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# Finding the radian value of this angle which passes through a point

1. Oct 29, 2013

### Ace.

1. The problem statement, all variables and given/known data
The terminal arm of an angle in standard position passes through (-7,8). Find the radian value of the angle in the interval [0,2∏], to the nearest hundredth.

2. Relevant equations
sinθ = $\frac{y}{r}$
cosθ = $\frac{x}{r}$
tanθ = $\frac{y}{x}$

3. The attempt at a solution
The terminal arm is in quadrant 2, and I found the side lengths to be -7,8, and $\sqrt{113}$ (hypotenuse). When I tried to find the value of θ I get different answers for different ratios.

θ = sin$^{-1}$$\frac{8}{\sqrt{113}}$
= 0.85

θ = cos$^{-1}$$\frac{-7}{\sqrt{113}}$
= 2.29

θ = tan$^{-1}$$\frac{8}{-7}$
= -0.85

The correct one is θ=2.29. Why is this correct and not the others?

2. Oct 29, 2013

### Staff: Mentor

The point -7,8 is in the second quadrant. So θ is less than π and greater than π/2.

3. Oct 29, 2013

### Staff: Mentor

Because, as you pointed out, the angle is in the 2nd quadrant. This means that θ is between $\pi/2$ and $\pi$ (roughly, between 1.5 and 3).

The sin-1 function returns an angle between $-\pi/2$ and $\pi/2$, as does tan-1. The cos-1 function returns an angle between 0 $\pi$, which was the correct interval for your angle.


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# What is the purpose of torque in projectile motion?

Torque, $\tau$ is defined as:

$$\vec{{\tau}} = \vec{r}\times\vec{F}$$

Intuitively, it's the rotational effect of force. However, using the above definition of torque we can calculate the torque due to a force acting on any particle about a reference point.

Consider projectile motion for instance. A particle was projected at an angle $\theta$ with a speed $u$. I calculated the torque due to weight at the highest point of the particle's trajectory about the point of projection to be $-mu^2\sin^2\theta$.

The thing that confuses me is that here there's torque but no rotation. What is the purpose of torque in projectile motion then?

• The angular position of the projectile with respect to the selected center around which rotation is measured changes, so there is 'rotation', and the question is founded on an misapprehension. Not that there is any point in working the problem in that way unless you are about to generalize to orbital motion but it is an internally consistent way to approach the problem. – dmckee Sep 25 '17 at 15:54

The equations that you have obtained for the trajectory of a projectile can be obtained by considering the torque which is acting on a mass $m$ and equating the torque to the rate of change of angular momentum.

Consider a mass which is projected at an angle $\theta$ to the horizontal from location $O$ at position $\vec r_{\rm O} = 0 \hat i+0 \hat j$ with velocity $\vec v_0 = \dot x_0 \hat i + \dot y_0 \hat j$ at time $t=0$.

At time $t$ the new position of the mass is $\vec r = x \hat i + y \hat j$ and it has a velocity $\vec v = \dot x \hat i + \dot y \hat j$.

The constant force acting on the mass is $-mg \hat j$.

At time $t$ the torque acting on the mass about location $O$ is

$\tau_{\rm O} = (x \hat i + y \hat j) \times (-mg\hat j) = -mgx\,\hat k$

where $\hat k$ is the unit vector coming out of the screen.

At time $t$ the angular momentum of the mass about location $O$ is

$L_{\rm O} = (x \hat i + y \hat j) \times m(\dot x \hat i + \dot y \hat j) = m(x \dot y - \dot x y)\, \hat z \Rightarrow \dfrac{d\vec L_{\rm O}}{dt} = m(x \ddot y - \ddot xy)\,\hat z$

$\vec \tau_{\rm O} = \dfrac{d\vec L_{\rm O}}{dt} \Rightarrow -gx = x \ddot y - \ddot xy$.

As the last equation must be true for all $x$ and $y$ this means that

$\ddot x =0$ and $\ddot y = -g$

which are the equations that you get by applying Newton's second law to the horizontal and vertical motions of a projectile.

In this case you could say that the torque has caused a change of direction, "rotation", of the velocity vector or you could say that the position vector $\vec r$ has been "rotated" but the key idea is that the torque has changed the angular momentum.

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States Of Matter : Gases And Liquids

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At 27^@C, the ratio of RMS vel...

Updated On: 27-06-2022

Get Answer to any question, just click a photo and upload the photo and get the answer completely free,

Text Solution

sqrt(3//5)sqrt(4//3)sqrt(2//3)0.25.

Solution : V_(RMS)=sqrt((3RT)/M)

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Transcript

at 27 degree Celsius the ratio of RMS velocity is ozone to oxygen is so we know that we are MS equals to square root of 3 Aarti on May I am now molecular mass of ozone is equal to 48 and Molecular mass of oxygen 32 not the values 3rt constant values

only variable here is molar mass RMS velocity of ozone and RMS velocity of oxygen the only variable thing here is molecular mass of oxygen upon molecular mass of ozone will be equal to 32 and 48 the final value will be under root ok 2 upon 3 Udaipur ratio of RMS velocities of ozone to aux

square root of 2 by 3 match with the option so the option will be the right answer thank you


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## anonymous one year ago ques

1. anonymous

For any level surface S given by $\phi(x,y,z)=c$ Then the region R that is it's projection on xy plane is always given by $\phi(x,y,0)=c$ ?

2. anonymous

That seems true .

3. anonymous

Is this a true false question?

4. anonymous

Nope, it's a doubt that I have

5. anonymous

we can do an example

6. anonymous

Idk, can you think of a counter example??

7. anonymous

can't think of a counterexample

8. anonymous

hmmm let's ask someone else @IrishBoy123

9. anonymous

maybe he can tell :P

10. IrishBoy123

let's test it first with a simple surface, a plane, which i shall just make uo $\pi: 2x + 3y + 4z = 5$ $\phi(x,y,z) = 2x + 3y + 4z [= const]$ $\phi(x,y,0) = 2x + 3y =5$ you might think of this in terms of the line of intersection of plane $$\pi$$ and the xy plane. in practise the projection is everywhere. the projection can be found from the surface integral formulation $\iint_{R} \frac{1}{|\hat n \bullet \hat k|} \ dx \ dy$ $= \frac{1}{4} \iint_{R} \ dx \ dy$

11. IrishBoy123

maybe we need to understand what projection means "intersection"? then yes. just set z to zero for intersection projection, no

12. IrishBoy123

another example paraboloid $$z = x^2 + y^2 - 4$$ $$\phi(x,y,z) = x^2 + y^2 - z = const \ \ [ie \ 4]$$ $$\phi(x,y,0) = x^2 + y^2 = 4$$ that's the circle on the xy plane, ie the *intersection*

13. anonymous

$\frac{1}{4}\int\limits_{0}^{\frac{2}{5}} \int\limits_{0}^{\frac{5-2x}{3}}dydx=\frac{1}{12}\int\limits_{0}^{\frac{2}{5}}(5-2x)dx=\frac{1}{12}[5x-x^2]_{0}^{\frac{2}{5}}$ Yeh sorry my bad, what I mean is that the equation we can form from $\phi(x,y,0)$ is what we will use as a limit when we transform from S to R

14. anonymous

What I'm trying to say is something like $\phi(x,y,0)$ will always give a curve in xy plane that bounds R

15. IrishBoy123

that looks good to me!

16. IrishBoy123

i think that's right

17. anonymous

So there's no reason to draw a graph every single time, we can just set z=0 and get our curve's equation, then we can double integrate within the limits

18. IrishBoy123

well, i always draw. always. seeing is believing. thing is you can get the line of intersection but you might still want to know if it is above or below the xy plane loads of people on here don't draw and they seem fine. to me, it's madness :-) and by draw, i mean rough sketch with key numbers. i use drawings to get the limits right, for example. you know, switch you integral from dydx to dxdy? i'll do that with my sketch. the integral http://www.wolframalpha.com/input/?i=%5Cfrac%7B1%7D%7B4%7D%5Cint%5Climits_%7B0%7D%5E%7B%5Cfrac%7B2%7D%7B5%7D%7D+%5Cint%5Climits_%7B0%7D%5E%7B%5Cfrac%7B5-2x%7D%7B3%7D%7Ddydx

19. anonymous

Good insight, this time I'll have less questions for my exams because it's university exams, in schools we had like 30 questions, but this time there will be around 7 questions of high weightage and each question would require more depth and explanation, so it might just help to draw a graph actually !!

20. IrishBoy123

ah! 2/5 in limit should be 5/2 ?!?! i spotted that from my sketch :-))

21. anonymous

Oops, $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ Not $ax$!!

22. anonymous

25/48

23. anonymous

Once again thanks!!

24. IrishBoy123


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# zbMATH — the first resource for mathematics

Study of a low Mach nuclear core model for single-phase flows. (English. French summary) Zbl 1329.76248
Summary: This paper deals with the modelling of the coolant (water) in a nuclear reactor core. This study is based on a monophasic low Mach number model (Lmnc model) coupled to the stiffened gas law for a single-phase flow. Some analytical steady and unsteady solutions are presented for the 1D case. We then introduce a numerical scheme to simulate the 1D model in order to assess its relevance. Finally, we carry out a normal mode perturbation analysis in order to approximate 2D solutions around the 1D steady solutions.

