ukmuiik

Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite Let $omega$ be an exact $1$-form on a path-connected manifold $M$. For a fixed $x in M$ we can define for all $y in M$. $$ g(y) := int_gamma gamma^*(omega), $$ where $gamma$ is a path from $x$ to $y$. Because $omega$ is exact, it is independent of $gamma$. What I want to show is that $dg = omega$. My thought is that $$ dleft( int_gamma gamma^*(omega) right) = int d(omega(gamma)), $$ because $d$ commutes with integration, and $gamma^*(omega)$ is defined to be $omega(gamma(t))$. Then $$ d(omega(gamma)) = domega(gamma) cdot dgamma $$ by chain rule. By change of variable, we get $$ int d omega = omega. $$ But, someone pointed out that $domega = 0$ by exactness. My thought is that this is only true when integrating over closed paths. What's going on? Thanks. Edit: Another thought is that $d(omega)$ and $domega$ are two different things. So, that $d(omega) = 0$, but $domega$ is a element of the cotangent bundle. Thu

Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite I try to upper-bound an error term $X$ with two random variables $A$ and $B$, but I only have closed form for $mathbfE[A] - mathbfE[B] = c$, and I derived $X$ into the following term: $$ mathbfE[X] leq n - dfrac11+mathbfEleft[ left(dfracBA right)^2 right] $$ where $n$ is a constant. I wonder if there are any tricks or inequalities to use to push this further? So far I know that $A>0$, $B>0$, and $mathbfE[A]>mathbfE[B]$ Edit: Thanks for Robert pointing out that the information is not enough, so let me add in all the information I have: Given $s$ probabilities $pi_1, pi_2, dots pi_s$, which are sampled from the distribution $Beta(alpha_s, beta_s)$ respectively. And both $alpha_s$ and $beta_s$ are sampled from the same distribution $mathcalP$ for all $s$. For each probability $pi_t$, we sample $n$ random variables with Binomial: $$ Y_i,t = Bin(2, pi_t) $$ Therefore, there are total $ns$ Binomial random varia