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# PhsyX Actor From X

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hi, i have a problem. I have added some static objects to my scene with ageia and now i want to create a sphere and move it around. This is the code to add the static stuff: NxTriangleMeshDesc TriMeshDesc; TriMeshDesc.numVertices = NumVerticies; TriMeshDesc.numTriangles = NumTriangles; TriMeshDesc.pointStrideBytes = sizeof(NxVec3); TriMeshDesc.triangleStrideBytes = 3*sizeof(short); TriMeshDesc.points = &verts[0].x; TriMeshDesc.triangles = &tris[0]; TriMeshDesc.flags = NX_MF_16_BIT_INDICES; MemoryWriteBuffer buf; bool status = NxCookTriangleMesh(TriMeshDesc, buf); ShapeDesc.meshData = g_CAgeia->m_PhysicsSDK->createTriangleMesh(MemoryReadBuffer(buf.data)); i do not add a bodydesc and then create the actor. Now if i use the ageia code:: NxBodyDesc SphereBodyDesc; SphereBodyDesc.angularDamping = 0.5f; SphereBodyDesc.linearVelocity = NxVec3(0.0f, 0.0f, 0.0f); NxSphereShapeDesc SphereDesc; SphereDesc.radius = 1.0f; NxActorDesc SphereActorDesc; SphereActorDesc.shapes.pushBack(&SphereDesc); SphereActorDesc.body = &SphereBodyDesc; SphereActorDesc.density = 10.0f; SphereActorDesc.globalPose.t = NxVec3(-70.0f, 32.0f, -190.0f); and create a sphere with a bodydesc so it can move it works perfect it collides with the scene. But if i want to create a NxActor from a ID3DXMesh* and move around with it i have no idea what to do. if i do it like in step1 and add a body desc it falls down but it never collides with anything it just falls invinite. So do i need another MeshDesc? I tryed NxConvexMeshDesc convexDesc; convexDesc.numVertices = NumVerticies; convexDesc.pointStrideBytes = sizeof(NxVec3); convexDesc.points = &verts[0].x; convexDesc.numTriangles = NumTriangles; convexDesc.triangles = &tris[0]; convexDesc.triangleStrideBytes = 3*sizeof(short); convexDesc.flags = NX_CF_COMPUTE_CONVEX | NX_CF_16_BIT_INDICES; but if i do this with:: NxInitCooking(); MemoryWriteBuffer buf2; bool status2 = NxCookConvexMesh(convexDesc, buf2); convexShapeDesc.meshData = g_CAgeia->m_PhysicsSDK->createConvexMesh(MemoryReadBuffer(buf2.data)); NxCookConvexMesh returns false and createConvexMesh fails. Any idea? thx

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I'm assuming, since you had success with the static triangle meshes, that the data structure types that you are passing into NxConvexMeshDesc are correct. I would recommend doing the following:

convexDesc.points = verts;

convexDesc.points = &verts[0].x

Since the latter makes an assumption about the internal implementation of NxVec3, and you cannot be 100% certain that the implementation will be the same in the future, or that the member 'x' will always remain public. That's just a general coding comment. (Oh, and if you happen to be using pointers to DirectX data structures here instead of copying to Ageia data structures...it might work fine, but be aware that you are getting lucky in a way. No guarantee that DX and Ageia will always define their vectors in a way that you can cast between them like this, e.g., what if Ageia's vector class goes double precision even while keeping .xyz in the same order?)

So, I see two possible things that might be causing NxCookConvexMesh to fail.

First, you are using the flag NX_CF_COMPUTE_CONVEX and also passing in triangles via convexDesc.triangles and convexDesc.triangleStrideBytes. The flag
tells the API to generate triangles by using a convex hull algorithm on a point cloud. If you tell it to do this but you are already giving it triangles---suggesting that you want it to use your triangles, it may be failing because you are giving it conflicting instructions.

Second, it might be failing because your triangles don't make up a convex mesh...maybe it is in reality concave.

Here's what I'd try, in this order:

1) Remove the NX_CF_COMPUTE_CONVEX flag and see what happens. If no failure, go with that. Your own triangles are a good, convex mesh.

2) If (1) fails, add back the NX_CF_COMPUTE_CONVEX flag and assign ConvexDesc.triangles = 0 and ConvexDesc.triangleStrideBytes = 0. See what happens.

My guess, assuming all else is good, is that one of these things will fix it.

A third option would be to treat your movable object using NxCookTriangleMesh, as you did with the static objects, just adding a body also to make it dynamic. But I would not do this at all unless you really have a concave mesh and need it to collide as concave. This will be your most expensive option.

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1) am i right that i cannot create a actor
with a NxTriangleMeshDesc that performs dynamic collision detection?
If i create one with such a struct and a body struct it falls throught
everything and never collides with anything :S

2) no NX_CF_COMPUTE_CONVEX flag is still failure

3) now i set it like this::

NxConvexMeshDesc convexDesc;
convexDesc.numVertices = NumVerticies;
convexDesc.pointStrideBytes = sizeof(NxVec3);
convexDesc.points = verts;
convexDesc.numTriangles = 0; //NumTriangles;
convexDesc.triangles = 0; //&tris[0];
convexDesc.triangleStrideBytes = 0; //3*sizeof(short);
convexDesc.flags = NX_CF_COMPUTE_CONVEX |NX_CF_16_BIT_INDICES;

NxConvexShapeDesc convexShapeDesc;
convexShapeDesc.localPose.t = NxVec3(0,0,0);

NxInitCooking();

after this the NxCookConvexMesh still fails

4) if i use a TriangleMesh and assign a body NxCookTriangleMesh it does not
fail but the mesh never collides

5) i dont really get what you mean with convace and convex but i can send
you the file if you want ;)

thank you

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Look in the SDK docs, under collisions, there is a matrix that shows what works and doesn't considering the type of mesh you setup with. IIRC forget the triangle mesh, it was almost useless to me for almost everything of than a static object, and it will not work with terrain mesh.

BTW you set to short? should be unsigned short, shouldn't matter since they should be both 16bit, but isn't correct.

TriMeshDesc.triangleStrideBytes = 3*sizeof(short);

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of_ownage,

I would agree with MARS_999...actually, regardless of whether technically would work for you, if you can avoid them, I would. They are going to be more expensive than the standard collision shapes. General triangle meshes are worse than convex meshes. The standard collision shapes use highly optimized, shape-specific code and can be very fast.

Now...you said something in your PM to me that gave me another clue about what might be going wrong. You said that your shapes had around 600 triangles. That is a problem! Ageia supports convex meshes only up to 256 triangles! Look at the NxConvexMesh documentation. It clearly states this triangle/polygon limit.

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Quote:
 Original post by MARS_999Look in the SDK docs, under collisions, there is a matrix that shows what works and doesn't considering the type of mesh you setup with. IIRC forget the triangle mesh, it was almost useless to me for almost everything of than a static object, and it will not work with terrain mesh. BTW you set to short? should be unsigned short, shouldn't matter since they should be both 16bit, but isn't correct.TriMeshDesc.triangleStrideBytes = 3*sizeof(short);

The size of a short is the same as the size of an unsigned short, so this line isn't a proble...though for clarity I personally would have made it sizeof(unsigned short).

The bigger issue is, of course, that of_ownage neds to make sure that the "tris" array is an array of unsigned shorts.

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Quote:
Original post by grhodes_at_work
Quote:
 Original post by MARS_999Look in the SDK docs, under collisions, there is a matrix that shows what works and doesn't considering the type of mesh you setup with. IIRC forget the triangle mesh, it was almost useless to me for almost everything of than a static object, and it will not work with terrain mesh. BTW you set to short? should be unsigned short, shouldn't matter since they should be both 16bit, but isn't correct.TriMeshDesc.triangleStrideBytes = 3*sizeof(short);

The size of a short is the same as the size of an unsigned short, so this line isn't a proble...though for clarity I personally would have made it sizeof(unsigned short).

The bigger issue is, of course, that of_ownage neds to make sure that the "tris" array is an array of unsigned shorts.

I agree as it's not an issue, like I stated before both are 2 bytes, but for obvious reasons its sloppy coding practices.

On the topic of 256 triangles I know that is in the SDK also, but not sure if that relfects 2.7.3 SDK, as I have a mesh with 2000k polygons and it works fine with it... I have more issues with the mesh being concave and not convex enough, then cooking it blows up.

If you're using concave meshes break up the mesh into chunks and try to make each one a NxConvexMeshDesc type.

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MARS*, your 2000 polygon meshes are being cooked as convex? Interesting.

I'll also point to John Ratcliff's CreateDynamics source code, built to work with PhysX (and partly while he was employed by Ageia), which can do the convex decomposition for you, if you have a concave mesh to begin with. It is SLOW so more a preprocess than a runtime thing.

John Ratcliff's Source Code (Look for "CreateDynamics")

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ok i use boxes now they work ok for me. But sometimes the behavior of them is really strange. Very small boxes stay above the ground while other fall onto it.
Maybe im confusing something with the creation of a Ageia Actor as far as
i know i need two things. the dimension and the world pos.

I have a ID3DXMesh* with a translate a rotate and a scale matrix
i have a local box with a min and a max point

so first i compute length/height/width of the new box which is::

D3DXVECTOR3 newStats;
newStats.x = box.maxPoint.x - box.centerPoint.x
newStats.y = box.maxPoint.y - box.centerPoint.y
newStats.z = box.maxPoint.z - box.centerPoint.z

now i have a vector with the half length half height half width and ageia
wants that.

Now since my mesh is scaled and rotated i need to do that with my local
box too so i do this:

D3DXMATRIX combined = rotate * scale;
D3DXVec3TransformCoord(&newStats, &newStats, &combined);

now i have a scaled rotated new local box and pash it to ageia
boxDesc.dimensions.set(NxVec3(newStats.x, newStats.y, newStats.z);

after this i set the actorDesc.globalPose.t to pos which is the position of
my ID3DXMesh* in worldSpace.

Is this correct or am i doing something wrong?

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Quote:
 ok i use boxes now they work ok for me. But sometimes the behavior of them is really strange. Very small boxes stay above the ground while other fall onto it.

Not sure if you're already doing this but using the PhysX remote visual debugger is VERY helpful as it shows you exactly what your actors are doing. to use it just add this line to your program after you create your physX SDK instance:

gPhysicsSDK->getFoundationSDK().getRemoteDebugger()->connect("localhost");

n.b. you can replace localhost with the IP address of another machine for remote debugging.

Then open the remote debugger tool at:

Start->All Programs->AGEIA PhysX SDK 2.7.3->Tools->Remote Debugger

It will run concurantly with your program and show you exactly what's going on in the PhysX simulation. This has gotten me out of a few scrapes in the past so I hope you find it as usefull.


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Is there a switch to turn off LaTeX counters from inside a box?

I know that amsmath typesets at least twice its material, the first one for measurement and that it takes measures (sic) to avoid some counters being updated twice.

