Suppose 500 individuals were asked one question having possible answers ”yes” or “no”. Denote by p the proportion of “yes” in the whole population. Estimate the probability that the proportion of “yes” in the sample (out of 500) is greater than in the whole population (p) by more than 5 %.  I was thinking of using the normal approx of the binomial where.. n = 500 p = 0.5 q = 0.5       Formula: \begin{align*}  P\left(\frac{N - np}{\sqrt{npq}}\right)\\ \end{align*} However, I'm not sure what the N value would be in this case.
Suppose $53$ percent of the population prefer red socks to green. If $100$ random people are asked, what is the probability that most (at least $50$) will say prefer GREEN:  So I set \begin{align*} N &amp; = 49.5\\   n &amp; = 100\\   p &amp; = 0.47 \end{align*}  Then.. \begin{align*}    P(N \geq 49.5) &amp; = P\left(\frac{N - np}{\sqrt{npq}}\right)\\                &amp; \geq \frac{49.5 - 47}{\sqrt{100*.47*.53}}\\                   &amp; = \varphi(0.50)\\                   &amp; = 0.3085   \end{align*} Except, that's not the right answer. What did I do wrong in my calculations?
Let $p$ be an odd prime number. Prove that $$\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \left( \frac{3 \cdot 4}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \right) = -1$$ where $\left( \frac{a}{p}\right)$ is the Legendre symbol.  This seems to be a tricky one! I've tried using the property $\left( \frac{ab}{p} \right)=\left( \frac{a}{p}\right) \left( \frac{b}{p} \right)$ and prime factoring all the non-primes but to no avail. I had a quick thought of induction haha, but that was silly. I tried factoring the common Legendre symbols like $\left( \frac{3}{p}\right) \left[ \left( \frac{2}{p} \right) + \left( \frac{4}{p} \right) \right]$ but that didn't bring anything either. And I've been looking for pairwise cancellation with $1$ term leftover, but that does not seem to work.  Can you help? Do positive semidefinite matrices have to be symmetric? Can you have a non-symmetric matrix that is positive definite? I can't seem to figure out why you wouldn't be able to have such a matrix, but all my notes specify positive definite matrices as "symmetric $n \times n$ matrices."  Can anyone help me with an example of a non-symmetric positive definite matrix, or some insight into a proof for why it would need to be symmetric should that be the case? Thanks! Steps for going from Polyakov action to low-energy effective action (SUGRA?) in String Theory A lot of string-theory questions have been asked here. This one is, hopefully, different in that this inquiry is specific and the expected answer would be more mathematical than philosophical in character.  To give some context, I'm familiar with LQG, QFT etc. I'm a beginner in CFT and string-theory. On reading papers on AdS/CFT and String Theory the following sentence often appears: "The low-energy effective action for $x$ dimensional ST of type $y$ reduces to that of supergravity (SUGRA) in $z$ dimensions with matter fields of types $a$, $b$ and $c$".  The $64,000 dollar question any noob, such as me, wants answered is what are the mathematical/technical steps or arguments which lead us from the fundamental action of string-theory (the Polyakov/Nambu-Goto action) to a low-energy effective description. Of course, I might be asking a question for which the only concise answer is: "go read GSW". Here's hoping otherwise ;) Make Javascript do List Comprehension What happen if I install VirtualBox Guest Additions on a host? I have a bootable USB drive with Ubuntu installed.   This morning I Installed VirtualBox and booted my Ubuntu drive in a VM. I then installed the guest additions in Ubuntu.   My question/curiosity is, what would happen if I boot the drive again outside of the VirtualBox VM, using a physical computer?  Will my Ubuntu work normally or is something going to change? Maximal ideals of polynomial rings in infinitely many variables Let $k$ be an algebraically closed field. Nullstellensatz states that the maximal ideals of the polynomial ring $R=k[X_1,\dots,X_n]$ are precisely those of the form $\langle X_1-a_1,\dots,X_n-a_n\rangle$, with $(a_1,\dots,a_n)\in k^n$.     