I have use jQuery Drill Down Menu From .  but this plugin is only work with given jQuery Version 1.4. If i am integrate with my project where jQuery 1.8.3, it produce following error into console.  Uncaught Error: Syntax error, unrecognized expression: [object Object][rel="2"]    Any solution?  I don't want to add another jQuery Version just for this menu effect, it increase page sige.
My company has purchased a product that renders an ASP.NET control on the page. This control uses jQuery 1.2.3 and adds a script tag to the page to reference it. The developers of the control will not support use of the control if it modified in any way (including modification to reference a different version of jQuery).  I'm about to start development of my own control and would like to use the features and speed improvements of jQuery 1.3. Both of these controls will need to exist on the same page.  How can I allow the purchased control to use jQuery 1.2.3 and new custom development to use jQuery 1.3? Also out of curiosity, what if we were to use an additional control that needed to reference yet another version of jQuery?
What is the difference between Newtonian and Lagrangian mechanics in a nutshell? What is Lagrangian mechanics, and what's the difference compared to Newtonian mechanics? I'm a mathematician/computer scientist, not a physicist, so I'm kind of looking for something like the explanation of the Lagrangian formulation of mechanics you'd give to someone who just finished a semester of college physics.  Things I'm hoping to have explained to me:   What's the overall difference in layman's terms? From what I've read so far, it sounds like Newtonian mechanics takes a more local "cause-and-effect"/"apply a force, get a reaction" view, while Lagrangian mechanics takes a more global "minimize this quantity" view. Or, to put it more axiomatically, Newtonian mechanics starts with Newton's three laws of motion, while Lagrangian mechanics starts with the Principle of Least Action. How do the approaches differ mathematically/when you're trying to solve a problem? Kind of similar to above, I'm guessing that Newtonian solutions start with drawing a bunch of force vectors, while Lagrangian solutions start with defining some function (calculating the Lagrangian...?) you want to minimize, but I really have no idea. What are the pros/cons of each approach? What questions are more naturally solved in each? For example, I believe Fermat's Principle of Least Time is something that's very naturally explained in Lagrangian mechanics ("minimize the time it takes to get between these two points"), but more difficult to explain in Newtonian mechanics since it requires knowing your endpoint. Problems updating Python from the repositories today (Feb 25, 2021) NOTE: This is a Master Canonical Post for the issues observed on February 25, 2020.  All issues on Feb 25 relating to Python updates not found should be dupe-linked here.  Errors have been seen such as this today, February 25, 2021: Failed to fetch http://security.ubuntu.com/ubuntu/pool/main/p/python2.7/libpython2.7-minimal_2.7.17-1~18.04ubuntu1.3_amd64.deb 404 Not Found  Is there something up with the repositories? How do I know if food left at room temperature is still safe to eat? Formerly List of all badges with full descriptions.  What are badge name's requirements? Why didn't I get badge name ? Which badges can I earn multiple times?    Jump to:                    Note: Some badges are awarded based on score. The term score means the total number of upvotes minus the total number of downvotes.  Any badge with a *** next to it is one of 20 badges that count towards the  displayed on a moderator candidate in an election. The candidate score is a total of 40 points: the first 20 are awarded based on the user's rep divided by 1000 and rounded down; the second 20 represents the total number of unique badges earned of the 20 that count. Note that if the same badge is earned multiple times, it will only count once towards the candidate score.  Any badge with a &quot;(retired)&quot; next to it is no longer awarded, but is retained by users who previously earned them. See  for more info on what it means for a badge to be retired.       Visit the  of the  on any site to see a complete list of badges that you can filter by earned, unearned, or type. Different seed gives different LASSO output Covering ten dots on a table with ten equal-sized coins: explanation of proof Note: . I have moved it here because:   I am curious about the answer The OP has not shown any interest in moving it himself     In the Communications of the ACM, , Peter Winkler asked the following question:     On the table before us are 10 dots,   and in our pocket are 10 $1 coins.   