The universal covering group of a symmetry group
Why exactly do sometimes universal covers, and sometimes central extensions feature in the application of a symmetry group to quantum physics?
Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map measure zero sets to measure zero sets? Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map measure zero sets to measure zero sets?   It is easy to prove this statement if the domain is bounded. Is there any way to extend the argument to unbounded domains? Can anyone give me some hint, or show me a simple proof sketch? I've installed phpmyadmin sudo apt-get install phpmyadmin. Of course, I have also mysql and a web server running properly.  Now how can I reach it? change caption in list of tables I would like to change the caption in the List of Tables in a way it is different from the caption in the corresponding table. How can I do that? Does there exist a command to do that? Prove $\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$ I am trying to prove  $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$  This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below.   $$I_n := \int\cos^n x \ dx = \int\cos^{n-1} x \cos x \ dx \tag{1}$$  First question: why rewrite the original instead of immediately integrating by parts of $\int \cos^n x \ dx$?  Integrate by parts with $$u = \cos^{n-1} x, dv = \cos x \ dx \implies du = (n-1)\cos^{n-2} x \cdot -\sin x, v = \sin x$$  which leads to  $$I_n = \sin x \ \cos^{n-1} x +\int\sin^2 x (n-1) \ \cos^{n-2} x \ dx \tag{2}$$  Since $(n-1)$ is a constant, we can throw it out front of the integral:  $$I_n = \sin x \ \cos^{n-1} x +(n-1)\int\sin^2 x \ \cos^{n-2} x \ dx\tag{3}$$  I can transform the integral a bit because $\sin^2 x + cos^2 x = 1 \implies \sin^2 x = 1-\cos^2 x$  $$I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}$$  According to Wikipedia as noted , this simplifies to:  $$I_n = \sin x \ \cos^{n-1} x + (n-1) \int \cos^{n-2} x \ dx - (n-1)\int(\cos^n x) \ dx \tag{5}$$  Question 2: How did they simplify the integral of $\int(1-\cos^2 x) \ dx$ to $\int(\cos^n x) \ dx$?  Assuming knowledge of equation 5, I see how to rewrite it as  $$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x  - (n-1) I_{n} \tag{6}$$  and solve for $I_n$. I had tried exploiting the fact that  $$\cos^2 x = \frac{1}{2} \cos(2x) + \frac{1}{2} $$  and trying to deal with $\int 1 \ dx - \int \frac{1}{2} \cos (2x) + \frac{1}{2} \ dx$  which left me with $\frac{x}{2} - \frac{1}{4} \sin(2x)$ after integrating those pieces. Putting it all together I have:  $$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x \left(-(n-1) (\frac{x}{2} - \frac{1}{4} \sin 2x) \right) \tag{7}$$  but I'm unsure how to write the last few terms as an expression of $I_{something}$ to get it to match the usual reduction formula of   $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$ I received an undesirable status for overstaying a SA visa while hospitalized for injuries sustained in Lesotho. I had to cross into SA from Lesotho. Now I am banned entry into SA until Feb 2016 and though I have submitted a waiver request to remove the undesirable status, I hear from some the waiver request may take 4-5 months or even take longer than the time of banishment.  I have a pre-paid safari planned in October in which I need to fly through Johannesburg to get to Botswana. The Jburg border control requires tourists to pass through SA immigration even when simply changing planes.  1) Will SA grant me "transit" access if I don't leave the airport?  