So I just learned: $$K_{max} = hf - \phi$$($\phi$ for ionization energy)  I wonder what will happen if $K_max$ is zero. Will it be attracted back to the ionized atom it has come from since it doesn't have any energy to move away from it?
So I was taught that:  Kinetic Energy (of electron) = Energy (of photon) - Ionization Energy  If E(photon) = IE, then KE=0 of the electron.   What does this physically/theoretically mean?   My current thoughts/interpretation is that enough energy/force is applied to ionize the electron so it is 'sufficiently far' from the atom, and then I guess it just moves with whatever speed it is moving at with natural kinetic laws, since no more energy/force is being applied....?  Any clarifications would be greatly appreciated, thanks in advance.
I just created my new character 2 hours ago and I've reached a town. Sometimes random guards will comment that I look ill, or that I should rest. "You look sick, honey. Do you want to lie down?", isn't uncommon.  Did I unexpectedly get vampirism? I didn't fight any vampires, only a couple of wolves and bandits.  I'm a dark elf, so maybe my race isn't seen much around Skyrim and the remarks are more roleplay than actual illness.  Any ideas? So, in my physics textbook it's written that the kinetic energy possessed by a body is equal to the amount of work done by an opposing force to stop the body, now that got me thinking that when a ball hits a wall, the wall which is the opposing force does no work  because there is no displacement, so to say isn't the kinetic energy possessed by the ball zero? Find the limit $$\lim_{x\to 0^{+}}\dfrac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$$  My attempt: Since $$\sin{x}=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+o(x^7)$$ $$\tan{x}=x+\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+\dfrac{1}{63}x^7+o(x^3)$$ So $$\sin{(\tan{x})}=\tan{x}-\dfrac{1}{3!}(\tan{x})^3+\dfrac{1}{5!}(\tan{x})^5-\dfrac{1}{7!}(\tan{x})^7+o(x^7)$$ Though this method might solve, I think this problem has nicer methods. Thanks. Show that $\left(1+\frac xk\right)^k$ is monotonically increasing Prove the inequality $\,\frac{1}{\sqrt{1}+ \sqrt{3}} +\frac{1}{\sqrt{5}+ \sqrt{7} }+\ldots+\frac{1}{\sqrt{9997}+\sqrt{9999}}\gt 24$ Prove the inequality $$\frac{1}{\sqrt{1}+ \sqrt{3}}  +\frac{1}{\sqrt{5}+ \sqrt{7} }+......... +\frac{1}{\sqrt{9997}+\sqrt{9999}} &gt; 24$$   My work:   Rationalizing the denominator gives   $$\frac{\sqrt{3}-1}{2}  +\frac{\sqrt{7}-\sqrt{5}}{2}+......+\frac{\sqrt{9999}-\sqrt{9997}}{2} .$$  Now by taking two as common and separating the positive and negative terms gives  $$\frac{1}{2} [ \{\sqrt{3} +\sqrt{7}+\dots +\sqrt{9999}\} - \{1+\sqrt{5} +\dots+\sqrt{9997}\}].$$  Can we do like this please suggest. Thanks. Confusion with floating point numbers On every loading of a lib, I get the error:     no version information available   This lib has been compiled on another PC (ubuntu 10.04) than the one running it (mandriva 2010.2).  $ ldd /usr/local/gnu-eabi-4.5.2/i686-pc-linux-gnu/arm-eabi/lib/libbfd-2.21.so /usr/local/gnu-eabi-4.5.2/i686-pc-linux-gnu/arm-eabi/lib/libbfd-2.21.so: /lib/libz.so.1: no version information available (required by /usr/local/gnu-eabi-4.5.2/i686-pc-linux-gnu/arm-eabi/lib/libbfd-2.21.so)  linux-gate.so.1 =&gt;  (0xffffe000)  libz.so.1 =&gt; /lib/libz.so.1 (0xb77a7000)  libc.so.6 =&gt; /lib/i686/libc.so.6 (0xb7655000)  /lib/ld-linux.so.2 (0xb787a000)   Edit: the workaround didn't work. A game called  is making rounds on social media. I am trying to determine the maximum score attainable for this game.  Let's assume WLOG that only 2s are returned (if 4s are possible the max score is doubled).  First, I'll relax some of the rules to create a game, $\mathcal{G}$, that can achieve a score at least as high as the original, 2048 game.  A run $R$  of $\mathcal{G}_n$ consists of a sequence of moves $m_1,\ldots,m_k$ which manipulate a multiset, $\mathcal{S}$ of maximum size n.  