text:
### Prime cube partnership题号：131 难度： 40 中英对照

There are some prime values, p, for which there exists a positive integer, n, such that the expression n3 + n2p is a perfect cube.

For example, when p = 19, 83 + 82×19 = 123.

What is perhaps most surprising is that for each prime with this property the value of n is unique, and there are only four such primes below one-hundred.

How many primes below one million have this remarkable property?

### Code

public final class p131 {
public static void main(String[] args) {
long start=System.nanoTime();
String result = run();
long end=System.nanoTime();
System.out.println(result);
System.out.println( (end-start)/1000000 + "ms" );
}
static public final int N=1000000;
static public boolean isPrime[]=new boolean[N+1];
static public String run(){
isPrime[2]=true;
for(int i=3;i<=N;i++){
isPrime[i]=true;
for(int j=3;j<=i/j && isPrime[i];j+=2)
if(i%j==0)
isPrime[i]=false;
}
int ans=0;
for(int x=1;x*x*3+x*3+1<N;x++)
if(isPrime[x*x*3+x*3+1])
ans++;
return ""+ans;
}
}
173
376ms

format: plain


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# Prove the following equation

#### Albert

##### Well-known member
$\triangle ABC (with \,\,side \,\,length \,\, a,b,c), \,\, given :\angle A+\angle C =2 \angle B$

prove :

$\dfrac {1}{a+b}+\dfrac{1}{b+c}=\dfrac {3}{a+b+c}$

#### Opalg

##### MHB Oldtimer
Staff member
\begin{array}{rcl}\dfrac {1}{a+b}+\dfrac{1}{b+c}=\dfrac {3}{a+b+c} & \Longleftrightarrow & (a+2b+c)(a+b+c) = 3(a+b)(a+c) \\ & \Longleftrightarrow & a^2+b^2+c^2 + 3ab + 3bc + 2ac = 3b^2 + 3ab + 3bc + 3ac \\ & \Longleftrightarrow & b^2 = a^2 + c^2 - ac. \end{array}
Next, by the sine rule, $a,b,c$ are proportional to $\sin A,\sin B,\sin C$, so it will be sufficient to prove that $\sin^2B = \sin^2A + \sin^2C - \sin A\sin C.$

Also, we are told that $\angle A+\angle C =2 \angle B .$ But $\angle A+\angle C = 180^\circ - \angle B$, so that $3\angle B = 180^\circ$, $\angle B = 60^\circ$, and so $\sin^2 B = 3/4.$ And $\angle C = 120^\circ - \angle A$, so $\sin C = \sin120^\circ \cos A - \cos120^\circ \sin A = \frac{\sqrt3}2\cos A + \frac12\sin A.$ Therefore \begin{aligned}\sin^2A + \sin^2C - \sin A\sin C &= \sin^2A + \tfrac14(\sqrt3\cos A + \sin A)^2 - \tfrac12\sin A(\sqrt3\cos A + \sin A) \\ &= \sin^2A + \tfrac14(3\cos^2 A + 2\sqrt3\sin A\cos A + \sin^2 A) - \tfrac12\sin A(\sqrt3\cos A + \sin A) \\ &= \tfrac34\sin^2 A + \tfrac34\cos^2 A = \tfrac34 = \sin^2 B,\end{aligned} as required.

#### Jester

##### Well-known member
MHB Math Helper
Couldn't you use use the law of Cosine with $\angle B = 60^\circ$ to get to the same answer quicker?

#### Opalg

##### MHB Oldtimer
Staff member
Couldn't you use use the law of Cosine with $\angle B = 60^\circ$ to get to the same answer quicker?

#### Albert

##### Well-known member
from Opalg's solution :
$b^2 = a^2 + c^2 - ac.$--------(1)
as jester said "Couldn't you use the law of Cosine with [FONT=MathJax_Main]∠[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]60[/FONT][FONT=MathJax_Main]∘[/FONT] to get to the same answer quicker?"
using law of Cosine :
$b^2=a^2+c^2-2ac\,\, cos B-------(2)$
compare (1) abd (2)
if (1)=(2) then $cos B=\dfrac {1}{2},\,\, \,\,now\,\, we\,\,\,only\,\, have\,\, to \,\, prove : \angle B=60^o$
$using :\angle A+\angle C=2\angle B,\therefore\,\, 3\angle B=180^o \Longleftrightarrow \angle B=60^0$
and the proof is finished


text:
QUESTION

# BUS 308 Week 4 Problem Set Week Four

This document of BUS 308 Week 4 Problem Set Week Four includes answers to the next questions:

1. Is the probability of having a graduate degree independent of the grade the employee is in? 2. Construct a 95% confidence interval on the mean service for each gender? Do they intersect?3. Are males and females distributed across grades in a similar pattern? 4. Do 95% confidence intervals on the mean length of service for each gender intersect?5. How do you interpret these results in light of our equity question?

• @
Tutor has posted answer for $6.49. See answer's preview$6.49


text:
# Troubleshooting¶

Use this section to read about known issues and for common troubleshooting steps.