##### MSC:
 76M25 Other numerical methods (fluid mechanics) (MSC2010) 76Dxx Incompressible viscous fluids 65M25 Numerical aspects of the method of characteristics for initial value and initial-boundary value problems involving PDEs
Full Text:
##### References:
 [1] M. Bernard, S. Dellacherie, G. Faccanoni, B. Grec and Y. Penel. Study of a low Mach nuclear core model for two-phase flows with phase transition I: stiffened gas law. (in preparation). · Zbl 1304.35529 [2] S. Dellacherie, G. Faccanoni, B. Grec and Y. Penel. Study of a low Mach nuclear core model for two-phase flows with phase transition II: tabulated EOS. (in preparation). · Zbl 1304.35529 [3] C. Calgaro, E. Creus\'{}e and T. Goudon. A hybrid finite volume – finite element method for variable density incompressible flows. J. Comput. Phys., 227(9), 4671–4696 (2008). · Zbl 1137.76037 [4] S. Dellacherie. On a low Mach nuclear core model. ESAIM: Proceedings, (to appear). · Zbl 1357.76074 [5] S. Dellacherie. On a diphasic low Mach number system. Math. Model. and Num. Anal., 39(3), 487–514 (2005). · Zbl 1075.35038 [6] D. Durran. Numerical Methods for Fluid Dynamics: with Application to Geophysics. Springer, Texts in Applied Mathematics 32, (2010). · Zbl 1214.76001 [7] O. Le M\'{}etayer, J. Massoni and R. Saurel. Elaborating equations of state of a liquid and its vapor for two-phase flow models. Int. J. Therm. Sci., 43(3), 265–276 (2004). [8] O. Le M\'{}etayer, J. Massoni and R. Saurel. Modelling evaporation fronts with reactive Riemann solvers. J. Comput. Phys., 205(2), 567–610 (2005). · Zbl 1088.76051 [9] E. W. Lemmon, M. O. McLinden and D. G. Friend. Thermophysical properties of fluid systems. In WebBook de Chimie NIST, Base de Donn\'{}ees Standard de R\'{}ef\'{}erence NIST Num\'{}ero 69, Eds. P.J. Linstrom and W.G. Mallard, National Institute of Standards and Technology, Gaithersburg MD, 20899, [10] A. Majda & J. Sethian. The derivation and numerical solution of the equations for zero mach number combustion. Combust. Sci. Technol., 42, 185–205 (1985). [11] Y. Penel. \'{}Etude th\'{}eorique et num\'{}erique de la d\'{}eformation d’une interface s\'{}eparant deux fluides non-miscibles ‘a bas nombre de Mach. PhD Thesis, Universit\'{}e Paris 13, France (December 2010). [12] N. Todreas & M. Kazimi. Nuclear Systems I: Thermal Hydraulic Fundamentals. Taylor & Francis Group, 2nd edition, (2011). ESAIM: PROCEEDINGS131 Figure 5. Linear case. Above: numerical (red dashed line), exact (black circles) and asymptotic (blue solid line) solutions to Eq. (5b) at different times. Below : relative error between numerical and exact solutions (black circles) and between numerical and asymptotic solutions (blue solid line) at different times. Figure 6. Linear case. Convergence rate (logarithmic scale) in norms L\inftyand L2. Figure 7. Linear case. Comparison at time t = 1 s between numerical (red dashed line) and asymptotic solutions (blue solid line) to Syst. (5). Figure 8. Linear case (with singular charge loss). Comparison at time t = 1 s between numerical (red dashed line) solutions to Syst. (5) with (5c) replaced by (5c’) and asymptotic solutions (blue solid line) to Syst. (5). ESAIM: PROCEEDINGS133 Figure 9. Negative {$$\Phi$$} case. Comparison at time t = 1 s between numerical (red dashed line) and asymptotic solutions (blue solid line) to Syst. (5). Figure 10. Sine case. Numerical (red dashed line) and asymptotic (blue solid line) solutions to Eq. (5b) at different times. Figure 11. Alternating {$$\Phi$$} case. Numerical (red dashed line) and asymptotic (associated to {$$\Phi$$}0– blue dotted line and to {$$\Phi$$}0/5 – black dotted line) solutions to Eq. (5b) at different times. Figure 12. Alternating {$$\Phi$$} case. Evolution with respect to time of the error between numerical solution to Eq. (5b) and asymptotic solution associated to {$$\Phi$$}0/5.
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# generating pilot sub-carriers in matlab

I wanted to simulate OFDM system in MATLAB. I'm using Fixed WiMAX OFDM parameters in my simulation, where the number of data sub-carriers are 192, number of Pilot sub-carriers are 8, number Null sub-carriers are 56. Null tones can be obtained in MATLAB by generating zero matrix with the required rows and columns.

% Null subcarrier:
msg(1,1:100)=0;

similarly, for data subcarriers, digitally modulated data is sent.

% Data subcarrier:
msg(2,1:100)= 1*sign(randn(1,100)) + i*sign(randn(1,100));

my question is, what type of data could I send for Pilot tones?? all books and literature I have read says that: Pilot tones are just known data. What type of Known data are the Pilot tones? What kind of data should I send in my specific problem? Please, clarify.

• Do I fail to see the question here? You calculate their value according to the standard and set the eight subcarriers of the OFDM symbol where they need to go to those values.
– jan
Apr 6, 2014 at 10:10

## Principle

The pilot tone/subcarrier (short: pilot) modulation for WiMax is specified in section 8.3.3.4.2 of IEEE standard 802.16-2004, which is available on the IEEE site. The data bits transmitted on the pilots are defined by a pseudo-random binary sequence (PRBS) $w_k$ of which the polynomial is given by $X^{11}+X^9+1$. The index $k$ stands for the OFDM symbol number and the the pilot tones are modulated using binary phase shift keying (BPSK), i.e. the subcarrier value is either $-A$ or $A$. The detailed allocation of BPSK symbols can be taken from the following equations that I have copied from the above-cited standard

As the energy of all subcarriers shall be equal and as you're using the symbol alphabet $\{1+j, 1-j, -1+j, -1-j\}$ with symbol energy $\sqrt{1^2+1^2}=\sqrt 2$ it must hold $A=\sqrt 2$.

## Example

The sequence for pilots in the uplink is $w_k=1,\ldots$ and the modulation of subcarriers in the first OFDM symbol ($k=0$) is given by

subcarrier index  value
----------------  -----
-88               -A
-63                A
-38               -A
-13                A
13               -A
38               -A
63               -A
88               -A

## Implementation

Create the sequence $w_k$ and create the BPSK symbols according to the above principle for every $k$. Then assign these values to the pilot subcarriers of every OFDM symbol. If rigorous accordance to the standard is not required you can just make up any random sequence of BPSK symbols and allocate them to the pilot subcarriers. Note that the pilot subcarrier allocation might be different depending on the version of the standard, uplink/downlink and PUSC/FUSC.

In your case, each OFDM symbol contains 256 sub-carriers. Store the pilot values for a OFDM symbol at time instant $n$ in an $8 \times 1$ array say V. Also, you have the positions of the pilot carriers stored in a array say, P. Let the OF(:,n) represent the $256 \times 1$ OFDM symbol at time n.

Then you can assign the pilot values V to the corresponding pilot positions P in the OFDM symbol at time n as

OF(P,n)=V

in MATLAB. I hope this helps you a bit.


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Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

# ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D.

Question:

ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D. Prove that ∠D = 1/2∠A.

Solution:

Let ∠ABE = 2x and ∠ACB = 2y

∠ABC = 180° − 2x [Linear pair]

∴ ∠A = 180° − ∠ABC − ∠ACB [Angle sum property]

= 180° − 180° + 2x + 2y

= 2(x − y) ..... (i)

Now, ∠D = 180° − ∠DBC − ∠DCB

⇒ ∠D = 180° − (x + 180° − 2x) − y

⇒ ∠D = 180° − x − 180° + 2x − y

= (x − y)

$=\frac{1}{2} \angle \mathrm{A} \ldots .$ from(i)

Hence, $\angle \mathrm{D}=\frac{1}{2} \angle \mathrm{A}$.


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# No change in entropy of the system+surroundings in reversible process.Really?

1. Sep 24, 2010

### kntsy

No change in entropy of the system+surroundings in "reversible" process.Really?

This is a challenge to the sentense from some basic text.
I think reversible is not enough.
It has to be ISOTHERMAL, right?
Lets me be sure that i am correct!

2. Sep 24, 2010

### Andrew Mason

Re: No change in entropy of the system+surroundings in "reversible" process.Really?

No. An adiabatic process (eg. expansion or compression) can be reversible. In such a case, the temperature is not constant.

AM

3. Sep 25, 2010

### kntsy

Re: No change in entropy of the system+surroundings in "reversible" process.Really?

Thanks! But what about REVERSIBLE isobaric, isovolume,? The total system+environment change in entropy still can remain zero????
I discover an important fact:
$$\triangle S_{total}=\triangle S_{in} +\triangle S_{out}=0$$
iff
$$\int\frac{dq_{in}}{T}=\int\frac{dq_{out}}{T}$$
iff


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# Unity 64-bit compiling (again)

This topic is 4184 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

I was going to post a reply to my last thread, but it's been retired. I'm still having issues with 64-bit compiling, this time on a much simpler project (I gave up with x64 on my larger project). I get the exact same error message: DruinkEdit.obj : error LNK2001: unresolved external symbol __CxxFrameHandler3 I can't disable exceptions, since then I get a bunch of errors about exceptions being disabled comming from the STL. Even this simple program doesn't link:
#include <windows.h>
#include <string>

static void ErrorBox(const std::wstring& strError)
{
MessageBox(NULL, strError.c_str(), L"Error", MB_OK | MB_ICONERROR);
}

int WINAPI WinMain(HINSTANCE, HINSTANCE, LPSTR, int)
{
ErrorBox(L"Test");
}

With the following errors:
------ Build started: Project: bar, Configuration: Release Win32 ------
Compiling...
Main.cpp
Main.obj : error LNK2001: unresolved external symbol "__declspec(dllimport) protected: wchar_t const * __cdecl std::basic_string<wchar_t,struct std::char_traits<wchar_t>,class std::allocator<wchar_t> >::_Myptr(void)const " (__imp_?_Myptr@?$basic_string@_WU?$char_traits@_W@std@@V?$allocator@_W@2@@std@@IEBAPEB_WXZ) Main.obj : error LNK2001: unresolved external symbol "__declspec(dllimport) public: wchar_t const * __cdecl std::basic_string<wchar_t,struct std::char_traits<wchar_t>,class std::allocator<wchar_t> >::c_str(void)const " (__imp_?c_str@?$basic_string@_WU?$char_traits@_W@std@@V?$allocator@_W@2@@std@@QEBAPEB_WXZ)
Main.obj : error LNK2001: unresolved external symbol "__declspec(dllimport) public: __cdecl std::basic_string<wchar_t,struct std::char_traits<wchar_t>,class std::allocator<wchar_t> >::~basic_string<wchar_t,struct std::char_traits<wchar_t>,class std::allocator<wchar_t> >(void)" (__imp_??1?$basic_string@_WU?$char_traits@_W@std@@V?$allocator@_W@2@@std@@QEAA@XZ) Main.obj : error LNK2001: unresolved external symbol "__declspec(dllimport) public: __cdecl std::basic_string<wchar_t,struct std::char_traits<wchar_t>,class std::allocator<wchar_t> >::basic_string<wchar_t,struct std::char_traits<wchar_t>,class std::allocator<wchar_t> >(wchar_t const *)" (__imp_??0?$basic_string@_WU?$char_traits@_W@std@@V?$allocator@_W@2@@std@@QEAA@PEB_W@Z)
c:\foo\bar\Release\bar.exe : fatal error LNK1120: 4 unresolved externals
Build log was saved at "file://c:\foo\bar\bar\Release\BuildLog.htm"
bar - 5 error(s), 0 warning(s)
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

Anyone got any [more] ideas? EDIT: It links fine if I remove the STL stuff and don't use any exceptions. But that's not very useful...

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It looks to me like it simply isn't linking the libs into the rest of the code. Which seems strange if it can include standard libs. I am in no way meaning to insult you but just thought I'd try a memory test, did you check and insure the libs you're using are included in the project?

Since STL deals with templates perhaps that has something to do with it. Have you tried compiling using a library that doesn't use templates? Not that this is a long term solution but it may help isolate or define the problem.

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Yup, it's only linking in the standard libs. I tried creating an application not using the STL, and that works fine. An empty application with exceptions (try, catch, throw) in it also won't link.

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Bump. Anyone had any problems with x64 and exception stuff?

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Try including string before windows.h?

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That won't affect anything, it's a link problem rather than a compile problem.

Tried it, no difference.

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Quote:
 Original post by Evil SteveThat won't affect anything, it's a link problem rather than a compile problem.Tried it, no difference.