Is there a switch a macro writer could use to inform LaTeX not to update counters (which are global in LaTeX) when it is needed to typeset some unknown contents #1 inside a box, which will be discarded later?

update from info gathered from egreg's answer and a (belated) look in the sources. It is not as I said above that amsmath avoids for some counters to be updated; it avoids it for all, because it uses a list of all counter names which is kept up-to-date by the latex kernel, and designed to allow anyone using it to do things with all these names. The conclusion is thus that amsmath defines the macro pair \savecounters@/\restorecounters@ because it needs the functionality, but that anyone can copy their definitions (\savecounters@ defines what \restorecounters@ will do) and use them with the same effect without having to load amsmath or even any portion of it. Just replace \@nx by \noexpand.

• \ifmeasuring@ is true in the first step and false during actual typesetting. However subtle problems can arise, if the counter's value changes from one to two digits. Can you give an example? – egreg Sep 27 '13 at 20:12
• @egreg I mentioned amsmath for background, I am asking for a switch provided by the LaTeX kernel. I have not looked at what exactly amsmath does to avoid counters being updated twice, I am asking if LaTeX provides now a way to do that. – user4686 Sep 27 '13 at 20:18
• @egreg the reason I am asking is that I just worked on this anwer and my method uses two preliminary renderings of the unknown argument, which may contain macros updating counters. – user4686 Sep 27 '13 at 20:20

This is basically what amsmath does for avoiding the problem of stepping counters twice when measuring. The changes to the counters' values should take place during the measuring phase, or there's the risk that a change from one to two digits ruins the measures made.

\documentclass{article}

\usepackage{amsmath,xcolor}

\newcounter{xyz}
\newcommand{\doxyz}{\stepcounter{xyz}--\thexyz--}

\makeatletter
\newdimen\jfbu@width
\newcommand{\processtwice}[1]{%
\begingroup
\savecounters@
% here we do the measurement
\sbox\z@{#1}%
\global\jfbu@width=2\wd\z@
% and we restore the counters' values
\restorecounters@
\endgroup
\colorbox{yellow}{\makebox[\jfbu@width]{#1}}%
}
\makeatother

\begin{document}

\processtwice{\doxyz}

\processtwice{\doxyz}

\end{document}

As you see, #1 is evaluated twice, but the counter is stepped only the right number of times.

• ok, so I ended up looking in amsmath and latex and I am very surprised that amsmath just uses the \cl@@ckpt thing in a simple way; it didn't have to tap into \newcounter etc... surprising really that \savecounters@ and restorecounters@ are provided by amsmath, and not from the latex kernel itself. The good news is thus that a macro writer only has to steal the macros from amsmath. – user4686 Sep 27 '13 at 20:57

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# Tag Info

7

$\mathbb{Z}$-modules are precisely abelian groups. As every ring is an abelian group, it is a $\mathbb{Z}$-module. It is entirely possible to be a module over more than one ring. For example, if $M$ is an $R$-module then it is also an $S$-module for any subring $S$ of $R$ (you seem to be interested in the case where $M = R$). Another example is given by ...

2

Let $A = \mathbb Z$ and $M = \mathbb Z/2\mathbb Z$.

1

I think the following should work, but take it with a grain of salt. Let $\mathbb Z(p^\infty)$ denote the $p$-Prüfer group, that is, the injective envelope of $\mathbb Z(p)$. The module $I=\prod_p \mathbb Z(p^\infty)$ is an injective module containing your module $M=\prod_p \mathbb Z(p)$ as a submodule. Now consider the module  E = M + \bigoplus_p \mathbb ...

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


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Describes the incast problem in storage networks. The incast problem is first described in the following two literatures:

@inproceedings{
author = {G. A. Gibson and D. F. Nagle and K. Amiri and J. Butler and F. W. Chang and H. Gobioff and C. Hardin and E. Riedel and D. Rochberg and J. Zelenka},
title = {A Cost-Effective, High-Bandwidth Storage Architecture},
booktitle = {Proc. 8th Int. Conf. on Architectural Support for Programming Languages and Operating Systems (ASPLOS)},
month = {Oct},
year = 1998,
}
@inproceedings{
author = {D. Nagle and D. Serenyi and A. Matthews},
title = {The Panasas ActiveScale Storage Cluster: Delivering Scalable High Bandwidth Storage},
booktitle = {Proc 2004 ACM/IEEE Conf on Supercomputing (SC'04)},
year = 2004,
}

Incast is caused by multiple senders overwhelming a fixed-size switch buffer. This is an important problem in storage network because, for example, parallel NFS (pNFS) is implemented as a component in NFS v4.1 to support parallel data transfer and data striping across multiple servers.

This paper reproduces incast in a setup using different switches:

• HP Procurve 2828 Ethernet switches in a tree hierarchy, connect up to 64 servers
• Force10 S50 switch connects 48 machines
• Force10 E1200 switch with 672 ports with at least 1MB output buffer per port, connecting 87 servers to 1 client

It found that incast occurs in the first two but not the last one, due to the large buffer in E1200 switch. The per-port buffer size in the experiment is not unknown because most switches allocate output buffer from a memory pool dynamically. The role of buffer in incast can be further confirmed with QoS enabled. As QoS enabled, the output buffer would be partitioned to support different classes of services, results in a smaller effective size of output queues. Hence the incast is more severe.

The paper also validated the result with simulation in ns2. It found that TCP RTO is the primary cause of incast. While only 2% of traffic are retransmitted packets, the degradation in goodput is due to underutilization during timeout. Moreover, the servers do not receive next job until the client received all the responses.

Three types of timeout are observed during incast:

• Full window loss: The entire window of data is lost and thus no feedback is available
• Last packet loss: The last packet of the SRU is dropped and no further data available to trigger the recovery
• Lost retransmit: The retransmitted packet is also dropped

Base on these, TCP-level solutions are proposed:

• use NewReno to catch partial ACKs and recover from multiple loss in the same window
• use limited transmit to ensure enough packets are sent to trigger the 3 DUPACK recovery, we can also reduce the DUPACK threshold from 3 to 1, then fast retransmit is triggered on every duplicated ACK
• disabling exponential slow start: Because the flows are mostly short flows, using more conservative linear increase of window can alleviate the situation
• reducing timeout from the popular value of min RTO of 200ms to typical RTT of gigabit ethernet of 100us or even 10us for Infiniband

Reducing timeout can lead to premature timeouts, which causes spurious retransmission and initiates slow-start from smallest window size, even if no packets are really lost. Moreover, small timeout needs fine-grained timer. BSD provides only coarse-grained software timer of 200ms and 500ms. Linux clock has a granularity of 1 to 10ms. Also, a small min RTO may be harmful for wide area networks as it causes premature timeout.

Ethernet flow control can also be a solution. Ethernet can issue pause frames to pause the forwarding of data for a designated period of time. But using Ethernet flow control can subject to head-of-line blocking problems.

## Bibliographic data

@inproceedings{
title = "Measurement and Analysis of TCP Throughput Collapse in Cluster-based Storage Systems",
author = "Amar Phanishayee and Elie Krevat and Vijay Vasudevan and David G. Andersen and Gregory R. Ganger and Garth A. Gibson and Srinivasan Seshan",
howpublished = "FAST'08",
booktitle = "Proc. 6th USENIX Conference on File and Storage Technologies (FAST'08)",
month = "Feb",
year = "2008",
}


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## help!!!!! Group Title Use basic identities to simplify the expression. one divided by cotangent of theta to the second power. + sec θ cos θ 6 months ago 6 months ago

1. help!!!!! Group Title

2. hartnn Group Title

so its $$\huge \dfrac{1}{\cot^2 \theta + \sec \theta \cos \theta}$$

3. hartnn Group Title

firstly, use the identity that sec theta = 1/cos theta so, sec theta cos theta = 1 and do you know what is $$\large 1+\cot^2 \theta = ... ?$$

4. help!!!!! Group Title

only the cot^2theta is the denominator

5. help!!!!! Group Title

no I don't know, I am COMPLETELY lost in this subject

6. hartnn Group Title

okk., so anyways you can use sec theta cos theta = 1 $$\large \dfrac{1}{\cot^2 \theta }+1$$

7. hartnn Group Title

now 1/ cot theta = tan theta so, $$\large \tan^2 \theta+1$$

8. hartnn Group Title

and ther's a standard pythagorean identity that $$\large 1+\tan^2 \theta = \sec^2 \theta$$ thats it! :)

9. help!!!!! Group Title

okay thank you!

10. help!!!!! Group Title

can you walk me through another please?

11. hartnn Group Title

sure

12. help!!!!! Group Title

Simplify the expression. quantity cosecant of x to the power of two times secant of x to the power of two divided by quantity secant of x to the power of two plus cosecant of x to the power of two.

13. help!!!!! Group Title

|dw:1389469428352:dw|

14. help!!!!! Group Title

yes

15. hartnn Group Title

$$\large \dfrac{\csc^2 x \sec^2 x }{\sec^2 x + \csc^2 x }$$

16. hartnn Group Title

ok, we use the identity that csc x = 1/ sin x sec x = 1/cos x

17. help!!!!! Group Title

okay

18. hartnn Group Title

$$\large \dfrac{\csc^2 x \sec^2 x }{\sec^2 x + \csc^2 x } = \large \dfrac{\sin^2 x \cos^2 x }{\sin^2 x + \cos^2 x }$$

19. hartnn Group Title

now we can use the famous pythagorean identity $$\Large \color{red}{\sin^2x + \cos^2 x =1 }$$

20. hartnn Group Title

21. help!!!!! Group Title

so the answer is just 1? or am I wrong?

22. hartnn Group Title

denominator = 1 yes what remains is numerator sin^2 x cos^2 x

23. hartnn Group Title

you have choices for this too ?

24. help!!!!! Group Title

yes I do the choices are: sin^2x cos^2x -1 1

25. help!!!!! Group Title

im sorry im so stupid. im usually really good at math but precalc just makes no sense to me

26. hartnn Group Title

oops, i did a simplification error $$\large \dfrac{\csc^2 x \sec^2 x }{\sec^2 x + \csc^2 x } = \large \dfrac{1 }{\sin^2 x + \cos^2 x } = 1$$

27. help!!!!! Group Title

okay and i have a question. next year I planned on taking ap stat. is that anything like precalc because i hope not. is it easier or harder?

28. hartnn Group Title

if you practice less, math is hard. if you practice more, math is easy. irrespective of any topic/subject

29. help!!!!! Group Title

well are the concepts of ap stat very similar to or very different to those of pre-calculus?

30. hartnn Group Title

statistics is a different branch of math altogether, so very different.

31. help!!!!! Group Title

i just think I'm not good at online class taking because I learn better from an actual person rather than a computer screen. I need the physical classroom environment and a teacher to give me instruction.

32. help!!!!! Group Title

what kind of problems does ap stat entail? word problems? graphs?

33. hartnn Group Title

both.

34. help!!!!! Group Title

oh okay thank you soooooo much! is there anything else that i should know?

35. hartnn Group Title

try to take up easy questions before going to harder ones. if you get stuck anytime, OpenStudy is always Open for you :)


text:
Mathematics

# $\displaystyle \int\dfrac{t^{2}}{t^{3}+1} dt =$

##### SOLUTION

Consider the following Equation.

