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# Basic elements of statistical decision theory and statistical  (Page 3/5)

 Page 3 / 5

The CRLB states that under some mild regularity assumptions about the conditional density function ${p}_{X|Y}\left(x|y\right)$ , the variance of any unbiased estimator is bounded from below by the inverse of the $I\left({y}^{*}\right)$ [link] , [link] , [link] . Recall that an unbiased estimator is any estimator $\stackrel{^}{Y}$ that satisfies $E\left[\stackrel{^}{Y}\right]={y}^{*}$ . The CRLB tells us is that

$\text{var}\left(\stackrel{^}{Y}\right)\phantom{\rule{4pt}{0ex}}\ge \phantom{\rule{4pt}{0ex}}\frac{1}{I\left({y}^{*}\right)}.$

If $Y$ is a vector-valued quantity, then the expected negative Hessian matrix(matrix of partial second derivatives) of the log-likelihood function is called the Fisher Information Matrix (FIM), and a similar inequality tells us that the varianceof each component of any unbiased estimator of ${y}^{*}$ is bounded below by the corresponding diagonal element of the inverse of the FIM.Since the MSE of an unbiased estimator is equal to its variance, we see that the CRLB provides a very useful lower bound on the best MSEperformance that we can hope to achieve. Thus, the CRLB is often used as a comparison point for evaluating estimators. It may or may not bepossible to achieve the CRLB, but if we find a decision rule that does, we know that it also minimizes the MSE risk among all possibleunbiased estimators. In general, it may be difficult to compute the CRLB, but in certain important cases it is possible to findclosed-form or computational solutions.

## Bayesian decision theory

Bayesian Decision Theory provides a formal system for integrating prior knowledge and observed observations. Forthe purposes of illustration we will focus on problems involving continuous variables and observations, but extensions to discretecases are straightforward (simple replace probability densities with probability mass functions, and integrals with summations). The keyelements of Bayesian methods are:

1. a prior probability density function ${p}_{Y}\left(y\right)$ describing a priori knowledge of probable states for the quantity $Y$ ;
2. the likelihood function ${p}_{X|Y}\left(x|y\right)$ , as described above;
3. the posterior density function ${p}_{Y|X}\left(y|x\right)$ .

The posterior density is a function of the prior and likelihood, obtained according to Bayes rule:

${p}_{Y|X}\left(y|x\right)=\frac{{p}_{X|Y}\left(x|y\right){p}_{Y}\left(y\right)}{\int {p}_{X|Y}\left(x|y\right){p}_{Y}\left(y\right)dy}.$

The posterior is an indicator of probable values for $Y$ , based on the prior knowledge and the observation. Several options exist for deriving a specific estimateof $Y$ using the posterior. The mean value of the posterior density is one common choice (commonly called the posterior mean ). The posterior mean is the decision rule that minimizes the expectedsquared error loss (MSE risk) function. The value $y$ where the posterior density is maximized is another popular estimator (commonlycalled the Maximum A Posteriori (MAP) estimator). Note that the denominator of the posterior is independent of $y$ , so the MAP estimator is simply the maximizer of the product of the likelihood andthe prior. Therefore, if the prior is a constant function, the MAP estimator and MLE coincide.

## Statistical learning

In all of the methods described above, we assumed some amount of knowledge about the distributions of the observation $X$ and quantity of interest $Y$ . Such knowledge can come from a careful analysis of the physical characteristics of the problem at hand, or it can begleaned from previous experience. However, there are situations where it is difficult to model the physics of the problem and we may nothave enough experience to develop complete and accurate probability models. In such cases, it is natural to adopt a statistical learning approach [link] , [link] .

#### Questions & Answers

I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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# Metrics learning

TODO : more detail

# Personally thinking

## Can we use object proposal algorithm (such as BING, DeepBox) to perform board segmentation

-> huhm, this is nonsense anyway

## One shot learning for semantic segmentation or even object detection

one-shot learning : Matching Networks Input: a. anchor : base image b. other test image One-Shot : object detection Output : which image is match

One-Shot Instance Segmentation Given a query image and a reference image showing an object of a novel category, we seek to detect and segment all instances of the corresponding category (in the image above ‘person’ on the left, ‘car’ on the right). https://github.com/bethgelab/siamese-mask-rcnn : 2018 https://github.com/timy90022/One-Shot-Object-Detection : 2019

-> very promising , but it's hard to implemented

## One shot object detection

Found the following 2 promissing approach

1. RepMet: Representative-based metric learning for classification and one-shot object detection https://github.com/jshtok/RepMet : 2018
2. One-Shot Object Detection with Co-Attention and Co-Excitation
https://github.com/timy90022/One-Shot-Object-Detection : 2019

Let’s compare the 1. and 2.

1. DML : only learn the baseline
2. Re-write the entire network architecture

-> We will use 2 for this task

apple has very great turi examples at: https://apple.github.io/turicreate/docs/userguide/one_shot_object_detection/ -> if we need to re-train an one-shot-object-detection then this is a good resource -> don’t know which algorithm are they using

Other

1. https://github.com/bingykang/Fewshot_Detection -> Few-shot Object Detection via Feature Reweighting -> few-shot problems : have the base model, then input some random images, -> we-erighting -> the model will have a very good feature nows -> very strange, maybe we need to read this deeper

# few-shot-learning

https://blog.floydhub.com/n-shot-learning/

N-shot learning = N sample for training few-shot learning : few = 1~5 -> zero shot learning -> one shot learning -> few-shot learning

training set = support set -> K classes, N examples per class testing set = query set -> K classes, a lot of example per class

-> reduce the size of training set as small as possible -> don’t know why they changed the name

sub categories:

zero-shot learning : do the task on unseed object
one-shot learning : do the task while using only 1 samples per class to learn
-> take the advantage of siamese (metrics learning)
few-shot learning : do the task while using only few samples per class to learn

How to measure performance: n-shot, k-way (n-example, k-class)

-> let’s dive a little bit

Zero-shot learning Learning to Compare: Relation Network for Few-Shot Learning Learning Deep Representations of Fine-Grained Visual Descriptions Improving zero-shot learning by mitigating the hubness problem

One-Shot Learning : siamese network , matching networks Model-Agnostic Meta-Learning for Fast Adaptation of Deep Networks One-shot Learning with Memory-Augmented Neural Networks Prototypical Networks for Few-shot Learning

Few-Shot learning -> just extend version of one-shot learning

-> famouse network : prototypical networks -> learn the mapping of an image in metric space (not vector space). -> metrics space : does not has origin -> vector space : has origin

-> now understand what is few-shots learning

—> bat dau khong hieu van de roi, doc lai cho ro nao

# meta-learning

https://lilianweng.github.io/lil-log/2018/11/30/meta-learning.html -> very good writing

https://www.sicara.ai/blog/2019-07-30-image-classification-few-shot-meta-learning

https://en.wikipedia.org/wiki/Meta_learning_(computer_science) -> a good warm up

meta-learning : learning to learn learning new concepts and skills fast with a few training examples? That’s essentially what meta-learning aims to solve.

A classifier trained on non-cat images can tell whether a given image contains a cat after seeing a handful of cat pictures. A game bot is able to quickly master a new game. A mini robot completes the desired task on an uphill surface during test even through it was only trained in a flat surface environment.

