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## A family of monotone quantities along the inverse curvature flow in the Euclidean space

Monotone quantities along hypersurfaces evolving under the inverse curvature flow have many applications in geometry and relativity. In [HI], Huisken and Ilmanen applied the monotone increasing property of Hawking mass to give a proof of the Riemannian Penrose Inequality. In a recent paper [BHW], Brendle, Hung and Wang discovered a monotone decreasing quantity along the inverse mean curvature flow in Anti-Desitter-Schwarzschild manifolds and used it to establish a Minkowski-type inequality for star-shaped hypersurfaces.

In this note, which is a sequel to one of our previous posts, we provide a series of new monotone increasing quantities along smooth solutions to the inverse curvature flow [Gerhardt] in ${ \mathbb{R}^n}$:

 Theorem 1 Let ${ \Sigma }$ be a smooth, closed, embedded hypersurface in ${ \mathbb{R}^n}$ with positive normalized ${k}$-th mean curvature ${\sigma_k>0}$ (${k\ge 2}$), and is star-shaped with respect to a fixed point ${O}$. Let ${ I }$ be an open interval and ${ X: \Sigma \times I \rightarrow \mathbb{R}^n }$ be a smooth map satisfying $\displaystyle \frac{\partial X}{\partial t} = \frac{\sigma_{k-2}}{\sigma_{k-1}} \nu, \ \ \ \ \ (1)$ where ${\sigma_l}$ is the normalized ${l}$-th mean curvature of the surface ${ \Sigma_t = X (\Sigma, t) }$ and ${ \nu }$ is the outward unit normal vector to ${ \Sigma_t}$. Let ${ r = r (x)}$ be the distance from ${ x }$ to ${ O}$. Then the function $\displaystyle Q(t)=\left(\int_{\Sigma_t}\sigma_{k-2}d\mu\right)^{-\frac{n-k-1}{n-k+1}}\left(\int _{\Sigma_t}\sigma_{k-2}d\mu- \int_{\Sigma_t} \sigma_k r^2d\mu\right) \ \ \ \ \ (2)$ is monotone increasing and ${Q(t) }$ is a constant function if and only if ${\Sigma_t }$ is a round sphere for each ${t}$. Here ${ d \mu }$ denotes the volume form on a hypersurface.

The precise definition of ${\sigma_k}$ will be given in Section 1. Let us just remark that ${\displaystyle \int_{\Sigma}\sigma_0 d\mu= \mathrm{Area}(\Sigma)}$. So for example, when ${k=2}$, then our result states that if ${\Sigma}$ has positive scalar curvature, then

$\displaystyle \textrm{Area}(\Sigma_t)^{-\frac{n-3}{n-1}}\left(\textrm{Area}(\Sigma_t)- \int_{\Sigma_t} \sigma_2 r^2d\mu\right)$

is monotone increasing along the (normalized) mean curvature flow (cf. also [LWW]).

Our motivation is as follows. In [K], the author used the Minkowski type integral formulas to obtain a series of inequalities relating the weighted ${k}$-th mean curvature of a hypersurface. Indeed, it was shown in [K] that:

 Theorem 2 Suppose ${\Sigma}$ is a closed hypersurface embedded in ${\mathbb{R}^n}$ with ${\sigma_{k}>0}$ for some ${1\le k\le n-1}$. Then $\displaystyle \mathrm{Area}(\Sigma)\le \int_\Sigma \sigma_1 r\le \cdots\le \int_\Sigma\sigma_{k} r^{k}$ and $\displaystyle n\mathrm{Vol}(\Omega)\le \int_\Sigma r\le \int_\Sigma \sigma_1 r^2 \le \cdots \le \int_\Sigma \sigma_{k}r^{k+1}. \ \ \ \ \ (3)$ Here ${r=|X|}$ and ${\Omega}$ is the region enclosed by ${\Sigma}$. The equality occurs if and only if ${\Sigma}$ is a sphere centered at ${O}$.

On the other hand, in [KM], it was shown that a special case of the inequalities in (3) can be obtained by using the monotonic property of a related geometric quantity along the inverse mean curvature flow. The approach of using monotone quantities along various inverse curvature flow to derive geometric inequalities is a very fruitful and has attracted a lot of attention recently, see e.g. [BHW], [LG], and [LWW]. In view of the series of inequalities in Theorem 2, it is natural to ask: is there an associated series of geometric quantities which are monotone increasing along some inverse curvature flow in ${\mathbb{R}^n}$? In this note, we give an affirmative answer to this question by showing that for each ${k\ge 2}$, there is a quantity ${Q(t)}$ which is monotone along the ${\frac{\sigma_{k-2}}{\sigma_{k-1}}}$ flow, as long as the initial hypersurface is star-shaped and ${k}$-convex, i.e. ${\sigma_k>0}$.

As an application of Theorem 1, we derive a series of sharp inequalities for star-shaped hypersurfaces in ${ \mathbb{R}^n}$, which partially generalizes Theorem 2:

 Theorem 3 Let ${\Sigma }$ be a smooth, closed hypersurface embedded in ${\mathbb R^n}$ with positive ${\sigma_k}$ (${k\ge 2}$) and is star-shaped with respect to ${O}$. Then $\displaystyle \int_\Sigma \sigma_{k-2}d\mu\le\int_\Sigma \sigma_k r^2d\mu \ \ \ \ \ (4)$ where ${r }$ is the distance to a fixed point ${O }$ and ${\sigma_k}$ is the normalized ${k}$-th mean curvature of ${ \Sigma}$. Furthermore, equality in (4) holds if and only if ${\Sigma}$ is a sphere centered at ${O}$.

The rest of this note is organized as follows: in Section 1, we will collect all the necessary definitions and preliminary results, and we will prove our main theorem in Section 2. I would also like to thank Prof. Pengzi Miao for discussion.

1. Preliminaries

Let us fix some notations. Let ${\Sigma\subset \mathbb R^n}$ be a hypersurface and ${m=n-1}$. Let ${\lbrace e_i\rbrace_{i=1}^{m}}$ be a local orthonormal frame on ${\Sigma}$ and let ${\nu=e_n}$ be the unit outward normal of ${\Sigma}$. Let ${\overline \nabla }$ and ${\nabla }$ be the connections on ${\mathbb R^n}$ and ${\Sigma}$ respectively. We define the the shape operator by ${A=\overline \nabla \nu: T\Sigma \rightarrow T\Sigma}$ and ${A_i^j}$ is defined by ${A(e_i)=\sum_{j=1}^m A_i^j e_j}$. By abusing of notation, we will also denote the second fundamental form ${\langle A(u), v\rangle}$ by ${A(u,v)}$. For simplicity, we will denote the dot product in ${\mathbb{R}^n}$ by ${\cdot}$.

We define the ${k}$-th mean curvature ${H_k}$ and the normalized ${k}$-th mean curvature ${\sigma_k}$ of ${\Sigma}$ by

$\displaystyle H _k = \frac 1{k!}\sum_{\substack{1\le i_1,\cdots, i_k\le m\\ 1\le j_1, \cdots, j_k\le m}} \epsilon_{j_1\cdots j_k}^{i_1\cdots i_k}A_{i_1}^{j_1}\cdots A_{i_k}^{j_k}\textrm{ and } \sigma_k= \frac{H_k}{{m\choose k}}$

respectively. Here ${\epsilon_{i_1 \cdots i_k}^{j_1\cdots j_k}}$ is zero if ${i_k=i_l}$ or ${j_k=j_l}$ for some ${k\ne l}$, or if ${\lbrace i_1, \cdots, i_k\rbrace \ne \lbrace j_1, \cdots, j_k\rbrace}$ as sets, otherwise it is defined as the sign of the permutation ${(i_1, \cdots, i_k)\mapsto (j_1, \cdots, j_k)}$. For convenience, we also define ${\sigma_0=1}$ and ${\sigma_{-1}=X\cdot \nu}$. The reason for this definition is that the Hsiung-Minkowski formula (Theorem 7) holds for ${\sigma_{-1}}$ (trivially).

Following [Reilly], we define the ${k}$-th Newton transformation ${T_k: T\Sigma\rightarrow T\Sigma}$ of ${A}$ as

$\displaystyle {(T_k)}_j^{\,i}= \frac 1 {k!} \sum_{\substack{1\le i_1,\cdots, i_k\le m\\ 1\le j_1, \cdots, j_k\le m}} \epsilon^{i i_1 \ldots i_k}_{j j_1 \ldots j_k} A_{i_1}^{j_1}\cdots A_{i_k}^{j_k}.$

For ${k=0}$, ${T_0= \textrm{I}}$ is the identity. It is not hard to see that if ${\lbrace e_i\rbrace_{i=1}^m}$ are the principal directions, then

$\displaystyle T_k(e_i)=\sum_{\substack{i_1<\cdots

where ${\{\lambda_i\}_{i=1}^m}$ are the principal curvatures.

From now on, we use the convention that repeated indices are summed from ${1}$ to ${m}$. We collect some basic properties of ${T_k}$ and ${H_k}$ (e.g. [BC] Lemma 2.1):

 Lemma 4 Let ${m=n-1}$, then on a hypersurface ${\Sigma}$ in ${\mathbb{R}^n}$, ${\mathrm{tr}(T_k)= (m-k)H_k=(m-k){m\choose k}\sigma_k}$. ${\mathrm{div}(T_k)=0}$. i.e. ${\displaystyle \nabla _{e_i}(T_{k})^i_{j}=0.}$ ${\displaystyle (T_k)_i^jA_j^i=(k+1) H_{k+1}. }$ ${ \displaystyle (T_{k-1})^l_j A^j_{i}A^{i}_l = H_k H_1-(k+1)H_{k+1} .}$

Let ${\Sigma}$ be a smooth hypersurface in ${\mathbb{R}^n}$ given by the embedding ${X_0}$. Let ${\Sigma_t}$ be the surface evolved by

$\displaystyle \begin{cases} \frac{\partial X}{\partial t}=F\nu\\ X(\cdot, 0)= X_0 \end{cases} \ \ \ \ \ (5)$

where ${F}$ is a function on ${\Sigma_t}$ which is independent of ${t}$. We have

 Lemma 5 Under the flow in (5), $\displaystyle \begin{array}{rl} g_{ij}'=&2FA_{ij}, \quad (g^{ij})'=-2FA^{ij}, \quad \nu'=-\nabla F, \quad \\ A_{ij}'=&-\nabla ^2_{ij} F+FA_{ik}A^k_j, \quad {A_i^j}'=-F_i^j-FA_{i}^kA_k^j, \quad d\mu'=FH_1d\mu,\\ H_m'=&(T_{m-1})_j^i\left(-(\nabla ^2F)_i^j-FA_{i}^kA_k^j\right)\\ =&-\nabla ^j\left((T_{m-1})_j^iF_i\right)-F\left(H_1H_m-(m+1)H_{m+1}\right). \end{array}$

Proof: These equations are standard, and we only prove some of them for illustration. We have

$\displaystyle \begin{array}{rl} \nu'=g^{ij}\langle \nu', \partial _i \rangle\partial _j =-g^{ij}\langle \nu, \overline \nabla _t\partial _i\rangle\partial _j =-g^{ij}\langle \nu, (F\nu)_i\rangle\partial _j =&-g^{ij}F_i\partial _j =-\nabla F. \end{array}$

