text:
Balaje Kalyanaraman

PhD Student at the University of Newcastle, Australia

25 July 2021

# Mixed DG methods for Biharmonic Equation

Let us take the biharmonic problem on $$\Omega$$ along with the clamped conditions on the boundary $$\partial\Omega$$.

\begin{align} \Delta^2 u = f \quad &\text{in }\Omega\\ u = \nabla u\cdot \mathbf{n} = 0 \quad &\text{on } \partial\Omega \end{align}

We now split the biharmonic equation into two equations by introducing an auxiliary variable $$v = \Delta u$$. We then have a system of equations

\begin{align} \begin{cases} \Delta u &= v \\ \Delta v &= f \end{cases}\quad &\text{in }\Omega\\ u = \nabla u\cdot \mathbf{n} = 0 \quad &\text{on } \partial\Omega \end{align}

The system is a little tricky to solve since, we do not have a boundary condition on $$v$$. It is possible to come up with a discontinuous Galerkin weak formulation, as discussed by Gudi, T., Nataraj, N. & Pani, A.K in their article. I will briefly discuss the construction of their scheme and how a simple modification leads to an increased rate of convergence.

## Mixed DG Method

Let us divide the domain $$\Omega$$ into shape regular finite elements (consider triangles). Let

$\mathcal{T}_h = \cup_{i=1}^{N_T} K_i$

denote the triangulation of the domain, with $$K_i$$ denoting the $$i$$-th triangle on the domain. Let us denote by $$\Gamma_\text{I}$$ the set of all interior edges, and $$\Gamma_\text{D}$$ the set of all boundary edges. Let $$\Gamma = \Gamma_\text{I} \cup \Gamma_\text{D}$$. For each edge $$e_k = \partial K_i \cap \partial K_j$$, associate a unit normal $$n_k$$. Note that the triangulation can contain hanging nodes along each side of the elements.

Since we are dealing with discontinuous functions, we require functions from the “broken” Sobolev space,

$H^s(\Omega, \mathcal{T}_h) = \left\{ v \in L^2(\Omega) \,:\, v |_{K_i} \in H^s(K_i) \; \forall K_i \in \mathcal{T}_h\right\}$

Let $$V = H^2(\Omega, \mathcal{T}_h)$$. The jump and average of a function $$v$$ across the edge $$e_k = \partial K_i \cap \partial K_j$$ are defined as

\begin{align} [v] &= v|_{K_i} - v|_{K_j}\\ \{v\} &= \frac{v|_{K_i} + v|_{K_j}}{2} \end{align}

There are other intricate assumptions on the triangulation which can be found in the article. But we have all the necessary ingredients to define the mixed DG method.

1. Let us take the first equation,

$\Delta u = v$

multiply by a function $$w \in V$$ and integrate over the domain

$\int_\Omega (- \Delta u + v)w\,dx = 0$

As usual proceeding with integration by parts, we obtain

\begin{align} \sum_{K\in \mathcal{T}_h}\int_K \nabla u\cdot \nabla w\,dx - \sum_{e_k \in \Gamma_\text{I}} \int_{e_k} \left\{\frac{\partial u}{\partial n_k}\right\}[w]\,ds + \int_\Omega u\,v\,dx = 0 \end{align}

Here we have used the fact that $$\nabla u \cdot \mathbf{n} = 0$$ on $$\Gamma_\text{D}$$. For a sufficiently smooth function $$[u] = 0$$ on $$\Gamma_{\text{I}}$$ and also since $$u = 0$$ on the boundary, adding the term

\begin{align} \sum_{K\in \mathcal{T}_h}\int_K \nabla u\cdot \nabla w\,dx - \sum_{e_k \in \Gamma_\text{I}} \int_{e_k} \left\{\frac{\partial u}{\partial n_k}\right\}[w]\,ds - \color{red}{\sum_{e_k \in \Gamma} \int_{e_k} \left\{\frac{\partial w}{\partial n_k}\right\}[u]\,ds} + \int_\Omega u\,v\,dx = 0 \end{align}

does not affect the equation. If $$u = g_1(x) \ne 0$$ on the boundary, then the term containing the weakly imposed DBC will appear on the RHS.

2. Similarly with the second equation,

$\Delta v = f$

multiply a function $$q \in V$$ and integrate over the domain

$\int_\Omega (-\Delta v + f)q \,dx = 0$

Applying integration by parts,

\begin{align} \sum_{K\in \mathcal{T}_h} \int_K \nabla v\cdot \nabla q\,dx - \sum_{e_k \in \Gamma} \int_{e_k} \left\{\frac{\partial v}{\partial n_k}\right\}[q]\,ds = -\int_\Omega fq\,dx \end{align}

For a sufficiently smooth function $$v$$, we have $$[v]=0$$ on $$\Gamma_\text{I}$$. We can add the term shown in red. Also like before, for a sufficiently smooth function $$[u] = 0$$ on $$\Gamma_{\text{I}}$$. Since $$u=0$$ on the boundary also, we can add the term shown in blue. The latter imposes the DBC weakly.

\begin{align} \small \sum_{K\in \mathcal{T}_h} \int_K \nabla v\cdot \nabla q\,dx - \sum_{e_k \in \Gamma} \int_{e_k} \left\{\frac{\partial v}{\partial n_k}\right\}[q]\,ds - \color{red}{\sum_{e_k \in \Gamma_\text{I}} \int_{e_k} \left\{\frac{\partial q}{\partial n_k}\right\}[v]\,ds} - \color{blue}{\sum_{e_k \in \Gamma} \int_{e_k} \alpha \,[u][q]\,ds} = -\int_\Omega fq\,dx \end{align}

To clean things up a bit, let us define the bilinear forms

\begin{align} B(w,q) &= \sum_{K\in \mathcal{T}_h} \int_K \nabla w\cdot \nabla q\,dx - \sum_{e_k \in \Gamma} \int_{e_k} \left\{\frac{\partial w}{\partial n_k}\right\}[q]\,ds - \sum_{e_k \in \Gamma_\text{I}} \int_{e_k} \left\{\frac{\partial q}{\partial n_k}\right\}[w]\,ds\\ J(w,q) &= \sum_{e_k \in \Gamma} \int_{e_k} \alpha \,[w][q]\,ds \end{align}

A mixed weak formulation of the biharmonic problem is to find $$(u,v) \in V\times V$$

\begin{align} B(w, u) + (u,v) &= 0,\\ B(v, q) - J(v, q) &= (f,q), \end{align}

for all $$(w,q) \in V\times V$$. Thus the finite dimensional version is to find $$(u_h,v_h) \in V_h\times V_h$$

\begin{align} B(w_h, u_h) + (u_h,v_h) &= 0,\\ B(v_h, q_h) - J(v_h, q_h) &= (f,q_h), \end{align}

for all $$(w_h,q_h) \in V_h\times V_h$$. The discrete space $$V_h$$ can be defined to be

$V_h = \left\{ v \in H^s(\mathcal{T}_h) \; : \; v|_K \in \mathbb{P}_k(K)\;\; \text{for all}\;\; K \in \mathcal{T}_h \right\},$

where $$\mathbb{P}_k(K)$$ denotes the space of polynomials of degree $$\le k$$. I want to talk about the $$h$$-rate of convergence as a function of the penalty term $$\alpha$$:

$\alpha |_{e_k} = \sigma_0 |e_k|^{-i}p^2$

where $$i=1$$ or $$i=3$$. It is well known in literature that for the case of the Non–symmetric interior penalty Galerkin (NIPG) method for second order elliptic problems, the choice of the penalty parameter plays a crucial role on the convergence of the solution. If $$i=1$$, under normal penalization, the convergence in the $$L^2-$$norm for the piecewise quadratic case is suboptimal. However, for the case when $$i=3$$, under super penalization, the convergence is optimal.

For the mixed Discontinuous Galerkin Method considered here, we observe that the penalization term with $$i=1$$ produces optimal convergence in all cases, as opposed to $$i=3$$ where optimal convergence rates are observed only for piecewise cubic elements or higher. The latter case was well studied by Gudi et al.. I did some tests with FreeFem a long time ago, and decided to test this again with Gridap.jl. I was able to test only upto piecewise cubic elements using FreeFem, but with Gridap.jl, I was able to go until order 6! The results are summarized below.

## Results

For conducting the numerical experiment, I chose the following parameters.

\begin{align} \Omega = [0,1]^2, \quad \sigma_0 = \frac{2}{p^2}, \quad i=1 \;(or)\; 3 \end{align}

### Example 1

I assume the exact solution $$u(x,y) = x^2(1-x)^2y^2(1-y)^2$$ and derive the function $$f(x,y)$$ accordingly. For this choice, fortunately $$u = \nabla u\cdot \mathbf{n} = 0$$ on the boundary. Here I show the results up until fifth order DG approximation. The codes are given below:

$$i=3$$ $$i=1$$
 Polynomial order Rate Polynomial order Rate 1 [0.3852, -0.1930 -0.0916] 1 [2.3182, 2.6385, 2.7471] 2 [3.4888, 2.8130, 2.4539] 2 [3.1978, 3.5151, 3.5987] 3 [5.1583, 3.8862, 3.9024] 3 [4.1920, 4.3205, 4.3889] 4 [5.6954, 4.6425, 4.7674] 4 [5.2752, 5.3111, 5.3818] 5 [6.3395, 5.8784, 5.9436] 5 [6.7739, 7.4219, 5.8162]

In the left-hand side, I show the results of the scheme discussed by Gudi et al. for $$i=3$$. In the right-hand side, I show the results of the scheme with the modified penalty parameter $$i=1$$. In the figures $$k$$ denotes the polynomial order of approximation.

• One thing that stands out immediately is the rate of convergence for piece-wise linear approximation. While setting $$i=3$$ does not yield convergence for linear elements, $$i=1$$ yields optimal (super?) convergence here.

• For $$i=3$$ we observe sub-optimal convergence in the $$L^2$$ norm for piece-wise quadratic elements, while $$i=1$$ yields optimal convergence. Although the magnitude of the error seems to be higher in $$i=1$$ case. Interesting.

• Higher order elements seem to yield optimal estimates (atleast close to) in both cases here. This is probably expected, as seen from the original article by Gudi et al..

So far the results are consistent with the observations of Gudi et al. for $$i=3$$. There is something interesting happening when $$i=1$$ in the penalty term.

### Example 2

Let us run the tests for one more example, $$u(x,y) = sin(\pi x)\sin(\pi y)$$. It is one of my favourite examples to test for convergence in elliptic FEM problems.

$$i=3$$ $$i=1$$
 Polynomial order Rate Polynomial order Rate 1 [1.4976, -0.3135, -0.1474] 1 [1.9061, 1.8918, 1.9492] 2 [3.1732, 2.9442, 2.7414] 2 [3.6448, 3.4870, 3.4528] 3 [4.0097, 3.8762, 3.9568] 3 [4.3743, 4.4074, 4.4296] 4 [4.7994, 4.7530, 4.9161] 4 [5.3946, 5.3610, 5.4243] 5 [6.3664, 5.9323, 5.9673] 5 [6.5199, 6.6304, 7.5057]

Mostly the same, except now for $$k=1$$ i.e., linear polynomials we get the optimal rate of convergence when $$i=1$$. It might be because the exact solution is not a polynomial. Again, the results seems consistent with Gudi et al. for $$k=3$$.