-> common approaches

1. Model-Based Model-based meta-learning models updates its parameters rapidly with a few training steps, which can be achieved by its internal architecture or controlled by another meta-learner model -> Memory-Augmented Neural Networks -> Meta Networks
2. Metrics-Based (metrics learning) The core idea in metric-based meta-learning is similar to nearest neighbors algorithms, which weight is generated by a kernel function. It aims to learn a metric or distance function over objects. The notion of a good metric is problem-dependent. It should represent the relationship between inputs in the task space and facilitate problem solving[ -> Siamese network -> Matching Network -> Relation network -> Prototypical Networks
3. Optimization-Based What optimization-based approach meta-learning algorithms intend for is to adjust the optimization algorithm so that the model can be good at learning with a few examples[7].

-> LSTM Meta-Learner -> Temporal Discreteness -> Reptile

Few-shot object recognition became a hot topic recently (from 4 few-shot papers in CVPR18 to around 20 in CVPR19). The usual setup is that you have categories with many examples you can use at training time; then at test time, you are given novel categories (usually 5) with only a few examples per category (usually 1 or 5; called “support-set”) and query images from the same categories. Next, I’ll try to break the few-shot methods to different families. Although these families are not well defined and many methods belong to more than one family. “Older” suggested methods were based on metric learning, where the objective is to learn a mapping from images to embedding space in which images from the same category are closed together and from different categories are far apart. Hoping that it will hold true for unseen categories. Following that, came meta-learning methods. These are models which are conditioned on the current task, so a different classifier is used as a function of the support-set. The idea is to find model hyper-parameters and parameters such that it will be easy to adapt to a new task without over-fitting to the few shots available. In parallel, data augmentation methods are also very popular. The idea is to learn data augmentation so we can generate more examples out of the few examples available. Finally, semantics-based methods are on the rise. It is inspired by zero-shot learning where classification is done based solely on the category name, textual descriptions, or attributes. Those extra semantic ques can also be of help when visual examples are scarce.

# Few Shot Semantic Segmentation Papers

One-Shot Learning for Semantic Segmentation -> 2017

https://github.com/xiaomengyc/Few-Shot-Semantic-Segmentation-Papers

# Incremental learning (continuous learning)

Continual Learning (CL) is built on the idea of learning continuously and adaptively about the external world and enabling the autonomous incremental development of ever more complex skills and knowledge.

In the context of Machine Learning it means being able to smoothly update the prediction model to take into account different tasks and data distributions but still being able to re-use and retain useful knowledge and skills during time.

# A Primer in BERTology: What we know about how BERT works

https://arxiv.org/pdf/2002.12327.pdf

# Camera calibration

matlab camera calibration

extrinsic : world points -> camera coordinates : extrinsics parameters intrinsic : camera coordinates -> image plane : intrinsics parameters

matlab one camera calibration

reason : re-calculate the extrinsics, intrinsic How to evaluate : re-projector error: +/- ??? (in pixel), lower = better examining camera extrinsics : see matlab page for detail View Undistorted Image : see matlab page for detail

opencv camera calibration and 3d reconstruction

Camera calibration With OpenCV cpp

camera calibration with opencv python

format: plain


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# Tag Info

17

You are correct. Position is a vector when you are working in a vector space, since, well, it is a vector space. Even then, if you use a nonlinear coordinate system, the coordinates of a point expressed in that coordinate system will not behave as a vector, since a nonlinear coordinate system is basically a nonlinear map from the vector space to ...

4

Great reasoning: as in Uldreth's fantastic answer but I would add one more thing that may help cement your good understanding in place. Co-ordinates are absolutely not vectors, they are labels on charts and are no more vectors than your street address is a vector. Almost certainly the reason people make the implication that you have correctly identified as ...

4

To my mind, Newton's equation makes the most sense as an equation of vector fields. Let $(M,g)$ be a (Pseudo-)Riemannian manifold with Riemannian connection $\nabla$. Then the equations of motion for a position-dependent conservative force $F$ are given by $$m {}^{\gamma}\nabla_{\frac{d}{dt}} \frac{d\gamma}{dt}=F\circ \gamma=-(\nabla V) \circ \gamma,$$ where ...

3

TL;DR: It is the wedge product $\wedge$ and the exterior derivative/differential $d$ (which squares to zero $d^2=0$) that give rise to Grassmann-odd elements and supersymmetry. More concretely, Ref. 1 first writes down a (non-relativistic) SUSY algebra ${\cal A}$ $$\tag{10} \{Q_1,Q_1\}_+~=~2H~=~\{Q_2,Q_2\}_+, \qquad \{Q_1,Q_2\}_+~=~0,$$ spanned by two ...

2

Let $\mathcal{M}$ be our spacetime. Then, a gauge theory is given by a connection form $A$ on a principal bundle over it (that locally projects onto the spacetime in a way compatible with gauge transformations), which is the gauge potential. Maxwell's equations1 (in vacuum) are the equations of motion for the gauge field for the Yang-Mills action coupled to ...

2

Here is a bare bones easy way to see that coordinate tuples are not 4-vectors. Start in an inertial coordinate system in flat spacetime. Change the coordinate system with a constant translation: $x' = x + A$ $y' = y$ $z' = z$ $t' = t$ Even in this idealistic case, 4-vectors and coordinate tuples transform differently. The components of the 4-vectors ...

2

I believe you and the author are referring to different "gradients". You mean the gradient along the manifold defined by the particle trajectory; since this is a one dimensional manifold, the gradient would indeed have only one component. The authors are referring to the full three-space gradient. Using the formulae $$r = \frac{1}{\kappa}+x \\ \phi = \kappa ... 2 I do not agree with the answer given by @ACuriousMind. @Scardenalli has asked for a compact Ricci-flat Riemannian manifold M having as isometry group U(1)\times SU(2)\times SU(3). This does not imply that M must be a symmetric Riemannian manifold. However, the answer to @Scardenalli's question is still no, and it follows from a classical result in ... 2 Well, the very notion of curvature relies on the Riemann curvature tensor, in fact, the curvature tensor arises from the fact that in a general curved space, the second covariant derivatives may not commute: A_{v;ps} - A_{v;sp} = A_{b}R^{b}_{vps}. That being said, however, I think at minimum to determine whether something is curved or not, in a general ... 1 Take your drawings, and imagine moving a vector around the loops you draw. Specifically, move the vector using parallel transport i.e. during each infinitesimal movement keep the vector parallel to its previous direction. If the space(time) is curved then when you have travelled around the loop you'll find the direction of the vector has changed. If we take ... 1 Yes, especially in research-level topics. There are several research groups that work with finding ways to apply differential geometry concepts to solid state systems (although condensed matter seems to be the preferred term nowadays). See for example the book by Altland and Simons, Condensed Matter Field Theory, Chapter 9 "Topology". This book is suitable ... 1 You can write$$\text dV = - m v \text d v$$where v is the velocity. 1 I think your limiting procedure should work. For the homogeneous Maxwell equations, when we restrict to a surface of constant x, the pull back of the field strength looks something like$$\iota^* F = E^\perp_i dt \wedge dx^i + B_x dy \wedge dz, where $E^\perp$ is the component of $E$ in the $yz$-plane. Now your timelike pillbox integral will have ...