$\displaystyle \begin{array}{rl} A_{ij}' =\langle \overline \nabla _i \nu, \partial _j\rangle' =\langle \overline \nabla _t\overline \nabla _i \nu, \partial _j\rangle+\langle \overline \nabla _i \nu, \overline \nabla _t\partial _j\rangle =&\langle \overline \nabla _i\overline \nabla _t \nu, \partial _j\rangle+\langle \overline \nabla _i \nu, \overline \nabla _j(F\nu)\rangle\\ =&-\langle \nabla _i\nabla F, \partial _j\rangle+F\langle \overline \nabla _i \nu, \overline \nabla _j\nu\rangle\\ =&-\nabla ^2_{ij} F+FA_{ik}A^k_j. \end{array}$

The formula ${ d\mu'=FH_1 d\mu}$ follows by the first variation formula for area. Finally, by Lemma 4,

$\displaystyle \begin{array}{rl} H_m'= \frac{\partial H_m}{\partial A_i^j}\partial _t A_i^j =&(T_{m-1})_j^i\left(-(\nabla ^2F)_i^j-FA_{i}^kA_k^j\right)\\ =&-\nabla ^j\left((T_{m-1})_j^iF_i\right)-F\left(H_1H_m-(m+1)H_{m+1}\right). \end{array}$

Here we have used the fact that

$\displaystyle \begin{array}{rl} \frac{\partial }{\partial A_i^j} H_m = \frac{\partial }{\partial A_i^j} \left(\frac{1}{m!}\epsilon^{i_1\cdots i_m}_{j_1\cdots j_m}A_{i_1}^{j_1}\cdots A_{i_m}^{j_m}\right) = \frac{m}{m!}\epsilon^{ii_1\cdots i_{m-1}}_{jj_1\cdots j_{m-1}}A_{i_1}^{j_1}\cdots A_{i_{m-1}}^{j_{m-1}} =(T_{m-1})^i_j. \end{array}$

$\Box$

 Proposition 6 Under the flow in (5), for any function ${f}$ defined locally around ${\Sigma_t}$, $\displaystyle \begin{array}{rl} \left(\int _\Sigma H_k f d\mu\right)'=\int _\Sigma(-(T_{k-1})_j^iF(\nabla ^2 f)_i^j +(k+1)H_{k+1}Ff+ H_k \langle \overline \nabla f , F\nu\rangle)d\mu. \end{array}$

Proof: By Lemma 5,

$\displaystyle \begin{array}{rl} \left(\int_\Sigma H_k f d\mu \right)' =& \int_\Sigma \left(H_k ' f + H_k f' +H_k fFH_1\right)d\mu\\ =&\int_\Sigma (-\nabla ^j((T_{k-1})_j^iF_i)f-Ff\left(H_1H_k-(k+1)H_{k+1}\right)\\ &+ H_k \langle \overline \nabla f , F\nu\rangle +H_k fFH_1)d\mu\\ =&\int_\Sigma \left((T_{k-1})_j^iF_i f^j +(k+1)H_{k+1}Ff+ H_k \langle \overline \nabla f , F\nu\rangle\right)d\mu\\ =&\int_\Sigma \left(-(T_{k-1})_j^iF (\nabla ^2f)_i^j +(k+1)H_{k+1}Ff+ H_k \langle \overline \nabla f , F\nu\rangle\right)d\mu. \end{array}$

$\Box$

The inverse curvature flow described in Theorem 1 actually exists for all time, given that the initial hypersurface is star-shaped:

 Theorem 6 ([Gerhardt] and [Urbas]) } Let ${\Sigma }$ be a smooth, closed hypersurface in ${ \mathbb{R}^{n}}$ with positive ${\sigma_k}$, given by a smooth embedding ${X_0 : \mathbb{S}^{n-1} \rightarrow \mathbb{R}^{n} }$. Suppose ${ \Sigma}$ is star-shaped with respect to a point ${ P }$. Then the initial value problem $\displaystyle \begin{cases} \frac{ \partial X}{ \partial t} = \frac{\sigma_{k-2}}{ \sigma_{k-1} } \nu \\ X (\cdot, 0) = X_0 ( \cdot) \end{cases} \ \ \ \ \ (6)$ has a unique smooth solution ${ X : \mathbb{S}^{n-1} \times [0, \infty) \rightarrow \mathbb{R}^{n} }$, where ${ \nu }$ is the unit outer normal vector to ${\Sigma_t = X (\mathbb{S}^{n-1}, t) }$ and ${\sigma_l}$ is the normalized ${l}$-th mean curvature of ${\Sigma_t}$. Moreover, ${\Sigma_t}$ is star-shaped with respect to ${P}$ and the rescaled hypersurface ${ \widetilde{\Sigma_t }}$, parametrized by ${ \widetilde{X}(\cdot , t) = e^{- t } X( \cdot, t) }$, converges to a sphere centered at ${P}$ in the ${ C^\infty}$ topology as ${ t \rightarrow \infty}$.

Finally, we also need the Hsiung-Minkowski formula ([Hsiung], [K]):

 Theorem 7 ([K] Cor. 3.1) Corollary 3.1} Suppose ${\Sigma}$ is a closed hypersurface in ${\mathbb{R}^n}$, and ${f}$ is a smooth function on ${\Sigma}$. Then $\displaystyle \int_\Sigma f \sigma_k d\mu =\int_\Sigma f\sigma_{k+1}\nu\cdot X d\mu- \frac{1}{(n-1-k){{n-1}\choose k}}\int_\Sigma \langle T_k(\nabla f), X^T\rangle d\mu.$ Here ${X}$ is the position vector, ${X^T}$ is its tangential component and ${\nu}$ is the unit outward normal of ${\Sigma}$.

2. Proof of the main results

Proof of Theorem 1: By Lemma 5, under the flow in (5),

$\displaystyle \left(\int_\Sigma H_k fd\mu\right)' =\int_\Sigma \left(-(T_{k-1})_j^iF (\nabla ^2f)_i^j +(k+1)H_{k+1}Ff+ H_k \langle \overline \nabla f , F\nu\rangle\right)d\mu. \ \ \ \ \ (7)$

For any function ${f}$, we have

$\displaystyle \nabla ^2 _{ij}f= \overline \nabla ^2_{ij}f- A_{ij}\overline \nabla f\cdot \nu.$

Now we fix ${f= \frac{r^2 }{2}}$, then ${\overline \nabla f=X}$ and ${\overline \nabla ^2 f = g}$ (the Euclidean metric). i.e.

$\displaystyle \nabla ^2_{ij}f=g_{ij}-A_{ij}u$

where ${u =X\cdot \nu\ge 0}$ as ${\Sigma_t}$ is star-shaped. Putting this into (7), and using Proposition 1, for this ${f}$, we have

$\displaystyle \begin{array}{rl} \left(\int_\Sigma H_k fd\mu\right)' =&\int_\Sigma \left(-(T_{k-1})_j^iF (\nabla ^2f)_i^j +(k+1)H_{k+1}Ff+ H_k \langle \overline \nabla f , F\nu\rangle\right)d\mu\\ =&\int_\Sigma \left(F \left((T_{k-1})_i^jA_j^iu- \textrm{tr}\left(T_{k-1}\right)\right) +(k+1)H_{k+1}Ff+ FH_k u\right)d\mu\\ =&\int_\Sigma \left(F \left(k H_k u- (n-k)H_{k-1}\right) +(k+1)H_{k+1}Ff+ FH_k u\right)d\mu\\ =&\int_\Sigma \left((k+1)FH_k u -(n-k)H_{k-1}F+(k+1)H_{k+1}Ff\right)d\mu. \end{array}$

This is equivalent to

$\displaystyle \begin{array}{rl} \left(\int_\Sigma \sigma_k fd\mu\right)' =&\int_\Sigma \left((k+1)F\sigma_k u -k \sigma_{k-1}F+(m-k)\sigma_{k+1}Ff\right)d\mu \end{array}$

where ${m=n-1}$. Let ${F=\frac{\sigma_{k-2}}{\sigma_{k-1}}}$, then the above equation becomes

$\displaystyle \begin{array}{rl} \left(\int_\Sigma \sigma_k fd\mu\right)' =&\int_\Sigma \left((k+1)\frac{\sigma_k \sigma_{k-2}}{\sigma_{k-1}} u -k \sigma_{k-2}+(m-k)\frac{\sigma_{k+1}\sigma_{k-2}}{\sigma_{k-1}}f\right)d\mu. \end{array}$

By the fact that ${\sigma_k>0}$ implies ${\sigma_l>0}$ for ${l\le k}$ (cf. e.g. [BC] Proposition 3.2), the Hsiung-Minkowski formula (Theorem 7), and by the Maclaurin’s inequality ${\sigma_{m-1}\sigma_{m+1}\le \sigma_m^2}$, which implies ${\frac{\sigma_{k+1}\sigma_{k-2}}{\sigma_{k-1}}\le \sigma_k}$, we have

$\displaystyle \begin{array}{rl} \left(\int_\Sigma \sigma_k fd\mu\right)' \le&\int_\Sigma \left((k+1)\sigma_{k-1}u -k \sigma_{k-2}+(m-k)\sigma_k f\right)d\mu\\ =&\int_\Sigma \left((k+1)\sigma_{k-2} -k \sigma_{k-2}+(m-k)\sigma_k f\right)d\mu\\ =&\int_\Sigma \left(\sigma_{k-2} +(m-k)\sigma_k f\right)d\mu. \end{array}$

i.e.

$\displaystyle \left(\int_\Sigma \sigma_k r^2 d\mu\right)'\le \int_\Sigma \left(2\sigma_{k-2} +(m-k)\sigma_k r^2\right)d\mu. \ \ \ \ \ (8)$

On the other hand, by (7), we have

$\displaystyle \left(\int _\Sigma\sigma_{k-2}d\mu\right)'= (n-k+1)\int_{\Sigma}\sigma_{k-1}Fd\mu=(n-k+1)\int_{\Sigma}\sigma_{k-2}d\mu. \ \ \ \ \ (9)$

Combining this equation and (8), we have

$\displaystyle \left(\int _\Sigma\sigma_{k-2}d\mu- \int_\Sigma \sigma_k r^2 d\mu\right)'\ge (n-1-k) \left(\int _\Sigma\sigma_{k-2}d\mu- \int_\Sigma \sigma_k r^2 d\mu\right).$

Equivalently,

$\displaystyle \left[e^{-(n-1-k)t}\left(\int _\Sigma\sigma_{k-2}d\mu- \int_\Sigma \sigma_k r^2 d\mu\right)\right]'\ge 0.$

In view of (9), this can also be written as

$\displaystyle \left[\left(\int_\Sigma\sigma_{k-2}d\mu\right)^{-\frac{n-k-1}{n-k+1}}\left(\int _\Sigma\sigma_{k-2}d\mu- \int_\Sigma \sigma_k r^2 d\mu\right)\right]'\ge 0.$

Suppose the equality holds, then by the equality case of the Newton’s inequality, ${\Sigma_t}$ must be umbilical, and therefore ${\Sigma}$ is a sphere. $\Box$

Proof of Theorem 3: By Theorem 6, there exists a smooth solution ${\{ \Sigma_t \}}$ to the inverse mean curvature flow with initial condition ${ \Sigma}$. Moreover, the rescaled hypersurface ${ \widetilde \Sigma_t = \{ e^{-\frac t{n-1}}x \ | \ x \in \Sigma_t \} }$ converges exponentially fast in the ${C^\infty}$ topology to a sphere. In particular, ${\widetilde \Sigma_t}$ and hence ${\Sigma_t}$, must be convex for large ${t}$.

