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 ...
Only top voted, non community-wiki answers of a minimum length are eligible