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


text: Calcium reacts slowly with water. This is due to Calcium being oxidation state +2 and Oxygen oxidation state is -2, this means when formed in a compound they cancel each other out to give CaO. Calcium metal slowly oxidizes in air, becoming encrusted with white $$CaO$$ and $$CaCO_3$$, which protect from attack by air. The reactions with fluorine and chlorine are exothermic while the reactions with bromine and iodine are endothermic. Reactions with Group 2 Elements. The reaction is catalysed by the oxygen-evolving complex (OEC) — an inorganic cluster of four manganese ions and one calcium ion bridged by oxide ligands 2, buried in the photosynthetic protein complex known as Photosystem II (PSII). The digest is aspirated into an 8,000 K argon plasma where resulting light emission is quantified for 30 elements simultaneously. It also reacts with the oxygen and nitrogen in the air to form a mixture of calcium oxide and calcium nitride. Calcium can be ignited and will when burning react with both oxygen and nitrogen forming calcium oxide, CaO, and calcium nitride, Ca. In bulk, calcium is less reactive: it quickly forms a hydration coating in moist air, but below 30% relative humidity it may be stored indefinitely at room temperature. Why Will An Egg Turn To Rubber If Put In Vinegar? Reaction of Calcium and Oxygen. 2Ca(s) + O 2 (g) → 2CaO(s) 3Ca(s) + N 2 (g) → Ca 3 N 2 (s) Reaction of calcium with water. Answered Feb 28, 2018 When calcium reacts with oxygen calcium oxide is formed by a highly exothermic reaction: 2Ca (s) + O2 (g) → 2CaO (s) The reaction will be favored by enthalpy because of its … Calcium ions are precipitated by phosphate ions in alkaline solutions. Reaction of Calcium with Oxygen. Calcium ions are precipitated by sulfate ions in neutral to alkaline solutions. Calcium reacts with hydrogen, forming calcium hydride [6]. Method 3500-Ca C Inductively Coupled Plasma Method [1]. Calcium ions are precipitated by carbonate ions in neutral to alkaline solutions [6]: Calcium reacts readily halogens forming the corresponding dihalides. At room temperature, calcium reacts with oxygen, forming a thin layer of CaO, that protects the metal from further oxidation. Calcium ions are precipitated by oxalate ions. How Does Iron React With Oxygen And Water? When finely divided, it spontaneously burns in air to produce the nitride. Explain Why The Reaction Of Oxygen With Iron Is A Problem But The Reaction Of Oxygen With Aluminum Is Not? How Many Atoms Of Oxygen Are There In 5.79moles Of Calcium Carbonate? Basically there are 10 rules that show which elements and their oxidation numbers take priority in a reaction. Copyright, ownership information and cookies- and privacy policy. When ignited, Calcium burns to give calcium xxide. Calcium metal dissolves readily in dilute or concentrated hydrochloric acid forming Ca(II) ions and hydrogen gas, H, At room temperature, calcium reacts with oxygen, forming a thin layer of CaO, that protects the metal from further oxidation. Alkaline earth metals also react with oxygen, though not as rapidly as Group 1 metals; these reactions also require heating. Calcium oxide is more normally made by heating calcium carbonate. A portion of the sample is digested in a combination of acids. The elements of Group 2 are beryllium, magnesium, calcium, strontium, barium, and radioactive radium. Calcium ions can be precipitated by strong bases provided the concentration of calcium is sufficient. Calcium, immediately below magnesium in the periodic table is more reactive with air than magnesium. Will Oxygen Turn Lime Water Cloudy? 2 Ca (s) + O 2 (g) 2 CaO (s) 3 Ca (s) + N 2 (g) Ca 3 N 2 (s) Calcium reacts slowly with water, forming calcium hydroxide and hydrogen. Calcium ions can not be precipitated by sulfide. Where Does Calcium Carbonate Come From? Calcium can be ignited and will when burning react with both oxygen and nitrogen forming calcium oxide, CaO, and calcium nitride, Ca 3 N 2. Because calcium is essential for life, it can be found in all organisms, living or dead. Similarly to Group 1 oxides, most group 2 oxides and hydroxides are only slightly soluble in water and form basic, or alkaline solutions. what is the symbol equation for the reaction of calcium with oxygen? $2Ca \,(s)+ O_2 \,(g)\rightarrow 2CaO\,(s)$ Calcium in living organisms.


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시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 256 MB 0 0 0 0.000%

문제

In Spring of 2007, we were a little late in scheduling the contest, so that it ended up being on a weekend during study days. To “help” students make up for the time they were wasting at the contest, we called it the “Final Exams” Contest, and had all problems have study and exam themes. One of the problems asked students to figure out optimal ways to allocate study time across different courses.

Given that study time is rather limited, it is crucial to divide it up well for studying for various exams. In the end, you just want to maximize your GPA, so maybe it’s worth to put up with a C in one course (where a better grade would require a lot of studying) to secure A’s (which can be achieved with relatively little studying each) in all other courses. Clearly, this is a crucial optimization problem, and solving it well will probably benefit you tremendously.

입력

The first line contains a number K ≥ 1, which is the number of input data sets in the file. This is followed by K data sets of the following form:

The first line of a data set contains two numbers n, H, the number of courses you are taking (an integer n between 1 and 10) and the number of hours available for studying (an integer H ≤ 100).

This is followed by n lines, each containing 10 integers. These integers in line i will describe, in the order given, the number of hours you need to study in order to get an A (4.0), A- (3.7), B+ (3.3), B (3.0), B- (2.7), C+ (2.3), C (2.0), C- (1.7), D+ (1.3), D (1.0) in the ith of your courses. The numbers will be non-increasing. The interpretation is that if you study at least that many hours in the particular subject, you will get the given grade. If you don’t even study enough for a D, your grade will be F (0.0).

출력

For each data set, first output “Data Set x:” on a line by itself, where x is its number. Then, output the maximum possible GPA (rounded to two decimals) you can obtain by dividing your H hours of studying between the subjects.

Each data set should be followed by a blank line.

예제 입력

1
3 60
40 37 35 33 30 26 20 10 5 1
10 10 10 10 10 10 10 10 10 1
24 23 22 21 20 20 20 20 20 20

예제 출력

Data Set 1:
3.43


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## Some formulas for Apostol-Euler polynomials associated with Hurwitz zeta function at rational arguments.(English)Zbl 1189.11012

In [Pac. J. Math. 1, 161–167 (1951; Zbl 0043.07103)] T. M. Apostol introduced a generalization of the classical Euler polynomials as analogous definition of Apostol type for the so-called Apostol-Euler numbers and polynomials of higher order. The generalization, the Apostol-Euler polynomials $$\mathcal E_n^{(\alpha)}(x;\lambda)$$, is defined by means of the following generating function: $\left(\frac 2{\lambda e^z+1}\right)^\alpha e^{xz} =\sum_{n=0}^\infty \mathcal E_n^{(\alpha)}(x;\lambda)\frac{z^n}{n!}\quad(| z| <| \log(-\lambda)| ),$ with: $$E_n^{(\alpha)}(x)=\mathcal E_n^{(\alpha)}(x;1)$$ and $$\mathcal E_n^{(\alpha)}(\lambda):=2^n\mathcal{E}_n^{(\alpha)}(\alpha/2;\lambda)$$; $$\mathcal E_n(x;\lambda)=\mathcal{E}_n^{(1)}(x;\lambda)$$ and $$\mathcal E_n(\lambda):=2^n\mathcal{E}_n(\alpha/2;\lambda)$$, where $$\mathcal E_n(\lambda)$$, $$\mathcal{E}_n^{(\alpha)}(\lambda)$$ and $$\mathcal E_n(x;\lambda)$$ denote the so-called Apostol-Euler numbers, Apostol-Euler numbers of order $$\alpha$$ and Apostol-Euler polynomials, respectively. In a similar manner the Apostol-Bernoulli polynomials of order $$\alpha$$, a generalization of the classical Bernoulli polynomials, is introduced. The author derives some relationships between these polynomials and the generalized Hurwitz–Lerch zeta-function. He also gives an explicit series representations for the polynomials involving the Hurwitz zeta-function and the Riemann zeta-function.