$I=\int\limits_{{}}^{{}}{\dfrac{{{t}^{2}}}{{{t}^{3}}+1}}dt$

$x={{t}^{3}}+1$

$\dfrac{dx}{dt}=2{{t}^{2}}$

$=\int\limits_{{}}^{{}}{\dfrac{{{t}^{2}}}{x}*\dfrac{1}{2{{t}^{2}}}}dx$

$=\dfrac{1}{2}\ln x+C$

$=\dfrac{1}{2}\ln \left( {{t}^{3}}+1 \right)+C$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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lf f(a$+$b-x) $=\mathrm{f}(\mathrm{x})$ then $\displaystyle \int_{\mathrm{a}}^{\mathrm{b}}\mathrm{x}.\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}=$
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Q5 Passage Hard
In calculating a number of integrals we had to use the method of integration by parts several times in succession.
The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts
$\displaystyle \int u\left ( x \right )v\left ( x \right )dx=u\left ( x \right )v_{1}-u'\left ( x \right )v_{2}\left ( x \right )+u''\left ( x \right )v_{3}\left ( x \right )+...+\left ( -1 \right )^{n-1}u^{n-1}\left ( x \right )V_{n}\left ( x \right ) \\ -\left ( -1 \right )^{n-1}\int u^{n}\left ( x \right )V_{n}\left ( x \right )dx$
where  $\displaystyle v_{1}\left ( x \right )=\int v\left ( x \right )dx,v_{2}\left ( x \right )=\int v_{1}\left ( x \right )dx ..., v_{n}\left ( x \right )= \int v_{n-1}\left ( x \right )dx$
Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration by parts is especially useful when  calculating $\displaystyle \int P_{n}\left ( x \right )Q\left ( x \right )dx,$ where $\displaystyle P_{n}\left ( x \right )$ is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n+1 times.


text:
# proof of Neumann series in Banach algebras

Let x be an element of a Banach algebra with identity, $\|x\|<1$. By applying the properties of the Norm in a Banach algebra, we see that the partial sums form a Cauchy sequence: $\|\sum_{n=l}^{m}x^{n}\|\leq\sum_{n=l}^{m}\|x\|^{n}\to 0$ for $l,m\to\infty$ (as is well known from real analysis), so by completeness of the Banach Algebra, the series converges to some element $y=\sum_{n=0}^{\infty}x^{n}$.

We observe that for any $m\in\mathbb{N}$,

 $(1-x)\sum_{n=0}^{m}x^{n}=\sum_{n=0}^{m}x^{n}-\sum_{n=1}^{m+1}x^{n}=1-x^{m+1}$ (1)

Furthermore, $\|x^{m+1}\|\leq\|x\|^{m+1}$, so ${\lim}_{m}x^{m+1}=0$.

Thus, by taking the limit $m\to\infty$ on both sides of (1), we get

 $(1-x)y=1$

(We can exchange the limit with the multiplication by $(1-x)$, since the multiplication in Banach algebras is continuous)

Since the Banach algebra generated by a single element is commutative and $(1-x)$ and $y$ are both in the Banach algebra generated by $x$, we also get $y(1-x)=1$. Hence, $y=(1-x)^{-1}$.

As in the first paragraph, the last claim $y\leq\frac{1}{1-\|y\|}$ again follows by applying the geometric series for real numbers.

Title proof of Neumann series in Banach algebras ProofOfNeumannSeriesInBanachAlgebras 2013-03-22 17:32:40 2013-03-22 17:32:40 FunctorSalad (18100) FunctorSalad (18100) 5 FunctorSalad (18100) Proof msc 46H05


text:
###### Question:

please help solve the part that says "show all your calculations" im confused on how to do this. I have attached the table i used below. my m1, b1, and correlation are correct.
this measurement and analysis on perior two solenoids. The length of both solenoids is 11 cm. Using your favorite math tool or package (graphing calculator, excel, etc) to analyze the data for Coil 1 included below in table 1. Do a linear fit (y = mx + b) on the data. Record your best fit result here: ml 3.7599 b1 = 0.0538 : Correlation 0.9977 Now, if we assume the relationship for the magnetic field at the center of a solenoid as given in many texts, Β = η μο Ι, where n is the number of turns per unit length, Ho is the magnetic permeability of free space (47 x 10-7 SI), and I is the current (amps), determine n and the total number of turns in the solenoid, N. Remember that in the equation above B is in units of Tesla (T). Show all your calculations here. nl N1 =
Table 1: Data for Solenoid 1. Voltage (V) 0 0.2 0.6 1.0 1.3 1.6 1.9 2.5 Current (A) 0 0.11 0.27 0.51 0.75 0.96 1.25 1.48 B-Field (T) 0 0.39 0.95 1.86 2.71 3.5 4.5 5.7

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# 🕸 Common Applications of CNNs

 Application Description 🧒👧 Face Verification Recognizing if that the given image and ID are belonging to the same person 👸 Face Recognition Assigning ID to the input face image 🌠 Neural Style Transfer Converting an image to another by learning the style from a specific image

# 🧒👧 Face Verification

## 🙌 Comparison

 Term Question Input Output Problem Class 🧒👧 Face Verification Is this the claimed person? 🕵♂ Face image / ID True / False 1:1 👸 Face Recognition Who is this person? 🧐 Face image ID of K faces in DB 1:K

# 🤸♀ Solving Approach

## 🤳 One Shot Learning

Learning from one example (that we have in the database) to recognize the person again

## 🖇 The Process

• Get input image

• Check if it belongs to the faces you have in the DB

## 👓 How to Check?

We have to calculate the similarity between the input image and the image in the database, so:

• ⭕ Use some function that

• similarity(img_in, img_db) = some_val

• 👷♀ Specifiy a threshold value

• 🕵♀ Check the threshold and specify the output

## 🤔 What can the similarity function be?

### 🔷 Siamese Network

A CNN which is used in face verification context, it recievs two images as input, after applying convolutions it calculates a feature vector from each image and, calculates the difference between them and then gives outputs decision.

In other words: it encodes the given images

Architecture:

## 👩🏫 How to Train?

We can train the network by taking an anchor (basic) image A and comparing it with both a positive sample P and a negative sample N. So that:

• 🚧 The dissimilarity between the anchor image and positive image must low

• 🚧 The dissimilarity between the anchor image and the negative image must be high

So:

$L=max(d(a,p)-d(a,n)+margin, 0)$

Another variable called margin, which is a hyperparameter is added to the loss equation. Margin defines how far away the dissimilarities should be, i.e if margin = 0.2 and d(a,p) = 0.5 then d(a,n) should at least be equal to 0.7. Margin helps us distinguish the two images better 🤸♀

Therefore, by using this loss function we:

• 👩🏫 Calculate the gradients and with the help of the gradients

• 👩🔧 We update the weights and biases of the Siamese network.

For training the network, we:

• 👩🏫 Take an anchor image and randomly sample positive and negative images and compute its loss function

# 🌠 Neural Style Transfer

Generating an image G by giving a content image C and a style image S

## 👀 Visualization

So to generate G, our NN has to learn features from S and apply suitable filters on C

## 👩🎓 Methodology

Usually we optimize the parameters -weights and biases- of the NN to get the wanted performance, here in Neural Style Transfer we start from a blank image composed of random pixel values, and we optimize a cost function by changing the pixel values of the image 🧐

In other words, we:

• 👩🏫 Define some cost function J

• 👩🔧 Iteratively modify each pixel so as to minimize our cost function

Long story short: While training NNs we update our weights and biases, but in style transfer, we keep the weights and biases constant, and instead update our image itself 🙌

### ⌚ Cost Function

We can define J as

$J(G)=\alpha J_{Content}(C,G)+\beta J_{Style}(S,G)$

Which:

• $J_{Content}$ denotes the similarity between G and C

• $J_{Style}$ denotes the similarity between G and S

• α and β hyperparameters


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The identity element of a group is unique

Suppose two elements $e$ and $e'$ are both identities. Then $$ee'=e'$$ because $e$ is an identity, and also $$ee'=e$$ because $e'$ is an identity too. Therefore, $e=e'$


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# 3.2 Lemniscates, Capacity, and Measure

The purpose of this section is not to give a detailed and rigorous account of the foundations of capacity and measure, but rather to indicate why the results of the previous section are much stronger if rephrased in terms of capacity or Hausdorff measure.

and M is sufficiently large,

Hence, provided

The basis of Section 6.5 is the result related to Cartan's lemma [Cartan. 1928; Nuttall, 1970b] that, for a polynomial qm{z) with leading coefficient unity, | qm(z)\ >t]m except on a set & of measure at most 7ttj2. The boundary on which | qm(z)| =fjm is usually called a lemniscate, and so all the results of

Section 6.5 apply except on lemniscatic regions which are arbitrarily small. We will see that the natural measure of the size of a lemniscatic region is its capacity.

To begin with, let us recall Tchebycheff's result for the minimax polynomial on an interval The problem is to find the polynomial Tn(x) in the class Pn of all polynomials p„(x) of degree n and leading coefficient unity for which the limit

is attained. The solution is well known. For «>1,

is a polynomial of degree n of leading coefficient unity, and

A linear change of variable

leads to the result that

To generalize these ideas to an arbitrary compact set in the z-plane, we have the following theorem.

Theorem 6.6.1. Let & be a compact set in the complex plane (containing infinitely many points). Then there is a unique Tchebycheff polynomial for which

may be written as

is attained. The minimax polynomial,

The zeros

is achieved at least n times on the boundary of &.

lie in the convex hull of \$, and the maximum value Mn of

Discussion. We shall not prove this important theorem, [Hille, 1962, p. 265], but elaborate it with a few remarks.

is attained at z = z', we find that

It is easy to see that all the z, lie within the convex hull of &. For suppose not, and let z[ lie outside the convex hull DC of S and the points z2, z3,..., z„. By considering a point z| nearer to %, and the point z' for which the

which contradicts the minimum property of Mn with respect to variation of the zr Hence the zeros of Tn(z) lie in the convex hull of S.

The proof of the existence of a minimax polynomial is based on "tracing" the roots to extremal positions in the closed convex hull. We omit the proofs of existence and uniqueness of Tn(z) in the general case. Since & is compact, it is obvious that the maximum of \Tn(z)\ is attained in S. Further, the maximum-modulus theorem shows that the maximum is attained on the boundary of &. That \Tn{z)\ should equal Mn at n distinct points on the boundary is another significant result we state without proof.

If \$ is a finite point set, a possibility excluded by the hypothesis of the theorem, then the Tchebycheff polynomial of order n is zero on & for n>N. Furthermore, it is not unique for n>N, and so this degenerate case is naturally excluded by the hypothesis.

Figure 1. A set S = S, US2. The union of the shaded region and the set £ comprises the convex hull of &.

Corollary. The Tchebycheff polynomial Tn{z) for a set & defines a lemniscate by \Tn{z)\ = Mn and a lemniscatic region £n by

Then Sc£„, and the boundary of £„ has at least n points in common with S.

Discussion. The important idea is that & becomes a subset of the lemniscatic region £n defined by \Tn(z)\<Mn, and the proof follows immediately from the maximum-modulus theorem and the main theorem.

