text:
## CryptoDB

### Benjamin Wesolowski

#### Publications

Year
Venue
Title
2022
EUROCRYPT
We study two important families of problems in isogeny-based cryptography and how they relate to each other: computing the endomorphism ring of supersingular elliptic curves, and inverting the action of class groups on oriented supersingular curves. We prove that these two families of problems are closely related through polynomial-time reductions, assuming the generalized Riemann hypothesis. We identify two classes of essentially equivalent problems. The first class corresponds to the problem of computing the endomorphism ring of oriented curves. The security of a large family of cryptosystems (such as CSIDH) reduces to (and sometimes from) this class, for which there are heuristic quantum algorithms running in subexponential time. The second class corresponds to computing the endomorphism ring of orientable curves. The security of essentially all isogeny-based cryptosystems reduces to (and sometimes from) this second class, for which the best known algorithms are still exponential. Some of our reductions not only generalise, but also strengthen previously known results. For instance, it was known that in the particular case of curves defined over $\mathbb F_p$, the security of CSIDH reduces to the endomorphism ring problem in subexponential time. Our reductions imply that the security of CSIDH is actually equivalent to the endomorphism ring problem, under polynomial time reductions (circumventing arguments that proved such reductions unlikely).
2021
ASIACRYPT
We present Séta, a new family of public-key encryption schemes with post-quantum security based on isogenies of supersingular elliptic curves. It is constructed from a new family of trapdoor one-way functions, where the inversion algorithm uses Petit's so called \emph{torsion attacks} on SIDH to compute an isogeny between supersingular elliptic curves given an endomorphism of the starting curve and images of torsion points. We prove the OW-CPA security of S\'eta and present an IND-CCA variant using the post-quantum OAEP transformation. Several variants for key generation are explored together with their impact on the selection of parameters, such as the base prime of the scheme. We furthermore formalise an uber'' isogeny assumption framework which aims to generalize computational isogeny problems encountered in schemes including SIDH, CSDIH, OSIDH and ours. Finally, we carefully select parameters to achieve a balance between security and run-times and present experimental results from our implementation.
2020
CRYPTO
Fixing a number field, the space of all ideal lattices, up to isometry, is naturally an Abelian group, called the *Arakelov class group*. This fact, well known to number theorists, has so far not been explicitly used in the literature on lattice-based cryptography. Remarkably, the Arakelov class group is a combination of two groups that have already led to significant cryptanalytic advances: the class group and the unit torus. In the present article, we show that the Arakelov class group has more to offer. We start with the development of a new versatile tool: we prove that, subject to the Riemann Hypothesis for Hecke L-functions, certain random walks on the Arakelov class group have a rapid mixing property. We then exploit this result to relate the average-case and the worst-case of the Shortest Vector Problem in ideal lattices. Our reduction appears particularly sharp: for Hermite-SVP in ideal lattices of certain cyclotomic number fields, it loses no more than a $\tilde O(\sqrt n)$ factor on the Hermite approximation factor. Furthermore, we suggest that this rapid-mixing theorem should find other applications in cryptography and in algorithmic number theory.
2020
ASIACRYPT
We introduce a new signature scheme, \emph{SQISign}, (for \emph{Short Quaternion and Isogeny Signature}) from isogeny graphs of supersingular elliptic curves. The signature scheme is derived from a new one-round, high soundness, interactive identification protocol. Targeting the post-quantum NIST-1 level of security, our implementation results in signatures of $204$ bytes, secret keys of $16$ bytes and public keys of $64$ bytes. In particular, the signature and public key sizes combined are an order of magnitude smaller than all other post-quantum signature schemes. On a modern workstation, our implementation in C takes 0.6s for key generation, 2.5s for signing, and 50ms for