Telemetry Streaming general troubleshooting tips:

• Examine the restnoded failure log at /var/log/restnoded/restnoded.log (this is where Telemetry Streaming records error messages)
• Examine the REST response:
• A 400-level response will carry an error message with it
• If this message is missing, incorrect, or misleading, please let us know by filing an issue on Github.
• Use Telemetry’s trace option to create a detailed trace of the configuration process for subsequent analysis. Telemetry’s trace option can be a powerful tool to learn about its working details and to review Telemetry’s operations in detail.

Troubleshooting

I’m receiving a path not registered error when I try to post a declaration

If you are receiving this error, it means either you did not install Telemetry Streaming, or it did not install properly. The error contains the following message:

{
"code":404,
"message": "Public URI path no registered. Please see /var/log/restjavad.0.log and /var/log/restnoded/restnoded.log for details.".
...
}


text:
You’ve reached the end of your free Videos limit.
#13 | Image Formed by Lens
(Physics) > Ray Optics and Optical Instruments
Unable to watch the video, please try another server
Related Practice Questions :

A concave lens of focal length 25 cm produces image $\frac{1}{10}th$ the size of the object. The distance of the object from the lens is

(1) 225 cm

(2) 250 cm

(3) 150 cm

(4) 175 cm

High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

A converging beam of light is incident on a concave lens of focal length 30cm, in such a way that their point of intersection would lie at a height 1cm above the principal axis and at a distance of 24cm from the position of the lens if there was no lens. Find the nature of the image formed when the beam passes through the lens.

1.  Real at a distance 60cm from the lens

2.  Real at a distance 120cm from the lens

3.  Virtual at a distance 60cm from the lens

4.  Virtual at a distance 120 cm from the lens

High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The focal length of a convex lens is f. An object is placed at a distance x from the focal point and a real image is formed. The linear magnification has a magnitude

1.

2.

3.

4.  $\frac{x}{f}$

High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

A thin equiconvex lens of power P is cut into three parts A, B, and C as shown in the figure. If  are powers of the three parts respectively, then

1.

2.

3.

4.

High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The slab of a material of refractive index 2 shown in figure has curved surface APB of radius of curvature 10 cm and a plane surface CD. On the left of APB is air and on the right of CD is water with refractive indices as given in figure. An object O is placed at a distance of 15 cm from pole P as shown. The distance of the final image of O from P, as viewed from the left is

(a)   20 cm

(b)   30 cm

(c)   40 cm

(d)   50 cm


text:
# Could something theoretically observe the first light in the Universe?

Light travels at 299,792,458 meters a second. If you look at something that is 299,792,458 meters away, you're observing light from 1 second in the past.

If an observer's distance from a light source exceeds the distance required for light to travel to the observer, and we have the advanced technology required to see something that far away, and also know the exact location in space to look (although data transfer speeds are bottle-necked by light-speed, but let's imagine that bottle-neck doesn't exist...), could we theoretically see the start of that light source?

You would also need to know the location of the Universe's first light source, but if we did, could we theoretically go one step further and go outside the radius of first light omitted from the Universe's centre and watch the birth of the Universe?

• cosmic microwave background is what you are looking for?en.wikipedia.org/wiki/Cosmic_microwave_background – Ch.Siva Ram Kishore Mar 25 at 12:23
• @Ch.SivaRamKishore Like how when light from an omitter moving away from us enters our hubble sphere the wavelength appears differently and more stretched out? So it becomes more faint and disappears into the background noise... Thanks for the link! – Lewis Harris Mar 25 at 12:44
• The first light source was the whole universe. See physics.stackexchange.com/q/136860/123208 – PM 2Ring Mar 25 at 16:18
• @PM2Ring I suppose there is easier ways to simulate the beginning of the universe – Lewis Harris Mar 25 at 18:36

If an observer's distance from a light source exceeds the distance required for light to travel to the observer, and we have the advanced technology required to see something that far away, and also know the exact location in space to look (although data transfer speeds are bottle-necked by light-speed, but let's imagine that bottle-neck doesn't exist...), could we theoretically see the start of that light source

Assume that the sun exploded right now. We would see the effect of it after 8 minute later, since that is the time it takes for the light travel between the sun and the earth.

This is true for any event. Even when you look at the mirror you are seeing the past. We always see the past. How past, well it depends on the distance between the source and the event (At this point things can be affected due to expansion of the universe, which is an important factor in large distances)

You would also need to know the location of the Universe's first light source

There is no such location in the universe. In other words, there's no direction which you can point at and say "that's where the universe started". Think about the surface of a ball. You are a 2D creature living on the surface. There's no physical center for this surface. The center lies inside the ball which a 2D creature can never point/reach.

but if we did, could we theoretically go one step further and go outside the radius of first light emitted from the Universe's center and watch the birth of the Universe?

We cannot because of there's no such place. There's no center for the universe.