Let ${ t_1 }$ be a time when ${ \Sigma_{t_1}}$ becomes convex. By using Theorem 7 twice, first with ${f=1}$ and then with ${f=u=X\cdot \nu}$, we have

$\displaystyle \begin{array}{rl} \int_{\Sigma_{t_1}} \sigma_{k-2} d\mu =&\int_{\Sigma_{t_1}} \sigma_{k-1}ud\mu\\ =&\int_{\Sigma_{t_1}} \sigma_k u^2 d\mu - \frac{1}{(m-k+1){m\choose {k-1}}}\int_{\Sigma_{t_1}} \langle T_{k-1}A(X^T), X^T\rangle d\mu\\ \le &\int_{\Sigma_{t_1}} \sigma_k r^2 d\mu - \frac{1}{(m-k+1){m\choose {k-1}}}\int_{\Sigma_{t_1}} \langle T_{k-1}A(X^T), X^T\rangle d\mu \end{array}$

Here we have used the fact that ${\nabla u=A(X^T)}$, which can be easily verified. As ${\Sigma_{t_1}}$ is convex, we have ${A>0}$ and ${T_{k-1}>0}$, and hence ${T_{k-1}A>0}$ (note that ${T_{k-1}}$ and ${A}$ are simultaneously diagonalizable), therefore

$\displaystyle \int_{\Sigma_{t_1}} \sigma_{k-2} d\mu\le\int_{\Sigma_{t_1}} \sigma_k r^2d\mu.$

In other words, ${Q(t_1)\le 0}$ for ${Q}$ defined in Theorem 1.

By Theorem 1, we know ${Q(t)}$ is monotone increasing, hence

$\displaystyle Q(0) \le Q(t_1) \le 0$

which proves (4).

If the equality in (4) holds, then ${ Q(0) = 0 }$. It follows from the monotonicity of ${ Q(t)}$ and the fact ${ Q(t) \le 0 }$ for large ${t}$ that

$\displaystyle Q(t) = 0, \ \forall \ t .$

By Theorem 1, this implies that ${ \Sigma_t }$ is a sphere for each ${t}$. By the above argument, we have ${X\cdot \nu= r}$ on ${\Sigma_{t_1}}$, from which we can deduce that ${\Sigma_{t_1}}$ is a sphere centered at ${O}$. Therefore, we conclude that the initial hypersurface ${ \Sigma}$ is a sphere centered at ${O}$. $\Box$

 Remark 1 We remark that the inequality in Theorem 3 also holds when ${k=1}$ by [KM] (using the convention that ${\sigma_{-1}=X\cdot \nu}$). When ${k=2}$, the assumption that ${\Sigma}$ is star-shaped can be dropped, by [K] Corollary 3.3.


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# neural network to find a very simple linear model (scikit-learn)

I'm trying to test different machine learning algorithm to try to find correlation between various data on MRI scans. Since I'm dealing with medical data, I don't have access to many events, but still I'm trying to see what a simple fully connected NN while provides me. By debugging, it looks like even if I enter the output almost as is in the input, a simple NN using the default scikit-learn NN regressor is unable to find a satisfactory model. I tried to reproduce the effect and I'd like to share with you the following code. I guess I'm just doing something wrong, because, basically, I'd like the NN to produce a simple model where label=X2-X1 and it doesn't work to me which is really surprising me.

here is the code:

import sklearn.neural_network as NN
import matplotlib.pyplot as plt
import numpy as np
N_events=100000 # number of rows of dataset
N_features = 2 # number of features
X = np.random.rand(N_events,N_features) # chose the data randomly
labels = X[:,1]-X[:,0] # label = X2-X1

C = NN.MLPRegressor(hidden_layer_sizes=(2,2),max_iter=500000,random_state=1) # very simple scikit-learn NN regressor
n = N_events // 2 # half data for training, half for test

print(X[0:10,1]-X[0:10,0]) # check that indeed label=X2-X1
print(labels[0:10])
C.fit( X[0:n,:] , labels[0:n]) # train the model

# plot y_predicted vs y_true for the test set...
plt.figure()
y_real = labels[n:] #
y_pred = C.predict(X[n:,:])
plt.plot(y_real,y_pred,'.')

# plot y_predicted vs y_true for the training set... even this doesn't work
plt.figure()
y_real = labels[0:n] #
y_pred = C.predict(X[0:n,:])
plt.plot(y_real,y_pred,'.')

I was expecting the NN with 2x2 = 4 degrees of freedom to easily find such a simple model (note that I even don't add unused features which could disturb the fit, since N_features is 2 here), so I guess I'm just not using the code correctly. Can anyone helps me ?

Many thanks

• Why would you expect a 2x2 to work? – gented Mar 18 '19 at 11:57
• since the exact model is label Y is X2-X1, I mean the model have one degree of freedom, and with 2x2 hidden layers (4 free parameters) and a lot of data points (50000 here) I was expecting the NN to find it. I mean there is more free parameters than actual degree of freedom, and enough data to constrain the model right ? – david guez Mar 18 '19 at 12:17
• sorry I meant 8 free parameters (2 input *2 cell in 1st layer = 4 , and 2 cell in first layer*2 cell in 2nd layer = 4 , so 4+4=8 weights, not counting for the bias term) – david guez Mar 18 '19 at 13:06
• What do you mean "expecting to find it"? What is that "doesn't work"? Are you seeing a coding error (unrelated to NN) or are you saying that the NN don't perform well? – gented Mar 18 '19 at 13:06
• no, actually the code is running, but the produced y_pred are very far from the y_true... so it performs very poorly (actually it the code predict always the same y_pred, no matter what the X is) . Indeed, by adding the number of cell in the model (even 3x3) the code produce a better (still not perfect) model, but I was expecting it to perform better. Maybe I'm just wrong, but I thought that fully connected NN would be general enough for such a simple fit...) – david guez Mar 18 '19 at 13:10

The problem is relu which is the default activation function. It zeros the inputs smaller than 0. Problem will be solved by using other functions. For example:
C = NN.MLPRegressor(activation="identity", hidden_layer_sizes=(1,1),max_iter=25,random_state=1)

Note that even 1 hidden layer with 1 neuron is enough to find the $$X_2 = X_1$$ decision boundary.

format: plain


text:
# Foncteurs en Grassmanniennes filtration de Krull et cohomologie des foncteurs

Let V denote a vector space over a finite field k and let Gr(V) be the Grassmannian of subspaces in V. Basic objects of study in this book are various categories of functors (called grassmannian functor categories) from the category of couples (V,W), where W belongs to Gr(V), to the category ₣ of vector spaces over k. The book contains a study of finite objects in these categories and their homological properties. General vanishing properties are proved, together with an application of grassmannian functor categories to the Krull filtration of the category of functors of the category ₣. Special attention is given to the case of the basic field k with two elements (in this case it is possible to prove Noetherian properties of studied functors).

Reviewer:
vs
Book details
Author:  Publisher:
Published:
2007
ISBN:
978-2-85629-248-8
Price:
EUR 37
Categorisation


text:
# Localization and chiral symmetry in three flavor domain wall QCD

David Antonio, Kenneth Bowler, Peter Boyle, Norman Christ, Michael Clark, Saul Cohen, Chris Dawson, Alistair Hart, Balint Joó, Chulwoo Jung, Richard Kenway, Shu Li, Meifeng Lin, Robert Mawhinney, Christopher Maynard, Shigemi Ohta, Robert Tweedie, Azusa Yamaguchi

Research output: Contribution to journalArticlepeer-review

## Abstract

We present results for the dependence of the residual mass of domain wall fermions (DWF) on the size of the fifth dimension and its relation to the density and localization properties of low-lying eigenvectors of the corresponding hermitian Wilson Dirac operator relevant to simulations of 2+1 flavor domain wall QCD. Using the DBW2 and Iwasaki gauge actions, we generate ensembles of configurations with a $16^3\times 32$ space-time volume and an extent of 8 in the fifth dimension for the sea quarks. We demonstrate the existence of a regime where the degree of locality, the size of chiral symmetry breaking and the rate of topology change can be acceptable for inverse lattice spacings $a^{-1} \ge 1.6$ GeV.
Original language English 014509 21 Physical Review D 77 1 https://doi.org/10.1103/PhysRevD.77.014509 Published - 22 Jan 2008

• hep-lat

## Fingerprint

Dive into the research topics of 'Localization and chiral symmetry in three flavor domain wall QCD'. Together they form a unique fingerprint.


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# A Bullet of Mass 0.012 Kg and Horizontal Speed 70 M S–1 Strikes a Block of Wood of Mass 0.4 Kg and Instantly Comes to Rest with Respect to the Block. the Block is Suspended from the Ceiling by Means of Thin Wires. Calculate the Height to Which the Block Rises. Also, Estimate the Amount of Heat Produced in the Block. - Physics

A bullet of mass 0.012 kg and horizontal speed 70 m s–1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

#### Solution 1

Mass of the bullet, m = 0.012 kg

Initial speed of the bullet, ub = 70 m/s

Mass of the wooden block, M = 0.4 kg

Initial speed of the wooden block, uB = 0

Final speed of the system of the bullet and the block = ν

Applying the law of conservation of momentum:

mub + MuB = (m + Mv

0.012 × 70 + 0.4 × 0 = (0.012 + 0.4) v

∴ v = 0.84 / 0.412 = 2.04 m/s

For the system of the bullet and the wooden block:

Mass of the system, m' = 0.412 kg

Velocity of the system = 2.04 m/s

Height up to which the system rises = h

Applying the law of conservation of energy to this system:

Potential energy at the highest point = Kinetic energy at the lowest point

m'gh = (1/2)m'v2

∴ h = (1/2)(v2 / g)

= (1/2) × (2.04)2 / 9.8

= 0.2123 m

The wooden block will rise to a height of 0.2123 m.

Heat produced = Kinetic energy of the bullet – Kinetic energy of the system

= (1/2) mu2 - (1/2) m'v2 = (1/2) × 0.012 × (70)2 - (1/2) × 0.412 × (2.04)2

= 29.4 - 0.857 = 28.54 J

#### Solution 2

Here, m1 = 0.012 kg, u1 = 70 m/s

m2 = 0.4 kg, u2 = 0

As the bullet comes to rest with respect to the block, the two behave as one body. Let v be the velocity acquired by the combination.

Applying principle of conservation of linear momentum, (m1 + m2) v = m1H1 + m2u2 = m1u1

v = (m_1u_1)/(m_1+m_2) = (0.012xx70)/(0.012+0.4) = (0.84)/0.412 = 2.04 ms^(-1)

Let the block rise to a height h

P.E. of the combination  =  K.E. of the combination

(m_1+m_2)gh = 1/2(m_1+m_2)v^2

h = v^2/(2g) = (2.04xx2.04)/(2xx9.8) = 0.212 m

For calculating heat produced, we calculate energy lost (W), where

W = intial K.E. of bullet - final K.E of combination

= 1/2m_1u_1^2 - 1/2(m_1+m_2)v^2

1/2xx0.012(70)^2 - 1/2(0.412)(2.04)

W = 29.4 - 0.86 = 28.54 joule

:. Heat produced H = W/J = (28.54)/4.2 = 6.8 cal

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 6 Work, Energy and Power
Q 24 | Page 137


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# Weird Bacteria-Procon Junior 2017

Can someone please tell me the reason of the following losing condition in problem WEIRD of PROCON Junior 2017

when both a and b are even and their difference is 2

It is clear that when a=b it is always possible for Second Player to reach same condition a=b irrespective of the first player move.