verification. While the soundness of the identification protocol follows from classical assumptions, the zero-knowledge property relies on the second main contribution of this paper. We introduce a new algorithm to find an isogeny path connecting two given supersingular elliptic curves of known endomorphism rings. A previous algorithm to solve this problem, due to Kohel, Lauter, Petit and Tignol, systematically reveals paths from the input curves to a `special' curve. This leakage would break the zero-knowledge property of the protocol. Our algorithm does not directly reveal such a path, and subject to a new computational assumption, we prove that the resulting identification protocol is zero-knowledge.
2020
JOFC
We construct a verifiable delay function (VDF). A VDF is a function whose evaluation requires running a given number of sequential steps, yet the result can be efficiently verified. They have applications in decentralised systems, such as the generation of trustworthy public randomness in a trustless environment, or resource-efficient blockchains. To construct our VDF, we actually build a trapdoor VDF. A trapdoor VDF is essentially a VDF which can be evaluated efficiently by parties who know a secret (the trapdoor). By setting up this scheme in a way that the trapdoor is unknown (not even by the party running the setup, so that there is no need for a trusted setup environment), we obtain a simple VDF. Our construction is based on groups of unknown order such as an RSA group or the class group of an imaginary quadratic field. The output of our construction is very short (the result and the proof of correctness are each a single element of the group), and the verification of correctness is very efficient.
2019
EUROCRYPT
We construct a verifiable delay function (VDF). A VDF is a function whose evaluation requires running a given number of sequential steps, yet the result can be efficiently verified. They have applications in decentralised systems, such as the generation of trustworthy public randomness in a trustless environment, or resource-efficient blockchains. To construct our VDF, we actually build a trapdoor VDF. A trapdoor VDF is essentially a VDF which can be evaluated efficiently by parties who know a secret (the trapdoor). By setting up this scheme in a way that the trapdoor is unknown (not even by the party running the setup, so that there is no need for a trusted setup environment), we obtain a simple VDF. Our construction is based on groups of unknown order such as an RSA group, or the class group of an imaginary quadratic field. The output of our construction is very short (the result and the proof of correctness are each a single element of the group), and the verification of correctness is very efficient.
2019
CRYPTO
The hardness of finding short vectors in ideals of cyclotomic number fields (hereafter, Ideal-SVP) can serve as a worst-case assumption for numerous efficient cryptosystems, via the average-case problems Ring-SIS and Ring-LWE. For a while, it could be assumed the Ideal-SVP problem was as hard as the analog problem for general lattices (SVP), even when considering quantum algorithms.But in the last few years, a series of works has lead to a quantum algorithm for Ideal-SVP that outperforms what can be done for general SVP in certain regimes. More precisely, it was demonstrated (under certain hypotheses) that one can find in quantum polynomial time a vector longer by a factor at most $\alpha = \exp ({\widetilde{O}(n^{1/2})})$ than the shortest non-zero vector in a cyclotomic ideal lattice, where n is the dimension.In this work, we explore the constants hidden behind this asymptotic claim. While these algorithms have quantum steps, the steps that impact the approximation factor $\alpha$ are entirely classical, which allows us to estimate it experimentally using only classical computing. Moreover, we design heuristic improvements for those steps that significantly decrease the hidden factors in practice. Finally, we derive new provable effective lower bounds based on volumetric arguments.This study allows to predict the crossover point with classical lattice reduction algorithms, and thereby determine the relevance of this quantum algorithm in any cryptanalytic context. For example we predict that this quantum algorithm provides shorter vectors than BKZ-300 (roughly the weakest security level of NIST lattice-based candidates) for cyclotomic rings of rank larger than about 24000.
2017
EUROCRYPT