Stupid question, but are you including the x64 versions of the standard libs? I'm new to the x64 platform myself, however I do know that a x64 application cannot link to an x86 dll (and vice versa). If I understand it correctly, it is because the 32 bit applications must be executed by the WOW64 service. In order to provide linking between the two, it would be quite a costly operation. I would imagine that an x64 app could not link to previously compiled x86 lib's because of the differences in code generation between the two. Not sure if this helps, you may even have looked into this already.

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can you please post you complete compilation and ofcourse linking line(s) ...

may be he links to the rwong libraries?
im just guessing that the standard setup links to the 32bit version and thus its crashing.

(and yes i have no visual studio 2005 avail, to try it myself)

did you set up the correct library and include directories in your settings ? ( i remember, that there are some subdirectorys with "x64" in it.)

... why is there a penguin at the title (without this penguin i probably would have never whatched this one :lol:)

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All the compiler directories are set to the /win64/amd64 folders where possible (E.g. E:\Microsoft Platform SDK\Bin\win64\AMD64).

My compiler line is:
/Od /I "E:\Source Control\DruinkClient\/" /D "WIN32" /D "_DEBUG" /D "_WINDOWS" /D "_CRT_SECURE_NO_DEPRECATE" /D "_CRT_NON_CONFORMING_SWPRINTFS" /D "_UNICODE" /D "UNICODE" /FD /EHsc /MDd /GS- /GR- /Fo"Debug x64\\" /Fd"Debug x64\vc80.pdb" /W4 /WX /nologo /c /Wp64 /Zi /TP /errorReport:prompt

/OUT:"E:\Source Control\DruinkClient\Debug x64\DruinkClient.exe" /INCREMENTAL /NOLOGO /MANIFEST /MANIFESTFILE:"Debug x64\DruinkClient.exe.intermediate.manifest" /DEBUG /PDB:"e:\Source Control\DruinkClient\Debug x64\DruinkClient.pdb" /SUBSYSTEM:WINDOWS /MACHINE:X64 /ERRORREPORT:PROMPT kernel32.lib user32.lib gdi32.lib winspool.lib comdlg32.lib advapi32.lib shell32.lib ole32.lib oleaut32.lib uuid.lib odbc32.lib odbccp32.lib

I'm using VC2005 Professional if it makes a difference (Which does support x64).

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well i dont understand much of this, but i doubt that this is everything ...
e.g. where is the linking of this crt stuff?

can you enable verbose output or sth like that?
or what does you build log say? (post it)

and did you try it without unicode? (at least i think that this "w"-stuff is unicode)

e.g. my gcc (amd64 ubuntu) says:
--------------------------------

#include <string>

int main( int argc, char **argv ) {

std::string s("asdf");

return 0;
}

---------------------------------------------
--------Compile------------------------------
---------------------------------------------
$g++ -v -c test.cpp Ziel: x86_64-linux-gnu Konfiguriert mit: ../src/configure -v --enable-languages=c,c++,java,f95,objc,ada,treelang --prefix=/usr --enable-shared --with-system-zlib --libexecdir=/usr/lib --without-included-gettext --enable-threads=posix --enable-nls --program-suffix=-4.0 --enable-__cxa_atexit --enable-clocale=gnu --enable-libstdcxx-debug --enable-java-awt=gtk-default --enable-gtk-cairo --with-java-home=/usr/lib/jvm/java-1.4.2-gcj-4.0-1.4.2.0/jre --enable-mpfr --disable-werror --enable-checking=release x86_64-linux-gnu Thread-Modell: posix gcc-Version 4.0.3 (Ubuntu 4.0.3-1ubuntu5) /usr/lib/gcc/x86_64-linux-gnu/4.0.3/cc1plus -quiet -v -D_GNU_SOURCE test.cpp -quiet -dumpbase test.cpp -mtune=k8 -auxbase test -version -o /tmp/ccPTwlkQ.s nicht vorhandenes Verzeichnis Â»/usr/local/include/x86_64-linux-gnuÂ« wird ignoriert nicht vorhandenes Verzeichnis Â»/usr/include/x86_64-linux-gnuÂ« wird ignoriert #include "..." - Suche beginnt hier: #include <...> - Suche beginnt hier: /usr/lib/gcc/x86_64-linux-gnu/4.0.3/../../../../include/c++/4.0.3 /usr/lib/gcc/x86_64-linux-gnu/4.0.3/../../../../include/c++/4.0.3/x86_64-linux-gnu /usr/lib/gcc/x86_64-linux-gnu/4.0.3/../../../../include/c++/4.0.3/backward /usr/local/include /usr/lib/gcc/x86_64-linux-gnu/4.0.3/include /usr/include Ende der Suchliste. GNU C++ version 4.0.3 (Ubuntu 4.0.3-1ubuntu5) (x86_64-linux-gnu) compiled by GNU C version 4.0.3 (Ubuntu 4.0.3-1ubuntu5). GGC-Heuristik: --param ggc-min-expand=100 --param ggc-min-heapsize=131072 as -V -Qy --64 -o test.o /tmp/ccPTwlkQ.s GNU assembler version 2.16.91 (x86_64-linux-gnu) using BFD version 2.16.91 20060118 Debian GNU/Linux --------------------------------------------- --------Link--------------------------------- ---------------------------------------------$ g++ -v -o test test.o

Ziel: x86_64-linux-gnu
Konfiguriert mit: ../src/configure -v --enable-languages=c,c++,java,f95,objc,ada,treelang --prefix=/usr --enable-shared --with-system-zlib --libexecdir=/usr/lib --without-included-gettext --enable-threads=posix --enable-nls --program-suffix=-4.0 --enable-__cxa_atexit --enable-clocale=gnu --enable-libstdcxx-debug --enable-java-awt=gtk-default --enable-gtk-cairo --with-java-home=/usr/lib/jvm/java-1.4.2-gcj-4.0-1.4.2.0/jre --enable-mpfr --disable-werror --enable-checking=release x86_64-linux-gnu
gcc-Version 4.0.3 (Ubuntu 4.0.3-1ubuntu5)
/usr/lib/gcc/x86_64-linux-gnu/4.0.3/collect2 --eh-frame-hdr -m elf_x86_64 -dynamic-linker /lib64/ld-linux-x86-64.so.2 -o test /usr/lib/gcc/x86_64-linux-gnu/4.0.3/../../../../lib64/crt1.o /usr/lib/gcc/x86_64-linux-gnu/4.0.3/../../../../lib64/crti.o /usr/lib/gcc/x86_64-linux-gnu/4.0.3/crtbegin.o -L/usr/lib/gcc/x86_64-linux-gnu/4.0.3 -L/usr/lib/gcc/x86_64-linux-gnu/4.0.3 -L/usr/lib/gcc/x86_64-linux-gnu/4.0.3/../../../../lib64 -L/usr/lib/gcc/x86_64-linux-gnu/4.0.3/../../.. -L/lib/../lib64 -L/usr/lib/../lib64 test.o -lstdc++ -lm -lgcc_s -lgcc -lc -lgcc_s -lgcc /usr/lib/gcc/x86_64-linux-gnu/4.0.3/crtend.o /usr/lib/gcc/x86_64-linux-gnu/4.0.3/../../../../lib64/crtn.o

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Is the project set as Win32? If so, have you tried setting it as x64 in the configuration manager?

Look around page 5 for the correct VS 2005 settings:
This may be useful to you

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Quote:
 Original post by Billr17Is the project set as Win32? If so, have you tried setting it as x64 in the configuration manager?Look around page 5 for the correct VS 2005 settings:This may be useful to you
Ah ha! I didn't think of configuration from Win32, it seems to be working great now, thanks [smile]

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I have a different question. what is this 64-bit compiling? and what is 64x and 86x.

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64-bit compiling is compiling applications for 64-bit operating systems like Windows XP x64. x64 refers to 64-bit CPUs and OSs, x86 refers to the 32-bit 386, 486, 586, etc family of CPUs.

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Quote:
 Original post by Evil Steve64-bit compiling is compiling applications for 64-bit operating systems like Windows XP x64. x64 refers to 64-bit CPUs and OSs, x86 refers to the 32-bit 386, 486, 586, etc family of CPUs.

To be correct, x86 refers to the "86" line of intel products (+ clones and derivative products, such as the 8088). The list includes:

* 8086, 8088 (first generation, 16 bit processor)
* 80186, 80188 (never really hit the shelves)
* 80286
* 80386 (first 32 bit CPU in this line)
* 80486
* Pentium (80586)
* Pentium II (80686)
* Pentium III
* Pentium IV
* AMD processors (Duron, Athlon, ...)

Current dual core processors and 64 bits processors are also part of this product line.

Regards,

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when you say 32 bit and 64 bit, are you refering to the register size of the cpu?

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It's a reference to the native word size of the platform. IA64 and IA32 have very different register mappings. In fact, IA32 technically has provision for registers of many sizes which are not 32 bits (floating point registers, in particular, as well as stuff for SIMD instruction sets).

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• ### Similar Content

• Hello fellow devs!
Once again I started working on an 2D adventure game and right now I'm doing the character-movement/animation. I'm not a big math guy and I was happy about my solution, but soon I realized that it's flawed.
My player has 5 walking-animations, mirrored for the left side: up, upright, right, downright, down. With the atan2 function I get the angle between player and destination. To get an index from 0 to 4, I divide PI by 5 and see how many times it goes into the player-destination angle.

In Pseudo-Code:
angle = atan2(destination.x - player.x, destination.y - player.y) //swapped y and x to get mirrored angle around the y axis
index = (int) (angle / (PI / 5));
PlayAnimation(index); //0 = up, 1 = up_right, 2 = right, 3 = down_right, 4 = down

Besides the fact that when angle is equal to PI it produces an index of 5, this works like a charm. Or at least I thought so at first. When I tested it, I realized that the up and down animation is playing more often than the others, which is pretty logical, since they have double the angle.

What I'm trying to achieve is something like this, but with equal angles, so that up and down has the same range as all other directions.

I can't get my head around it. Any suggestions? Is the whole approach doomed?

Thank you in advance for any input!

• By devbyskc
Hi Everyone,
Like most here, I'm a newbie but have been dabbling with game development for a few years. I am currently working full-time overseas and learning the craft in my spare time. It's been a long but highly rewarding adventure. Much of my time has been spent working through tutorials. In all of them, as well as my own attempts at development, I used the audio files supplied by the tutorial author, or obtained from one of the numerous sites online. I am working solo, and will be for a while, so I don't want to get too wrapped up with any one skill set. Regarding audio, the files I've found and used are good for what I was doing at the time. However I would now like to try my hand at customizing the audio more. My game engine of choice is Unity and it has an audio mixer built in that I have experimented with following their tutorials. I have obtained a great book called Game Audio Development with Unity 5.x that I am working through. Half way through the book it introduces using FMOD to supplement the Unity Audio Mixer. Later in the book, the author introduces Reaper (a very popular DAW) as an external program to compose and mix music to be integrated with Unity. I did some research on DAWs and quickly became overwhelmed. Much of what I found was geared toward professional sound engineers and sound designers. I am in no way trying or even thinking about getting to that level. All I want to be able to do is take a music file, and tweak it some to get the sound I want for my game. I've played with Audacity as well, but it didn't seem to fit the bill. So that is why I am looking at a better quality DAW. Since being solo, I am also under a budget contraint. So of all the DAW software out there, I am considering Reaper or Presonus Studio One due to their pricing. My question is, is investing the time to learn about using a DAW to tweak a sound file worth it? Are there any solo developers currently using a DAW as part of their overall workflow? If so, which one? I've also come across Fabric which is a Unity plug-in that enhances the built-in audio mixer. Would that be a better alternative?
I know this is long, and maybe I haven't communicated well in trying to be brief. But any advice from the gurus/vets would be greatly appreciated. I've leaned so much and had a lot of fun in the process. BTW, I am also a senior citizen (I cut my programming teeth back using punch cards and Structured Basic when it first came out). If anyone needs more clarification of what I am trying to accomplish please let me know.  Thanks in advance for any assistance/advice.

