In Hamill's "A Student's Guide to Lagrangians and Hamiltonians", section 5.2, the equations for a canonical transformation $(q,p) \to (Q,P)$, induced by the generating function $F(q,Q,t)$ are derived using the principle of stationary action and the fact that the equations of motion are indifferent to the addition of a complete time derivative $d{\Lambda}(q,t)/dt$ (not a function of $\dot{q}$!) to the Lagrangian; this indifference being because the small variations $\delta q$ to the extremal path vanish at the endpoints.  In particular, the author states that:  $\delta \displaystyle\int_{t_1}^{t_2} \left( P\dot{Q} -K(Q,P) \, + dF(q,Q,t)/dt \right)dt = \delta \int_{t_1}^{t_2} \left( P\dot{Q} -K(Q,P)\right)dt$  (where $K$ is the new Hamiltonian) because supposedly we have: $$\delta\int_{t_1}^{t_2} (dF(q,Q,t)/dt)dt=\delta \left(F\left(q(t_2),Q(t_2),t_2\right)-F\left(q(t_1),Q(t_1),t_1\right)\right)=0$$  for the reasons stated above for $\Lambda$.  I don't understand why this can be assumed. In general, $q$ may be a function of both $Q$ and $P$, and the small variations of $P$ at the endpoints need not necessarily vanish, and therefore $F$ may obtain different values at the point $t_1$ (and of course at $t_2$) for different variations of $Q$..  All this can also be stated in terms of $q$ and $\dot{q}$, in which case I would say that since $Q$ is a function of $q$ and $p$ (and $p$ is a function of $\dot{q}$), we get $F(q,\dot{q},t)$ which may vary at the endpoints since $\delta \dot{q}$ need not vanish at the endpoints.  This is an important step in the derivation of a canonical transformation using a generating function... What am I missing?
This might be a stupid question, but I just don't get it. In Hamiltonian mechanics when examining conditions for a $(\boldsymbol{q},\boldsymbol{p})\rightarrow(\boldsymbol{Q},\boldsymbol{P})$ transformation to be canonical one starts with $$ \dot{q}_ip^i-H(\boldsymbol{q},\boldsymbol{p},t)= \dot{Q}_iP^i-\bar{H}(\boldsymbol{Q},\boldsymbol{P},t)+\frac{d}{dt}W(\boldsymbol{q},\boldsymbol{Q},t)$$ where $\bar{H}$ is the transformed Hamiltonian, and $W$ is the generating function (now a function of $\boldsymbol{q}$ and $\boldsymbol{Q}$). This term shouldn't break Hamilton's principle, since $$ \delta\int_{t_1}^{t_2} dt\frac{d}{dt}W(\boldsymbol{q},\boldsymbol{Q},t)=\delta W(\boldsymbol{q},\boldsymbol{Q},t)|_{t_2}-\delta W(\boldsymbol{q},\boldsymbol{Q},t)|_{t_1}=0-0=0 .$$ But I don't see why the variation of $W$ should disappear at the endpoints (say at $t_1$). Expanding leads to: $$ \delta W(\boldsymbol{q},\boldsymbol{Q},t)|_{t_1}=\left(\frac{\partial W}{\partial q_i}\right)_{t_1}\underbrace{\delta q_i(t_1)}_{=0}+ \left(\frac{\partial W}{\partial Q_i}\right)_{t_1}\delta Q_i(t_1)=\left(\frac{\partial W}{\partial Q_i}\right)_{t_1}\delta Q_i(t_1).$$ $\boldsymbol{Q}$ is itself a function of $\boldsymbol{q}$ and $\boldsymbol{p}$, so $$ \delta Q_i(t_1)=\left(\frac{\partial Q_i}{\partial q_k}\right)_{t_1}\underbrace{\delta q_k(t_1)}_{=0}+\left(\frac{\partial Q_i}{\partial p_k}\right)_{t_1}\delta p_k(t_1)=\left(\frac{\partial Q_i}{\partial p_k}\right)_{t_1}\delta p_k(t_1). $$ It seems as if we also needed the variation of $\boldsymbol{p}$ to vanish at the endpoints, and I don't get this because (at least in cartesian coordinates) $\boldsymbol{p}=m\dot{\boldsymbol{q}}$ and the velocity can be different along the original and the varied orbitals even at the endpoints (they can point in totally different directions), so in general $\delta \dot{\boldsymbol{q}}(t_1)\neq 0$. What am I doing wrong? Can someone help me with this, please?