PKC 2022


text:
# The Keras Functional API

The Keras API makes creating deep learning models fast and easy. While the sequential API allows you to create models layer-by-layer it is limited in that it does not allow you to create models that share layers or have multiple inputs or outputs. The functional API is an alternate way of creating models that offers a lot more flexibility, including creating more complex models.

## Setup

The following installs pydot and graphviz. You can use brew install graphviz on Mac as well.

We need Tensorflow, of course. The clear_session ensures that the state is reset (e,,g, when using this code in a notebook):

## Example

The advantages of the functional API are:

• re-use of layers across multiple models
• enabling models with complex (non-linear) topologies
• enabling models with multiple inputs and/or outputs

Let’s see how a basic model can created with the functional API:

The input layer defines how the 16-dimensional vectors are digested:

The batch size is always omitted, we only specify the shape of each sample. You can check the input layer like so:

The functional aspect shows when adding a new layer which calls the input layer:

The “layer call” action is like drawing an arrow from “inputs” to this layer we created.
We’re “passing” the inputs to the dense layer, and out we get x.

Let’s add a few more layers to our graph of layers:

At this point, we can create a Model by specifying its inputs and outputs in the graph of layers:

The model’s summary can be checked:

We can also plot the model as a graph (this is where graphviz is used):

And optionally display the input and output shapes of each layer in the plotted graph:

## Training, evaluation, serialization and inference

The functional API only affects how you define your mode. All other aspects remain precisely the same. Including serialization. For example, this saves and loads the model and it would be the same for a model defined via the sequence API:

## Using the same graph of layers to define multiple models

In the functional API, models are created by specifying their inputs
and outputs in a graph of layers. That means that a single graph of layers
can be used to generate multiple models.

In the example below, we use the same stack of layers to instantiate two models:
an encoder model that turns image inputs into 16-dimensional vectors,
and an end-to-end autoencoder model for training.

Note that we make the decoding architecture strictly symmetrical to the encoding architecture,
so that we get an output shape that is the same as the input shape (28, 28, 1).
The reverse of a Conv2D layer is a Conv2DTranspose layer, and the reverse of a MaxPooling2D
layer is an UpSampling2D layer.

## All models are callable, just like layers

You can treat any model as if it were a layer, by calling it on an Input or on the output of another layer.
Note that by calling a model you aren’t just reusing the architecture of the model, you’re also reusing its weights.

Let’s see this in action. Here’s a different take on the autoencoder example that creates an encoder model, a decoder model,
and chain them in two calls to obtain the autoencoder model:

As you can see, model can be nested: a model can contain submodels (since a model is just like a layer).

A common use case for model nesting is ensembling.
As an example, here’s how to ensemble a set of models into a single model that averages their predictions:

## Manipulating complex graph topologies

### Models with multiple inputs and outputs

The functional API makes it easy to manipulate multiple inputs and outputs.
This cannot be handled with the Sequential API.

Here’s a simple example. Let’s say you’re building a system for ranking custom issue tickets by priority and routing them to the right department.

You model will have 3 inputs:

• Title of the ticket (text input)
• Text body of the ticket (text input)
• Any tags added by the user (categorical input)

It will have two outputs:

• Priority score between 0 and 1 (scalar sigmoid output)
• The department that should handle the ticket (softmax output over the set of departments)

Let’s built this model in a few lines with the Functional API.

Let’s plot the model:

When compiling this model, we can assign different losses to each output.
You can even assign different weights to each loss, to modulate their
contribution to the total training loss.

Since we gave names to our output layers, we could also specify the loss like this:

We can train the model by passing lists of Numpy arrays of inputs and targets:

When calling fit with a Dataset object, it should yield either a
tuple of lists like ([title_data, body_data, tags_data], [priority_targets, dept_targets])
or a tuple of dictionaries like
({'title': title_data, 'body': body_data, 'tags': tags_data}, {'priority': priority_targets, 'department': dept_targets}).

## A toy resnet model

In addition to models with multiple inputs and outputs, the Functional API makes it easy to manipulate non-linear connectivity topologies. That is, models where layers are not connected sequentially. This also cannot be handled with the Sequential API.

A common use case for this is residual connections.

Let’s build a toy ResNet model for CIFAR10 to demonstrate this.

Let’s plot the model:

It can be trained like any other neural topology:

## Sharing layers

Another good use for the functional API are models that use shared layers. Shared layers are layer instances that get reused multiple times in a same model: they learn features that correspond to multiple paths in the graph-of-layers.

Shared layers are often used to encode inputs that come from similar spaces (say, two different pieces of text that feature similar vocabulary), since they enable sharing of information across these different inputs, and they make it possible to train such a model on less data. If a given word is seen in one of the inputs, that will benefit the processing of all inputs that go through the shared layer.

To share a layer in the Functional API, just call the same layer instance multiple times. For instance, here’s an Embedding layer shared across two different text inputs:

## Extracting and reusing nodes in the graph of layers

Because the graph of layers you are manipulating in the Functional API is a static datastructure, it can be accessed and inspected. This is how we are able to plot Functional models as images, for instance.

This also means that we can access the activations of intermediate layers (“nodes” in the graph) and reuse them elsewhere. Let’s look at an example. This is a VGG19 model with weights pre-trained on ImageNet:

And these are the intermediate activations of the model, obtained by querying the graph datastructure:

We can use these features to create a new feature-extraction model, that returns the values of the intermediate layer activations — and we can do all of this in 3 lines.

