I tried to define (my first! - I'm a LateX newbie) command to insert notes in the margins of my text, but it introduces gaps wherever I insert the command :-( The code for my command is:  \newcommand{\mnote}[1] {   \mbox{}\marginnote{\color{red}\footnotesize#1} }   And I use it for example like this:  \subsection{Differentiable Manifold} \mnote{This is the central object in differential geometry.}A differentiable manifold is   but lo and behold, this has the unpleasant effect of adding white space before the letter of the paragraph:    How can I restore proper alignment for the initial letter of the paragraph?
I have  experience where empty spaces cause unwanted effects. It is not easy to trace the cause of these unwanted effects. In order to eliminate any doubt, I often overuse % as follows.   \usepackage% [% left=3cm, right=3cm% ]% {geometry}   or  \newcommand{\mycommand}% {% This is my command.% }   I got an extreme example that will break what Leo Liu said     In fact, only spaces which would be   output have to be removed by comment.   The following code does not produce output, but we cannot remove the % and leave a blank line between two elements of a list below.  \newpsstyle{gridstyle} { gridwidth=0.4pt, % griddots=0 }   Shortly speaking, where are the necessary places to be appended with % to remove unwanted spaces?
Proving by induction that $ \sum_{k=0}^n{n \choose k} = 2^n$ Real analysis derivate problem. How can I prove this? Let $f$ be differentiable in $[a,b]$ and $f(a)=0$. If $\exists M \in R$ such that $\vert f'(x) \vert \leq M \vert f(x)\vert \ $,  $\forall x \in [a,b]$, then $f(x)=0, \ \forall x \in [a,b]$.  $\\$ I have an idea, but I can´t see it clear. I have arrived to the inequality $\vert f(x)\vert \leq  K\vert f(y)\vert $ for $y \in (a,x)$, supposing $k&gt;0$ and repeating the process $\vert f(x)\vert \leq  K^n\vert f(z)\vert$ for $a&lt;z&lt;y&lt;x$. If $K&lt;1$, then if $n \rightarrow \infty \ $, $f(x)=0$. That´s a sort of intuitive idea. What is the proper way to change the DNS IP? ERROR running force:source:pull: Entity of type 'CustomApplication' named 'Contact' cannot be found Making an MFC application international I've got several large MFC applications here, and converting them into any other format is out of the question.  We're expanding into other markets, and would like to make the apps work in other languages and cultures.  So far, I've found occasional references as to what to do with Visual C++ version 6, with one mention that later versions of MFC have additional relevant features.  Searching MSDN gives me instructions on how to convert the apps to Unicode, which we already did.  I found nothing on MSDN on how to make languages multilingual once they're in Unicode, only a few older things using Google, and one book about internationalization using VC++6.  (We're using Visual Studio 2008 now, on XP and Vista.)    I make no claims for the strength of my Google-fu, and would be happy to be directed to things I've missed.  Is it reasonable to use the methods of VC++6, or should I use later features?  If I should use features later than that, where can I find some reference to them?  Is there something I should worry about other than setting the locale, converting all strings to resources, and duplicating resources in different languages?  I can find more things about .NET internationalization, but I'm inexperienced in .NET.  What will I find there that's applicable to VC++ and MFC?  Edit:  I just ran into difficulties trying to put Unicode strings into the String Table resource.  (As far as I can tell, Unicode strings need to be entered into the .rc file with a text editor, in L"0x0034" form.  I haven't found a resource editor way to do that.)  Any tips on that?  Any other resource I can use in VS2008 using VC++ and MFC?  Edit:  Somebody on a Microsoft forum suggested that I open the offending .rc file in Notepad, and save it in Unicode.  Unfortunately, this doesn't seem to fix the dialogs, although the stringtable seems to be working in Japanese. Let $b>0.$ Evaluate $\lim_{n \to \infty} \int^{b}_{0} \frac{\sin nx}{nx}dx$ I'm stumped at the last part of this problem. Could anyone advise, please? Thank you.  Here is  my  proof.   Define  $f: \mathbb{R} \to \mathbb{R}$ by $$ f(x) = \left\{\begin{aligned} &amp;\frac{\sin nx}{nx}&amp;&amp;, x \neq0 \\ &amp;1 &amp;&amp;, x=0 \end{aligned} \right.$$   Hence, $$f(x) = \sum^{\infty}_{k=0}(-1)^k\frac{(nx)^{2k}}{(2k+1)!} , \forall x \in \mathbb{R} \mbox{ and } f \in {\mathscr R[0,b]}.$$  $$\int^{b}_{0} \frac{\sin nx}{nx}dx = \int^{b}_{0}f(x)dx=\sum^{\infty}_{k=0}\int^{b}_{0}(-1)^k\frac{(nx)^{2k}}{(2k+1)!}dx= \sum^{\infty}_{n=0}(-1)^k\frac{(n)^{2k}b^{2k+1}}{(2k+1)(2k+1)!}$$  $$\lim_{n \to \infty} \sum^{\infty}_{k=0}(-1)^k\frac{(n)^{2k}b^{2k+1}}{(2k+1)(2k+1)!}=  ?