### MSC:

 11B68 Bernoulli and Euler numbers and polynomials 11M35 Hurwitz and Lerch zeta functions 11B73 Bell and Stirling numbers 33C45 Orthogonal polynomials and functions of hypergeometric type (Jacobi, Laguerre, Hermite, Askey scheme, etc.)

Zbl 0043.07103
Full Text:


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# Homework Help: Questions on 3 simultaneous equations (3D vector planes)

1. Jan 31, 2010

### Charismaztex

1. The problem statement, all variables and given/known data

1) Determine, by elimination, value of a (if any) such that the given system will have a unique solution

$x+2y+3z=2, 2x+2y+az=0, 3x+2y+z=0$

2) Determine, by elimination, values of a (if any) such that the given system:

a) Is consistent with and infinity of solutions;
b) has a unique solution;
c) is inconsistent, with no solution

$x+2y+3z=a, x+y+z=0, 3x+2y+z=0$

3) consider the following system of 3 equations in x,y, and z

$2x+2y+2z=9, x+3y+4z=5, Ax+5y+6z=B$

Give possibly values of A and B in the third equation which make this system:

a) inconsistent
b) consistent but with an infinite number of solutions

2. Relevant equations

N/A

3. The attempt at a solution

1) By a unique solution I'm presuming that all three planes meet at the same point. Would this be to solve and get a so that x,y, and z have a unique value?

2) Consistent with infinity of solutions, is that when we get a situation 0=0? Possibly with at least 2 of the plane equations the same, or intersecting like the "spine" of a book. So a value of a to get some sort of 0=0?

would inconsistent be a situation when we get 0=a number (a nonsense statement)?

To be honest, I have no idea how to approach the question which as to determine values of a or A and B. What sort of method would be suitable?

Charismaztex

2. Jan 31, 2010

### HallsofIvy

Yes, just go ahead and solve the equations with the "a" in there. At some point, perhaps more than once, you will have to divide by an expression involving a. The system has a unique solution if, each time you have to divide by an expression involving a, that expression is NOT 0. a can be any number that does not make any of those expressions equal to 0.

Yes. Here, a will be such that, at some point, you have to "divide by 0" but the number you are dividing into is also 0.

Right. Again, a must be such that it makes some expression you divide by 0 but now the value you are dividing into is NOT 0.

The best way to do this would be to actually go ahead and solve for x, y, and z, equal to fractions with a in numerator and denominator. There will be a unique solution if a is such that none of those denominators is 0. There will be an infinite number of solutions if a is such that at least one of the denominators is 0 but so are the corresponding numerators. There will be no solution if a is such that there exist at least one denominator equal to 0 and the corresponding numerator is not equal to 0.

Last edited by a moderator: Jan 31, 2010
3. Jan 31, 2010

### Charismaztex

Thanks for the reply. Using your method, I've successfully found for Q1) that a is anything except 2 (confirmed by inputting any number except 2 to get a unique solution on the calculator).

For Q2) I've found that no matter how I manipulate the equations, it always comes out to be something like, for example, $y+2z=a, y+2z=0$. Hence I can see that a must be 0 for infinite solutions, and a is any number other than 0 to be inconsistent with no solutions. I am stymied, however, at part b) where "a" should be a number where there is a unique solution. I can only see that a is either consistent with a=0 or inconsistent with a=any number, with not third option.

For Q3) I've worked out A=3 (in my equation, I got A=3 to make the denominator 0) and B=14 (I made the numerator =0 and solved for B) for infinite solutions and B=any number except 14 for inconsistent solution. I am wondering if this is the correct way to approach this.

Thanks,
Charismaztex


text:
## Co/lax functors without apriori comparison cells

A profunctor is functorial iff every object of has a reflection in within . Fixing (arbitrarily) one reflection arrow for each object yields a functor such that .
Dually, is co-functorial iff every object of has a coreflection in , yielding a functor such that .
If both are satisfied, then above is left adjoint to .

This analogy continues in dimension 2 (for bicategories or double categories), yielding the colax functors as double profunctors with the reflection property and the lax functors as double profunctors with the coreflection property. If a double profunctor has both property, it determines a colax/lax adjunction.


text:
# TrigonometryFinding the greatest angle

#### Pranav

##### Well-known member
Problem:

In $\Delta$ABC, $\displaystyle \cos \left( \frac{A}{2} \right) \cos \left( \frac{B}{2} \right) \cos \left( \frac{C}{2} \right)\leq \frac{1}{4}$, then greatest angle of triangle

A)lies in $\left(0,\frac{\pi}{2}\right)$

B)lies in $\left(\frac{2\pi}{3},\frac{5\pi}{6}\right)$

C)lies in $\left(\frac{5\pi}{6},\pi\right)$

D)lies in $\left(\frac{\pi}{2},\frac{2\pi}{3}\right)$

Attempt:

I haven't been able to proceed anywhere with this problem. I could only simplify the given inequality to
$$\sin A+\sin B+\sin C\leq 1$$

(The above can be proved by using $C=\pi-(A+B)$)

But I am not sure if the above helps.

Any help is appreciated. Thanks!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Problem:

In $\Delta$ABC, $\displaystyle \cos \left( \frac{A}{2} \right) \cos \left( \frac{B}{2} \right) \cos \left( \frac{C}{2} \right)\leq \frac{1}{4}$, then greatest angle of triangle

A)lies in $\left(0,\frac{\pi}{2}\right)$

B)lies in $\left(\frac{2\pi}{3},\frac{5\pi}{6}\right)$

C)lies in $\left(\frac{5\pi}{6},\pi\right)$

D)lies in $\left(\frac{\pi}{2},\frac{2\pi}{3}\right)$

Attempt:

I haven't been able to proceed anywhere with this problem. I could only simplify the given inequality to
$$\sin A+\sin B+\sin C\leq 1$$

(The above can be proved by using $C=\pi-(A+B)$)

But I am not sure if the above helps.

Any help is appreciated. Thanks!
Hi!

Suppose A is the greatest angle.
That means A is at least 60 degrees and at most 180.
$$\frac \pi 3 \le A < \pi \qquad \qquad (1)$$

You've got:
$$\sin A+\sin B+\sin C\leq 1$$
That means that:
$$-1 \le \sin A \le \frac 1 3 \qquad \qquad (2)$$

Combine (1) and (2) to find:
$$0 < \sin A \le \frac 1 3 \wedge A \ge \frac \pi 3$$

With $\sin(\frac {5\pi}{6}) = \frac 1 2$, we can conclude that...

#### Pranav

##### Well-known member
Hi ILS!

With $\sin(\frac {5\pi}{6}) = \frac 1 2$, we can conclude that...
I understand what you have done before this but isn't 1/2 out of the range of $\sin A$?