Theorem 6.6.2. Let & be a compact set in the complex z-plane (containing infinitely many points). Let Tn(z) be the Tchebycheff polynomials defined on \$, and let

Then the capacity of & is uniquely defined by

Hence

We now proceed to three theorems which we prove, because the proofs illustrate the structure of lemniscates and the idea of capacity. Repeated use is made of the maximum-modulus theorem, which states that the maximum modulus of an analytic function in a compact region is achieved on the boundary.

Proof. The only proof required is that the limit (6.7) is well defined. To do this, let

If 8 is the maximum diameter of the compact set S in the usual sense, then from (6.4),

Given e>0, we may find N such that

and

For any positive integers m, k but with m<N,

where

is a constant independent of k. Recognising zm[Tn{z)]k as a polynomial of order m + nk, we see that

and by taking the limit as k^> oo,

Equation (6.8) holds for any positive m<N, and any e>0. Hence /8 = a, and cap(S) is well defined by (6.7).

The quantity cap(S) is a measure of the magnitude of the set &. It is called the logarithmic capacity, colloquially abbreviated to capacity or transfinite diameter in different contexts.

is given by

defined on 8, so that

le the «th-order Tchebycheff polynomial

The next theorem shows the key role of lemniscates in the theory of capacity.

Theorem 6.6.3. If & is a lemniscatic region given by

then

Proof. The maximum modulus theorem show that 3S (the boundary of S)

then

and consequently

By definition of

Recall

RoucHfi's theorem. If /(z) and g(z) are analytic inside and on a closed contour C and |g(z)|<|/(z)| on C, then /(z) and /(z) + g(z) have the same number of zeros inside C.

In this case f(z) = Tn(z), g(z)= —/>„(z), and the theorem implies that a polynomial of degree n— 1 has n zeros in \$. This is impossible, hence tj'=tj and cap(S)=rj, proving the theorem.

Before giving examples of the capacity of a set, we prove the major result which improves the theorems of the previous section.

where

We will consider the family of lemniscates

given by \p(z)\ = p for all positive p. This consists of at most n disjoint closed curves K>, K2,..., K . The maximum-modulus theorem shows that

Theorem 6.6.4. Let & be a compact set. Then

Proof. Let Tn( z) be the «th Chebychev polynomial denned on &. ror any e>0, (6.7) implies that a sufficiently large n exists such that

This inequality defines a lemniscatic region £„, of capacity -q = cap( &) + e, which has & as a subset. If we prove that

it follows that

Hence the theorem is true provided we prove (6.10). For brevity, let the lemniscate be defined by

Figure 2. The mapping of K, and K2 onto the circle |f| = p, showing two branch points in the f-plane.

the curves lie outside each other. There is at least one root of p{z) in each

curve K. • 1ft thfirp hp nrprisplv ii . rnnts in K. As thp r>r>int 7 mnvps arnnnrl

moves once around the circlt

and

moves a,, times around the circle

mapping

and the inverse

Herep(z) is clearly a single-valued function of z, but

Now consider the mappings

the inverse map z=z(f) is not, and z=z(J) has n— 1 branch points occurring at p'(z) = 0. However, assuming that |?| = p avoids these branch points, one of the branches of z=/(f) is regular on |f'| = pl/llk, and has the Laurent expansion

i.e.,

(6.12)

To find the area of Kk,

Hence

Therefore

. (6.13)

Equation (6.13) is an increasing and continuous function of p, and so is bounded by its value at p= oc. For p sufficiently large, Kl contains all n roots of d(z\ and therefore a,=n and m = 1.

(6.14)

which makes (6.12) more explicit, and (6.13) becomes

The result is proved, but it is interesting to note from (6.15) that equality is attained when a0=fli = ■ ■ ■ =0 and (6.14) shows that this locus is the circle z" = p.

We now turn to a few examples.

Example 1. The capacity of the interval a<x<b is (b—a)/4.

Discussion. We assume the result (6.1) [Cheney, 1966, p. 61; Rivlin, 1969, Chapter 1] for the proof. Then result (6.3) follows from the substitution (6.2), and from (6.16) it follows that

and the theorem is proved.

Example 2. The capacity of the disk | z | < R and the circle | z | = R is each equal to R.

Proof. The circle |z| = i? is a lemniscate, given by \zm\ = Rm, and so the results follow from Theorem 6.6.3 (and also from an inspection of the proof of Theorem 6.6.4).

Example 3. If & is a countable compact set, cap(S) = 0. We leave the proof as an exercise.

We next state two very important theorems about capacity, without proof. Each gives considerable insight into the magnitude of the capacity of an arbitrary point set.

Discussion. The points

ire chosen so as to maximize

Theorem 6.6.5 [Hille, 1962, p. 268-273]. Let & be a compact set. Then

which contains n(n— l)/2 terms in its expansion. Hence it is clear that cap(S) is bounded by the maximum diameter of 6. Furthermore, cap(S) is some sort of geometric mean of the distance apart of the points of \$, and so it is called the transfinite diameter.

Two corollaries of the theorem are self-evident:

Corollary 1 (Monotonicity). If <3) and & are compact sets with йі!С&,

Corollary 2 (Homogeneity).

Theorem,6.6.6 [Hille, 1962, p. 280-289]. Let ju(z) be a normalized measure defined on &, and define

Let

Then

Discussion. We have in mind that ju(z) is a charge distribution on a two-dimensional surface S, so that I[ji] is the self-energy associated with the distribution ji. In physical equilibrium, this functional is a minimum, and so the potential due to the physical distribution of unit charge on \$ is

The further definition that ln(cap(S)) = - V{&) explains the name of logarithmic capacity for cap(S).

Finally, we remark that we have emphasized the role of capacity as a point-set measure of the region of inaccurate approximation of Pade approximants to meromorphic functions. Another popular point-set measure is Hausdorff measure, and the results of the previous section generalize to convergence in a-dimensional Hausdorff measure [Wallin, 1974; Lubinsky, 1980]. In this case, the measure is defined by

where the minimum is taken over all possible denumerable families of circular disks Dt which cover the set S, and S( Dt ) is the diameter of the disk Dr Two-dimensional Hausdorff measure is similar to area in its properties, and one-dimensional Hausdorff measure is more like length.

The following Boutroux-Cartan lemma is most useful in the context of one-dimensional Hausdorff measure.

satisfies

Theorem 6.6.7. For any v>0, the polynomial

the inequality

We do not prove this theorem, and we refer to EPA (Chapter 14) for a more complete account of the role of Hausdorff measure in convergence of Pade approximants. We merely note that (6.16) is the basic type of inequality required in this context. We also note that many of these result, also extend, mutatis mutandis, to the rational interpolation problem [Walsh, 1969, 1970; Karlsson, 1976].

We conclude this section by repeating that an inspection of the proofs of the theorems of Section 6.5 shows that they prove convergence in capacity, and we have shown that this is a substantially stronger result than convergence in measure. In particular, an interval has zero measure, but nonzero capacity (and nonzero one-dimensional Hausdorff measure), and so a finite interval, or any set containing a finite interval, is not a permissible exceptional set for the theorems of Section 6.5. Finally, we draw attention to a result of Nuttall's, possibly generalizable in the future, that for certain functions with branch points connected by branch cuts, the Pade approximants converge in capacity in the z-plane except in a region which minimizes the capacity of the cuts in the z ~ 1 -plane [Nuttall, 1977; Nuttall and Singh, 1977]. This conjecture should be contrasted with the conjecture of Baker, Gammel, and Wills.


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# Course Description

This course enables students to broaden their understanding of relationships and extend their problem-solving and algebraic skills through investigation, the effective use of technology, and abstract reasoning. Students will explore quadratic relations and their applications; solve and apply linear systems; verify properties of geometric figures using analytic geometry; and investigate the trigonometry of right and acute triangles. Students will reason mathematically and communicate their thinking as they solve multi-step problems.

# Overall Provincial Curriculum Expectations

See Ontario math curriculum for grades 9 and 10 for details.

### A: Quadratic Relations of the Form $y=ax^2+bx+c$

1. determine the basic properties of quadratic relations;
2. relate transformations of the graph of $y = x^2$ to the algebraic representation $y = a(x - h)^2 + k$;
3. solve quadratic equations and interpret the solutions with respect to the corresponding relations;

### B: Analytic Geometry

1. model and solve problems involving the intersection of two straight lines;
2. solve problems using analytic geometry involving properties of lines and line segments;
3. verify geometric properties of triangles and quadrilaterals, using analytic geometry.

### C: Trigonometry

1. use their knowledge of ratio and proportion to investigate similar triangles and solve problems related to similarity;
2. solve problems involving right triangles, using the primary trigonometric ratios and the Pythagorean theorem;
3. solve problems involving acute triangles, using the sine law and the cosine law.

## Specific Provincial Curriculum Expectations

### A1: Investigating the Basic Properties of Quadratic Relations

• – collect data that can be represented as a quadratic relation, from experiments using appropriate equipment and technology (e.g., concrete materials, scientific probes, graphing calculators), or from secondary sources (e.g., the Internet, Statistics Canada); graph the data and draw a curve of best fit, if appropriate, with or without the use of technology
• – determine, through investigation with and without the use of technology, that a quadratic relation of the form $y = ax^2 + bx + c, a\neq 0$ can be graphically represented as a parabola, and that the table of values yields a constant second difference
• – identify the key features of a graph of a parabola (i.e., the equation of the axis of symmetry, the coordinates of the vertex, the y-intercept, the zeros, and the maximum or minimum value), and use the appropriate terminology to describe them;
• – compare, through investigation using technology, the features of the graph of $y = x^2$ and the graph of $y = 2^x$, and determine the meaning of a negative exponent and of zero as an exponent (e.g., by examining patterns in a table of values for $y = 2^x$; by applying the exponent rules for multiplication and division)

### A2: Relating the Graph of $y=x^2$ And Its Transformations

• – identify, through investigation using technology, the effect on the graph of $y = x^2$ of transformations (i.e., translations, reflections in the x-axis, vertical stretches or compressions) by considering separately each parameter $a, h, k$ [i.e., investigate the effect on the graph of $y = x^2$ of $a, h, k$ in $y = x^2 + k, y = (x – h)^2$ and $y = ax^2$];
• – explain the roles of $a, h, k$ in $y = a(x – h )^2 + k$, using the appropriate terminology to describe the transformations, and identify the vertex and the equation of the axis of symmetry;
• – sketch, by hand, the graph of $y = a(x – h )^2 + k$ by applying transformations to the graph of $y = x^2$
• – determine the equation, in the form $y = a(x – h)^2 + k$, of a given graph of a parabola

