What is the best practice for testing a function with a list of inputs and outputs with python's unittest module?
How do you generate dynamic (parameterized) unit tests in Python?
Theorem 13.9 in these  is the polynomial spectral mapping theorem:     Theorem 13.9 For a polynomial $p$ we have $\sigma(p(T)) = p(\sigma(T))$.     My questions:   First direction: Why would $q(T) (p(T) - p(\lambda) I)$ be an inverse for $T - \lambda I$? Update: @RobertIsrael said that instead $(p(T) - p(\lambda) I)^{-1} q(T)$ should be $(p(T) - p(\lambda) I)^{-1}$ be? According to the lines in the proof above, we have $$ (p(T) - p(\lambda) I)^{-1} = ((T - \lambda I) q(T))^{-1} \overset{?!}{=} q(T)^{-1} (T - \lambda I)^{-1}, $$ but as $\lambda \in \sigma(T)$ by assumption, $(T - \lambda I)^{-1}$ doesn't exist, right? For the second direction: Why does $x_i \in \sigma(T)$ imply that $p(x_i) - \mu = 0$? I thought that $x_i \in \sigma(T)$ only implies that $T - x_i I$ is not boundedly invertible. Since $\mu \in \sigma(p(T))$ we have that $p(T) - \mu I = c \prod_{k = 1}^{n} (T - x_i I)$ is not boundedly invertible but how does that help?   In both statements there seem to be a notion of "a product of operators is not invertible iff all factors aren't invertible", but how can that be true? Consider i.e. shift operators on a sequence spaces such that $ST = I$, i.e. the left and right shift operators, then the above mentioned notion doesn't hold, right? "The antipodal map of $S^2$ is the isometry $\bar m$ sending each point $P$ to its opposite point $-P$" (from textbook)  Show that antipodal points remain antipodal under any isometry of $S^2$. In other words, for each isometry f of $S^2$, if $P$ and $P'$ are antipodal, then so are $f(P)$ and $f(P')$  I am unsure how I can define the $P$ and $P'$ so that they are opposites of each other in the antipodal map, and am I supposed to apply the traditional isometries (translation, reflection and rotation) to the points? I don't understand this problem. Performance of Non Clustered Indexes on Heaps vs Clustered Indexes Is there some way to PUSH data from web server to browser? Of course I am aware of Ajax, but the problem with Ajax is that the browser should poll the server frequently to find whether there is new data. This increases server load.  Is there any better method (even using Ajax) other than polling the server frequently? Let's get rid of the 10K flag queue are pretty cool... You get a birds-eye view of activity on the site, a &quot;dashboard&quot; view of what's happening. Some of the individual tools haven't scaled particularly well with Stack Overflow's growth, but the concept behind them is still sound: we trust you to enough to be a bee watcher now.  ...and then there's . What a let-down! Once upon a time, this queue contained spam and offensive flags, which one might reasonably assume were important enough to put in front of the more trusted members of the site. Re-flagging them brought them that much closer to deletion, while editing them offered salvation to some hapless author. Nowadays, it's a bunch of Not an Answer flags and a smattering of assorted cruft. 10K users can't even vote to delete these; only 20K users have that privilege. Re-flagging them does nothing but increase their priority in the moderator flag queue, where they frequently outrank more pressing issues;  has . Also, it's , and the behavior has diverged far enough from that of the moderator flag queue on which it is based that it has become an active hindrance to further development of the tooling there. Worst of all, we have  that's available to anyone with the editing privilege. It even has logic built in to prioritize answers likely to be deleted for users with . It's like we gave you a car for your 10th birthday, and then replaced it with a rusty bicycle when you turned 16. It was a nice idea, but it has outlived its usefulness. Proposed changes The /tools/flagged route goes away. Period. No replacement. The rest of the 10K tools stick around as informational pages. Not an Answer flags go into /review/low-quality, . Then we beef up the Low Quality review process to make better use of more experienced reviewers and solve this whole &quot;declined / helpful / disputed&quot; flag debate once and for all:  Effective # of reviews required == ReviewsRequired + # of applicable flags (where ReviewsRequired is 2 on Stack Overflow, 1 everywhere else). So 1 VLQ or NAA flag means EffectiveReviewsRequired=3. LQ tasks are not dequeued until one of the following conditions is met:  Post is edited from within review. Outcome: flags are marked &quot;helpful&quot; (current behavior). Post accumulates 3 Delete votes (can only happen when post scores &lt;= 0 and reviewers have &gt;= 20K rep). Outcome: post is deleted, flags are marked &quot;helpful&quot;. Task accumulates EffectiveReviewsRequired &quot;Looks Good&quot; reviews. Outcome: if the number of (Recommend)Delete reviews is &gt;= the number of Looks Good reviews, mark flags &quot;disputed&quot; and raise DisputedLowQuality mod flag. Otherwise, mark flags &quot;declined&quot;. Task accumulates 6* RecommendDelete + Delete reviews. Outcome: mark flags &quot;helpful&quot;. If the post scores &gt; 0 then raise DisputedLowQuality mod flag, else just delete post (current behavior).   As under the current system, flags on posts that've already completed one full review cycle without being deleted should skip /review and go directly into the mod queue.  End result  Flags get handled faster, more accurately, and with less wasted effort from 10K reviewers. Moderators are free to focus on situations that can't be handled by the community - the ! Developers are free to make enhancements to the moderator tooling without having to work around 10K user restrictions.  