If $H \le G$ and $H$ has index 2 in $G$ then $a^2 \in H, \forall a\in G$
If $[G:H]=n$, then $g^{n!}\in H$ for all $g\in G$.
So my teacher gave a pretty tough (for me) problem in class today: Does the series $$ \sum_{n=1}^{\infty}\frac {\sin{\frac 1 n}} {\sqrt n} $$ converge or diverge?  So far I've thought of trying the Comparison Test and Ratio Test but couldn't really get anywhere. Intuitively I think it should converge, but how do I actually show this? Proving the given quadratic form is chi-squared $k$ Suppose $\underline{X}$ is an $m$-dimensional vector following multivariate Normal distribution i.e. $\underline{X}$~$N_m(\underline{\mu},\Sigma)$ where $\Sigma$ is positive definite. Let $B$ be a symmetric $m$ by $m$ matrix, having the following properties:  $B\Sigma B=B$  $\underline{\mu}'B\underline{\mu}=0$  $trace(B\Sigma)=k$, where $k&lt;m$   Prove that $\underline{X}'B\underline{X}$ follows chi-squared ($k$).  I proceeded a lot but got stuck in one last step where I think linear algebra is troubling me. Kindly have  a look at my &quot;solution&quot;.  The second condition can be written as $\underline{\mu}'B\Sigma B\underline{\mu}=0\implies B\underline{\mu}=\underline{0}$ as $\Sigma$ is Positive definite. Then after a little algebra and considering condition 2, we can write $\underline{X}'B\underline{X}=(\underline{X}-\underline{\mu})'B(\underline{X}-\underline{\mu})=(\underline{X}-\underline{\mu})'B\Sigma B(\underline{X}-\underline{\mu})$ But $\Sigma$ is positive definite hence we can have a symmetric matrix $\sqrt{\Sigma}$ defined such that $\sqrt{\Sigma}\sqrt{\Sigma}=\Sigma$. Using this, we can write $(\underline{X}-\underline{\mu})'B\Sigma B(\underline{X}-\underline{\mu})=(\sqrt{\Sigma}B(\underline{X}-\underline{\mu}))'(\sqrt{\Sigma}B(\underline{X}-\underline{\mu}))$ We want to show that this follows chi-squared ($k$). Now, it is obvious that as $(\underline{X}-\underline{\mu})$~$N(\underline{0},\Sigma)$, $\sqrt{\Sigma}B(\underline{X}-\underline{\mu})$~$N(\underline{0},\sqrt{\Sigma}B\Sigma B\sqrt{\Sigma})=N(\underline{0},\sqrt{\Sigma}B\sqrt{\Sigma})$ Now, $trace(\sqrt{\Sigma}B\sqrt{\Sigma})=trace(B\Sigma)=k$ and $\sqrt{\Sigma}B\sqrt{\Sigma}$ is idempotent (verified). So its eigenvalues are only $0$ or $1$. Since its trace is $k$, there are $k$ eigenvalues with value $1$ and other $m-k$ eigenvalues are $0$. Now here is the thing I want to prove: I want to show that$\sqrt{\Sigma}B\sqrt{\Sigma}$, which is the covariance matrix, is diagonal and has only $k$ $1$'s and remaining $m-k$ diagonal entries $0$. The reason is, then I will be able to say that $k$ components of $\sqrt{\Sigma}B(\underline{X}-\underline{\mu})$ are independent standard normals whose sum of squares is our quadratic form and thus it follows chi-squared ($k$). But I really could not show that $\sqrt{\Sigma}B\sqrt{\Sigma}$ has the desired property i.e. it is diagonal with rank $k$. But this is crucial, it is the finale. If this does not happen, then some components will be correlated with each other and we cannot say that our quadratic form has chi-squared ($k$). I would really appreciate if someone helped me to prove this final part. Early propagation of carries in the BLAKE hash function Prove that if the function $f$ is defined on the set of positive real numbers, its values are real, and $f$ satisfies the equation $$f\left( \frac{x+y}{2}\right) + f\left(\frac{2xy}{x+y} \right) =f(x)+f(y)$$ for all positive $x,y$, then $$2f(\sqrt{xy})=f(x)+f(y)$$ for every pair $x,y$ of positive numbers.    Source: Miklos Schweitzer Memorial Competition 2001    I can see how the repeated application of the functional equation condition upon itself forms a bound, but how can I formally prove this? How to get only one vertically centered equation number in align environment with two equations How do you unpin and move the form in the Visual Studio designer? Random bits of my chat die and come back to life during the day I was trying to accomplish this through a trigger, but its not really working. Is there a way to get this done through Process Builder? If so, what is that process?  I'm trying to have ProductB being added to the opportunity automatically once ProductA is manually added to the opportunity.  Also, is there a way to have the $ amount of ProductB being 0.5% of ProductA?  