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# Dyson Sphere only ever possible theoretically

Assume technology more advanced than current levels, let us theorise for this question that we are capable of reaching and if needed extracting material from any planet or moon / body in our solar system.

Question simply is would there actually be enough material available allowing for the above ( and what we could extract from other planets via huge automated mining operations perhaps ) in our solar system to build such a creation.

Given the size it would have to likely be , there does not seem to be enough erm "stuff" around to make it any more than a theoretical dream ?

I was not 100% sure which S.E to put this in , sorry if I have chosen the wrong one! :) I do not know how to move if if it is wrong.

• It's not clear to me how you are computing the amount of material which needs to go into a Dyson sphere... Hint: Have you ever seen a really large mirror? Let's say a mirror 5 meters wide by 3 meters tall, 15 square meters (160 square feet) total area; how much silver do you think is in that mirror? How much silver would there be in a mirror with an area equal to the United States of America? Is that a lot of silver? – AlexP Jan 22 '20 at 17:19
• Hi AndyF. It's not clear to me what question you're actually asking. Can you maybe edit your post to clearly specify what your question is and what kind of answer you're looking for? – Morris The Cat Jan 22 '20 at 17:20
• I'm not sure how else I can ask it sorry :( The dead simple summary would be " I'm not sure there is enough material available in our solar system to build one even if we were capable " , Second paragraph . – AndyF Jan 22 '20 at 17:23
• Niven Rings and Dyson spheres have more problems than just the amount of mass/material available. They probably require ultra-exotic substances... low-grade (or even high-grade) steel just isn't going to cut it. The sphere builders will also probably need construction techniques sophisticated beyond anything you or I can imagine. I have to be honest, I struggle with getting the corner of the sheet on the mattress without the other side pulling off... imagine doing that with a partial sphere somewhere around the orbit of Jupiter. Ouch. – John O Jan 22 '20 at 18:26

Let's do some math to find out.

Let's say each "panel" of the sphere is 0.01 m thick, at a distance of 1 AU (149597871000 m) we would need to cover $$4\pi r^2=2.8 \cdot 10^{23} \ m^2$$, resulting in a volume of $$2.8 \cdot 10^{21} \ m^3$$.

Assuming a density equal to that of water, we get that we need $$2.8 \cdot 10^{24} \ kg$$ of matter to complete the Dyson sphere, which is the same order of magnitude of the mass of Earth ($$5.972 \cdot 10^{24} \ kg$$.)

So, being Earth not the biggest guy in the Solar system, we have plenty of mass to build a Dyson sphere.

• This proves only that there's enough matter in the Solar System to build a shell around the Sun. I'd be very surprised if it turns out that Dyson spheres only need to be 1 centimeter thick. – Halfthawed Jan 22 '20 at 20:48
• Agree with @Halfthawed; not only is 1cm a vanishingly thin layer, you're talking about needing about half Earth's mass worth of material suitable to build it. That means, in part, that you're looking for material strong enough in that thickness to support the weight of the polar regions of the sphere, whose gravitational pull toward the Sun is going to be significant and which isn't being counteracted by centripetal force of tangential orbital motion. – KeithS Jan 23 '20 at 0:41
• Dyson sphere was originally envisioned as gauze or a swarm of loosely associated satellites. So 1x1x1m satellites, 10 meters apart and viola. – pjp Jan 23 '20 at 11:25

You raise legitimate concerns. Even if there was simply enough mass, it may not be the right atoms (we don't build anything out of helium for instance).

Which is why some people already came up with the dyson swarm, as well as other variants.

A Dyson swarm is a swarm of smaller constructs which gather and send back the energy they collect. It is more "practical" because:

• it can be operational even with a few constructs
• once a part of the swarm is operational, the energy gathered can be used to complete the swarm

Once you have consequent operational dyson swarm, building a dyson sphere seems more feasible. Also, regarding the type of materials needed, let me remind you that with enough energy, we are already able to change atoms (by fusion or particle bombardment). So if you somehow run short of a rare element, you can just create more.

There are qualified people who already spent more time than you and me "planning" how to build a swarm. Maybe their plan will convince you.

• It's worse than just not having the right types of atoms... you might need something that's not even atomic in nature. Transmutation of elements is probably trivial by the time you have the technology to build one. – John O Jan 22 '20 at 18:28
• The swarm is quite an interesting idea , needing much less resources initially too. It was no a thought I had considered. – AndyF Jan 23 '20 at 10:35


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# How to find weight of neural network layer mlp classifier?

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Content

FAQ

Those who are looking for an answer to the question «How to find weight of neural network layer mlp classifier?» often ask the following questions:

### 💻 Is neural network classifier a nonlinear classifier?

I've just read an article talking about neural nets being a non-linear classification model. This can be shown to any number of layers, since linear combination of any number of weights is again linear.

### 💻 What is neural network classifier?

Summary. Neural networks are complex models, which try to mimic the way the human brain develops classification rules. A neural net consists of many different layers of neurons, with each layer receiving inputs from previous layers, and passing outputs to further layers.

### 💻 Are neural network a linear classifier?

Perhaps the simplest neural network we can define for binary classification is the single-layer perceptron. Given an input, the output neuron fires (produces an output of 1) only if the data point belongs to the target class. Otherwise, it does not fire (it produces an output of -1). The network looks something like this:

Predict using the multi-layer perceptron classifier. predict_log_proba (X) Return the log of probability estimates. predict_proba (X) Probability estimates. score (X, y[, sample_weight]) Return the mean accuracy on the given test data and labels. set_params (**params) Set the parameters of this estimator.

So, the weight change from the input layer unit i to hidden layer unit j is: ∆wij =η⋅δj ⋅ xi k k where δj = oj (1 −oj)∑wjk ⋅δ The weight change from the hidden layer unit j to the output layer unit k is: ∆wjk =η⋅δk ⋅oj where δk = (ytarg et,k − yk )yk (1 − yk)

Solution : A working solution is to inherit from MLPClassifier and override the _init_coef method. In the _init_coef write the code to set the initial weights. Then use the new class "MLPClassifierOverride" as in the example below instead of "MLPClassifier".

1) Choose your classifier. from sklearn.neural_network import MLPClassifier mlp = MLPClassifier(max_iter=100) 2) Define a hyper-parameter space to search. (All the values that you want to try out.)

The standard multilayer perceptron (MLP) is a cascade of single-layer perceptrons. There is a layer of input nodes, a layer of output nodes, and one or more intermediate layers. The interior layers are sometimes called “hidden layers” because they are not directly observable from the systems inputs and outputs.

To see this let us see the example we took above but now the weights are initialized with very large values instead of 0 : W[l] = np.random.randn(l-1,l)*10. Neural network is the same as earlier, using this initialization on the dataset “make circles” from sklearn.datasets, the result obtained as the following :

The single-layer perceptron classifiers discussed previously can only deal with linearly separable sets of patterns. The multilayer networks to be introduced here are the most widespread neural network architecture – Made useful until the 1980s, because of lack of efficient training algorithms (McClelland and Rumelhart 1986) – The ...

Around 2^n (where n is the number of neurons in the architecture) slightly-unique neural networks are generated during the training process, and ensembled together to make predictions. A good dropout rate is between 0.1 to 0.5; 0.3 for RNNs, and 0.5 for CNNs. Use larger rates for bigger layers.

Each neuron of the hidden layers receives the output from every neuron of the previous layers and transforms these values with a weighted linear summation $$\sum_{i=0}^{n-1}w_ix_i = w_0x_0 + w_1x_1 + ... + w_{n-1}x_{n-1}$$ into an output value, where n is the number of neurons of the layer and $w_i$ corresponds to the i th component of the weight vector. The output layer receives the values from the last hidden layer.

Consider a neural network having an l layer, which has l-1 hidden layers and 1 output layer. Then, the parameters i.e, weights and biases of the layer l are represented as, Image Source: link. In addition to weights and biases, some more intermediate variables are also computed during the training process, Image Source: link. Steps of Training ...

We've handpicked 22 related questions for you, similar to «How to find weight of neural network layer mlp classifier?» so you can surely find the answer!

What is neural network hidden layer?

What is a Hidden Layer? In neural networks, a hidden layer is located between the input and output of the algorithm, in which the function applies weights to the inputs and directs them through an activation function as the output. In short, the hidden layers perform nonlinear transformations of the inputs entered into the network.

What layer is hte input layer in neural network?

On paper the network has input neurons. On implementation level you have to organize this data (usually using arrays/vectors) which is why you speak of an input vector: An input vector holds the input neuron values (representing the input layer). If you're familiar with basics of graph theory or image processing - it's the same principle.

A generic three layer neural network octave?

Neural network with one hidden layer consisting of three nodes. We are going to build a simple network with only one hidden layer containing three nodes. The input will equal the transpose of the ...

How many layer in a neural network?

#### Traditionally, neural networks only had three types of layers: hidden, input and output....Table: Determining the Number of Hidden Layers.