@vijju123 please use this one : https://www.codechef.com/PROJ2017/problems/WEIRD

i also want to know the solution but does not have enough karma to ask!

Okay, i will work for a solution

Thanks for dhruv and alax for their views. I seriously felt that answer was not upto the mark after reading that. Your approaches were really awesome

I think I have an answer. Here, we define a winning position to be when the player who will play next has a confirmed strategy to win.

I assume you know why (k, k) is a losing position. So Bran tries to convert the two testtubes into a (k, k) form. So, let him add x bactera from TT B to TT A. Here, for simplicity bacteria in TT B is larger than bacteria in TT A. So,

a + {x/2} = b - x
b - a = x + {x/2}

Now, look at the values of x + {x/2}, you’ll observe they are always 0, 1 \pmod{3}. Hence, if a - b \equiv 0, 1 \pmod{3} we are done.

Now if a-b = 2 \pmod {3} for x > 2, we’ll try to prove that we can turn it into a set of form (k, k+2) which is also a losing position for other. This is always possible as b-a-2 \equiv 0 \pmod{3} and we can find an x for this and simply remove x from TT B (try to think why this works).

Now, here is the heart of the question, why is the exception when b - a = 2?
The answer: because here b - a - 2 = 0, the minimum we can take is 1.

If you have any doubts, feel free to ask.

1 Like

@vijju123 I’ve added a solution. Could you review it?

I think you might need to expand this line a bit -

Now, look at the values of x+x/2, you'll observe they are always 0,1(mod3). Hence, if a−b≡0,1(mod3)a−b≡0,1(mod3) we are done.


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# Math Help - what is the appropriate approach this time!?

1. ## what is the appropriate approach this time!?

hi all and especially mr_fantastic! remember the post "what is the...". that one was the case of Bayesian theorem. now what about this one?

Consider two players A and B who are accompanying in a sequence of independent games. Probability of A win is p^2, probability of B win is q^2, and probability of equality (neither A nor B wins) is 2pq (p+q=1 and 0<p<1). At each run, the winner gets 2 points and in the situation of equality each player gets 1 point. Suppose X and Y are the total points of relatively A and B after the end of n runs of game. What is COV(X, Y)?
ps. power is denoted by ^.
{hence my absolute unfamiliarity with posting rules!!}

2. Originally Posted by alv
hi all and especially mr_fantastic! remember the post "what is the...". that one was the case of Bayesian theorem. now what about this one?

Consider two players A and B who are accompanying in a sequence of independent games. Probability of A win is p^2, probability of B win is q^2, and probability of equality (neither A nor B wins) is 2pq (p+q=1 and 0<p<1). At each run, the winner gets 2 points and in the situation of equality each player gets 1 point. Suppose X and Y are the total points of relatively A and B after the end of n runs of game. What is COV(X, Y)?

ps. power is denoted by ^.
{hence my absolute unfamiliarity with posting rules!!}
Cov(X, Y) = E(XY) - E(X) E(Y).

Define the following random variables:

$W_A$ is the number of games won by player A. Then $W_A$ ~ Binomial (n, $p^2$).

$W_B$ is the number of games won by player B. Then $W_B$ ~ Binomial (n, $q^2$).

$D_A = D_B$ is the number of games drawn by both players.

Note that:

1. $W_A + W_B + D_A = n$.

2. The joint distribution of $W_A$, $W_B$ and $D_A$ is multinomial.

From 1:

$X = 2W_A + D_A = W_A - W_B + n$.

$Y = 2W_B + D_B = W_B - W_A + n$.

Note that:

3. $X + Y = 2n \Rightarrow (X + Y)^2 = 4n^2 \Rightarrow X^2 + 2XY + Y^2 = 4n^2$.

$E(X) = E(W_A - W_B + n) = E(W_A) - E(W_B) + n = np^2 - nq^2 + n = 2np$.

$E(Y) = E(W_B - W_A + n) = E(W_B) - E(W_A) + n = 2nq$.

From 3: $E(XY) = 2n^2 - \frac{1}{2} E(X^2) - \frac{1}{2} E(Y^2)$.

But $E(X^2) = Var(X) + [E(X)]^2$ and:

* $Var(X) = Var(W_A - W_B + n) = Var(W_A) + Var(W_B) + 2 Cov(W_A, W_B)$.

* $Cov(W_A, W_B) = -np^2 q^2$ (this is a standard result following from 2).

etc. (the details are left left for you).

3. ## thanks but a general question?

by the way where can i find help on how to use math tags like what you guys do? it's really hard to explain the problems in plain text...those math tags are necessary. i have no information on how to use those tags...where can i look?

4. Hello
Originally Posted by alv
by the way where can i find help on how to use math tags like what you guys do? it's really hard to explain the problems in plain text...those math tags are necessary. i have no information on how to use those tags...where can i look?
Take a look at this thread in the Latex Help forum.

5. Originally Posted by mr fantastic
Cov(X, Y) = E(XY) - E(X) E(Y).

Define the following random variables:

$W_A$ is the number of games won by player A. Then $W_A$ ~ Binomial (n, $p^2$).

$W_B$ is the number of games won by player B. Then $W_B$ ~ Binomial (n, $q^2$).

$D_A = D_B$ is the number of games drawn by both players.

Note that:

1. $W_A + W_B + D_A = n$.

2. The joint distribution of $W_A$, $W_B$ and $D_A$ is multinomial.

From 1:

$X = 2W_A + D_A = W_A - W_B + n$.

$Y = 2W_B + D_B = W_B - W_A + n$.

Note that:

3. $X + Y = 2n$

[snip]
A shorter approach might now be to note that $X + Y = 2n \Rightarrow Var(X + Y) = Var(2n) = 0$.

But $Var(X + Y) = Var(X) + Var(Y) + 2 Cov(X, Y)$.

Therefore:

$Var(X) + Var(Y) + 2 Cov(X, Y) = 0$

$\Rightarrow Cov(X, Y) = -\frac{1}{2} (Var(X) + Var(Y))$.

Now note that:

(a) $Var(X) = Var(W_A - W_B + n) = Var(W_A) + Var(W_B) + 2 Cov(W_A, W_B)$

$= n p^2 (1 - p^2) + n q^2 (1 - q^2) + 2 Cov(W_A, W_B)$

$= {\color{red}2npq(1-pq)} + 2 Cov(W_A, W_B)$.

(You might want to check the simplification in red).

(b) $Var(Y) = Var(W_B - W_A + n) = Var(W_A) + Var(W_B) + 2 Cov(W_A, W_B) = Var(X)$.

(c) $Cov(W_A, W_B) = -np^2 q^2$ (this is a standard result following from the fact that the joint distribution of $W_A$, $W_B$ and $D_A$ is multinomial).

Therefore Cov(X, Y) = .........

6. Originally Posted by mr fantastic
Cov(X, Y) = E(XY) - E(X) E(Y).

Define the following random variables:

$W_A$ is the number of games won by player A. Then $W_A$ ~ Binomial (n, $p^2$).

$W_B$ is the number of games won by player B. Then $W_B$ ~ Binomial (n, $q^2$).

$D_A = D_B$ is the number of games drawn by both players.

Note that:

1. $W_A + W_B + D_A = n$.

2. The joint distribution of $W_A$, $W_B$ and $D_A$ is multinomial.

From 1:

$X = 2W_A + D_A = W_A - W_B + n$.

$Y = 2W_B + D_B = W_B - W_A + n$.

Note that:

3. $X + Y = 2n$
[snip]
One more thought:

Therefore Cov(X, Y) = Cov(X, 2n - X) = .......


text:
## Learning and Clustering Plan Abstractions to Improve Hierarchical Planning

• Hierachical planning can be improved by explanation-based learning (EBL) of abstract plans from detailed, successfully solved planning problems. Abstract plans, expressed in well-established terms of the domain, serve as useful problem decompositions which can drastically reduce the planning complexity. The learned plan abstraction must be valid for a class of planning cases rather than for a single case, to ensure their successful application in a larger spectrum of new situations. A hierarchical organization of the newly learned knowledge must be archieved to overcome the utility problem in EBL. This paper presents a new formal model of shared plan abstraction and the closely related explanation-based procedure S-PABS. Unlike other apporaches to plan abstraction, our model allows a total different terminology to be introduced at the abstract level. Finally, an unsupervised incremental procedure for constructing a hierachy of shared abstract plans is proposed, as a kind of concept formation over explanations.

$Rev: 13581$


text:
Journal article

### Observation of new states decaying into Λc+π-π+

Abstract:
The spectrum of charmed baryons consisting of a charmed quark and two light quarks denoted by Ac and σc baryons were investigated. The sample used in the investigation corresponds to an integrated luminosity 13.7/fb from the data taken on the Y(4S) resonance and in the continuum at energies just below the Y(4S). A broad state with an invariant mass roughly above 480 MeV above that of the Ac baryon and a narrow state with an invariant mass of 596 MeV above the Ac mass were found.

### Access Document

Publisher copy:
10.1103/PhysRevLett.86.4479

### Authors

Boulahouache, C More by this author
Dambasuren, E More by this author
Journal:
Physical Review Letters
Volume:
86
Issue:
20
Pages:
4479-4482
Publication date:
2001-05-14
DOI:
EISSN:
1079-7114
ISSN:
0031-9007
URN:
Source identifiers:
321868
Local pid:
pubs:321868
Language:
English


text:
How do you solve 6 - sqrt (3x) = -3?

Feb 17, 2016

$x = 27$

Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} 6 - \sqrt{3 x} = - 3$

Adding $\left(\sqrt{3 x} + 3\right)$ to both sides (and exchanging the sides)
$\textcolor{w h i t e}{\text{XXX}} \sqrt{3 x} = 9$

Squaring both sides:
$\textcolor{w h i t e}{\text{XXX}} 3 x = 81$

Dividing both sides by $3$
$\textcolor{w h i t e}{\text{XXX}} x = 27$

Feb 17, 2016

$x = 27$

Explanation:

First, you isolate the root in one side of the equation by subtracting 6 to both sides:

$6 - \sqrt{3 x} = - 3$

$- \sqrt{3 x} = - 3 - 6 = - 9$

You can eliminate the minus sines by multiplicating both side by -1:

$\sqrt{3 x} = 9$

The next step is to square both sides of the equation. This can add false solutions to the equation, so be aware that the dominion is x>0.

${\left(\sqrt{3 x}\right)}^{2} = {9}^{2}$

$3 x = 81$

divide both sides by 3, you will get the answer:

$x = 27$ (is larger than zero, so is a valid solution)

Feb 17, 2016

You must isolate the radical to one side of the equation before squaring to get rid of it.

Explanation:

It would probably be easier to send the radical to the other side and the 3 to be with the 6, because it is simpler to work with a positive radical.

$6 + 3 = \sqrt{3 x}$

$9 = \sqrt{3 x}$

Now we square to eliminate the square root:

${\left(9\right)}^{2} = {\left(\sqrt{3 x}\right)}^{2}$

$81 = 3 x$

$27 = x$

So, x = 27.

N.B: Always check your solutions in the original equation. When dealing with more complex questions, usually the type where there are two solutions, oftentimes only one or neither solution will work. Solutions that don't work in the original equation are called extraneous solutions.