This comes in handy when implementing neural style transfer, among other things.

## Extending the API by writing custom layers

Keras has a wide range of layers but if you don’t find what you need, it’s easy to extend the API by creating your own layers.

All layers subclass the Layer class and implement:
– A call method, that specifies the computation done by the layer.
– A build method, that creates the weights of the layer (note that this is just a style convention; you could create weights in __init__ as well).

Here’s a simple implementation of a Dense layer:

If you want your custom layer to support serialization, you should also define a get_config method,
that returns the constructor arguments of the layer instance:

Optionally, you could also implement the classmethod from_config(cls, config), which is in charge of recreating a layer instance given its config dictionary. The default implementation of from_config is:

## When to use the Functional API

How to decide whether to use the Functional API to create a new model, or just subclass the Model class directly? In general, the Functional API is higher-level, easier & safer to use, and has a number of features that subclassed Models do not support. However, Model subclassing gives you greater flexibility when creating models that are not easily expressible as directed acyclic graphs of layers (for instance, you could not implement a Tree-RNN with the Functional API, you would have to subclass Model directly).

Here are the strengths of the Functional API:

The properties listed below are all true for Sequential models as well (which are also data structures), but they aren’t true for subclassed models (which are Python bytecode, not data structures).

#### It is less verbose.

No super(MyClass, self).__init__(...), no def call(self, ...):, etc.

Compare:

With the subclassed version:

#### It validates your model while you’re defining it.

In the Functional API, your input specification (shape and dtype) is created in advance (via Input), and every time you call a layer, the layer checks that the specification passed to it matches its assumptions, and it will raise a helpful error message if not.

This guarantees that any model you can build with the Functional API will run. All debugging (other than convergence-related debugging) will happen statically during the model construction, and not at execution time. This is similar to typechecking in a compiler.

#### Your Functional model is plottable and inspectable.

You can plot the model as a graph, and you can easily access intermediate nodes in this graph — for instance, to extract and reuse the activations of intermediate layers, as we saw in a previous example:

#### Your Functional model can be serialized or cloned.

Because a Functional model is a data structure rather than a piece of code, it is safely serializable and can be saved as a single file that allows you to recreate the exact same model without having access to any of the original code.

### Here are the weaknesses of the Functional API:

#### It does not support dynamic architectures.

The Functional API treats models as DAGs of layers. This is true for most deep learning architectures, but not all: for instance, recursive networks or Tree RNNs do not follow this assumption and cannot be implemented in the Functional API.

#### Sometimes, you just need to write everything from scratch.

When writing advanced achitectures, you may want to do things that are outside the scope of “defining a DAG of layers”: for instance, you may want to expose multiple custom training and inference methods on your model instance. This requires subclassing.

## Mix-and-matching different API styles

Choosing between the Functional API or Model subclassing isn’t a binary decision that restricts you to one category of models. All models in the tf.keras API can interact with each, whether they’re Sequential models, Functional models, or subclassed Models/Layers written from scratch.

You can always use a Functional model or Sequential model as part of a subclassed Model/Layer:

Inversely, you can use any subclassed Layer or Model in the Functional API as long as it implements a call method that follows one of the following patterns:

• call(self, inputs, **kwargs) where inputs is a tensor or a nested structure of tensors (e.g. a list of tensors), and where **kwargs are non-tensor arguments (non-inputs).
• call(self, inputs, training=None, **kwargs) where training is a boolean indicating whether the layer should behave in training mode and inference mode.
• call(self, inputs, mask=None, **kwargs) where mask is a boolean mask tensor (useful for RNNs, for instance).
• call(self, inputs, training=None, mask=None, **kwargs) — of course you can have both masking and training-specific behavior at the same time.

In addition, if you implement the get_config method on your custom Layer or Model, the Functional models you create with it will still be serializable and clonable.