Num Hidden LayersResult
noneOnly capable of representing linear separable functions or decisions.
How to do neural network multi layer?

Neural networks can also have multiple output units. For example, here is a network with two hidden layers layers L_2 and L_3 and two output units in layer L_4: To train this network, we would need training examples (x^{(i)}, y^{(i)}) where y^{(i)} \in \Re^2. This sort of network is useful if there’re multiple outputs that you’re interested in predicting.

What is a hidden layer neural network?

One hidden layer Neural Network Why do you need non-linear activation functions?

What is a multi-layer neural network?

A multi-layer neural network contains more than one layer of artificial neurons or nodes. They differ widely in design. It is important to note that while single-layer neural networks were useful early in the evolution of AI, the vast majority of networks used today have a multi-layer model.

What is a neural network dense layer?

Dense Layer = Fullyconnected Layer = topology, describes how the neurons are connected to the next layer of neurons (every neuron is connected to every neuron in the next layer), an intermediate layer (also called hidden layer see figure) Output Layer = Last layer of a Multilayer Perceptron

What is a single-layer neural network?

What Does Single-Layer Neural Network Mean? A single-layer neural network represents the most simple form of neural network, in which there is only one layer of input nodes that send weighted inputs to a subsequent layer of receiving nodes, or in some cases, one receiving node. This single-layer design was part of the foundation for systems which have now become much more complex.

What is fc layer in neural network?

These are the first layers in the network. The final layer(s), which are usually Fully Connected NNs, whose goal is to classify those features. The latter do have a typical equation (i.e. $f(W^T \cdot X + b)$), where $f$ is an activation function.

What is hidden layer in neural network?

In neural networks, a hidden layer is located between the input and output of the algorithm, in which the function applies weights to the inputs and directs them through an activation function as the output. In short, the hidden layers perform nonlinear transformations of the inputs entered into the network.

What is input layer in neural network?

The input layer of a neural network is composed of artificial input neurons, and brings the initial data into the system for further processing by subsequent layers of artificial neurons. The input layer is the very beginning of the workflow for the artificial neural network.

What is linear layer in neural network?

In the network, the hidden layer and output layer is composed made up of Dense layer. Hidden layer has 3 nodes and output layer has 1 node. All nodes in hidden layer are connected to all nodes in the input layer and all nodes in the output layer, in this case there is only one node, is also connected to all nodes in hidden layer.

What is multi layer neural network example?

Convolutional neural networks (CNNs), so useful for image processing and computer vision, as well as recurrent neural networks, deep networks and deep belief systems are all examples of multi-layer neural networks. CNNs, for example, can have dozens of layers that work sequentially on an image.

What is multi layer neural network model?

Multi-Layer Neural Network. Consider a supervised learning problem where we have access to labeled training examples (x^{(i)}, y^{(i)}). Neural networks give a way of defining a complex, non-linear form of hypotheses h_{W,b}(x), with parameters W,b that we can fit to our data. To describe neural networks, we will begin by describing the simplest possible neural network, one which comprises a single “neuron.” We will use the following diagram to denote a single neuron: This “neuron ...

What is multi layer neural network theory?

A multi-layer neural network contains more than one layer of artificial neurons or nodes. They differ widely in design. It is important to note that while single-layer neural networks were useful early in the evolution of AI, the vast majority of networks used today have a multi-layer model. Advertisement.

What is output layer in neural network?

What Is An Output Layer? The output layer is responsible for producing the final result. There must always be one output layer in a neural network. The output layer takes in the inputs which are passed in from the layers before it, performs the calculations via its neurons and then the output is computed.

Why need hidden layer in neural network?

#### Hidden layers and neurons

They allow you to model complex data thanks to their nodes/neurons. They are “hidden” because the true values of their nodes are unknown in the training dataset. In fact, we only know the input and output. Each neural network has at least one hidden layer. How to find number of weight terms neural network?

You can find the number of weights by counting the edges in that network. To address the original question: In a canonical neural network, the weights go on the edges between the input layer and the hidden layers, between all hidden layers, and between hidden layers and the output layer.

Are neural networks a type of classifier?
• Neural networks are also effective in categorizing data into identifiable groups or features. Classification neural networks used for feature categorization are very similar to fault-diagnosis networks, except that they only allow one output response for any input pattern, instead of allowing multiple faults to occur for a given set of operating conditions. The classification network selects the category based on which output response has the highest output value.
How neural networks work human classifier wrinkles?

The ANN classifier is designed by using two layer feedforward backpropagation neural networks. The proposed age classification framework is trained and tested with face images from PAL face database and shown considerable improvement in the age classification accuracy up to 94.17% and 93.75% for male and female respectively.

A neural network with three hidden layer linear?

Figure 6.1 is an example of a simple three layer neural network The neural network consists of: An input layer A hidden layer An output layer Each of the layers are interconnected by modifiable weights, which are represented by the links between layers Each layer consists of a number of units (neurons) that loosely mimic the

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# Question 06eb8

Mar 30, 2016

Here's what I got.

#### Explanation:

COMPLETE LAB SHEET

This answer is based on the solution provided here:

http://socratic.org/questions/2-what-volume-of-oxygen-must-react-with-a-full-tank-of-gas-for-the-following-at-

So, you know that the balanced chemical equation for this combustion reaction looks like this

$\textcolor{red}{2} {\text{C"_8"H"_text(18(l]) + color(blue)(25)"O"_text(2(g]) -> 16"CO"_text(2(g]) + 18"H"_2"O}}_{\textrm{\left(l\right]}}$

You also know that the reaction used up

18.5color(red)(cancel(color(black)("gal"))) * (3.79color(blue)(cancel(color(black)("L"))))/(1color(red)(cancel(color(black)("gal")))) * (1000color(green)(cancel(color(black)("mL"))))/(1color(blue)(cancel(color(black)("L")))) * "0.703 g"/(1color(green)(cancel(color(black)("mL")))) = 4.93 * 10^4"g"

of octane. This many grams of octane will contain

4.93 * 10^4color(red)(cancel(color(black)("g"))) * "1 mole octane"/(114.23color(red)(cancel(color(black)("g")))) = "431.6 moles octane"

Now, your job at this point is to use the $\textcolor{red}{2} : 16$ mole ratio that exists between octane and carbon dioxide to determine how many moles of carbon dioxide will be produced by the reaction.

431.6 color(red)(cancel(color(black)("moles octane"))) * "16 moles CO"_2/(color(red)(2)color(red)(cancel(color(black)("moles octane")))) = "3452.8 moles CO"_2

The problem provides you with the pressure and temperature inside the combustion chamber, which means that you can use the ideal gas law equation to find the volume occupied by the carbon dioxide

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

Convert the temperature from degrees Fahrenheit to degrees Celsius, then from degrees Celsius to Kelvin by using the conversion factors

color(purple)(|bar(ul(color(white)(a/a)color(black)(t[""^@"C"] = (t[""^@"F"] - 32) xx 5/9)color(white)(a/a)|))) " " and " "color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))

You should get

t[""^@"C"] = (1940^@"F" - 32) * 5/9 = 1060^@"C"

and

T["K"] = 1060^@"C" + 273.15 = "1333.15 K"#

Rearrange the ideal gas law equation to solve for $V$, the volume of the gas

$P V = n R T \implies V = \frac{n R T}{P}$

Plug in your values to get

$V = \left(3452.8 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 1333.15color(red)(cancel(color(black)("K"))))/(11.56color(red)(cancel(color(black)("atm}}}}\right)$

$V = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{32,700 L} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ rounded to three sig figs

Make sure to double-check my calculations!


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1
JEE Main 2022 (Online) 27th July Evening Shift
+4
-1

(A) The drift velocity of electrons decreases with the increase in the temperature of conductor.

(B) The drift velocity is inversely proportional to the area of cross-section of given conductor.

(C) The drift velocity does not depend on the applied potential difference to the conductor.

(D) The drift velocity of electron is inversely proportional to the length of the conductor.

(E) The drift velocity increases with the increase in the temperature of conductor.