2) Will SA allow me to spend a night in the airport as the layover is 12 hours>  Thanks for passing on any info. Overriding "+=" in Python? (__iadd__() method) Why doesn't running "sudo shutdown now" shut down? Discontinuous functions with closed graphs I tried looking up a question regarding graphs of continuous functions on this site, but all the ones I found consider functions from $\mathbb{R}$ into $\mathbb{R}$. I have been pondering the following question: given a general topological spaces $X, Y$, and a function $f: X\to Y$, when does $Graph(f)$ closed in $X\times Y$ imply that $f$ is continuous. By the closed graph theorem, this is true whenever $X$ and $Y$ are both Banach spaces.   Also, it is fairly easy to prove that whenever $Y$ is a Hausdorff space and $f$ is continuous, then $Graph(f)$ is closed, but I do not think that the converse is true, so I am trying to find an example where $X$ is some topological space, $Y$ a Hausdorff space, $f: X\to Y$ a function with a closed graph in $X\times Y$, but who fails to be continuous. As of yet I have not been able to find such a counterexample, partially because I have no clue where to look for such a counterexample. I would really appreciate getting some directions to go in. How to find sum of power series $\sum_{n=0}^\infty\frac{1}{n!(n+2)}$ by differentiation and integration? Let $$ S = \sum_{n=0}^\infty \frac{1}{n!(n+2)} $$  Integrate the Taylor Series of $xe^x$ to show that S = 1  Also, Differentiate the Taylor series of $\frac{e^x - 1}{x}$ to show that S = 1  For the integration one, I got $\int xe^x dx$ = $x^2\sum_{n=0}^\infty \frac{1}{n!(n+2)} x^n$ and pretty much stuck. I can see the S term but i have no idea how to move on from there Integral $\int_{0}^{\infty}e^{-ax}\cos (bx)\operatorname d\!x$ I want to evaluate the following integral via complex analysis $$\int\limits_{x=0}^{x=\infty}e^{-ax}\cos (bx)\operatorname d\!x \ \ ,\ \ a &gt;0$$  Which function/ contour should I consider ? This is fairly easy.  If I go to https://webserver.com/someurl?a=b, will the someurl?a=b part be protected, or just the contents of the site? If you could travel to the center of the Earth (or any planet), would you be weightless there? In my test class i have the creation of a dummy custom setting and a dummy lead:  Group g1 = new Group(Name='group name', type='Queue');              insert g1;             QueuesObject q1 = new QueueSObject(QueueID = g1.id, SobjectType = 'Lead');              insert q1;       SetOwner__c mycs = SetOwner__c.getValues('standardset');               if(mycs == null) {                               mycs = new SetOwner__c(Name= 'CustomValues');                  mycs.OwnerId__c ='00520000000z3PrAAI';                  mycs.RecordtypeID__c='012L00000004QsrIAE';                  insert mycs;      Lead lead = createNewLead(u2.id);                           insert lead;   ERROR:  MIXED_DML_OPERATION, DML operation on setup object is not permitted after you have updated a non-setup object (or vice versa): SetOwner__c, original object: QueueSobject: []   I get this error because i am trying to perform DML on setup and non-setup objects in the same transaction.  Any solution for this problem? Let $A,B$ be real symmetric matrices, and for any $n\in \Bbb N^{+}$, and for all $x,y\in \Bbb R$, we have $$tr(xA+yB)^n=x^ntr(A^n)+y^ntr(B^n).$$  Show that $AB=0$.  My try:since $$(xA+yB)^n=x^nA^n+\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}+y^nB^n$$ so $$tr\left(x^nA^n+\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}+y^nB^n\right)=x^ntr(A^n)+y^ntr(B^n)$$  then  $$tr\left(\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}\right)=0$$ so $$\binom{n}{1}x^{n-1}ytr(A^{n-1}B)+\cdots+\binom{n}{n-1}xy^{n-1}tr(AB^{n-1})=0$$ then I can't  I found this is a nice result, maybe this is an odd problem, and I can't solve it. Thank you. Let $x*y=\gcd(x,y)$ on $D_n=\{x\in \mathbb N:x |n\}$. I have to prove that $(D_n,*)$ is a commutative monoid and I have a problem when finding the identity element.  $x*e=x$  So $gcd(x,e)=x \to e=xk,k\in \mathbb N^*$ and $e \leq n$  A good $e$ is $0$ but it's not in $D_n$ Is floating point math broken? Consider the following code:   0.1 + 0.2 == 0.3  -&gt;  false   0.1 + 0.2         -&gt;  0.30000000000000004   Why do these inaccuracies happen? Proof of an inequality by induction: $(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + x_1 + x_2 + ... + x_n$ Let $n \in \mathbb N^+$. Show that if $x_1, x_2, ... , x_n$ are $n$ real numbers such that $-1 \le x_i \le 0$ for each $1 \le i \le n$, then $$(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + x_1 + x_2 + ... + x_n$$  I am using a proof by induction, but I am unable to complete it. I would appreciate any tips on how I could finish the proof.   Incomplete Proof Let $P(n)$ be the proposition that $(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + \sum_{i = 1} ^ n x_i$. When $n = 1$, LHS = RHS = $1 + x_i$. Clearly $1 + x_i \ge 1 + x_i$ and so $P(1)$ is true. Assume that $P(k)$ is true, i.e. $(1 + x_1)(1 + x_2)...(1 + x_k) \ge 1 + \sum_{i=1}^k x_i$. Case 1 Suppose $x_{k + 1}  = -1$. Then $1 + x_{k + 1} = 0$, hence LHS = $(1 + x_1)(1 + x_2)...(1 + x_k)(1 + x_{k+1}) = 0$. RHS = $1 + x_1 + x_2 + ... + x_k - 1 = x_1 + x_2 + ... + x_k$ which is always nonpositive since each $x_i$ is nonpositive. Hence RHS $\le 0$ and so LHS $\ge$ RHS. Case 2 Suppose $x_{k+1} = 0$. Then $1 + x_{k+1} = 1$, hence multiplying it to LHS does not change anything. Also, adding $x_{k+1}$ to RHS does not change anything. Therefore, LHS $\ge$ RHS still holds. Case 3 Suppose $-1 &lt; x_{k+1} &lt; 0$. Then $0 &lt; 1 + x_{k+1} &lt; 1$. Adding $x_{k+1}$ to RHS makes the RHS smaller since $x_{k+1}$ is negative. If LHS $= 0$, then multiplying $1 + x_{k+1}$ to LHS does not change anything, and so LHS $\ge$ RHS holds. If LHS > $0$, then multiplying $1 + x_{k+1}$ to LHS makes the LHS smaller since $0 &lt; 1 + x_{k+1} &lt; 1$. In this case LHS $\ge$ RHS does not necessarily hold? ... Kuratowski-Zorn Lemma with pre-order (quasi-order, proset) instead poset? What would happen if we would use pre-order (there is no weak-antisymmetry, only reflexivity and transitivity) in  Kuratowski-Zorn Lemma instead of partial order?      Suppose that we have set P which is preordered and  has the property   that every chain (i.e. totally ordered subset) has an upper bound in   P. Then the set P contains at least one maximal element.   Is this statement true in ZFC? If not, what can we postulate to make it true? I think it can be false, or at least we need to postulate something more (or not?), but I am just missing something and don't see it.  P.S.      I wanted to put tags with combination of words like: Kuratowski Zorn Lemma Pre-order Proset, but it seems that none of them exist and I need to have 300+ reputation to create new ones, thus I only typed (set-theory) :( If someone with bigger reputation could make this topics' tags more precise, I would be grateful :) If $f$ is measurable and $fg$ is in $L^1$ for all $g \in L^q$, must $f \in L^p$? Let $f$ be a measurable function on a measure space $X$ and suppose that $fg \in L^1$ for all $g\in L^q$. Must $f$ be in $L^p$, for $p$ the conjugate of $q$? If we assume that $\|fg\|_1 \leq C\|g\|_q$ for some constant $C$, this follows from the Riesz Representation theorem. But what if we aren't given that such a $C$ exists? "1 out of 100 chickens is" or "1 out of 100 chickens are"? I'm in an argument. To me "are" makes more sense. I understand the rationale for is because it's only one chicken, but chickens itself is plural. Help?