Prove the coins can be placed on the   table (no two overlapping) in such a   way that all dots are covered. Figure   2 shows a valid placement of the coins   for this particular set of dots; they   are transparent so we can see them.   The three coins at the bottom are not   needed.   In the , he presented his proof:     We had to show that any 10 dots on a   table can be covered by   non-overlapping &#36;1 coins, in a problem   devised by Naoki Inaba and sent to me   by his friend, Hirokazu Iwasawa, both   puzzle mavens in Japan.      The key is to note that packing disks   arranged in a honeycomb pattern cover   more than 90% of the plane. But how do   we know they do? A disk of radius one   fits inside a regular hexagon made up   of six equilateral triangles of   altitude one. Since each such triangle   has area $\frac{\sqrt{3}}{3}$, the hexagon   itself has area $2 \sqrt{3}$; since the   hexagons tile the plane in a honeycomb   pattern, the disks, each with area $\pi$,   cover $\frac{\pi}{2\sqrt{3}}\approx .9069$ of the   plane's surface.      It follows that if the disks are   placed randomly on the plane, the   probability that any particular point   is covered is .9069. Therefore, if we   randomly place lots of &#36;1 coins   (borrowed) on the table in a hexagonal   pattern, on average, 9.069 of our 10   points will be covered, meaning at   least some of the time all 10 will be   covered. (We need at most only 10   coins so give back the rest.)      What does it mean that the disks cover   90.69% of the infinite plane? The easiest way to answer is to say,   perhaps, that the percentage of any   large square covered by the disks   approaches this value as the square   expands. What is "random" about the   placement of the disks? One way to   think it through is to fix any packing   and any disk within it, then pick a   point uniformly at random from the   honeycomb hexagon containing the disk   and move the disk so its center is at   the chosen point.   I don't understand. Doesn't the probabilistic nature of this proof simply mean that in the majority of configurations, all 10 dots can be covered. Can't we still come up with a configuration involving 10 (or less) dots where one of the dots can't be covered? What are the chances of getting $K$ identical outcomes in a row in a series of $Y$ independent events? Each event has two possible outcomes: a and b.  Each event is independent.  Outcome a has $X\%$ chance of happening.  If we run $Y$ number of such events, what are the chances of getting $K$ outcomes a in a row, at least once?  To put it in practical terms imagine a series of coin tosses. Each coin toss has two possible outcomes: a = heads and b = tails. Outcome a = heads has a chance of $50\%$. If we run $Y = 100$ number of coin tosses, what are the chances of getting $K = 10$ outcomes a=heads in a row, at least once?   I'm using the practical example only so I can express the idea more clearly. I'm interested in finding out the general formulas and computations so that I can run the calculations for any $X$, $Y$, or $K$.  I would also be very interested in knowing how the probabilities would change if the events would not be independent. What if the more outcomes a we have in a row the bigger the chance of getting an outcome a in the next event? What if it's the other way around, and the more outcomes a in a row we have the lower the chance of getting another a? Total of attribute field of point layer per polygon layer I have a polygon shapefile of many lakes, I also have a point shapefile of locations on those lakes where fish were caught.  The point shapefile contains the attributes: Fisherman_iD, Fish_Caught, Lat, Long.  The polygon shapefile contains the attributes Lake_name, Points_per_polygon(PNTCNT)  Each point obviously has differing entries under the fish_caught field.   What I want to get is the average fish caught per lake (polygon).  So I need to have a total number of fish caught, per individual polygon, and then I can divide that by the PNTCNT file I create using the Points in a Polygon tool.  Ideas? I am relatively new to QGIS. which vectors are perpendicular to each other?  $\vec a = (1, -2, 3)$, $\vec b = (5, 4, 1)$, $\vec c = (1, 0, -5)$  Do i just take the dot product of 2 of them. If the dot product they are at $90^\circ$? But how do i know if there perpendicular? What is the relationship between a chi squared test and test of equal proportions? Suppose that I have three populations with four, mutually exclusive characteristics.  