in my blender for some reason it isint working right like i cant see the entire thing it cuts out
I'm having an odd issue - When I try to zoom in on a detail using my MMB, the "camera" in the 3D view seems to be cutting off the exterior surface of my object and showing me the inside instead. I'd expect this behavior, but only when I was much more zoomed in.  If I zoom out a bit (one click on the scroll wheel) then I can see the exterior as I intend.  Is there any way to adjust this behavior?  Thanks!
While I was reading the book Exploratory examples for real analysis, I came across this:     to show that : $$\lim_{x\rightarrow x_o}\frac{1}{h(x)} = \frac{1}{\lim_{x\rightarrow x_o}{h(x)}} = \frac{1}{M}$$      Discussion:Let $\epsilon \gt 0$.we want to find $\delta \gt 0$ such that       if  $0 \lt |x-x_0| \lt \delta $,then $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|\lt \epsilon$.      In order to get expression involving $|h(x)-M|$,we first find the common denominator:        $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|=\Big|\frac{M-h(x)}{M~h(x)}\Big|$      $~=\Big|\frac{h(x)-M}{|M|~|h(x)|}\Big|$.      As we have to deal with variable term $|h(x)|$ in denominator.We can construct bound .Let $\epsilon =\frac{|M|}{2}$ ,since lim$_{x\to x_o}h(x)=M$,there exists $\delta_1\gt 0~~$ s.t.       if $0 \lt |x-x_0| \lt \delta_1 $,then $|h(x)-{M}|\lt \frac{|M|}{2}$.      As you study steps below,identify that step where $h(x)\neq 0$ for all $0 \lt |x-x_0| \lt \delta_1 $ plays a critical role:      $|M|=|M-h(x)+h(x)|$      $~~~~~~\leq |M-h(x)|+|h(x)|$      $~~~~~~= |h(x)-M|+|h(x)|$      $~~~~~~\lt \frac{M}{2}+|h(x)| \implies$      $\frac{|M|}{2} \lt |h(x)| \implies$      $\frac{1}{h(x)}\lt \frac{2}{|M|}$,where $0 \lt |x-x_0| \lt \delta_1 $.      Now that we've bound for $1/h(x)$,we can apply the definition to term $|h(x)-M|$.      Since     lim$_{x\rightarrow x_o}h(x)=M$,we can find $\delta_2 \gt 0$ such that,       $~~~~~~~~~~~~~~~~~~~~~~~~~~~~$if  $0 \lt |x-x_0| \lt \delta_2$, then $|h(x)-M|\lt \frac{|M|^2 \epsilon }{2}$   The thing I can't understand is for what purpose did the author work with such complicated expression involving $\epsilon $ in the end of the discussion? Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = \gcd(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$. Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = hcf(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$.   We know that if $a|c$ and $b|c$ then $a\cdot b\cdot s=c$ (for some positive integer $s$). $(ab|c)$  Then doesn't $ab|dc$ since $ab|c$?  I feel like I'm misunderstanding my givens.    Can we say $\operatorname{lcm}(a,b)=c$,  $\operatorname{hcf}(a,b)=d$, and $\operatorname{lcd}(a,b) \operatorname{hcf}(a*b)=a*b$? Thus, $ab|dc$  as $dc = ab$. Find an example where Markov's inequality is tight in the following sense: For each positive integer $a$, find a non-negative random variable $X$ such that $P(X\ge a)=E(X)/a$.   How to do this problem, I am really confused. Also, what is the definition of Markov's inequality? VBox on 14.04, Kernel driver not installed (rc=-1908) On Installing Virtual box on Ubuntu 14.04, I got following message:     Kernel driver not installed (rc=-1908)      The VirtualBox Linux kernel driver (vboxdrv) is either not loaded or there is a >permission problem with /dev/vboxdrv. Please reinstall the kernel module by executing      '/etc/init.d/vboxdrv setup'      as root. If it is available in your distribution, you should install the DKMS package >first. This package keeps track of Linux kernel changes and recompiles the vboxdrv >kernel module if necessary.   Then on following the   I entered these commands:  sudo apt-get install linux-headers-generic build-essential dkms sudo apt-get remove --purge virtualbox-dkms sudo apt-get install virtualbox-dkms   On Running the last command   sudo apt-get install virtualbox-dkms   I get an error:  Reading package lists... Done Building dependency tree        Reading state information... Done Some packages could not be installed. This may mean that you have requested an impossible situation or if you are using the unstable  distribution that some required packages have not yet been created or been moved out of Incoming. The following information may help to resolve the situation:  The following packages have unmet dependencies: virtualbox-dkms : Depends: virtualbox (&gt;= 4.3.10-dfsg-1) E: Unable to correct problems, you have held broken packages.   