Why and what for launchd listens on TFTP port 69? How to prevent?  $ sudo lsof -nP -i:69 COMMAND PID USER   FD   TYPE             DEVICE SIZE/OFF NODE NAME launchd   1 root   23u  IPv6 0x55f5dbd8d897c847      0t0  UDP *:69 launchd   1 root   57u  IPv4 0x55f5dbd8d897e687      0t0  UDP *:69 launchd   1 root   59u  IPv6 0x55f5dbd8d897c847      0t0  UDP *:69 launchd   1 root   60u  IPv4 0x55f5dbd8d897e687      0t0  UDP *:69   I think this is a root problem why tftpd cannot listen on port 69 – port is already in use – and exits with status 1.
Enabling tftpd  sudo launchctl load -F /System/Library/LaunchDaemons/tftp.plist   Start:  sudo launchctl start com.apple.tftpd   See logs syslog -w  com.apple.xpc.launchd[1] (com.apple.tftpd[30687]) &lt;Warning&gt;: Service exited with abnormal code: 1   Or with log stream --process tftp --level debug  Filtering the log data using "process BEGINSWITH[cd] "tftp"" Timestamp                       Thread     Type        Activity             PID    TTL 2020-01-21 12:35:44.903240+0200 0x5292a    Activity    0x6afa0              31479  0    tftpd: (libsystem_info.dylib) Retrieve User by Name 2020-01-21 12:35:44.903949+0200 0x5292a    Default     0x0                  31479  0    tftpd: recvfrom: Resource temporarily unavailable   I also tried to run tftpd directly without launchctl sudo /usr/libexec/tftpd -d -i /private/tftpboot/ with the same results.  Earlier I used this configuration and it worked fine. But after some changes (may be OS update) it became broken. Current MacOS version is 10.14.6 (18G2022)    How to fix tftpd?    Appendix. Configuration file /System/Library/LaunchDaemons/tftp.plist has default content:  &lt;?xml version="1.0" encoding="UTF-8"?&gt; &lt;!DOCTYPE plist PUBLIC "-//Apple Computer//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd"&gt; &lt;plist version="1.0"&gt; &lt;dict&gt;     &lt;key&gt;Disabled&lt;/key&gt;     &lt;true/&gt;     &lt;key&gt;Label&lt;/key&gt;     &lt;string&gt;com.apple.tftpd&lt;/string&gt;     &lt;key&gt;ProgramArguments&lt;/key&gt;     &lt;array&gt;         &lt;string&gt;/usr/libexec/tftpd&lt;/string&gt;         &lt;string&gt;-i&lt;/string&gt;         &lt;string&gt;/private/tftpboot&lt;/string&gt;     &lt;/array&gt;     &lt;key&gt;inetdCompatibility&lt;/key&gt;     &lt;dict&gt;         &lt;key&gt;Wait&lt;/key&gt;         &lt;true/&gt;     &lt;/dict&gt;     &lt;key&gt;InitGroups&lt;/key&gt;     &lt;true/&gt;     &lt;key&gt;Sockets&lt;/key&gt;     &lt;dict&gt;         &lt;key&gt;Listeners&lt;/key&gt;         &lt;dict&gt;             &lt;key&gt;SockServiceName&lt;/key&gt;             &lt;string&gt;tftp&lt;/string&gt;             &lt;key&gt;SockType&lt;/key&gt;             &lt;string&gt;dgram&lt;/string&gt;         &lt;/dict&gt;     &lt;/dict&gt; &lt;/dict&gt; &lt;/plist&gt;   Update1 It seems port is in use. Why? Who?  $ sudo lsof -i:69 COMMAND PID USER   FD   TYPE             DEVICE SIZE/OFF NODE NAME launchd   1 root   23u  IPv6 0x55f5dbd8d897c847      0t0  UDP *:tftp launchd   1 root   57u  IPv4 0x55f5dbd8d897e687      0t0  UDP *:tftp launchd   1 root   59u  IPv6 0x55f5dbd8d897c847      0t0  UDP *:tftp launchd   1 root   60u  IPv4 0x55f5dbd8d897e687      0t0  UDP *:tftp   launchd uses UDP port 69. Why?  $ ps -ef | grep tftp 501 94727 48073   0  4:16PM ttys007    0:00.00 grep --color=auto tftp
Windows registry entries for default explorer view Show that $\lim\limits_{y\downarrow 0} y\mathbb{E}[\frac{1}{X};X>y]=0$. Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and let $X$ be a nonnegative random defined on this space. For any $A\in \mathcal{F}$, let $\mathbb{E}[X;A]:=\mathbb{E}[X\mathbb{1}_{A}]$, where $\mathbb{1}_{A}$ denotes the indicator random variable on $A$. Without assuming $\mathbb{E}\left[\frac{1}{X}\right]&lt;\infty$, show that  \begin{eqnarray} \lim\limits_{y\downarrow 0} ~~y~\mathbb{E}\left[\frac{1}{X};X&gt;y\right]=0. \end{eqnarray} Any initial ideas will be greatly appreciated. triggered my curiosity for this question. Why most of manga are black and white? Is this because of more cost if they add colors to it? how to replace multiple strings together in Oracle I have a string coming from a table like "can no pay{1},as your payment{2}due on {3}". I want to replace {1} with some value , {2} with some value and {3} with some value .  