• Hi , I was considering this start up http://adshir.com/, for investment and i would like a little bit of feedback on what the developers community think about the technology.
So far what they have is a demo that runs in real time on a Tablet at over 60FPS, it runs locally on the  integrated GPU of the i7 . They have a 20 000 triangles  dinosaur that looks impressive,  better than anything i saw on a mobile device, with reflections and shadows looking very close to what they would look in the real world. They achieved this thanks to a  new algorithm of a rendering technique called Path tracing/Ray tracing, that  is very demanding and so far it is done mostly for static images.
From what i checked around there is no real option for real time ray tracing (60 FPS on consumer devices). There was imagination technologies that were supposed to release a chip that supports real time ray tracing, but i did not found they had a product in the market or even if the technology is finished as their last demo  i found was with a PC.  The other one is OTOY with their brigade engine that is still not released and if i understand well is more a cloud solution than in hardware solution .
Would there  be a sizable  interest in the developers community in having such a product as a plug-in for existing game engines?  How important  is Ray tracing to the  future of high end real time graphics?

• Good day,

I just wanted to share our casual game that is available for android.

Description: Fight your way from the ravenous plant monster for survival through flips. The rules are simple, drag and release your phone screen. Improve your skills and show it to your friends with the games quirky ranks. Select an array of characters using the orb you acquire throughout the game.


text:
# poj 3278 Catch That Cow （BFS+剪枝）

Catch That Cow
 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 58474 Accepted: 18173

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points - 1 or + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

4

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

Source

#include <cstdio>
#include <queue>
using namespace std;
const int maxn=200010;
int m,n;
int vis1[maxn];
struct node
{
int x,step;
};
int bfs()
{
node sd;
sd.x=n;
sd.step=0;
queue<node>q;
q.push(sd);
while(!q.empty())
{
node nd=q.front();
if(nd.x==m)
return nd.step;
q.pop();
for(int i=0;i<3;i++)//枚举三种变换方法
{
if(i==0)
{
sd.x=nd.x-1;
sd.step=nd.step+1;
if(sd.x>=0&&!vis1[sd.x])  //该数字没有访问过，可加入队列
{
vis1[sd.x]=1;
q.push(sd);
}
}
else if(i==1)
{
sd.x=nd.x+1;
sd.step=nd.step+1;
if(sd.x>maxn)         //超出最大范围的数字舍去
continue;
if(sd.x>=0&&!vis1[sd.x])
{
vis1[sd.x]=1;
q.push(sd);
}
}
else
{
sd.x=nd.x*2;
sd.step=nd.step+1;
if(sd.x>maxn)
continue;
if(sd.x>=0&&!vis1[sd.x])
{
vis1[sd.x]=1;
q.push(sd);
}
}
}
}

}
int main()
{
scanf("%d%d",&n,&m);
printf("%d\n",bfs() );
return 0;
}

• 本文已收录于以下专栏：

## poj 3278 Catch That Cow（BFS，剪枝）

Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 43761  ...

## poj3278——Catch That Cow（BFS，剪枝）

DescriptionFarmer John has been informed of the location of a fugitive cow and wants to catch her im...

## POJ-3278 Catch That Cow(广搜+剪枝)

Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 39887   Accepted: ...

## poj 3278 Catch That Cow 【bfs】

Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 52335  ...

BFS 模板题 &&符号机制

## POJ 3278 Catch That Cow (BFS)

Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 46662  ...

## POJ - 3278 - Catch That Cow （BFS）

举报原因： 您举报文章：深度学习：神经网络中的前向传播和反向传播算法推导 色情 政治 抄袭 广告 招聘 骂人 其他 (最多只允许输入30个字)


text:
Given a list of triangle indices, how exactly does one go about converting that to a list of indices with adjacency for a geometry shader?

Note that we're strictly talking about indices here - vertices are present, but we're going to focus solely on indices, because we can use them to match duplicate vertices without having to get into floating-point comparisons and epsilons - that work has already been done.

I know that for any given triangle in the list, indices {0, 1}, {1, 2} and {2, 0} (or {n, n+1}, {n+1, n+2}, {n+2, n} if you prefer) of it form it's edges; the index list is well-formed and respects winding order correctly.

I know that for any given such edge, we can search the entire list for another triangle that uses two of those indices, and the third index of that triangle is the one to use for completing the adjacent triangle for that edge.

I know that in the adjacency list each original triangle is represented by 6 indices, the original indices go into slots 0, 2, 4; the new indices to complete the adjacency go into slots 1, 3, 5. The index to complete for edge {0, 1} goes into slot 1, the index to complete for edge {1, 2} goes into slot 3, the index to complete for edge {2, 1} goes into slot 5.

What have I tried?

I've tried brute-forcing it, and yes, that will work, but I'm after a more elegant approach.

I've tried Eric Lengyel's edge list builder, but (1) it doesn't seem to respect the original triangle order, (2) it doesn't seem to respect winding order, (3) it's as clear as mud where to go next after you've built the edge list, and (4) I've a suspicion of sample code that has such obvious glaring errors as "triangleIndex" versus "faceIndex" - did the author even compile the code, never mind run it to verify it?

So -- any suggestions or pointers from here on?

• Jimmy, I edited out the stuff about shadow volumes and changed the title, since it didn't seem relevant and was potentially confusing - the question is actually just about building adjacency data, even though your ultimate purpose is to use it for shadow volumes. – Nathan Reed Sep 12 '13 at 17:32

I would try using a hash table for this (for instance, std::unordered_map if you're in C++). Build a hash table that maps from a half-edge (expressed as a pair of indices, in order) to the third index of the triangle the half-edge belongs to.

This can be built by simply iterating over the triangle list and adding each triangle's three half-edges to the hash table. If your initial index list had a pair of adjacent triangles, (0, 1, 2, 2, 1, 3), you'd end up with a hash table containing:

(0, 1) -> 2
(1, 2) -> 0
(2, 0) -> 1
(2, 1) -> 3
(1, 3) -> 2
(3, 2) -> 1

Note that edges (1, 2) and (2, 1) both appear in the table, representing the two sides of the edge, and pointing to the third vertex of each of the two triangles.

Then, to build the adjacency data all you have to do is iterate over the triangle list again, and query each triangle's edges with the opposite winding. So when processing triangle (0, 1, 2) you'd query edges (1, 0), (2, 1), and (0, 2). This will find the opposite vertex of each edge, if it exists.

• Cool, searching for edges with the opposite order was a key piece of missing information; works a champ; +1 and accepted. – Maximus Minimus Sep 12 '13 at 18:37

Look at the half-edge data structure.

The idea roughly is that you store a list of half edges, e.g. the half of a particular edge attached to a particular face. This structure then also stores information about the corresponding half-edge for the adjoining face.

The data structure makes queries about connectivity, adjacency, and so on very efficient. There are algorithms to generate the data efficiently, though not necessarily "game run-time" efficiently (you'll probably want to do this offline in your asset pipeline).

See also these blog posts. I believe the structure and algorithms are also in Real-Time Collision Detection, but I don't recall for sure and dont' have a copy at the office.

• -1, sorry, but that didn't give any info I didn't already have, and was oriented around usage on the CPU rather than building a list with adjacency for use in a GS. – Maximus Minimus Sep 17 '13 at 16:41


text:
# More fun in 2015, Part 46

Find

$S=\sum\omega^{1729},$

where the sum is taken over all primitive $$2015^\text{th}$$ roots of unity $$\omega$$.

×


text:
Tutorial on Dualities

These are the slides of my tutorial on Dualities at the $16^{th}$ Latin American Symposium on Mathematical Logic. 28th July – 1st August 2014. Buenos Aires, Argentina.  A shorter version can be found here.

Slides on Duality (SLALM 2014)

Dualities and geometry

Finally I wrote some slides about the long-waiting article I am writing together with Olivia Caramello and Vincenzo Marra on adjunctions, dualities, and Nullstellensätze .  These slides where presented at the AILA meeting in Pisa and at the Apllied Logic seminar in Delft.

Dualities and geometry


text:
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# If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is th

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If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is th  [#permalink]

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29 Aug 2018, 23:32
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If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is the average of a, b, c, and 13?

(A) 8
(B) 8.5
(C) 9
(D) 9.5
(E) It cannot be determined from the information given.

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Re: If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is th  [#permalink]

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29 Aug 2018, 23:54
a+b+c+5+6=5*6
-> a+b+c = 19
-> a+b+c+13 = 32
-> (a+b+c+13)/ 4 = 32/4 = 8, hence option (A)
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Re: If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is th  [#permalink]

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30 Aug 2018, 00:04
very basic question.

(a+b+c+5+6)/5=6, so a+b+c=19.

(a+b+c+13)/4=(19+13)/4=8

ANS.A
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Re: If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is th  [#permalink]

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30 Aug 2018, 00:09
Bunuel wrote:
If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is the average of a, b, c, and 13?

(A) 8
(B) 8.5
(C) 9
(D) 9.5
(E) It cannot be determined from the information given.

Given, a+b+c+5+6=5*6=30
Or, a+b+c+5+6+2=30+2=32
Or, $$\frac{a+b+c+13}{4}=\frac{32}{4}=8$$
Or, average of a, b, c, and 13 is 8

Ans. (A)
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Re: If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is th  [#permalink]

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30 Aug 2018, 05:14
Bunuel wrote:
If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is the average of a, b, c, and 13?

(A) 8
(B) 8.5
(C) 9
(D) 9.5
(E) It cannot be determined from the information given.

$$\frac{a + b + c + d + 5 + 6}{5 = 6}$$

a + b + c + d + 11 = 30

a + b + c = 19

a + b + c + 13 / 4

$$\frac{19 + 13}{4}$$

= $$\frac{32}{4}$$

= 8

The best answer is A.
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Re: If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is th  [#permalink]

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30 Aug 2018, 06:12
Bunuel wrote:
If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is the average of a, b, c, and 13?

(A) 8
(B) 8.5
(C) 9
(D) 9.5
(E) It cannot be determined from the information given.

$$a + b + c + 5 + 6 = 30$$

So, $$a + b + c = 19$$

So, the average of a, b, c, and 13 = $$\frac{19 + 13}{4} = 8$$ Answer must be (A)
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Re: If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is th  [#permalink]

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04 Sep 2018, 03:38
Bunuel wrote:
If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is the average of a, b, c, and 13?