Just for reference, I checked the version of the Virtual Box, I got 4.3.16 r95972  It worked using the solution  sudo apt-get install build-essential module-assistant sudo m-a prepare   Second as mentioned in the error reinstall kernal module.  sudo /etc/init.d/vboxdrv setup   More Details are from I want to do an , then cross-validate model prediction accuracy with 80% data for training and 20% for validation, and calculate  for predictions.  Consider this dataset:    X     Y ----------  2.3    1  3.1    2  3.5    2  10.0   5  6.8    4  5.0    3  5.4    2  3.2    1   I did this:  x=c(2.3,3.1,3.5,10.0,6.8,5.0,5.4,3.2) y=c(1,2,2,5,4,3,2,1) myData=data.frame(cbind(x,y))  library("MASS") reg=polr(as.factor(myData$y)~myData$x,data=myData,method="probit")   I saw , but I couldn't fully understand. Suppose myValidationData contains 20% of data which I want to use for validation. So, I would do:  fit=predict(reg,type="probs") x=c(5.6, 5.1) y=c(3,3) myValidationData=data.frame(cbind(x,y))   This is how I tried to predict, but is it correct, when I want to cross-validate?  fit=predict(reg,data=myValidationData,type="probs")   How should I measure RMSE? And, how can I plot the prediction? I'm too spoiled by Windows utilities that take a digital snapshot of your entire drive, which you can then restore from in the event of a drive crash.  (e.g. like Time Machine for Mac OS X).  Is there a similar way of doing this in Linux? Inverse function of $x^x$ How can I find the inverse function of $f(x) = x^x$? I cannot seem to find the inverse of this function, or any function in which there is both an $x$ in the exponent as well as the base. I have tried using logs, differentiating, etc, etc, but to no avail. This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$ let $0&lt;a\le b\le c\le d$, and such $abcd=1$,show that   $$a+b^2+c^3+d^4\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}+\dfrac{1}{d^4}$$   it seems harder than Show that $\textbf{ZF}$ shows $\phi$ is $\Delta_1$ What are the proper uses of:   static_cast dynamic_cast const_cast reinterpret_cast C-style cast (type)value Function-style cast type(value)   How does one decide which to use in which specific cases? Reasons why not to update an Ubuntu version to a newer/older kernel version I had several friends ask me to update their Ubuntu to a newer kernel without changing the version of Ubuntu. So for example, staying in 10.04 with the 3.3 Kernel version.  When having an older or present Ubuntu version, many users want to apply the newer stable/unstable kernel version to the Ubuntu version they are using thinking that it is similar to upgrading a program. So what I am looking for, are some simple answers to explain the dangers (that are pretty much bigger than the advantages) of forcing a newer kernel version in a Ubuntu release that was not meant for it.  What can I tell them or explain to them of why they should NOT update to a newer/older kernel when having an Ubuntu version that was not made for that kernel version. Ball motion with air resistance coupled differential equation for fourth-order Runge-Kutta I've created a MATLAB function for solving coupled differential equation with the fourth-order Runge-Kutta method based on what is provided  (Simultaneous Equations of First Order). Here the function:  function [y] = rk4_c(f, g, h, x, y, z, n) % Runge-Kutta % Implementation of the fourth-order method for coupled equations % h = dt % x is the time here  for ii=1:(n-1)     k1 = h * f(x(ii), y(ii), z(ii));     l1 = h * g(x(ii), y(ii), z(ii));     k2 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1);     l2 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1);     k3 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2);     l3 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2);     k4 = h * f(x(ii) + h, y(ii) + k3, z(ii) + l3);     l4 = h * g(x(ii) + h, y(ii) + k3, z(ii) + l3);     y(ii+1) = y(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4);     z(ii+1) = z(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end   I would like to use this function for solving motion equations for a ball with air resistance. So starting with the basic equations (I hope they are correct): $$md^2x/dt^2 = F_{D_x}$$ and $$md^2y/dt^2 = F_{D_y}-mg$$ with $$F_{D_x} = -Dvvcos(\theta)$$ and $$F_{D_y} = -Dvvsin(\theta)$$ where $D$ is a drag coefficient. From my understanding, and from what I read, both second-order equation can be converted to two coupled equations that could be used with my coupled Runge-Kutta method function. Can someone help me understand how to convert my equation to a set of coupled equations that could be used with my MATLAB function?  