And btw CMB is not the "first light". There was always light (photons). In the early universe, these photons were coupled to the baryonic matter. Which implies that the photons were always interacting with the particles. When they decoupled from the matter, these photons became free and that is the CMBR. It's just an event that occurred in the early universe. If we had some sort of a superpower to detect very low energy neutrinos (about $$1.9K$$) we would see the Cosmic Neutrino Background which occurred around $$1s$$ after the big bang. Meanwhile, the photon decoupling (Event that created CMBR) occurred around $$300,000$$ years later.

http://www.astro.ucla.edu/~wright/nocenter.html

Theoretically, you can go outside of our observable universe. But thats a completely different story from what you are describing.

• Great information just the kind of answer I was looking for. I suppose I meant anything between observable wavelengths that can be observed by our instrumentation. Theoretically could someone observe old waves is what I'm asking, but know that those waves are so old that the reason that they haven't got to them yet is because the source of the wave is so far away. – Lewis Harris Mar 31 at 16:46

The cosmic microwave background is the farthest back we can see due to before that the entire universe was composed of opaque plasma, if someone somehow went far enough to observe the birth of the universe they would not see anything, until the universe was cold enough to allow light to pass through, that is the cosmic microwave background (it is in microwave radiation due to the extreme Doppler shift).

These articles provide the basic ideas:

https://www.scientificamerican.com/article/what-is-the-cosmic-microw/

https://www.popularmechanics.com/space/deep-space/a9632/how-close-to-the-big-bang-can-we-see-16105099/


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# Fisher-Yates shuffle in JavaScript

This is my first time making a function just from reading about it on Wikipedia, so I'm sure there's room for a lot of improvement.

function fisherYates(str) {
let result = '';
for (let i = 0, len = str.length; i < len; i++) {
let rand = (Math.random()*str.length)|0;
result += str[rand];
str = str.slice(0, rand) + str.slice(rand+1);
}
return result;
}

• I know it is 10 years old and pertains to arrays instead of strings, but you mind find this post interesting – Sᴀᴍ Onᴇᴌᴀ Oct 17 '17 at 17:49

## Mostly just style

Why add the variable len inside the for loop. It is a constant so put it outside the loop as one.

The variable result is in function scope. Use var rather than let.

The variable rand does not change inside the loop so it should be a const.

The expression to get the random number does not need the (...) as operator precedence ensures that the | is applied correctly.

Add some spaces between operators for readability. eg (Math.random()*str.length)|0; is better as (Math.random() * str.length) | 0

### One issue

The only problem is that you have one to many iterations. The last random value will always be to select from a 1 character string, so it is not needed. You need only iterate the string length minus one.

### The rewrite

How I would rewrite your code. I personally don't like putting the let inside the for loop, so I put the declaration in the scope above. This is because for loops can have a lot of noise (clutter) and that reduces it a bit. Also the function is so short and simple variable names can be shortened without problem, I would make rand just the r.

function fisherYates(str) {
var i, result = '';
const len = str.length - 1;
for (i = 0; i < len; i ++) {
const r = Math.random() * str.length | 0;
result += str[r];
str = str.slice(0, r) + str.slice(r + 1);
}
return result + str;
}

• I'd go the other way and change rand to randomIndex, but each to his own, I guess. – Tamoghna Chowdhury Oct 18 '17 at 4:20
• I'd also declare each variable in the smallest lexical scope, using let. – Tamoghna Chowdhury Oct 18 '17 at 4:21
• @TamoghnaChowdhury Block scope has its place but without a clear need its should not be used. You use scope to protect a variable's name, but good function design (no rambling 100+ line functions) eliminates that as an issue (no risk in a 8 line function of using the same variable name by mistake). So i go on the side of efficiency, each scope declaration requires a heap assignment and additional context stack work. In nested loops this can contribute considerable overhead for no good reason but that of "protecting against incompetent programmers" – Blindman67 Oct 18 '17 at 5:04
• I didn't know that let was that inefficient in JavaScript... Anyway, your comment should IMO be a part of your answer. Especially given that almost every article about ES6+ online is preaching about how var is outdated and let should be used everywhere. – Tamoghna Chowdhury Oct 18 '17 at 8:22
• @TamoghnaChowdhury That is standard in the language since day one (1995). Implementation method is not part of the language spec, V8 uses a "context" object which carries the scope chain. The scope is mutable so I guess it is similar to Python (though I am not familiar with python's inner workings) – Blindman67 Oct 18 '17 at 15:27


text:
# Linked Questions

12 questions linked to/from Using sets of arguments from a list
4answers
178 views

### @ and Sequence ( and /@ ) [duplicate]

list = { {a,b}, {c,d}, {e,g} } How to efficiently get { f[a,b], f[c,d], f[e,g] } I tried various combinations using ...
4answers
228 views

### How do I pass a “list of lists” as the argument to a function of the form F[x,y]? [duplicate]

I have a function F[x,y] and a list of the form: {{a1,b1},{a2,b2},{a3,b3},{an,bn}} Basically, I want to pass the list ...
1answer
123 views