## Conclusions

In Symmetric Interior Penalty Galerkin (SIPG) Method, the choice of penalty parameter $$\sigma_0$$ has mostly been an arbitrary real number that is sufficiently large. A definite lower bound to my knowledge has not yet been established. In this case for mixed DG method, however, an upper bound for the penalty term $$\alpha|_{e_k} = \sigma_0 |e_k|^{-i} p^2$$ may exist. A high value (proportional to $$|e_k|^{-3}$$) may erode the method into producing sub-optimal rates. A low value (proportional to $$|e_k|^{-1}$$), and hence a low value of $$\sigma_0$$ could explain the good rate of convergence.

While an improvement in the convergence is observed for linear elements when the penalty term is reduced (taken proportional to $$|e_k|^-1$$), the magnitude of the error for higher order elements seems to actually be higher when the penalty term is proportional to $$|e_k|^{-1}$$ than when it is proportional to $$|e_k|^{-3}$$. This also needs some more tests.

## Update (14 Aug 2021)

I did some test for 3D problems and I observed similar results as in 2D. The exact solution I assumed is $$u(x,y,z) = x^4(1-x)^4y^4(1-y)^4z^4(1-z)^4$$ on a unit cube.

 $$i=3$$ $$i=1$$ Polynomial order Rate Polynomial order Rate 1 [2.8811, 1.3726, -0.0986] 1 [2.0525, 2.2049, 2.4382] 2 [5.1508, 3.4932, 2.5361] 2 [3.8400, 3.6734, 4.7405]

## References

Gudi, T., Nataraj, N. & Pani, A.K. Mixed Discontinuous Galerkin Finite Element Method for the Biharmonic Equation. J Sci Comput 37, 139–161 (2008). https://doi.org/10.1007/s10915-008-9200-1

tags: wip - discontinuous - Galerkin - gridap


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# Intensity

Intensities are extremely useful in the insurance industry. In this post I will introduce intensities in the setting of pricing out a simple life insurance.

## Life insurance

Take the most simple life insurance: We have to pay $1$ to the customer if he/she dies before the age of $n$ years old.

When tasked with coming up with a price for this insurance, we want to know how likely the customer is to die before reaching the age of $n$ years. When we know that, and a few other things like how much interest we can accumulate in the time span and our expenses etc. we are ready to price the insurance.

We hopefully have access to some statistics that will tell us at what age people normally die, allowing us to figure out a density function for the distribution of life lengths of new-borns. Lets call it $f(t)$. Let $X$ be the time of death.

$$Pr[X\leq n] = \int_0^n f(t) dt$$

This is all well and good, but often the customer will not be $0$ years old when taking up the insurance, and as such should expect a lower price since he/she can promise not to die before xers current age, call it $a$. So really we then want to know:

$$Pr[X\leq n | X>a] =\frac {Pr[a < X\leq n]}{Pr[X>a]} = {\int_a^n f(t) dt}/{\int_0^a 1 - f(t) dt}$$

This is not too bad, but it will be a lot cleaner when using intensities.

## Cumulative functions

To avoid messing with too many integrals lets define $F(t)=\Pr[X\leq t]$ and $\bar F(t)=\Pr[X>t]=1-F(t)$. Actuarians call these the mortality and survival functions. We can restate the probability above as

$$\Pr[X\leq n | X>a]=\int_a^n \bar F(t|X>a) f(t|X>t)dt=\frac{\bar{F}(a+t)}{\bar F(a)}\frac{f(t)}{\bar(F)(t)}$$

## Intensities

Lets start with defining the intensity $\mu(t)$:

$$\mu(t)=\frac{f(t)}{\bar F(t)}$$

This definition implies that $\mu(t)=\frac{-\frac d{dt}\bar F(t)}{\bar F(t)}=\frac d{dt} -\ln(\bar F(t))$ and so:

$$\bar F (t) = e^{-\int_0^t \mu(\tau) d\tau}$$

So our expression reduces to

$$\Pr[X\leq n | X>a]=\int_a^n e^{-\int_a^{a+t}\mu(\tau) d\tau}\mu(t) dt$$

If we redefine $n$ so instead of saying we want an insurance until age $n$, we say that we want $n$ years of insurance it becomes even simpler:

$$\Pr[X\leq a+n | X>a]=\int_0^n e^{-\int_0^t\mu(a+\tau) d\tau}\mu(t) dt$$

## Why does it work?

What we need to figure out when selling insurances to a person of $a>0$ years of age really is a new density function, for the distribution of remaining life length of a $a$-year old.

$f(t|a) = \frac {f(a+t)}{\bar F(a)}$.

The probabiliy of living $t$ years after reaching age $a$ should be the same as a newborn living $a+t$ years, conditioned on him/her surviving $a$ years. However conditioning $\mu(t)$ is a lot simpler, consider first what happens to the survival function:

$$\bar F(t| a) = \frac {\bar F(a+t)} {\bar F(a)}$$

Plug these in to our definition of $\mu$ and we see that

$$\mu(t|a)=\frac {f(a+t)} {\bar F(a+t)}=\mu(a+t)$$

Which is why intensities are so handy.

format: plain


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## Review of Differential Equations in Circuits (5 minutes)

• The differential equations defining charge (or current, if you wish) in basic circuits as a function of time are found using Kirchoff's Current Law and Kirchoff's Voltage Law. Using Kirchoff's Voltage law, we can find that:

$$V_{L}=L\frac{dI}{dt} \; \; ,$$

and

$$V_{C}=\frac{Q}{C} \; \; .$$

Now, as an example of Kirchoff's Voltage Law, if we combine a capacitor and an inductor in series as shown below,

we find that

$$V_{C}+V_{L}=0 \; \; ,$$

which implies that

$$\frac{Q}{C}+L\frac{d^{2}Q}{dt^{2}}= 0 \; \; .$$

This differential equation will define what the charge in the circuit will do.

So, the main strategy for solving for $Q(t)$ in a simple circuit made from inductors, capacitors, and resistors is to look for the differential equation and solve it.

NOTE: Resistors, capacitors, and inductors in series and parallel all have different properties from one another. Remind students to find review these properties again.

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# Math Help - Write Using Rectangular Coordinates

1. ## Write Using Rectangular Coordinates

The letters r and theta represent polar coordinates. Write each equation using rectangular coordinates (x,y).

(1) y = 4

(2) r = [3]/[3 - cos(theta)]

2. Originally Posted by magentarita
The letters r and theta represent polar coordinates. Write each equation using rectangular coordinates (x,y).

(1) y = 4

(2) r = [3]/[3 - cos(theta)]
(1) is already expressed in rectangular coordinates.

(2) $\Rightarrow 3r - r \cos \theta = 3$.

Substitute $x = r \cos \theta$ and $r = \sqrt{x^2 + y^2}$.

3. ## sorry

Originally Posted by mr fantastic
(1) is already expressed in rectangular coordinates.

(2) $\Rightarrow 3r - r \cos \theta = 3$.

Substitute $x = r \cos \theta$ and $r = \sqrt{x^2 + y^2}$.
Sorry, the first question should be r = 4 not y = 4.

Can you show me now?

4. ## Also...

The letters r and theta represent polar coordinates. Write each equation using rectangular coordinates (x,y).

The second question is a fraction.

(2) r = [3] divided by [3 - cos(theta)]

5. Originally Posted by magentarita
The letters r and theta represent polar coordinates. Write each equation using rectangular coordinates (x,y).

The second question is a fraction.

(2) r = [3] divided by [3 - cos(theta)]
I realise that. The symbol => means 'it follows that'. Do you see how what I posted follows from the second question.

6. Originally Posted by magentarita
Sorry, the first question should be r = 4 not y = 4.

Can you show me now?
r = 4 => r^2 = 16. Make the obvious substitution.

7. ## I don't....

Originally Posted by mr fantastic
r = 4 => r^2 = 16. Make the obvious substitution.
I don't see the obvious substitution.

8. You should recall that $r^2 = x^2 + y^2$ which should remind you of a circle

9. ## ok

Originally Posted by o_O
You should recall that $r^2 = x^2 + y^2$ which should remind you of a circle
Our teacher did not teach this to the class.


text:
# An error has occurred. See the log file C:\Program Files (x86)\Justinmind\Justinmind Prototyper7.2.2

seetharam shared this problem 2 years ago
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I get this error while i access justinmind after updataing from 7.2.2 to 7.5, please help as its very urgent.

An error has occurred. See the log file - C:\Program Files (x86)\Justinmind\Justinmind Prototyper7.2.2\workspace\.metadat.log.

1

Hi Seetharam,

Your request is being dealt with in our Customer Support Portal.

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text:
Hi Jean Guillaume,

I guess Marco Patzer and Wolfgang Schuster gave you already the appropriate
Nevertheless, I wanted to share with you the definitions I use for the kind of
work you want to do:

%%% begin example-exercise.tex
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% defining \startexo, \stopexo as an enumeration:
\startsetups style:simple
\defineenumeration[exo]
[alternative=hanging,
width=fit,
stopper={.},
text=Exercise,
%       between=,
before=,
after=\blank]
\stopsetups % style:simple

\startsetups style:textrule
\definenumber[MyExoNumber][way=bysection,sectionumber=yes]
\setuptextrules[rulecolor=darkred]
\define\ExoCommand{\incrementnumber[MyExoNumber]
\textrule[top]{Exercise \getnumber[MyExoNumber]}
\startbackground[frame=off,leftframe=on,backgroundcolor=white,
framecolor=darkred]}

\defineenumeration[exo]
[alternative=hanging,
width=fit,
text={},
number=hide,
number=no,
before={\ExoCommand},
after={\stopbackground\blank}]
\stopsetups % style:textrule

% end defining \startexo, \stopexo as an enumeration
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%defining  numbered questions
\defineconversion[exercise][\numbers]
%\setupitemize[packed]
\def\StartQuestions{%
\startitemize[exercise][width=2em,packed,style=bold,stopper=,right=)]}
\def\StopQuestions{\stopitemize}
\def\q{\item}

% trye each of the following
\setups[style:simple]
%\setups[style:textrule]

\starttext
\startexo
Prove that for all $a,b\in {\Bbb K}$, a field of characteristic 2, one has
$(a+b)^2 = a^2 + b^2$.
\stopexo

\startexo
Prove that for all $a,b\in {\Bbb K}$, a field of characteristic 2, one has
$(a+b)^4 = a^4 + b^4$.
\stopexo

\startexo
\StartQuestions
\q Prove that
\startformula
\sum_{n=1}^\infty{1 \over n^2} = {\pi^2 \over 6}.
\stopformula

\q Prove that for any $n \geq 1$ integer one has
\startformula
\sum_{k=1}^n k^3 = \left({n(n+1) \over 2}\right)^2.
\stopformula
\StopQuestions
\stopexo

\stoptext
%%% end example-exercise.tex

Best regards: OK

On 9 janv. 2014, at 14:23, Jean-Guillaume <jng...@gmail.com> wrote:

> But this:
>
>  [Exercise] [subsection]
>  [after=]
> \starttext
>  \startExercise [title=This is the exercise's title]
>    The exercise's content.
>  \stopExercise
> \stoptext
>
> results to:
>
> 1 This is the exercise's title
> The exercise's content.
>
> and I would like it to be:
>
> 1 This is the exercise's title The exercise's content.
>
> (When I can typeset that, I will see about formatting the number and so on.)
>
> I'm sorry it wasn't very clear. Now I understand that I misunderstood
> \startExercise{...} for \startExercise[title=...] and that's why I was
> confused about which newline I want to get rid of.
>
> Jean-Guillaume
> ___________________________________________________________________________________
> If your question is of interest to others as well, please add an entry to the
> Wiki!
>
> maillist : ntg-context@ntg.nl / http://www.ntg.nl/mailman/listinfo/ntg-context
> webpage  : http://www.pragma-ade.nl / http://tex.aanhet.net
> archive  : http://foundry.supelec.fr/projects/contextrev/
> wiki     : http://contextgarden.net
> ___________________________________________________________________________________

___________________________________________________________________________________


text:
Combination theorems for discrete subgroups of isometry groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:37:42Z http://mathoverflow.net/feeds/question/109068 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109068/combination-theorems-for-discrete-subgroups-of-isometry-groups Combination theorems for discrete subgroups of isometry groups anonymous 2012-10-07T15:23:39Z 2012-10-07T20:56:30Z
Maskit's combination theorem says: if $M=M_1\cup_\Sigma M_2$ is a union of hyperbolic 3-manifolds $M_1=\Gamma_1\backslash H^3, M_2=\Gamma_2\backslash H^3$ along a surface $\Sigma$, and if the limit set of $H:=\pi_1\Sigma$ is a codimension 1 submanifold of $\partial H^3$ dividing $\partial H^3$ into domains $\Omega_1,\Omega_2$ which are precisely $H$-invariant in $\Gamma_i$ (meaning that $\Omega_i$ is not preserved by any $g\in \Gamma_i\setminus H$) then the amalgamated product $\Gamma=\Gamma_1*_H\Gamma_2$ is a discrete subgroup of $Isom(H^3)$ and in particular $M$ carries a hyperbolic metric, i.e., $\widetilde{M}$ is isometric to $H^3$.