(A) 8
(B) 8.5
(C) 9
(D) 9.5
(E) It cannot be determined from the information given.

We can create the equation:

(a + b + c + 5 + 6)/5 = 6

a + b + c + 11 = 30

a + b + c = 19

So the average of a, b, c, and 13 is:

(a + b + c + 13)/4

(19 + 13)/4

32/4 = 8

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Re: If the average (arithmetic mean) of a, b, c, 5, and 6 is 6, what is th   [#permalink] 04 Sep 2018, 03:38
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text:
Question

# A block of mass m moving with speed v collides with another block of mass $$2$$m at rest. The lighter block comes to rest after the collision. The coefficient of restitution is?

Solution

## Linear momentum will be conserved: $$mv+2m\times0=2mu+m\times0$$$$\Rightarrow u = \dfrac{v}{2}$$Coefficient of restitution $$e=\dfrac { relative\quad velocity\quad of\quad separation }{ relative\quad velocity\quad of\quad approach\quad }$$$$\\ =\dfrac { v/2 }{ v } =\dfrac { 1 }{ 2 }$$Physics

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• # question_answer 15) If Anika finds that she is 10th from the right in a line of girls and fourth from the left. How many girls should be added to the line such that there are 28 girls in the line? A) 11                          B) 12           C) 13                          D) 15

Number of girls in the line $=\left( 10+4 \right)-1=13$ Number of girls to be added $=28-13=15$girls.


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Air pump physics problem

by Yegor
Tags: pump
 P: 120 Air pump physics problem Each cycle corresponds to a T=const. process. (in another version of this problem the pumping process is very fast and a the transformation would correspond to an adiabatic process - all you have to do is to add the exponent $$\gamma$$ to the equations below). After the first cycle the air volume increases from V to V+dV and the pressure decreases at $$p_1$$. You have $$p_0\cdot V=p_1\cdot (V+dV)$$ After the first cycle, the pump takes out the air from dV, and you get again $$p_1=p_0 \frac{V}{V+dv}$$ In the second cycle, $$p_1\cdot V=p_2\cdot (V+dV)$$ and $$p_2=p_1 \frac{V}{V+dv}=p_0 \frac{V^2}{(V+dv)^2}$$ and so on..... $$p_n=p_{n-1} \frac{V}{V+dv}=p_0 \frac{V^n}{(V+dv)^n}$$


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# Zariski topology in the complex plane: an example

I want to find the closure under the zariski topology, of this set $\left\{ {\left( {x,y} \right) \in {\Bbb C}^2 ;\left| x \right| + \left| y \right| = 1} \right\}$ I have no idea what I can do

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Let $S$ be your set, and let $f\in\mathbb C[x,y]$ a polynomial vanishing on $S$.

We prove $f=0$ as follows.

Let $(a,b)$ be a nonzero vector in $\mathbb C^2$.

The restriction of $f$ to the line $$L:=\{(ta,tb)\ |\ t\in\mathbb C\}$$ is a polynomial in $t$ vanishing on $S\cap L$, which is the set of those $(ta,tb)$ with $$|t|=\frac{1}{|a|+|b|}\quad.$$ This set being infinite, $f$ vanishes on $L$. As $\mathbb C^2$ is covered by such lines, we have indeed $f=0$.

Variation: $f(ru,(1-r)v)$ is a polynomial in $r,u,v$ vanishing on $$0\le r\le1,\quad |u|=1=|v|.$$ From this it is easy to deduce $f=0$.

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I proved in a different way, thanks for all! All the answers was useful –  Arkj Mar 28 '12 at 2:28

Your set $S$ has closure $\bar S=\mathbb C^2$.
To prove that it is enough to prove that no non-zero polynomial $P(x,y)$ vanishes on $S$.
Here is a proof using real manifolds:

The set $S$ is a real submanifold of dimension $3$ of $\mathbb C^2=\mathbb R^4$ since its equation is $f(x,y)=(x_1^2+x_2^2)^{1/2}+(y_1^2+y_2^2)^{1/2}=1$ and $f$ has non-zero gradient on $S$.
But except at finitely many points $V(P)$, the variety defined by $P$, is smooth and thus a real manifold of dimension $2$, so that you can't have $S\subset V(P)$.

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(expanding my comment to an answer)

Let $S$ be your set $S=\{(x,y)\in\mathbb{C}^2\mid |x|+|y|=1\}$. If we can show that the only bivariate polynomial that vanishes on all the points of $S$ is the constant zero, then the ideal $\cal{I}(S)$ consists of the zero polynomial alone, and hence $\overline{S}=\cal{V}(\cal{I}(S))=\mathbb{C}^2$.

Let $f(x,y)$ be a polynomial that vanishes at all the point of $S$. Then all the complex numbers $x$ on the circle $|x|=1/2$ must be zeros of the univariate polynomial $g(x):=f(x,1/2)$. There are infinitely many numbers $x$ with this property, so we must have $g(x)=0$. Therefore $y-\frac12$ must be a factor of $f(x,y)$. But we can repeat the same argument for any complex number $y_0$ such that $|y_0|<1$. So all such polynomials $y-y_0$ must be factors of $f(x,y)$. But the polynomial $f(x,y)$ has a finite degree, so this is impossible by unique factorization.

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# Crazy Calendar(LightOJ-1393)

## 题面

2011 was a crazy year. Many people all over the world proposed on 11-11-11, married on 11-11-11, some even went through surgery only to have 11-11-11 as their child’s birth date. How crazy people can be! Don’t they see there is a “20” hidden? Then what to do? A very elegant solution came from ARR, a very famous and funny character - why do we need to follow Christian (or some calls it Gregorian) calendar? Why don’t we start our own calendar on the day of marriage? And those who like to celebrate their marriage ceremony too frequent, why don’t they declare only 1 day per year. In that fashion they can celebrate their anniversary every day. And may be one minute a year or a second or … Uh.. getting complex. Let’s back to the title. From now, we start to have a new calendar system, “Kisu Pari Na”. And we hope to update this calendar on every national contest.

The purpose of this calendar is - we all will try our best to learn something new in every year. For this first year let’s learn some combinatory. It reminds me of my first year in college. I faced this problem but could not solve this then. But see how easy it is:

(1, 1)(1, 2)(1, 3)(1, 4)
(2, 1)(2, 2)(2, 3)(2, 4)
(3, 1)(3, 2)(3, 3)(3, 4)

Say you start from upper left cell and want to go to lower right cell. The only restriction is you can only move downward or rightward. How many ways are there? How to solve it? Not that difficult. You have to go two times Down and three times Right (whichever way you try) to reach the goal from the starting cell, right? So the answer is number of ways you can arrange two D (represents Down) and three R (represent Right). 2 same characters and 3 same characters, total 5 characters. So it is:

$\frac{5!}{2!3!}$ Or = $\frac{(D+R)!}{D!R!}$ D+RCR. Easy isn’t it?

Ok enough with learning. Now back to problem, given a grid and at each cell there are some coins. Inky and Pinky are playing a game getting inspiration from the above problem. At each turn, a player chooses a non empty cell and then removes one or more coins from that cell and put them to the cell exactly right of it or exactly beneath it. A player can’t divide the coins and put one part to right and others to down. Note that, for the cells at the right column the player can’t move it to more right, and same for the bottom-most row. So a player can’t move coins from the lower right cell. The game will finish when no moves are available and the player who moved last will win. Now inky being very modest asked Pinky to move first. Can you say if Pinky will win if both play perfectly?

## 输入

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing two integers R C (1 ≤ R * C ≤ 50000), where R denotes the number of rows and C denotes the number of columns of the grid respectively. Each of the next R lines contains C space separated integers denoting the grid. These integers lie in the range [0, 109].

## 输出

For every test case, output case number followed by “win” if Pinky can win or “lose”.

## 提示

Dataset is huge, use faster I/O methods.

## 代码

_/_/_/_/_/ EOF _/_/_/_/_/


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# 1. Classify the following numbers as rational or irrational: (iv)     $\frac{1}{\sqrt{2}}$

Given number is $\frac{1}{\sqrt{2}}$
$\Rightarrow \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}=\frac{1.41421356..}{2} =0.7071068...$


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# zbMATH — the first resource for mathematics

Linear systems of hyperplane sections on varieties of low codimension. (English. Russian original) Zbl 0577.14032
Funct. Anal. Appl. 19, 165-173 (1985); translation from Funkts. Anal. Prilozh. 19, No. 3, 1-10 (1985).
Let $$X^ n\subset {\mathbb{P}}^ N$$ be a non-degenerate smooth projective variety over an algebraically closed field, and suppose that X can be isomorphically projected to $${\mathbb{P}}^{2n+1-\delta}$$, where $$\delta >0$$. Then it turns out that $$N\leq f(Ent(n/\delta))$$, where f is a quadratic function whose explicit form is computed in the paper. In other words, if $$X^ n\subset {\mathbb{P}}^ r$$ has ”small codimension”, i.e. $$r\leq 2n$$, then $$h^ 0(X,{\mathcal O}_ X(1))\leq f(Ent(n/2n+1-r))+1.$$
The above result follows from a study of higher secant varieties; in particular, if $$S^ kX$$ denotes the variety of k-dimensional secant subspaces of X (such that $$S^ 0X=X$$ and $$S^ 1X=SX$$ is the usual secant variety), then it is shown that $$S^{Ent(n/\delta)}={\mathbb{P}}^ N$$ [for $$\delta >n$$ this implies Hartshorne’s conjecture on linear normality; cf. the author Mat. Sb., Nov. Ser. 116(158), 593-602 (1981; Zbl 0484.14016)].
It should be noted that all varieties X for which the above inequalities turn into equalities have been classified; they are called the Scorza varieties, and for $$\delta =n$$ this notion coincides with the notion of Severi varieties as defined in the author’s paper in Mat. Sbornik 126(168), No.1, 115-132 (1985; see the preceding review).

##### MSC:
 14J40 $$n$$-folds ($$n>4$$) 14N05 Projective techniques in algebraic geometry 14M07 Low codimension problems in algebraic geometry 14C20 Divisors, linear systems, invertible sheaves
Full Text:
##### References:
 [1] T. Fujita and J. Roberts, ”Varieties with small secant varieties: the extremal case,” Am. J. Math.,103, No. 5, 953-976 (1981). · Zbl 0475.14046 · doi:10.2307/2374254 [2] W. Fulton and R. Lazarsfeld, ”Connectivity and its applications in algebraic geometry,” in: Proc. Midwest Conf. on Algebraic Geometry, Univ. of Illinois, 1980, Lecture Notes in Math., Vol. 862, Springer-Verlag, Berlin?Heidelberg?New York (1981), pp. 26-92. · Zbl 0484.14005 [3] Ph. Griffiths and J. Harris, ”Algebraic geometry and local differential geometry,” Ann. Sci. Ecole Norm. Sup.,12, No. 3, 355-452 (1979). · Zbl 0426.14019 [4] R. Lazarsfeld and A. Van der Ven, ”Topics in the geometry of projective space: Recent work of F. L. Zak,” Birkhäuser Verlag, Basel?Boston?Stuttgart (1984). [5] F. L. Zak, ”Projections of algebraic varieties,” Mat. Sb.,116, No. 4, 593-602 (1981). · Zbl 0484.14016 [6] F. L. Zak, ”Severi varieties,” Mat. Sb.,126, No. 1, 115-132 (1985). · Zbl 0577.14031
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Are all pseudo-random number generators ultimately periodic?