### How to use a table with two variables as the values for a function with two variables (x and y)? [duplicate]

I'm sorry if this is a dumb question. I am a student and have 0 prior experience with Mathematica or any sort of coding and I have looked at various resources trying to figure it out myself but still ...
2answers
46 views

### How to use an imported table data in a calculation of f[x,y,z], where each row of the column correspond to x,y,z? [duplicate]

I have a table with desired inputs of a given function efvc[n,m,u]. I want to calculate for different values of n, ...
1answer
40 views

### How to use Apply and Map properly in this situation? [duplicate]

So this does what I want: #1 <> #2 &["♠", "A"] ♠A The output is fine. How do I quickly apply this to the list ...
0answers
97 views

### Applying a function of two variables to a the elements in a list of pairs [duplicate]

I have a function: f = # - ((1 - #2)*a + #2*b) &; and I would like to apply it to a list like this: {{1,2},{2,3}....} I ...
0answers
43 views

### Apply all possible permutations into a function [duplicate]

I need to create a function that returns all possible trebles of integers that sum up to a given number. For example, is n=2 then I need something like this: ...
0answers
29 views

### Apply a function to each array in a list of arrays [duplicate]

Given a function f[x_,y_], and an array {{x1,y1},{x2,y2},...}, how would one go about applying the function to each {xn,yn}? I attempted using Map[f[#,#]&,array], but that applied the function ...
0answers
11 views

### Evaluating a function over the rows of a matrix [duplicate]

I performed quite a search, and I am unable to find a quick solution to the following problem. I have a $p \times q$ matrix $c$, say c = {{1 , 3}, {3, 4}} and ...
3answers
7k views

### How do I convert an argument list to a sequence of arguments?

I think this is a basic question, but I am having difficulty finding the answer in the documentation. Thread is not what I am looking for, I think. Suppose that I ...
9answers
688 views

### How to substitute values from a list into a function?

I was hoping to take pairs of numbers from a list and substitute them into a function. So if my list was list = {{1,2}, {3,4}, {5,6}} and my function was <...
3answers
440 views

### Map and Pure Function

Trying to take the first item of a member and raising it to the power of the second item of the same member and then subtracting the output from the third item of the same member, the list: ...


text:
# What's the idea behind Carleman estimates?

A standard Carleman-type estimate is of the form $$\sum_{|\alpha| where $$\phi$$ is some weight function. This formula turns to be very useful in the study of uniqueness of Cauchy problems, and many mathematicians have considered this (such as Calderon, Hormander, Kenig, Sogge, and Tataru...)

For a first look at this inequality, I'm wondering whether the weight fuction has an essential role, and furthermore, what's the original idea of it? Are there some very simple but illuminating examples that show the reasonability of the Carleman estimates?

One example in my mind is the first order operator $$P=D+ix$$, where $$D = \frac{1}{i} \frac{d}{dx}$$. It's easy to see that $$P^*=D-ix$$, and $$P^*P-I=PP^*+I=-\frac{d^2}{dx^2}+x^2,$$ which is the so-called harmonic oscillator. Here we have $$2\|u\|_{L^2}\leq \|Pu\|_{L^2},\quad u\in C_{0}^{\infty}.$$ But in this simple example, there is no need to use a weight function. From the proof I guess the decomposition $$P=\frac{P+P^*}{2}+\frac{P-P^*}{2}$$ may be one of the general ideas.

• Besides the good answer bellow you can get a good overview over some relevant results and the philosophy behind Carleman estimates from these notes of Lerner: math.jussieu.fr/~lerner/m2carl.pdf. – CPJ May 21 '14 at 15:20

The weight function $\phi$ indeed plays an (I would even say the) essential role in Carleman estimates, and constructing one such that the estimate holds is one of the major challenges in proving and applying Carleman estimates. I will give a very informal overview, and refer to the references below for the technicalities.

The original motivation for introducing Carleman estimates was to prove unique continuation theorems, which can informally be stated as follows: Given a partial differential operator $P$, an oriented hypersurface $\Sigma$, and a function $u$ satisfying $Pu=0$, then if $u=0$ (locally) on one side $\Sigma^+$ of $\Sigma$, it vanishes on the other side $\Sigma^-$ as well. This can be interpreted in the sense that the complete information about $u$ in $\Sigma^-$ can be retrieved from information in $\Sigma^+$, i.e., the complete information flows across $\Sigma$. For linear hyperbolic equations, such information flow is described by the characteristics, and the idea behind Carleman estimates is to extend this concept to general (pseudo-)differential operators via microlocal analysis. This leads to considering so-called bicharacteristic rays, which need to satisfy the geometric property that all such rays passing near $\Sigma$ in $\Sigma^+$ must cross into $\Sigma^-$. Furthermore, solutions to $Pu=0$ should not decay exponentially along such rays towards $\Sigma$ (otherwise the information gets lost before it crosses $\Sigma$). A surface satisfying these conditions is called strongly pseudo-convex (with respect to $P$).