#### Opalg

##### MHB Oldtimer
Staff member
... isn't 1/2 out of the range of $\sin A$?
If $\Delta$ is the area of the triangle then $\Delta = \frac12bc\sin A$ and so $\sin A = \frac{2\Delta}{bc}$, and similarly $\sin B = \frac{2\Delta}{ca}$ and $\sin C = \frac{2\Delta}{ab}$. The inequality $\sin A + \sin B + \sin C \leqslant 1$ then becomes $2\Delta(a+b+c) \leqslant abc.$ But (see here for example) $abc = 4R\Delta$, where $R$ is the radius of the circumcircle. Therefore $2\Delta(a+b+c) \leqslant 4R\Delta$, or $s\leqslant R$, where $s = \frac12(a+b+c)$ is the semi-perimeter of the triangle. In particular, we must have $a\leqslant R$, where $a = |BC|$ is the longest side. It follows that the angle $BOC$ in the diagram below ($O$ being the centre of the circumcircle) must be at most $\pi/3$, and hence the angle $A$ must be at least $5\pi/6$. Therefore $\sin A \leqslant \frac12.$

#### Pranav

##### Well-known member
Hi Opalg!

...... $a\leqslant R$, where $a = |BC|$ is the longest side.
Sorry if this is obvious but I can't quite follow this. How do you get $a\leq R$? Why not $a\leq 2R$?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hi ILS!

I understand what you have done before this but isn't 1/2 out of the range of $\sin A$?
Yep. It's out of range. I'll get to that in a sec.

What is the solution of:
$$0 < \sin A \le \frac 1 3$$

#### Opalg

##### MHB Oldtimer
Staff member
Hi Opalg!

Sorry if this is obvious but I can't quite follow this. How do you get $a\leq R$? Why not $a\leq 2R$?
$a$ must be less than $s$ because $a\leqslant b+c$ (any side of a triangle must be less than half the perimeter, because the sum of the other two sides is larger then it). So $a = \frac12(a+a) \leqslant \frac12(a+b+c) = s \leqslant R.$

#### Pranav

##### Well-known member
$a$ must be less than $s$ because $a\leqslant b+c$ (any side of a triangle must be less than half the perimeter, because the sum of the other two sides is larger then it). So $a = \frac12(a+a) \leqslant \frac12(a+b+c) = s \leqslant R.$
Thanks Opalg but I am still stuck at your previous post.

Why do you say the least angle is $5\pi/6$ when it is clearly greater than $\pi/3$.

Yep. It's out of range. I'll get to that in a sec.

What is the solution of:
$$0 < \sin A \le \frac 1 3$$
$$A \in \left(0,\arcsin\left(\frac{1}{3}\right)\right] \cup \left[\pi-\arcsin\left(\frac{1}{3}\right),\pi \right)$$

Correct?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
$$A \in \left(0,\arcsin\left(\frac{1}{3}\right)\right] \cup \left[\pi-\arcsin\left(\frac{1}{3}\right),\pi \right)$$

Correct?
Yes. Correct.

Since, $$\displaystyle \sin(\pi/6) = 1/2 > 1/3$$, we can conclude that the angle corresponding to $\arcsin(1/3) < \pi/6$.

Therefore a weaker, but still true, condition is:
$$A \in \left(0,\pi/6 \right) \cup \left(\pi-\pi/6,\pi \right)$$

Combine with the fact that $A > \pi /3$, and we find that $$\displaystyle A \in \left(\frac {5\pi}{6},\pi \right)$$.

#### Pranav

##### Well-known member
Yes. Correct.

Since, $$\displaystyle \sin(\pi/6) = 1/2 > 1/3$$, we can conclude that the angle corresponding to $\arcsin(1/3) < \pi/6$.

Therefore a weaker, but still true, condition is:
$$A \in \left(0,\pi/6 \right) \cup \left(\pi-\pi/6,\pi \right)$$

Combine with the fact that $A > \pi /3$, and we find that $$\displaystyle A \in \left(\frac {5\pi}{6},\pi \right)$$.
Thanks.

But I am again confused. I was looking at your previous and found the following confusing.

You've got:
$$\sin A+\sin B+\sin C\leq 1$$
That means that:
$$-1 \le \sin A \le \frac 1 3 \qquad \qquad (2)$$

$1/3=0.33$. I can have $\sin A=0.9$ and $\sin B=\sin C=0.05$ which satisfies the inequality but not the range you have mentioned for $\sin A$. Sorry if I am being dumb.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
$1/3=0.33$. I can have $\sin A=0.9$ and $\sin B=\sin C=0.05$ which satisfies the inequality but not the range you have mentioned for $\sin A$. Sorry if I am being dumb.
Good point. I made a mistake there.

#### Pranav

##### Well-known member
Good point. I made a mistake there.
I am sorry, the values I mentioned won't satisfy A+B+C=$\pi$.

But I still don't see how you get $-1\leq \sin A \leq \frac{1}{3}$, can you please elaborate that?

And I am still stuck at Opalg's post. I understand how $a\leq R$. When a=R, I see that A is $5\pi /6$. I don't see how A=$\pi/6$ is achieved.

EDIT: I think I have arrived at the answer but I am still interested to know about Opalg's method. Can you please explain a bit more, it would really help. Many thanks!

From triangle inequality, b+c>a $\Rightarrow \sin B+\sin C>\sin A \Rightarrow \sin A+\sin B+\sin C>2\sin A$. Hence $\sin A<1/2$. This gives the answer. Thanks a lot both of you.

Last edited:

#### Opalg

##### MHB Oldtimer
Staff member
And I am still stuck at Opalg's post. I understand how $a\leq R$. When a=R, I see that A is $5\pi /6$. I don't see how A=$\pi/6$ is achieved.
Given that $a\leqslant R$, it follows that either $A\geqslant 5\pi/6$ or $A\leqslant \pi/6$. But $A$ is supposed to be the greatest angle in the triangle, so it cannot be $< \pi/6$. If the greatest angle in the triangle is $\pi/6$ then the triangle would have to be equilateral, in which case $\sin A + \sin B + \sin C = 3\sqrt3/2 >1$, so we can certainly rule out that case.

#### Pranav

##### Well-known member
Given that $a\leqslant R$, it follows that either $A\geqslant 5\pi/6$ or $A\leqslant \pi/6$. But $A$ is supposed to be the greatest angle in the triangle, so it cannot be $< \pi/6$. If the greatest angle in the triangle is $\pi/6$ then the triangle would have to be equilateral, in which case $\sin A + \sin B + \sin C = 3\sqrt3/2 >1$, so we can certainly rule out that case.
I realise my mistake, thanks a lot Opalg!