Practice exercises:

1. Solve for x. Beware of extraneous solutions.

a) $\sqrt{2 x + 3} = 5$

b) $\sqrt{3 x + 1} = x - 1$

c) $\sqrt{3 x} - \sqrt{2 x + 1} = 1$

d) $\sqrt{12 x + 1} + \sqrt{17 x - 4} = 15$

Good luck!


text:
Calculus: substitution for multiple variables

Problem
I have a question about variable substitution for a multivariate integral. Let me first pose the question and then provide some context, since I think the context is not necessarily very important.

Say we have (for example) the following integral $$\int_0^1 \int_0^{1-\varepsilon_1} \int_0^{1-\varepsilon_1 -\varepsilon_2} 24 (1- \varepsilon_1 - \varepsilon_2 - \varepsilon_3) d\varepsilon_3 d\varepsilon_2 d\varepsilon_1= 1$$

Furthermore, let $z = \varepsilon_1 + \varepsilon_2 + \varepsilon_3$. I'm trying to find $f(z)$ satisfying $$\int_0^1 f(z) dz = 1$$ corresponding to the integral above. At first glance, this seems like an easy problem, but whatever I do, I cannot seem to eliminate all the epsilon!

I suppose there are three cases:

$$\text{substituting } \varepsilon_1 = z - \varepsilon_2 - \varepsilon_3: \qquad \int_{\varepsilon_2 + \varepsilon_3}^{1+\varepsilon_2 + \varepsilon_3} \int_0^{1-\varepsilon_1} \int_{0}^{1- \varepsilon_1 - \varepsilon_2} 24 (1- z) d\varepsilon_3 d\varepsilon_2 dz$$

$$\text{substituting } \varepsilon_2 = z - \varepsilon_1 - \varepsilon_3: \qquad \int_0^1 \int_{\varepsilon_1 + \varepsilon_3}^{1+\varepsilon_3} \int_{0}^{1- \varepsilon_1 - \varepsilon_2} 24 (1- z) d\varepsilon_3 dz d\varepsilon_1$$

$$\text{substituting } \varepsilon_3 = z - \varepsilon_1 - \varepsilon_2: \qquad \int_0^1 \int_0^{1-\varepsilon_1} \int_{\varepsilon_1 + \varepsilon_2}^{1} 24 (1- z) dz d\varepsilon_2 d\varepsilon_1$$

The first two integrals will actually not evaluate to a numerical value. Only the last one will, but $z$ is eliminated in the first step (which is not what I want of course). So my question is: can I partially solve this integral in a way that leaves me with a function $f(z)$?

Context
I am trying to compute the convolution of a linear combination of three dependent variables $\varepsilon_1$, $\varepsilon_2$ and $\varepsilon_3$. So I am trying to determine the distribution of $\alpha_1 \varepsilon_1 + \alpha_2 \varepsilon_2 + \alpha_3 \varepsilon_3 + \cdots + \alpha_n \varepsilon_n$. I was able to determine the joint distribution $f(\alpha_1 \varepsilon_1, \ldots, \alpha_n \varepsilon_n)$, so I really only have the integration problem left to solve.

• Really confusing. What means "corresponding to the integral above"? – Martín-Blas Pérez Pinilla May 12 '18 at 11:11

The expression $$\int_{\varepsilon_2+\varepsilon_3}^{1+\varepsilon_2+\varepsilon_3} \int_0^{1-\varepsilon_1}\int_0^{1-\varepsilon_1-\varepsilon_2} 24(1 - z)\,d\varepsilon_3\,d\varepsilon_2\,dz$$ is utterly nonsensical. The external integration limits must be constant.
If you do the change of variable $$\epsilon_1 = z - \epsilon_2 - \epsilon_3,$$ $$\epsilon_2 = \epsilon_2,$$ $$\epsilon_3 = \epsilon_3,$$ in the original integral, as the Jacobian is 1, after the change the new integral in the order $d\varepsilon_3\,d\varepsilon_2\,dz$ is $$\int_0^1\int_0^z\int_0^{z-\varepsilon_2}24(1 - z)\,d\varepsilon_3\, d\varepsilon_2\,dz = \int_0^1 12(z^2 - z^3)\,dz = 1.$$
• Thanks for your answer! I understand that the expression with non-constant external limits is nonsense, but I hoped it made it easier to spot my (faulty) rationale. The solution you provide makes sense, but I still don't understand how you obtained the integration bounds. For example, if I use your change of variable, when I compute the new bounds for $\varepsilon_1$ then I get: (lower bound) $z = 0 + \varepsilon_2 + \varepsilon_3 = \varepsilon_2 + \varepsilon_3$ and (upper bound) $z = 0 + \varepsilon_2 + \varepsilon_3 = \varepsilon_2 + \varepsilon_3$. – Lerak May 14 '18 at 9:36
• @Karel, $0\le z = (\cdots)\le 1$ is obvious. Fixed $z$, use that $0\le\epsilon_2\le z$. Finally, $\epsilon_ 2 + \epsilon_3\le z$. – Martín-Blas Pérez Pinilla May 14 '18 at 10:02
• @Karel, $\epsilon_1$ isn't among the new variables. – Martín-Blas Pérez Pinilla May 14 '18 at 10:04


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## Floodfilling a texture in HLSL Compute shader

I have a very large texture which I want to fill with values representing "distance in units from a river tile". The texture is already seeded with these origin river points (meaning distance/value = 0 in them).

From these points I want to floodfill outwards, incrementing the value with each step from these origin points.

Doing this on the CPU is no problem using a stack or similiar structure but ideally I want to do this in the middle of a Compute shader execution which runs over the entire texture.

I have read this which sound similiar to what I want to do but it mentions there might be a smarter way to do this with Compute shaders – which is what I am using.

Any ideas on how to solve this on compute shaders?

## How to compute matrices in Einstein notation:

I would like to compute $$r_{ii’}=\sum_{kk’}A_{ii’}^{kk’}r_{kk’}$$ where all indices vary from 0 to 1 to yield a vector / matrix.

## Evaluate the limit of a subsequence: Compute the limit $\lim_{n\to\infty} \cos\left( \pi \sqrt{4n^2 + 5n + 1} \right)$ Integer $n$ on Mathematica

I want to compute the limit $$\lim \limits_{n\to\infty} \cos\left( \pi \sqrt{4n^2 + 5n + 1} \right)$$for integer $$n$$. By completing the square, we can determine that this limit is equal to $$– \tfrac1{\sqrt2} \approx -0.7071$$.

But if we don’t restrict $$n$$ to an integer, then the limit is indeterminate / does not exist. And can be easily found by typing it on WolframAlpha. Or in Mathematica:

However, I do not know how to compute the limit (on Mathematica) with the original constraint that $$n$$ must be an integer.

I know that we can plot a graph on Mathematica:

The graph suggests that the limit is equal to $$-\tfrac1{\sqrt2}$$. However, this doesn’t look like a convincing result because we can’t know that the limit is exactly equal to $$-\tfrac1{\sqrt2}$$.

Question: Is there a way to compute this limit in Mathematica where it spits out a single numerical value (of $$-1/{\sqrt2}$$)?

## How to compute a vector V when N points are given and V satisfies given properties

We are given N points P1,P2,…,PN in a 2D plane(All points are distinct and N is as large as 10^5). For each valid i, the coordinates of the point Pi are (xi,yi). Help me to find a vector V = (a, b) ( where |a|, |b| <= 1e9) such that the following holds:

For each i (1 ≤ i ≤ N), let Si= dot(V, G(Pi, Pi+1)). lets assume PN+1=P1. where G(v1, v2) = ((v2(x) – v1(x)), (v2(y) – v1(y)) and dot(V1, V2) denotes dot product of two vectors

How to choose V such that It is possible to find two integers l and r (1 ≤ l ≤ r ≤ N) such that:

Si < 0 if(i <= r and i >= l) and Si > 0 otherwise

(or)

Si > 0 if(i <= r and i >= l) and Si < 0 otherwise

I need to know if there is a way of choosing vector (a, b) to satisfy the above conditions(If the solution is possible)

## Compute duration of an event based on consecutive values in another column

I need to compute the sum of durations of multiple events based on consecutive values in another column.

             ts             | w ----------------------------+---  2020-07-27 15:40:04.045+00 | t  2020-07-27 15:41:04.045+00 | t  2020-07-27 15:41:14.045+00 | f  2020-07-27 15:42:14.045+00 | t  2020-07-27 15:43:14.045+00 | t

The event is considered being active as long as the column w has a consecutive value of true.

The duration of the first event would be 60 seconds. '2020-07-27 15:41:04.045+00' - 2020-07-27 15:40:04.045+00. The second event has the same duration. The sum of both would be 120 seconds.

What would be the best/most performant approach to computing these values? The longest time range we’ll be looking at will probably be half a year involving about 30 million rows.

I wrote a custom aggregate function that computes the duration but it takes about 16 seconds for only 1.5 million rows.

 Aggregate  (cost=444090.03..444090.04 rows=1 width=4) (actual time=16290.826..16290.828 rows=1 loops=1)    ->  Seq Scan on discriminator0  (cost=0.00..57289.03 rows=1547203 width=9) (actual time=0.016..1723.178 rows=1547229 loops=1)  Planning Time: 0.196 ms  JIT:    Functions: 3    Options: Inlining false, Optimization false, Expressions true, Deforming true    Timing: Generation 0.889 ms, Inlining 0.000 ms, Optimization 0.508 ms, Emission 6.472 ms, Total 7.870 ms  Execution Time: 16291.836 ms

I’m new to SQL and basically just got this working through trial and error, so I’m sure there is room for improvement. Maybe even a whole different approach. Here’s a fiddle with the aggregate function.

I’m not sure if i should include the code because it’s quite long

## Given $n$ unique items and an $m^{th}$ normalised value, compute $m^{th}$ permutation without factorial expansion

We know that the number of permutations possible for $$n$$ unique items is $$n!$$. We can uniquely label each permutation with a number from $$0$$ to $$(n!-1)$$.

Suppose if $$n=4$$, the possible permutations with their labels are,

0:  1234 1:  1243 2:  1324 3:  1342 4:  1432 5:  1423 6:  2134 7:  2143 8:  2314 9:  2341 10: 2431 11: 2413 12: 3214 13: 3241 14: 3124 15: 3142 16: 3412 17: 3421 18: 4231 19: 4213 20: 4321 21: 4312 22: 4132 23: 4123

With any well defined labelling scheme, given a number $$m, 0 \leq m < n!$$, we can get back the permutation sequence. Further, these labels can be normalised to be between $$0$$ and $$1$$. The above labels can be transformed into,

0:       1234 0.0434:  1243 0.0869:  1324 0.1304:  1342 0.1739:  1432 0.2173:  1423 0.2608:  2134 0.3043:  2143 0.3478:  2314 0.3913:  2341 0.4347:  2431 0.4782:  2413 0.5217:  3214 0.5652:  3241 0.6086:  3124 0.6521:  3142 0.6956:  3412 0.7391:  3421 0.7826:  4231 0.8260:  4213 0.8695:  4321 0.9130:  4312 0.9565:  4132 1:       4123

Now, given $$n$$ and $$m^{th}$$ normalised label, can we get the $$m^{th}$$ permutation while avoiding the expansion of $$n!$$ ? For example, in the above set of permutations, if we were given the $$m^{th}$$ normalised label to be $$0.9$$, is it possible to get the closest sequence 4312 as the answer without computing $$4!$$ ?