Choose the correct answer from the options given below :

A
(A) and (B) only
B
(A) and (D) only
C
(B) and (E) only
D
(B) and (C) only
2
JEE Main 2022 (Online) 27th July Morning Shift
+4
-1

Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are $$r_{1}$$ and $$r_{2}$$ $$\left(r_{1}>r_{2}\right)$$. If the potential difference across the source of internal resistance $$r_{1}$$ is zero, then the value of R will be :

A
$$r_{1}-r_{2}$$
B
$$\frac{r_{1} r_{2}}{r_{1}+r_{2}}$$
C
$$\frac{r_{1}+r_{2}}{2}$$
D
$$r_{2}-r_{1}$$
3
JEE Main 2022 (Online) 26th July Evening Shift
+4
-1

A battery of $$6 \mathrm{~V}$$ is connected to the circuit as shown below. The current I drawn from the battery is :

A
1A
B
2A
C
$$\frac{6}{11}$$ A
D
$$\frac{4}{3}$$ A
4
JEE Main 2022 (Online) 26th July Morning Shift
+4
-1

The current I in the given circuit will be :

A
10 A
B
20 A
C
4 A
D
40 A
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text: # Python Challenge Answers Challenges for 10-26-18 - python answers  # cumulative sum x=[3,10,4,12,55] cs=[0]*5 for i in range(0,len(x)): cs[i]=sum(x[0:(i+1)]) print(cs) x=[3,10,4,12,55] cs=list() for i in range(0,len(x)): cs.append(sum(x[0:(i+1)])) print(cs) #how long to 5 5's x=[5,3,2,5,5,1,2,5,3,5,1,5,1] count=0 i=0 while count < 5: if x[i]==5: count+=1 i+=1 print(i) # answers for Lecture 11 extra practice import pandas as pd # 0 - calculate sum of female and male wages in wages.csv wages=pandas.read_csv("wages.csv",header=0,sep=",") femaleSum=0 maleSum=0 for i in range(0,len(wages),1): if wages.gender[i]=="female": femaleSum=femaleSum+wages.wage[i] else: maleSum=maleSum+wages.wage[i] femaleSum maleSum sum(wages.gender=="female") sum(wages.gender=="male") wages.tail() # find runs # load file findRuns=pd.read_csv("findRuns.txt",header=None,sep="\t") # create a variable out that is currently undefined out=pd.DataFrame(columns=['startIndex','runLength']) # I will use this variable cur to hold onto the previous number in the vector; # this is analagous to using findRuns[i-1] cur=findRuns.iloc[0,0] #cur=findRuns[i-1] # this is a counter that I use to keep track of how long a run of repeated values is; # if there are not repeated values than this count equals 1 count=1 # loop through each entry of our vector (except the 1st one, which we set to cur above) for i in range(1,50,1): # test if the ith value in the vector findRuns equals the previous (stored in cur) if findRuns.iloc[i,0]==cur: # test whether count is 1 (we aren't in the middle of a run) or >1 (in the middle of a run) if count==1: # if the ith value in the vector equals the previous (stored in cur) and count is 1, we # are at the beginning of a run and we want to store this value (we temporarily store it in 'start') start=(i-1) # we add one to count because the run continued based on the ith value of findRuns being equal to # the previous (stored in cur) count=count+1 # if the ith value in findRuns is not the same as the previous (stored in cur) we either are not in a run # or we are ending a run else: # if count is greater than 1 it means we were in a run and must be exiting one if count>1: # add a row to 'out' that will hold the starting positions in the first column and the length # of runs in the second column; this appends rows to out after finding and counting each run out.loc[len(out)]=[start,count] # reset count to 1 because we just exited a run count=1 # remember cur holds the previous element in findRuns, so we need to update this after each time # we go through the for loop cur=findRuns.iloc[i,0] cur out


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On the energy-momentum tensor in Moyal space

Abstract : After reviewing the known results for the definition and properties of the energy-momentum tensor(s) in Minkowski space, we study the properties of the energy-momentum tensor of gauge fields coupled to matter in non-commutative (Moyal) space. In general, the non-commutativity affects the usual conservation law of the tensor as well as its transformation properties (gauge covariance instead of gauge invariance). It is known that the conservation of the energy-momentum tensor can be achieved by a redefinition involving another star product. Furthermore, for a pure gauge theory it is always possible to define a gauge invariant energy-momentum tensor by means of a gauge invariant Wilson line. We show that the latter two procedures are incompatible with each other if couplings of gauge fields to matter fields (scalars or fermions) are considered: The gauge invariant tensor (constructed via Wilson line) does not allow for a redefinition assuring its conservation, and vice-versa the introduction of another star product does not allow for gauge invariance by means of a Wilson line.
Type de document :
Article dans une revue
European Physical Journal C: Particles and Fields, Springer Verlag (Germany), 2015, 75, pp.284. 〈10.1140/epjc/s10052-015-3492-8〉
Domaine :

http://hal.in2p3.fr/in2p3-01142855
Contributeur : Sylvie Flores <>
Soumis le : jeudi 16 avril 2015 - 10:38:09
Dernière modification le : jeudi 15 mars 2018 - 11:40:08

Citation

H. Balasin, D.N. Blaschke, F. Gieres, M. Schweda. On the energy-momentum tensor in Moyal space. European Physical Journal C: Particles and Fields, Springer Verlag (Germany), 2015, 75, pp.284. 〈10.1140/epjc/s10052-015-3492-8〉. 〈in2p3-01142855〉

Métriques

Consultations de la notice


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# Function Notation

Function notation is challenging for many students yet we teachers overlook the reasons for the challenges. For example, students see the parentheses and rely on the rule they were taught previously, y(5) is “y times 5”. Because this is overlooked by teachers we often skim over the concept of notation and delve into the steps. This document is a means of presenting the concept and the use of function notation in a meaningful way. Feel free to use or revise and use the document as you wish.


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### Home > CCA > Chapter 5 > Lesson 5.2.2 > Problem5-66

5-66.

Choose one of the sequences in problem 5-65 for which you determined that $447$ is not a term. Write a clear explanation describing how you can be sure that $447$ is not a term of the sequence.

Justification will vary, but $n$ must be a positive integer.


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# Equal sign with a “c”

I am not sure whether this is common sense, but I really don't know what this $c$ means in an equivalent sign, from Lindell's How to Simulate It:

$$\big\{(S_1(1^n, x, f_1(x,y)), f(x,y))\big\}\stackrel{c}{\equiv} \big\{(\mathsf{view}^\pi_1(x,y,n),\mathsf{output}^\pi(x,y,n))\big\}_{x,y,n}$$ $$\big\{(S_2(1^n, x, f_2(x,y)), f(x,y))\big\}\stackrel{c}{\equiv} \big\{(\mathsf{view}^\pi_2(x,y,n),\mathsf{output}^\pi(x,y,n))\big\}_{x,y,n}$$

• The paragraph introducing the notation (bottom of page 3) talks about computationally indistinguishability, so I assume that's what it means. – CodesInChaos Oct 23 '17 at 13:37

Two probability ensembles... are said to be computationally indistinguishable, denoted $X\stackrel{c}{\equiv}Y$, if...


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# $\cos(n\vartheta)=\frac{a_n}{3^n}$

I want to show,

if i know that $$\cos(\vartheta)=\frac{1}{3}$$ than $$\cos(n\vartheta)=\frac{a_n}{3^n}$$ for $$n\in \mathbb{N}$$, where $$a_n \in \mathbb{Z}$$,$$3 \nmid a_n$$

My approach was to do it by induction. For n=1 it is clear. Than i used $$\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$$ with $$\alpha=n\vartheta$$ and $$\beta=\pm \vartheta$$. I added the results and got $$\cos((n+1)\vartheta)=2\cos(\vartheta)\cos(n\vartheta)-\cos((n-1)\vartheta)$$.

I get then

$$\cos((n+1)\vartheta)=2\frac{1}{3}\frac{a_n}{3^n}-\cos((n-1)\vartheta) = \frac{2a_n-\cos((n-1)\vartheta)3^{n+1}}{3^{n+1}}$$

I don't know how to argue that $$2a_n-\cos((n-1)\vartheta)3^{n+1} \in \mathbb{Z}$$ and that $$3 \nmid 2a_n-\cos((n-1)\vartheta)3^{n+1}$$.

If $$a_n = 3^n \cos(n \vartheta)$$, your equation $$\cos((n+1) \vartheta) = 2 \cos(\vartheta) \cos(n \vartheta) - \cos((n-1) \vartheta)$$ can be written as $$a_{n+1} = 2 a_n - 9 a_{n-1}$$ and then it's obvious that if $$a_n$$ and $$a_{n-1}$$ are integers, so is $$a_{n+1}$$, and if $$a_{n} \not\equiv 0 \bmod 3$$ then $$a_{n+1} \not\equiv 0 \bmod 3$$.