text:
#### Can any one tell me the properties and uses of parabola & elipes

$Properties\;of\;Parabola-\\*The\;eccentricity\;of\;any\;parabola\;is\;1.\\* The\;parabola\;is\;symmetric\;about\;its\;axis.\\*The\;axis\;is\;perpendicular to\;the\;directrix.\\*The\;axis\;passes\;through\;the\;vertex\;and\;the focus.\\*The\;tangent\;at\;vertex\;is\;parallel\;to\;the\;directrix.\;etc.\\* Uses\;of\;Parabola-\\*Parabolas\;are\;frequently\;used\;in\;physics\;and\\* engineering\;for\;things\;such\;as\;the\;design\;of\;automobile\;headlight\\* reflectors\;and\;the\;paths\;of\;ballistic\;missiles.\\*Properties\;of\;Ellipse-\\* An\;ellipse\;is\;formed\;by\;a\;plane\;intersecting\;a\;cone\;at\;an\;angle\;to\;its\;base.\\* All\;ellipses\;have\;two\;focal\;points\\*All\;ellipses\;have\;a\;center\;and\;a\;major and\;minor\;axis.\\*All\;ellipses\;have\;eccentricity\;values\;greater\;than\;or equal\;to\;zero,\;and\;less\;than\;one.\\*$


text:
# zbMATH — the first resource for mathematics

Connection and curvature in crystals with non-constant dislocation density. (English) Zbl 1425.74121
Summary: Given a smooth defective solid crystalline structure defined by linearly independent ‘lattice’ vector fields, the Burgers vector construction characterizes some aspect of the ‘defectiveness’ of the crystal by virtue of its interpretation in terms of the closure failure of appropriately defined paths in the material and this construction partly determines the distribution of dislocations in the crystal. In the case that the topology of the body manifold $$M$$ is trivial (e.g., a smooth crystal defined on an open set in $$\mathbb{R}^2)$$, it would seem at first glance that there is no corresponding construction that leads to the notion of a distribution of disclinations, that is, defects with some kind of ‘rotational’ closure failure, even though the existence of such discrete defects seems to be accepted in the physical literature. For if one chooses to parallel transport a vector, given at some point $$P$$ in the crystal, by requiring that the components of the transported vector on the lattice vector fields are constant, there is no change in the vector after parallel transport along any circuit based at $$P$$. So the corresponding curvature is zero. However, we show that one can define a certain (generally non-zero) curvature in this context, in a natural way. In fact, we show (subject to some technical assumptions) that given a smooth solid crystalline structure, there is a Lie group acting on the body manifold $$M$$ that has dimension greater or equal to that of $$M$$. When the dislocation density is non-constant in $$M$$ the group generally has a non-trivial topology, and so there may be an associated curvature. Using standard geometric methods in this context, we show that there is a linear connection invariant with respect to the said Lie group, and give examples of structures where the corresponding torsion and curvature may be non-zero even when the topology of $$M$$ is trivial. So we show that there is a ‘rotational’ closure failure associated with the group structure – however, we do not claim, as yet, that this leads to the notion of a distribution of disclinations in the material, since we do not provide a physical interpretation of these ideas. We hope to provide a convincing interpretation in future work. The theory of fibre bundles, in particular the theory of homogeneous spaces, is central to the discussion.

##### MSC:
 74E15 Crystalline structure 53Z05 Applications of differential geometry to physics 22E70 Applications of Lie groups to the sciences; explicit representations
Full Text:
##### References:
 [1] [1] Davini, C. A proposal for a continuum theory of defective crystals. Arch Ration Mech Anal 1986; 96: 295-317. · Zbl 0623.73002 [2] [2] Elżanowski, M, Preston, S. On continuously defective elastic crystals. Miskolc Math Notes 2013; 14(2): 659-670. · Zbl 1299.53161 [3] [3] Kobayashi, S, Nomizu, K. Foundations of differential geometry, vols. I & II. New York: John Wiley & Sons, Inc., 1996. [4] [4] Davini, C, Parry, GP. A complete list of invariants for defective crystals. Proc R Soc London, Ser A 1991; 432: 341-365. · Zbl 0726.73032 [5] [5] Elżanowski, M, Parry, GP. Material symmetry in a theory of continuously defective crystals. J Elast 2004; 74: 215-237. · Zbl 1058.74028 [6] [6] Olver, PJ. Equivalence, invariants, and symmetry. Cambridge: Cambridge University Press, 1995. [7] [7] Parry, GP. Discrete structures in continuum descriptions of defective crystals. Philos Trans R Soc London, Ser A 2016; 374(2066): DOI: 10.1098/rsta.20150172. · Zbl 1353.82073 [8] [8] Parry, GP, Silhavý, M. Elastic invariants in the theory of defective crystals. Proc R Soc London, Ser A 1999; 455: 4333-4346. DOI: 10.1098/rspa.19990503. · Zbl 0954.74013 [9] [9] Gorbatsevich, VV, Onishchik, AL, Vinberg, EB. Lie groups and Lie algebras I (Encyclopedia of Mathematical Sciences, vol. 20). Berlin: Springer-Verlag, 1993. [10] [10] Palais, RS. A global formulation of the Lie theory of transformation groups (Memoirs of the American Mathematical Society, vol. 22). Providence, RI: American Mathematical Society, 1957. [11] [11] Parry, GP. Group properties of defective crystal structures. Math Mech Solids 2003; 8: 515-538. · Zbl 1072.74016 [12] [12] Poor, WA. Differential geometric structures. New York: McGraw-Hill, 1981. [13] [13] Čap, A, Slovàk, J. Parabolic geometries I (Mathematical Surveys and Monographs, vol. 154) Providence, RI: American Mathematical Society, 2008. [14] [14] Baker, A. Matrix groups. London, Springer-Verlag, 2003.
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# RSA Cryptosystem

This question is regarding the RSA cryptosystem.

Alice gives Bob her public key, $N$ and $e$, on a USB flash drive. Unfortunately, Bob loses the flash drive on the train.

Question 1: Does Alice need to generate a new key pair? Justify your answer.

After hearing of Bob’s carelessness, Alice sends her public key to Bob in an unencrypted email.

Question 2: Explain how Eve, with the ability to intercept and modify messages, could eavesdrop on their conversations without either of them knowing. Why would this not be a problem if Bob hadn’t lost the flash drive?

Could I have help in answering these questions.

This is my thought on question 1: I would think Alice does not need to generate a new Public Key, since if anyone picks up the flash drive they still do not know the Private key which is difficult to determine with just $N$ and $e$. They can send Alice messages, but can't decrypt the ciphertext sent to Alice.

• – mikeazo Jan 17 '17 at 20:39


text:
step_kpca_rbf a specification of a recipe step that will convert numeric data into one or more principal components using a radial basis function kernel basis expansion.

step_kpca_rbf(
recipe,
...,
role = "predictor",
trained = FALSE,
num_comp = 5,
res = NULL,
sigma = 0.2,
prefix = "kPC",
skip = FALSE,
id = rand_id("kpca_rbf")
)

# S3 method for step_kpca_rbf
tidy(x, ...)

## Arguments

recipe A recipe object. The step will be added to the sequence of operations for this recipe. One or more selector functions to choose which variables will be used to compute the components. See selections() for more details. For the tidy method, these are not currently used. For model terms created by this step, what analysis role should they be assigned? By default, the function assumes that the new principal component columns created by the original variables will be used as predictors in a model. A logical to indicate if the quantities for preprocessing have been estimated. The number of PCA components to retain as new predictors. If num_comp is greater than the number of columns or the number of possible components, a smaller value will be used. An S4 kernlab::kpca() object is stored here once this preprocessing step has be trained by prep.recipe(). A numeric value for the radial basis function parameter. A character string that will be the prefix to the resulting new variables. See notes below. A logical. Should the step be skipped when the recipe is baked by bake.recipe()? While all operations are baked when prep.recipe() is run, some operations may not be able to be conducted on new data (e.g. processing the outcome variable(s)). Care should be taken when using skip = TRUE as it may affect the computations for subsequent operations A character string that is unique to this step to identify it. A step_kpca_rbf object

## Value

An updated version of recipe with the new step added to the sequence of existing steps (if any). For the tidy method, a tibble with columns terms (the selectors or variables selected).

