How to accept case-insensitive inputs?
Case-insensitive string comparison in C++
I have a local user on my laptop. My company is migrating to a domain setup and I want to migrate the local user to a domain user, with all the settings (desktop, Start menu, programs) intact. How can I do that?  Also, is it possible to have a local user with the same settings as the domain, for when I'm working on my laptop while not connected to the domain? If $\mathbf A$ and $\mathbf B$ are Hermitian, when does $\det(\mathbf A-\lambda\mathbf B)=0$ have only real roots? Let $\mathbf A, \mathbf B$ be Hermitian matrices of the same size. What is the characterization of $\mathbf A, \mathbf B$ such that $p(\lambda)=\det(\mathbf A-\lambda\mathbf B)=0$ has only real roots?   If $\mathbf B$ is positive definite (I corrected this), it is easy to see $p(\lambda)$ has only real roots. I have 3 tables and I have to inner join Table A with Table B but a left outer between Table A and Table C.   Can I combine outer and inner join in the same query? I can nest queries and attain the desired result but I am unable to do both the joins in the same query. It appears that in other SQL languages, the order of joining is important. Is this the case in SQL Server as well?    Ok, here's the scenario.  Consider 3 tables. Table A, Table F, Table D.  I will need the recordset to contain all rows in D irrespective of whether it exists in F (after it's inner joined with A). So, a outer join comes to mind. What I would need is:   First do a inner join between A and F to get a set (this may be a null set) Then do a outer join with the recordset in (1) with D Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$ Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$.   Attempt:  I've tried proving it but it's not equating to $\frac{\pi}{4}$. Please someone should help try to prove it. Is anything wrong with the equation? If there is, please let me know. Doubling the tails in two-sample permutation test Prediction interval for binomial random variable What is the formula (approximate or exact) for a prediction interval for a Binomial random variable?  Assume $Y \sim \mathsf{Binom}(n, p)$, and we observe $y$ (drawn from $Y$). The $n$ is known.  Our goal is to obtain a 95% prediction interval for a new draw from $Y$.   The point estimate is $n\hat{p}$, where $\hat{p}=\frac{y}{n}$. A confidence interval for $\hat{p}$ is straightforward, but I cannot find a formula for a prediction interval for $Y$. If we knew $p$ (rather than $\hat{p}$), then a 95% prediction interval just involves finding the quantiles of a binomial. Is there something obvious I'm overlooking? How can I get better results when doing camera motion tracking? Whenever I do camera tracking in Blender the reconstruction is always zooming out or not on the axis. I have a great quality camera, but the end results of the camera tracking always has the model sliding. If you have a solution please tell me. How do I determine if a number is prime?  I'm writing a program where a user inputs any integer and from that the program determines whether the number is prime, but how do I go about that? Input field for decimal property I sometimes have long running processes that I want to kick off before going home, so I create a SSH session to the server to start the process, but then I want to close my laptop and go home and later, after dinner, I want to check on the process that I started before leaving work. How can I do that with SSH? My understanding is that if you break your SSH connection you will also break your login session on the server, therefore killing the long running process. "All of these things are not beneficial and (they) do not help her" Should I mention the pronoun again in this case?     "All of these things are not beneficial and (they?) do not   help her"   I know that there is already answer about the question "if it's needed to repeat on the pronoun in case that they are in the same position of grammar tense". But in this case it's different because the question deals with case of two different  tenses (in bold). Why is electric field of an infinite plate constant at all points? I know from Gauss law, it is $\vec{E}=\dfrac{\sigma}{2 \epsilon_0}(\hat{n})$ at all points. But it doesn't make sense because of the inverse square nature of electric field which suggests if you move further away from the plane, electric field must reduce.   Then why is electric field of an infinite plate constant at all points? Let $G$ be a group and $c\in G$. Define $*$ on $G$ by: $x*y=xdy$ for all $x,y\in G$, where $d$ is the inverse of $c$. Prove $(G,*)$ is a group. Let $G$ be a group and $c$ be a fixed element of $G$. Define a new operation $*$ on G by:   $$x*y=xdy$$   for all $x$ and $y$ in $G$, where $d$ is the inverse of $c$. Prove that $G$ is a group under the operation $*$.    