I take random samples from each population and construct a crosstab or frequency table for the characteristics that I am measuring.  Am I correct in saying that:   If I wanted to test whether there is any relationship between the populations and the characteristics (e.g. whether one population has a higher frequency of one of the characteristics), I should run a chi-squared test and see whether the result is significant. If the chi-squared test is significant, it only shows me that there is some relationship between the populations and characteristics, but not how they are related. Furthermore, not all of the characteristics need to be related to the population.  For example, if the different populations have significantly different distributions of characteristics A and B, but not of C and D, then the chi-squared test may still come back as significant. If I wanted to measure whether or not a specific characteristic is affected by the population, then I can run a test for equal proportions (I have seen this called a z-test, or as prop.test() in R) on just that characteristic.   In other words, is it appropriate to use the prop.test() to more accurately determine the nature of a relationship between two sets of categories when the chi-squared test says that there is a significant relationship? Extra }, or forgotten \right My MWE:  \documentclass{report}  \usepackage{amsmath}  \begin{document}  \begin{align} y &amp;= a \\ =&amp; -2c_1\bar{z}_1^2 + 2\bar{z}_1\bar{z}_2 + 2\bar{z}_2 \left ( -\bar{z}_1 -c_2z_2 - \frac{3\epsilon}{2}\bar{z}_2 -\sigma_{2,1}\cos(x_1)\Delta x_1 \notag \right. \\ \left. &amp;- \sigma_{2,2}\Delta x_2 -\sigma_{2,3} \Delta u +c_2 \chi_2 \vphantom{\frac{1}{1}} \right ) - 2\gamma \sigma_{2,1}^2\cos^2(x_1) \Delta x_1^2 - \gamma \sigma_{2,2}^2 \Delta x_2^2   - 2\gamma \sigma_{2,3}^2 \Delta u^2 \end{align}  \end{document}   The error I get:  Extra }, or forgotten \right.   The expected result: Foo* set = new Foo[100]; // ... delete [] set;   You don't pass the array's boundaries to delete[]. But where is that information stored? Is it standardised? Double booked - User bookmarking question recorded twice In this  one user is shown as having bookmarked  twice:  This doesn't appear to be the same question as:  or  though it's certainly possible that the cause is the same. How can I merge properties of two JavaScript objects dynamically? I need to be able to merge two (very simple) JavaScript objects at runtime.  For example I'd like to:  var obj1 = { food: 'pizza', car: 'ford' } var obj2 = { animal: 'dog' }  obj1.merge(obj2);  //obj1 now has three properties: food, car, and animal   Does anyone have a script for this or know of a built in way to do this?  I do not need recursion, and I do not need to merge functions, just methods on flat objects. Consider the following MWE:  \documentclass{article} \usepackage{pstricks} \begin{document} A \end{document}   It produces the error  Runaway argument? ] {pgffor.code.tex}  ! Paragraph ended before \ProvidesPackageRCS@i was complete. &lt;to be read again&gt;                     \par  l.11   How do I fix this? Cannot start XAMPP on Ubuntu 11.10 Lat Long to X Y Z position in JS .. not working Convergence of sequence given by $x_1=1$ and $x_{n+1}=x_n+\sqrt{x_n^2+1}$ Be $(x_n)_n\ge 1 $  such that $x_1=1$ and $x_{n+1}=x_n+\sqrt{x_n^2+1}$ for every $n\ge 1$. Prove that the sequence $y_n=(2^n/x_n)_n \ge 1$ is convergent and find it s limit.  Being positive and decreasing, $y_n$ is clearly convergent. But finding it's limit really put me intro trouble. Any help?  Thank you! Suppose f is a twice differentiable function with $f(0)=0$ and $f(1)=1$, $f'(0)=f'(1)=0$. Then show that $|f''(x)| > 4$ for some $x$ in $(0,1)$ Any help would​ be appreciated.  My initial attempts were using Lagrange's Mean Value Theorem first between $f(0)$ and $f(1)$ to show that $f'(t)=1$ for some $t$ in $(0,1)$.  Now applying LMV between $f'(t)$ and $f'(0)$, and $f'(t)$ and $f'(1)$ I could prove it if t lies in either $(0,1/4]$ or in $[3/4,1)$. Don't really know where to go next. Using something like this:  try:    # Something... except Exception as excep:    logger = logging.getLogger("component")    logger.warning("something raised an exception: " + excep)    logger.info("something raised an exception: " + excep)   I would rather not have it on the error-level cause in my special case it is not an error.