## Details

Kernel principal component analysis (kPCA) is an extension of a PCA analysis that conducts the calculations in a broader dimensionality defined by a kernel function. For example, if a quadratic kernel function were used, each variable would be represented by its original values as well as its square. This nonlinear mapping is used during the PCA analysis and can potentially help find better representations of the original data.

This step requires the dimRed and kernlab packages. If not installed, the step will stop with a note about installing these packages.

As with ordinary PCA, it is important to standardize the variables prior to running PCA (step_center and step_scale can be used for this purpose).

The argument num_comp controls the number of components that will be retained (the original variables that are used to derive the components are removed from the data). The new components will have names that begin with prefix and a sequence of numbers. The variable names are padded with zeros. For example, if num_comp < 10, their names will be kPC1 - kPC9. If num_comp = 101, the names would be kPC001 - kPC101.

## References

Scholkopf, B., Smola, A., and Muller, K. (1997). Kernel principal component analysis. Lecture Notes in Computer Science, 1327, 583-588.

Karatzoglou, K., Smola, A., Hornik, K., and Zeileis, A. (2004). kernlab - An S4 package for kernel methods in R. Journal of Statistical Software, 11(1), 1-20.

step_pca() step_ica() step_isomap() recipe() prep.recipe() bake.recipe()

## Examples

library(modeldata)
data(biomass)

biomass_tr <- biomass[biomass$dataset == "Training",] biomass_te <- biomass[biomass$dataset == "Testing",]

rec <- recipe(HHV ~ carbon + hydrogen + oxygen + nitrogen + sulfur,
data = biomass_tr)

kpca_trans <- rec %>%
step_YeoJohnson(all_predictors()) %>%
step_normalize(all_predictors()) %>%
step_kpca_rbf(all_predictors())

if (require(dimRed) & require(kernlab)) {
kpca_estimates <- prep(kpca_trans, training = biomass_tr)

kpca_te <- bake(kpca_estimates, biomass_te)

rng <- extendrange(c(kpca_te$kPC1, kpca_te$kPC2))
plot(kpca_te$kPC1, kpca_te$kPC2,
xlim = rng, ylim = rng)

tidy(kpca_trans, number = 3)
tidy(kpca_estimates, number = 3)
}
#> 2020-10-19 00:09:02: Calculating kernel PCA#> 2020-10-19 00:09:02: Trying to calculate reverse#> 2020-10-19 00:09:02: DONE#> # A tibble: 5 x 2
#>   terms    id
#>   <chr>    <chr>
#> 1 carbon   kpca_rbf_68zye
#> 2 hydrogen kpca_rbf_68zye
#> 3 oxygen   kpca_rbf_68zye
#> 4 nitrogen kpca_rbf_68zye
#> 5 sulfur   kpca_rbf_68zye


text:
## Precalculus (6th Edition) Blitzer

The magnitude of the force is $335\text{ lb}$.
The magnitude of the barrel against the ramp is calculated by using the law of sine’s because in this case we take only vertical component into account and ignore the horizontal components. Therefore, we get $3500\sin 5.5{}^\circ =335\text{ lb}$


text:
# Arduino serial.read to int

I am trying to read a serial string which comes through as "R0123" for example then I need the 0123 to be in an int. to send out through another method

Here is my code it is not working the way that I think it should

void loop()
{
if(Serial.available())
{
delay(100);
{
int r1 = Serial.read();
int r2 = Serial.read();
int r3 = Serial.read();
int r4 = Serial.read();
int red = ((int)r1 * 1000) + ((int)r2 * 100) + ((int)r3 * 10) + (int)r4;

sb.sendColour(red,0,0);

Serial.print(r1,0);
Serial.print(r2,0);
Serial.print(r3,0);
Serial.print(r4,0);
Serial.print(red);
}
}
}

-

What about using atoi()? You will still need to trim of the 'R', and you might have to pad it with a NULL character at the end ('\0'). But then it should work as:

int rInteger = atoi(s);

when s is a pointer to a null terminated string.

-
So if i do Serial.read into a string for the four characters then add \o and run in it atoi it should work – Ashley Hughes Aug 26 '12 at 9:55
I am getting an error ArduinoShiftyWireless.cpp: In function 'void loop()': ArduinoShiftyWireless:45: error: cannot convert 'String' to 'const char*' for argument '1' to 'int atoi(const char*)' – Ashley Hughes Aug 26 '12 at 10:04
Like I said, atoi() doesn't take a string, it takes a pointer to a string, or to an array of chars, hence the error :) I am not sure if arduino uses ANSI C(?) Never the less, this should be accurate enough: acm.uiuc.edu/webmonkeys/book/c_guide/2.13.html#atoi – Bladt Aug 26 '12 at 10:13
Frankly, I'm a little rusty on my C, bit something like //code to read the 'R' character char* readString; for(int i=0; i<4; i++) { readString[i]=Serial.Read(); } int a = atoi(readString); I imagine would do the trick. – Bladt Aug 26 '12 at 10:43
Ehh formatting? See it here: pastebin.com/Q4GCLh75 – Bladt Aug 26 '12 at 10:49

You are confusing integer values and ascii character values. When your

  int r1 = Serial.read();

reads a 0 it will read the ASCII character 0. The integer value of the character 0 is (decimal) 48 (check for instance this ascii table). So instead that line should read

  int r1 = Serial.read() - 48;

or even better

  int r1 = Serial.read() - '0';

-

This reads 4 characters after the "R", places them in a char array and appends a null character. Then converts to integer. The code doesn't check if there are at least 4 chars following the "R", or that they are digits, though.

#define numberOfDigits 4
char theNumberString[numberOfDigits + 1];
int theNumber;

{
for (int i = 0; i < numberOfDigits; theNumberString[i++] = Serial.Read());
theNumberString[numberOfDigits] = 0x00;
theNumber = atoi(theNumberString);
}

-

You can try:

int intParse(byte length) //return parsed byte from serial
{
char BUFFER[length + 1];
delay(5); //need for Baud 9600, less delay for faster speed
for(byte i = 0; i < length; i++)
BUFFER[length] = 0; //I had this earlier, but not necessary for Arduino.
return atoi(BUFFER);
}

For example: R1234 You would write:

void loop()
{
char input = Serial.read();
if(input == 'R')
{
int output = intParse(4);
}
}

-
You should also set BUFFER[length] = 0;, as C doesn't guarantee this by default. – Ben Voigt May 2 '13 at 1:10

more simple code using parseInt()

void loop()
{
if (Serial.available())
{
int f = Serial.parseInt();
if (f > 0)
{
Serial.println(f);
}
}
}

-

Why not just use the "readString()" function? Then you can have your code as:

void setup() {
Serial.begin(9600);
Serial.println("Waiting for intstructions...\n");
}

void loop() {
if(Serial.available()) {
`


text:
# How to find the coefficient of heat equation solution with non-odd initial condition

I've got a situation where I'm supposed to solve the 1-D heat equation $$v_t = c^2 u_{xx}$$ under boundary conditions and an initial condition, and so after some run-of-the-mill process, I attained the general solution $$v(x,t) = \sum_{n = 1}^{\infty} A_n\sin(n\pi x)e^{-\lambda_n^2 t}$$ where $$\lambda_n = cn\pi$$ and $$n \in \mathbb{N}$$.