What if we are working with infinitely many variables, say $\{X_i\}_{i\in I}$, being $I$ an infinite set? Again, the ideals of the form $\bigl\langle\{X_i-a_i: i\in I\}\bigr\rangle$, with $(a_i)_{i\in I}\subseteq k$ are maximal (even if $k$ is not algebraically closed), but what about the converse? Identifying Unused Stored Procedures This next year, I am helping an effort to clean several SQL Server environments.    We have about 10,000 stored procedures and estimate that only about 1000 of them are used on a regular basis, and another 200 or so are used on a rare occasion, meaning we have a lot of work to do.    Since we have multiple departments and teams that can access these databases and procedures, we are not always the ones calling the procedures, meaning that we must determine what procedures are being called.  On top of that, we want to determine this over a few months, not in a few days (which eliminates some possibilities).  One approach to this is to use the SQL Server Profiler and track what procedures are being called and compare them to the list of what procedures we have, while marking whether the procedures are used or not.  From then, we could move the procedures to a different schema in case a department comes screaming.  Is using the Profiler the most effective approach here?  And/Or have any of you done something similar and found another way/better way to do this? How to read a line from the console in C? What is the simplest way to read a full line in a C console program The text entered might have a variable length and we can't make any assumption about its content. My question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.  Page 57. Exercise 12. Show that the following are polynomials by converting them to the form $\sum_{k=0}^{m}a_kx^k$ for a suitable $m$. In each case $n$ is a positive integer. $a)$ $(1+x)^{2n}.$ $b)$ $\frac{1-x^{n+1}}{1-x}, x\not=1.$ $c)$ $\prod_{k=0}^{n}(1+x^{2^k}).$  The attempt at a solution: a) part of the problem is pretty easy I guess, it is example of binomial theorem, so the answer would be $(1+x)^{2n}=\sum_{k=0}^{2n}(^{2n}_{k})x^k.$ Answer to part b) would be the following: $$\frac{1-x^{n+1}}{1-x}=\frac{(1-x)(1+x+x^2+\cdots+x^n)}{1-x}=1+x+x^2+\cdots+x^n=‌​\sum_{k=0}^{n}x^k.$$thanks to @DiegoMath's hint. As for part c), we have $$\prod_{k=0}^{n}(1+x^{2^k})=(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots(1+x^{2^n})$$ and I have trouble &quot;converting&quot; this to sum which would be of a form of polynomial. Passive voice phrasing of "The bottle contains two liters" PhD in Europe without scholarship from non-European homeland Why does the series $\sum_{n=1}^\infty\frac1n$ not converge? 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. Exclusive differences Profiles vs permission sets? I understand, the differences between Profiles and Permission Sets. I just wanna ask are there some configs that make Profile or Permission Set unique.  Are there things that could only be done via permission set and not via Profiles and vice versa ? Let $R$ be a commutative ring with 1 then why does $a\in N(R) \Rightarrow 1+a\in U(R)$? Unable to uninstall Universal Apps through PowerShell What is the JavaScript >>> operator and how do you use it? This is a problem from Introduction to Representation Theory by Etingof et al. I have no idea how to approach it. My friend wrote a program that uses Gaussian elimination to find a basis, which starts with x, y, [x, y], [y, [x, y]], ... but freezes once it gets to elements with 9 or more occurrences of x, y.  I expect that there is some way to characterize the nonzero elements of this Lie algebra (i.e., elements that are "immune" to being shown to be zero through repeated application of the Jacobi identity) which could be turned into a way to enumerate them. I don't see what it is, though.  It's also totally unclear to me how to show that some set we know spans the Lie algebra (proven using some criterion as described in the previous paragraph) is actually linearly independent, unless we find a faithful representation... but I have no idea how to find a faithful representation for an infinite-dimensional Lie algebra.  