Is it Possible to replace all 3 in one replace function ? or is there any way I can directly write query and get replaced value ?   I want to replace these strings in Oracle stored procedure the original string is coming from one of my table I am just doing select on that table   and then I want to replace {1},{2},{3} values from that string to the other value that I have from another table Adding hyperlinks to pop-up data in qgis2web? Does anyone know if it is possible to add hyperlinks to image fields etc. in the pop-up windows of Qgis2web?  I've had a look in the layers javascript file, but the image source path isn't rendered using html, so I can't simply append &lt;a href" to it.  The image is coming in via the usual method of having a field that contains the path of the image file.  I'd like to achieve linking of the image, like in the example below. Alternatively, I could just populate the 'more info' field with the URL, but if I did that I'd like to avoid just doing it by having a field that contains the entire URL, as some may be quite long (these will only be internal links). Why is pruning not needed for random forest trees? Breiman says that the trees are grown with out pruning. Why? I mean to say that there must be a solid reason why the trees in random forest are not pruned. On the other hand it is considered very important to prune a single decision tree to avoid over fitting. Is there some literature available to read for this reason? Of course the trees may not be correlated but there still will be a possibility of over fitting. Is there a way to combine custom settings with custom labels ? So that the output from a setting value would be a custom label ?   To explain:  In example, we have an international organisation, and say countries are stored as an ISO code. As this is a limited and stable set of data, using custom settings would be the appropriate way to deal with say the full country name.   The issue arises when I do not just want to store the English country name, but want to have it translatable, presenting the right translation depending on the user's locale. Now, this is the point where I'd like to integrate custom settings with custom labels.  So that say the custom setting with name 'some iso code' doesn't return the full country name, but Label.Country'some iso code', which is than automatically translated in the right version.   Sounds good right. But as custom settings are fixed to certain types, as far as I've gotten is put the label name in the settings as a string value. And then .. i'm stuck :).  No way to cast a string to a Custom label, or have it be interpreted (and translated) as such, or is there ?   I'm aware this can be done in a custom object &amp; apex, I was hoping for a more automated, "cleaner", less resource consuming way, better integrated, configurable way.  Additional clarrification: *The goal is to extend the use of custom labels (translation workbench), by adding a mapping from any data-based value to a custom label.*   While we could do a long matching process in apex, it would be faster &amp; neater if the custom settings mechanism could be used. There are disadvantaged to custom objects, they require soql &amp; apex, while custom settings &amp; custom labels are cached. They too can be used in many other places. Countries were an example, I'd use this mechanism for every variable value that needs to be translatable. How can I get distribution name and version number in a simple shell script? I have video tutorials that I want to show in my presentation with beamer class. I don't want the video files get embedded to the PDF because they will make the PDF size very very huge.  If you know how the videos can be played on the HTML with a reference rather than embedding the video file to the HTML file, I want something like this to be applied to my PDF presentation. I just want to reference the video URL from my PDF presentation.   Shortly speaking, can we use media9 to show a video on a PDF but the video is not embedded to the PDF?  