# Tag Info

5

The "only if" is fine, but the "if" requires connectedness. For consider the disjoint union of two spheres, and let $\omega$ be a form that integrates to $1$ on the first sphere and to $-1$ on the second. Obviously, such an $\omega$ cannot be exact. (Apply Stokes's Theorem on each sphere separately.) Indeed, the argument I just gave ...

2

Let us compute $f'(u)$ for $u > 0$: $$f'(u) = -e^{\alpha u}\log(u)-\alpha u e^{\alpha u}\log(u) - e^{\alpha u} = -e^{\alpha u}(\log(u)+\alpha u \log(u) +1)$$ This is continuous on $(0,+\infty)$, so it is locally bounded on $(0,+\infty)$. Hence, $f$ is locally Lipschitz on $(0,+\infty)$. However, $$\lim_{u \to 0⁺} f'(u) = +\infty,$$ hence $f$ is not ...

1

First, the series are $$\log(1+x)=\sum_{k=1}^\infty(-1)^{k-1}\frac{x^k}k$$ and $$-\log(1-x)=\sum_{k=1}^\infty\frac{x^k}k$$ The series is $2\pi$-periodic and the sum for $-\pi\le x\le\pi$ is  \begin{align} \sum_{k=1}^\infty\frac{\sin(kx)}k &=\frac1{2i}\sum_{k=1}^\infty\frac{e^{ikx}-e^{-ikx}}k\\ &=-\frac1{2i}\left(\log(1-e^{ix})-\log(1-e^{-ix})\...

1

\begin{gather*} y=-( 1+\mid \ln( -1-x) \mid )\\ The\ domain\ is\ ( -\infty ,-1)\\ and\ the\ range\ is\ ( -\infty ,-1) . \end{gather*} Does this answer your question?

1

I think this is true. As Greg Martin says, it's only reliable to prove it from the definitions. Throughout the proof I'll only write one $\delta >0$, because we can take the minimum of all $\delta_i$'s and call it $\delta$. \ Let $M>1$ there exists $\delta >0$ such that $|\frac{f(x)}{g(x)}|>\sqrt{M}$ and $|g(x)|>\sqrt{M}$ whenever $0<|x-a|&... 1 First, note that$X_n\xrightarrow{p}X$iff every subsequence of$\{X_n\}$has a sub-subsequence converging to$X$a.s. Take a subsequence$\{\xi_{n_k}\}$. We know that$\xi_{n_k}\xrightarrow{p}\xi$and, thus, there is a subsequence$\{\xi_{n_{k_j}}\}$converging a.s. to$\xi$. Since$\{f \text{ is discontinuous at }\xi\}$is a null set,$f(\xi_{n_{k_j}})\to ...

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