Here’s a quick example where we use a custom RNN written from scratch in a Functional model:

Tags:

format: plain


text:
# Revision history [back]

### access data of binary image and display coordinates of some points

Hello everyone,

I want access image data of a binary image in order to determine the white point located in the up of the image(having minimal y), down(maximum y), right(maximum x), and in the left(having minimal X) of the image.

here is my code:

cvThreshold(gray, Treshold_img, 150, 255, CV_THRESH_BINARY);
IplConvKernel *element = cvCreateStructuringElementEx(3, 3, 1,1, CV_SHAPE_RECT, NULL );
cvMorphologyEx(Treshold_img, Morphology_img, Temp_img, element, CV_MOP_CLOSE, 2);// remove noises

for (int i = 0; i < Morphology_img->height; i++){
for (int j = 0; j < Morphology_img->width; j++)
{

if ((Morphology_img->imageData[i* Morphology_img->widthStep + j] = 255) && (i<min_y))
{
min_y = i;
min_y_x = j;

}

else if (Morphology_img->imageData[i* Morphology_img->widthStep + j] = 255 && j < min_x)
{
min_x = j;
min_x_y = i;

}
else if (Morphology_img->imageData[i* Morphology_img->widthStep + j] = 255 && i > max_y)
{
max_y = i;
max_y_x = j;

}
else if (Morphology_img->imageData[i* Morphology_img->widthStep + j] = 255 && j >max_x)
{
max_x = j;
max_x_y = i;

}

}
}

The problem is that when i run the code , my Morphology_img become a white image and the coordinate of my points are not correct. Any help? any Idea?
Thanks


text:
### Who's the boss?

Time Limit: 1000 ms Memory Limit: 65536 KiB

#### Problem Description

Several surveys indicate that the taller you are, the higher you can climb the corporate ladder. At TALL
Enterprises Inc. this de facto standard has been properly formalized: your boss is always at least as tall as
you are. Furthermore, you can safely assume that your boss earns a bit more than you do. In fact, you can be
absolutely sure that your immediate boss is the person who earns the least among all the employees that earn
more than you and are at least as tall as you are. Furthermore, if you are the immediate boss of someone, that
person is your subordinate, and all his subordinates are your subordinates as well. If you are nobody s boss,
then you have no subordinates. As simple as these rules are, many people working for TALL are unsure of to
whom they should be turning in their weekly progress report and how many subordinates they have. Write a
program that will help in determining for any employee who the immediate boss of that employee is and how
many subordinates they have. Quality Assurance at TALL have devised a series of tests to ensure that your
program is correct. These test are described below.

#### Input

On the first line of the input is a single positive integer n, telling the number of test scenarios to follow. Each
scenario begins with a line containing two positive integers m and q, where m (at most 30000) is the number
of employees and q (at most 200) is the number of queries. The following m lines each list an employee by
three integers on the same line: employee ID number (six decimal digits, the first one of which is not zero),
yearly salary in Euros and finally height in μm (1 μm = 10-6 meters -- accuracy is important at TALL). The
chairperson is the employee that earns more than anyone else and is also the tallest person in the company.
Then there are q lines listing queries. Each query is a single legal employee ID.
The salary is a positive integer which is at most 10 000 000. No two employees have the same ID, and no two
employees have the same salary. The height of an employee is at least 1 000 000 μm and at most 2 500 000
μm.

#### Output

For each employee ID x in a query output a single line with two integers y k, separated by one space
character, where y is the ID of x\'s boss, and k is the number of subordinates of x. If the query is the ID of the
chairperson, then you should output 0 as the ID of his or her boss (since the chairperson has no immediate
boss except, possibly, God).

#### Sample Input

2
3 3
123456 14323 1700000
123458 41412 1900000
123457 15221 1800000
123456
123458
123457
4 4
200002 12234 1832001
200003 15002 1745201
200004 18745 1883410
200001 24834 1921313
200004
200002
200003
200001

#### Sample Output

123457 0
0 2
123458 1
200001 2
200004 0
200004 0
0 3

#### Source

Northwestern 2003-2004


text:
GCSE Maths Geometry and Measure

3D Shapes

Cuboid

Volume of A Cube

# Volume Of A Cube

Here we will learn about the volume of a cube, including how to calculate the volume of a cube and how to find missing lengths of a cube given its volume.

There are also volume of a cube worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

## What is the volume of a cube?

The volume of a cube is the amount of space there is within a cube.

A cube is a three-dimensional shape with 6 square faces.

To find the volume of a cube, with side length a, we can use the volume of a cube formula,

\text {Volume }=a^{3}.

Volume is measured in cubic units for example, mm^{3}, \ cm^{3} or m^{3}.