Uploading those videos to a web server might be difficult for me because internet connection is very slow and sometimes not available in the presentation room. :-) Cyclic orbit animation I'm trying to make one object orbit another. But I want it to repeat, rather than stop at 100 frames. It seems there is no option for this. Is there a way to make this happen? I have seen many camera review sites illustrate the effect of using different focal lengths on the same photograph using frames. There are also similar illustrations to compare different sensors (). I find this kind of visualization to be very useful in comparing the actual effect of the longer focal length\different sensor size. Of course the ability to do a wider focal length visualization would also be equally useful, but that's not possible.  Is there any ready-made software or plugin that enables this? If not, is there any simple technique to visualize the effect of longer focal lengths on an image (one would of course have to use the zoom information in the image EXIF data to accurately represent this)? All I can think of is to use some trigonometry to do the necessary calculations for the frame sizes for cropping. Can QGIS export Geospatial PDF for Avenza PDF map app? Since QGIS uses GDAL, does any map exported as a PDF become a Geospatial PDF? I am hoping to import maps from QGIS into the Avenza PDF Map app on my phone. Previous posts mentioned exporting as Geospatial PDF's and GeoTiff's, but none had an exact answer or process. warning message at command line `sudo` Everytime I try to sudo something at the terminal, I get a warning message:  alexandre@XPS-15Z $ sudo mount file.iso /mnt/ -o loop [sudo] password for alexandre:  no talloc stackframe at ../source3/param/loadparm.c:4864, leaking memory   It's just annoying to get that all the time. How do I solve that? I'm on Ubuntu 14.04 (upgraded from 13.10) Is there a string math evaluator in .NET? If I have a string with a valid math expression such as:  String s = "1 + 2 * 7";   Is there a built in library/function in .NET that will parse and evaluate that expression for me and return the result? In this case 15. Ramanujan log-trigonometric integrals I discovered the following conjectured identity numerically while studying a family of related integrals.  Let's set $$ R^{+}:=  \frac{2}{\pi}\int_{0}^{\pi/2}\sqrt[\normalsize{8}]{x^2 + \ln^2\!\cos x} \sqrt{ \frac{1}{2}+\frac{1}{2}\sqrt{ \frac{1}{2}+ \frac{1}{2}  \sqrt{ \frac{\ln^{2}\!\cos x}{ x^2 + \ln^2\! \cos x}}}}\,\mathrm{d}x, \tag1 $$ $$ R^{-}:=  \frac{2}{\pi}\int_{0}^{\pi/2}\frac{1}{\sqrt[\normalsize{8}]{x^2 + \ln^2\!\cos x}} \sqrt{ \frac{1}{2}+\frac{1}{2}\sqrt{ \frac{1}{2}+ \frac{1}{2} \sqrt{ \frac{\ln^{2}\!\cos x}{ x^2 + \ln^2\! \cos x}}}}\,\mathrm{d}x. \tag2 $$ We may numerically observe with at least 500 digits of precision that $$  \begin{align} &amp; R^{+}R^{-} \stackrel{?}{=}1 \tag 3 \\\\ &amp; R^{+} \stackrel{?}{=} \sqrt[\normalsize{4}]{\ln 2} \tag4 \\\\ &amp; R^{-} \stackrel{?}{=}\frac{1}{\sqrt[\normalsize{4}]{\ln 2}}. \tag5 \end{align} $$  How can we prove it?  A version of this has been sent to Eric Weisstein, these integrals are on Mathworld as . In Section 9 of Chapter 6, Lang constructrs for the first time a non-abelian extension as a splitting field $K$ of $X^n - a$ when the ground field $k$ does not contain primitive $n$-th roots of unity $\zeta_n$. He constructs an injective homomorphism of the Galois group $G(K/k)$ into the matrices of type   $$ M=\begin{pmatrix} 1 &amp; 0 \\ b &amp; d \end{pmatrix} \quad \hbox{with}\ b\in \mathbb{Z}/n\mathbb{Z} \quad \hbox{and} \quad d\in (\mathbb{Z}/n\mathbb{Z})^*,$$  where $(\mathbb{Z}/n\mathbb{Z})^*$ the multiplicative group of integers modulo n.  Afterwards, he states in Theorem 9.4 that if $n$ is an odd positive integer prime to the characterstic and $[k(\zeta_n):k]=\varphi(n)$ (i.e, the primitive roots of unity are not in the ground field), then the above homomorphism is, in fact, an isomorphism. Besides, the commutator group is $G(K/k(\zeta_n))$, so $k(\zeta_n)$ is the maximal abelian subextension of $K$.  Lang proves the theorem inductively over the number of prime factors in the decomposition of $n$ (i.e. if $n=9= 3\cdot 3$, then the proof "would take two steps" to complete). The induction start with $n$ prime is written clearly and I agree with it.   