If you want to check, this problem is in the book “A Course In Group Theory”, by John F. Humphreys (problem $3$, page $28$)   I know what are the axioms that $(G,*)$ must obey in order to be a group and I did show that all the axioms where obeyed except the inverse axiom.  My problem is that, whenever I try to verify what would the inverse be for some element $x$, I get an equation where the $x$ is cancelled in both sides.  Can please someone guide me?  Thank you!  (P.S.: This is NOT a homework assignment, I’m studying Group Theory by this book and got problems solving this question) What is the !! (not not) operator in JavaScript? Does applying the Collatz conjecture algorithm to negative integers always result in sequences ending with $-1$ and then repeating themselves an infinite number of times? It is so for the first several seed values.  After hours of thinking and research, I don't know how to answer the question. How to use a string column with pgfplotstable to label datapoints in a pgfplot? I have a comma-separated list of data points. They should be plotted as a scatter plot, and should be labeled using data from the .csv file. Having specified the first column of my file as string type, I still keep running into errors saying Could not parse input [...] as a floating point number, sorry.  Given a CSV file of the following format, how can I read Name and Z as labels?  Name,X,Y,Z ODIN,250,45,2001   Here is my MWE:  \documentclass{article}  \usepackage{tikz,pgfplots,pgfplotstable}     \pgfplotsset{compat=1.5.1}  \begin{document}  \pgfplotstabletypeset[columns/Name/.style={string type}]{table.csv};  \begin{tikzpicture}     \begin{axis}[     visualization depends on=\thisrow{Name} \as \labela,     visualization depends on=\thisrow{Z} \as \labelb,     every node near coord/.append style={font={\tiny}},     nodes near coords=\labela/\labelb,     nodes near coords align={horizontal},]      \addplot[only marks] table [x=X,y=Y,col sep=comma] {table.csv};      \end{axis} \end{tikzpicture}  \end{document}   Note that if the data in column Name is just floats it should work fine. A sum of series problem: $\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$ I have a question regarding the sum of this series: $$\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$$ My approach: I found that this sum is equal to: $$\sum_{n=3}^{2008}\frac{n}{(n-2)!+(n-1)!+(n)!}$$  I reduced it to : $$\sum_{n=3}^{2008}\frac{1}{n(n-2)!}$$  Please suggest how to proceed further. Top of answer covered by sticky top bar after clicking direct link Evaluating $\int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x$ I was generalizing the following integral: $$\int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x \hspace{40pt} n\geq 0$$ We can start by noting that $\sin x =\dfrac{e^{ix}-e^{-ix}}{2i}$ and thus $\displaystyle \sin^{2n+1}x=\frac{(-1)^n}{2^{2n}}\sum_{r=0}^n (-1)^r \binom{2n+1}{r}\sin(2r+1)x$. This implies $$\begin{aligned}\displaystyle \int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x &amp;=\frac{(-1)^n}{2^{2n}}\sum_{r=0}^n (-1)^r \binom{2n+1}{r}\int_{0}^{\infty}\frac{\sin(2r+1)x}{x}\mathrm{d}x \\ &amp;=\frac{(-1)^n\pi}{2^{2n+1}}\sum_{r=0}^n(-1)^r\binom{2n+1}{r} \\ &amp;=\frac{(-1)^n\pi}{2^{2n+1}}\sum_{r=0}^n\left((-1)^r\binom{2n}{r}-(-1)^{r-1}\binom{2n}{r-1}\right)\end{aligned}$$ Where I have used the well known  $\displaystyle \int_{0}^{\infty} \frac{\sin(2r+1)x}{x}\mathrm{d}x=\int_{0}^{\infty}\frac{\sin x}{x}\mathrm{d}x=\frac{\pi}{2}$ and the property of binomial coefficients that $\displaystyle \binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$ Since the above sum telescopes, we have $$\displaystyle \int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x=\frac{(-1)^n\pi}{2^{2n+1}}(-1)^n\binom{2n}{n}=\frac{\pi}{2^{2n+1}}\binom{2n}{n} ~\forall ~ n\in \mathbb{Z^{+}}$$ I would like to know other methods for evaluating this integral. I've been trying a simple recipe I found on the internet, which is a simple recipe consisting mostly of gelatin.  The problem is that the end result is basically just a fruit-flavoured jelly (Jell-O?), rather than the more chewy gummy-bears (Haribo) style of jelly confectionery.  Can anyone suggest what I need to do to make them less springy and more chewy.  It's all difficult to articulate!  Here's what I used for the first test-batch:   12 tbsp sugar 12/3 cup fruit juice 8 tbsp Golden Syrup 8 tbsp gelatin   soften the gelatin in 1/2 cup of cold water. place sugar, syrup and the juice in a pan and gently heat until sugar dissolves. Stir in the gelatin and stir until dissolved. Pour into moulds and wait until set.    Thanks