$$ I am working on the following problem asked in an examination: "What is the maximum number of cylindrical pencils of 0.5 cm diameter that can be stood in a square shaped stand of 5cm × 5cm inner cross section? 1. 99    2. 121   3. 100    4. 105." A simple calculation yielded me the answer 100.  But the answer in the key is 105. I don't how such an answer is possible. Any hint please.....Thanks in advance! Redirect the output using `sudo` I have a small question regarding using sudo with output redirect &gt;. To enable IP forwrding, someone can use the command:  echo 1 &gt; /proc/sys/net/ipv4/ip_forward   Executing this command will give permission denied as it requires root privileges. However, executing the same command with sudo gives also permission denied error! It seems that output redirect &gt; does not inherit the permissions of the preceding command echo. Is this right?  As a workaround I do:  echo 1 | sudo tee /proc/sys/net/ipv4/ip_forward   Is this the best way to do it? Am I missing something?  Please, note that this is an example and it applies to all commands that use output redirect. How to use JACK and Pulseaudio/ALSA at the same time on the same audio device? Whenever I launch hydrogen it breaks my audio in all other applications which use Pulse, for example, I can no longer play videos in Chrome after having launched Hydrogen once, until I restart my PC or do sudo alsa force-reload. I have found people with similar issues, but no actual guide to fix this, often there are links to  (which has moved to  ) I am using Ubuntu Studio 14.04 64bit.  ALTERNATIVELY if this is truly impossible to achieve, can someone tell me how to use Jack on one soundcard and Pulse on another? This way I could plug in a USB sound card only when I need JACK.  Please don't make me use windows. Prove $[A,B^n] = nB^{n-1}[A,B]$ I am trying to show that $[A,B^n] = nB^{n-1}[A,B]$ where A and B are two Hermitian operators that commute with their commutator. However, I am running into a little problem and would like a hint of how to proceed.  If A and B commute then $[A,B] = ABA^{-1}B^{-1} = e$ where e is the identity element of the group.  $$\therefore AB=BA$$  $$n=1; [A,B^1] = (1)B^0[A,B] = e$$ This statement is certainly true. however moving on to $n = 2$ I find...  $$[A,B^2] = AB^2A^{-1}B^{-2} = ABBA^{-1}B^{-1}B^{-1} = BBAA^{-1}B^{-1}B^{-1}$$  Where in the last step I have used the fact that A and B commute to rearange the terms. However, it is plain to see that this last term simply reduces to the identity as well and for the n = 2 case we have:  $$[A,B^2] = e \ne 2B[A,B] = 2Be = 2B$$  Clearly I have assumed something I shouldn't have. The fact that there is a multiplicative factor of n implies I should be adding things, but I thought if I kept it as general as possible, the answer should just fall out naturally. I don't want an answer please, only guidance. Why does the Cauchy distribution have no mean? I am working on binary classification problem, I try to evaluate the performance of some classification algorithms (LR,Decission Tree , Random forest ...). I am using  a cross validation technique (to avoid over-fitting) with AUC ROC as scoring function to compare the performance of the algorithms, but I am getting a weird results with Random forest and AdbBoost, I have a perfect AUC_ROC score (i.e. =1) despite the fact that the recall(TPR) and FPR of this algorithms are different from 1 and 0 respectively .  def FPR(y_true, y_pred):     tn, fp, fn, tp = confusion_matrix(y_true, y_pred).ravel()     result = fp / (fp+tn)     return result def FNR(y_true, y_pred):     tn, fp, fn, tp = confusion_matrix(y_true, y_pred).ravel()     result = fn / (tp+fn)     return result FPR_scorer = make_scorer(FPR) FNR_scorer = make_scorer(FNR)     def get_CrossValResults2(model,cv_rst,bestIndx):     best=pd.DataFrame.from_dict(cv_rst).iloc[bestIndx]     roc=&quot;{:.12f}&quot;.format(best['mean_test_roc_auc'])     acc =&quot;{:.0%}&quot;.format(best['mean_test_accuracy'])     prec =&quot;{:.0%}&quot;.format(best['mean_test_precision'])     rec =&quot;{:.0%}&quot;.format( best['mean_test_recall'])     f1 =&quot;{:.0%}&quot;.format(best['mean_test_f1'])     r2=&quot;{:.2f}&quot;.format(best['mean_test_r2'])     g_mean=&quot;{:.2f}&quot;.format(best['mean_test_gmean'])     pr_auc=&quot;{:.8f}&quot;.format(best['mean_test_pr'])     fnr=&quot;{:.0%}&quot;.format(best['mean_test_fnr'])     fpr=&quot;{:.0%}&quot;.format(best['mean_test_fpr'])     rst = pd.DataFrame([[ model, acc,prec,rec,fpr,fnr,f1,roc,pr_auc,g_mean,r2]],columns = ['Model', 'Accuracy', 'Precision', 'Recall','FPR','FNR', 'F1-Score','ROC_auc','PR_auc','gmean','r2'])     return rst        cross_val_rst = pd.DataFrame(columns = ['Model', 'Accuracy', 'Precision', 'Recall','FPR','FNR', 'F1-Score','ROC_auc','PR_auc','gmean','r2'])              scoring = {'accuracy':'accuracy','recall':'recall','precision':'precision','fpr':FPR_scorer,'fnr':FNR_scorer,'f1':'f1' ,'roc_auc':'roc_auc','pr':'average_precision','gmean':Gmean_scorer,'r2':'r2'}     