Question: what is known about generalizations in the setting of, say, nonpositively curved manifolds, i.e., replacing the universal cover $H^3$ by any simply connected manifold of nonpositive curvature?

Li-Ohshika-Wang in http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.ojm/1260892842 handled the case of $H^n$. (The statement of their main result Theorem 4.2. is somewhat involved but, if I am not mistaken, it implies the above statement for $H^n$ instead $H^3$.) Are some other cases known?
 http://mathoverflow.net/questions/109068/combination-theorems-for-discrete-subgroups-of-isometry-groups/109086#109086 Answer by Agol for Combination theorems for discrete subgroups of isometry groups Agol 2012-10-07T18:14:43Z 2012-10-07T20:56:30Z
There's a combination theorem of Bestvina-Feighn for hyperbolic groups.

There are gluing theorems for $CAT(\kappa)$ spaces in Chapter 11 of Bridson-Haefliger which can give rise to combination theorems for groups.

Another combination theorem is due to Baker-Cooper for groups acting on hyperbolic space.

A similar combination theorem for hyperbolic groups is due to Gitik.

A similar combination theorem but applying to groups acting on cube complexes appears in Lemma 4.14 of Haglund-Wise.
 http://mathoverflow.net/questions/109068/combination-theorems-for-discrete-subgroups-of-isometry-groups/109090#109090 Answer by Igor Rivin for Combination theorems for discrete subgroups of isometry groups Igor Rivin 2012-10-07T19:12:05Z 2012-10-07T19:12:05Z
There is also this thesis (Combination of quasiconvex subgroups in relatively hyperbolic groups by Martinez Pedroza, Eduardo, Ph.D., THE UNIVERSITY OF OKLAHOMA, 2008,) , and references therein.


text:
Adjusting R8 to Increase HP-97 Printer Intensity
04-01-2021, 11:39 PM (This post was last modified: 04-01-2021 11:41 PM by James Linder.)
Post: #21
 James Linder Junior Member Posts: 27 Joined: Jul 2020
RE: Adjusting R8 to Increase HP-97 Printer Intensity
They only see occasional use as collector items, and I would rather have pristine printouts when I show them off.

How many desktop HP calculators do you actually own?
04-02-2021, 12:06 AM (This post was last modified: 04-02-2021 12:09 AM by teenix.)
Post: #22
 teenix Senior Member Posts: 1,419 Joined: May 2016
RE: Adjusting R8 to Increase HP-97 Printer Intensity
(04-01-2021 11:39 PM)James Linder Wrote:  They only see occasional use as collector items, and I would rather have pristine printouts when I show them off.

How many desktop HP calculators do you actually own?

That's fair enough. I am a bit over cautious at times :-)

None, I have one on loan while I develop the 97 replacement CPU board.

Maybe one day....

cheers

Tony
04-02-2021, 12:50 AM
Post: #23
 James Linder Junior Member Posts: 27 Joined: Jul 2020
RE: Adjusting R8 to Increase HP-97 Printer Intensity
If you do not mind the outer case having become darker over time, I could put a fully
functional 97 together for you for about $275 with soft case, or$325 if you want the
adapter/charger and a rebuilt NiCad battery pack.

It would be the best of the four I have left using the best case with the best keyboard
and after I restore the card reader and the printer page advance spool. The printers
I have remaining provide readable print, however the print is not as dark as the ones
I own.

If you want to do the work on the card reader and the printer page advance yourself,
I would take $75 off these prices. You would need to pay shipping, and in USA First Class mail for this would cost about$25.

If you also want the user's manual and Standard Pac software with manual for the 97,
I would include good but not excellent manuals and software for another $50. Please understand, these prices are considerably lower than typical to show you my appreciation for your assistance provided on this forum. 04-02-2021, 01:24 AM Post: #24  teenix Senior Member Posts: 1,419 Joined: May 2016 RE: Adjusting R8 to Increase HP-97 Printer Intensity Hi James, A nice offer indeed and I appreciate the thought. Sadly I'll have to think about it as I am on the verge of retirement and am still looking for an old country home to settle down in. Unfortunately with prices rapidly going up here, it's making things more and more difficult. What would be nice is a HP-91 loaner (working or not) so I can create a CPU replacement for it and the 92. The keyboard layout is different than the 97, so I guess the CPU board layout will be also. With the replacement CPU boards, I have control of the printer so I can change print intensity via software settings. cheers Tony 04-02-2021, 01:53 AM Post: #25  James Linder Junior Member Posts: 27 Joined: Jul 2020 RE: Adjusting R8 to Increase HP-97 Printer Intensity I think you are in luck. I have a dead HP-91 that when I bought it a year ago it had a mint exterior and a display so dim that the room had to be dark to see it. Then I bought a 91 that worked but had a cracked top case so I swapped cases so I would have a nice one that worked and looked nice. The leftover parts are what you would receive: top & bottom outer case, keyboard with display, and motherboard mounted on center support shelf. The printer has been removed and the ribbon connector from the keyboard to the display has been ruined. If you pay for shipping both directions I will loan it to you for as long as you need it. I think First Class Mail will run about$15.

You may keep it as long as you need to, however when finished please return to me so I may use it for parts. How does that sound?

I will see if I can come up with a 92 for you after you are done with the 91. If you prefer having a 92 first, then let me check into that to see what I can come up with.
04-02-2021, 02:08 AM
Post: #26
 Don Williams Junior Member Posts: 30 Joined: Apr 2014
RE: Adjusting R8 to Increase HP-97 Printer Intensity
Just a heads up concerning mailing costs.

I think you may be discussing an overseas shipment. You should check the price's before you make a promise you might not be willing to keep.

I have some devices I would like to ship to Panamatik in Germany, but overseas shipping charges are just staggeringly expensive right now.

Hope the world gets back to normal soon.

Don't ask me I just lurk here.
04-02-2021, 03:58 AM
Post: #27
 teenix Senior Member Posts: 1,419 Joined: May 2016
RE: Adjusting R8 to Increase HP-97 Printer Intensity
On TAS currently, HP-97's shipped to Australia from the US vary from about $Aus40 to$Aus120 so might depend on dispatch location.

cheers

Tony
04-02-2021, 12:02 PM
Post: #28
 James Linder Junior Member Posts: 27 Joined: Jul 2020
RE: Adjusting R8 to Increase HP-97 Printer Intensity
Do you want me to check the mailing costs for you? Typically First Class
runs one third the cost of priority mail.
04-02-2021, 01:02 PM
Post: #29
 teenix Senior Member Posts: 1,419 Joined: May 2016
RE: Adjusting R8 to Increase HP-97 Printer Intensity
(04-02-2021 12:02 PM)James Linder Wrote:  Do you want me to check the mailing costs for you? Typically First Class
runs one third the cost of priority mail.

Yes thanks James, I would appreciate it, thank you.

cheers

Tony
04-02-2021, 01:15 PM
Post: #30
 James Linder Junior Member Posts: 27 Joined: Jul 2020
RE: Adjusting R8 to Increase HP-97 Printer Intensity
Please email me your mailing address. Jim_Linder@sbcglobal.net
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text:
# An introduction to Eclipse for Visual Studio users

I’m seeing more and more people moving to Eclipse as a development platform, even those Windows users who have traditionally used Visual Studio. As an Eclipse user for quite a while now I’m often asked how good it is, or how to use it. Of course, …

I’m seeing more and more people moving to Eclipse as a development platform, even those Windows users who have traditionally used Visual Studio. As an Eclipse user for quite a while now I’m often asked how good it is, or how to use it. Of course, telling people to simply try it out isn’t enough. Many people just don’t get Eclipse and cannot understand or translate the skills and experience they already have to the Eclipse environment. That’s where An introduction to Eclipse for Visual Studio users can help. It’s a quick overview of the fundamentals of Eclipse from the perspective of a Visual Studio user. For a more in depth examination, there’s a tutorial Eclipse for Visual Studio developers, and another on migrating your applications from VS to Eclipse: Migrate Visual Studio C and C++ projects to Eclipse CDT.I can recommend any (or indeed all) of these.


text:
Language:   Search:   Contact
Zentralblatt MATH has released its new interface!
For an improved author identification, see the new author database of ZBMATH.

Query:
Fill in the form and click »Search«...
Format:
Display: entries per page entries
Zbl 0955.37040
Tanaka, Kazunaga
Periodic solutions for singular Hamiltonian systems and closed geodesics on non-compact Riemannian manifolds.
(English)
[J] Ann. Inst. Henri Poincaré, Anal. Non Linéaire 17, No.1, 1-33 (2000). ISSN 0294-1449

The author deals with the so-called prescribed energy problem: $$\ddot q+ \nabla V(q)= 0, \qquad \tfrac 12 |\dot q|^2+ V(q)= H,$$ where $q(t): \bbfR\to \bbfR^N \setminus \{0\}$, $N\geq 2$, $V: \bbfR^n \setminus \{0\}\to \bbfR$ and $H\in \bbfR$. The author studies the existence of periodic solutions of similar Hamiltonian systems as well as the existence of closed geodesics on noncompact Riemannian manifolds in a related situation.
[Messoud Efendiev (Berlin)]
MSC 2000:
*37J45 Periodic, homoclinic and heteroclinic orbits, etc.
58E10 Appl. of bifurcation theory to geodesics
34C25 Periodic solutions of ODE

Keywords: energy problem; Hamiltonian systems; closed geodesics; Riemannian manifold

Cited in: Zbl 0992.53029

Highlights
Master Server


text:
## Journal of Symbolic Logic

### Large Cardinals and Large Dilators

Andy Lewis

#### Abstract

Applying Woodin's non-stationary tower notion of forcing, I prove that the existence of a supercompact cardinal $\kappa$ in V and a Ramsey dilator in some small forcing extension V[G] implies the existence in V of a measurable dilator of size $\kappa$, measurable by $\kappa$-complete measures.