## How to compute the number of occurrences of a cron schedule over a time span?

For example, given a cron schedule such as: 0 * 1 7 6 and a time span, like the next year, can you directly compute the number of times the schedule will fire? It can be brute forced pretty trivially but I’d like to use this to sort a list so I’d rather not just brute force it. It’s pretty easy to compute the trivial cron patterns:

fn('* * * * *', 1 day) -> 60 * 24 fn('0,30 * * * *', 1 day) -> 2 * 24

It’s easy up until the calendar gets involved with different length months and then you have to worry about days or the week coinciding with days of the month.

## Point Subdomain in Google Domains to Google Compute Instance running WordPress

I have installed wordpress (the bitnami multi-site version) to a compute instance from the marketplace. I’ve also recently moved my domain registration to domains.google.com. Everything is working fine with the top-level domain. I also set up a subdomain, wp.mydomain.com in the domains interface pointed to the static external IP I configured for the compute instance running wordpress. When I go to wp.mydomain.com it redirects me to http://.xip.io/.

How do I get this set up so that I stay on wp.mydomain.com instead of going to my ip address as the url?

## Write a program that accepts the radius of a circle in the client side of a program and compute the area and perimeter in the server side program

Write a program that accepts the radius of a circle in the client side of a program and compute the area and perimeter in the server side program and return the values to the client program.

// Client   import java.io.*; import java.net.*;  public class Client {             public static void main(String[] args) throws IOException{                 Socket s = new Socket("localhost",1234);                 BufferedReader br;                 PrintStream ps;   String str;                System.out.println("Enter Radius  :");                 br=new BufferedReader(new InputStreamReader(System.in));    ps=new PrintStream(s.getOutputStream());                ps.println(br.readLine());    br=new BufferedReader(new InputStreamReader(s.getInputStream()));                str=br.readLine();                System.out.println("Area of the circle is : "+str);                 br.close();                 ps.close();            } }
// Server  import java.io.*; import java.net.*;  class Server {       public static void main(String[] args){           try {           ServerSocket ss = new ServerSocket(1234);                  System.out.println("Waiting for Client Request");                   Socket s=ss.accept();                   BufferedReader br;                   PrintStream ps;                   String str;                   br=new BufferedReader(new InputStreamReader(s.getInputStream()));                   str=br.readLine();                   double r=Double.parseDouble(str);                   double area=3.14*r*r;                   ps=new PrintStream(s.getOutputStream());                   ps.println(String.valueOf(area));                   br.close();                   ps.close();                   s.close();                   ss.close();        }           catch(Exception e)           {                       System.out.println(e);           }  } }

how to add perimeter of a circle

## Compute Structure Tensor

As i know, Structure Tensor is:

How i compute $$I_x^2, I_y^2, I_xI_y$$. I used Hadamard product for compute $$I_x^2, I_y^2, I_xI_y$$. But if i use Hadamard product i will get $$det(M) = 0$$. I’m stuck in here for too long. Can anyone explain for me?. Thank you.


text:
## Tuesday, June 21, 2016

### DEFINITIONS – ALGORITHM & LEMMA

DEFINITIONS – ALGORITHM & LEMMA

ALGORITHM
An algorithm is a series of well defined steps which gives a procedure for solving a type of problem. The word algorithm comes from the name of the 9th century Persian mathematician al-Khwarizmi. In fact, even the word ‘algebra’ is derived from a book, he wrote, called Hisab al-jabr w’al-muqabala.

LEMMA

A lemma is a proven statement used for proving another statement.


text:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically

### Ratio Sudoku 2

##### Stage: 3 and 4 Challenge Level:

A Sudoku with clues as ratios.

### Intersection Sudoku 1

##### Stage: 3 and 4 Challenge Level:

A Sudoku with a twist.

##### Stage: 3 and 4 Challenge Level:

Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.

### Ratio Sudoku 1

##### Stage: 3 and 4 Challenge Level:

A Sudoku with clues as ratios.

### Tetrahedra Tester

##### Stage: 3 Challenge Level:

An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?

### Wallpaper Sudoku

##### Stage: 3 and 4 Challenge Level:

A Sudoku that uses transformations as supporting clues.

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##### Stage: 2, 3 and 4 Challenge Level:

Solve this Sudoku puzzle whose clues are in the form of sums of the numbers which should appear in diagonal opposite cells.

### Twin Corresponding Sudoku

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This sudoku requires you to have "double vision" - two Sudoku's for the price of one

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This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it?

### Intersection Sudoku 2

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A Sudoku with a twist.

### Twin Corresponding Sudoku III

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Two sudokus in one. Challenge yourself to make the necessary connections.

### Instant Insanity

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Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.

### Intersection Sums Sudoku

##### Stage: 2, 3 and 4 Challenge Level:

A Sudoku with clues given as sums of entries.

### Ratio Sudoku 3

##### Stage: 3 and 4 Challenge Level:

A Sudoku with clues as ratios or fractions.

### When Will You Pay Me? Say the Bells of Old Bailey

##### Stage: 3 Challenge Level:

Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?

### You Owe Me Five Farthings, Say the Bells of St Martin's

##### Stage: 3 Challenge Level:

Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?

### Corresponding Sudokus

##### Stage: 3, 4 and 5

This second Sudoku article discusses "Corresponding Sudokus" which are pairs of Sudokus with terms that can be matched using a substitution rule.

### Twin Corresponding Sudokus II

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Two sudokus in one. Challenge yourself to make the necessary connections.

### Product Doubles Sudoku

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Each clue number in this sudoku is the product of the two numbers in adjacent cells.

### Integrated Sums Sudoku

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The puzzle can be solved with the help of small clue-numbers which are either placed on the border lines between selected pairs of neighbouring squares of the grid or placed after slash marks on. . . .

### Rectangle Outline Sudoku

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Each of the main diagonals of this sudoku must contain the numbers 1 to 9 and each rectangle width the numbers 1 to 4.

### Colour Islands Sudoku 2

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In this Sudoku, there are three coloured "islands" in the 9x9 grid. Within each "island" EVERY group of nine cells that form a 3x3 square must contain the numbers 1 through 9.

### Gr8 Coach

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Can you coach your rowing eight to win?

### Bochap Sudoku

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This Sudoku combines all four arithmetic operations.

### Making Maths: Double-sided Magic Square

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Make your own double-sided magic square. But can you complete both sides once you've made the pieces?

### 1 to 8

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Place the numbers 1 to 8 in the circles so that no consecutive numbers are joined by a line.

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A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour.

### Cayley

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The letters in the following addition sum represent the digits 1 ... 9. If A=3 and D=2, what number is represented by "CAYLEY"?

### Ones Only

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Find the smallest whole number which, when mutiplied by 7, gives a product consisting entirely of ones.

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A few extra challenges set by some young NRICH members.

### Counting on Letters

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The letters of the word ABACUS have been arranged in the shape of a triangle. How many different ways can you find to read the word ABACUS from this triangular pattern?

### Counters

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Hover your mouse over the counters to see which ones will be removed. Click to remover them. The winner is the last one to remove a counter. How you can make sure you win?

### One to Fifteen

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Can you put the numbers from 1 to 15 on the circles so that no consecutive numbers lie anywhere along a continuous straight line?

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Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?

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You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?

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If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?

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Four small numbers give the clue to the contents of the four surrounding cells.

### Nine-pin Triangles

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How many different triangles can you make on a circular pegboard that has nine pegs?

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Can you put the 25 coloured tiles into the 5 x 5 square so that no column, no row and no diagonal line have tiles of the same colour in them?

### Masterclass Ideas: Working Systematically

##### Stage: 2 and 3 Challenge Level:

A package contains a set of resources designed to develop students’ mathematical thinking. This package places a particular emphasis on “being systematic” and is designed to meet. . . .

### Oranges and Lemons, Say the Bells of St Clement's

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Bellringers have a special way to write down the patterns they ring. Learn about these patterns and draw some of your own.

### More on Mazes

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There is a long tradition of creating mazes throughout history and across the world. This article gives details of mazes you can visit and those that you can tackle on paper.

### A Square of Numbers

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Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?

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Can you find all the different triangles on these peg boards, and find their angles?

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Can you find all the different ways of lining up these Cuisenaire rods?

### Fault-free Rectangles

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Find out what a "fault-free" rectangle is and try to make some of your own.

### Triangles to Tetrahedra

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Starting with four different triangles, imagine you have an unlimited number of each type. How many different tetrahedra can you make? Convince us you have found them all.

### Isosceles Triangles

##### Stage: 3 Challenge Level:

Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?


text:
# A Future of the electron neutrino mass limits

#### exponent137

Summary
Upper bound of the electron neutrino mass was calculated, 0,086 eV. https://arxiv.org/abs/1811.02578 This team also plans to calculate the lower bound of the electron neutrino mass. It is interesting what is the future of these calculations and measurements.
In five years also experiment KATRIN will give either the upper bound of electron neutrino mass (0,2 eV) or even the mass of the electron neutrino. https://www.katrin.kit.edu/

My question is, what we can expect from the astronomical and non-astronomical measurements to improve these data? I suppose that measurements of gravitational waves will give new data, LIGO, VIRGO, KAGRA, LISA? I suppose that James Webb telescope will give new data, useful for neutrino mass? Then, IceCube neutrino observatory is useful? There are also neutrino oscillations measurements, as NOvA and Hyper-Kamiokande.

Which measurements are the most promising and which are less promising, according to neutrinos rest masses?

Last edited:
Related Astronomy and Astrophysics News on Phys.org

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
There is no such thing as the "mass of an electron neutrino". The electron neutrino is not a mass eigenstate. Instead what these experiments measure is an effective neutrino mass, which is a particular combination of the masses of the neutrino mass eigenstates. The exact combination depends on the type of experiment performed. For example, cosmology is typically sensitive to the sum of neutrino masses, whereas beta decay experiments target a particular combination involving the lepton mixing matrix. Neutrino oscillations are only sensitive to the mass squared differences, not to the masses themselves. Neutrinoless double beta decay experiments target yet another combination of the masses.

Neutrino telescopes such as IceCube are sensitive mainly to high-energy neutrinos, where the masses are negligible for kinematical purposes.

Edit: grammar

Last edited:

Staff Emeritus
\Upper bound of the electron neutrino mass was calculated, 0,086 eV. https://arxiv.org/abs/1811.02578
You are misrepresenting what they wrote. Orodruin is right, there is no such thing as the "mass of an electron neutrino". Furthermore, the author of that paper never claim that is what they are calculating, and indeed, the words "electron neutrino" don't even appear.

#### exponent137

You are misrepresenting what they wrote. Orodruin is right, there is no such thing as the "mass of an electron neutrino". Furthermore, the author of that paper never claim that is what they are calculating, and indeed, the words "electron neutrino" don't even appear.
As I look now, does $m^\nu_0$ mean either $m^\nu_1$ or $m^\nu_3$, dependent on normal hierarchy (NH) or inverted hierarchy (IH)?

Staff Emeritus
Itr is defined on page 2, line 8.

You simply have to put more effort. I don't think it's PF's job to read the paper for you.

#### berkeman

Mentor

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# Tag Info

61

It has to be so common a question that the answer is actually given in various places on Dupont's own website (Dupont are the makers of Teflon): “If nothing sticks to Teflon®, then how does Teflon® stick to a pan?" Nonstick coatings are applied in layers, just like paint. The first layer is the primer—and it's the special chemistry in the primer that ...

41

First off, may I say that I applaud your decision to test this through an experiment. It's rare to see that than I would like. Now, on to the matter at hand. It's fairly well known from industrial chemistry that non-polar solvents degrade latex quite heavily. I work with latex seals a lot, and the hexanes we use routinely break the seals down in under a ...