To make these conditions precise, one considers $\Sigma$ as the level set $\{\phi(x)=0\}$ (and $\Sigma^+$ as $\{\phi(x)>0\}$). The pseudo-convexity conditions above then become positivity conditions for the derivative of $\phi$ along the bicharacteristic flow (expressed using Poisson brackets). The connection is now that if $\Sigma=\{\phi(x)=0\}$ is a strongly pseudo-convex surface, then $e^{\lambda\phi}$ is strongly pseudo-convex function (in the sense of these positivity conditions) for sufficiently large $\lambda$. The parameter $\lambda$ is tied to the decay rate along the bicharacteristic rays, and hence the Carleman weight can be seen as accounting for this decay: The weight vanishes as the information flow across $\Sigma$ decays. The weight thus localizes the estimate to that part of $\Sigma^-$ where you still have usable information from $\Sigma^+$.

For example, for elliptic operators of second order, every smooth surface is strongly pseudo-convex (the decay condition is always satisfied); similarly, for hyperbolic operators with constant (real) coefficients, every non-characteristic surface is strongly pseudo-convex. In particular, for the wave operator $Pu=u_{{{tt}}} - c^2\Delta u$, any convex surface is strongly pseudo-convex, as well as the zero level sets for $\phi = x^2 - \beta t^2$ for $\beta < c^2$ (this bound is sharp).

[1] A good overview of unique continuation using Carleman estimates (along the lines I've given) is Daniel Tataru, Unique Continuation Problems for Partial Differential Equations, The IMA Volumes in Mathematics and its Applications Volume 137, 2004, pp 239-255.

[2] Carleman estimates are also useful for inverse problems for partial differential equations; this side is discussed in Victor Isakov, Inverse Problems for Partial Differential Equations, 2nd ed., 2006, Springer. (The example for the wave equation is Theorem 3.4.1.)

[3] David Dos Santos Ferreira has written a habilitation thesis on Carleman estimates.

Just a try: I can imagine, that the function $w\equiv 1$ as a weight function would not be sufficient to conclude unique continuation. You should rather think if Carleman holds true for a weight function which is approximately the dirac delta function. Thats why typical carleman estimates have a weight function which have a pole at zero. For example $$\int w^{-1-2\alpha} (x) f^2 (x) \leq C \int w^{-2-2\alpha} (\Delta f)^2,$$ (I dont know if the exponents are correct, see Bourgain Kenig 2005) where $w (x) = \phi (\lvert x \rvert)$ with $$\phi (s) = s \exp \left(-\int_0^s \frac{1-e^{-1}}{t} dt \right) .$$ In this case $w^{-1}$ has a pole at zero and I guess this is more useful for application in unique continuation.


text:
# Overview¶

This section describes the basic working principles of the cuTensorNet library. For a general introduction to quantum circuits, please refer to Introduction to quantum computing.

## Introduction to tensor networks¶

Tensor networks emerge in many mathematical and scientific domains, ranging from quantum circuit simulation, quantum many-body physics, quantum chemistry, to machine learning. As network sizes scale up exponentially, there is an increasing need for a high-performance tensor network library in order to perform tensor network contractions efficiently, which cuTensorNet aims to serve.

A tensor network is a collection of tensors contracted together to form a tensor of arbitrary rank. The contractions between the constituent tensors fully determines the network topology. For example, the tensor $$T$$ below is given by contracting the tensors $$A$$, $$B$$, $$C$$, and $$D$$:

$T_{abcd} = A_{aij} B_{bjk} C_{klc} D_{lid},$

where the modes with the same label are implicitly summed over following the Einstein summation convention. In this example, the mode label (index) $$i$$ connects the tensors $$D$$ and $$A$$, the mode label $$j$$ connects the tensors $$A$$ and $$B$$, the mode label $$k$$ connects the tensors $$B$$ and $$C$$, the mode label $$l$$ connects the tensors $$C$$ and $$D$$. The four uncontracted modes with labels $$a$$, $$b$$, $$c$$, and $$d$$ refer to free modes (sometimes also referred to as external modes), indicating the resulting tensor $$T$$ is of rank 4.

### Description of tensor networks¶

In the cuTensorNet library, we follow cuTENSOR’s nomenclature:

• A rank (or order) $$N$$ tensor has $$N$$ modes

• Each mode has an extent (the size of the mode), so a $$3\times 3$$ matrix has two modes, each of extent 3

• Each mode has a stride accounting for the distance in physical memory between two logically consecutive elements along that mode, in unit of elements

Note

For NumPy/CuPy users, rank/order translates to the ndim attribute, the sequence of extents translates to shape, and the sequence of strides has the identical meaning as strides.

A tensor network in the cuTensorNet library is represented by the cutensornetNetworkDescriptor_t descriptor that effectively encodes the topology and data type of the network. To be precise, this descriptor specifies the number of input tensors in numInputs and the number of modes for each tensor in the array numModesIn, along with each tensor’s modes, extents, and strides in the arrays of pointers modesIn, extentsIn, and stridesIn, respectively.