For example,

The volume of this cube is

\begin{aligned} \text{Volume} &=a^{3} \\\\ \text{Volume} &=8^{3} \\\\ \text{Volume} &=512 \mathrm{~cm}^{3} \end{aligned}

## How to calculate the volume of a cube

In order to calculate the volume of a cube:

1. Write down the formula.

\text{Volume }=a^{3}

2. Substitute the values into the formula.
3. Work out the calculation.
4. Write the answer and include the units.

## Volume of a cube examples

### Example 1: volume of a cube

Work out the volume of the cube.

1. Write down the formula.

\text{Volume }=a^{3}

2Substitute the values into the formula.

Here the sides of the cube are 6cm.

\text{Volume }=6^{3}

3Work out the calculation.

\begin{aligned} \text{Volume} &=6 \times 6 \times 6\\\\ &=216 \end{aligned}

4Write the answer and include the units.

The measurements are in cm therefore the volume will be in cm^{3}.

\text{Volume }=216 \mathrm{~cm}^{3}

### Example 2: volume of a cube

Work out the volume of this cube.

\text{Volume }=a^{3}

Here the edges are 7cm.

\text{Volume }=7^{3}

\begin{aligned} \text{Volume} &=7 \times 7 \times 7\\\\ &=343 \end{aligned}

The measurements are in cm therefore the volume will be in cm^{3}.

\text{Volume }=343 \mathrm{~cm}^{3}

### Example 3: volume of a cube – different units

Work out the volume of this cube.

\text{Volume }=a^{3}

Notice here that one of the units is in cm whilst the other is in m. We need all the units to be the same to calculate the volume. This is a cube so we know all the edges are the same length, so we can easily change m to cm, \ 0.6 \ m = 60 \ cm.

\text{Volume }=60^{3}

\begin{aligned} \text {Volume }&=60 \times 60 \times 60 \\\\ &=216000 \end{aligned}

The measurements are in cm therefore the volume will be in cm^{3}.

\text{Volume }=216000 \mathrm{~cm}^{3}

## How to work out a missing length given the volume

Sometimes we are given the volume of a cube and need to work out the length of an edge.

We can do this by substituting the volume into the formula for the volume of a cube and then solving the equation that is formed. This will give us the length of the sides of a cube.

1. Write down the formula.
\text{Volume }=a^{3}
2. Substitute the values that you do know into the formula.
3. Solve the equation.
4. Write the answer and include the units.

## Finding the length given the volume examples

### Example 4: find the length of a cube given the volume

The volume of the cube is 64 \ cm^{3}. Work out the length of the cube.

\text{Volume }=a^{3}

a^{3}=64

\begin{aligned} a^{3}&=64 \\\\ a&=\sqrt[3]{64} \\\\ a&=4 \end{aligned}

Since the measurement for the volume in this question was in cm^{3}, the length will be in cm.

a=4 \ \mathrm{~cm}

### Example 5: find the length of a cube given the volume

The volume of the cube is 1000 \ m^{3}. Work out the length of each side, x.

\text{Volume }=x^{3}

x^{3}=1000

\begin{aligned} x^{3}&=1000 \\\\ x&=\sqrt[3]{1000} \\\\ x&=10 \end{aligned}

Since the measurement for the volume in this question was in m^{3}, the length will be in m.

x=10 \ \mathrm{~m}

### Common misconceptions

• Missing/incorrect units

Volume is measured in units cubed (e.g. mm^{3}, \ cm^{3}, \ m^3 etc).

• Calculating with different units

You need to make sure all measurements are in the same units before calculating volume.

For example, you can’t have some in cm and some in m.

• Dividing by three rather than cube rooting

If you are given the volume of a cube and you need to find the side length, remember the inverse of cube is cube root, not divide by 3.

For example, if the volume of a cube is 27 \ cm^{3}, the side length is \sqrt[3]{27}=3 \ \mathrm{cm} (not 27 \div 3 ).