Are all pseudo-random number generators ultimately periodic? Or are they periodic at all in the end?

By periodic I mean that, like rational numbers, they in the end generate a periodic subsequence...

And pseudo-random means algorithmic/mathematical generation of random numbers...

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This is a pedantic point to make, but on a finite-memory computer, every non-halting program is ultimately periodic. You could analyse the algorithm as running on a Turing machine, but any PRNG whose memory use is unbounded with time wouldn't be very practical. –  Peter May 5 '14 at 20:20
You should check out this video on youtube. I found this video during one of those times where you just click through the suggested videos list. It has a lot of information about random numbers and how they work. And yes, they are periodic as said in the video. –  matthapkidokarate May 5 '14 at 20:28
for an example of a PRNG-like construction that seems to be aperiodic see collatz problem (or other fractal-like problems). any proof of its periodicity would likely be a proof of the problem (still unsolved for ~¾ century). this question also would seem to have some connection to the busy beaver problem and undecidability. however this is an unconventional view. might attempt to work this into answer with some comment upvotes. –  vzn May 6 '14 at 16:16
Another unconventional view is based on real numbers: assume that the digits of some real number are considered random... If I have an algorithm to generate the digits, those digits will not be necessarily periodic...? –  user13675 May 11 '14 at 17:27

All pseudorandom generators that don't rely on outside randomness and use a bounded amount of memory are necessarily ultimately periodic since they have finite state. You can think of them as huge deterministic finite automata which have special "output" states in which they give their output. All finite automata are eventually periodic, and so all pseudorandom generators produce eventually periodic output.

However, the period length can be enormous. For example, a PRNG with a cryptographic state of 128 bits might only cycle once every $2^{128}$ bits of output, and so even if outputting one bit every nanosecond, the solar system will be dead ere the PRNG repeats.

If the PRNG is allowed to use an unbounded amount of memory (which isn't realistic) then it can, for example, output the binary expansion of $\sqrt{2}$, which we know isn't eventually periodic (since $\sqrt{2}$ is irrational).

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Why can't they be general Turing machines? –  Niklas B. May 5 '14 at 20:03
Ah I see, they would also need to use infinitly much space –  Niklas B. May 5 '14 at 20:25
Wouldn't $\sqrt 2$ have to be normal as well, for it to be a true PRNG? I don't believe this is known. –  WChargin May 6 '14 at 0:39
Could you use some sort of iterative algorithm with the BBP $\pi$-digit generation formula to generate PR numbers, assuming $\pi$ is normal? Or some other formula to extract a given digit from a normal irrational number? Seems like there might be a way to do this in finite memory (though maybe not for BBP due to precision)/ –  WChargin May 6 '14 at 0:41
@WChargin Note that what you are suggesting does take unbounded memory. If you have a function f(i) that returns the ith digit of pi, you must have unbounded memory to list all the digits, because i does take growing memory amounts, even though its grow is logarithmic. If you fix the amount of memory such a function can only compute the digits up to a certain I that fills all the available memory (and proably less, since some space is required to perform the calculations). –  Bakuriu May 6 '14 at 11:12

PRNGs are state-machines. If they're based only in internal input (in contrast to Poker Stars RNG which is a combination of hardware and software, seeding itself continuously from... sound samples) you'll get (C, S1,...) where C is the current (or previous) value and S1,... could be a set of states:

If there are possible N values (since the memory is bounded) of C and you iterate N+1 times, you will hit the same value for C at least twice. If you iterate 2N+1 times, you will hit the same value for C at least 3 times.

Let T = (Ct, S1t, S2t) be a certain state (current value and other states).
Let M = #{values for S1}X{values for S2}X{...} be the cardinal of possible states combinations (again: since the memory is bounded).

If you iterate NM+1 times the algoritm, you'll reach at least twice the same state (Ct, S1t, S2t, ...), thus generating the same output value and the same following state sequence than the first time, and so becoming periodic.

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# Show that $\int_\Omega f \, dA = \frac{3\pi}{2} \, f(0) + \frac{\pi}{2} \, f'(0)$

Let $\Omega = \{ z + z^2/2 : |z| < 1\}$ be the interior of a shape in the complex plane, and let $f(z)$ be an analytic function. Show the following formula:

$$\int_\Omega f \, dA = \frac{3\pi}{2} \, f(0) + \frac{\pi}{2} \, f'(0)$$

One strategy is to notice that both sides are linear in $f$ so we can assume a power series expansion, $f(x)=\sum a_n \, z^n$. This is not very geometric. Stokes theorem is another possibility.

It's slightly odd to integrate complex function with area? Perhaps I'm a bit rusty, I calculated the area $dA$ should be:

$$(dx + idy)\,(dx - idy) = 2i\, dx \,dy$$

I plotted an image of the interior region. It's simply connected but it's not quite a circle.

As stated in the comments, the map $z \mapsto z + z^2/2$ seems to be injective on the unit disk $\mathbb{D} = \{ |z| < 1 \}$:

Originally I had posted $z \mapsto z/2 + z^2$ which has another interesting shape and analogous question could be posed.

• $dA$ should be $dx\wedge dy = \frac{1}{2i} d\overline{z} \wedge dz$. Since $g \colon z \mapsto z + \frac{z^2}{2}$ is injective on the unit disk, a change-of-variables suggests itself. – Daniel Fischer Dec 6 '17 at 14:20
• @DanielFischer No it's not, the image curve is clearly overlapping. – cactus314 Dec 6 '17 at 14:24
• We have $\operatorname{Re} g'(z) = \operatorname{Re} (1+z) > 0$ on the unit disk, so $g$ is injective there. I'm not sure what you plotted, I agree that that doesn't look injective. – Daniel Fischer Dec 6 '17 at 14:31
• @DanielFischer I have the corret picture now. Could you explain where the $f'(0)$ term happens? – cactus314 Dec 6 '17 at 14:41
• From the change of variables, you get $$\frac{1}{2i}\int_{\mathbb{D}} f(g(z)) (1+z)(1+\overline{z})\,d\overline{z}\wedge dz.$$ Change to polar coordinates, and all terms containing an $e^{ik\varphi}$ with $k\neq 0$ integrate to $0$. From $\overline{g'(z)}$ you have a term $e^{-i\varphi}$, and that gets you the $f'(0)$ term by the grouping $f'(0)\cdot re^{i\varphi}\cdot (1 + \dotsc)(\dotsc + e^{-i\varphi})\, rd\varphi\,dr$. – Daniel Fischer Dec 6 '17 at 14:48

We must of course assume that $f$ is integrable over $\Omega$. Then, since $g$ is injective, we have

\begin{align} \int_{\Omega} f\,dA &= \frac{1}{2i} \int_{\Omega} f(w) \, d\overline{w} \wedge dw \\ &= \frac{1}{2i} \int_{\mathbb{D}} f(g(z))\, d\overline{g(z)} \wedge dg(z) \\ &= \frac{1}{2i} \int_{\mathbb{D}} d\bigl(\overline{g(z)}\cdot f(g(z))\,dg(z)\bigr). \end{align}

If $f\circ g$ is continuous on $\overline{\mathbb{D}}$, we can directly apply Stokes' theorem to that last integral, otherwise we apply it to the integral over $r\cdot \mathbb{D}$ for $0 < r < 1$ and let $r\to 1$. Let's do that anyway. Then Stokes' theorem yields

\begin{align} \frac{1}{2i} \int_{r\mathbb{D}}&\, d\bigl(\overline{g(z)}\cdot f(g(z))\,dg(z)\bigr) \\ &= \frac{1}{2i} \int_{\lvert z\rvert = r} \overline{g(z)}\bigl(f(g(z))g'(z)\bigr)\,dz \\ &= \frac{1}{2i} \int_{\lvert z\rvert = r} \biggl(\overline{z} + \frac{\overline{z}^2}{2}\biggr) f(g(z))(1+z)\,dz \\ &= \frac{1}{2i} \int_{\lvert z\rvert = r} \biggl(\frac{r^2}{z} + \frac{r^4}{2z^2}\biggr)f(g(z))(1+z)\,dz \tag{$\overline{z} = r^2/z$} \\ &= \pi r^2 \frac{1}{2\pi i} \int_{\lvert z\rvert = r} \frac{f(g(z))(1+z)}{z}\,dz + \frac{\pi r^4}{2}\cdot \frac{1}{2\pi i} \int_{\lvert z\rvert = r} \frac{f(g(z))(1+z)}{z^2}\,dz\\ &= \pi r^2 f(g(0))(1+0) + \frac{\pi r^4}{2} \frac{d}{dz}\biggr\rvert_{z = 0}\bigl(f(g(z))(1+z)\bigr) \\ &= \pi r^2 f(0) + \frac{\pi r^4}{2}\bigl(f(0) + f'(0)\cdot g'(0)^2\bigr) \\ &= \pi f(0)\biggl(r^2 + \frac{r^4}{2}\biggr) + \frac{\pi r^4}{2} f'(0)\,. \end{align}

Taking the limit $r \to 1$ then gets us to

$$\int_{\Omega} f\,dA = \frac{3\pi}{2} f(0) + \frac{\pi}{2} f'(0).$$

• Right. It doesn't just "fall out" of the change of variables. I think the $f'(0)$ term has to do with the cusp of the shape at $g(-1)=-1/2$. Because of that one point, the interior of the cardioid is not diffeomorphic to the disk. – cactus314 Dec 6 '17 at 16:11


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# Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined

I appreciate the help.

My attempt:

\begin{align} \tan(x) + \cot(x) &= \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \\ &= \frac{\sin^2(x)}{\cos(x) \sin(x)}+\frac{\cos^2(x)}{\cos(x) \sin(x)} \\ &= \frac{\sin^2(x)+\cos^2(x)}{\cos(x) \sin(x)}\\ &= \frac{1}{\cos(x) \sin(x)}\\ &= \frac{1}{\frac{1}{\sec(x)}\frac{1}{\csc(x)}}\\ &=\frac{1}{\frac{1}{\sec \csc}}\\ &=\frac{1}{1}\cdot \frac{\sec(x) \csc(x)}{1}\\ &= \sec(x) \csc(x) \end{align}

• OK! If you have to do this for an exam, however, I suggest you write in all of the " $\ \theta \$ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.) – colormegone Feb 3 '14 at 0:48
• yup. It's quicker to go from $\frac1{cos\cdot{sin}}$ to $\frac1{cos}\frac1{sin}=sec\cdot{csc}$. – Eleven-Eleven Feb 3 '14 at 0:49

That is exactly correct! Just two things: First, $\tan,\sin,\cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $\tan x$, $\cos x$ etc rather than just $\tan$ or $\cos$.