• – expand and simplify second-degree polynomial expressions [e.g., $(2x + 5)^2$, (2x – y)(x + 3y)], using a variety of tools (e.g., algebra tiles, diagrams, computer algebra systems, paper and pencil) and strategies (e.g., patterning);
• – factor polynomial expressions involving common factors, trinomials, and differences of squares [e.g., $2x^2 + 4x$, $2x – 2y + ax – ay$, $x2 – x – 6$, $2a^2 + 11a + 5$, $4x^2 – 25$], using a variety of tools (e.g., concrete materials, computer algebra systems, paper and pencil) and strategies (e.g., patterning);
• – determine, through investigation, and describe the connection between the factors of a quadratic expression and the x-intercepts (i.e., the zeros) of the graph of the corresponding quadratic relation, expressed in the form y = a(x – r)(x – s);
• – interpret real and non-real roots of quadratic equations, through investigation using graphing technology, and relate the roots to the x-intercepts of the corresponding relations;
• – express $y = ax^2 + bx + c$ in the form $y = a(x – h)^2 + k$ by completing the square in situations involving no fractions, using a variety of tools (e.g. concrete materials, diagrams, paper and pencil);
• – sketch or graph a quadratic relation whose equation is given in the form $y = ax^2 + bx + c$, using a variety of methods (e.g., sketching $y = x^2 – 2x – 8$ using intercepts and symmetry; sketching $y = 3x^2 – 12x + 1$ by completing the square and applying transformations; graphing $h = –4.9t^2 + 50t + 1.5$ using technology);
• – explore the algebraic development of the quadratic formula (e.g., given the algebraic development, connect the steps to a numerical example; follow a demonstration of the algebraic development [student reproduction of the development of the general case is not required]);
• – solve quadratic equations that have real roots, using a variety of methods (i.e., factoring, using the quadratic formula, graphing)

### A4: Solving Problems Involving Quadratic Equations

• – determine the zeros and the maximum or minimum value of a quadratic relation from its graph (i.e., using graphing calculators or graphing software) or from its defining equation (i.e., by applying algebraic techniques);
• – solve problems arising from a realistic situation represented by a graph or an equation of a quadratic relation, with and without the use of technology (e.g., given the graph or the equation of a quadratic relation representing the height of a ball over elapsed time, answer questions such as the following: What is the maximum height of the ball? After what length of time will the ball hit the ground? Over what time interval is the height of the ball greater than 3 m?)

### B1: Using Linear Systems to Solve Problems

• – solve systems of two linear equations involving two variables, using the algebraic method of substitution or elimination
• – solve problems that arise from realistic situations described in words or represented by linear systems of two equations involving two variables, by choosing an appropriate algebraic or graphical method

### B2: Solving Problems Involving Properties of Line Segments

• – develop the formula for the midpoint of a line segment, and use this formula to solve problems (e.g., determine the coordinates of the midpoints of the sides of a triangle, given the coordinates of the vertices, and verify concretely or by using dynamic geometry software);
• – develop the formula for the length of a line segment, and use this formula to solve problems (e.g., determine the lengths of the line segments joining the midpoints of the sides of a triangle, given the coordinates of the vertices of the triangle, and verify using dynamic geometry software);
• – develop the equation for a circle with centre (0, 0) and radius r, by applying the formula for the length of a line segment;
• – determine the radius of a circle with centre (0, 0), given its equation; write the equation of a circle with centre (0, 0), given the radius; and sketch the circle, given the equation in the form $x^2 + y^2 = r^2$;
• – solve problems involving the slope, length, and midpoint of a line segment (e.g., determine the equation of the right bisector of a line segment, given the coordinates of the endpoints; determine the distance from a given point to a line whose equation is given, and verify using dynamic geometry software)

### B3: Using Analytic Geometry to Verify Geometric Properties

• – determine, through investigation (e.g., using dynamic geometry software, by paper folding), some characteristics and properties of geometric figures (e.g., medians in a triangle, similar figures constructed on the sides of a right triangle);
• – verify, using algebraic techniques and analytic geometry, some characteristics of geometric figures (e.g., verify that two lines are perpendicular, given the coordinates of two points on each line; verify, by determining side length, that a triangle is equilateral, given the coordinates of the vertices);
• – plan and implement a multi-step strategy that uses analytic geometry and algebraic techniques to verify a geometric property (e.g., given the coordinates of the vertices of a triangle, verify that the line segment joining the midpoints of two sides of the triangle is parallel to the third side and half its length, and check using dynamic geometry software; given the coordinates of the vertices of a rectangle, verify that the diagonals of the rectangle bisect each other)

### C1: Investigating Similarity and Solving Problems Involving Similar Triangles

• – verify, through investigation (e.g., using dynamic geometry software, concrete materials), the properties of similar triangles (e.g., given similar triangles, verify the equality of corresponding angles and the proportionality of corresponding sides);
• – describe and compare the concepts of similarity and congruence;
• – solve problems involving similar triangles in realistic situations (e.g., shadows, reflections, scale models, surveying)

### C2: Solving Problems Involving the Trigonometry of Right Triangles

• – determine, through investigation (e.g., using dynamic geometry software, concrete materials), the relationship between the ratio of two sides in a right triangle and the ratio of the two corresponding sides in a similar right triangle, and define the sine, cosine, and tangent ratios
• – determine the measures of the sides and angles in right triangles, using the primary trigonometric ratios and the Pythagorean theorem;
• – solve problems involving the measures of sides and angles in right triangles in reallife applications (e.g., in surveying, in navigating, in determining the height of an inaccessible object around the school), using the primary trigonometric ratios and the Pythagorean theorem

### C3: Solving Problems Involving the Trigonometry of Acute Triangles

• – explore the development of the sine law within acute triangles (e.g., use dynamic geometry software to determine that the ratio of the side lengths equals the ratio of the sines of the opposite angles; follow the algebraic development of the sine law and identify the application of solving systems of equations [student reproduction of the development of the formula is not required]);
• – explore the development of the cosine law within acute triangles (e.g., use dynamic geometry software to verify the cosine law; follow the algebraic development of the cosine law and identify its relationship to the Pythagorean theorem and the
• cosine ratio [student reproduction of the development of the formula is not required]);
• – determine the measures of sides and angles in acute triangles, using the sine law and the cosine law
• – solve problems involving the measures of sides and angles in acute triangles.


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# 4.13: Simplify Variable Expressions Involving Integer Division

Difficulty Level: At Grade Created by: CK-12
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Practice Simplify Variable Expressions Involving Integer Division
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Have you ever had to do homework while on vacation?

Molly was working on her homework in the ski lodge. She had one more problem to do before she could go back out for another run on the mountain.

Here is the problem.

12x3$\frac{-12x}{-3}$

Molly isn't sure how to work with this problem. She knows that it involves integers, variables and division.

This Concept will teach you how to help Molly with her dilemma.

### Guidance

Previously we used variable expressions with addition, subtraction and multiplication. Now we are going to apply division of integers with variable expressions. Remember that a variable expression is a math phrase that uses numbers, variables and operations.

Find the value of this expression 18x÷(2)$-18x \div (-2)$

It may help you to rewrite the problem like this using a fraction bar to divide. Now you can see which values can be divided.

18x2$\frac{-18x}{-2}$

Then separate out the integers like this.

18x2=18x2=182x$\frac{-18x}{-2} = \frac{-18 \cdot x}{-2} = \frac{-18}{-2} \cdot x$

Notice that we can divide the integers. The x$x$ remains alone because there isn’t another x$x$ . We separate it out. Then we divide the integer part and add the x$x$ to the answer.

Since 18÷(2)=18÷2=9$-18 \div (-2) = 18 \div 2 = 9$ , we know that 182x=9x=9x$\frac{-18}{-2} \cdot x= 9 \cdot x = 9x$

So, the value of the expression is 9x$9x$ .

Let’s look at another problem where there is a matching variable too.

24y÷2y$-24y \div 2y$

Next, we rewrite the expression using a fraction bar.

24y2y$\frac{-24y}{2y}$

Now, we can separate the terms.

&\frac{-24}{2} \cdot \frac{y}{y}\\
&-24 \div 2 = -12

y÷y=1$y \div y =1$ because the y$y$ ’s cancel each other out

12(1)=12$-12(1) = -12$

The value of the expression is -12.

Now it's time for you to try a few on your own.

#### Example A

14a÷7$-14a \div -7$

Solution: 2a$2a$

#### Example B

28ab÷7b$28ab \div -7b$

Solution: 4a$-4a$

#### Example C

6x÷2$-6x \div -2$

Solution: 3x$3x$

Here is the original problem once again.

Molly was working on her homework in the ski lodge. She had one more problem to do before she could go back out for another run on the mountain.

Here is the problem.

12x3$\frac{-12x}{-3}$

Molly isn't sure how to work with this problem. She knows that it involves integers, variables and division.

Molly is correct.

First, let's complete the division of the numbers. We can follow the rules for dividing integers and divide negative 12 by negative 3. That answer is 4.

Because there isn't another x$x$ to work with, we can simply add that to our quotient.

4x$4x$

### Guided Practice

Here is one for you to try on your own.

18ab÷9b$-18ab \div 9b$

First, rewrite the expression using a fraction bar.

18ab9b$\frac{-18ab}{-9b}$

Next, separate out the terms.

&\frac{-18}{-9} \cdot \frac{ab}{b}\\
&-18 \div -9 = 2

ab÷b=a$ab \div b= a$ Notice that the b$b$ ’s cancel, but the a doesn’t. It is left as part of the final expression.

Our final answer is 2a$2a$ .

### Explore More

Directions: Find each quotient with variable expressions.

1. 36t÷(9)$36t \div (-9)$

2. 56n÷(7)$-56n \div (-7)$

3. 22n÷11n$-22n \div -11n$

4. 28n÷7$-28n \div 7$

5. 18xy÷2x$18xy \div 2x$

6. 72t÷(9t)$72t \div (-9t)$

7. 48xy÷(8y)$48xy \div (-8y)$

8. 54xy÷(9xy)$54xy \div (-9xy)$

9. 16a÷(4a)$16a \div (4a)$

10. 16ab÷(4b)$-16ab \div (-4b)$

11. 99xy÷(9x)$-99xy \div (-9x)$

12. 121a÷(11b)$121a \div (11b)$

13. 144xy÷(12)$-144xy \div (-12)$

14. 169y÷(13x)$-169y \div (-13x)$

15. 225xy÷(5z)$-225xy \div (5z)$

### Vocabulary Language: English

Quotient

Quotient

The quotient is the result after two amounts have been divided.
Undefined

Undefined

An expression in mathematics is undefined when it has no meaning. Division by zero is undefined.
Variable Expression

Variable Expression

A variable expression is a mathematical phrase that contains at least one variable or unknown quantity.

## Date Created:

Nov 30, 2012

Jun 09, 2015
You can only attach files to Modality which belong to you
If you would like to associate files with this Modality, please make a copy first.


text:
# Geometry at Infinity

Priority programme of the DFG

## Welcome to SPP 2026

This is the platform of a coordinated research programme in mathematics, funded by the German Research Foundation (DFG).  It comprises 33 research projects in the fields of differential geometry, geometric topology, and global analysis. More than 80 researchers from doctoral to professorial level and based at more than 20 German and Swiss universities are represented in this programme.

## Latest publications

#### Asymptotic geometry of the moduli space of parabolic $SL(2,\mathbb{C})$-Higgs bundlesLaura Fredrickson, Rafe Mazzeo, Jan Swoboda, Hartmut Weiß

Given a generic stable strongly parabolic $SL(2,\mathbb{C})$-Higgs bundle
$(\mathcal{E}, \varphi)$, we describe the family of harmonic metrics $h_t$...