#### Article information

Source
J. Symbolic Logic, Volume 63, Issue 4 (1998), 1496-1510.

Dates
First available in Project Euclid: 6 July 2007

https://projecteuclid.org/euclid.jsl/1183745644

Mathematical Reviews number (MathSciNet)
MR1665763

Zentralblatt MATH identifier
0926.03070

JSTOR


text:
# Does sum−1^{n}ln 2n^{frac{1}{n}}) converge or diverge?

Question
Functions
Does $$\sum−1^{n}\ln 2n^{\frac{1}{n}})$$ converge or diverge?

2021-02-14
Let $$a_{n}= (-1)^{n}\log(2n^{\frac{1}{2}}).$$ Here
$$\lim_{n \rightarrow \infty}|an| = \lim_{n \rightarrow \infty} \log(2n^{\frac{1}{n}})| =\log(\lim 2n^{\frac{1}{n}}) (log is continuous function in [1,∞) ) =\log2 (Since \lim_{n \rightarrow \infty} n^{\frac{1}{n}}=1$$
$$\sum_{n=1}^\infty (-1)^{n}\log(2n^{\frac{1}{n}})$$ is not convergent.

### Relevant Questions

Does $$\displaystyle\sum−{1}^{{{n}}}{\ln{{2}}}{n}^{{{\frac{{{1}}}{{{n}}}}}}{)}$$ converge or diverge?
Does the attached series converge or diverge? Note: it is a complex series and not a real series. I've been told the limit comparison test is not a valid approach to solving this problem.
$$\sum_{n=1}^\infty\frac{1}{n-i^n}$$
Does the series (showing the picture) converge or diverge?
1) The integral test shows that the series converges
2) The nth-term test shows that the series converges
3) The series diverges because the series is a geometric series with |r|>=1
4) The nth-term test shows that the series diverges
$$\sum_{n=1}^\infty\frac{4^n}{n+1}$$
Given series:
$$\sum_{k=3}^\infty[\frac{1}{k}-\frac{1}{k+1}]$$
does this series converge or diverge?
If the series converges, find the sum of the series.
Determine whether the sequence $$\displaystyle{\left\lbrace{a}{n}\right\rbrace}\infty$$
n=2 with $$\displaystyle{a}{n}={\frac{{{\ln}}}{{{n}}}}{n}{2}$$ converges. Show all steps.
Question 1
The function $$N(t)=15t\sqrt{t^{2}+1}+2000$$ denotes the number of members of a club t months after the club's opening. You can use the following
$$N'(t)=\frac{30t^{2}+15}{\sqrt {t^{2}+1}}$$ / and/or $$\int N(t)dt = 2000t+5(t^{2}+1)^{\frac{3}{2})}+ C$$
1.1 How many members were in the club after 12 months?
Question 1
The function $$\displaystyle{N}{\left({t}\right)}={15}{t}\sqrt{{{t}^{{{2}}}+{1}}}+{2000}$$ denotes the number of members of a club t months after the club's opening. You can use the following
$$\displaystyle{N}'{\left({t}\right)}={\frac{{{30}{t}^{{{2}}}+{15}}}{{\sqrt{{{t}^{{{2}}}+{1}}}}}}$$ / and/or $$\displaystyle\int{N}{\left({t}\right)}{\left.{d}{t}\right.}={2000}{t}+{5}{\left({t}^{{{2}}}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}\rbrace+{C}$$
1.1 How many members were in the club after 12 months?
Which series converge, and which diverge? Give reasons for your answers. If a series converges, find its sum.
$$\sum_{n=0}^\infty(\ln(4e^n-1)-\ln(2e^n+1))$$
\lim_{x \rightarrow \infty} x^{\frac{1}{ln x}}
Select one:
a 1
b 0
c ∞
d e


text:
Featured

The ASSESS team

At IAASARS/NOA I am part of the ASSESS project – Episodic MAss LoSS in Evolved MaSsive Stars. This project, funded by an ERC Consolidator grant (2018-2023, PI Bonanos), investigates the impact of episodic mass loss on the evolution of massive stars, and its role in the early Universe. The team is led by dr. Alceste Bonanos, and currently consists of three post-docs (dr. Grigoris Maravelias, dr. Ming Yang, and myself), and one PhD student (Stephan de Wit).

Reconstructing the EUV Spectrum of Star-forming Regions from Millimeter Recombination Lines of H I, He I, and He II

Murchikova, Lena; Murphy, Eric J.; Lis, Dariusz C.; Armus, Lee; de Mink, Selma; Sheth, Kartik; Zakamska, Nadia; Tramper, Frank; Bongiorno, Angela; Elvis, Martin; Kewley, Lisa; Sana, Hugues

The Astrophysical Journal, 903 29 (2020)

Abstract

The extreme ultraviolet (EUV) spectra of distant star-forming regions cannot be probed directly using either ground- or space-based telescopes due to the high cross section for interaction of EUV photons with the interstellar medium. This makes EUV spectra poorly constrained. The millimeter/submillimeter recombination lines of H and He, which can be observed from the ground, can serve as a reliable probe of the EUV. Here we present a study based on ALMA observations of three Galactic ultracompact H II regions and the starburst region Sgr B2(M), in which we reconstruct the key parameters of the EUV spectra using millimeter recombination lines of H I, He I, and He II. We find that in all cases the EUV spectra between 13.6 and 54.4 eV have similar frequency dependence: ${L}_{\nu }\propto {\nu }^{-4.5\pm 0.4}.$ We compare the inferred values of the EUV spectral slopes with the values expected for a purely single stellar evolution model (Starburst99) and the Binary Population and Spectral Synthesis code (BPASS). We find that the observed spectral slope differs from the model predictions. This may imply that the fraction of interacting binaries in H II regions is substantially lower than assumed in BPASS. The technique demonstrated here allows one to deduce the EUV spectra of star-forming regions, providing critical insight into photon production rates at $\lambda \leqslant 912\mathring{{\rm{A}}}$ , and can serve as calibration to starburst synthesis models, improving our understanding of star formation in the distant universe and the properties of ionizing flux during reionization.

Keywords: Astrophysicists;Interstellar medium;Star-forming regions;Star formation;Radiative recombination;Compact H II region;105;847;1565;1569;2057;286;Astrophysics – Astrophysics of Galaxies;Astrophysics – Solar and Stellar Astrophysics

Evolved massive stars at low metallicity. II. Red supergiant stars in the Small Magellanic Cloud

Yang, Ming; Bonanos, Alceste Z.; Jiang, Bi-Wei; Gao, Jian; Gavras, Panagiotis; Maravelias, Grigoris; Wang, Shu; Chen, Xiao-Dian; Tramper, Frank; Ren, Yi; Spetsieri, Zoi T.; Xue, Meng-Yao

Astronomy & Astrophysics, Volume 639, A116 (2020)

Abstract

We present the most comprehensive red supergiant (RSG) sample for the Small Magellanic Cloud (SMC) to date, including 1239 RSG candidates. The initial sample was derived based on a source catalog for the SMC with conservative ranking. Additional spectroscopic RSGs were retrieved from the literature, and RSG candidates were selected based on the inspection of Gaia and 2MASS color-magnitude diagrams (CMDs). We estimate that there are in total ∼1800 or more RSGs in the SMC. We purify the sample by studying the infrared CMDs and the variability of the objects, though there is still an ambiguity between asymptotic giant branch stars (AGBs) and RSGs at the red end of our sample. One heavily obscured target was identified based on multiple near-IR and mid-IR (MIR) CMDs. The investigation of color-color diagrams shows that there are fewer RSGs candidates (∼4%) showing PAH emission features compared to the Milky Way and LMC (∼15%). The MIR variability of RSG sample increases with luminosity. We separate the RSG sample into two subsamples (risky and safe), and identify one M5e AGB star in the risky subsample based on simultaneous inspection of variabilities, luminosities, and colors. The degeneracy of mass loss rate (MLR), variability, and luminosity of the RSG sample is discussed, indicating that most of the targets with high variability are also the bright ones with high MLR. Some targets show excessive dust emission, which may be related to previous episodic mass loss events. We also roughly estimate the total gas and dust budget produced by entire RSG population as ∼1.9-1.1+2.4 × 10-6 M yr-1 in the most conservative case, according to the derived MLR from IRAC1-IRAC4 color. Based on the MIST models, we derive a linear relation between Teff and observed J – KS color with reddening correction for the RSG sample. By using a constant bolometric correction and this relation, the Geneva evolutionary model is compared with our RSG sample, showing a good agreement and a lower initial mass limit of ∼7 M for the RSG population. Finally, we compare the RSG sample in the SMC and the LMC. Despite the incompleteness of LMC sample in the faint end, the result indicates that the LMC sample always shows redder color (except for the IRAC1-IRAC2 and WISE1-WISE2 colors due to CO absorption) and higher variability than the SMC sample, which is likely due to a positive relation between MLR, variability and the metallicity.

Keywords: infrared: stars;Magellanic Clouds;stars: late-type;stars: massive;stars: mass-loss;stars: variables: general;Astrophysics – Solar and Stellar Astrophysics;Astrophysics – Astrophysics of Galaxies

Properties of OB star-black hole systems derived from detailed binary evolution models

Langer, N.; Schürmann, C.; Stoll, K.; Marchant, P.; Lennon, D. J.; Mahy, L.; de Mink, S. E.; Quast, M.; Riedel, W.; Sana, H.; Schneider, P.; Schootemeijer, A.; Wang, C.; Almeida, L. A.; Bestenlehner, J. M.; Bodensteiner, J.; Castro, N.; Clark, S.; Crowther, P. A.; Dufton, P. Evans, C. J.; Fossati, L.; Gräfener, G.; Grassitelli, L.; Grin, N.; Hastings, B.; Herrero, A.; de Koter, A.; Menon, A.; Patrick, L.; Puls, J.; Renzo, M.; Sander, A. A. C.; Schneider, F. R. N.; Sen, K.; Shenar, T.; Simón-Días, S.; Tauris, T. M.; Tramper, F.; Vink, J. S.; Xu, X. -T.

Astronomy & Astrophysics, Volume 638, A39 (2020)

Abstract

Context. The recent gravitational wave measurements have demonstrated the existence of stellar mass black hole binaries. It is essential for our understanding of massive star evolution to identify the contribution of binary evolution to the formation of double black holes.
Aims: A promising way to progress is investigating the progenitors of double black hole systems and comparing predictions with local massive star samples, such as the population in 30 Doradus in the Large Magellanic Cloud (LMC).
Methods: With this purpose in mind, we analysed a large grid of detailed binary evolution models at LMC metallicity with initial primary masses between 10 and 40 M, and identified the model systems that potentially evolve into a binary consisting of a black hole and a massive main-sequence star. We then derived the observable properties of such systems, as well as peculiarities of the OB star component.
Results: We find that ∼3% of the LMC late-O and early-B stars in binaries are expected to possess a black hole companion when stars with a final helium core mass above 6.6 M are assumed to form black holes. While the vast majority of them may be X-ray quiet, our models suggest that these black holes may be identified in spectroscopic binaries, either by large amplitude radial velocity variations (≳50 km s-1) and simultaneous nitrogen surface enrichment, or through a moderate radial velocity (≳10 km s-1) and simultaneous rapid rotation of the OB star. The predicted mass ratios are such that main-sequence companions can be excluded in most cases. A comparison to the observed OB+WR binaries in the LMC, Be and X-ray binaries, and known massive black hole binaries supports our conclusion.
Conclusions: We expect spectroscopic observations to be able to test key assumptions in our models, with important implications for massive star evolution in general and for the formation of double black hole mergers in particular.