Finally, you could save time on your proof by noticing on the fourth step that $$\frac{1}{\cos x\sin x}=\frac{1}{\cos x}\frac{1}{\sin x}=\sec x \csc x$$

Your steps are correct, but just keep in mind that robotically converting everying into $\sin$s and $\cos$s isn't the only option available to you.

Note that $$\cot\theta = \frac{\cos\theta}{\sin\theta}=\frac{\frac{1}{\sin\theta}}{\frac{1}{\cos\theta}}=\frac{\csc\theta}{\sec\theta}$$ that $$\cot\theta\tan\theta=\frac{1}{\tan\theta}\cdot\tan\theta=1$$ and that $$\sec^2\theta=\tan^2+1$$ then $$\begin{array}{lll} \tan\theta+\cot\theta&=&1\cdot(\tan\theta+\cot\theta)\\ &=&(\cot\theta\tan\theta)(\tan\theta+\cot\theta)\\ &=&(\cot\theta)(\tan\theta(\tan\theta+\cot\theta))\\ &=&\frac{\csc\theta}{\sec\theta}(\tan^2\theta+1)\\ &=&\frac{\csc\theta}{\sec\theta}\sec^2\theta\\ &=&\sec\theta\csc\theta \end{array}$$

An alternative approach writes $$t=\tan x/2$$ so $$\tan x+\cot x=\frac{2t}{1-t^2}+\frac{1-t^2}{2t}=\frac{1+t^2}{1-t^2}\frac{1+t^2}{2t}=\sec x\csc x.$$

Equations last but one, last but two and last but three can be deleted and you can straight away jump to last result because you already know $$\sec,\csc$$ are inverse trig functions of $$\cos, \sin$$.
For acute $$\theta$$, there's this trigonograph:
$$\sec\theta \cdot \csc\theta \;=\; 2\,|\triangle OPQ| \;=\; 1\cdot( \tan\theta +\cot\theta)$$


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# Find the volume of the parallelepiped whose conterminous edges are represented by the vectors

Question:

Find the volume of the parallelepiped whose conterminous edges are represented by the vectors+

i. $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$

ii. $\overrightarrow{\mathrm{a}}=-3 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=-5 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{c}}=7 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$

iii. $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, \overrightarrow{\mathrm{c}}=\hat{\mathrm{j}}+\hat{\mathrm{k}}$

iv. $\bar{a}=6 \hat{l}, \bar{b}=2 \hat{\jmath}, \bar{c}=5 \hat{k}$

Solution:


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# PowerShell chrgp script

## PowerShell

how can I assign permissions to user specific folders using powershell?

Hi

We have a Windows 2012 R2 servers and run the Microsoft NFS Server on it. As Windows doesn't currently expose the file ACL's via NFS the group ownership of files is important to allow permissions to work. The GUI doesn't allow you to display file group ownership or change it, probably cause it's not really an issue for Windows. The old solution was to use "chgrp" in Services For Unix, but SFU has been removed from 2012 R2. So I looked to PowerShell to write a replacement. I was surprised no one else has already done this, from what I tell.

I found a simple chown (Change Ownership) example and modified it to be a chgrp (Change Group), it's below. It does work properly but is limited to single file or directory names. Does anyone have a framework of a script that I can plug these chgrp works into that would allow this script to accept: multiple filename/directory arguments, file/directory globs/wildcards and accept a recursive command line flag.

I can view the group ownership with "Get-Acl".

My very basic chgrp version for a single filename has syntax chgrp.ps1 groupname filename. I'd like:

chgrp.ps1 /r groupname dirname dirname

chgrp.ps1 groupname filename dirname

chgrp.ps1 groupname c:\dirname\*.* dirname filename

etc etc (where say /r would recurse through these dirs).

Here is my simple version that takes only a single file.

Param (

$group,$filename

)

$acct = New-Object System.Security.Principal.NTAccount($group)

$file=Get-Item$filename

$acl=$file.GetAccessControl()

$acl.SetGroup($acct)

$file.SetAccessControl($acl)

Thanks

Colin

Mace
OP
##### cduff Jun 19, 2014 at 3:37 UTC

Powershell
Param (

$group, [Parameter(ValueFromPipeline=$True)]
$filesystemobject ) Begin{$acct = New-Object System.Security.Principal.NTAccount($group) } Process{$acl=$filesystemobject.GetAccessControl()$acl.SetGroup($acct)$filesystemobject.SetAccessControl($acl) }  Something like this. I didn't test this, but that is the gist of it. Popular Topics in PowerShell TEST YOUR SMARTS Which of the following retains the information it's storing when the system power is turned off? • GPU • RAM • CPU • ROM 88% of IT pros got this right. ## 5 Replies · · · Mace OP ##### cduff Jun 19, 2014 at 3:33 UTC As for the recursion and globbing you would use the builtin Get-ChildItem cmdlet to do that, then pipe it into your script that only processes a single directory. So it would look like Get-ChildItem -Recuse -Path c:\ -Filter *.txt | cgrp.ps1 -group groupname To run your script on all text files on the system. Just an example. Your only task is to set it up to accept pipline input. 0 · · · Mace OP Best Answer ##### cduff Jun 19, 2014 at 3:37 UTC Powershell Param ($group,

[Parameter(ValueFromPipeline=$True)]$filesystemobject

)
Begin{
$acct = New-Object System.Security.Principal.NTAccount($group)
}
Process{
$acl=$filesystemobject.GetAccessControl()

$acl.SetGroup($acct)

$filesystemobject.SetAccessControl($acl)
}

Something like this. I didn't test this, but that is the gist of it.

0
· · ·
Pimiento
OP
##### colinsimpson Jun 19, 2014 at 4:58 UTC

Looks good, but sadly doesn't run. Do you not have to have something to convert file paths into file objects:

$file=Get-Item$filename

Text
PS G:\test> Get-ChildItem -Recurse -Path g:\test | c:\it\chgrppipe.ps1 ion\itgrp
You cannot call a method on a null-valued expression.
At C:\it\chgrppipe.ps1:17 char:5
+     $file.SetAccessControl($acl)
+     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+ CategoryInfo          : InvalidOperation: (:) [], RuntimeException
+ FullyQualifiedErrorId : InvokeMethodOnNull

You cannot call a method on a null-valued expression.
At C:\it\chgrppipe.ps1:17 char:5
+     $file.SetAccessControl($acl)
+     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+ CategoryInfo          : InvalidOperation: (:) [], RuntimeException
+ FullyQualifiedErrorId : InvokeMethodOnNull

You cannot call a method on a null-valued expression.
At C:\it\chgrppipe.ps1:17 char:5
+     $file.SetAccessControl($acl)
+     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+ CategoryInfo          : InvalidOperation: (:) [], RuntimeException
+ FullyQualifiedErrorId : InvokeMethodOnNull

You cannot call a method on a null-valued expression.
At C:\it\chgrppipe.ps1:17 char:5
+     $file.SetAccessControl($acl)
+     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+ CategoryInfo          : InvalidOperation: (:) [], RuntimeException
+ FullyQualifiedErrorId : InvokeMethodOnNull

0
· · ·
Mace
OP
##### cduff Jun 19, 2014 at 5:34 UTC

Get-ChildItem passes an object, you give the path to Get-ChildItem and it does the conversion.  I did it that way to take advantage of the recursion and globbing already built into that cmdlet.

Change $file.SetAccessControl($acl) to $filesystemobject.SetAccessControl($acl)

0
· · ·
Pimiento
OP
##### colinsimpson Jun 20, 2014 at 10:05 UTC

That looks good. Thanks for all your help.

0

This discussion has been inactive for over a year.

You may get a better answer to your question by starting a new discussion.


text:
# Proof that there are infinitely many primes

The proof that there are infinitely many prime numbers is a proof by contradiction.

First, we note that there is indeed a prime: namely $$2$$. We will also state a lemma: that every number greater than $$1$$ has a prime which divides it. (This is the easier half of the Fundamental Theorem of Arithmetic, and the slightly stronger statement that “every number may be written as a product of primes” is proved there.)

Now we can proceed to the meat of the proof. Suppose that there were finitely many prime numbers $$p_1, p_2, \ldots, p_n$$. Since we know $$2$$ is prime, we know $$2$$ appears in that list.

Then consider the number $$P = p_1p_2\ldots p_n + 1$$ — that is, the product of all primes plus 1. Since $$2$$ appeared in the list, we know $$P \geq 2+1 = 3$$, and in particular $$P > 1$$.

The number $$P$$ can’t be divided by any of the primes in our list, because it’s 1 more than a multiple of them. But there is a prime which divides $$P$$, because $$P>1$$; we stated this as a lemma. This is immediately a contradiction: $$P$$ cannot be divided by any prime, even though all integers greater than $$1$$ can be divided by a prime.

# Example

There is a common misconception that $$p_1 p_2 \dots p_n+1$$ must be prime if $$p_1, \dots, p_n$$ are all primes. This isn’t actually the case: if we let $$p_1, \dots, p_6$$ be the first six primes $$2,3,5,7,11,13$$, then you can check by hand that $$p_1 \dots p_6 + 1 = 30031$$; but $$30031 = 59 \times 509$$. (These are all somewhat ad-hoc; there is no particular reason I knew that taking the first six primes would give me a composite number at the end.) However, we have discovered a new prime this way (in fact, two new primes!): namely $$59$$ and $$509$$.

In general, this is a terrible way to discover new primes. The proof tells us that there must be some new prime dividing $$30031$$, without telling us anything about what those primes are, or even if there is more than one of them (that is, it doesn’t tell us whether $$30031$$ is prime or composite). Without knowing in advance that $$30031$$ is equal to $$59 \times 509$$, it is in general very difficult to discover those two prime factors. In fact, it’s an open problem whether or not prime factorisation is “easy” in the specific technical sense of there being a polynomial-time algorithm to do it, though most people believe that prime factorisation is “hard” in this sense.

Parents:

• Prime number

The prime numbers are the “building blocks” of the counting numbers.

• How does this prove that P isn’t divisible by any non-prime number? This could be clearer.


text:
#### Chapter 4 System of Linear Equations

Section 4.1 What are Systems of Linear Equations?

# 4.1.2 Contents

Before we can really start, let us clarify the terminology.
##### Info 4.1.2

Several equations relating a specific number of variables at the same time form a so-called system of equations. If the variables in every equation occur only linearly, i.e. at most to the power of $1$, and are only multiplied by (constant) numbers, the system is called a system of linear equations, or LS (linear system).

The two equations in the first example 4.1.1 form a system of linear equations in the variables $x$ and $y$. In contrast, the three equations

$x+y+z=3,\mathrm{ }\text{}\mathrm{ }x+y-z=1,\mathrm{ }\text{and}\mathrm{ }x·y+z=2$

do form a system of equations in the variables $x$, $y$, and $z$, but the system is not linear since in the third equation the term $x·y$ occurs, which is bilinear in the variables $x$ and $y$ and hence violates the condition of linearity.
By the way, in a system of equations the number of equations need not be equal to the number of variables; we will return to this later on.
##### Info 4.1.3

If the number of equations in a system equals the number of variables, the system of equations is called a square.
##### Exercise 4.1.4
Which of the following systems is a system of linear equations?
 $x+y-3z=0$, $2x-3=y$, and $1.5x-z=22+y$, $\mathrm{sin}\left(x\right)+\mathrm{cos}\left(y\right)=1$ and $x-y=0$, $2z-3y+4x=5$ and $z+y-{x}^{2}=25$.