#### Orientation data for moduli spaces of coherent sheaves over Calabi-Yau 3-foldsDominic Joyce, Markus Upmeier

Let X be a compact Calabi-Yau 3-fold, and write $$\mathcal{M}, \overline{\mathcal{M}}$$ for the moduli stacks of objects in coh(X) and the derived...

#### On the equivariant cohomology of cohomogeneity one Alexandrov spacesManuel Amann, Masoumeh Zarei

We give a characterization of those Alexandrov spaces admitting a cohomogeneity one action of a compact connected Lie group G for which the action is...

#### Area Estimates for High genus Lawson surfaces via DPWLynn Heller, Sebastian Heller, Martin Traizet

Starting at a saddle tower surface, we give a new existence proof of the Lawson surfaces ξm,k of high genus by deforming the corresponding DPW...

#### Isothermic constrained Willmore tori in 3-spaceLynn Heller, Sebastian Heller, Cheikh Birahim Ndiaye

We show that the homogeneous and the 2-lobe Delaunay tori in the 3-sphere provide the only isothermic constrained Willmore tori in 3-space with...


text:
Fermi energy/temperature

1. Oct 7, 2011

AriAstronomer

Hey everyone,
So unfortunately I haven't taken Stat Mech yet (and my GRE is in a week), but through casual reading and forums I've gathered what the concept of fermi temperature and energy is...or thought I did. From what I understand, below the fermi energy (the fermi energy within an atom corresponding to some highest orbital) all orbitals below it are completely filled, and all orbitals above it are completely empty. I assumed however, that this happens at the fermi temperature, which I thought was going to be close to T=0K, but apparently for most elements is like 10,000K? What gives? Also, what's this about below the fermi temperature things get degenerate (as in, more than one electron configuration corresponding to the same energy), but above there is no degeneracy? I was under the impression that at fermi temperature (which I thought was close to 0Kelvin), all electrons fill the lowest possible orbitals without violating the Pauli Exclusion Principle, and then as you raise temperature as things get more energy they gain more freedom and thus degeneracy starts....

The GRE is in a week, any chance someone can resolve this for me?!! I'd really appreciate it!

Ari

2. Oct 9, 2011

xiyangxixia

If Ec-Ef>>0, they are under degerancy. If 0<Ec-Ef<2KT, they are under weak degenerancy. If Ec-Ef>2KT, they are under nondegenerancy. The energy of electron became larger with temperature increase. Many electrons escaped from feimi energy level.

3. Oct 10, 2011

genneth

Fermi energy is only defined at 0K. At finite temperatures the relevant quantity is the chemical potential; as T -> 0, this converges to the Fermi energy. The Fermi temperature is the scale one gets from converting the Fermi energy (via k_B) to a temperature. The relevance is that the Fermi distribution has a "width" which is of the order of the Fermi energy, and at high temperatures it converges to a classical Boltzmann distribution; so the Fermi temperature sets the scale between quantum effects dominating (things like degeneracy) and being able to treat the fermions as a classical gas.


text:
# Show $\lim_{n \rightarrow \infty} \int_{\Omega} f_n\,d\mu_n = \int_{\Omega} f\,d\mu$

Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of measures on $A$ satisfying $\mu_n \nearrow \mu$ and $f,f_n \in M^+$ ( $n \in \mathbb{N}$ ) satisfying $f_n \nearrow f$. Show that $$\lim_{n \rightarrow \infty} \int_{\Omega} f_n\, d\mu_n = \int_{\Omega} f\, d\mu.$$

So my idea is to show this claim for a fixed $f_n$. Maybe we could use this then for the claim itself. Moreover I think that monotone convergence theorem could be helpful.

Remark: $M^+$ := {$f | f$ is measurable and non-negative }

So $\mu_{n}$ is absolutely continuous with respect to $\mu$, so $\dfrac{d\mu_{n}}{d\mu}$ exists, and we see that \begin{align*} \mu_{n}(A)=\int_{A}\dfrac{d\mu_{n}}{d\mu}d\mu\leq\mu_{n+1}(A)=\int_{A}\dfrac{d\mu_{n+1}}{d\mu}d\mu, \end{align*} so \begin{align*} \int_{A}\left(\dfrac{d\mu_{n+1}}{d\mu}-\dfrac{d\mu_{n}}{d\mu}\right)d\mu\geq 0 \end{align*} holds for every $A$, so \begin{align*} \dfrac{d\mu_{n+1}}{d\mu}\geq\dfrac{d\mu_{n}}{d\mu},~~~~\text{a.e.}. \end{align*} Since \begin{align*} \mu_{n}(A)=\int_{A}\dfrac{d\mu_{n}}{d\mu}d\mu, \end{align*} we take $n\rightarrow\infty$ and using Monotone Convergence Theorem to deduce that \begin{align*} \mu(A)=\int_{A}\lim_{n\rightarrow\infty}\dfrac{d\mu_{n}}{d\mu}d\mu, \end{align*} this shows that \begin{align*} \lim_{n\rightarrow\infty}\dfrac{d\mu_{n}}{d\mu}=1,~~~~\text{a.e}. \end{align*} Now we write that \begin{align*} \int_{\Omega}f_{n}d\mu_{n}=\int_{\Omega}f_{n}\dfrac{d\mu_{n}}{d\mu}d\mu, \end{align*} and use the fact that $f_{n}\dfrac{d\mu_{n}}{d\mu}\uparrow f$ a.e. to get the result.
• Thank you for your answer user284331, but I have two questions: what do you mean with $\frac{d\mu_n}{d\mu}$? and why is $\mu_{n}(A)=\int_{A}\dfrac{d\mu_{n}}{d\mu}d\mu$ ? I think the rest is clear then. – RukiaKuchiki Apr 25 '18 at 18:07
• This is our definition of absolutely continuous: Let µ and ν be measures on a measurable space . We call $ν$ absolutely continuous with respect to $µ$ if for all A ∈ $A$, $µ$[A] = 0 ⇒ $ν$[A] = 0 . – RukiaKuchiki Apr 25 '18 at 18:10


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# Indeterminism in Newtonian mechanics?

1. May 22, 2010

### Blue_Jaunte

This paper, written by a University of Pittsburgh professor, John Norton, describes a simple situation in which Newtonian mechanics allows for purportedly non-deterministic equations of motion:
http://www.pitt.edu/~jdnorton/papers/DomePSA2006.pdf

In the first part of the paper, he describes the problem and finds the equations of motion. In the problem, a ball rests atop a dome in a constant, downward, gravitational field. The dome is given by the equation
h = 2/(3g)r^(3/2)
(fudging the units a bit for simplicity).
Where
h = distance traveled in the vertical direction
r = distance traveled along the surface of the dome
The net force on the ball (frictionless surface) is
F = a = g*sinθ = g(dh/dr) = r^(1/2)
(assuming unit mass)
Where
θ = angle between direction of motion and the horizontal

The resulting differential equation,
r''(t) = r^(1/2)
has the expected solution, r(t)=0, as well as a class of solutions
r(t) =(1/144)[t - T]^4, for t ≥ T
for some arbitrary time T.

In the later parts of the paper, he addresses possible objections to his conclusions.

I was just wondering what other people thought of this.

-Mike

Last edited: May 22, 2010
2. May 22, 2010

### Q_Goest

Hi Mike. I read through parts of the paper and skimmed others so I'm not claiming to know everything Norton wrote. If you've read it over in detail, and assuming I miss Norton's point somewhere, please point out where in the paper I've overlooked it.

I disagree with Norton's conclusion that spontaneous motion at t=T is allowed by Newtonian mechanics, regardless of whether we are talking about a "dome" case as suggested by the paper or not. In fact, I think that's the key here. Consider the case where r approaches infinity. In other words, consider the case in which the "dome" becomes flatter and flatter, until the radius of curvature approaches infinity. Any force causing motion of the ball down the dome becomes vanishingly small for any t>T. For r = infinity, the dome is flat and the force completely disappears. Certainly, we wouldn't say that on a flat surface, there is any propensity for the ball to suddenly begin rolling. Obviously, this would require a force, and there is no force available on a flat surface, regardless of where the ball is.

Now let's compare the ball on a flat surface to a ball on a dome. Is the contact between the ball on the top of the dome the same as the contact between the ball on a flat surface? I think it is. As long as the ball is indeed at the location on the dome where the surface is perpendicular to the gravitational force, then in theory there's no difference between the two. So if a ball can't be allowed to move on a flat surface because there is no force available to move it, then how can the same ball move on a domed surface if the contact is identical?

3. May 22, 2010

### gabbagabbahey

I haven't looked at it too in-depth, but at first glance it looks as though the extraneous solutions (with spontaneous acceleration) constrain the particle to move along the dome. I'm not sure that the normal force and gravity alone would account for such motion, given the shape of the dome. Certainly, at the top of the dome, any acceleration or initial velocity would cause the particle to lose contact with the dome, and I haven't checked whether the same is true at other points on the dome, but I suspect it is.

4. Aug 14, 2010

### Petr Mugver

I think It's a question about unstable equilibrium. The point r = 0 on the dome has horizontal tangent plane, but the second derivatives of h with respect to x and y form e negatively defined matrix. I think the same problems arise in all examples of surfaces with this property, not only the dome. I'm not able to prove this assertion right now, so I go "heuristically". If the initial velocity is not zero but \epsilon, no matter how small, then the solution is unique. Letting \epsilon go to zero you find other solutions, which in my opinion are unphysical. In the real world we cannot rule out every small perturbation to the motion. In this case of unstable equilibrium, a small perturbation can cause big changes in the motion, as is the case, for example, of an egg in "equilibrium" with the longer axis in vertical position. The fact that we cannot predict the direction the ball will fall is due to the fact that we don't know the exact initial conditions at t = T.

5. Aug 14, 2010

### Petr Mugver

I thought it a bit better. Let's stay in 2D (in that article there was no need to be in 3D). So let's suppose the particle is linked to a curve y = f(x). Suppose f has a point of unstable equilibrium. By shifting the origin we can suppose:

f(0) = 0
f'(0) = 0 (equilibrium point)
f''(0) < 0 (unstable equilibrium point)

The lagrangian is

$$L=\frac{1}{2}(\dot x^2+\dot y^2)-gf=\frac{1}{2}(1+f'^2)\dot x^2-gf$$

The hamiltonian is

$$H=p\dot x-L=\frac{1}{2}(1+f'^2)\dot x^2+gf$$

This is independent of time, so it's conserved. At t = 0 we have H = 0, so

$$\dot x=\left(\frac{-2gf}{1+f'^2}\right)^{1/2}$$

This equation has (at least) two solutions: x(t) = 0 and

$$\int_{0}^{x(t)}\left(\frac{-2gf}{1+f'^2}\right)^{-1/2}dx=t>0$$

Note that this last solution exists only if f < 0 in a neighbourhood of x = 0, i.e. only if the origin is an unstable equilibrium point. Note also that the solution is of the form x(t) = x(t)u(t), where u is the step function. So x(t) and x(t - T) with T > 0 have the same initial condition, thus we get a whole family of solutions. You can add a +\dot z^2 term if you really want 3D, all you get is an extra angular momentum term, which is zero for our initial conditions, so the conclusions don't change.