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The science case for LIGO-India
Abstract The global network of gravitational-wave detectors has completed three observing runs with ∼50 detections of merging compact binaries. A third LIGO detector, with comparable astrophysical reach, is to be built in India (LIGO-Aundha) and expected to be operational during the latter part of this decade. Such additions to the network increase the number of baselines and the network SNR of GW events. These enhancements help improve the sky-localization of those events. Multiple detectors simultaneously in operation will also increase the baseline duty factor, thereby, leading to an improvement in the detection rates and, hence, the completeness of surveys. In this paper, we quantify the improvements due to the expansion of the LIGO global network in the precision with which source properties will be measured. We also present examples of how this expansion will give a boost to tests of fundamental physics.
Authors:
; ; ; ; ; ; ; ; ; ; ; ; ; ;
Award ID(s):
Publication Date:
NSF-PAR ID:
10314085
Journal Name:
Classical and Quantum Gravity
Volume:
39
Issue:
2
ISSN:
0264-9381
National Science Foundation
##### More Like this
1. Abstract

The waveform of a compact binary coalescence is predicted by general relativity. It is therefore possible to directly constrain the response of a gravitational-wave (GW) detector by analyzing a signal’s observed amplitude and phase evolution as a function of frequency. GW signals alone constrain the relative amplitude and phase between different frequencies within the same detector and between different detectors. Furthermore, if the source’s distance and inclination can be determined independently, for example from an electromagnetic (EM) counterpart, one can calibrate the absolute amplitude response of the detector network. We analyze GW170817’s ability to calibrate the LIGO/Virgo detectors, findingmore »

2. Abstract

Based on the prior O1–O2 observing runs, about 30% of the data collected by Advanced LIGO and Virgo in the next observing runs are expected to be single-interferometer data, i.e. they will be collected at times when only one detector in the network is operating in observing mode. Searches for gravitational-wave signals from supernova events do not rely on matched filtering techniques because of the stochastic nature of the signals. If a Galactic supernova occurs during single-interferometer times, separation of its unmodelled gravitational-wave signal from noise will be even more difficult due to lack of coherence between detectors. Wemore »

3. Abstract

Dark matter exists in our Universe, but its nature remains mysterious. The remarkable sensitivity of the Laser Interferometer Gravitational-Wave Observatory (LIGO) may be able to solve this mystery. A good dark matter candidate is the ultralight dark photon. Because of its interaction with ordinary matter, it induces displacements on LIGO mirrors that can lead to an observable signal. In a study that bridges gravitational wave science and particle physics, we perform a direct dark matter search using data from LIGO’s first (O1) data run, as opposed to an indirect search for dark matter via its production of gravitational waves.more »

4. ABSTRACT

GW170817 showed that neutron star mergers not only emit gravitational waves but also can release electromagnetic signatures in multiple wavelengths. Within the first half of the third observing run of the Advanced LIGO and Virgo detectors, there have been a number of gravitational wave candidates of compact binary systems for which at least one component is potentially a neutron star. In this article, we look at the candidates S190425z, S190426c, S190510g, S190901ap, and S190910h, predicted to have potentially a non-zero remnant mass, in more detail. All these triggers have been followed up with extensive campaigns by the astronomical communitymore »

5. The detection of gravitational-wave signals by the LIGO and Virgo observatories during the past few years has ushered us into the era of gravitational-wave astronomy, shifting our focus from detection to source parameter estimation. This has imposed stringent requirements on calibration in order to maximize the astrophysical information extracted from these detected signals. Current detectors rely on photon radiation pressure from auxiliary lasers to achieve required calibration accuracy. These photon calibrators have made significant improvements over the last few years, realizing fiducials displacements with sub-percent accuracy. This achieved accuracy is directly dependent on the laser power calibration. For the nextmore »


text:
## Engineer's Book

Engineers – Telecoms – Physics – Teaching – Surveying

## HP 05 – How Do You Do When You Are a Gifted Person ?

Written By: Jean-Paul Cipria - Juin• 22•17

3D Ascii Probabilities Table – ©J.P. Cipria 2017.

Gifted ? Talented ? Intelligent ? Cleverness. Have you got problems with … life ?

# First Rules to Survive Efficiently

## Think and Stop and Do

Do it when you think it … and then stop to think further !

We knows methods are generaly toxic ! Then don’t use it in engineering !

First we ask us : – « How to browse a text and read ONLY ONE TIME a data and find its presence probability ? »

Immediate thinking was to store it in a table with ascii value as indice. Then if you want to store $i \longrightarrow j \longrightarrow k$ you store $Table(i,j,k)=NumberOfOccurences$.

Then forgot the following very, very, very slow thought : – « Try to do a recursive method : Find ‘a’, browse text, count, find b, browse text … etc ». Very bad algorithm not so methodic but surely very, verrrry, slooooow.

Stop to think and construct immediately you table …

3D Ascii Probabilities Table – ©J.P. Cipria 2017.

… Then table dimension is : $256*256*256=16777216$ 16 MO ! You use big memory but you get  instantaneous calculation ! Let’s choose our constraints « Minimum Time Calculation » by sensations.

When we choose to find order five for text probabilities we’ve got $256*256*256*256*256=1.0995e+12$ then Matlab worked a long time and said us : – « Lack of memories. »

## Needless to do some Industrial Stupid Jobs ?

Yes it does ! … and think about … nothing. « Be Aware » high potential ! Open your chacras ! … Let’s Play Solitaire game or return to a big society. You have some time to think about nothing.

After the 50th bugs we returned to a recursive method and we waited a long time to play with « solitaire » without thinking or « Mind in the air » to find another solution.

Let’s your Mind in the Air. Play Solitaire ! »

Solitaire – Open Your Chacras ?

And – « How to reduct ascii table only to ‘ab…z » characters ? Then $Table_{dim}=26^5=11881376$ «  Then … do it ==> Instantaneous calculation for order 5 texts probabilities. Yeah. … But you got only 90% of right statistics because you forgot informations when you limit text values. – « But are those values contains informations ? » … « Think about entropy. »  … etc.

Physics, inventions, discovers are the result of long chainS of knowledgeS. Let’s go fishing and pulling the net. J. P. Cipria.

Ishikawa – Telecom – SMS Trouble Shooting

.

Jean-Paul Cipria
22/06/2017.

You can follow any responses to this entry through the RSS 2.0 feed. You can skip to the end and leave a response. Pinging is currently not allowed.


text:
Having chosen your $4$ consecutive numbers you place $+$ and $-$ signs in between them.
You have $3$ empty spaces, so how many possible combinations are there for arranging the $+$ and $-$ signs?


text:
liu.seSearch for publications in DiVA
Change search

BETA
##### Publications (10 of 36) Show all publications
Fowler, S., Baravdish, G. & Baravdish, G. (2019). Compressed Sensing of Wireless Signals for Image Tensor Reconstruction. In: IEEE Global Communications Conference: Signal Processing for Communications (Globecom2019 SPC). Paper presented at IEEE Global Communications Conference, Waikoloa, HI, USA, 9-13 December 2019.
Open this publication in new window or tab >>Compressed Sensing of Wireless Signals for Image Tensor Reconstruction
2019 (English)In: IEEE Global Communications Conference: Signal Processing for Communications (Globecom2019 SPC), 2019Conference paper, Published paper (Refereed)
##### Abstract [en]

Use of wireless signal for identification of unknown object, or technology to see-through a wall to form an image, is gaining growing interest from various fields including law enforcement and military sectors, disaster management, or even in civilian sectors such as construction sites. The great challenge in the implementation of such technology is the stochastic disturbances on wireless signal which will result in a signal with missing samples. Compressive Sensing (CS) is a powerful tool for estimating the missing samples since it can find accurate solution to largely underdetermined linear wireless signals. However, sparse models like CS can also suffer from information loss dues to stochastic lossy nature of wireless, making CS not to have accurate information for reconstruction of a signal. In this paper, we developed a theoretical and experimental framework for the mapping of obstacles by reconstructing the wireless signal based on a sparse signal. We apply tensor format to perform the computations along each mode by relaxing the tensor constraints to obtain accurate results. The proposed framework demonstrates how to take 2D signals, formulate estimate signals and produce a 3D image location in a completely unknown area inside of the obstacle (wall).

##### Keywords
Compressed Sensing; Basis Pursuit; Ill-posed; Tensors; Lagrangian; Wireless Signals;Augmented-Lagrangian, Alternating Direction Method of Multipliers, Basis Pursuit, Ill-posed and Inverse Problems, ImagingAugmented-Lagrangian, Alternating Direction Method of Multipliers, Basis Pursuit, Ill-posed and Inverse Problems, Imaging
##### National Category
Engineering and Technology Signal Processing
##### Identifiers
urn:nbn:se:liu:diva-163576 (URN)
##### Conference
IEEE Global Communications Conference, Waikoloa, HI, USA, 9-13 December 2019
Available from: 2020-02-14 Created: 2020-02-14 Last updated: 2020-02-17Bibliographically approved
Zhang, X., Zhang, X., Fu, Z., Yu, B. & Fowler, S. (2019). Joint Distortion Estimation and Layer Selection of Unequal Error Protection for SVC Video Transmission over FSO Networks. In: 2019 IEEE 21st International Conference on High Performance Computing and Communications; IEEE 17th International Conference on Smart City; IEEE 5th International Conference on Data Science and Systems (HPCC/SmartCity/DSS): . Paper presented at 2019 IEEE 21st International Conference on High Performance Computing and Communications; IEEE 17th International Conference on Smart City; IEEE 5th International Conference on Data Science and Systems (HPCC/SmartCity/DSS) (pp. 726-733). Piscataway, New Jersey, US
Open this publication in new window or tab >>Joint Distortion Estimation and Layer Selection of Unequal Error Protection for SVC Video Transmission over FSO Networks
2019 (English)In: 2019 IEEE 21st International Conference on High Performance Computing and Communications; IEEE 17th International Conference on Smart City; IEEE 5th International Conference on Data Science and Systems (HPCC/SmartCity/DSS), Piscataway, New Jersey, US, 2019, p. 726-733Conference paper, Published paper (Refereed)
##### Abstract [en]

To meet the needs of high throughput, low delay and high resolution video services, the free space optical network (FSO) has recently received increasing attention due to its large bandwidth and high transmission rate. However, atmospheric turbulence is a major challenge, which can lead to severe BER performance degradation or even make the link ineffective. To mitigate the significant degradation of the video streaming quality on FSO networks, in this paper, an unequal error protection (UEP) scheme for scalable video coding is designed to minimize the video distortion. By adaptively choosing the number of video layers and the redundancy rate of the channel coding, unequal error protection of Scalable video coding is achieved. In addition, to solve the optimization model, a heuristic-based fast algorithm is proposed. A large number of simulation experiments in an NS3 network simulator show that the proposed scheme offers significantly higher quality in the reconstructed video sequences.

