I'm trying to understand the proof by induction of: $$ (a+b)^n = \sum \limits_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k} $$ I'm at the point of deriving the inductive step and am getting next: $$ (a+b)^{n+1} = (a+b)(a+b)^n=\dotsb = a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\binom{n}{k}a^{n+1-k}b^{k} + \sum_{k=1}^{n}\binom{n}{k-1}a^{n-1-k}b^{k} $$  Now, I also have a solution to this problem, and in the solution the two summations are turned into exactly the middle elements of the series (since $a^{n+1}$ and $b^{n+1}$ are the first and the last elements of the series correspondingly) and we get a perfect proof. However, I don't understand how next was achieved:  $$ \sum_{k=1}^{n}\binom{n}{k}a^{n+1-k}b^{k} + \sum_{k=1}^{n}\binom{n}{k-1}a^{n-1-k}b^{k} = \sum_{k=1}^{n}\binom{n+1}{k}a^{n+1-k}b^{k} $$  Any advises and hints are welcome.
I have to do proof of $(a+b)^n$ and $(a+b)^{n+1}$ with mathematical induction.  I finished the first one $(a+b)^n = a^n + na^{n-1}b+\dots+b^n$.  I however have trouble with the second one, I don't know what to start exactly. Any help/hints?
Inner automorphisms form a normal subgroup of $\operatorname{Aut}(G)$ For an arbitrary group $(G,\cdot)$ let $\operatorname{Aut}(G) = \{f: G \to G \mid f \text{ is an isomorphism}\}$ be the set of all automorphisms of the group $G$. We assume that $(\operatorname{Aut}(G),\circ)$ where $\circ$ is composition of mappings is a group.  1) Prove that for arbitrary $a \in G$ there is an automorphism $p_a: G \to G;\quad p_a(x) = a^{-1}xa$.  2) Prove, that $\operatorname{Inn}(G) = \{p_a \mid a \in G\}$ is a normal subgroup of $(\operatorname{Aut}(G),\circ)$. How to configure an exposed filter in a view using a date picker? Complex zeros of the polynomials $\sum_{k=0}^{n} z^k/k!$, inside balls this is a question from a Temple prelim exam, and i'm trapped in it! We have $p_n(z)=\sum_{k=0}^n\frac{z^k}{k!}$ and we have to prove that $\forall r&gt;0 \quad \exists N\in\mathbb{N}$ s.t. $p_n(z)$ has no zeros in $B_r(0)$ for $n&gt;N$.  I tried to use Rouche's theorem observing that $1+\frac{z^n}{n!}$ has the nth roots of $n!$ as complex zeros, which are outside a fixed ball for n sufficiently large, but i'm having trouble proving that $|1+\frac{z^n}{n!}|&gt;|z+\cdots+\frac{z^{n-1}}{(n-1)!}|$ in $|z|=r$ (if this is true). If anyone could help.. How do you join or create a server in multiplayer? I just got Minecraft Pocket Edition on my iPhone, but I'm having trouble figuring out how to start multiplayer. When I go to menu, there is only one button that says Play. How do I join or create servers? I had a problem with disappearing marker as the data sum up to more than 100 while xmax was set to 100. See .  As written in the answer to the cited question I tried to avoid the problem by normalizing the value to 100. But this lead to (much smaller) rounding errors in previously correct rows and now the marker disappear there. See the following example. Before the normalisation the column F ist affected, after the column B.   How can I in a reliable way avoid this rounding problem? Or at least get a clear error message. I have quite a number of this plots and I don't want to have to check them all the time for missing markers.    \documentclass{scrreprt} \usepackage{pgfplots} \pgfplotsset{compat=1.13} \usepackage{pgfplotstable} \pgfplotsset{  my stackbar plot/.style={              xbar stacked,              xmin=0,xmax=100,              symbolic y coords={A,B,C,D,E,F,G},              ytick=data,              nodes near coords={xxx},}}  \begin{document}  \pgfplotstableread[col sep=space]{  text   --  -    +       ++  A      0.0 1.7 13.8    84.5  B      0.0 0.6 20.1    79.3  C      0.0 1.9 13.2    84.9  D      0.0 1.6 27.9    70.5  E      1.3 3.9 19.5    75.3  F      0.0 1.4 15.0    83.7  G      0.3 1.7 24.8    73.2  }\data   \pgfplotstablecreatecol[create col/expr={\thisrow{--}+\thisrow{-}+\thisrow{+}+\thisrow{++}}]{sum}\data %  \pgfplotstablecreatecol[create col/copy=--]{--o}\data  \pgfplotstablecreatecol[create col/copy=-]{-o}\data  \pgfplotstablecreatecol[create col/copy=+]{+o}\data  \pgfplotstablecreatecol[create col/copy=++]{++o}\data  \pgfplotstablecreatecol[create col/expr={100/\thisrow{sum}*\thisrow{--o}}]{--}\data  \pgfplotstablecreatecol[create col/expr={100/\thisrow{sum}*\thisrow{-o}}]{-}\data  \pgfplotstablecreatecol[create col/expr={100/\thisrow{sum}*\thisrow{+o}}]{+}\data  \pgfplotstablecreatecol[create col/expr={100/\thisrow{sum}*\thisrow{++o}}]{++}\data  \pgfplotstablecreatecol[create col/expr={\thisrow{--}+\thisrow{-}+\thisrow{+}+\thisrow{++}}]{sumnew}\data  \pgfplotstabletypeset[columns={text,sum}   ,precision=10,columns/text/.style={string type}]\data \quad \pgfplotstabletypeset[columns={text,sumnew},precision=10,columns/text/.style={string type}]\data   %\pgfplotstablesave{\data}{pgfplotstempout.dat} \begin{tikzpicture} \begin{axis}[my stackbar plot]  \addplot table [x expr = \thisrow{--o},y=text] {\data};  \addplot table [x expr = \thisrow{-o}, y=text] {\data};  \addplot table [x expr = \thisrow{+o}, y=text] {\data};  \addplot table [x expr = \thisrow{++o}, y=text] {\data}; \end{axis} \end{tikzpicture} \quad \begin{tikzpicture} \begin{axis}[my stackbar plot]  \addplot table [x expr = \thisrow{--},y=text] {\data};  \addplot table [x expr = \thisrow{-}, y=text] {\data};  \addplot table [x expr = \thisrow{+}, y=text] {\data};  \addplot table [x expr = \thisrow{++}, y=text] {\data}; \end{axis} \end{tikzpicture}  \end{document} Allow users to hide connections between accounts This feature request is inspired by the question on Programmers.SE on .   Stack Overflow is a site aimed at professional programmers, and it is used by some employers and some employees to examine or showcase the candidates abilities. I don't see any problem with that, but there is one aspect I'm a bit concerned about. Certain SE sites are far more private than the professional sites like SO. Anyone that participates in Christianity, Judaism, Islam or Biblical Hermeneutics is likely giving out a significant amount of information about their personal belief or world view to anyone who can connect the user name to the actual person. Another example would be controversial topics on Skeptics or Politics; not everyone would like to give their employer access to that kind of information.  The fundamental problem is that the SE platform does not allow for any separation inside one account. Of course one could create separate accounts for private and professional topics, but that requires a significant amount of clairvoyance and a rather annoying user experience juggling separate browsers to log into different accounts at once. An additional problem is that SE has automatic user merge features that can be triggered by logging into two accounts in the same session. I'm not sure how they work exactly, but I suspect that having your two accounts accidentally merged is a real danger if you are not careful.  There should be a mechanism to hide the connection between SE accounts, without having to use separate OpenIDs. For example, if I would like to use my SO account in job applications, I should be able to separate it from all my other accounts. The separation would only be for the publicly accessible part, moderators could still be allowed to see the connection between accounts. And behind the scenes it would just work as it is now, just without any links between the accounts.   The aspects that would need to be changed that I can think of are   the accounts tab in the user profile, that shouldn't show any connections for separated profiles copying the user profile to other sites should not work for separated profiles the network profile should not show the connections between separated profiles Area 51 should also not show the profile connections   This separation would not have to be bulletproof, but it should prevent users googling a specific person to make connections the user does not want to make public. When creating the id attributes for HTML elements, what rules are there for the value? Dynamic Server Name in SQL Statements (from clause) Will grammar errors become correct after enough people use them for enough time? I searched for links__system_main_menu, but just find usages of this theme. Where is it defined? Set of branch points isn't discrete, but branch points are isolated? I refer to , which I understand says that the branch points of a nonconstant holomorphic map $F: X \to Y$ between Riemann surfaces $X$ and $Y$, both of which are not necessarily compact, form a discrete subset of range $Y$ of the map $F$. (Not sure if 'connected' is part of the definition of Riemann surfaces in other textbooks, but it is here.) Question 1: Do I understand correctly, and is this indeed true? What else I understand:  $A$ is a discrete subspace/subset (I just assume everything here is the usual subspace topology) of $B$ if and only if every singleton subset of $A$ is open in $A$ if and only if every point of $A$ is isolated.  This overflow question () seems to suggest this is not true but is true if $F$ is proper. However, I'm not sure the definitions of Riemann surface are the same (maybe some answers use definitions where Riemann surfaces are not necessarily connected or something).  2.1. I know $F$ is 'discrete' (meaning that its fibres are discrete; here, i refer to a ): see definition before Lemma 3.1, on p. 7), by Proposition II.3.12, and open, by open mapping theorem, which is Proposition II.3.8, but I'm not sure these help show $F$ is proper or are otherwise helpful. Ostensibly 'discrete' maps don't map discrete subsets to discrete subsets.   This stackexchange question () seems to suggest this is true.  