User:Johnlemmas

In Helmholtz decomposition - why are the second components of the decomposition zero when $V$ is $\mathbb{R}^3$?

> $\varphi(\mathbf{r})=\frac{1}{4\pi}\int_{V}\frac{\nabla'\cdot\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'-\frac{1}{4\pi}\int_{S}\frac{\mathbf{F}(\mathbf{r}')\cdot\mathbf{\mathrm{d}S}'}{\left|\mathbf{r}-\mathbf{r}'\right|},$ > > $\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\int_{V}\frac{\nabla'\times\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'+\frac{1}{4\pi}\int_{S}\frac{\mathbf{F}(\mathbf{r}')\times\mathbf{\mathrm{d}S}'}{\left|\mathbf{r}-\mathbf{r}'\right|}.$ > > > If $V$ is $\mathbb{R}^3$ itself (unbounded), and $F$ vanishes sufficiently fast > at infinity, then the second component of both scalar and vector > potential are zero. That is, > > $\varphi(\mathbf{r})=\frac{1}{4\pi}\int_{V}\frac{\nabla'\cdot\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V',$ > > $\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\int_{V}\frac{\nabla'\times\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'.$

Why is it true that second components of these two potentials are zero?

multivariable-calculus

Analysis of function $f$ in real setting to hyperreal setting: can all "if"s be changed to "iff"?

> A function $f$ on an interval $I$ is uniformly continuous if its natural extension $f^{*}$ in $I^{*}$ has the following property: > > for every pair of hyperreals $x$ and $y$ in $I^{*}$, if $x\approx y$ then $f^*(x)\approx f^*(y)$

Question is, for the first "if": Can this be changed into "if and only if"?

The same question for the quote below:

> A real function $f$ is continuous at a standard real number $x$ if for every hyperreal $x'$ infinitely close to $x$, the value $f(x')$ is also infinitely close to $f(x)$. This captures Cauchy's definition of continuity.