Keywords: stars: evolution;stars: massive;binaries: close;stars: black holes;stars: early-type;stars: rotation;Astrophysics – Solar and Stellar Astrophysics;Astrophysics – High Energy Astrophysical Phenomena

The Tarantula Massive Binary Monitoring. IV. Double-lined photometric binaries

Mahy, L.; Almeida, L. A.; Sana, H.; Clark, J. S.; de Koter, A.; de Mink, S. E.; Evans, C. J.; Grin, N. J.; Langer, N.; Moffat, A. F. J.; Schneider, F. R. N.; Shenar, T.; Tramper, F.

Astronomy & Astrophysics, Volume 634, A119 (2020)

Abstract

Context. A high fraction of massive stars are found to be binaries but only a few of them are reported as photometrically variable. By studying the populations of double-lined spectroscopic binaries in the 30 Doradus region, we found a subset of them that have photometry from the OGLE project and that display variations in their light curves related to orbital motions.
Aims: The goal of this study is to determine the dynamical masses and radii of the 26 binary components in order to investigate the mass-discrepancy problem and to provide an empirical mass-luminosity relation for the Large Magellanic Cloud (LMC).
Methods: We use the PHOEBE programme to perform a systematic analysis of the OGLE V and I light curves obtained for 13 binary systems in the 30 Doradus region. We adopt the effective temperatures, and orbital parameters derived previously to obtain the inclinations of the systems and the parameters of the individual components.
Results: Three systems display eclipses in their light curves, while the others only display ellipsoidal variations. We classify two systems as over-contact, five as semi-detached, and four as detached. The two remaining systems have uncertain configurations due to large uncertainties on their inclinations. The fact that systems display ellipsoidal variations has a significant impact on the inclination errors. From the dynamical masses, luminosities, and radii, we provide LMC-based empirical mass-luminosity and mass-radius relations, and we compare them to other relations given for the Galaxy, the LMC, and the Small Magellanic Cloud. These relations differ for different mass ranges, but do not seem to depend on the metallicity regimes. We also compare the dynamical, spectroscopic, and evolutionary masses of the stars in our sample. While the dynamical and spectroscopic masses agree with each other, the evolutionary masses are systematically higher, at least for stars in semi-detached systems. This suggests that the mass discrepancy can be partly explained by past or ongoing interactions between the stars.

Keywords: stars: early-type;binaries: eclipsing;open clusters and associations: individual: 30 Doradus;stars: fundamental parameters;Astrophysics – Solar and Stellar Astrophysics

The Tarantula Massive Binary Monitoring. III. Atmosphere analysis of double-lined spectroscopic systems

Mahy, L.; Sana, H.; Abdul-Masih, M.; Almeida, L. A.; Langer, N.; Shenar, T.; de Koter, A.; de Mink, S. E.; de Wit, S.; Grin, N. J.; Evans, C. J.; Moffat, A. F. J.; Schneider, F. R. N.; Barbá, R.; Clark, J. S.; Crowther, P.; Gräfener, G.; Lennon, D. J.; Tramper, F.; Vink, J. S.

Astronomy & Astrophysics, Volume 634, A118 (2020)

Abstract

Context. Accurate stellar parameters of individual objects in binary systems are essential to constrain the effects of binarity on stellar evolution. These parameters serve as a prerequisite to probing existing and future theoretical evolutionary models.
Aims: We aim to derive the atmospheric parameters of the 31 double-lined spectroscopic binaries in the Tarantula Massive Binary Monitoring sample. This sample, composed of detached, semi-detached and contact systems with at least one of the components classified as an O-type star, is an excellent test-bed to study how binarity can impact our knowledge of the evolution of massive stars.
Methods: In the present paper, 32 epochs of FLAMES/GIRAFFE spectra are analysed by using spectral disentangling to construct the individual spectra of 62 components. We then apply the CMFGEN atmosphere code to determine their stellar parameters and their helium, carbon, and nitrogen surface abundances.
Results: Among the 31 systems that we study in the present paper, we identify between 48 and 77% of them as detached, likely pre-interacting systems, 16% as semi-detached systems, and between 5 and 35% as systems in or close to contact phase. Based on the properties of their components, we show that the effects of tides on chemical mixing are limited. Components on longer-period orbits show higher nitrogen enrichment at their surface than those on shorter-period orbits, in contrast to expectations of rotational or tidal mixing, implying that other mechanisms play a role in this process. For semi-detached systems, components that fill their Roche lobe are mass donors. They exhibit higher nitrogen content at their surface and rotate more slowly than their companions. By accreting new material, their companions spin faster and are likely rejuvenated. Their locations in the N – v sin i diagram tend to show that binary products are good candidates to populate the two groups of stars (slowly rotating, nitrogen-enriched objects and rapidly rotating non-enriched objects) that cannot be reproduced through single-star population synthesis. Finally, we find no peculiar surface abundances for the components in (over-)contact systems, as has been suggested by evolutionary models for tidal mixing.
Conclusions: This sample, consisting of 31 massive binary systems, is the largest sample of binaries composed of at least one O-type star to be studied in such a homogeneous way by applying spectral disentangling and atmosphere modelling. The study of these objects gives us strong observational constraints to test theoretical binary evolutionary tracks.

Keywords: stars: early-type;binaries: spectroscopic;open clusters and associations: individual: 30 Doradus;stars: fundamental parameters;Astrophysics – Solar and Stellar Astrophysics;Astrophysics – Astrophysics of Galaxies


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# Speed Distance Time Calculator

Formula: Speed = DistanceTime

Enter the unknown value as 'x'

Speed : meters/second

Distance: meters

Time: seconds

x =

The Speed Distance Time Calculator an online tool which shows Speed Distance Time for the given input. Byju's Speed Distance Time Calculator is a tool
which makes calculations very simple and interesting. If an input is given then it can easily show the result for the given number.

#### Practise This Question

Dimensional formula ML1T2 does not represent the physical quantity


text:
news

# Heart failure market worth $16 billion by 2026 as Novartis’ drug triggers growth 56 SHARES Share via The heart failure space across the seven key markets of the US, France, Germany, Italy, Spain, the UK and Japan is set to grow from$3.7 billion in 2016 to around \$16.1 billion by 2026, representing an impressive compound annual growth rate of 15.7%, according to research and consulting firm GlobalData.

The company’s latest report states that the strongest driver of this rise in market value will be the growing uptake of Novartis’ Entresto over the forecast period, despite initial modest sales. Other drivers will include the launch of several chronic heart failure therapies, including Amgen and Cytokinetics’ omecamtiv mecarbil, and an increase in the global prevalence of chronic heart failure and incidence of acute heart failure.

Elizabeth Hamson, PhD, Healthcare Analyst for GlobalData, explains: “Over the past two decades, chronic heart failure therapies have demonstrated success in slowing the progression of the disease and in reducing both mortality and morbidity in large-scale clinical trials. However, these successes have been limited to heart failure with reduced ejection fraction (HF-REF), showing only moderate benefits in heart failure with preserved ejection fraction (HF-PEF). Despite the lack of strong clinical evidence, guideline-recommended HF-REF therapies are widely used to treat HF-PEF.

“Based on this, drug developers have historically only targeted HF-REF in their drug strategies. With HF-PEF on the rise, however, this patient cohort represents a lucrative opportunity for pharmaceutical companies such as Novartis, which recently launched its first-in-class angiotensin receptor-neprilysin inhibitor Entresto in the US for HF-REF, and is currently conducting late-stage trials of the drug in HF-PEF patients.”

GlobalData anticipates Entresto’s label expansions to HF-PEF to be approved in 2020, which will boost the drug’s uptake dramatically. Due to the lack of evidence-based therapies for HF-PEF, if Entresto proves to be efficacious in this patient, it will help Novartis further penetrate the heart failure market, undoubtedly benefiting the company immensely.

Hamson concludes: “Although it is thought that Entresto will fulfill a major unmet need over the forecast period, it is important to acknowledge that others will remain. For example, effective treatment of patients with multiple comorbidities, particularly those with renal impairment, will remain elusive. GlobalData does not expect this unmet need to be fulfilled during the forecast period, although the recent FDA approval of several potassium-binding agents to treat hyperkalemia may relieve the burden of this unmet need to a slight extent.”

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### All Issues

Vol.10, 2020
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 Volume 5, Number 1, 2015, Pages 64-76                                                                DOI：10.11948/2015006 Traveling wavefronts of a delayed lattice reaction-diffusion model Li Shu,Peixuan Weng,Yanling Tian Keywords:Pioneer-climax model, lattice differential system, harmless delay, traveling wave solution, minimal wave speed. Abstract: We investigate a system of delayed lattice differential system which is a model of pioneer-climax species distributed on one dimensional discrete space. We show that there exists a constant $c^*>0$, such that the model has traveling wave solutions connecting a boundary equilibrium to a co-existence equilibrium for $c\geq c^*$. We also argue that $c^*$ is the minimal wave speed and the delay is harmless. The Schauder's fixed point theorem combining with upper-lower solution technique is used for showing the existence of wave solution. PDF      Download reader


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# solve this diophantine equation: ${\rm lcm}[x,y]+{\rm lcm}[y,z]+{\rm lcm}[z,x]=3(x+y+z)$

I am almost certain it is a duplicate question but I am looking for a reference regarding how solve the diophantine equation，Find the postive integer $x,y,z$ such $${\rm lcm}[x,y]+{\rm lcm}[y,z]+{\rm lcm}[z,x]=3(x+y+z)$$

Now that's a nice little problem. I didn't expect it to be solvable completely, but... wait a minute.

Obviously, a multiple of any solution is also a solution, so we may just as well divide it by $\gcd(x,y,z)$, if any, and look for the primitive triples. Now, being coprime as a triple does not mean being pairwise coprime, so we may assume $\gcd(x,y)=d_1, \gcd(x,z)=d_2,$ and $\gcd(y,z)=d_3$. Then $x=a\cdot d_1d_2, y=b\cdot d_1d_3$, and $z=c\cdot d_2d_3$, where $a,b,c,d_1,d_2,\text{ and }d_3$ are all pairwise coprime. (Upd. Not quite all of them, as explained in a comment by Litho, but $d_3$ is coprime to each of $d_1, d_2$, and $a$, which suffices for our purpose.) With that in mind, the equation becomes $$(ab+ac+bc)\cdot d_1d_2d_3=3a\cdot d_1d_2+3b\cdot d_1d_3+3c\cdot d_2d_3$$

Now, the LHS is apparently divisible by $d_3$, and so are two of the three summands in the RHS. Then $3ad_1d_2$ must be also divisible by $d_3$, which is only possible if $d_3$ is either $1$ or $3$. The same reasoning applies to $d_1$ and $d_2$. Being pairwise coprime, they can't equal 3 all at once, or any two of them. So we have two possible cases:

1. All gcd's equal 1;
2. One gcd equals 3 and the rest are 1.

In the first case the equation reduces to $$ab+ac+bc=3(a+b+c)$$ Considering the three sub-cases ($c=1,c=2,\text{ and }c\geqslant3$), we arrive at the solutions (1,3,9), (2,2,8), and (3,3,3), which are no good for us, since they violate the requirement of coprimality.