Trigger that I have so far:  trigger Attempt1 on OpportunityLineItem (after insert) { List&lt;OpportunityLineItem&gt; oliList = new List&lt;OpportunityLineItem&gt;();  for( OpportunityLineItem item: [SELECT  OpportunityId,                                          Name                                  FROM    OpportunityLineItem                                 WHERE   Id = :Trigger.new AND                                          Name = 'NAME OF PRODUCTA'])  {   oliList.add(new OpportunityLineItem(     OpportunityId = item.OpportunityId,      PricebookEntryId ='01uA000000QPDldEAt',     Quantity = 1,     UnitPrice = 25)   );  }  Insert oliList; } This is from Fraleigh's First Course in Abstract Algebra (page 82, Theorem 8.16) and I keep having hard time understanding its proof. I understand only until they mention the map $\lambda_x (g) = xg$. Can someone explain this proof step by step? Thank you so much! Here is the proof:  Let $G$ be a group. We show that $G$ is isomorphic to a subgroup of $S_G$. Define a one-to-one function $\phi: G \to S_G$ such that $\phi(xy)=\phi(x) \phi(y)$ for all $x,y \in G$. For $x \in G$, let $\lambda_x: G \to G$ be defined by $\lambda_x (g) = xg$ for all $g \in G$. The equation $\lambda_x (x^{-1} c) = x(x^{-1} c) = c$ for all $c \in G$ shows that $\lambda_x$ maps $G$ onto $G$. If $\lambda_x (a) = \lambda_x (b)$, then $xa=xb$ so $a=b$ by cancellation. Thus $\lambda_x$ is also one to one, and is a permutation of $G$. We now define $\phi: G \to S_G$ by defining $\phi(x) = \lambda_x$ for all $x \in G$. To show that $\phi$ is one to one, suppose that $\phi(x) = \phi(y)$. Then $\lambda_x = \lambda_y$ as functions mapping $G$ into $G$. In particular $\lambda_x (e) = \lambda_y (e)$, so $xe=ye$ and $x=y$. Thus $\phi$ is one to one. It only remains to show that $\phi(xy) = \phi(x) \phi(y)$, that is, that $\phi_{xy} = \lambda_x \lambda_y$. Now for any $g \in G$, we have $\lambda_{xy} (g) = (xy)g$. Permutation multiplication is function composition, so $(\lambda_x \lambda_y)(g) = \lambda_x (\lambda_y (g)) = \lambda_x (yg) = x(yg)$. Thus by associativity, $\lambda_{xy} = \lambda_x \lambda_y$. Suppose the following dialogue:  ... &quot;I accept only one notion of land property. Namely, 'I am doing my stuff here, therefore I am here&quot;. &quot;But this means,&quot; he responded, &quot;you can break into any place and stay there, using this rule.&quot; &quot;There is another rule, though. Right to act in its most natural sense, 'I am here, therefore I am doing my stuff here.&quot; &quot;But do not these rules put together become circular reasoning?&quot; &quot;Of course, they do,&quot; I put my head closer to him, &quot;and this is exactly that rare case when circular reasoning is not only sufficient, but is necessary.&quot; &quot;But why is that? Circular reasoning never was a correct argument.&quot; &quot;There are many circular things happening in nature. Life forms, for example, exist only in order to produce new life forms or to prolongate their own lifetime. Is there any reason to say this can't be the case for reason itself? Why must reasoning not be circular?&quot; ...  Can circular reasoning ever be justified? For example, we know all the words in our languages can't be defined using other words without the use of circularity. We know logic can't prove itself without circularity. Are these arguments rational? Are there any other arguments where circular reasoning is still rational? Russian babel and chemformula How to prove that a number is irrational With all the various races in the federation, how do the humans manage to dominate it? I accidentally removed my /etc/apt/sources.list file. Is it hosted by Canonical somewhere so I can re-add it to the system? It's a bit of an emergency, so I hope someone can provide a quick answer. Wall of Sand on Difficult Terrain Looking into  prompted this inquiry:  What happens when you move through a wall of sand on difficult terrain?  This involves the understanding that wall of sand changes the cost of moving 1 foot to 3 feet and that difficult terrain typically changes the cost of moving 1 foot to 2 feet. How do I remove the file suffix and path portion from a path string in Bash? Given a string file path such as /foo/fizzbuzz.bar, how would I use bash to extract just the fizzbuzz portion of said string? Assume that $f(x),g(x)$ are positive and are in $L^1$. Moreover, they are differentiable and their derivative is integrable. Let $h(x)=f(x)*g(x)$, the convolution of $f$ and $g$. Does the derivative of $h(x)$ exist? If yes, how can we prove that $$ \frac{d}{dx}(f(x)*g(x)) = \left(\frac{d}{dx}f(x)\right)*g(x)$$  Thanks Evaluate $\int \frac{1}{(x^2+1)^2}dx$ $$\int \frac{1}{(x^2+1)^2}\mathrm dx$$ Look simple, but I stuck a little bit. What method is better in this situation? Please, help. Rotating an object continously with OnMouseDown I have a script that works but it does not do what I want it to do. My script makes an object rotate on when another is clicked. However, I want the object to continuously rotate until I click on the other object again. Right now when I click it rotates just a bit and stops.  Here is my script:   public class Rotate : MonoBehaviour {     public float speed =50f;     public GameObject target;      public void OnMouseDown(){        target.transform.Rotate (new Vector3(0, Time.deltaTime, 0) * speed);     }  }   How can I make the object rotate continuously until I click it again? How do I read and parse an XML file in C#?