Deployed web parts don't show up in Web Part Gallery SciFi movie (or miniseries) of the afterlife I am trying to find the name of a direct to TV movie, possibly a 2-3 part miniseries which I believe aired on the SciFi channel in the 90s. I don't remember who starred in it, but the movie opened with an astronaut attempting reentry whose shuttle ends up exploding. He awakes in a large bubble of air in a river and after reaching the surface and swimming to shore, meets several other people, from seemingly many different times throughout history (including one alien, who died on Earth).  Over the course of the movie, we learn the world is the afterlife, where everyone goes when they die. The major plot involves launching a steamboat (built by Mark Twain, no less) to travel up a river to the mountains, where the beings responsible for bringing everyone to this world are.  Does anyone know what the title of this movie/miniseries is? Is it ever a good idea to close credit cards? 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My attempt:  let $x,y \in \mathbb{Z}$, $y\neq 0$.  $$\begin{align*} f(x/y)&amp;=f(x)f(1/y) &amp;&amp;\text{because }f\text{ is a ring homomorphism}\\  &amp;=g(x)f(1/y) &amp;&amp;\text{because }f\text{ and }g\text{ agree on the integers} \end{align*}$$  I do not know how to rewrite $f(1/y)$ so that I can replace it with $g(1/y)$. Any hints will be much appreciated. Thanks. What is the right way to indent your source code? How do I add a Pokestop to my area? How to favorite a question from Android app? I recently upgraded Windows 7 to Windows 10. The biggest issue I have now is that the Windows Search service no longer works, making the Start menu mostly unusable.  If I go to Indexing Options in the control panel I see that Indexing is not running and the list of paths to index is empty.  Going to advanced options, the current index path is empty, and the new path is empty too. This is true even if I try to use the "New Index" button.  Using the troubleshooter won't help since the service doesn't start. If I try to manually start the "Windows Search" service, I get the error  "Error 3: Unable to find specified path" (this is the rough translation, the original message is in my language).  What can I do to find out which path/file is missing ? I tried to check with Process Monitor, but the only file (beside system libraries) the SearchIndexer.exe try to access seem to be c:\ProgramData\Microsoft\Search\Data\Application (everytime the service is started the whole folder c:\ProgramData\Microsoft\Search is deleted and recreated) and C:\Windows\System32\config\systemprofile (which exists).  Edit: all available updates are installed, and issuing      SFC /scannow   Doesn't reveal any issue No homomorphism from $Z_{16}\oplus Z_{2}$ onto $Z_{4}\oplus Z_{4}$. For some reasons, homomorphism is a very hard area for me to make improvements. I've been hitting the brick wall for almost 2 hours.   Prove that no homomorphism exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$.  Assume a homomorphism $\phi$ exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$. Then $\phi$ is an isomorphism.  Note that the $\left | \ker \phi \right |=2$ The $\ker \phi$ is also a normal subgroup for $Z_{16}\oplus Z_{2}$. We want possible normal subgroup of order 2. I.e., 2 elements in each normal subgroups.  By Lagrange's theorem, the order of each element in a group divides the order of a group so the possible order of the elements are 1 or 2. If the elements has order 1, then the Ker \phi cannot have 2 elements.  Hence, $\ker\phi$ = $\left \{ (0,0),(8,0) \right \}$ ,$\left \{ (0,0),(0,8) \right \}$,$\left \{ (0,0),(8,1) \right \}$, possibly.  By the First isomorphism theorem:  $\phi: Z_{16}\oplus Z_2 \rightarrow Z_2\oplus Z_2$  $\Psi: G/\ker \phi \rightarrow \phi\left ( Z_16\oplus Z_2 \right )=Z_2\oplus Z_2$  $g\ker \phi  \mapsto \phi(g)=\Psi(g\ker \phi)$  Any help is appreciated.  Edit: I know that $(2,0)$ has order $8$ in $Z_{16}\oplus Z_{2}$ but any elements in $Z_{4}\oplus Z_{4}$ does not have order $8$ which would have solved the question at the outset. But I would like a different route. Parse XML with Namespace using SimpleXML I have a canon 760D. It has a EF-S lens mount. Am I able to use a EF lens with my camera? NOTE: this question is sorta like a follow-on question to  I asked. In that question, the main suggestion was that I was trying to show two polys. Now, I'm asking about a SINGLE poly.  I have the following GEOGRPAHY of Los Angeles in my Sql Server 2008 :-     (or in )  It's bascially ..  -- Shape of Los Angeles.  DECLARE @WKT VARCHAR(MAX) = 'POLYGON ((-118.66815700000011 34.181235000000029, ... )) SELECT GEOGRAPHY::STGeomFromText(@WKT, 4326)   Now when I just try and render this out, It now looks like this (NOTE: this is different, but still not correct .. compared to my previous GIS Question).    Now, in the previous question people were suggesting that the previous data was a MULTIPOLYGON so there was some bad stuff I must have been doing, trying to render out holes in polys.  This time, this is a single polygon. So i'm assuming the hole theory still doesn't work. Now, I understand that by looking at this image, there's a number of holes ... but the Well Known Text data which Sql Server gives me is a POLYGON which to me means .. one single shape. So i'm not sure how I can fix this :(  This is why i'm asking the a similar question to my previous one.  Can anyone help? Can they see the data I've posted and see what happens in other systems? (I don't have any other software besides Sql Server 2008).  Maybe I need to extract all my data from Sql Server 2008 .. and clean it up (whatever that means) through another system. For example. 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For example,   "Salary tax" instead of "Salaries tax" "Office supplies" instead of "Offices supplies" "Product demo" instead of "Product demo"   In the above cases, the word "Salary" is used as an adjective to describe the type of "tax", so it will be in singular form. But I saw people using salaries tax and not sure if it is grammatical.  So my question is if there are any general rules on how to use the singular/plural form of an adjective? How to fix the weird surface in fluid object Currently using ubuntu 15.10 on azure and installed the gnome and vnc4server latest version on ubuntu. Please find log file below:  Current VNC client screen:    $ cat .vnc/myhost1\:1.log    --cut-- VNCSConnST:  Server default pixel format depth 16 (16bpp) little-endian rgb565 VNCSConnST:  Client pixel format depth 8 (8bpp) colour-map Window manager warning: Invalid WM_TRANSIENT_FOR window 0x20000c specified for    0x20000b (I        nitialization error). 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