In the second case it is $$3ab+3ac+3bc=3(3a+3b+c)$$ which can be simplified to $$ab+ac+bc=3a+3b+c$$ Analyzing it in the similar manner, we discover some solutions which translate nicely to those of the original equation.

Finally, the primitive triples are: $$(x,y,z)=(3,3,5)\text{ or }(1,9,21)$$ The rest of solutions are the permutations of these two and multiples thereof.

• A minor correction: you say that if $x,y,z$ are coprime, then $a,b,c,d_1,d_2,\text{ and }d_3$ are all pairwise coprime. That's not true: $a,b$ and $c$ are pairwise coprime; $d_1, d_2$ and $d_3$ are pairwise coprime; in addition, $\gcd(a,d_3) = \gcd(b, d_2) = \gcd(c, d_1) = 1$. But,for example, $a$ doesn't have to be coprime with $d_1$: if $x=4, y=2, z=1$, then $a=d_1=2$. Fortunately, it doesn't affect the rest of your solution. – Litho Mar 29 '17 at 8:06
• That's right; I stand corrected. – Ivan Neretin Mar 29 '17 at 8:14


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Search: MSC category 54D45 ( Local compactness, $\sigma$-compactness )
 On the Hereditary Paracompactness of Locally Compact, Hereditarily Normal Spaces We establish that if it is consistent that there is a supercompact cardinal, then it is consistent that every locally compact, hereditarily normal space which does not include a perfect pre-image of $\omega_1$ is hereditarily paracompact. Keywords:locally compact, hereditarily normal, paracompact, Axiom R, PFA$^{++}$Categories:54D35, 54D15, 54D20, 54D45, 03E65, 03E35


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# Homework Help: Force and acceleration on sodium atom if laser bundle shines on atom

1. May 25, 2014

### jennyjones

1. The problem statement, all variables and given/known data
I want to calculate the average force and acceleration on an sodium atom, if a laser bundle shines on it
which has I>>>> I(sat). I(sat) is the saturation intensity of the 589nm 3s-3p transition in sodium.
3p lives 16 ns

2. Relevant equations

c = 3 * 10 ^ 8
m sodium = 3.8 * 10^-26 kg
planck = 6.6* 10 ^-34 Js
tua = 16 ns

3. The attempt at a solution

F = - h(bar)/(2*λ*(tau)) ????

not sure which formula to use

thanx,

jenny

2. May 26, 2014

### haruspex

Not an area I know anything about, but the formula you quoted looks almost right to me.
The h/λ part gives you the momentum of each photon. 1/tau, where tau is the remanence time in the excited state, gives you the max rate of excitation. The reradiation is in a random direction, so you should only count the momentum of the absorbed photons. Force equals rate of change of momentum, so momentum per excitation multiplied by rate of excitation events.

What I don't understand is the constant factor of 1/4π. Maybe someone who knows this area will comment.

3. May 26, 2014

### BvU

Hello Jenny,

Could you reveal a bit more of the context of your post ?
I am intrigued by the I >> Isat. I take it the laser is tuned to the 589 nm, but the spectral width isn't indicated: is it sharper than the D-line or much broader?

When you irradiate Na atoms, you get resonance fluorescence (in all directions) and you also get stimulated emission (in the same direction as the incoming photon). When the intensity in increased, the latter shortens the residence time in the upper level (power broadening, page 90). Hope this link works, otherwise: Laser Spectroscopy: Vol. 1: Basic Principles
By Wolfgang Demtröder

I seem to recall that with a broadly tuned laser you get line broadening and with a very sharply tune laser you can get line splitting, but I'm not that sure - long time ago.

However, momentum conservation goes: Cohen-Tannoudji and Dalibard -- real experts in this field, the former a Nobel laureate -- mention a recoil of the atom of around 3 cm/s on p 16 . ( $p = \hbar{\bf k} = h/\lambda$. No $1/(4\pi)$. The $1/\tau$ comes in as the number of photons that can be absorbed per second, as Haruspex already wrote).

CT & D work out radiation pressure on p 17; more or less what you are asking.

4. May 27, 2014

### jennyjones

thanks so much for the help!

i think i know how to solve the problem now!

F = dp/dt = - h/λτ
F = - 6.6* 10 ^-34/(589 * 10^-9 * 16 * 10^-9) = -7 * 10^-20

a = F/m = -7 *10^-20/ 3.8 * 10^-26 kg = -1.8 *10^6


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# Club sets and diagonal intersection

Let $$\kappa$$ be any regular cardinal, and let $$$$ be a sequence of club sets. Define their diagonal intersection $$\Delta C_i$$ as follows:

$$\Delta C_i = \{\alpha<\kappa \mid \forall i<\alpha:\alpha\in C_i\}$$

I must prove that this set is also club. Now, I would like to consider the following supposed counterexample. Clearly, $$\omega$$ is regular, so if I define the following sequence of sets:

$$C_i =\{p_i ^n \mid n\in\mathbb N\}$$

Where $$p_i$$ is the $$i$$th prime number, we have a sequence of infinite (trivially) club sets with no pairwise intersection. If I understand correctly, this means that every number larger than $$1$$ is an element of $$0$$ or $$1$$ sets, and is not an element of the diagonal intersection. In particular, the diagonal intersection contains only $$2$$ elements, and so it is not unbounded and not club.

I assume this counterexample fails somewhere, as I was supposed to prove the claim. Can anyone please help me locate my error?

• I looked at a few set theory books on my shelves (because I mainly know about the situation for $\kappa = {\omega}_1$ and wasn't sure of the exact hypothesis used for other cardinals), and the assumption for this diagonal intersection result is that $\kappa$ is a regular cardinal with uncountable cofinality. Perhaps you overlooked this aspect, or maybe what you're looking at overlooked saying this (or made this assumption earlier in the discussion and you didn't notice)? – Dave L. Renfro Dec 9 '18 at 17:22
• Ok, so there really was some confusion. Thank you for the clarification! – Uri George Peterzil Dec 9 '18 at 17:29
• This result (when $\kappa=cf (\kappa)\geq \omega_1$) is used in proving an important tool known as Fodor's Lemma, or the Pressing-Down Lemma. – DanielWainfleet Dec 9 '18 at 20:37

When talking about clubs/stationary sets/etc., we have to restrict to uncountable regular cardinals - or at least, ordinals of uncountable cofinality - to get a nontrivial theory. For example, both $$\{$$evens$$\}$$ and $$\{$$odds$$\}$$ are club in $$\omega$$, but their intersection is empty; so it's not even diagonal intersection that fails, but regular (hehe) intersection!


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# Counting the numbers between $1$ and $1,000,000$ whose digits sum to $30$

What's the number of numbers between $1$ and $1,000,000$ whose digits sum is $30$?

So I thought of this as a stars and sticks problem, so in the case you have $35\choose 5$ numbers whose sum is $30$. But obviously this number of numbers includes many overlaps (if for instance the requested sum was $8$, then $80=800=8000=134=143=431$ etc.). How do I get rid of the overlaps (inclusion-exclusion)?

Thanks in advance for any assistance!

-
Just a note: If each digit would be random from $\{0, \ldots, 9\}$ then the sum of $6$ digits would approximately be distributed as $\mathcal{N}(27, 99/2)$ leading to about $\approx 51,750$ out of every $1,000,000$ numbers with sum $30$. This is pretty close to the actual number. –  TMM May 11 '13 at 18:29
(The lazy Mathematica-answer is $50,877$.) –  TMM May 11 '13 at 19:53

The problem isn’t overlaps: it’s restricting the values to the range from $0$ through $9$. You’re looking for the number of solutions in non-negative integers to the equation

$$x_1+x_2+x_3+x_4+x_5+x_6=30\tag{1}$$

with the added requirement that $x_k\le 9$ for $k=1,\dots,6$: each of those solutions defines one of the numbers that you’re trying to count, and vice versa. The number of solutions without the upper bound restriction is, as you say,

$$\binom{30+6-1}{6-1}=\binom{35}5\;.$$

However, some of those solutions require ‘digits’ greater than $9$. Start by counting the solutions in non-negative integers to $(1)$ that have $x_1\ge 10$. Those are clearly in one-to-one correspondence with solutions in non-negative integers to

$$y_1+x_2+x_3+x_4+x_5+x_6=20\;,$$

where $y_1=x_1-10$, and you know how to compute their number. Call that number $n_1$. There are just as many solutions to $(1)$ that violate the upper bound on $x_2$, just as many again that violate the upper bound on $x_3$, and so on, so a better approximation to the solution to the original problem is

$$\binom{35}5-6n_1\;.$$

Of course this overcorrects, since it’s possible for a solution to $(1)$ to violate two of the upper bound constraints at once. You’ll have to calculate the number $n_2$ of solutions that have $x_1\ge 10$ and $x_2\ge 10$ and then add back the appropriate multiple of $n_2$. And since it’s actually possible to violate as many as three of the constraints at once, you’ll have to make one further inclusion-exclusion correction.

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Hi, thank you for the insight. I didn't think of the upper boundry! Now, as for you corrections. Would $n_1={20\choose 5}$? Also, I didn't understand your last paragraph. How can two of the upper bound constraints be violated at once? Or in short, I didn't understand the last paragraph idea :/ –  ohad May 11 '13 at 17:53
@ohad: Calculating $n_1$ is a stars-and-bars problem: $n_1=\binom{20+6-1}{6-1}=\binom{25}5$. The solution $x_1=x_2=15$, $x_3=x_4=x_5=x_6=0$ violates the upper bound constraints on both $x_1$ and $x_2$. The $6$-‘digit’ number $(11)23(12)11$ violates the constraints on $x_1$ and $x_4$. –  Brian M. Scott May 11 '13 at 18:03
Yes, my mistake. I understand now why there can be two and even three violations on that constraint, but how does that reflect on the number of numbers? Thanks and sorry to bother you so much. –  ohad May 11 '13 at 18:15
@ohad: It means that when we subtract $6\binom{25}5$ to get rid of the solutions that exceed one of the bounds, each solution that exceeds two bounds get subtracted twice and therefore counted $-1$ times in the total $\binom{35}5-6\binom{25}5$. We want to count it $0$ times, so we have to add it back in. There are $\binom{15}5$ solutions to $(1)$ that exceed the bounds on $x_1$ and $x_2$ (why?), and there are $\binom62$ pairs of variables, so there are $\binom62\binom{15}5$ solutions to $(1)$ that violate at least two constraints and have to be added back in. But if you check, ... –  Brian M. Scott May 11 '13 at 18:19
... you’ll find that every solution that violates three constraints (and there are some) has been counted once in $\binom{35}5$, $3$ times in $6\binom{25}5$, and $3$ times in $\binom62\binom{15}5$ (why?), for a total of $1-3+3=1$ time. But we don’t want to count those solutions, so we need to subtract them from $\binom{35}5-6\binom{25}5+\binom62\binom{15}5$. And that’s the last such correction that’s needed, since it’s impossible to get a total of $30$ with $4$ or more variables violating the constraint. –  Brian M. Scott May 11 '13 at 18:21

One looks for the number $N$ of integer solutions of $$\sum_{i=1}^6n_i=30,\qquad 0\leqslant n_i\leqslant9.$$ Thus, $N$ is the coefficient of $x^{30}$ in the polynomial $$\prod_{i=1}^6\sum_{n_i=0}^9x^{n_i}=\left(\frac{1-x^{10}}{1-x}\right)^6.$$ To compute $N$, note that $$(1-x^{10})^6=\color{red}{1}-\color{red}{6}x^{10}+\color{red}{15}x^{20}-\color{red}{20}x^{30}+o(30),$$ and that $$\frac{1}{(1-x)^6}=\sum_{k=0}^{30}\color{green}{{5+k\choose5}}x^k+o(30),$$ where $o(30)$ denotes terms of order at least $31$, hence $$N=\color{red}{1}\cdot\color{green}{{5+30\choose5}}-\color{red}{6}\cdot\color{green}{{5+20\choose5}}+\color{red}{15}\cdot\color{green}{{5+10\choose5}}-\color{red}{20}\cdot\color{green}{{5+0\choose5}}=50877.$$

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If I remember correctly, this is the first exercise in Polya-Szego's treatise. –  Did May 11 '13 at 19:43

HINT: Can you find the number of integer soloutions of the equation $$x_1+x_2+x_3+x_4+x_5+x_6=30$$ such that $0\leq x_i\leq 9$ for all $i$. Use generating functions!