Systems of linear equations are distinguished from general systems of equations by their relative simplicity. Nevertheless, they play an important role in fields as diverse as medical science (e.g. for CT scans), engineering (e.g. for describing sound propagation in complex designed spaces), or physics (e.g. concerning the question of which wave lengths excited atoms can emit). It is, without doubt, worth dealing with systems of linear equations intensely.
For systems of equations generally, the question focuses on which values the variables must take such that all equations of the system are simultaneously satisfied. Such a set of values for the variables is called a solution of a system of equations.
Before we solve systems of equations a detail should be noted: depending on the problem, it may not be useful to accept all variable values. In the first example 4.1.1 the variables $x$ and $y$ are the numbers of unicycles and bicycles the group of stuntmen owns. Such numbers can only be non-negative integers, i.e. elements of $ℕ{}_{0}$. Hence, in this case the number range for the solutions has to be restricted to $ℕ{}_{0}$ in advance (namely, for both $x$ and $y$).
##### Info 4.1.5

The possible number range for the solutions of a system of equations is called the base set of the system. The domain is the subset of the base set for which all the terms of the equations of the system are defined. For systems of linear equations, base set and domain coincide. Finally, the solution set is the subset of the domain which merges the solutions of the system. The solution set is denoted by $L$.
If the base set is not explicitly specified - and cannot be derived from the problem description - we will assume implicitly that the base set is $ℝ$, the number range of the real numbers.


text:
ID 683847
Date 1/20/2023
Public

## 6.3. FPGA Partial Reconfiguration from Linux

The Partial Reconfiguration started from Linux relies on the device tree overlay mechanism which allows you to modify the original device tree at run time.

The components needed to perform the Platform Reconfiguration from Linux are listed next. These items are already provided as part of the Partial Reconfiguration example provided in the following RocketBoards page:

• dtbt—This is a script used to manage the overlay. This script provides the dtbt command that can be called from the Linux shell providing the ability to add or remove an overlay to the device tree and also list the overlays currently applied. This is located under Linux file system at /sbin directory. The source code for this script can be seen from the altera-opensource / dtbt web page.
• pr_fpga_static_region.dtbo—This is the overlay for the static region. This is generated from the pr_fpga_static_region.dts file and creates the new device tree nodes that describe the PR region and the freeze controller. The attributes for the PR region node are expected to be updated by the PR persona overlays. This overlay needs to be applied before persona overlays. This file is located under Linux file system at the /boot/devicetree directory. Next is shown the structure of the static region device tree overlay DTS.
Note: The static region and PR region can also be defined directly from the initial Linux device tree as alternative to use an overlay for this purpose.
fragment@0 {
target-path = "/soc/base_fpga_region";  /* Device tree hierarchy */
:
__overlay__ {
external-fpga-config;
:
/* PR region node */
fpga_pr_regionN{
compatible = "fpga-region";
fpga-bridges = <&freeze_controller_0>;
:
};
/* PR freeze Controller node */
freeze_controller@offset {
compatible = "altr,freeze_controller-ver", "altr,freeze-bridge-controller";
:
};
};
};

• pr_personaX.dtbo—This is the overlay for the PR Persona X and extends the fpga_pr_regionN node created in the PR Static Region overlay. This overlay is generated from the pr_personaX.dts file. Here is where it is indicated which is the RBF file that corresponds to the FPGA design to be loaded in the PR region (referred to as the personaX.rbf). This overlay also defines additional nodes and specific attributes that applies for this persona in the PR region node in the device tree. There is a PR Persona overlay file for each persona FPGA design to be loaded in the PR region. These overlay files are located under Linux file system at the /boot/devicetree directory. Next is shown the structure of the PR Persona Overlay DTS.
fragment@1 {
target-path = "/soc/base_fpga_region/fpga_pr_regionN";
:
__overlay__ {
partial-fpga-config;
firmware-name = "personaX.rbf";  /* RBF file for this persona */
:
/* Specific attributes for this PR Persona in fpga_pr_regionN  node*/
:
pr_region_N_nodeA:{
/* Specific attributes for this PR Persona in sub node A */

};
pr_region_N_nodeB:{
/* Specific attributes for this PR Persona in sub node B */

};
};
};

• personaX.rbf—This is the bitstream used by personaX overlay and corresponds to the FPGA design to be loaded into the PR region.

During the partial reconfiguration started by Linux, this accesses the PR RBF files under its file system at the /lib/firmware directory.
Note: In the hardware design, it is not allowed enabling both SGMII and PR at the same time. Because of this, the hardware design needs to be recompiled with SGMII disabled and PR enabled to allow PR to be used.
The following figure shows the components involved in Partial Reconfiguration for 2 Persona using the Linux device tree overlays mechanism:
Figure 34. Components Involved
As indicated before, to perform the partial reconfiguration from Linux, we make use of the dtbt command which allows the device tree overlays handling. At the time that the overlay is applied, the PR is performed automatically, and the Linux driver associated to the overlay is loaded. The following table lists some of the functionality that this command provides:
Description Command Usage
List overlays applied dtbt -l
Add an overlay dtbt -a <.dtbo to apply> -p <.dtbo search path>
Remove an overlay dtbt -r <.dtbo to remove > -p <.dtbo search path>
Note: When the PR overlay is removed, the corresponding Linux driver is removed.
To perform the PR successfully, the following sequence should be followed from the Linux shell:
1. Load static region overlay and confirm that this was applied by listing the overlays.
# dtbt -a pr_fpga_static_region.dtbo -p /boot/devicetree
# dtbt -l

Only one static region overlay should be shown.

2. Perform PR by loading the desired overlay and confirm that this was applied by listing the overlays applied.
# dtbt -a pr_persona0.dtbo -p /boot/devicetree
# dtbt -l

Static region and persona0 overlays should be shown.

3. If another PR persona wants to be loaded, the current PR persona overlay needs to be removed and then the new PR Persona overlay should be applied.
root@agilex:~# dtbt -r pr_persona0.dtbo -p /boot/devicetree
root@agilex:~# dtbt -a pr_persona1.dtbo -p /boot/devicetree
root@agilex:~# dtbt -l

Static region and persona1 overlays should be shown.


text:
# Testing for useful variables in a “net lift model”

I am often involved in modeling the Net lift, aka Uplift, aka incremental response of direct marketing campaigns. In a nutshell, this approach looks to model and thus select for marketing those individuals who require promotion in order to take the desired action (e.g. order a product). Given prospect A and B, if A will buy if we send a direct mail letter and B will buy anyway, we target A. Think of "swing voters" in political elections - you put resources against the persuadable undecided.

I am looking for affirmation or suggestions on the following approach to help understand which variables are important. Here is an example of a binary predictor variable. Lets say that 20,000 direct mail letters were sent (10,000 to those with predictor1 =0 and 10,000 with predictor1 =1) and 2,000 control customers were selected and held out to not receive the letter. In this case, the difference in response rate (treated group - control) is 1.5% for predictor1 =0 and -0.3% for predictor1=1. We would conclude that those with predictor1=0 are the better candidates to send a letter to - as the letter actually decreased the point estimate of response for those with a value of 1 for predictor1.

The difference in the odds ratio is 1.413 (treated - control).

Here is an example where predictor 1 is not useful.

Is it accurate to test the value of a given predictor on this difference in response rates (for any type of predictor, nit just a simple binary one) using a standard logistic regression? For example,the first scenario shown above would result in a significant interaction variable of treated*predictor1:

dat<-data.frame(treated=c(1,1,0,0),predictor=c(0,1,0,1), responded=c(250,122,10,15), notresponded = c(9750,9879,990,985))
mod<-glm(cbind(responded,notresponded)~predictor*treated, data=dat,family=binomial)
summary(mod)

Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept)        -4.5951     0.3178 -14.458  < 2e-16 ***
predictor           0.4105     0.4107   1.000  0.31754
treated             0.9316     0.3242   2.873  0.00406 **
predictor:treated  -1.1411     0.4255  -2.682  0.00733 **

While the second scenario does not:

dat<-data.frame(treated=c(1,1,0,0),predictor=c(0,1,0,1), responded=c(130,122,10,15), notresponded = c(9870,9879,990,985))
mod<-glm(cbind(responded,notresponded)~predictor*treated, data=dat,family=binomial)
summary(mod)

Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept)        -4.5951     0.3178 -14.458   <2e-16 ***
predictor           0.4105     0.4107   1.000    0.318
treated             0.2654     0.3299   0.805    0.421
predictor:treated  -0.4750     0.4299  -1.105    0.269

Question:

Issues with multiple testing aside, does this not represent a viable method for testing variables for net lift modeling?

Should I exclude the main effects and only have the interaction term? This is a principle question and speaks to the interpretation of an interaction - given that I am looking for "significant" variables which could be used to select customers to market to in order to maximize the difference between treated and control.

Can one also use this framework to test for deeper interactions (treated*predictor1*predictor2) to see if we need to really look at predictor1 and predictor2 combinations?


text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
 any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
On quasi-periodic perturbations of elliptic equilibrium points. (English) Zbl 0863.34043

From the authors’ summary: This work focuses on quasi-periodic time-dependent perturbations of ordinary differential equations near elliptic equilibrium points. This means studying

$\stackrel{˙}{x}=\left(A+\epsilon Q\left(t,\epsilon \right)\right)x+\epsilon g\left(t,\epsilon \right)+h\left(x,t,\epsilon \right),$

where $A$ is elliptic and $h$ is $𝒪\left({x}^{2}\right)$. It is shown that, under suitable hypothesis of analyticity, nonresonance and nondegeneracy with respect to $\epsilon$, there exists a Cantorian set $ℰ$ such that for all $\epsilon \in ℰ$ there exists a quasi-periodic solution such that it goes to zero when $\epsilon$ does. This quasi-periodic solution has the same set of basic frequencies as the perturbation. Moreover, the relative measure of the set $\left[0,{\epsilon }_{0}\right]\setminus ℰ$ in $\left[0,{\epsilon }_{0}\right]$ is exponentially small in ${\epsilon }_{0}$. The case $g\equiv 0$, $h\equiv 0$ (quasi-periodic Floquet theorem) is also considered. Finally, the Hamiltonian case is studied. In this situation, most of the invariant tori that are near the equilibrium point are not destroyed but only slightly deformed and “shaken” in a quasi-periodic way. This quasi-periodic “shaking” has the same basic frequencies as the perturbation.

Reviewer: P.Smith (Keele)
##### MSC:
 34C27 Almost and pseudo-almost periodic solutions of ODE 34D10 Stability perturbations of ODE 34C99 Qualitative theory of solutions of ODE 37C55 Periodic and quasiperiodic flows and diffeomorphisms 37J40 Perturbations, normal forms, small divisors, KAM theory, Arnol’d diffusion 37J99 Finite-dimensional Hamiltonian, Lagrangian, contact, and nonholonomic systems
##### Keywords:
quasi periodic solutions; Floquet theory; KAM theory