I think this is true if $F$ is injective or at least something like locally injective. (Please don't make me type up all the stuff I did for this part.) However based on the question in (3) and based on the , which I think uses the term 'branch points' to mean the same thing as what Miranda means by 'ramification points' (and thus is different from Miranda's 'branch points'), it seems that no neighbourhood $U$ of a ramification point $p \in U \subseteq X$ of $F$ is such that the restriction $F|_U$ is injective.  Kinda related:    This page () says    Now, the ramification and branch points must form a discrete set  I don't think the page assumes anything like $F$ is proper.  The  text in (2.1) actually seems to assume proper based on the '3.7. Proper holomorphic maps' part, but said part refers to 'Lemma 3.17', so I think proper can be replaced with closed. Maybe $F$ is closed or something.  Kinda related (see Question 2 below):     says    If $F$ is proper then the image $\Delta = F(R)$ is discrete in $Y$.   This seems to have all the same assumptions as with Miranda. Ostensibly, either Miranda is wrong or there's some higher level machinery that allows us to omit $F$ proper.  Question 2: To possibly generalise this, what are the minimum requirements on $F, X$ and $Y$ to make $F$ map discrete subsets to discrete subsets? I mean, I'm not sure we really need 'holomorphic' here. I have a feeling this applies perhaps to maps that are just open/closed, continuous and discrete and spaces that are just Hausdorff and locally compact or something.  (again) Kinda related: Can someone give a simple explanation as to why the       $$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$   doesn't converge, on the other hand it grows very slowly?   I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks. Rank correlation statistics comparison I am trying to understand the relative behavior of the following rank correlation statistics:   Spearman coefficient Kendall Tau / Concordance percentage Normalized Gini coefficient (area under curve of percentage captured versus percentage observations) Normalized Area under ROC curve (for binary classifiers)   I don't believe any of these are functionally related to the others.  The accepted answer  references  and Spearman and Kendall are highly correlated (as one would expect).  Are there good intuitions behind/discussions of relative (across datasets) or absolute (for a given dataset) differences for (any pair of) these measures? Suppose we have $n$ covariates $x_1, \dots, x_n$ and a binary outcome variable $y$. Some of these covariates are categorical with multiple levels. Others are continuous. How would you choose the "best" model? In other words, how do you choose which covariates to include in the model?  Would you model $y$ with each of the covariates individually using simple logistic regression and choose the ones with a significant association? Benchmarking hosting providers IO with Bonnie Why does default auto.arima stop at (5,2,5)? The function auto.arima in the forecast package of R is a powerful tool to identify the best ARIMA(p,d,q) model for a given data series. In the   of the function, they report the following:  auto.arima(y, d=NA, max.p=5, max.q=5, max.order=5, max.d=2, start.p=2, start.q=2, ...)   According to the above function it seems that, by default, auto.arima tries to find the best model among all the combination having:   p equal or lower than 5 q equal or lower than 5 d equal or lower than 2 order equal or lower to 5, i.e. p + q + d  ≤ 5.   However, why does the default code stops at these conditions? Is there any particular reason why p, q and d should be equal or lower than 5, 5 and 2 and their sum lower than 5? In Ubuntu, I want to scroll up while a command is running in my terminal window and generates new output (the command that is running is rspec). I can scroll up, but each time rspec generates a new line of console output, it throws me back down to the bottom. Is there a way to scroll all the way up? I'm using terminator on Linux Mint/Ubuntu 12.04 LTS. I found a possible workaround.  says that with stty ixon  you can turn on &quot;XON/XOFF flow control for terminal output. Being able to pause output by typing ^S and resume it by typing ^Q [...]&quot; This it not totally what I wanted though, without having to pause the output (and maybe wondering when forgetting to turn it on again), I thought there might be a simple modifier key like shft-alt with mouse wheel scroll-up. Are there better solutions? I thought the new Apple font was called San Francisco, but I cannot find it in the font list of the Keynote.app for my Mac. Is it called another way or it is not available for Keynote?  I am currently using OS X El Capitan. TeX editor with line-spacing preferences Why $\sqrt{-1 \times -1} \neq \sqrt{-1}^2$?