archive.ubuntu.com/ubuntu zesty/universe amd64 putty-tools amd64 0.67-3   503  Service Unavailable Err:2 http://il.archive.ubuntu.com/ubuntu zesty/universe amd64 putty amd64 0.67-3   503  Service Unavailable E: Failed to fetch http://il.archive.ubuntu.com/ubuntu/pool/universe/p/putty/putty-tools_0.67-3_amd64.deb  503  Service Unavailable E: Failed to fetch http://il.archive.ubuntu.com/ubuntu/pool/universe/p/putty/putty_0.67-3_amd64.deb  503  Service Unavailable E: Unable to fetch some archives, maybe run apt-get update or try with --fix-missing? guy@LG-LAPTOP:~$ Markdown for ordered lists, starting with zero? I see that you can create an ordered list using Markdown notation like the following.  1. line 1 2. line 2 3. line 3 ... N. line N   The Markdown content supports starting each line with any arbitrary number and reorders the list starting with 1. I'd like to start an ordered list with zero. My use case for zero would be in describing capture groups from regular expressions where group zero always returns the entire string. So I'd like to display something like this:  # Groups 0. Entire string 1. The first capture group you were looking for 2. The second capture group 3. The third group with the string I heart Kittens   Markdown instead reorders the group and forces the ordered list to start at 1:    Trying to force the issue with inline HTML: &lt;ol start="0"&gt; or &lt;ol&gt;&lt;li value="0"&gt;entire string&lt;/li&gt;&lt;/ol&gt; doesn't work either. Any ideas on how I can create ordered list starting with zero?  If it's not possible to do so, then this is a feature-request. I'm trying to use the Bash parameter expansions to modify the output of a command substitution or another parameter expansion. The following nested expansions work quite well in Zsh; but result in a &quot;bad substitution&quot; error in Bash: ${${PWD##*/}//trunk/latest}  or ${$(basename $PWD)//trunk/latest}  the output should be the last folder of the $PWD, replaced by latest when my current directory is trunk so /home/user/trunk should become latest Is there a Bash equivalent allowing to chain expansions without relying on variables or pipes? Or do Bash expansions only allows the input to be a string or a plain variable? Cutting strands of wire to fit plug I am a hvac tech, but my question is I have 2awg stranded and my plug is 6awg, will it matter much if I cut some strands to fit my plug? the plug is 60amp 250vac for a piece of kitchen equipment if the matters. Proof verification: column rank = row rank $\newcommand{\span}{\operatorname{span}}\newcommand{\rank}{\operatorname{rank}}$I would like to prove that for every martix, column rank = row rank Let $A\in M_{m\times n}$ be some matrix. fix $R_A$ = vector space of the rows of $A$ , $C_A$ = vector space of the columns of $A$  $$\rank (R_A) = dim \ (\span \{R_1, R_2, \ldots, R_m\})$$ $$\rank (C_A) = dim \ (\span \{C_1, C_2, \ldots, C_n\})$$  let $x$ be some vector. $$x\in \operatorname{Null}(A) \Leftrightarrow \forall i  \ \ (1 \leq i \leq m) : \langle x,R_i\rangle = 0$$ (to be clear - I'm referring to the inner product of $x$ with each row of $A$) Using the rank nullity theorem, $\dim \operatorname{Null}(A) =  n - \rank(R_A)$ as $n$ =  number of columns. what I would like to do is to say: $$\dim \operatorname{Null}(A) =  n - \rank(R_A)$$ $$\dim \operatorname{Null}(A) = n - \rank(C_A)$$ therefore,  $\dim \operatorname{Null}(A) = n - \rank(R_A)= n - \rank(C_A)  \Longrightarrow \rank(R_A) = \rank(C_A) $ it that false? is using the rank- nullity  theorem this way is cheating? or just doesn't prove what needs to be proven formally? "This site is all about getting answers. It's not a discussion forum. There's no chit-chat." Simple PHP Pagination script How was Nearly Headless Nick unpetrified? If $a,b,c \in \mathbb{R+}$ and $abc=1$ Prove That  $$S=\sum \frac{a}{a+b^4+c^4} \le 1$$  My approach:  we have $$S=\sum \frac{\frac{1}{bc}}{\frac{1}{bc}+b^4+c^4}=\sum \frac{1}{1+b^5c+bc^5}$$  Now by $AM \ge GM$ we have  $$\frac{1+b^5c+bc^5}{3} \ge b^2c^2$$ $\implies$  $$\frac{1}{1+b^5c+bc^5} \le \frac{1}{3b^2c^2}=\frac{a^2}{3}$$ Hence  $$S \le \frac{a^2+b^2+c^2}{3}$$  Can we proceed with this? $X\sim\mathrm{Poisson}(\lambda_1 = 5)$ and $Y\sim\mathrm{Poisson}(\lambda_2 = 15)$. Let $Z = X + Y$ . Compute $\mathrm{Corr}(X, Z)$. Error installing Apache2 Server How is effect precedence determined? I was using  and my opponent has a . I attacked Sylvanas, fearing that the Deathrattle will take control of one Hyena. I feel lucky, because it did not happen.  I want to know if there are some general rules of effect precedence.   Has Blizzard published something about this?  Are there articles teaching about precedence?   Please note, I want guidance about effect precedence in general, not just about Deathrattle. "Thought you'd never ask" is "I thought you'd never ask" with "I" omitted. "Hope this helps" is "I hope this helps" with "I" omitted.   In English grammar, normally every sentence should have a subject, right?   My first thought is that these two examples are so often used that they are like set phrases. But these are not really set phrases. You can alter the words after "thought" and "hope".  Another possible explanation is the tendency to drop the subject if it is the first person pronoun. It seems that in many languages, such as Spanish, Italian and Japanese, the first person subject is usually omitted. Maybe English is going the same way? (Not exactly the same, since in Italian, verb forms change according to the person, so the subject is not necessary to understand who one's referring to.)  And, apparently, such omission is more common in spoken English than in written English.  Are there more examples of such first person subject omission? How frequent is it? Clicking community promotion ad renders an instance of SO within the frame of that ad Today I came across an ad on this question on SO:  (I don't expect it to stay very long).   There was a community promotion ad on the right hand side of the site, I think for one of our English sites. It had a sentence that I only partially remember:     Idiom ... poor or even poorer   I found it interesting, so I clicked that ad. Which caused a new instance of SO to open within the ad space. Unfortunately I failed to take a screenshot. I've been refreshing the page like crazy, but can't get the ad to appear again.  Edit: I managed to get a new community add, for the photography site, and captured a screenshot:   Edit2: This bug still persists, I saw it today with an OOP add from CodeReview, and an area 51 add. A correct preposition - account closed due/by/at 12th of May