Now my question is (unless I misunderstood everything, which is EXTREMELY possible): why the dome? I don't really get the point...

Last edited: Aug 14, 2010
6. Jan 24, 2011

### LuisVela

Hi there, I read the paper. I think he is jumping to deeper and more complicated explanations too soon.

Here is some curious result, that helps understand my point of view:

If the ball on the top of the dome has already been set in motion, i.e. is already falling to a given direction, we can calculate its velocity at any given time via using:
$$0=\frac{1}{2}mv^2-mgh\Rightarrow v(h)=\sqrt{2gh}$$

Now, lets imagine the opposite situation. What if we throw the ball upwards with a given speed from the position r(h) for a given h?
Well, if the speed is less than v(h), the ball wont make it to the top, and if its bigger, then it will go to the other side, and start falling there.
Whats interesting, and not too many people realise it, is this: Given that we can set the ball in motion, upwards, with the exact speed v(h). What is the time it will take the ball to reach the top?

Well, it is infinite.

I guess form here you can derive the rest, and light a bit the problem, before jumping to make any deeper assertions.


text:
How find this limit$\lim_{n\to\infty}\frac{1}{n}\left(\frac{n}{\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1}}\right)^n$

How find this limit $$\lim_{n\to\infty}\dfrac{1}{n}\left(\dfrac{n}{\dfrac{1}{2}+\dfrac{2}{3}+\cdots+\dfrac{n}{n+1}}\right)^n$$

My try: since $$\dfrac{1}{2}+\dfrac{2}{3}+\cdots+\dfrac{n}{n+1}=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+\cdots+\left(1-\dfrac{1}{n+1}\right)=(n+1)-H_{n+1}$$

where $$H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$

then I can't.Thank you

this problem is from a book,and only give this answer $$e^{\gamma-1}$$ where $\gamma$ is denotes the Euler–Mascheroni constant.

• What book is it from? – chs21259 Jan 23 '14 at 3:56

Hint: You're on the right track! Start by noting that if $a_n$ is the $n$-term of your sequence, then \begin{align*} \ln a_n&=\ln\left(\frac{1}{n}\right)+n\ln\left(\frac{n}{n+1-H_{n+1}}\right)\\ &=n\ln(n)-n\ln(n+1-H_{n+1})-\ln(n)\\ &=(n-1)\ln(n)-n\ln(n+1-H_{n+1}) \end{align*} Next, note that $$\ln(n+1-H_{n+1})=\ln\left[n\left(1+\frac{1}{n}-\frac{H_{n+1}}{n}\right)\right]=\ln n+\ln\left(1+\frac{1}{n}-\frac{H_{n+1}}{n}\right),$$ so that \begin{align*} \ln(a_n)&=(n-1)\ln(n)-n\ln(n)-n\ln\left(1+\frac{1}{n}-\frac{H_{n+1}}{n}\right)\\ &=-\ln(n)-n\ln\left(1+\frac{1}{n}-\frac{H_{n+1}}{n}\right). \end{align*} Now, what do you know about asymptotics for $\frac{H_{n+1}}{n}$? How about for $\ln(1+w)$ as $w\to0$?

I thing that Nicholas R. Peterson gave you almost all the tricks required for your problem. I shall try to help you for the last details (since this is not homework - at least, no tagges as).

For sufficiently large values of $n$, the ratio of H(n+1)/n behaves as
(EulerGamma - Log[1/n]) / n
So, Log[1 + 1/ n - H(n+1) /n] behaves as 1 + (1 - EulerGamma + Log[1/n]) / n
So, - Log[n] - n Log[1 + 1/ n - H(n+1) /n] behaves as (-1 + EulerGamma) and, finally, $a(n)$ behaves such as Exp[EulerGamma - 1] when $n$ goes to infinite values.

One point I would like to undeline here is that all the above has been obtained on the basis of Taylor expansions limited to first order.


text:
The data templating language
Jsonnet

# Standard Library

This page describes the functions available in Jsonnet's standard library, i.e. the object implicitly bound to the std variable. Some of the standard library functions can be implemented in Jsonnet. Their code can be found in the std.jsonnet file. The behavior of some of the other functions, i.e. the ones that expose extra functionality not otherwise available to programmers, is described formally in the specification.

The standard library is implicitly added to all Jsonnet programs by enclosing them in a local construct. For example, if the program given by the user is {x: "foo"}, then the actual code executed would be local std = { ... }; {x: "foo"}. The functions in the standard library are all hidden fields of the std object.

### External Variables

#### std.extVar(x)

If an external variable with the given name was defined, return its string value. Otherwise, raise an error.

### Types and Reflection

#### std.thisFile

Note that this is a field. It contains the current Jsonnet filename as a string.

#### std.type(x)

Return a string that indicates the type of the value. The possible return values are: "array", "boolean", "function", "null", "number", "object", and "string".

#### std.length(x)

Depending on the type of the value given, either returns the number of elements in the array, the number of codepoints in the string, the number of parameters in the function, or the number of fields in the object. Raises an error if given a primitive value, i.e. null, true or false.

#### std.objectHas(o, f)

Returns true if the given object has the field (given as a string), otherwise false. Raises an error if the arguments are not object and string respectively. Returns false if the field is hidden..

#### std.objectFields(o)

Returns an array of strings, each element being a field from the given object. Does not include hidden fields.

#### std.objectHasAll(o, f)

As std.objectHas but also includes hidden fields.

#### std.objectFieldsAll(o)

As std.objectFields but also includes hidden fields.

#### std.prune(x)

Recursively remove all "empty" members of x. "Empty" is defined as zero length arrays, zero length objects, or null values.

### Mathematical Utilities

The following mathematical functions are available: std.abs(n), std.max(a, b), std.min(a, b), std.pow(x, n), std.exp(x), std.exponent(x), std.mantissa(x), std.floor(x), std.ceil(x), std.sqrt(x), std.sin(x), std.cos(x), std.tan(x), std.asin(x), std.acos(x), and std.atan(x).

The function std.mod(a, b) is what the % operator is desugared to. It performs modulo arithmetic if the left hand side is a number, or if the left hand side is a string, it does Python-style string formatting with std.format().

### Assertions and Debugging

#### std.assertEqual(a, b)

Ensure that a == b. Returns true or throws an error message.

### String Manipulation

#### std.toString(a)

Convert the given argument to a string.

#### std.codepoint(str)

Returns the positive integer representing the unicode codepoint of the character in the given single-character string. This function is the inverse of std.char(n).

#### std.char(n)

Returns a string of length one whose only unicode codepoint has integer id n. This function is the inverse of std.codepoint(str).

#### std.substr(s, from, len)

Returns a string that is the part of s that starts at offset from and is len codepoints long.

#### std.startsWith(a, b)

Returns whether the string a is prefixed by the string b.

#### std.endsWith(a, b)

Returns whether the string a is suffixed by the string b.

#### std.substr(s, from, len)

Returns a string that is the part of s that starts at offset from and is len codepoints long.

#### std.split(str, c)

Split the string str into an array of strings, divided by the single character c.

Example: std.split("foo/bar", "/") yields ["foo", "bar"].

Example: std.split("/foo/", "/") yields ["", "foo", ""].

#### std.splitLimit(str, c, maxsplits)

As std.split(str, c) but will stop after maxsplits splits, thereby the largest array it will return has length maxsplits + 1. A limit of -1 means unlimited.

Example: std.splitLimit("foo/bar", "/", 1) yields ["foo", "bar"].

Example: std.splitLimit("/foo/", "/", 1) yields ["", "foo/"].

#### std.stringChars(str)

Split the string str into an array of strings, each containing a single codepoint.

Example: std.stringChars("foo") yields ["f", "o", "o"].

#### std.format(str, vals)

Format the string str using the values in vals. The values can be an array, an object, or in other cases are treated as if they were provided in a singleton array. The string formatting follows the same rules as Python. The % operator can be used as a shorthand for this function.

Example: std.format("Hello %03d", 12) yields "Hello 012".

Example: "Hello %03d" % 12 yields "Hello 012".

Example: "Hello %s, age %d" % ["Foo", 25] yields "Hello Foo, age 25".

Example: "Hello %(name)s, age %(age)d" % {age: 25, name: "Foo"} yields "Hello Foo, age 25".

#### std.parseInt(str)

Parses a signed decimal integer from the input string

Example: std.parseInt("123") yields 123.

Example: std.parseInt("-123") yields -123.

#### std.escapeStringBash(str)

Wrap str in single quotes, and escape any single quotes within str by changing them to a sequence '"'"'. This allows injection of arbitrary strings as arguments of commands in bash scripts.

#### std.escapeStringDollars(str)

Convert $to$$in str. This allows injection of arbitrary strings into systems that use$ for string interpolation (like Terraform).

#### std.escapeStringJson(str)

Convert str to allow it to be embedded in a JSON representation, within a string. This adds quotes, escapes backslashes, and escapes unprintable characters.

Example:
{ local description = "Multiline\nc:\\path", json: "{name: %s}" % std.escapeStringJson(description) }
yields: {"json": "{name: \"Multiline\\nc:\\\\path\"}"}

#### std.escapeStringPython(str)

Convert str to allow it to be embedded in Python. This is an alias for std.escapeStringJson.

### Manifestation

#### std.manifestIni(v)

Convert the given structure to a string in INI format. This allows using Jsonnet's object model to build a configuration to be consumed by an application expecting an INI file. The data is in the form of a set of sections, each containing a key/value mapping. These examples should make it clear:

{ main: { a: "1", b: "2" }, sections: { s1: {x: "11", y: "22", z: "33"}, s2: {p: "yes", q: ""}, empty: {}, } }

Yields a string containing this INI file:

a = 1 b = 2 [empty] [s1] x = 11 y = 22 z = 33 [s2] p = yes q =

#### std.manifestPython(v)

Convert the given value to a JSON-like form that is compatible with Python. The chief differences are True / False / None instead of true / false / null.

{ b: ["foo", "bar"], c: true, d: null, e: { f1: false, f2: 42 }, }

Yields a string containing this Python code:

{"b": ["foo", "bar"], "c": True, "d": None, "e": {"f1": False, "f2": 42}}

#### std.manifestPythonVars(v)

Convert the given object to a JSON-like form that is compatible with Python. The key difference to std.manifestPython is that the top level is represented as a list of Python global variables.

{ b: ["foo", "bar"], c: true, d: null, e: { f1: false, f2: 42 }, }

Yields a string containing this Python code:

b = ["foo", "bar"] c = True d = None e = {"f1": False, "f2": 42}

### Arrays

#### std.makeArray(sz, func)

Create a new array of sz elements by calling func(i) to initialize each element. Func is expected to be a function that takes a single parameter, the index of the element it should initialize.

Example: std.makeArray(3,function(x) x * x) yields [0, 1, 4].

#### std.count(arr, x)

Return the number of times that x occurs in arr.

#### std.map(func, arr)

Apply the given function to every element of the array to form a new array.

#### std.filterMap(filter_func, map_func, arr)

This is primarily used to desugar array comprehension syntax. It first filters, then maps thte given array, using the two functions provided.