Now come the questions:   I do not understand Lang's argument to prove that the commutator subgroup is the subgroup $\begin{pmatrix}1 &amp; 0 \\ b &amp; 1\end{pmatrix}$, $b\in \mathbb{Z}/n\mathbb{Z}$. What is this projection on the diagonal and how does one see that the factor group of the commutator group is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^*$?   By the way, direct computation of $x\, y\, x^{-1}\, y^{-1}$ for two arbitrary matrices $x=\begin{pmatrix} 1 &amp; 0 \\ b &amp; d \end{pmatrix}$, $y=\begin{pmatrix} 1 &amp; 0 \\ c &amp; e \end{pmatrix}$ yields a matrix $\begin{pmatrix} 1 &amp; 0 \\ b(1-e)+c(d-1) &amp; 1 \end{pmatrix}$. As $n$ is odd, $2 \in (\mathbb{Z}/n\mathbb{Z})^*$, so taking $e=1$ and $d=2$, we indeed reconstruct the above-mentioned group.   The second (inductive) part of the proof is a complete mystery to me with virtually every sentence being a Theorem which needs an extra proof. Can someone give me some hints?   2.a. Lang writes $n=p\,m$ and takes $\alpha$ to be the root of $X^n-a$. It is clear that $\beta=\alpha^p$ is a root of $X^m -a$. Now comes the argument "By induction we can apply the theorem to $X^m-a$". I agree that we can apply it here but the result is that $K=k(\beta,\zeta_m)$ is the splitting field of $X^m-a$. In the diagram below, there appears $k(\beta,\zeta_n)$ out of nowhere.   2.b. Now Lang proceeds to apply the theorem on $X^p-\beta$ over $k(\beta)$. Bergman's "Companion" reminds to check that $[k(\beta,\zeta_p):k(\beta)]=\varphi(p)$ before applying the theorem and gives a hint "$[k(\zeta_p):k]$ and $[k(\beta):k]$ are relatively prime". I disagree with this hint. Take $n=21= m p$ where $m=3$ and $p=7$. Then $[k(\zeta_p):k]=\varphi(p)=p-1=6$ but $[k(\beta):k]=m=3$ and both are not co-prime. Can we apply the theorem here at all? Or if so what is the argument? If Bergman's argument is valid, where is the fault in my logic?  2.c. Even assuming that we can apply the theorem in 2.b., then we can say something about $k(\alpha, \zeta_p)$ and not $k(\alpha,\zeta_n)$. How do we proceed to checking everything Lang states in the last section of the proof (basically, everything after the diagram)?  EDIT: I have found the essentially same question Tokenizing strings in C I have been trying to tokenize a string using SPACE as delimiter but it doesn't work. Does any one have suggestion on why it doesn't work?  Edit: tokenizing using:  strtok(string, " ");   The code is like the following  pch = strtok (str," "); while (pch != NULL) {   printf ("%s\n",pch);   pch = strtok (NULL, " "); } Ghost Schedulable Classes Blocking Deployment I look for similar questions to this one but my problem is different. I want to join this community:    so I click in     but I got this error:    but then, I found that the community is already in list of communities:    However, if I click on the community, I'm an anonymous member     and even I have the option to join again     but if I click on it, the problem starts again :( what can I do? Prove that $\displaystyle \sum^{n}_{r=0}(-1)^r\cdot \large\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{3!}{2(n+3)}$   $\bf{My\; Try::}$ We can write $$\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{n!}{r!\times (n-r)!}\times \frac{r!\times 3!}{(r+3)!} = \frac{3!\times n!}{(n-r)!\times (r+3)!}$$  So our sum is $$ 3!\sum^{n}_{r=0}(-1)^r\frac{n!}{(n-r)!\times (r+3)!} = \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\frac{(n+3)!}{(r+3)!\times (n-r)!}$$  So we get Sum $$ \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\binom{n+3}{r+3} = -\frac{3!}{(n+1)(n+2)}\sum^{n}_{r=0}(-1)^{r+3}\binom{n+3}{r+3}$$  $$  = -\frac{3!}{(n+1)(n+2)(n+3)}\left[-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]$$  $$=-\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]+\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}\right]$$  So we get $$ = \frac{3!}{(n+1)(n+2)(n+3)}\left[1-(n+3)+\frac{(n+3)(n+2)}{2}\right] = \frac{3!}{(n+1)(n+2)}\times \frac{n^2+3n+2}{2} = \frac{3!}{(n+1)(n+2)(n+3)}\times \frac{(n+1)(n+2)}{2} = \frac{3!}{2(n+3)}$$  Can we prove it Using Combinatorial way,bcz above method is very lengthy  If yes then plz explain here  Thanks