EDIT  I was able to use my rk4_c function, however it only works for solving the velocities. I'm now trying to see how I could solve for the positions x and y. I'm kind of learning about Runge-Kutta methods at the same time, so sorry for the questions that may sound trivials.  % Runge Kutta code to solve projectile motion with quadratic drag % dVx/dt = -(D/m)*vx*sqrt(vx^2+vy^2) % dVy/dt = -(D/m)*vy*sqrt(vx^2+vy^2) - g  clc clear all  % Constant D = 0.24; % m = 2; % kg g = 9.80665; % m/s^2  % Define function handles fVx = @(t,vx,vy) -(D/m)*vx*sqrt(vx^2+vy^2); fVy = @(t,vx,vy) -(D/m)*vy*sqrt(vx^2+vy^2) - g;  % Initial conditions v0 = 200; % m/s theta = 30*pi/180; % rad t(1) = 0; vx(1) = v0*cos(theta); vy(1) = v0*sin(theta);  % Step size h = 0.01; % s tFinal = 2; N = ceil(tFinal/h);  % RK4 simultaneous coupled loop for ii = 1:N     % Update time     t(ii+1) = t(ii) + h;      % Update vx and vy     k1 = h * fVx(t(ii), vx(ii), vy(ii));     l1 = h * fVy(t(ii), vx(ii), vy(ii));     k2 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1);     l2 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1);     k3 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2);     l3 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2);     k4 = h * fVx(t(ii) + h, vx(ii) + k3, vy(ii) + l3);     l4 = h * fVy(t(ii) + h, vx(ii) + k3, vy(ii) + l3);     vx(ii+1) = vx(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4);     vy(ii+1) = vy(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end  % Plot the solution figure(1) plot(t, vx, t,vy) xlabel('Time (s)') ylabel('Velocities (m/s)') legend('Vx', 'Vy') Is there a word for the sadness over "What might have been"? How to find all graph isomorphisms in FindGraphIsomorphism Does anyone know of any good tools (I'm looking for IDEs) to write assembly on the Mac. Xcode is a little cumbersome to me.  Also, on the Intel Macs, can I use generic x86 asm?  Or is there a modified instruction set? Any information about post Intel.  Also: I know that on windows, asm can run in an emulated environment created by the OS to let the code think it's running on its own dedicated machine. Does OS X provide the same thing? Who turned on the lightsaber? Spoilers for The Last Jedi:      Just before Snoke is slain,   he narrates Kylo Ren's thoughts/feelings (saying things like "he turns the lightsaber towards his enemy" and so on) we see the Skywalker lightsaber being manipulated through the Force to point towards and then slay Snoke. My question is, do we know for sure who was using to Force to use the Skywalker lightsaber in this scene?   It could have been   A) Kylo Ren was turning igniting the Skywalker lightsaber with the Force. Snoke assumed the thoughts/feelings he was sensing were directed at Rey but they weren't.   Or... B) Snoke was so preoccupied with Kylo's thoughts that he wasn't paying any attention to Rey's thoughts. Rey takes advantage of this and uses the Force to turn and ignite the Skywalker lightsaber to slay Snoke. This option is reinforced by the fact that  1) the Skywalker lightsaber then zooms into Rey's hand, not Kylo's.  2) after the ensuing scuffle, Kylo Ren reveals that he hasn't switched sides; he is still a believer in the First Order.  3) Kylo Ren even tells Hux that Rey slew Snoke. Note: If you're looking for a simple explanation as to why comments aren't required on downvotes, see .   I used to get "upset" (though that is too strong a term) when I got downvoted without comment. If my answer isn't good enough then I'd like to know why. Not only does it improve the answer for the OP but it improves my knowledge too.  Where the down-vote has been explained I've found it useful &amp; it has improved my answer, or forced me to delete the answer if it was totally wrong.  So is there any way we can encourage people to leave a comment? Perhaps they don't lose rep if they explain their down-vote?  I must admit though that I haven't always explained my down-vote either so you could call me a hypocrite. I've also grown a thicker skin over the months of using SO (it seems to have come with the higher rep score ;-)), so I'm less bothered about this now. Whenever both my roommate and I use the wifi in a certain room in our apartment, it slows dramatically for me. It doesn't slow for him. If he uses it in another room, it's fine; if I use it in another room together with him, it's fine; it's just in this particular room. Does anyone know of a reason this could be? Can $\frac{\sqrt{3}}{\sin20^{\circ}}-\frac{1}{\cos20^{\circ}}$ have two values? How to change an application icon programmatically in Android?