Hints would be greatly appreciated.  Edit: After looking at the partial output, we can form the "straightforward" conjecture that:   If $n \geq 4$ is even, then there is one basis element of grade $n$, namely $\mathrm{ad}([x, y])^{(n-4)/2}[y, [y, [y, x]]]$. If $n \geq 5$ is odd, then there are two basis elements of grade $n$, namely $[x, z]$ and $[y, z]$ where $z$ is the single basis element of grade $n-1$.   I also have a different conjecture, which is mutually exclusive with the above: that if we flatten out all nonzero elements into the form ${\rm ad}(e_1) {\rm ad}(e_2) \ldots {\rm ad}(e_{n-1}) e_n$ using the Jacobi identity, then in the sequences $e_1, \ldots, e_{n-1}, e_n$ and $e_1, \ldots, e_n, e_{n-1}$, any span containing two x's must contain at least two y's and any span containing five y's must contain at least two x's, otherwise we can use the Jacobi identity to rearrange the element somehow to show that it vanishes. If this is true, then the long-range behaviour must be:   For $n \geq 8$ there is exactly one basis element of grade $n$, namely $[... [y, [x, [y, [y, [x, y]]]]] ...]$ (the pattern $y x y$ repeats).   I still don't see an approach to proving that none of the elements in this sequence actually vanishes, though.  Edit 2: Here is the hint given by Etingof in an email:     No, there are no published solutions, since these problems are used as   a homework for classes. This is indeed a nontrivial problem.      This is the positive part of the affine Kac-Moody algebra of type   A_2^2, which is the twisted loop algebra for sl(2). So what you need   to do is to relate this Lie algebra to loops into sl(2). I see from   your post that you are on the right track. Prove that the power set of an $n$-element set contains $2^n$ elements Theorem. Let $X$ denote an arbitrary set such that $|X|=n$. Then $|\mathcal P(X)|=2^n$.      Proof. The proof is by induction on the numbers of elements of $X$.      For the base case, suppose $|X|=0$. Clearly, $X=\emptyset$. But the empty set is the only subset of itself, so $|\mathcal P(X)|=1=2^0$.      Now the induction step. Suppose $|X|=n$; by the induction hypotesis, we know that $|\mathcal P(X)|=2^n$. Let $Y$ be a set with $n+1$ elements, namely $Y=X\cup\{a\}$. There are two kinds of subsets of $Y$: those that include $a$ and those that don't. The first are exactly the subsets of $X$, and there are $2^n$ of them. The latter are sets of the form $Z\cup\{a\}$, where $Z\in\mathcal P(X)$; since there are $2^n$ possible choices for $Z$, there must be exactly $2^n$ subsets of $Y$ of which $a$ is an element. Therefore $|\mathcal P(Y)|=2^n+2^n=2^{n+1}$. $\square$      From the above explanation, I don't understand why the set that contains $\{a\}$ will contain $2^{|n|}$ elements when it should clearly be $2^{|1|}$.  The construction of a new set $S$ is the union of the old set with cardinality $n$ and an new element $\{a\}$, therefore the set that do not contain $\{a\}$ still has cardinality $n$ and the set that contains $\{a\}$ is just $\{a\}$, one element...  Can someone please elucidate? Consider the following Gaussian Integral $$I = \int_{-\infty}^{\infty} e^{-x^2} \ dx$$  The usual trick to calculate this is to consider $$I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2} \ dx \right) \left(\int_{-\infty}^{\infty} e^{-y^{2}} \ dy \right)$$  and convert to polar coordinates. We get $\sqrt{\pi}$ as the answer.   Is it possible to get the same answer by considering $I^{3}, I^{4}, \dots, I^{n}$? A uniformly continuous function maps bounded set to bounded sets I am trying to prove the following:  If $A\subset\mathbb R$ is bounded and $\,f:A\to \mathbb R\,$ is uniformly continuous, then $f[A]$ is bounded.  Could you check my proof?  Let $A \subseteq [-K,K]\subseteq \mathbb R$. Let $\varepsilon = 1$. If $f$ is uniformly continuous there is $\delta$ wuth $|x-y| &lt;\delta$ imply that $|f(x)-f(y)|&lt;1$ for all $x,y\in A$. Let $a \in A$. Then because $A$ is bounded there is a finite number of balls $B(a_n,\delta)$ that cover $A$. Let the number be $N$. Then $f(a)-N \le f(x) \le f(a) + N$ for all $x\in A$.