##### Place, publisher, year, edition, pages
Piscataway, New Jersey, US: , 2019
##### Keywords
channel coding;error statistics;free-space optical communication;image sequences;optical links;optimisation;redundancy;video coding;video streaming;FSO networks;unequal error protection scheme;scalable video coding;video distortion;video layers;redundancy rate;channel coding;NS3 network simulator show;reconstructed video sequences;layer selection;SVC video transmission;low delay;high resolution video services;free space optical network;high transmission rate;atmospheric turbulence;link ineffective;video streaming quality;BER performance degradation;Streaming media;Forward error correction;Error correction codes;Redundancy;Distortion;Video coding;Packet loss;Free space optical network;Forward error correction;Unequal error protection (UEP);Scalable video coding
##### National Category
Electrical Engineering, Electronic Engineering, Information Engineering
##### Identifiers
urn:nbn:se:liu:diva-162721 (URN)10.1109/HPCC/SmartCity/DSS.2019.00108 (DOI)
##### Conference
2019 IEEE 21st International Conference on High Performance Computing and Communications; IEEE 17th International Conference on Smart City; IEEE 5th International Conference on Data Science and Systems (HPCC/SmartCity/DSS)
Available from: 2019-12-17 Created: 2019-12-17 Last updated: 2020-01-24Bibliographically approved
Fowler, S., Baravdish, G. & Rudberg, M. (2019). Optimizing Compressed Sensing for seeing through walls based on Wireless Signals. In: IEEE Symposium on Computers and Communications (ISCC): . Paper presented at IEEE Symposium on Computers and Communications (ISCC), Barcelona, Spain, 29 June-3 July 2019 (pp. 1-6).
Open this publication in new window or tab >>Optimizing Compressed Sensing for seeing through walls based on Wireless Signals
2019 (English)In: IEEE Symposium on Computers and Communications (ISCC), 2019, p. 1-6Conference paper, Published paper (Refereed)
##### Abstract [en]

In this paper, we developed a theoretical and experimental framework for the mapping of obstacles using WiFi, based on a small number of wireless channel samples. This is very challenging due to the numerous channel coefficients to be estimated over the time-varying channel and the channel estimation of a wireless signal transmission to be considered for compressive sampling. In a typical communication system, the signal is sampled at least twice at the highest frequency contained in the signal. However, this limits efficient ways to compress the signal, as it places a huge burden on sampling the entire signal while only a small number of the transform coefficients are needed to represent the signal. To tackle this problem, we will focused on a mathematical optimization problem for the most efficient compressed sensing method called $\ell_1$-norm, known as Basis Pursuit. Before optimizing the problem, the noise was removed from the signal, namely, multipath fading. Our experimental results show the improved performance in the number of iterations for obtaining a framework for the mapping of obstacles.

##### Keywords
Compressed Sensing; Imaging; Signal Reconstruction; Basis Pursuit; Radio Tomographic; Digital Image Reconstruction
##### National Category
Electrical Engineering, Electronic Engineering, Information Engineering
##### Identifiers
urn:nbn:se:liu:diva-163563 (URN)10.1109/ISCC47284.2019.8969738 (DOI)
##### Conference
IEEE Symposium on Computers and Communications (ISCC), Barcelona, Spain, 29 June-3 July 2019
Available from: 2020-02-14 Created: 2020-02-14 Last updated: 2020-02-17Bibliographically approved
Al Haji, G., Fowler, S. & Andersson Granberg, T. (2018). Smart traffic calming measures for smart cities - a pre-study. Linköping: Linköping University Electronic Press
Open this publication in new window or tab >>Smart traffic calming measures for smart cities - a pre-study
##### Abstract [en]

Traffic calming measures, such as speed bumps and elevated crossing points, are used to reduce speed, to prevent overtaking and generally contribute to a safer traffic situation. However, they might also cause increased response times for rescue vehicles (e.g. ambulances or fire trucks). An alternative to the conventional traffic calming measures is so-called smart traffic calming measures. These can determine when a vehicle approaches, whose journey should not be hindered, and adjust to allow for free passage for this vehicle.

This report gives an overview of the problem, and some examples of smart  traffic calming measures are discussed. Special focus is put on the wireless communication necessary to detect emergency vehicles. Furthermore, existing challenges and possible solutions for traffic calming measures and the communication needed to make them smart are discussed.

##### Abstract [sv]

Trafiklugnande åtgärder, såsom vägbulor och förhöjda övergångsställen, används för att minska hastigheten, hindra farliga omkörningar och generellt bidra till en säkrare trafiksituation. Dock kan de också bidra till att insatstiderna för räddningsfordon (t.ex. ambulanser eller räddningstjänstens släckbilar) förlängs. Ett alternativ till de konventionella trafiklugnande åtgärderna, är s.k. smarta trafiklugnade åtgärder. Dessa kan avgöra när ett fordon närmar sig, vars färd inte bör hindras, och anpassa sig så att fri färd för detta fordon tillåts.

I denna rapport ges en översikt av problemet, och några exempel på smarta trafiklugnade åtgärder diskuteras, med fokus på sådana som hämtar information och styrs med hjälp av trådlös kommunikation. Vidare diskuteras existerande utmaningar och möjliga lösningar för trafiklugnande åtgärder och den kommunikation som krävs för att göra dem smarta.

##### Series
CARER Report ; 22
##### National Category
Transport Systems and Logistics Computer Engineering Infrastructure Engineering
##### Identifiers
urn:nbn:se:liu:diva-145386 (URN)
Available from: 2018-02-27 Created: 2018-02-27 Last updated: 2018-02-27Bibliographically approved
Mushtaq, M. S., Fowler, S., Augustin, B. & Mellouk, A. (2016). QoE in 5G Cloud Networks using Multimedia Services. In: 2016 IEEE WIRELESS COMMUNICATIONS AND NETWORKING CONFERENCE: . Paper presented at IEEE Wireless Communication and Networking Conference (WCNC’16). IEEE Computer Society
Open this publication in new window or tab >>QoE in 5G Cloud Networks using Multimedia Services
2016 (English)In: 2016 IEEE WIRELESS COMMUNICATIONS AND NETWORKING CONFERENCE, IEEE Computer Society, 2016Conference paper, Published paper (Refereed)
##### Abstract [en]

The 4G standard Long Term Evolution-Advanced(LTE-A) has been deployed in many countries. Now, technologyis evolving towards the 5G standard since it is expecting tostart its service in 2020. The 5G cellular networks will mainlycontain in cloud computing and primarily Quality of Service(QoS) parameters (e.g. delay, loss rate, etc.) influence thecloud network performance. The impact of user perceivedQuality of Experience (QoE) using multimedia services, andapplication significantly relies on the QoS parameters. The keychallenge of 5G technology is to reduce the delay less thanone millisecond. In this paper, we have described a methodthat minimizes the overall network delay for multimediaservices; which are constant bit rate (VoIP) and variablebit rate (video) traffic model. We also proposed a methodthat measures the user’s QoE for video streaming trafficusing the network QoS parameters, i.e. delay and packet lossrate. The performance of proposed QoE method is comparedwith QoV method, and our proposed QoE method performsbest by carefully handle the impact of QoS parameters. Theresults show that our described method successfully reduces theoverall network delays, which result to maximize the user’s QoE.