-


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Parallel Plate Capacitors Video Lessons

Concept

# Problem: Describe qualitatively but specifically how the circuit quantities change after the switch is closed in the circuit a) current b) voltage across the capicitor. Explain why these quantities change as they do, in terms of motion and presence of the capacitor.

###### FREE Expert Solution

a)

If the switch is closed, the capacitor becomes a short circuit.

The current in this circuit is given by:

$\overline{){\mathbf{i}}{\mathbf{=}}\frac{{\mathbf{V}}_{\mathbf{1}}\mathbf{+}{\mathbf{V}}_{\mathbf{2}}}{\mathbf{R}}}$

93% (211 ratings)
###### Problem Details

Describe qualitatively but specifically how the circuit quantities change after the switch is closed in the circuit a) current b) voltage across the capicitor. Explain why these quantities change as they do, in terms of motion and presence of the capacitor.


text:
# If alpha, beta and gamma are the roots of the equation x^(3)+x+2=0, then the equation whose roots are (alpha- beta)(alpha-gamma), (beta-gamma)(beta-gamma) and (gamma-alpha)(gamma-alpha) is

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Text Solution

x^(3)-6x^(2)+216=0x^(3)-3x^(2)+112=0x^(3)+6x^(2)-216=0x^(3)+3x^(2)-112=0

D

Transcript

TimeTranscript
00:00 - 00:59hello friend here we have a question that alpha beta gamma are the root of the equation x cube + X + 2 equal to zero then the equation whose roots are so we have to find the equation whose roots are this one right now if alpha beta if Alpha is the root of the equation that mean Alpha will satisfy in this in this equation and hear this Alpha minus beta into Alpha minus Gama is the root of the equation let's suppose that has why right so that will so that will lies in the in the equation right so hair Alpha minus beta into Alpha minus Gama this week and write it as Alpha square minus alpha + beta
01:00 - 01:59plus Bita Gama write this value will lies on that equations route that equation right now here the equation we have x cube + X + 2 now as we know that some of the route that is alpha + beta + Gamma equal to minus of coefficient of x square upon coefficient of x cube here we haven't any coefficient of x square so this value is equal to 0 right and the product of the route that is alpha beta gamma is equal to minus of coefficient of upon my upon coefficient of x cube Sohail this value is -2 right so here we can put hair if I want to find in Alpha so we will get alpha value is equal to
02:00 - 02:59- of Beta + Gamma in hair my alpha value that is alpha beta value equal to beta + Gamma value equal to minus alpha and beta gamma value equal to minus 2 by Alpha Sohail minus 2 by Alpha equal to beta gamma right I can put this value here so my bi we will get why will be Alpha square minus Alpha into minus alpha plus minus of to buy Alpha after solving this we will get Two Alpha cube minus 2
03:00 - 03:59Alpha now the equation that we have x cube + X + 2 equal to 0 9 Alpha is its root so I will satisfy this equation that is alpha Cube + alpha + 2 equal to zero so here are my alpha Q value equal to minus of Alpha minus of two so I can put here this is 2 into minus of Alpha minus of 2 minus 2 upon Alpha that is I will get my wind terms of Alpha so hair this will come as - 4 - 2 alpha minus 2 upon Alpha this will be minus x minus 2 alpha upon Alpha this is minus 6 by Alpha minus 2 equal to y
04:00 - 04:59write some Alpha value equal to minus of 2 upon this is - of 6 - of sex upon by plus two right now this Alpha lies on that equation that is here I am writing that Alpha lies on equation x cube + X + 2 equal to zero so we can put here that is minus x upon Y + 2 whole cube here we will get - of sex Y + 2 + 2 equal to zero the solving this we will get one
05:00 - 05:5928 - - of 3 into Y + 2 square + Y + 2 this is here we have 2 Y + 2 cube equal to zero so this we will get - of 108 - of 3 y square - 12 - 12 + y cube + 8 + 6 y square + 12 Y equal to zero write this term and this item will cancel out here we will have equal to y cube
06:00 - 06:59Plus 3 y square - 112 so this is the equation which we have to find thank


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Category:Definitions/Inverse Tangent

This category contains definitions related to Inverse Tangent.
Related results can be found in Category:Inverse Tangent.

The inverse tangent is a multifunction defined on $S$ as:

$\forall z \in S: \tan^{-1} \left({z}\right) := \left\{{w \in \C: \tan \left({w}\right) = z}\right\}$

where $\tan \left({w}\right)$ is the tangent of $w$.

Pages in category "Definitions/Inverse Tangent"

The following 14 pages are in this category, out of 14 total.


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${{\boldsymbol \Lambda}}$ BARYONS($\boldsymbol S$ = $-1$, $\boldsymbol I$ = 0) ${{\mathit \Lambda}^{0}}$ = ${\mathit {\mathit u}}$ ${\mathit {\mathit d}}$ ${\mathit {\mathit s}}$ INSPIRE search

# ${{\boldsymbol \Lambda}{(1380)}}$

See the related review on "Pole Structure of the ${{\mathit \Lambda}{(1405)}}$ Region."
${{\boldsymbol \Lambda}{(1380)}}$ POLE POSITION
 REAL PART
 $-2{\times }$IMAGINARY PART


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# Error when Adding a Python Script to a Driver

I seem to have a bit of a problem that I was wondering if someone could help me with. As the title says, for some reason I keep getting an error when I try to add my script to the driver, and I can’t seem to figure out why at all. Here is the screenshot of how I’m applying it to the driver:

and here is the code:

import bpy
import math

def ShiftMoonMonth(objectName):
obj = bpy.data.objects[objectName]
sourceRotation = obj.rotation_euler.z
finalRotation = (math.floor(sourceRotation/389.10)*29.10564039280042)
return finalRotation

bpy.app.driver_namespace['ShiftMoonMonth'] = ShiftMoonMonth

I have auto-run python scripts turned on, and when I look through the available namespaces in the blender console, ShiftMoonMonth is one of the options I can use. So I’m pretty sure that the code is linked to the driver, but for some reason it’s still giving this error.

The place I got the code template from was this tutorial here: https://youtu.be/rfEFU0Mdq3Q?t=21m50s at 21:50, where he is trying to solve a very similar problem to what I’m solving. As far as I can tell, it seems to be the same as what I have, although it is quite old, so there may be something else you have to do. I also don’t understand the language, so I may have missed something.

Anyway, any help would be very appreciated!

For a single property type driver variable you need also to set the path value. For example

import bpy

def x(name):
print(name)
return 0

bpy.app.driver_namespace["x"] = x

for example shown prints "Lamp", the name of the lamp object, to the console. Ok, string properties can be passed via driver variables. (personally don't recommend it) (note also for this case, the expression could simply be x("Lamp") with no need for a driver variable)

Not setting path for the single property driver variable will result in "None" being printed to console.

Ok, now that we know a name can be passed, it's best to check for errors with something like,

if name is None:
print("problem with name variable")
return 0

obj = bpy.context.scene.objects.get(name)

if obj is None:
print("no object named:", name)
return 0

as errors will mark the driver as "bad" which can result it in being turned off.

A more usual way would be, where the x Euler rotation of the Lamp object, is passed as a float variable, "var", to the expression floor(2 * var / pi).

PS also of note is that rotation in blender is in radians.

You're using an object reference, but your code expects a name.

You may want to use rotation_euler as data path setting, and use that as parameter value. It'll make your code easier (and probably solves this problem).

It also avoids accessing objects by name; name based access is usually a bad idea, since names are not unique. Objects linked from other blend files may have the same name.

(I had to put this here to have enough characters to reply)

Ah, I see now. bpy.data.objects[ ] is looking for the name of an object, and if I had gone the next step and set the Driver’s Data Path to rotation_euler, I would have been trying to pass the rotation into it when it was looking for the name. The reason it was giving the error was that it didn’t have any name being passed into it.

To help out anyone in the future, here is the way it’s set up so that it works:

and how the working code looks with the object reference omitted:

import bpy
import math

def ShiftMoonMonth(valueToGet):
#obj = bpy.data.objects[objectName]
#sourceRotation = obj.rotation_euler.z
finalRotation = (math.floor(valueToGet/389.10)*29.10564039280042)
return finalRotation

bpy.app.driver_namespace['ShiftMoonMonth'] = ShiftMoonMonth

That’s great to know that just adding floor to the expression area works as well, batFINGER. I had thought I would need to import the math library to get it to work, which is why I went the script route in the first place. Thanks for the tip of using radians, too. That would have been very difficult trying to figure out why the calculations weren’t working with degrees. Everything is working perfectly now. :)

Thanks so much to you both for taking the time to answer it!

Is there a way to mark both of your answers as correct, or do I need to choose just one?


text:
# Apostol's Calculus - Prove the set is open (inequality)

This is from Apostol's Calculus, Vol. II Section 8.3 #3(a).

Prove that $S=\{(x,y,z) \mid z^2 - x^2 - y^2 -1>0\}$ is open.

The only way to prove that a set is open which has been covered so far is to prove that for an arbitrary point $\mathop a\limits^{\small \to} \in S$ there is an open ball $B(\mathop a\limits^{\small \to} , r) \subset S$ by finding an explicit $r$ and showing that $\mathop x\limits^{\small \to} \in B(\mathop a\limits^{\small \to} , r) \implies \mathop x\limits^{\small \to} \in S$. Nothing topological (compactness, completeness, etc.) has been discussed. For some reason I just am stuck trying to figure out an $r$ which would work.

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Choose arbitrary $(a,b,c)\in S$. Assume without loss of generality that $a,b,c\geq 0$. Put $k=c^2-a^2-b^2-1$.

Then, for any $(a',b',c')\in B((a,b,c),r)$ (with $0<r<c$, $r$ to be specified later -- notice that $c>0$) we have that $$c'^2-a'^2-b'^2-1\geq (c-r)^2-(a+r)^2-(b+r)^2-1=k-2r(a+b+c)-r^2$$ Thus, we can see that if we choose $r>0$ small enough to have $r^2+2r(a+b+c)<k$ (and $r<c$), $B((a,b,c),r)$ will be contained in $S$.