Likewise, it holds similar information about the output tensor (e.g., numModesOut, modesOut, extentsOut, stridesOut). Note that there is only one output tensor per network, so there is no need to set numOutputs and the corresponding arguments are just plain arrays.

It is possible for all these network metadata to live on the host, since when constructing a tensor network only its topology and the data-access pattern matter; we do not need to know the actual content of the input tensors at network descriptor creation.

Internally, cuTensorNet utilizes cuTENSOR to create tensor objects and perform pairwise tensor contractions. cuTensorNet’s APIs are designed such that users can just focus on creating the network description without having to manage such “low-level” details by themselves. The tensor contraction can be computed in a different precision from the data type, given by a cutensornetComputeType_t constant.

Once a valid tensor network is created, one can

1. Find a low-cost contraction path, possibly with slicing and additional constraints

2. Access information concerning the contraction path

3. Get the needed workspace size to accommodate intermediate tensors

4. Create a contraction plan according to the info collected above

5. Auto tune the contraction plan to optimize the runtime of the network contraction

6. Perform the actual contraction to retrieve the output tensor

It is the users’ responsibility to manage device memory for the workspace (from Step 3) and input/output tensors (for Step 5). See API Reference for the cuTensorNet APIs (section Workspace Management API). Alternatively, the user can provide a stream-ordered memory pool to the library to facilitate workspace memory allocations, see Memory Management API for details.

## Contraction pathfinder¶

A contraction path is a sequence of pairwise contractions represented in the numpy.einsum_path() format. The role of a path optimizer is to find a contraction path that minimizes the cost of contracting the tensor network. The cuTensorNet pathfinder is based on a graph-partitioning approach (called phase 1), followed by slicing and reconfiguration (called phase 2). Practically, experience indicates that finding an optimal contraction path can be sensitive to the choice of configuration parameters. Therefore, many of these are available to be configured via cutensornetContractionOptimizerConfigSetAttribute().

### Slicing¶

In order to fit a tensor network contraction into available device memory, as specified by workspaceSizeConstraint, it may be necessary to use slicing (also known as variable projection or bond cutting). Each slice can be computed independently from the others. Thus, if we intend to run a parallel computation, slicing is also one of the best techniques as it creates independent work for each device. We may similarly use a sliced contraction in order to create work for all available nodes. Slicing means that we compute the contraction for only one particular position in a certain mode (or combination of modes), creating a number of slices equal to the product of the extents of the sliced modes. We can then sum over the individually computed values to reproduce the full tensor network contraction. Such a technique is useful for large tensor networks, in particular quantum circuits, where the memory footprint to perform the contraction could exceed any existing memory storage. Taking the above $$T$$ tensor as an example, if we slice over the mode i we obtain the following:

$T_{abcd} = A_{aij} B_{bjk} C_{klc} D_{lid} \longrightarrow \sum_{i_s} \left( A_{a {i_s} j} B_{bjk} C_{klc} D_{l {i_s} d} \right),$

where the sliced mode $$i_s$$ is no longer implicitly summed over as part of tensor contraction, but instead explicitly summed (potentially in parallel). Although slicing reduces the memory footprint, it usually worsens the flops count of the contraction, and there is no simple way to determine what set of sliced modes will yield the best performance.

The cuTensorNet library offers some controls to influence the slice-finding algorithm:

### Reconfiguration¶

At the end of each slice-finding iteration, the quality of the contraction tree has been diminished by the slicing. We can improve the contraction tree at this stage by performing reconfiguration. Reconfiguration considers a number of small subtrees within the overall contraction tree and attempts to improve their quality. Although the process is computationally expensive, a non-reconfigured sliced contraction tree may be orders of magnitude more expensive to execute than expected. The cuTensorNet library offers some controls to influence the reconfiguration algorithm:

• CUTENSORNET_CONTRACTION_OPTIMIZER_CONFIG_RECONFIG_NUM_ITERATIONS: Specifies the number of subtrees to consider during each reconfiguration. The amount of time spent in reconfiguration, which usually dominates the pathfinder run time, is linearly proportional to this value. Based on our experiments, values between 500 and 1000 provide very good results. Default is 500. Setting this to 0 will disable reconfiguration.

• CUTENSORNET_CONTRACTION_OPTIMIZER_CONFIG_RECONFIG_NUM_LEAVES: Specifies the maximum number of leaf nodes in each subtree considered by reconfiguration. Since the time spent is exponential in this quantity for optimal subtree reconfiguration, selecting large values will invoke faster non-optimal algorithms. Nonetheless, the time spent by reconfiguration increases very rapidly as this quantity is increased. Default is 8. Must be at least 2. While using the default value usually produces the best flops, setting it to 6 will speed up the pathfinder execution without significant increase in the flops count for many problems.