##### Place, publisher, year, edition, pages
IEEE Computer Society, 2016
##### Series
IEEE Wireless Communications and Networking Conference, ISSN 1525-3511
##### Keywords
QoE; QoS; 5G; Cloud; VoIP; Video; Multimedia Services.
##### National Category
Communication Systems
##### Identifiers
urn:nbn:se:liu:diva-128164 (URN)10.1109/WCNC.2016.7565173 (DOI)000388603103053 ()978-1-4673-9814-5 (ISBN)
##### Conference
IEEE Wireless Communication and Networking Conference (WCNC’16)
Available from: 2016-05-19 Created: 2016-05-19 Last updated: 2016-12-27Bibliographically approved
Mushtaq, M. S., Mellouk, A., Augustin, B. & Fowler, S. (2016). QoE Power-Efficient Multimedia Delivery Method for LTE-A. IEEE Systems Journal, 10(2), 749-760
Open this publication in new window or tab >>QoE Power-Efficient Multimedia Delivery Method for LTE-A
2016 (English)In: IEEE Systems Journal, ISSN 1932-8184, E-ISSN 1937-9234, Vol. 10, no 2, p. 749-760Article in journal (Refereed) Published
##### Abstract [en]

The fastest growing of multimedia services overfuture wireless communication system demand more networkresources, efficient delivery of multimedia service with highusers satisfaction, and power optimization of User Equipments(UEs). The resources and power optimization are significant infuture mobile computing systems, because emerging multimediaservices consume more resources and power. The 4G standard ofLTE-A wireless system has adopted the Discontinuous Reception(DRX) method to extend and optimize the UE battery life,while there is no standard scheduling method to distribute theradio resources among the UE. This paper presents a downlinkscheduler, i.e. Quality of Experience (QoE) Power EfficientMethod (QEPEM) for LTE-A, which efficiently allocates theradio resources and optimizes the use of UE power using theDRX mechanism. We investigates how the different durationof DRX Light and Deep Sleep cycle influences the QoS andQoE of end users, using VoIP over the LTE-A. The QEPEMis evaluated with the traditional methods, in terms of SystemThroughput, Fairness Index, Packet Loss Rate, and PacketDelay. The QEPEM measures the user’s QoE, and feedback toeNodeB for scheduling decisions along with other importantparameters. Our proposed method reduces the packet delay,packet loss, and increases the fairness and UE’s power savingwith high users satisfaction.

##### Place, publisher, year, edition, pages
IEEE Communications Society, 2016
##### Keywords
QoE; QoS; LTE; Power Saving; Scheduling; VoIP; DRX; Light Sleep; Deep Sleep; Multimedia
##### National Category
Communication Systems
##### Identifiers
urn:nbn:se:liu:diva-117941 (URN)10.1109/JSYST.2015.2435994 (DOI)000383258600033 ()
Available from: 2015-08-22 Created: 2015-05-18 Last updated: 2017-12-04Bibliographically approved
Fowler, S., Shahidullah, A. O., Osman, M., Karlsson, J. M. & Yuan, D. (2015). Analytical evaluation of extended DRX with additional active cycles for light traffic. The International Journal of Computer Networks (COMNET), Elsevier, 77, 90-102
Open this publication in new window or tab >>Analytical evaluation of extended DRX with additional active cycles for light traffic
2015 (English)In: The International Journal of Computer Networks (COMNET), Elsevier, ISSN 1389-1286, Vol. 77, p. 90-102Article in journal (Refereed) Published
##### Abstract [en]

Abstract LTE and LTE-Advanced mobile technologies have integrated discontinuous reception (DRX) power saving method to optimize the power consumption at the user equipment (UE). The DRX method was proposed by the 3rd Generation Partnership Project (3GPP), and since then, the traffic behavior has been analyzed in several studies with a standard 3-state DRX model to describe the trade-off between power saving and delay. In this paper, we presented a novel 4-state and 5-state 3GPP LTE DRX mechanisms. The proposed mechanisms were developed by augmenting (an) active state(s) to deep and/or light sleep cycle of standard 3-state DRX for handling a small burst of packets, thereby bypassing the process of returning to the timer-dependent active mode. We have generated analytical models using a semi-Markov process for bursty packet data traffic and evaluated these augmented DRX mechanisms against a standard 3-state DRX method. Overall, the analytical results from varying timing parameters showed that our augmented DRX (both 4-state and 5-state) improved power saving factor (ranging between 1% and 8%) and reduced delay (ranging between 20% and 60%) compared to the standard 3-state DRX. Furthermore, the magnitude of improvement for both delay and power-saving was somewhat greater in 5-state than 4-state.

Elsevier, 2015
##### Keywords
LTE, LTE-Advanced, DRX, Power saving, Delay, Semi-Markov process
##### National Category
Other Electrical Engineering, Electronic Engineering, Information Engineering
##### Identifiers
urn:nbn:se:liu:diva-114286 (URN)10.1016/j.comnet.2014.11.020 (DOI)000349502300006 ()
##### Projects
EC-FP7 Marie Curie CIG Grant, Proposal number: 294182
##### Funder
EU, FP7, Seventh Framework Programme Available from: 2015-02-17 Created: 2015-02-17 Last updated: 2016-03-02Bibliographically approved
Fowler, S., Sarfraz, J., Abbas, M. M. & Angelakis, V. (2015). Gaussian semi-Markov Model based on Real Video Multimedia Traffic. In: IEEE (Ed.), IEEE International Conference on Communications Communications (ICC) 2015: . Paper presented at IEEE ICC 2015 - Communications Software, Services and Multimedia Applications Symposium, 8-12 June 2015, London, UK (pp. 6971-6976). IEEE Press
Open this publication in new window or tab >>Gaussian semi-Markov Model based on Real Video Multimedia Traffic
2015 (English)In: IEEE International Conference on Communications Communications (ICC) 2015 / [ed] IEEE, IEEE Press, 2015, p. 6971-6976Conference paper, Published paper (Refereed)
##### Abstract [en]

The 3rd Generation Partnership Project (3GPP) introduced the new radio access technology, LTE (Long Term Evolution) and LTE-Advanced, which has the capability to provide larger bandwidth and low latencies on a wireless network in order to fulfill the demand of Users' Equipment (UEs) with acceptable Quality of Service (QoS). One of the data-heavy applications that has exploded on the market is Video. This calls for accurate modeling of video traffic. To ensure the quality and correctness of complex systems with Video traffic, it is important to evaluate the behavior of the traffic in the heterogeneous environment. This paper presents measurements on LTE Video traffic which were performed in a real environment. We derived a semi-Markov model which accurately reproduces the statistics of composite Video measurements over LTE network.

IEEE Press, 2015
##### Series
IEEE International Conference on Communications, ISSN 1550-3607
##### Keywords
LTE; semi-Markov; lognormal; Gaussian Mixture Model; Real Video Traffic; Empirical Data
##### National Category
Telecommunications
##### Identifiers
urn:nbn:se:liu:diva-117277 (URN)10.1109/ICC.2015.7249437 (DOI)000371708107033 ()
##### Conference
IEEE ICC 2015 - Communications Software, Services and Multimedia Applications Symposium, 8-12 June 2015, London, UK
##### Projects
EC-FP7 Marie Curie CIG grant, Proposal number: 294182;European Unions Seventh Framework Programme (FP7/2007-2013) under grant agreement no [609094 RERUM].
##### Funder
EU, FP7, Seventh Framework Programme, 609094EU, FP7, Seventh Framework Programme, 294182 Available from: 2015-08-22 Created: 2015-04-22 Last updated: 2016-04-20Bibliographically approved
Bitam, S., Mellouk, A. & Fowler, S. (2015). MQBV: multicast quality of service swarm bee routing for vehicular ad hoc networks. Wireless Communications & Mobile Computing, 15(9), 1391-1404
Open this publication in new window or tab >>MQBV: multicast quality of service swarm bee routing for vehicular ad hoc networks
2015 (English)In: Wireless Communications & Mobile Computing, ISSN 1530-8669, E-ISSN 1530-8677, Vol. 15, no 9, p. 1391-1404Article in journal (Refereed) Published
##### Abstract [en]

Vehicular ad hoc networks (VANETs) are witnessing in recent years a rapid development for road transmissions and are considered as one of the most important types of next generation networks, in which drivers can have access anywhere and anytime to information. However, vehicles have to deal with many challenges such as the links failures due to their frequent mobility as well as limited degrees of freedom in their mobility patterns. In this paper, we propose a new quality of service multicast and multipath routing protocol for VANETs, based on the paradigm of bees communication, called multicast quality of service swarm bee routing for VANETs (MQBV). The MQBV finds and maintains robust routes between the source node and all multicast group members. Therefore, the average end-to-end delay and the normalized overhead load should be reduced, while at the same time increasing the average bandwidth and the packet delivery ratio. Extensive simulation results were obtained using ns-2 simulator in a realistic VANET settings and demonstrated the efficiency of the proposed protocol. Copyright (c) 2013 John Wiley and Sons, Ltd.