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For the main inequality, do you use $c'^2\ge(c-r)^2$ and so on for each one? I'm trying to see how to get there from having $\sqrt{(a'-a)^2 + (b'-b)^2 + (c'-c)^2}<r$ –  process91 Aug 14 '12 at 1:51
@MichaelBoratko: Yes. This is obvious: for $(a',b',c')$ in this ball, $c'$ belongs to the interval $(c-r,c+r)$, and of course $(c-r)^2$ is smaller than the square of any number in that interval (because they're all positive). Similarly for $a',b'$. –  tomasz Aug 14 '12 at 2:00
Should we also have $r<a$ and $r<b$ to be sure that $a'$ and $b'$ are positive? –  process91 Aug 14 '12 at 2:06
@MichaelBoratko: You can do that, but that is not necessary. If $a\geq 0$, then $(a+r)^2$ is greater than the square of any number in the interval $(a-r,a+r)$, whether or not there are negative numbers within. –  tomasz Aug 14 '12 at 2:09
OK, good deal. Thank you very much! –  process91 Aug 14 '12 at 2:19

Note that $|x-x'| \leq \|(x,y,z)-(x',y',z')\|$. Furthermore, we have $|x^2-(x')^2| = |x-x'| |x+x'|$. Hence if $|x|, |x'| < L$, then $|x^2-(x')^2| \leq 2 L \|(x,y,z)-(x',y',z')\|$. The same applies to $y,y', z, z'$, of course.

Pick $(x,y,z) \in S$, let $M = \max\{|x|,|y|,|z|\}$ and $r = \min \{M, \frac{1}{6M}(z^2-(1+x^2+y^2))\}$ (which is $>0$, of course).

Suppose $\|(x,y,z)-(x',y',z')\| < r$. Then we have $|x^2-(x')^2| < 2M r$, and similarly for the other coordinates.

Finally, we have $(z')^2-(1+(x')^2+(y')^2) > z^2-2Mr - (1 +x^2+y^2) - 4Mr$. By choice of $r$, we have $z^2 - (1 +x^2+y^2) \geq 6Mr$, from which we obtain $(z')^2-(1+(x')^2+(y')^2) > 0$, hence $(x',y',z') \in S$. It follows that $S$ is open since $B((x,y,z), r) \subset S$.

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Oops - I had the inequality going the wrong way in my question - I will fix it –  process91 Aug 14 '12 at 1:28
My repair was interrupted by dinner! –  copper.hat Aug 14 '12 at 3:04

Let $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ be the function $f(x, y, z) = z^2 - x^2 - y^2 - 1$. Since $f$ is continuous and $(0, \infty)$ is open in $\mathbb{R}$, $S = f^{-1}((0, \infty))$ is open.

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He wrote that he it not using anything topological. –  copper.hat Aug 14 '12 at 1:19
I do not have the theorem to back this up yet - only what I put above. Although I suppose I could rip into the details of this in order to reach that conclusion, I doubt that is what was intended. –  process91 Aug 14 '12 at 1:19
A pity the OP can't use this since it is, by far, the easiest solution, though I believe it must be done with $\,(0,\infty)\,$ –  DonAntonio Aug 14 '12 at 2:33
@DonAntonio Someone edited the question. Originally it was $(-\infty, 0)$. –  Makoto Kato Aug 14 '12 at 3:36
Indeed @MakotoKato . –  DonAntonio Aug 14 '12 at 3:41


text:
# Cp(omega 1 + 1) is not normal

In this and subsequent posts, we consider $C_p(X)$ where $X$ is a compact space. Recall that $C_p(X)$ is the space of all continuous real-valued functions defined on $X$ and that it is endowed with the pointwise convergence topology. One of the compact spaces we consider is $\omega_1+1$, the first compact uncountable ordinal. There are many interesting results about the function space $C_p(\omega_1+1)$. In this post we show that $C_p(\omega_1+1)$ is not normal. An even more interesting fact about $C_p(\omega_1+1)$ is that $C_p(\omega_1+1)$ does not have any dense normal subspace [1].

Let $\omega_1$ be the first uncountable ordinal, and let $\omega_1+1$ be the successor ordinal to $\omega_1$. The set $\omega_1$ is the first uncountable ordinal. Furthermore consider these ordinals as topological spaces endowed with the order topology. As mentioned above, the space $\omega_1+1$ is the first compact uncountable ordinal. In proving that $C_p(\omega_1+1)$ is not normal, a theorem that is due to D. P. Baturov is utilized [2]. This theorem is also proved in this previous post.

For the basic working of function spaces with the pointwise convergence topology, see the post called Working with the function space Cp(X).

The fact that $C_p(\omega_1+1)$ is not normal is established by the following two points.

• If $C_p(\omega_1+1)$ is normal, then $C_p(\omega_1+1)$ has countable extent, i.e. every closed and discrete subspace of $C_p(\omega_1+1)$ is countable.
• There exists an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$.

We discuss each of the bullet points separately.

____________________________________________________________________

The function space $C_p(\omega_1+1)$ is a dense subspace of $\mathbb{R}^{\omega_1}$, the product of $\omega_1$ many copies of $\mathbb{R}$. According to a result of D. P. Baturov [2], any dense normal subspace of the product of $\omega_1$ many separable metric spaces has countable extent (also see Theorem 1a in this previous post). Thus $C_p(\omega_1+1)$ cannot be normal if the second bullet point above is established.

____________________________________________________________________

Now we show that there exists an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$. For each $\alpha$ with $0<\alpha<\omega_1$, define $h_\alpha:\omega_1 + 1 \rightarrow \left\{0,1 \right\}$ by:

$h_\alpha(\gamma) = \begin{cases} 1 & \mbox{if } \gamma \le \alpha \\ 0 & \mbox{if } \alpha<\gamma \le \omega_1 \end{cases}$

Clearly, $h_\alpha \in C_p(\omega_1 +1)$ for each $\alpha$. Let $H=\left\{h_\alpha: 0<\alpha<\omega_1 \right\}$. We show that $H$ is a closed and discrete subspace of $C_p(\omega_1 +1)$. The fact that $H$ is closed in $C_p(\omega_1 +1)$ is establish by the following claim.

Claim 1
Let $h \in C_p(\omega_1 +1) \backslash H$. There exists an open subset $U$ of $C_p(\omega_1 +1)$ such that $h \in U$ and $U \cap H=\varnothing$.

First we get some easy cases out of the way. Suppose that there exists some $\alpha<\omega_1$ such that $h(\alpha) \notin \left\{0,1 \right\}$. Then let $U=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in \mathbb{R} \backslash \left\{0,1 \right\} \right\}$. Clearly $h \in U$ and $U \cap H=\varnothing$.

Another easy case: If $h(\alpha)=0$ for all $\alpha \le \omega_1$, then consider the open set $U$ where $U=\left\{f \in C_p(\omega_1 +1): f(0) \in \mathbb{R} \backslash \left\{1 \right\} \right\}$. Clearly $h \in U$ and $U \cap H=\varnothing$.

From now on we can assume that $h(\omega_1+1) \subset \left\{0,1 \right\}$ and that $h$ is not identically the zero function. Suppose Claim 1 is not true. Then $h \in \overline{H}$. Next observe the following:

Observation.
If $h(\beta)=1$ for some $\beta \le \omega_1$, then $h(\alpha)=1$ for all $\alpha \le \beta$.

To see this, if $h(\alpha)=0$, $h(\beta)=1$ and $\alpha<\beta$, then define the open set $V$ by $V=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in (-0.1,0.1) \text{ and } f(\beta) \in (0.9,1.1) \right\}$. Note that $h \in V$ and $V \cap H=\varnothing$, contradicting that $h \in \overline{H}$. So the above observation is valid.

Now either $h(\omega_1)=1$ or $h(\omega_1)=0$. We claim that $h(\omega_1)=1$ is not possible. Suppose that $h(\omega_1)=1$. Let $V=\left\{f \in C_p(\omega_1 +1): f(\omega_1) \in (0.9,1.1) \right\}$. Then $h \in V$ and $V \cap H=\varnothing$, contradicting that $h \in \overline{H}$. It must be the case that $h(\omega_1)=0$.

Because of the continuity of $h$ at the point $x=\omega_1$, of all the $\gamma<\omega_1$ for which $h(\gamma)=1$, there is the largest one, say $\beta$. Now $h(\beta)=1$. According to the observation made above, $h(\alpha)=1$ for all $\alpha \le \beta$. This means that $h=h_\beta$. This is a contradiction since $h \notin H$. Thus Claim 1 must be true and the fact that $H$ is closed is established.

Next we show that $H$ is discrete in $C_p(\omega_1 +1)$. Fix $h_\alpha$ where $0<\alpha<\omega_1$. Let $W=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in (0.9,1.1) \text{ and } f(\alpha+1) \in (-0.1,0.1) \right\}$. It is clear that $h_\alpha \in W$. Furthermore, $h_\gamma \notin W$ for all $\alpha < \gamma$ and $h_\gamma \notin W$ for all $\gamma <\alpha$. Thus $W$ is open such that $\left\{h_\alpha \right\}=W \cap H$. This completes the proof that $H$ is discrete.

We have established that $H$ is an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$. This implies that $C_p(\omega_1 +1)$ is not normal.

____________________________________________________________________

Remarks

The set $H=\left\{h_\alpha: 0<\alpha<\omega_1 \right\}$ as defined above is closed and discrete in $C_p(\omega_1 +1)$. However, the set $H$ is not discrete in a larger subspace of the product space. The set $H$ is also a subset of the following $\Sigma$-product:

$\Sigma(\omega_1)=\left\{x \in \mathbb{R}^{\omega_1}: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \omega_1 \right\}$

Because $\Sigma(\omega_1)$ is the $\Sigma$-product of separable metric spaces, it is normal (see here). By Theorem 1a in this previous post, $\Sigma(\omega_1)$ would have countable extent. Thus the set $H$ cannot be closed and discrete in $\Sigma(\omega_1)$. We can actually see this directly. Let $\alpha<\omega_1$ be a limit ordinal. Define $t:\omega_1 + 1 \rightarrow \left\{0,1 \right\}$ by $t(\beta)=1$ for all $\beta<\alpha$ and $t(\beta)=0$ for all $\beta \ge \alpha$. Clearly $t \notin C_p(\omega_1 +1)$ and $t \in \Sigma(\omega_1)$. Furthermore, $t \in \overline{H}$ (the closure is taken in $\Sigma(\omega_1)$).

____________________________________________________________________

Reference

1. Arhangel’skii, A. V., Normality and Dense Subspaces, Proc. Amer. Math. Soc., 48, no. 2, 283-291, 2001.
2. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.

____________________________________________________________________
$\copyright \ 2014 \text{ by Dan Ma}$


text:
LLVM  8.0.0svn

#include "llvm/BinaryFormat/COFF.h"

[legend]

## Public Attributes

uint16_t Machine

int32_t NumberOfSections

uint32_t TimeDateStamp

uint32_t PointerToSymbolTable

uint32_t NumberOfSymbols

uint16_t Characteristics

## Detailed Description

Definition at line 66 of file COFF.h.

## ◆ Characteristics

Definition at line 73 of file COFF.h.

## ◆ Machine

Definition at line 67 of file COFF.h.

## ◆ NumberOfSections

Definition at line 68 of file COFF.h.

Referenced by encodeBase64StringEntry(), and isAssociative().

## ◆ NumberOfSymbols

Definition at line 71 of file COFF.h.

Referenced by encodeBase64StringEntry(), and isAssociative().

## ◆ PointerToSymbolTable

Definition at line 70 of file COFF.h.

Referenced by encodeBase64StringEntry(), and isAssociative().

Definition at line 72 of file COFF.h.

Referenced by encodeBase64StringEntry().