### Deferred rank simplification¶

Since the time taken by the path-finding algorithm increases quickly as the number of tensors increases, it is advantageous to minimize the number of tensors, if possible. Rank simplification removes trivial tensor contractions from the network in order to improve performance. These contractions are those where a tensor is only contracted with at most two neighbors, effectively making a matrix multiplication. The necessary contractions to perform the simplification are not immediately performed but rather are prepended to the contraction path returned. If, for some reason, such simplification is not desired, it can be disabled:

While simplification helps lower the FLOP count in most cases, it may sometimes (depending on the network topology and other factors) lead to a path with higher FLOP count. We recommend that users experiment with the impact of turning simplification off (using the option listed above) on the computed path.

## Hyper-optimizer¶

cuTensorNet provides a hyper-optimizer for the pathfinder that can automatically generate many instances of contraction path and return the best of them in terms of total flops. The number of instances is user controlled by CUTENSORNET_CONTRACTION_OPTIMIZER_CONFIG_HYPER_NUM_SAMPLES and is set to 0 by default. The idea here is that the hyper-optimizer will create CUTENSORNET_CONTRACTION_OPTIMIZER_CONFIG_HYPER_NUM_SAMPLES instances, each with different parameters of the pathfinder algorithm. Each instance will run the full pathfinder algorithm including reconfiguration and slicing (if requested). At the end of the hyper-optimizer loop, the best path (in term of flops) is returned.

The hyper-optimizer runs its instances in parallel. The desired number of threads can be set using CUTENSORNET_CONTRACTION_OPTIMIZER_CONFIG_HYPER_NUM_THREADS and is chosen to be half of the available logical cores by default. The number of threads is limited to the number of the available logical cores to avoid the resource contention that is likely with an unnecessarily large number of threads.

Currently, hyper-optimizer multithreading is implemented via OpenMP and:

• OpenMP environment variables (e.g., OMP_NUM_THREADS) are not used.

• Internal OpenMP configuration/settings are not affected, i.e. no omp_set_*() functions are called.

The configuration parameters that are varied by the hyper-optimizer are:

Some of these parameters may be fixed to a given value (via cutensornetContractionOptimizerConfigSetAttribute()). When a parameter is fixed, the hyper-optimizer will not randomize it. The randomness can be fixed by setting the seed via the attribute CUTENSORNET_CONTRACTION_OPTIMIZER_CONFIG_SEED.

## Supported data types¶

A valid combination of the data and compute types for tensor network contractions inherits straightforwardly from that of cuTENSOR. Please refer to cutensornetCreateNetworkDescriptor() and cuTENSOR’s User Guide for detail.

## References¶

For a technical introduction to cuTensorNet, please refer to the NVIDIA blog:

For further information about general tensor networks, please refer to the following:

For the application of tensor networks to quantum circuit simulations, please see:


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text:
# Multiplication properties in pseudo-differential calculus with small regularity assumptions on the symbols

Joachim Toft

given at  esi12 (16.10.12 15:45)
id:  2355
length:  45min
status:
type:
LINK-Presentation:  http://univie.ac.at/nuhag-php/dateien/talks/Toft_2012-09_Multiplication12.pdf
ABSTRACT:
There are four important products which appear on the symbol levels in
pseudo-differential calculus. The composition of two Weyl quantizations
corresponds on the symbol level on the so called Weyl product.
The convolution appears when putting the theory of Toeplitz
operators (the same as localization operators) on functions on R^d
into the context of pseudo-differential calculus, using the fact that the
Weyl symbol of a Toeplitz operator is a convolution by the Toeplitz
symbol and an other convenient function. Finally, the twisted
convolution and ordinary multiplication appear when
applying the (symplectic) Fourier transform on Weyl products and
ordinary convolutions, respectively.

In the talk we establish Young and Hölder relations for such products
on Schatten-von Neumann symbols, Lebesgue spaces and modulation
spaces. We use the results to extend the class of possible window functions in
the definition of modulation spaces,
and to prove that any Schatten-p symbol in the Weyl calculus gives rise to
a Schatten-p Toeplitz operator. The Schatten-von Neumann classes here are
of the form I_p(H_1,H_2), where
H_1,H_2 are Hilbert spaces of (ultra-)modulation space type
which stay between the Gelfand-Shilov space Sigma_1(R^d)
and its dual Sigma_1'(R^d). Furthermore, the symbol classes of pseudo-differential
operators stay between Sigma_1(R^2d)\$ and its dual Sigma_1'(R^2d).

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### Ball Bearings

If a is the radius of the axle, b the radius of each ball-bearing, and c the radius of the hub, why does the number of ball bearings n determine the ratio c/a? Find a formula for c/a in terms of n.

### Overarch 2

Bricks are 20cm long and 10cm high. How high could an arch be built without mortar on a flat horizontal surface, to overhang by 1 metre? How big an overhang is it possible to make like this?

### Cushion Ball

The shortest path between any two points on a snooker table is the straight line between them but what if the ball must bounce off one wall, or 2 walls, or 3 walls?