##### Place, publisher, year, edition, pages
Wiley: 12 months, 2015
##### Keywords
vehicular ad hoc network; routing; bee swarm; quality of service; road safety
##### National Category
Civil Engineering
##### Identifiers
urn:nbn:se:liu:diva-120040 (URN)10.1002/wcm.2420 (DOI)000355738200008 ()
Available from: 2015-07-06 Created: 2015-07-06 Last updated: 2017-12-04
Mushtaq, M. S., Fowler, S., Augustin, B. & Mellouk, A. (2015). QoE in 5G Wireless Cellular Networks based on Mobile Cloud Network. In: : . Paper presented at IEEE INFOCOM Workshop (MMCloudCom). IEEE conference proceedings
Open this publication in new window or tab >>QoE in 5G Wireless Cellular Networks based on Mobile Cloud Network
2015 (English)Conference paper, Published paper (Refereed)
##### Abstract [en]

The 4G standard Long Term Evolution-Advanced(LTE-A) has been deployed in many countries. Now, technologyis evolving towards the 5G standard since it is expecting tostart its services in 2020. The 5G cellular networks will mainlycontain in cloud computing and primarily Quality of Service(QoS) parameters (e.g. delay, loss rate, etc.) influence the cloudnetwork performance. The impact of user perceived Quality ofExperience (QoE) significantly relies on the QoS parameters. Thekey challenge of 5G technology is to reduce the delay less thanone millisecond. In this paper, we have described a method thatminimizes the overall network delay for constant bit rate (VoIP)and variable bit rate (video) traffic model. We also proposed amethod that measures the user’s QoE for video streaming trafficusing the network QoS parameters, i.e. delay and packet lossrate. The performance of proposed QoE method is comparedwith QoV method, and our proposed QoE method performsbest by carefully handle the impact of QoS parameters. Theresults show that our described method successfully reduces theoverall network delays, which result to maximize the user’s QoE.

##### Place, publisher, year, edition, pages
IEEE conference proceedings, 2015
##### Keywords
QoE; 5G; Mobile Cloud Computing; LTE; LTE-A: QoS; Video; VoIP;Cloud
##### National Category
Communication Systems
##### Identifiers
urn:nbn:se:liu:diva-117748 (URN)
##### Conference
IEEE INFOCOM Workshop (MMCloudCom)
Available from: 2015-08-22 Created: 2015-05-08 Last updated: 2015-10-21Bibliographically approved
##### Identifiers
ORCID iD: orcid.org/0000-0002-0019-8411

#### Search in DiVA

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text:
Free Version
Moderate

# Graphical Interpretation: Comparing Values

APCALC-M5RESL

Given the graph of $f(x)$ above (going through the listed points), which of the following statements is true?

A

$f''(-3) < f'(-3) < f(-3)$

B

$f(-3) < f''(-3) < f'(-3)$

C

$f'(-3) < f(-3) < f''(-3)$

D

$f"(-3) < f(-3) < f'(-3)$


text:
# Don't wait for luck

Geometry Level 2

Find area enclosed by $$2|x| + 3|y| \leq 6$$ in square units.

Details And Assumptions:

• $$|a|$$ is the distance of "$$a$$" from origin. For example , $$|9| = 9$$ and $$|-9| = 9$$ .
×


text:
You’ll find, most likely, a variety of posts, spanning many topics. They form, together, a compendium of my thoughts, observations, musings, scribbles, guides, memoirs, and other writings.

In fact, they form a better about me than this page can. However, in short here are some things you can expect in further reading:

Personal favourites are:

• $mathematics$ – in particular type and category theory,
• computer science,
• theology,
• music

to name a few.

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text:
# Thread: problems with logs and a proof.

1. ## problems with logs and a proof.

1. I used the intermediate value theorem for (square root x to the third)= x^2+2x-1 on (0,1), I plugged 0 and 1, and I got (left side)0=-1(right side) and 1=2 respectively. Since I got a negative(on 0), and positive on the other(on 1), there is at least one solution for (0,1), is this right? If I have the wrong idea, tell me. 2. 1/2 lnx/ln3=ln8/ln2. ans. 729. 3. (square root to the third 2)^x + 10=2^(x^2). ans. 2, -5/3 4.(ln x/ln 3)+(ln 2x+1/ln 3)=1 ans. 1 I do not know how to put the small numbers on logs so I put it in natural log form. Thanks in advance.

2. Originally Posted by driver327
1. I used the intermediate value theorem for (square root x to the third)= x^2+2x-1 on (0,1), I plugged 0 and 1, and I got (left side)0=-1(right side) and 1=2 respectively. Since I got a negative(on 0), and positive on the other(on 1), there is at least one solution for (0,1), is this right? If I have the wrong idea, tell me.
Wow. Ummmm... Let me break this down a bit. You've got the function:
$\sqrt{x^3} = x^2 + 2x - 1$
on the interval (0, 1).

Are you trying to find a solution for x?

If so then try moving everything to the left:
$\sqrt{x^3} - (x^2 + 2x - 1) = 0$

NOW plug in x = 0 and x = 1:
$\sqrt{0^3} - (0^2 + 2 \cdot 0 - 1) = 1$

$\sqrt{1^3} - (1^2 + 2 \cdot 1 - 1) = -1$

Because the first is positive and the second negative we know that it (the LHS) passes through 0 at some point in the interval.

-Dan

3. Originally Posted by driver327
2. 1/2 lnx/ln3=ln8/ln2. ans. 729.
Again, solve for x?

$\frac{1}{2} \frac{ln(x)}{ln(3)} = \frac{ln(8)}{ln(2)}$

Now,
$ln(8) = ln(2^3) = 3 \cdot ln(2)$, so

$\frac{1}{2} \frac{ln(x)}{ln(3)} = \frac{3 \cdot ln(2)}{ln(2)}$

$\frac{1}{2} \frac{ln(x)}{ln(3)} = 3$

$\frac{ln(x)}{ln(3)} = 6$

$ln(x) = 6 \cdot ln(3)$

$ln(x) = ln(3^6) = ln(729)$

$x = 729$

-Dan

4. Originally Posted by driver327
3. (square root to the third 2)^x + 10=2^(x^2). ans. 2, -5/3
Is this $(\sqrt{2}~^3)^x + 10 = 2^{x^2}$?

-Dan

5. Originally Posted by driver327
4.(ln x/ln 3)+(ln 2x+1/ln 3)=1 ans. 1
Is this supposed to be
$\frac{ln(x)}{ln(3)} + \frac{ln(2x + 1)}{ln(3)} = 1$?

$ln(x) + ln(2x + 1) = ln(3)$

$ln(x(2x + 1)) = ln(3)$

$x(2x + 1) = 3$

$2x^2 + x - 3 = 0$

$(2x + 3)(x - 1) = 0$

So
$x = -\frac{3}{2}$
or
$x = 1$

But a negative x is outside the domain of the expression on the LHS. Thus the only solution is x = 1.

-Dan

6. Originally Posted by topsquark
Is this $(\sqrt{2}~^3)^x + 10 = 2^{x^2}$?

-Dan
The 3 was on top of the square root, and the +10 was part of the exponent. Sorry for the confusion.

7. Originally Posted by driver327
The 3 was on top of the square root, and the +10 was part of the exponent. Sorry for the confusion.
So it's
$\sqrt[3]{2^{(x+10)}} = 2^{x^2}$ <-- This I can work with!

Well, $\sqrt[3]{x} = x^{1/3}$ so:
$(2^{(x+10)})^{1/3} = 2^{x^2}$

$(2^{(x+10)/3}) = 2^{x^2}$

Thus
$\frac{x + 10}{3} = x^2$

$x + 10 = 3x^2$

$3x^2 - x - 10 = 0$

$(3x + 5)(x - 2) = 0$

Thus
$x = -\frac{5}{3}$ or $x = 2$.

Both solutions are in the domain of the original expression, so both are solutions.

-Dan


text:
# regular languages – Simplify \$R:=0^*+0^*1left(1+000^*1right)^*0^*\$

I’m trying to simplify the following REGEX:

$$R:=0^*+0^*1left(1+000^*1right)^*0^*$$

$$R$$ is the result of transforming a GNFA that recognizes $$L:= {w in {0,1}^* | left(forall i in left(1,|w|-2right)cap mathbb{N}right) w_iw_{i+1}w_{i+2} neq 101}$$

I already know that there exists $$R’:=0^*left(1+000^*1right)^*0^*$$ such that $$L(R)=L(R’)=L$$

Any hints?


text:
# How to export a formula from a SAS model

I am a bit of a SAS novice so please forgive my ignorance.

I have a generalized Linear mixed model in SAS based on past data but I don't know how to export it to a pure mathematical formula that could be used to generates probabilities varying different inputs.

For example, I want to be able to input a values to different parameters and get a probability for that instance. Is there an easy way to export the mathematical model from SAS?

Yes there are various ways to export your model. The quickest and easiest method that comes to my mind is using ODS output tables

You should look specifically for ParameterEstimates output table. This table (or dataset) will contain all the estimated parameters of your model. From here you can exported to R, Excel or use it with Proc Score within SAS to score new observations.

So your code should look like this:

ods output ParameterEstimates=ParameterEstimates;

proc GLMMIX or GML;
Model ...;
<more options> ...;
run;

Here is a link to all the output tables available for GLMMIX.