However, I was given an initial condition of $$v(x, 0) = f(x) = -(1 + x)$$ with $$0 < x < 1$$, of which I understand is neither an odd or an even function. Thus, I'm not entirely sure that using Fourier series will be able to lead me to find values for $$A_n$$, and if it was able to be done then could someone explain how? I'm in a bit of a pickle now and so any help is much appreciated!

If you apply the initial condition to your solution you will be able to solve for the coefficient $$A_{n}$$.
$$v(x,0) = f(x) = \sum_{n=1}^{\infty} A_{n} \sin(n \pi x) = -(1+x)$$
$$A_{n} = \int_{0}^{1} f(x) \sin(n \pi x) \textrm{d}x$$
$$A_{n} = - \int_{0}^{1} (1+x) \sin(n \pi x) \textrm{d}x = - \frac{\pi n + \sin(\pi n) - 2\pi \cos( n \pi)}{\pi^{2} n^{2}}$$
• When the function is even or odd you get simpler representations I believe but this method for finding the coefficients works either way. The coefficients go to $0$ as $n \to \infty$ – Shogun May 10 at 17:44


text:
# Can I backpropagate different distributions at once using Policy Gradient?

Hi, my issue is that I use a set of different batches to compute one (multinomial) distribution and one sampling per each batch, in a single epoch. How can I backpropagate those different distributions? I’m thinking on just creating a single Multinomial containing all the distribution vectors as a matrix and also get all the sampling vectors at once (as a matrix, again). Does it have sense?

So, something like \pi(a_i = k | s) ~ p_i^k (1-p_i)^k up to a normalization for i=1…N
And you want to learn it as a single distribution: \pi(A=[k1..kn] | s) right?


text:
## Fornal Charge Exceptions?

$FC=V-(L+\frac{S}{2})$

Jose Lupian 1C
Posts: 27
Joined: Fri Apr 06, 2018 10:03 am

### Fornal Charge Exceptions?

Are there any elements in the periodic table that are exceptions to the formal charge equation or does the equation apply to all the elements?

Jimmy lira-1G
Posts: 61
Joined: Fri Apr 06, 2018 10:03 am

### Re: Fornal Charge Exceptions?

I think I may have found one check this link out

http://www.chem.ucla.edu/~harding/tutor ... charge.pdf

"(The rare exception is a pure
covalent bond between two identical atoms such as in molecular chlorine, Cl-Cl.)
Because of this uneven sharing atoms have fractional, rather than integer, charges.
Calculations to determine
exact electron distribution and atomic charges require complex
computer programs and often many hours of computer time. "
also chek out


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# 数学代写| 随机过程代考|Prediction and conditional expectation

A population starts with one individual at time $n=0: Z_{0}=1$.

After one unit of time (at time $n=1$ ) the sole individual produces $Z_{1}$ identical clones of itself and dies. $Z_{1}$ is an $\mathbb{N}_{0}$-valued random variable.

(a) If $Z_{1}$ happens to be equal to 0 the population is dead and nothing happens at any future time $n \geq 2$.

(b) If $Z_{1}>0$, a unit of time later, each of $Z_{1}$ individuals gives birth to a random number of children and dies. The first one has $Z_{1,1}$ children, the second one $Z_{1,2}$ children, etc. The last, $Z_{1}^{\text {th }}$ one, gives birth to $Z_{1, Z_{1}}$ children. We assume that the distribution of the number of children is the same for each individual in every generation and independent of either the number of individuals in the generation and of the number of children the others have. This distribution, shared by all $Z_{n, i}$ and $Z_{1}$, is called the offspring distribution. The total number of individuals in the second generation is now
$$Z_{2}=\sum_{k=1}^{Z_{1}} Z_{1, k}$$
(c) The third, fourth, etc. generations are produced in the same way. If it ever happens that $Z_{n}=0$, for some $n$, then $Z_{m}=0$ for all $m \geq n$ – the population is extinct. Otherwise,
$$Z_{n+1}=\sum_{k=1}^{Z_{n}} Z_{n, k}$$

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## 数学代写| 随机过程代考|mathematicians

Probability theory is not merely an intellectual pursuit, but provides also a framework for estimation and prediction. Practical men and women of ten need to make decisions based on quantities which are not easily measurable, either because they lie in the future or because of some intrinsic inaccessibility; in doing so they usually make use of some current or feasible observation. Economic examples are commonplace (business trends, inflation rates, and so on); other examples include weather prediction, the climate in prehistoric times, the state of the core of a nuclear reactor, the cause of a disease in an individual or a population, or the paths of celestial bodies. This last problem has the distinction of being amongst the first to be tackled by mathematicians using a modern approach to probability.

## 数学代写| 随机过程代考|triangle inequalit

At its least complicated, a question of prediction or estimation involves an unknown or unobserved random variable $Y$, about which we are provided with the value of some (observable) random variable $X$. The problem is to deduce information about the value of $Y$ from a knowledge of the value of $X$. Thus we seek a function $h(X)$ which is (in some sense) close to $Y$; we write $\widehat{Y}=h(X)$ and call $\widehat{Y}$ an ‘estimator’ of $Y$. As we saw in Section 7.1, there are many different ways in which two random variables may be said to be close to one anotherpointwise, in $r$ th mean, in probability, and so on. A particular way of especial convenience is to work with the norm given by
$$|U|_{2}=\sqrt{\mathbb{E}\left(U^{2}\right)},$$
so that the distance between two random variables $U$ and $V$ is
$$|U-V|_{2}=\sqrt{\mathbb{E}\left{(U-V)^{2}\right}}$$

## 数学代写| 随机过程代考|TRIANGLE INEQUALIT

|ü|2=和(ü2),

|UV|_{2}=\sqrt{\mathbb{E}\left{(UV)^{2}\right}}|UV|_{2}=\sqrt{\mathbb{E}\left{(UV)^{2}\right}}

## Matlab代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。


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# Closedness and boundedness in metrizable topological spaces

This is a quick question that I have not managed to answer myself: let $X$ be a metrizable topological space, let $A\subset X$ be a closed, bounded subset. $f:X\to Y$ is a homeomorphism, must $f(A)$ be closed and bounded?

I know that if $X=\mathbb{R}^n$ the answer is yes, because compactness is a topological property, and also that closedness and boundedness alone are not enough to guarantee that $f(A)$ is either closed or bounded (consider $[0,\infty)$ and $[0,1)$), but what happens if we have both?

For metric spaces: Take $X = (0..1)$ and take the inclusion $ι\colon X → ℝ$. Now take $A = X$. Or even take $f\colon X → ℝ,~x↦ \frac{x}{1-x}$ and $A = X$ if you want $f(A)$ to be neither bounded nor closed in $ℝ$.
• @Dr.: It's worth noting that, even in metrizable topological spaces, there needn't be a natural way to interpret boundedness. For example, $X=(0,1)$ is certainly bounded as a subspace of $\Bbb R$ under the usual metric. However, if we instead consider the function $\rho:X\times X\to\Bbb R$ given by $$\rho(x,y)=\left|\frac1x-\frac1y\right|,$$ which turns out to be a metric on $X$ topologically equivalent to the usual one, we find that $X$ is not bounded with respect to $\rho.$ – Cameron Buie Nov 24 '14 at 23:21


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