Each $m_i$ is either:   a put $p$, which inserts a 2 into $\mathcal{S}$, or a merge, $\alpha_i$, which replaces two $2^i$ elements from $\mathcal{S}$ with a single $2^{i+1}$ elt.   I allow sequential moves of the same type because 2048 allows multiple merges or no-merge turns.  The score of a position is the sum of $\mathcal{S}$.  What is the maximum score possible on a game of order n?  Note: The final position must have only unique values, else a merge and put would be possible in increase score.  I claim it is $\sum_{i=1}^{n} 2^i$, but couldn't come up with any good induction arguments.  It's pretty easy to prove that an element $2^k$ is producible in a game $\mathcal{G}_k$, but proving an upper bound for production got a bit hairy, and also doesn't cover the full space. Vieta Jumping: Related to IMO problem 6, 1988: If $ab + 1$ divides $a^2 + b^2$ then $ab + 1$ cannot be a perfect square. The famous IMO problem 6 states that if $a,b$ are positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $\frac{ a^2 + b^2}{ab + 1 }$ is a perfect square, namely, $gcd(a,b)^2$.  How about a modification of this problem:    If $a,b$ are (strictly) positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $ab + 1$ cannot be a perfect square.    I am looking for a proof of the claim above, or a counter-example.  One possible approach towards a proof is the following:   Suppose there is such a pair $(a,b)$ as above, then by the famous IMO problem 6 from 1988, $\frac{ a^2 + b^2}{ab + 1 } = g^2$ where $g = gcd(a,b)$. Since $ab + 1$ is a perfect square, $a^2 + b^2 = c^2$ for some integer $c$. So that $(a,b,c)$ is a Pythagorean tripple, therefore there exists positive integers $n,m,l$ with $n$ coprime to $m$, such that $a = l(n^2 - m^2)$ and $b = 2lnm$ - by Euclids formula.  Then by plugging in $a$ and $b$ in terms of $n,m,l$ into the original equation, and solving for $l$, it is possible to obtain the following:  There exists positive coprime integers $n,m$ such that   $\frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)}$ is a perfect square.  If we put the quotient above into a program, as in this python code snipet:  N = 1000  for n in range(1,N):     for m in range(n+1, N):         A = (n*n + m*m + 1)*(n*n + m*m - 1)         B = 2*m*n*(n+m)*(m-n)         if A % B == 0:             print(A/B)   The quotient always is 2, regardless of whether or not $n$ and $m$ are co prime! So if the very strong implication   $2mn(n+m)(n-m) \mid (n^2 + m^2 + 1)(n^2 + m^2 - 1) \implies \frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)} = 2$    holds, then the original problem will be solved. Problem of unbalanced data Is there a fast way (without looping over all edges/faces/verts) using ops.mesh or bmesh to get the lists of indices associated with mesh islands as shown below? These islands are part of one mesh but are disjoint from each other.    I know how to select the boundaries by running the operator Select boundary Loop or calling bpy.ops.mesh.region_to_loop() but can't find any API calls to return the vert/edge/face indices associated with the individual islands. An equality involving the mean value property I am trying to establish that  $$\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\cos^2\left(\frac{\theta}{2}\right)\;\mathrm d\theta=2f(0)+f^\prime(0)$$  for $f$ analytic on $|z| &lt; R,\; R &gt; 1$  I think I am making a mistake somewhere, please point it out to me.Here's what I've done unsing the mean value property and Cauchy's integral formula:  $$2f(0)+f&#39;(0)=2(\frac1{2\pi}\int_0^{2\pi} f(e^{i\theta})d\theta + \frac1{2\pi i}\int_0^{2\pi}\frac{f(e^{i\theta})ie^{i\theta}}{e^{i2\theta}}=\frac{1}{\pi}\int_0^{2\pi} f(e^{i\theta})d\theta\frac1{2\pi}\int_0^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta}}\quad\text{(1)}$$  But when I develop the expression $\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\cos^2(\frac{\theta}{2}) d\theta$ I get:  $$\frac1{\pi}\int_0^{2\pi} f(e^{i\theta})d\theta+\frac1{2\pi}\int_0^{2\pi}f(e^{i\theta})[{e^{i\theta}}+e^{i-\theta}]d\theta \quad\text{(2)}$$  The second integral in (1) differs from the second integral in (2). Thanks. Can't open snap-store I have several input and option elements on my page, each (well almost) have an event attached to update some text on the page once they change. I use jQuery which is really really cool :)   I also use Microsofts  framework, utilizing the UpdatePanel. The reason why I do that is that certain elements are created on the page based on some server-side logic. I don't really want to explain why I use the UpdatePanel - even if it could (it can with quite some effort) be rewritten to use only  I still want that UpdatePanel.  You probably guessed it - once I have a postback on the UpdatePanel the jQuery events stops working. I actually was expecting this, since the "postback" is not really a new postback so my code in document.ready that binds the events won't be fired again. I also confirmed my suspicion by reading up on it in the jQuery help libraries.  Anyway I'm left with the problem of rebinding my controls after the UpdatePanel is done updating the . I preferably need a solution that does not require adding more .js files (jQuery plug-ins) to the page but something as simple as being able to catch the UpdatePanel's 'afterupdating' where I can just call my method to rebind all the form elements. 1 Plot, 2 Scale/Axis How would you make a comma-separated string from a list of strings? Is "guy" gender-neutral? How to highlight (color/draw attention to) a particular snippet in Minted I'm currently doing some work on error messages in programming languages. As such, I'm showing faulty pieces of code, alongside the compiler-errors accompanying them.  I'm using minted for the code snippets, but ideally, in addition to normal syntax highlighting, I'd like to highlight the sections of the code at which the error is located, by underlining them, putting a red box around them, shading their background, etc.  Is this possible using Minted? 3 more votes from other users are needed to close this question.   This is the message I got hovering the mouse over close voting tab.   I wonder can we begin any sentence with a numeric value? Is it good practice? What is the right ffmpeg output format for a v4l2-loopback device? Following the guide in  I did:  $ sudo modprobe v4l2loopback $ ffmpeg -f x11grab -r 15 -s 1280x720 -i :0.0+0,0 -vcodec rawvideo -pix_fmt yuv420p -threads 0 -f v4l2 /dev/video0   The device /dev/video0 shows up as expected but the ffmpeg command fails with following output:  $ ffmpeg -f x11grab -r 15 -s 1280x720 -i :0.0+0,0 -vcodec rawvideo -pix_fmt yuv420p -threads 0 -f v4l2 /dev/video0 ffmpeg version 4.0 Copyright (c) 2000-2018 the FFmpeg developers ... [x11grab @ 0x559767794cc0] Stream #0: not enough frames to estimate rate; consider increasing probesize .... [v4l2 @ 0x55976779e6c0] ioctl(VIDIOC_G_FMT): Invalid argument Could not write header for output file #0 (incorrect codec parameters ?): Invalid argument Error initializing output stream 0:0 --  Conversion failed!   I already tried to follow the suggestions below the answer to no avail. I noticed that v4l2 is listed as demux only in ffmpeg -formats. Does that mean I can't use it for output to /dev/video0? If so, what should I use instead? I tried h264 but that crashes after ~1000 frames with following error message:  $ ffmpeg -f x11grab -r 15 -s 1280x720 -i :0.0+0,0 -vcodec rawvideo -pix_fmt yuv420p -threads 0 -f h264 /dev/video0 ffmpeg version 4.0 Copyright (c) 2000-2018 the FFmpeg developers ... [x11grab @ 0x55da61e29cc0] Stream #0: not enough frames to estimate rate; consider increasing probesize ... File '/dev/video0' already exists. Overwrite ? [y/N] y .... Error writing trailer of /dev/video0: Invalid argument .... Conversion failed!