Hope it helps

it depends on exactly what's needed but sas can actually provide the predicted values for you. I assume you're using genmod, thus, see 'predicted' in the output statement: https://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_genmod_sect023.htm, or "plots=predicted" on the genmod statement


text:
Publication
Title
Comparison of dynamic simulations with RBS measurements of low energy ion implantation of $Sb^{+}$ into $SiO_{2}/Si$ substrates
Author
Abstract
 Antimony ions were implanted into SiO2/Si substrates at energies of 30 and 50 keV at relatively low doses (max. 5 x 10(15) ions cm(-2)). The behaviour of the implanted ions was Simulated with a dynamic TRIM code, allowing treating alteration of the target in a dynamic mode, i.e. as a function of ion dose. For cases, where the projected range of the implant coincides with the interface, it has been observed that there is significant mixing between the SiO2 layer and the Si substrate together with a constant value for the O/Si ratio except for a pronounced oxygen surface peak. Whereas in the cases, where the implant is far from the interface, the O/Si ratio shows characteristic oscillations together with the surface peak. For high doses, above about 10(16) ions cm(-2), a gradual increase of the sputtering yield Of SiO2 is observed due to Sb incorporation in the substrate. The balance between the sputtering and swelling of the target is shown to be negative up to a fluence of 9.5 x 10(15) ions cm(-2), although the total sputtering yield is more than 1. RBS (Rutherford Backscattering Spectrometry) measurements confirm the mixing at the interface in cases where the latter coincides with the projected range of the implanted ions.
Language
English
Source (journal)
Microchimica acta
Publication
2004
Volume/pages
145(2004), p. 67-74
ISI
000221453500012
Full text (Publisher's DOI)
Full text (publisher's version - intranet only)
UAntwerpen
 Faculty/Department Research group Publication type Subject Affiliation Publications with a UAntwerp address


text:
# Finding the equation of a circle given the tangent line and a point on that line.

• March 15th 2014, 07:37 PM
RandomInquirer
Finding the equation of a circle given the tangent line and a point on that line.
I've solved some of them and the rest I don't really understand how to go about.
1. How do I find the equation of the circle if it is tangent to the line -3x+2y+5 = 0 at the point (1, -1)?

I'm stumped on this one since I don't know how I'd be able get any of the details through equations. I don't have the center point so I can't use the perpendicular distance formula/look for perpendicular line by finding the inverse of the slope (or can I? I'm not sure.)

I'm still quite confused about the circle. And I'm not really sure if the concepts I'm using are correct or not. I'm not sure if I can do this and that for this problem and that problem, but I'm slowly learning. I don't have a proper background on it since I'm just searching willy nilly.

--
If you guys don't mind I have follow-ups. You don't need to give me the answers, it would help if you can explain what I'm supposed to do and how to go about doing it (Beginner level formulas for this, I guess).

1. Given circle is tangent to the line -x+y+4 = 0 at point (3, -1) and the circle's center is on the line x + 2y -3 = 0, how will I find the equation of the circle?

2. Is there a faster way to find out the equation of the circle inscribed in the triangle?
Take this for example: SOLUTION: Find the equation of the circle that is inscribed in the triangle whose sides lie on the lines 3x+y=5,x-3y=1,and x+3y=-7.
Also in that page why is it that there are only four solutions to the system. Can't the equations on the left side be negative as well?
Here's another example which is shorter but I totally don't understand how she figured out the signs.
SOLUTION: find the equation of the circle inscribed in a triangle whose sides are 3x+y-5=0, x+3y-1=0, x-3y+7=0.

Thanks for the help guys. It's greatly appreciated.
• March 15th 2014, 08:47 PM
Soroban
Re: Finding the equation of a circle given the tangent line and a point on that line.
Hello, RandomInquirer!

Quote:

2. Is there a faster way to find out the equation of the circle inscribed in the triangle?

Example: Find the equation of the circle that is inscribed in the triangle whose sides
lie on the lines: . $3x+y\:=\:5,\;\;x-3y\:=\:1,\;\;x+3y\:=\:-7.$

The distance from a point $(x_1,y_1)$ to a line $Ax + By + C \:=\:0$

. . is given by: . $d \;=\;\frac{|Ax_1 + By_1+C|}{\sqrt{a^2+B^2}}$

We want a point $(h,k)$ which is equidistant from: $\begin{Bmatrix}3x+y-5 &=& 0 \\ x-3y - 1 &=& 0 \\ x+3y+7 &=& 0 \end{Bmatrix}$

$\text{We have: }\;d \;=\;\underbrace{\frac{3h+k-5}{\sqrt{10}}}_{[1]} \;=\;\underbrace{\frac{h-3k-1}{\sqrt{10}}}_{[2]} \;=\;\underbrace{\frac{h+3k+7}{\sqrt{10}}}_{[3]}$

$[2] = [3]\!:\;h-3k-1 \:=\:h+3k+7 \quad\Rightarrow\quad -6k \:=\:8 \quad\Rightarrow\quad k \,=\,\text{-}\tfrac{4}{3}\;\;[4]$

$[1] = [2]\!:\;\;3h+k-5 \:=\: h-3k-1 \quad\Rightarrow\quad h \:=\:2 - 2k\;\;[5]$

Substitute [4] into [5]: . $h \:=\:2-2\left(\text{-}\tfrac{4}{3}\right) \quad\Rightarrow\quad h \:=\:\tfrac{14}{3}$

The center of the circle is: . $\left(\frac{14}{3},\;\text{-}\frac{4}{3}\right)$

The radius is: . $[1]\;\;r \;=\;\frac{3(\tfrac{14}{3}) + (\text{-}\frac{4}{3}) - 5}{\sqrt{10}} \;=\;\frac{23}{3\sqrt{10}}$

The equation of the circle is: . $\left(x-\frac{14}{3}\right)^2 + \left(y + \frac{4}{3}\right)^2 \;=\;\frac{529}{90}$
• March 15th 2014, 09:09 PM
RandomInquirer
Re: Finding the equation of a circle given the tangent line and a point on that line.
I used an online circle graph application to compare your answer to the answer given on the hyperlinked page and your answer ended up being the red circle right next to the inscribed circle.
Attachment 30420
From what I gathered, the circle with the smallest radius/diameter is the inscribed circle. Is this true for all or is there some sort of special case where it's not? I can't think of a case where it is false but that doesn't mean that somebody else has not.
• March 16th 2014, 09:04 AM
Soroban
Re: Finding the equation of a circle given the tangent line and a point on that line.
Hello again, RandomInquirer!

Quote:

1. Given circle is tangent to the line $-x+y+4 \,=\, 0$ at point $P(3, -1)$
and the circle's center is on the line $x + 2y -3 \,=\, 0,$
find the equation of the circle.

A tangent to the circle is perpendicular to the radius at the point of tangency.
The tangent is: $y \,=\,x-4$; its slope is $1$.
The slope of the radius is $\text{-}1.$

Hence, the center lies on the line: . $y+1 \,=\,\text{-}1(x-3) \quad\Rightarrow\quad x+y \,=\,2$

The center also lies on the line: . $x + 2y \,=\,3$

Solve the system.
. . The center is $C(1,1).$

The radius is: . $CP \:=\:\sqrt{(3-1)^2 + (-1-1)^2} \:=\:\sqrt{8}$

Therefore: . $(x-1)^2 + (y-1)^2 \:=\:8$
• March 16th 2014, 10:57 AM
RandomInquirer
Re: Finding the equation of a circle given the tangent line and a point on that line.
Thank you for your help. It's much appreciated!
• March 17th 2014, 07:45 PM
johng
Re: Finding the equation of a circle given the tangent line and a point on that line.
Hi,
Rarely does Soroban make an error, but he did this time; nobody bats a thousand. He didn't choose the right sign for the distance of (h,k) to 3x+y-5=0. Your first link shows a correct solution, but the second link also has an error. I don't see any easy way to solve this problem The two attachments try to explain:

Attachment 30439

Attachment 30440
• March 17th 2014, 08:18 PM
SlipEternal
Re: Finding the equation of a circle given the tangent line and a point on that line.
Quote:

Originally Posted by RandomInquirer
I've solved some of them and the rest I don't really understand how to go about.
1. How do I find the equation of the circle if it is tangent to the line -3x+2y+5 = 0 at the point (1, -1)?

All circles have equations of the form $(x-h)^2 + (y-k)^2 = r^2$. Using implicit differentiation, we can find the derivative: $\dfrac{dy}{dx} = \dfrac{h-x}{y-k}$

The line $-3x+2y+5=0$ has slope $\dfrac{3}{2}$

Plugging that in, you get $\dfrac{3}{2} = \dfrac{h-1}{-1-k}$

Additionally, you know $(1,-1)$ is a point on the circle, so $(1-h)^2 + (-1-k)^2 = r^2$.

Solving for $k$ in the first equation, we get: $k = -\dfrac{2h+1}{3}$

Plugging into the second equation, we get:

$(1-h)^2 + \left(\dfrac{2h-2}{3}\right)^2 = r^2$

$r^2 = \dfrac{13}{9}(1-h)^2$

Thus, the circle's formula is

$(x-h)^2 + \left(y+\dfrac{2h+1}{3}\right)^2 = \dfrac{13}{9}(1-h)^2$ where $h \in \mathbb{R}$ (it gives you an infinite family of circles).