#### std.filter(func, arr)

Return a new array containing all the elements of arr for which the func function returns true.

#### std.foldl(func, arr, init)

Classic foldl function. Calls the function on the result of the previous function call and each array element, or init in the case of the initial element. Traverses the array from left to right.

#### std.foldr(func, arr, init)

Classic foldl function. Calls the function on each array element and the result of the previous function call, or init in the case of the initial element. Traverses the array from right to left.

#### std.range(from, to)

Return an array of ascending numbers between the two limits, inclusively.

#### std.join(sep, arr)

If sep is a string, then arr must be an array of strings, in which case they are concatenated with sep used as a delimiter. If sep is an array, then arr must be an array of arrays, in which case the arrays are concatenated in the same way, to produce a single array.

Example1: std.join(".", ["www", "google", "com"]) yields "www.google.com".

Example2: std.join([9, 9], [[1], [2, 3]]) yields [1, 9, 9, 2, 3].

#### std.lines(arr)

Concatenate an array of strings into a text file with newline characters after each string. This is suitable for constructing bash scripts and the like.

#### std.flattenArrays(arrs)

Concatenate an array of arrays into a single array.

#### std.sort(arr)

Sorts the array using the <= operator.

#### std.uniq(arr)

Removes successive duplicates. When given a sorted array, removes all duplicates.

### Sets

Sets are represented as ordered arrays without duplicates.

#### std.set(arr)

Syntax sugar for std.uniq(std.sort(arr)).

#### std.setInter(a, b)

Set intersection operation (values in both a and b).

#### std.setUnion(a, b)

Set union operation (values in either a or b). Note that + on sets will simply concatenate the arrays, possibly forming an array that is not a set (due to not bein ordered without duplicates).

#### std.setDiff(a, b)

Set difference operation (values in a but not b).

### Encoding

#### std.base64(v)

Encodes the given value into a base64 string. The encoding sequence is A-Za-z0-9+/ with = to pad the output to a multiple of 4 characters. The value can be a string or an array of numbers, but the codepoints / numbers must be in the 0 to 255 range. The resulting string has no line breaks.

#### std.base64DecodeBytes(s)

Decodes the given base64 string into an array of bytes (number values). Currently assumes the input string has no linebreaks and is padded to a multiple of 4 (with the = character). In other words, it consumes the output of std.base64().

#### std.base64Decode(s)

Behaves like std.base64DecodeBytes() except returns a string instead of an array of bytes.

#### std.md5(s)

Encodes the given value into an MD5 string.

### JSON Merge Patch

#### std.mergePatch(target, patch)

Applies patch to target according to RFC7396


text:
How do you simplify 12 - 2 {24 + 2 [3 (7 + 4) + (6)^(2)]}?

Aug 23, 2017

$= - 312$

Explanation:

There are only 2 terms. The first is already in its simplest form.
Simplify the second term:

$\textcolor{b l u e}{12} - 2 \left\{24 + 2 \left[3 \textcolor{red}{\left(7 + 4\right)} + \textcolor{red}{{\left(6\right)}^{2}}\right]\right\}$
$\textcolor{w h i t e}{w w w w w w w w w w w w} \downarrow \textcolor{w h i t e}{\times x} \downarrow$
$= \textcolor{b l u e}{12} - 2 \left\{24 + 2 \left[3 \textcolor{red}{\left(11\right)} + \textcolor{red}{\left(36\right)}\right]\right\}$
$\textcolor{w h i t e}{w w w w w w w w w w w} \downarrow$
$= \textcolor{b l u e}{12} - 2 \left\{24 + 2 \left[\textcolor{red}{33} + 36\right]\right\}$

=color(blue)(12)-2{24+2{color(lime)(33+36]}
$\textcolor{w h i t e}{w w w w w w w w w w w w w} \downarrow$
=color(blue)(12)-2{24" + "2color(lime)((69)}
$\textcolor{w h i t e}{w w w w w w w w w . w w w} \downarrow$
$= \textcolor{b l u e}{12} - 2 \left\{24 \text{ "+" } \textcolor{\lim e}{138}\right\}$

=color(blue)(12)-2{color(magenta)(24+138)]}
$\textcolor{w h i t e}{w w w w w w w} \downarrow$
$= \textcolor{b l u e}{12} - 2 \left\{\textcolor{m a \ge n t a}{162}\right\}$
$\textcolor{w h i t e}{w w w w w w w} \downarrow$
$= \textcolor{b l u e}{12} \text{ } - \textcolor{m a \ge n t a}{324}$

$= - 312$

Aug 23, 2017

$- 312$

Explanation:

We have: $12 - 2 \left\{24 + 2 \left[3 \left(7 + 4\right) + {\left(6\right)}^{2}\right]\right\}$

Let's apply the "PEMDAS" rules for operator precedence.

"PEMDAS" is an acronym for:

$\text{Parentheses} ,$ $\text{exponents} ,$ $\text{multiplication,}$ $\text{division} ,$ $\text{addition}$ $\mathmr{and}$ $\text{subtraction.}$

First, let's evaluate the operations within parentheses:

$= 12 - 2 \left\{24 + 2 \left[3 \times 11 + {6}^{2}\right]\right\}$

Next, we evaluate any exponents:

$= 12 - 2 \left\{24 + 2 \left[3 \times 11 + 36\right]\right\}$

Then, we perform any multiplication:

$= 12 - 2 \left\{24 + 2 \left[33 + 36\right]\right\}$

There is no division in this case, so let's perform any addition:

$= 12 - 2 \left\{24 + 2 \left[69\right]\right\}$

$= 12 - 2 \left\{24 + 138\right\}$

$= 12 - 2 \left\{162\right\}$

$= 12 - 324$

Finally, let's subtract the two numbers:

$= - 312$


text:
# Why do we take the closure of the support?

In topology and analysis we define the support of a continuous real function $f:X\rightarrow \mathbb R$ to be $\left\{ x\in X:f(x)\neq 0\right\}$. This is the complement of the fiber $f^{-1} \left\{0 \right\}$. So it looks like the support is always an open set. Why then do we take its closure?

In algebraic geometry, if we look at elements of a ring as regular functions, then it's tempting to define their support the same way, which yields $\operatorname{supp}f= \left\{\mathfrak p\in \operatorname{Spec}R:f\notin \mathfrak p \right\}$. But these are exactly the basic open sets of the Zariski topology. I'm just trying to understand whether this is not a healthy way to see things because I've been told "supports should be closed".

• In differential geometry, there is the notion of (smooth) functions (and more generally differential forms) with compact support. It's an important subset of all (smooth) functions, and since we're talking about manifolds here, the support cannot be compact if it's not closed. That's one reason for supports being closed, at least. Dec 18 '15 at 11:23
• We could also defined the support using germs : say a function is zero around $x$ if it is zero in a neighborhood of $x$. Then the support of a function $f$ is the set of $x$ where $f$ is not zero around $x$. Using this definition, it is automatically closed. Dec 18 '15 at 11:37
• I think it is a matter of technical convenience. For instance, when constructing partitions of unity, one often wants supports to be closed, not open. Dec 18 '15 at 11:40
• @Roland so is it kind of unhealthy to think of supports as a basis for the Zariski topology?
– user153312
Dec 18 '15 at 11:41
• Your first section is confusing: in topology and analysis the support of a continuous real function is defined as the closure of $\left\{ x\in X:f(x)\neq 0\right\}$. Dec 18 '15 at 13:21

Given a scheme $X$ there are two notions of support $f\in \mathcal O(X)$:
1) The first definition is the set of of points $$\operatorname {supp }(f)= \left\{ x\in X:f_x\neq 0_x\in \mathcal O_{X,x}\right\}$$ where the germ of $f$ at $x$ is not zero.
2) The second definition is the good old zero set of $f$ defined by $$V(f)=\{ x\in X:f[x]=\operatorname {class}(f_x)\neq 0\in \kappa (x)=\mathcal O_{X,x}/ \mathfrak m_x\}$$ It is also automatically closed.
3) The relation between these closed subsets is$$V(f)\subset \operatorname {supp }(f)$$with strict inclusion in general:
For a simple example, take $X=\mathbb A^1_\mathbb C=\operatorname {Spec}\mathbb C[T],\: f=T-17$ .
Then for $a\in \mathbb C$ and $x_a=(T-a)$ we have $f[a]=a-17\in \kappa(x_a)=\mathbb C$ and for the generic point $\eta=(0)$ we have $f[\eta]=T-a\in \kappa(\eta)=\operatorname {Frac}(\frac {\mathbb C[T]}{(0)})=\mathbb C(T)$.
Thus $f[x_{17}]=0$ and $f[P]\neq 0$ for all other $P\in \mathbb A^1_\mathbb C$ , so that $$V(f)=\{x_{17}\}\subsetneq \operatorname {supp }(f)=\mathbb A^1_\mathbb C$$


text:
# 2nd question from Chap.20: Combination, Permutation and Probability

Level pending

Remember that no number can be used more than once!

×


text:
HERE Routing API 7
In maintenance
Routing API v7 Developer's Guide

# Alternative Routes

If requested, the service can provide any available alternative routes. Alternative routes are available for all modes of transport. However, alternative routing for public transport takes some extra factors into account, such as timetables and number of transfers. Timetable routing can provide accurate time information but estimated routing cannot. This means that alternative routes are defined differently for each type of public transport routing:

In estimated routing, possible alternative routes are:

• Geometrically different from the original route
• Via different transport types
• Not significantly different from the original route in estimated time of arrival.

In timetable routing, possible alternative routes can be:

• Scheduled at different times
• Via different transport types
• Geometrically different, if not significantly different from the original route in estimated time of arrival.
Note: As timetable information is highly dynamic, we cannot give any particular guarantees on the number of and contents of alternative routes.

## Requesting Alternative Routes

To request multiple alternative routes in the response, set the alternatives request parameter to the number of additional routes that the service should return. For example, if you specify alternatives=3 in the request, the service can return up to 4 routes in the response: the original route and three alternatives. Note that the alternatives parameter specifies maximum number of additional routes returned. For any route, the service may or may not find admissible alternatives, so it is possible that even with several alternatives requested, the response will contain only one route.

## Alternative Route Ordering

Public transport alternatives are ordered by estimated preference. Simply, if no alternatives are requested, the best possible route is returned. Public transport routing algorithms take into account several criteria for estimating preference of resulting alternatives. Among these:
• Route departure and arrival time
• Duration of the trip
• Number of transfers (changes) involved
• Amount of walking distance involved
• User preferences (if specified in the request)

In addition to public transport alternatives, your request may result in a walking only route. This happens when walking is preferred to public transportation or walking makes a viable alternative.


text:
# No-hole $\lambda$-$L(k, k-1, \ldots, 2, 1)$-labeling for Square Grid

• Published 2016

#### Abstract

Given a fixed k ∈ Z and λ ∈ Z, the objective of a λ-L(k, k − 1, . . . , 2, 1)-labeling of a graph G is to assign nonnegative integers (known as labels) from the set {0, . . . , λ − 1} to the vertices of G such that the adjacent vertices receive values which differ by at least k, vertices connected by a path of length two receive values which differ by at… (More)