param_grid = {'n_estimators': [200],               'max_depth': [80,90],               'min_samples_leaf': [2,3, 4],             'min_samples_split': [2,5,12],             'criterion': [ 'gini'],              'class_weight' : [class_weights], 'n_jobs' : [-1]}  clf = GridSearchCV(RandomForestClassifier(class_weight=class_weights), param_grid, cv=kfold,scoring=scoring,refit=refit)#Fit the model bestmodel = clf.fit(X,Y)       cross_val_rst = cross_val_rst.append(get_CrossValResults2(model='Random Forrest',bestIndx=bestmodel.best_index_,cv_rst=bestmodel.cv_results_),ignore_index=True) To my knowledge there are three types of acceleration when a body (e.g. a rod) is moving in a circle about an axis. These are:   Angular acceleration : this is the rate of change of angular velocity. Tangential acceleration : this is the linear acceleration of the system in a tangential direction to the circle and equals the radius times the angular acceleration. Radial/centripetal acceleration : this is the linear acceleration of the system that is directed inwards towards the center of the circle.   I also think there are two types of velocity:   Linear velocity: this is its velocity in the tangential direction and is constantly changing Angular velocity or angular frequency: this is the rate of change of angle.   Is the above correct? And have I missed anything? Can't install wine from winehq.org on Ubuntu (actually Lubuntu) 18.04 LTS I have followed all the instructions on .  In particular, I have run the following commands: sudo dpkg --add-architecture i386  wget -nc https://dl.winehq.org/wine-builds/winehq.key sudo apt-key add winehq.key sudo apt-add-repository 'deb https://dl.winehq.org/wine-builds/ubuntu/ bionic main'   The first three of the above seemed to work fine.  However, the last command gives me the following error: 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:  winehq-stable : Depends: wine-stable (= 5.0.0~bionic) E: Unable to correct problems, you have held broken packages.  I'm running Lubuntu 18.04.3 on a USB live stick.  (However, if I remember correctly, a couple days ago I got the exact same error on a normal hard disk Lubuntu 18.04.3 installation.) I've read that a lot of people get similar errors.  Unfortunately I haven't found any solutions online that worked. Any help would be much appreciated.  Thanks! EDIT:  Thanks for the comments and suggestions everyone.  I'll try them out later today. In the mean time, here's something I don't understand.  Ubuntu 18.04 has been around for almost two years now.  In all that time, surely somebody has successfully installed wine, right?  What did they do to get it working? Why isn't $V= V_1 + V_2$? $V=V_a - V_c = V_a - V_b + V_b - V_c$, $V_a - V_b= V_1$ and $V_b - V_c = V_2$ Doesn't that prove that $V = V_1 + V_2$? Regardless of $V_3$, If i'm wrong , is there a way to obtain $V$ in terms of $V_1$, $V_2$ and $V_3$ This is an idea I have seen around, but I have not seen a feature request for it, so here it is.  Downvotes that have no comments are frustrating and difficult to manage.  This is a way to help improve that:   Have a downvote initially cost more to cast for the downvoter (5 points for example) When a comment (from the downvoter) on the downvoted question/answer gets an upvote, you get your points back.   All points (ie all 5) could be refunded or just up till the traditional -1 that is in place now (ie 4 points refunded).    This would encourage good comments and downvotes for real reasons.  (Because comments for questions/answers that are bias related will not get the upvotes as readily.)  This could/should be limited to non-subjective tags.  (ie tags where a downvote just because you "don't like it" is not appropriate.) bibtex and genitive / possessive 's: the proper way to obtain "Kuran's (1989) model" Is there an automatic way to attach a genitive apostrophe or " ' s " to an author's name?  Currently I use:  \citeauthor{kuran1989}'s \citeyear{kuran1989}   But since this comes up regularily I'd like to know whether there's a general solution. How to prove $\operatorname{Tr}(AB) = \operatorname{Tr}(BA)$? there is a similar thread here , but I'm only looking for a simple linear algebra proof. Prove: if $|z|=1$ than $\frac{z+1}{z-1}$ is an imaginary number Prove: if $|z|=1$ than $\frac{z+1}{z-1}$ is an imaginary number    I have tried to look at $$\frac{z+1}{z-1}=\frac{z+1}{z-1}\cdot \overline{\frac{z-1}{\overline{z-1}}}=\frac{z+1}{z-1}\cdot \frac{\overline{z}-1}{\overline{z}-1}$$  But did not get far This is something I don't particularly understand. I get that for Windows, activation exists to prevent software piracy, but you don't have to pay for iOS. Perhaps activation exists to prevent Android devices from using iOS or to prevent jailbreaks and downgrades? Anyone know the real reason why?