I came up with this line somewhere:  To properly experience $n$ dimensions the retina of the organism should be at least of $n-1$ dimensions. And human eye has $2$ dimensional retina, so we can experience at most 3 dimensions properly.  Now my question is:  Do more than 3 dimensions exist?  What is the proof that 'there exist only 3' or 'more dimensions do exist'?
As human beings, we observe the world in which we live in three dimensions. However, it is certainly theoretically possible that more dimensions exist.  Is there any direct or indirect evidence supporting a fourth dimension?
Let $a,b,m,n$ be positive integers such that g.c.d.$(a,b)=1$ and $a^m+b^m|a^n+b^n$ , then is it true that $m$ divides $n$ ? I'm trying to enable alsa on different prefix than default through winetricks.  More specifically, the prefix that has Rosetta Stone on it, as it cannot see my microphone.  When I open winetricks, it only lets me select default.  I tried anyway, on default, and it did not resolve the issue.  I saw some answers online, through the terminal, but that didn't work either.  Maybe I have the syntax wrong?  steve@stevelap:~$ WINEPREFIX=~/.PlayOnLinux/wineprefix/Rosetta sh winetricks alsa sh: 0: Can't open winetricks steve@stevelap:~$ sudo WINEPREFIX=~/.PlayOnLinux/wineprefix/Rosetta sh winetricks alsa [sudo] password for steve:  sh: 0: Can't open winetricks   Could anyone please offer assistance?  Thanks! I want to make a full focus on the body, for example: I need to have a clear focus on the each balls in the photo. I used just single point focus.    In this picture I used:   shutter speed: 8 aperture: F/4 ISO: 200 2 flashes, one in the right and one on the top of project camera: nikon d300s lens : 105mm micro 2.8   Thanks Access denied for user 'root@localhost' (using password:NO) What does 1x1 convolution mean in a neural network? I am currently doing the Udacity Deep Learning Tutorial. In Lesson 3, they talk about a 1x1 convolution. This 1x1 convolution is used in Google Inception Module. I'm having trouble understanding what is a 1x1 convolution.  I have also seen  by Yann Lecun.  Could someone kindly explain this to me? Why can't I suggest edits on SE meta sites? Word or phrase for mere coincidence that brings happiness Let $E$ be a metric $\mathbb R$-vector space, $I$ be a nonempty set and $(\mu_t)_{t\in I},(\nu_t)_{t\in I}$ be families of nonnegative finite measures on $\mathcal B(E)$.  that if $(\mu_t)_{t\in I}$ and $(\nu_t)_{t\in I}$ are tight$^1$ and bounded in total variation norm, then the family of convolutions $(\mu_t\ast\nu_t)_{t\in I}$ is tight and bounded in total variation norm as well.  Now I would like to show the converse: If $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, then $(\nu_t)_{t\in I}$ is tight. (Most probably we need to impose a boundedness assumption on $(\mu_t)_{t\in I}$ and on $(\mu_t\ast\nu_t)_{t\in I}$ or even a priori on $(\nu_t)_{t\in I}$.)  For the moment, I'm only able to prove the claim assuming that the $\mu_t,\nu_t$ are probability measures. Under this assumption, the proof goes as follows: Let $\varepsilon&gt;0$. Since $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, there is a compact $K\subseteq E$ with $$\sup\left(\sup_{t\in I}\mu_t(K^c),\sup_{t\in I}(\mu_t\ast\nu_t)(K^c)\right)&lt;\varepsilon\tag1.$$ Now $K-K$ is compact and \begin{equation}\begin{split}1-\varepsilon&lt;(\mu_t\ast\nu_t)(K)&amp;=\underbrace{\int_K\mu_t({\rm d}x)\nu_t(\underbrace{K-x}_{\subseteq\:K-K})}_{\le\:\nu_t(K-K)}+\underbrace{\int_{K^c}\mu_t({\rm d}x)\underbrace{\nu_t(K-x)}_{\le\:1}}_{&lt;\:\varepsilon}\\&amp;&lt;\nu_t(K-K)+\varepsilon\end{split}\tag2\end{equation} for all $t\in I$, which immediately yields $$\nu_t(K-K)&gt;1-2\varepsilon\;\;\;\text{for all }t\in I\tag3.$$  Can we drop the probability measure assumption?  EDIT I'm not able to obtain the claim, even with the help of the answer below. If $\varepsilon&gt;0$, the goal is to show that $$\nu_t(E)-\nu_t(K-K)=\nu_t((K-K)^c)&lt;\varepsilon\tag4.$$ If I'm choosing all the constants as generic as possible, we obtain $$\mu_t(E)\le c_1\tag5$$ $$\mu_t(K^c)\le k_1(\varepsilon)\tag6$$ $$(\mu_t\ast\nu_t)(K)&lt;c_1\nu_t(K-K)+c_2k_1(\varepsilon)\tag6$$ $$\mu_t(E)\ge\alpha_1\tag7$$ $$(\mu_t\ast\nu_t)(E)\le c_3\tag8$$ $$\nu_t(E)\le\frac{c_3}{\alpha_1}=:c_2\tag9$$ $$(\mu_t\ast\nu_t)(E)\ge\alpha_3\tag{10}$$ $$\nu_t(E)\ge\frac{\alpha_3}{c_1}\tag{11}$$ $$(\mu_t\ast\nu_t)(K^c)&lt;k_3(\varepsilon)$$ The assumptions that I'm imposing is that $(\mu_t)$ and $(\mu_t\ast\nu_t)$ are bounded above (by $c_1$ and $c_3$) and below (by $\alpha_1$ and $\alpha_3$) in total variation norm and are tight (so that we can find $K$ for arbitrary $k_1(\varepsilon)$ and $k_3(\varepsilon)$). In order to obtain the claim we need to choose $k_1(\varepsilon)$ and $k_3(\varepsilon)$ such that $(4)$ holds. From everything above, we obtain $$\nu_t(E)-\nu_t(K-K)&lt;\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)\tag{12}$$ Even if we simplify this by assuming $\alpha_1=\alpha_3$, $c_1=c_3$ and $k_1(\varepsilon)=k_3(\varepsilon)$, the equation $$\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)=\varepsilon\tag{13}$$ doesn't seem to admit a solution $(k_1(\varepsilon),k_3(\varepsilon))$ with positive $k_1(\varepsilon)$ and $k_3(\varepsilon)$; unless $\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}\ge0$; which clearly doesn't need to be the case (it is clearly the case for probability measures though ...)  $^1$ Remember that a family $\mathcal F$ of finite signed measures on $\mathcal B(E)$ is called tight if for all $\varepsilon&gt;0$, there is a compact $K\subseteq E$ such that $$\sup_{\mu\in\mathcal F}|\mu|(K^c)&lt;\varepsilon.$$ Any reason to have a strong user password? I recently decided to change my (very) strong user password to something trivial.  This was mostly to eliminate frustration with typing a long password each time I sudo or whatever.  My question is, is there any reason to have a strong user password? considering that most services which I run on my box do not use PAM for auth:   apache2 - uses .htpasswd for access   sshd - only private keys allowed   transmissiond - uses its own http auth   mpd - uses its own auth   samba - limited to LAN   Physical access to the box is out of the question. I have notifications in the OSX Notification Center, as for most of my apps I am using the setting, "Show in Notification Centre: 10 Recent Items" This is a good setting but rather than have a very full Notification Center, I would still like to clear it from time to time with 1 click.  I am using Yosemite 10.10.3 I asked this question at stackoverflow and have been kindly redirected here.  I'd like to align some equations in Latex using the AMS packages. Each equation has two equal signs that need to be aligned. So something in the line of  A = B = C D = E = F   I've tried using the align-environment like this  \begin{align} A &amp;= B &amp;= C \\ D &amp;= E &amp;= F \end{align}   This works in principle (it aligns), however it adds ridiculously large spaces before the second equal sign in each line. But I just want the line to runs as if there was no additional alignment tab. Only when I replace for example "B" by "BBB" I want the equal sign before "F" to shift to right the exact amount of space.  Could anyone help me out on that one? It's kind of driving me crazy since I don't get the idea of that strange behavior and I just can't find any solution. Maybe alignat could help, but I don't really get how that environment works.  Cheers, Oliver Show that the Sorgenfrey line does not have a countable basis. I have a small server which I use for testing and programming. Currently it runs Debian 9.4 stretch with 4.14.0-0.bpo.3-amd64 kernel.  Today I tried to connect through SSH but I couldn't then I tried to ping it and it was unreachable. Therefore I had to hard-restart it by unplugging power cable. Then I went to  /var/log/syslog  and I found a strange line containing exactly 6140 characters like the following  ^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@  then nothing else until new log entries of system restart. This is actually the first time it happens.  Does someone know what could it be? Remote desktop without a password I have an HTPC with a user that has no password and autologin.  I want to be able to connect to it via Remote Desktop without a password, so I can login with the original user. (I don't want to add another user just for the mstsc)  Is there a way of doing that?  The OS is Windows 7, 32bit. How can I use Photoshop to imitate a photo taken with a low-quality camera? How can I create an image in Photoshop that imitates a photo taken with a low-quality phone camera? Examples of qualities I want to simulate:   blurring in some areas caused by a dirty lens low quality exposure handling in the darkroom noise   Some examples: php - insert a variable in an echo string Why doesn't console.log work when passed as a parameter to forEach? What will result in better page ranking in search results: Page views or time spent on page? Is it possible to play D&D 5e solo? Is it at all possible to run a solo game where you act as the DM/GM for yourself? I know there was a 4th edition adventure published in an old issue of one of the D&amp;D magazines but recently I read somewhere (Reddit I think) that the games Outbreak: Undead and Eclipse phase were good games to play solo.  Can anyone with experience with running a game of D&amp;D 5e for themselves explain how to do so? Is "Michael here" correct?