Questions Am I forgetting anything here? I haven't really spent much time in the 10K flag queue since back when it was filled with spam flags and therefore useful - is there a use-case that you depend on that would be lost with its removal? Post 'em below. *: on Stack Overflow, only . What's the difference between a Long Term Support Release and a Normal Release? Now the cubs are growing, fed on mom's milk, but mom herself hasn't eaten in five months, and is getting thinner by the day.   What does in mean here, and how would the meaning of that piece change if for were used? I can't run Cisco packet tracer 7.1 on Ubuntu I have downloaded and extracted the packet tracer file, then I used this command as root  chmod +x install   then I ran (as root)  ./install    Then I used this command   ./set_ptenv.sh   but whenever I run this command   packettracer    it just doesn't start Removing empty lines in Notepad++ Wow, the iOS app feed sure showed me an interesting question! I want to read more. Uh, how do I get there? In the iOS app, pulling up a question from the feed gives the standard question view, which provides no means of navigating elsewhere on that site, either to browse or search. This should be changed for feed questions.  For example, I tapped on a highlighted Seasoned Advice question today, and found that I wanted to make a comment pointing to a related question on the site. The best way I could think of finding that post was to leave the app and use my browser. Then I came back with the link to copy-paste in.  Similarly, a question in the feed might be so intriguing that I want to look around for more cool stuff on the site. That requires two taps back and then a search via the "All Sites" menu item.  I'd suggest that questions selected from the feed should get a special header bar (or maybe stuck in between the questions and the answers) that provides a "More posts!" link at least to the site's top view; ideally there would also be a search field that could pop out of it.  Another possibility that would work pretty well, although it's a small tap target on a phone, would be to link the site name that appears under the question title to the site's top view. An idempotent operator is compact if and only if it is of finite rank Would you help me to solve this problem. Show that an idempotent operator on hilbert space is compact if and only if it has finite rank. Discrepancy in using adjective or adverb with “taste” Generalized RBF Kernels There is the notion of Generalized RBF Kernels, for example in "Towards Optimal Bag-of-Features for Object Categorization and Semantic Video Retrieval" from Jiang (1) or in formula (2.72) in  (2).  They define: Gen. RBF Kernel: $e^{-pd(x,y)}$, where (1) says d can be any distance function and (2) requires it to be a metric.  I now have been wondering for a long time: Is a function defined by $e^{-pd(x,y)}$ where d is a metric always a Mercer's Kernel? I think in (2) it sounds like it, but if it were true, not many people know of it, since they are still looking to prove that $e^{-EMD(x,y)}$ is a kernel, where EMD = Earth Mover's Distance, which is a metric.  If its not a Mercer's Kernel - do you know of any counter example? I have a normal door, and left to its own devices it will simply rest about 20-30 degrees. It won't swing shut on its own, nor stay open on its own. It hangs a bit in the middle. Convenient enough to catch you elbow when you're not looking.  I'd like the door to stay put wherever we leave it.  I've heard idea such as slipping cardboard in to the hinge to rebalance the door, etc. but no guidelines as to the thickness of that cardboard, or which hinge to change to move it in which direction.  I would hope you can appreciate how I'd rather not make this a task of trial and error and continually removing the hinge and putting it back on, etc.  I've also heard something about lightly bending one of the hinge pins, but not sure how I should go about that, or "how much" to bend it.  So, I'm looking for some tips on how to fix this (and other) door. Is CAT5 cable good enough for RS-485 vs "true" RS-485 cable Beamer: how to justify the whole text? How can I justify the text in the whole document in beamer without using \usepackage{ragged2e} and \justifying at the beginning of each frame?  \documentclass[8pt,handout,t]{beamer} \usepackage{pifont} %used inside itemize environment \item[\ding{226}] \usepackage{lipsum} \begin{document}  \begin{frame} \begin{columns}[T]  \begin{column}{0.3\textwidth} \includegraphics[width=\columnwidth]{example-image-1x1} \end{column} \begin{column}{0.65\textwidth} \begin{itemize} \item[\ding{226}] \lipsum[2] \item[\ding{226}] \lipsum[3] \end{itemize} \end{column}  \end{columns} \end{frame}  \begin{frame}  \lipsum[3]  \begin{itemize}    \item[\ding{226}] \lipsum[4]  \end{itemize} \end{frame} \begin{frame}   \begin{columns}[t] \begin{column}{0.4\textwidth} \lipsum[3] \end{column} \begin{column}{0.4\textwidth} \lipsum[5] \end{column} \end{columns} \end{frame} \end{document}     Edit: I had already checked  out but it does not work for both the columns and the itemize environment. What files should i put in a logo pack to give to clients? Should I put small, medium and large version of my logo as well as the vector file. DEF file syntax error in visual studio 2012 For any matrix $A$, $AA^T$ is symmetric, and thereby diagonalizable (Spectral Theorem). This implies that $AA^T$ has n linearly independent eigenvectors and therefore full rank.  Does this imply that for any matrix $A$, $AA^T$ is always invertible ? I know there is something wrong here because from the above chain we can conclude that a symmetric matrix is always invertible, which is definitely not true.  Could someone please point out where I am going wrong? Limit of divided integrals Prove that: $$\lim\limits_{m\to \infty} \frac{\int\limits_0^{\pi/2}(\sin(x))^{2m} dx}{\int\limits_0^{\pi/2}(\sin(x))^{2m+1} dx} = 1$$  I had tried to get it from inequalities like $$(\sin(x))^{2m+1} \le (\sin(x))^{2m} \le (\sin(x))^{2m-1}$$ But I always hit a wall in that regard (I integrate the sin functions and from there keep on going but there it ends).