32

Corrosion Resistant Products, Ltd., with the help of Dupont, has established this source of information on what can and cannot eat teflon. Here's a list: Sodium and potassium metal - these reduce and defluorinate PTFE, which finds use in etching PTFE Finely divided metal powders, like aluminum and and magnesium, cause PTFE to combust at high temperatures ...

32

Just to add a bit to Ben's excellent answer... A number of fluorinating agents also react with PTFE, $\ce{XeF2}$ and $\ce{CoF3}$ being examples Ben mentioned the reaction of magnesium metal. Typically with metals, they must be in intimate contact with the PTFE surface, so molten metals or metals dissolved in anhydrous solvents will react. The magnesium ...

26

Nitrile gloves are made of nitrile rubber, or poly(butadiene/acrylonitrile). This polymer is highly soluble in chloroform, with some papers I found indicating that one can dissolve up to 18% in mass of nitrile butadiene rubber in chloroform. Moreover, it permeates easily through NBR, meaning we can expect the dissolution to be fast in addition to ...

23

When you rip, tear or mechanically deform a polymer (for example a piece of plastic) you are putting energy into the material. The energy from this deformation causes the polymer chains in the vicinity of the deformation to attempt to align. To some degree this partial alignment makes continued deformation easier. To continue tearing the polymer apart you ...

23

I am aware that this answer does not describe a ‘chemical that can destroy PTFE’. However, since you are also asking ‘is there anything that can destroy Teflon through only chemical means?’, and in order to complete the other answers, I would like to add that PTFE can be easily destroyed by means of radiation chemistry. (Radiation chemistry is the study of ...

23

Kevlar is an A,B co-polymer where the monomers are terephthalic acid and 1,4-diaminobenzene. The amide linkage, together with the aromatic rings in the polymer, makes a very rigid polymeric structure that stacks the chains in an organized way, maximizing the effectiveness of hydrogen bonding between amide linkages, together with pi-stacking of the rings: ...

17

Agreed: a physical action can cause a chemical reaction. For example, any attempt to manipulate $\ce{NI3}$ physically causes it to decompose (detonate). For that matter, dividing metals into tiny pieces changes chemical behavior, because bonds that would have been hidden in the interior are exposed at the surface. Finely divided iron, for example, becomes ...

15

Based on appearance and extent of deformation, your bottle is likely made of PVC plastic which is not compatible with oils. Plastic bottles are made from blow molding. This process leaves residual stress in the polymer chains of the materials but creates a smooth and transparent surface. Lavender oil can diffuse into PVC and make it softer. Once it is ...

14

The International Chewing Gum Association tells us: Gum base is what gives chewing gum its “chew.” It is made of a combination of food-grade polymers, waxes, and softeners that give it the texture desired by consumers and enable it to effectively deliver sweetness, flavor, and various other benefits, including dental benefits. A more complete description ...

14

According to the Compendium of Polymer Terminology and Nomenclature (IUPAC Recommendations 2008, i.e. the ‘Purple Book’), three different types of names can be used for polymers: traditional names, structure-based names, and source-based names. When traditional names fit into the general pattern of systematic nomenclature, they are retained, in this case: ...

13

Polylactic acid polymers are creating a lot of interest these days because they're biodegradable. On the downside, corn starch is often one of the starting materials...and corn is also used as a food source, so there's some controversy over the wisdom of this approach. I think I've found a site that takes the approach you're basing your question on...but ...

13

You have a possible answer to your question in proteins, an example which includes some long polymer chains. Intramolecular interactions - while not necessarily the driving force for formation of a collapsed protein globule (usually argued to be due to the hydrophobic effect, requiring intermolecular interactions) - are the basis for higher order structure ...

12

Before putting Teflon on a pan, the pan is scratched hardly leaving some tiny holes where the hole opening is smaller than the hole inner size (like a bottle). When Teflon is cast into that, it physically can't get out. That's also why when you scratch a Teflon pan, things start sticking to it.

12

Teflon (PTFE, polytetrafluoroethylene) is prepared by the polymerization of tetrafluoroethylene as shown below. If we took polyethylene and replaced all of the hydrogens with fluorines, we would have PTFE. Here is a Table that compares the strengths of some single bonds involving a methyl group. \begin{array}\hline Experimental ~Bond~ Enthalpies for~ CH_3 ...

12

TL;DR I think you might need two more things to interpret this line: the first code part of the standard, and the standard itself - ASTM D2000 (needs to be purchased). These descriptors are not chemical elements, they seem to be the part of ASTM D2000 "Standard Classiﬁcation System for Rubber Products in Automotive Applications", so what you are looking at ...

12

Polyethylene is prepared by polymerizing ethylene. Ethylene has a double bond, thanks to which the polymerization goes on. You could not do that with a methylene group. So the name comes after the smallest unit which is actually used to prepare the polymer.

11

There are two ways in which the band gap of a material: using optical spectroscopy and using electrochemistry. The way you go about analyzing the data to determine the band gap is independent of material, but the process of collecting the data is very specific to your material, and you should consult papers where others have made similar measurements on ...

11

Starch is what plants primarily use as a glucose storage. As such, it is essential that they can break it back down into glucose otherwise it would defeat its own purpose and be removed by evolution very quickly. Think about it: Requiring a lot of energy to build up something that serves no further purpose — not exactly giving a plant an advantage in ...

10

There's no reason why the structural subunits of polymers could not exhibit resonance, provided there are conjugated $\pi$-bonds present. A polymer such as kevlar, for example, certainly benefits from resonance stabilization within each monomeric unit. Whether there exist any polymers in which the resonance is truly distributed over the entire molecule (if ...

10

The persistence length is a statistically defined property - in other words, it is defined by looking at many lengths of many polymer molecules over a long period of time. It is the length (L) at which a vector drawn tangent to the polymer is no longer correlated in space with another tangent vector at distance (L). In other words, it tells you how well the ...

10

Based on what I gathered from this Wikipedia article on super glue : Super glue is typically composed of a class of compounds called cyanoacrylates. Cyanoacrylates are capable of taking part in this process we call "Polymerization", where chains of cyanoacrylate molecules, called "polymers" ( Poly- Many, mere- parts) are formed. Each constituent ...

10

It is an equation useful for describing the behaviour of a large system, not predicting the fate of a few monomers. Polymerization is a stochastic process—if we only have 80 monomers, the resulting polymer molecules will vary wildly in length if we perform the experiment several times (and logically, could not ever have an average degree of polymerization of ...

10

First, it's important to define what you mean by "nylon", a term which is often used generically to refer to many different aliphatic polyamides. The most common are "nylon 6" and "nylon 6,6". Nylon 6 is the polymer of 6-aminohexanoic acid (aka 6-aminocaproic acid), while nylon 6,6 is from alternating units of 1,6-diaminohexane (aka hexamethylene diamine) ...

10

You can combine materials in two ways: chemically and mechanically. On average, if you want to the desired physical properties to add up, you want to go mechanical way, not the chemical one. Once the reaction takes place, the there is absolutely no reason to assume the product preserves properties of both precursors. It might sometimes, but more often than ...

9

This question appears to be about the preparation of polyacrylonitrile dendrimers. Each addition of acrylonitrile completely reacts away the NH2 groups previously present: assume that all reactions are complete. Your answer of 28 assumes that the reactions could be controllable to allow partial consumption of NH2 groups, which requires less addition of ...

9

This difference between living and non-living polymerizations is small in words but large in effects. In principle cationic and anionic polymerizations can be living. In practice, it is not as easy as it looks on paper. See below. A living polymerization is any chain (or addition) polymerization that is prevented from terminating, i.e. the ends of the ...

9

It's been my experience that initiators are chosen for: 1) initiation temperature, 2) solubility, and 3) cost. 1) Initiation temperature - Most radical initiators are stable at room temperature; they need to be heated to dissociate. For example, the half-life of benzoyl peroxide at room temperature is very long, but it is about 30 minutes at 92 °C and 1 ...

9

Polypropylene is hydrocarbon and does not adsorb polar and ionic compounds. So, several treatments with mild hydrochloric acid followed with several treatments with clean water should work. However, I wouldn't store drinking water there without a proven test. For example, leave water there for several weeks and then perform analysis of the water for metal ...

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


text:
## $$p$$-adic and real subanalytic sets.(English)Zbl 0693.14012

Denote by $${\mathbb{Z}}_ p\{X\}$$ the ring of restricted power series over $${\mathbb{Z}}_ p$$ in the set of variables $$X_ 1,\dots,X_ m$$. Let $$P_ n$$ be the subset of $${\mathbb{Q}}_ p^*$$ consisting of $$n$$-th powers. A basic subset of $${\mathbb{Z}}_ p^ m$$ is a subset of the type $$B=\{x\in {\mathbb{Z}}_ p^ m;\quad f(x)=0,\quad g_ 1(x)\in P_{n(1)},\dots,g_ r(x)\in P_{n(r)}\}$$ where $$f,g_ 1,\dots,g_ r\in {\mathbb{Z}}_ p\{X\}$$ and n(1),…,n(r) are positive integers. A D-function $$f:\quad {\mathbb{Z}}_ p^ m\to {\mathbb{Z}}_ p$$ is a function obtained by composing maps induced by restricted power series and the map $$D:\quad {\mathbb{Z}}_ p^ 2\to {\mathbb{Z}}_ p$$ which is defined by $$D(x,y)=x/y$$ if $$| x| \leq | y|$$ and $$y\neq 0$$ and $$D(x,y)=0$$ otherwise. Similarly as basic subsets one defines D-basic subsets. The central result of the paper is the
Elimination theorem: Let $$\pi:\quad {\mathbb{Z}}_ p^ m\to {\mathbb{Z}}_ p^ n$$ for $$m\geq n$$ be a projection. Then the image of a basic subset B in $${\mathbb{Z}}_ p^ m$$ is D-basic in $${\mathbb{Z}}_ p^ n.$$
A subset S of a $$p$$-adic manifold M is called semianalytic if each point x of M admits an open neighborhood U of x such that $$U\cap S$$ is a finite union of sets of the type $$\{y\in U;\quad f(y)=0,\quad g_ 1(y)\in P_{n(1)},\dots,g_ r(y)\in P_{n(r)}\}$$ where $$f,g_ 1,\dots,g_ r$$ are analytic functions on U and n(1),…,n(r) are positive integers. A subset S of M is called subanalytic if for each point x of M there is an open neighborhood U of x such that $$U\cap S$$ is an image of a semianalytic subset $$S'$$ of $$U\times {\mathbb{Z}}_ p^ m$$ under the projection to U.
Uniformization theorem: Let $$S\subset {\mathbb{Z}}_ p^ m$$ be subanalytic. Then there exists a compact $$p$$-adic manifold M of dimension m and an analytic map h: $$M\to {\mathbb{Z}}_ p^ m$$ such that $$h^{-1}(S)$$ is semianalytic and such that h is a composition of finitely many blowing- ups of closed submanifolds.
As applications of the uniformization theorem the authors obtain the following results:
Rationality of Poincaré series: Let $$S\subset {\mathbb{Z}}_ p^ m$$ be subanalytic and denote by N(n,S) the cardinality of the residue classes mod $$p^ n$$ which admit representatives in S. Then $$P_ S(T)=\sum_{n\in {\mathbb{N}}}N(n,S)T^ n$$ is rational.
Rationality of Łojasiewicz exponents: Let $$S\subset {\mathbb{Z}}_ p^ m$$ be closed and subanalytic and let f,g: $$S\to {\mathbb{Z}}_ p$$ be functions whose graphs are subanalytic such that $$| f|,| g|: S\to {\mathbb{R}}$$ are continuous. Suppose that $$g^{-1}(0)\subset f^{-1}(0)$$. For $$\alpha >0$$ consider the property $P(\alpha):\quad | f(x)|^{\alpha}\leq c| g(x)| \quad for\quad some\quad c\in {\mathbb{R}}\quad and\quad all\quad x\in S.$ Then there exists some $$\alpha >0$$ such that P($$\alpha)$$ is satisfied. Let $$\alpha_ 0$$ be the infimum over all $$\alpha >0$$ such that P($$\alpha)$$ is satisfied. Then $$\alpha_ 0$$ is a rational number and if $$\alpha_ 0>0$$ then $$P(\alpha_ 0)$$ is satisfied.
Reviewer: W.Lütkebohmert

### MSC:

 14Pxx Real algebraic and real-analytic geometry 14G20 Local ground fields in algebraic geometry
Full Text:


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# Setting $base_url without breaking non-production environments¶ Drupal 8 does not support $base_url. For more information, see this issue on Drupal.org.