# Population Dynamics

##### Stage: 5 Challenge Level:

A hive of bees, a colony of ants and a parliament of owls.

These are just a few examples of animal groups, or populations. A population is dynamic; this means it is constantly changing in size and demographics. New animals are born, old animals die and other factors such as drought, fire and lack of predators, all cause a change in the population.

The population growth is the change in the number of individuals in a population, per unit time. For example, if a population has ten births and five deaths per year, then the population growth is five individuals per year.

In the following pages, we aim to represent populations and changes in populations using mathematics. This involves using differential equations and even probability.

Links to pages on differential equations:

 Part 1 A First Model Part 2 Exponential and Geometric Models Part 3 The Logistic Equation Part 4 The Logistic Map Part 5 The Lotka-Volterra Equations Part 6 Modified L-V Equations

Links to pages on probability:

### Beginning the Model

We are able to describe population growth by making some generalizations and using simple differential equations:

The size, $N_t$, of a population depends upon:

• The initial number of individuals, $N_0$
• The number of births, B
• The number of deaths, D
• The number of immigrants, I
• The number of migrants, E

This gives us the equation: $$N_t=N_0+B-D+I-E$$

When a population is closed, there is no immigration or emigration. This often occurs on remote islands, such as the Galapagos Islands. Our equation then becomes $N_t=N_0+B-D$ , or equivalently $$N_{t+1}=N_t+B-D$$

Clearly the population will increase if $B> D$, and will decrease if $B< D$.

A population is in equilibrium if on average the population size remains constant over a long period of time. Mathematically, this means:   $N_t=N_{t+\Delta t}$

Question:

We can rewrite the equation $N_{t+1}=N_t+B-D$ , as: $$N_{t+1}-N_t=\Delta N_t=B-D$$ Intuitively, why does this make sense? Think of an example of a population to explain why.


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# Derive Definition of Exponential Function (Power Series) From Compound Interest

This example demonstrates how the formula for compound interest can be used to derive the power series definition of the exponential function. The power series of the exponential function is shown below.

From a high-level perspective, compound interest represents an iterative approach to modeling exponential growth and the exponential function is a natural limit to how fast something can “continuously” grow.

## Step 1

The variables and their meanings are summarized below.

Expression Description
Accumulated amount.
Principle amount.
Rate of interest expressed as a decimal value.
Frequency of investment per time period.
Time elapsed.

Then, let’s look at some examples of the formula. Specifically, we are interested in what happens as the frequency of investment, represented by the variable , increases.

Given an interest rate of , a principle amount of , total time of and a yearly investment strategy per year the formula calculates the accumulated amount shown below.

Next, given the same interest rate, principle amount and time elapsed, let’s calculate the accumulated amount if we switch quarterly investment strategy per year.

Finally, given the same interest rate, principle amount and time elapsed, let’s calculate the vaue for a monthly investment strategy per year.

These examples demonstrate that as the frequency of investment goes up so does the accumulated amount. The question we are trying to answer is “what happens as the frequency of the growth rate continues to increase?” Is there a natural limit, or does it grow in an unbounded manner.

## Step 2

We can pose this question mathematically by taking the limit of the function as goes to infinity. However, finding the limit at this stage poses some challenges and so we are first going to transform the function to be more simple and abstract.

Substitute the variable into the expression, where represents the total number of times the interest is applied and is given by . This changes the function to the equivalent form:

Then, we are going to make two observations. First, observe that scales the output (growth curve) of the function vertically. Second, observe that scales the input of the function horizontally. Because we are interested in modeling the growth curve (shape) of the function and it makes the math cleaner later on, we can set and .

To finish the transformation, from now on, let’s consider to represent the total number of times interest is applied and replace with the more generic variable .

## Step 3

Take the limit of the function as approaches infinity.

There are two reasonable approaches to taking the limit of this function. The first, shown on this page, is to expand the product as a series and look for patterns. The second, shown on another page, is to approximate a value for the limit.

Expand the first three cases of the limit.

Case Expression

This is tedious to do by hand and instead, we can use the binomial expansion, substituting and into each case of the expansion. This is shown below and aligned so that the powers of line up.

Case Binomial Expansion

The math becomes pretty hairy at this point, and while we won’t prove anything rigorously, hopefully you will be convinced that as the coefficients approach known values. Take, for example, the expansion of the expresions of where is equal to , and shown below. Clear patterns start to appear in the coefficients associated with each power. For example, the coefficient associated with approaches the fraction .

The generalized pattern that emerges as approaches infinity is the coefficient in front of the -th power of approaches over the factorial of . This pattern leaves us with the infinite power series below which represents the definition of the exponential function. This is the same definition that can derived using a taylor series[1].

We can give this function the abbreviated name and we have finished the derivation of power series definition of the exponential function.

If desired, the scalar values of the principle and interest rate can be substituted back into the formula to recover the ability to scale the function vertically and horizontally. This formula is the same as the population growth formula.