Occasionally, it may be necessary to set the $base_url value in a Drupal application’s settings.php file, because a contributed module may require this variable to be set, or an application may have specialized configurations. Unfortunately, setting this variable can cause issues on Acquia Cloud, where there are separate Development, Staging, and Production environments. If it is set unconditionally, the $base_url variable instructs Drupal to rewrite all requests for all environments, which breaks environments to which this variable doesn’t point.

The solution is to make use of the $_ENV['AH_SITE_ENVIRONMENT'] environment variable that is set by the Acquia Cloud environment. Note for multisite users Multisite installations should set $base_url in a dynamic fashion, as outlined here. Setting the $base_url to an empty string disables caching and is not recommended. Modify and use the following example code as necessary to fit your needs, and be sure to add your Remote Administration environment if your application has one: if (isset($_ENV['AH_SITE_ENVIRONMENT'])) {
switch ($_ENV['AH_SITE_ENVIRONMENT']) { case 'dev':$base_url = 'http://dev.example.com';
break;
case 'test': $base_url = 'http://test.example.com'; break; case 'prod':$base_url = 'http://www.example.com';
break;
}
}

You can further modify the preceding code if there are environments that don’t require $base_url to be explicitly set. For example, if the $base_url variable is required only for Production, you can use the following much shorter code snippet:

if (isset($_ENV['AH_SITE_ENVIRONMENT']) &&$_ENV['AH_SITE_ENVIRONMENT'] === 'prod') {
$base_url = 'http://www.example.com'; }  Note If you set a custom $base_url, be aware that the variable should not contain a trailing slash and should always start with either http:// or https://.


text:
## Files in this item

FilesDescriptionFormat

application/pdf

9904414.pdf (7MB)
(no description provided)PDF

## Description

 Title: Interface Structure Studies of Copper Thin Films on C-Plane Sapphire by Crystal Truncation Rod Analysis Using Synchrotron X-Ray Author(s): Chung, Ki-Sup Doctoral Committee Chair(s): Chen, Haydn Department / Program: Materials Science and Engineering Discipline: Materials Science and Engineering Degree Granting Institution: University of Illinois at Urbana-Champaign Degree: Ph.D. Genre: Dissertation Subject(s): Engineering, Electronics and Electrical Abstract: Thin films were grown by UHV-MBE technique. The substrates underwent two different heat treatments prior to Cu deposition. One substrate was annealed in air at $1500\sp\circ$C for 3 hours followed by an anneal in UHV at $1200\sp\circ$C for 30 minutes prior to Cu deposition. The other substrate was not subjected to any high-temperature annealing except for an intermediate temperature cleaning treatment at $600\sp\circ$C before deposition. The CTR from a clean sapphire substrate without Cu coverage and $\rm C\sb6\sb0/\alpha$-$\rm Al\sb2O\sb3$ were measured as a reference. The measured data were fitted to the theoretical equation of diffracted crystal truncation rod intensity profile based upon the kinematical approximation. The CTR analysis method provide us the nature of the terminating layer sequence of the substrate, the structural relaxations (interatomic distance change) of the layers near the surface and the interfacial roughness. Issue Date: 1998 Type: Text Language: English Description: 150 p.Thesis (Ph.D.)--University of Illinois at Urbana-Champaign, 1998. URI: http://hdl.handle.net/2142/82905 Other Identifier(s): (MiAaPQ)AAI9904414 Date Available in IDEALS: 2015-09-25 Date Deposited: 1998


text:
DXR is a code search and navigation tool aimed at making sense of large projects. It supports full-text and regex searches as well as structural queries.

#### Mercurial (d38398e5144e)

Line Code
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62
/*! \page usage_decode Decoding

    The vpx_codec_decode() function is at the core of the decode loop. It
    processes packets of compressed data passed by the application, producing
    decoded images. The decoder expects packets to comprise exactly one image
    frame of data. Packets \ref MUST be passed in decode order. If the
    application wishes to associate some data with the frame, the
    user_priv member may be set. The deadline
    parameter controls the amount of time in microseconds the decoder should
    spend working on the frame. This is typically used to support adaptive
    \ref usage_postproc based on the amount of free CPU time. For more
    information on the deadline parameter, see \ref usage_deadline.

    \ref samples

    \section usage_cb Callback Based Decoding
    There are two methods for the application to access decoded frame data. Some
    codecs support asynchronous (callback-based) decoding \ref usage_features
    that allow the application to register a callback to be invoked by the
    decoder when decoded data becomes available. Decoders are not required to
    support this feature, however. Like all \ref usage_features, support can be
    determined by calling vpx_codec_get_caps(). Callbacks are available in both
    frame-based and slice-based variants. Frame based callbacks conform to the
    signature of #vpx_codec_put_frame_cb_fn_t and are invoked once the entire
    frame has been decoded. Slice based callbacks conform to the signature of
    #vpx_codec_put_slice_cb_fn_t and are invoked after a subsection of the frame
    is decoded. For example, a slice callback could be issued for each
    macroblock row. However, the number and size of slices to return is
    implementation specific. Also, the image data passed in a slice callback is
    not necessarily in the same memory segment as the data will be when it is
    assembled into a full frame. For this reason, the application \ref MUST
    examine the rectangles that describe what data is valid to access and what
    data has been updated in this call. For all their additional complexity,
    slice based decoding callbacks provide substantial speed gains to the
    overall application in some cases, due to improved cache behavior.

    \section usage_frame_iter Frame Iterator Based Decoding
    If the codec does not support callback based decoding, or the application
    chooses not to make use of that feature, decoded frames are made available
    through the vpx_codec_get_frame() iterator. The application initializes the
    iterator storage (of type #vpx_codec_iter_t) to NULL, then calls
    vpx_codec_get_frame repeatedly until it returns NULL, indicating that all
    images have been returned. This process may result in zero, one, or many
    frames that are ready for display, depending on the codec.

    \section usage_postproc Postprocessing
    Postprocessing is a process that is applied after a frame is decoded to
    enhance the image's appearance by removing artifacts introduced in the
    compression process. It is not required to properly decode the frame, and
    is generally done only when there is enough spare CPU time to execute
    the required filters. Codecs may support a number of different
    postprocessing filters, and the available filters may differ from platform
    to platform. Embedded devices often do not have enough CPU to implement
    postprocessing in software. The filter selection is generally handled
    automatically by the codec, depending on the amount of time remaining before
    hitting the user-specified \ref usage_deadline after decoding the frame.

*/


text:
# Thread: [SOLVED] Equation to plane through 3 pts

1. ## [SOLVED] Equation to plane through 3 pts

I have points p, q, r,

p = (-4,-1,-1)
q = (-2,0,1)
r = (-1,-2,-3)

Okay so I need to find the equation to the plane using these three points.

Step 1. I find vector pq and vector qr

vector pq = (2,1,2)
vector qr = (1,-2,-4)

Step 2. Find the cross product pq x qr since they are parallel to the plane. I got (0,6,-5)

Step 3. I sub this into any point using (I forget what it's called form) and I get

0(x+4) + 6(y+1) + -5(z+1) = 0
6y - 5z +1 = 0

I think that's right, what do you think?

2. Originally Posted by thekrown
I have points p, q, r,

p = (-4,-1,-1)
q = (-2,0,1)
r = (-1,-2,-3)

Okay so I need to find the equation to the plane using these three points.

Step 1. I find vector pq and vector qr

vector pq = (2,1,2)
vector qr = (1,-2,-4)

Step 2. Find the cross product pq x qr since they are parallel to the plane. I got (0,6,-5)

Step 3. I sub this into any point using (I forget what it's called form) and I get

0(x+4) + 6(y+1) + -5(z+1) = 0
6y - 5z +1 = 0

I think that's right, what do you think?
Neither q nor r are on that plane...

Tonio

3. What am I doing wrong? Are my steps correct or are my calculations wrong?

4. Originally Posted by thekrown
What am I doing wrong? Are my steps correct or are my calculations wrong?
$\overrightarrow {PQ} = \left\langle {2,1,2} \right\rangle \,\& \,\overrightarrow {PR} = \left\langle {3, - 1, - 2} \right\rangle$
$\overrightarrow {PQ} \times \,\overrightarrow {PR} = \left\langle {0,2, - 1} \right\rangle$

5. Originally Posted by thekrown
I have points p, q, r,

p = (-4,-1,-1)
q = (-2,0,1)
r = (-1,-2,-3)

Okay so I need to find the equation to the plane using these three points.

Step 1. I find vector pq and vector qr

vector pq = (2,1,2)
vector qr = (1,-2,-4)

Step 2. Find the cross product pq x qr since they are parallel to the plane. I got (0,6,-5)
pq x qr= (0, 10, -5)

Step 3. I sub this into any point using (I forget what it's called form) and I get

0(x+4) + 6(y+1) + -5(z+1) = 0
6y - 5z +1 = 0

I think that's right, what do you think?
You should at least have put p, q, and r into that equation to see for yourself!


text:
Frequency of a Damped Oscillator (RLC Circuit)

In my last post I mentioned that the frequency of a damped oscillator is not actually the same as of an undamped oscillator.  That makes the assumptions for measuring inductance I used sadly incorrect.  I wanted to briefly run down the math on how to do the calculation correctly:

Frequency of an undamped (perfect) LC oscillator: $\omega_0=\sqrt{\frac{1}{LC}}$

Frequency of a damped oscillator:

Wikipedia page  informs us the formula for the frequency of a tank circuit taking into account the resistance of the coil is:

$\omega = \sqrt{\frac{1}{LC}-(\frac{R}{L})^2}$

I’m too lazy to solve equations since grad school.  So I typed this into my TI-89. I got:

$L=\frac{\sqrt{1-4C^2R^2\omega^2}+1}{2C\omega^2}$

In order to use this we can pulse through the resistor and measure how much voltage is dropped through the tank circuit.  At DC the capacitor is an open, so that will tell us the resistance of the inductor, which should be the only non-negligible part of this.  My R1 is 150 ohms.  I measured one of my 1/4 Watt inductors at 20 ohms.  So, I expect to measure around .6 mV, certainly not more than a volt or so.  That means I’ll do better using the internal voltage reference on the arduino.

I’ll code this up and modify the circuit and we’ll try again in the next post.