### Practice volume of a cube questions

1. Work out the volume of the cube.

9 \ \mathrm{cm}^{3}

27 \ \mathrm{cm}^{3}

54 \ \mathrm{cm}^{3}

36 \ \mathrm{cm}^{3}
\begin{aligned} \text{Volume } &=a^{3} \\\\ \text{Volume } &=3^{3} \\\\ &=3 \times 3 \times 3 \\\\ &=27 \mathrm{~cm}^{3} \end{aligned}

2. Work out the volume of the cube.

1.25 \ \mathrm{m}^{3}

150 \ \mathrm{m}^{3}

125 \ \mathrm{m}^{3}

0.125 \ \mathrm{m}^{3}
\begin{aligned} \text{Volume }&=a^{3}\\\\ \text{Volume }&=0.5^{3}\\\\ &=0.5 \times 0.5 \times 0.5\\\\ &=0.125 \mathrm{~m}^{3} \end{aligned}

3. Work out the volume of this cube.

64000 \ \mathrm{cm}^{3}

96 \ \mathrm{cm}^{3}

64 \ \mathrm{cm}^{3}

6.4 \ \mathrm{m}^{3}

Notice here that one of the units is in cm whilst the other is in m . We need all the units to be the same to calculate the volume.

This is a cube, we know all the edges are the same length, so we can easily change m to cm, \ 0.4 \ m = 40 \ cm.

\begin{aligned} \text{Volume }&=a^{3}\\\\ \text{Volume }&=40^{3}\\\\ &=40 \times 40 \times 40\\\\ &=64000 \end{aligned}

The measurements are in cm therefore the volume will be in cm^{3}.

\text{Volume }=64000 \ \mathrm{~cm}^{3}

4. The volume of a cube is 343 \ cm^{3}. Find the length of the side.

7 \ \mathrm{cm}^{3}

114.3 \ \mathrm{~cm}

7 \ \mathrm{~cm}

114.3 \ \mathrm{m}^{3}
\text{Volume }=a^{3}

\begin{aligned} a^{3}&=343 \\\\ a&=\sqrt[3]{343} \\\\ a&=7 \mathrm{~cm} \end{aligned}

5. The volume of the cube is 125 \ m^{3}. Work out the length of each side x.

5 \ \mathrm{~m}

41.6 \ \mathrm{m}^{3}

41.6 \ \mathrm{~m}

5 \ \mathrm{m}^{3}
\text{Volume }=x^{3}

\begin{aligned} x^{3}&=125 \\\\ x&=\sqrt[3]{125} \\\\ x&=5 \mathrm{~m} \end{aligned}

6. The volume of a cube is 64 \ m^{3}. Work out the length of each side. Give your answer in cm.

4 \ \mathrm{~m}

21.3 \ \mathrm{m}^{3}

21.3 \ \mathrm{~m}

400 \ \mathrm{~cm}
\text{Volume }=a^{3}

\begin{aligned} a^{3}&=64 \\\\ a&=\sqrt[3]{64} \\\\ a&=4 \mathrm{~m} \end{aligned}

4m = 400 \ cm

a=400 \mathrm{~cm}

### Volume of a cube GCSE questions

1. Here is a cube.

Work out the volume of the cube.

(2 marks)

\text{Volume}=2 \times 2 \times 2

(1)

\text{Volume}=8 \ \mathrm{m}^{3}

(1)

2. (a) This sculpture is formed by placing one cube on top of another. Work out the total volume of the sculpture.

(b) The density of the material used to make the sculpture is 2.7 \ g/cm^{3}. Work out the total mass of the sculpture. Give your answer in kg, to the nearest kg.

You may use \text{density }= \frac{\text{mass}}{\text{volume}} .

(5 marks)

(a)

Volume of bottom cube = 60 \times 60 \times 60=216000 \ cm^3

(1)

Volume of top cube = 35 \times 35 \times 35=42875 \ cm^3

(1)

Total volume = 216000+42875=258875 \ cm^3

(1)

(b)

2.7=\frac{m}{258875}

(1)

2.7 \times 258875=698962.5 \ \mathrm{g}

(1)

698962.5 \ g=698.9625 \ kg=699 \ kg (nearest kg )

(1)

3. A cube has a volume of 343 \ m^{3}. Find the length of the cube. Give your answer in cm.

(3 marks)

x^{3}=343 or x=\sqrt[3]{343}

(1)

x= 7 \ m

(1)

x= 700 \ cm

(1)

## Learning checklist

You have now learned how to:

• Understand and apply the formula to calculate the volume of cubes
• Use the properties of faces, surfaces, edges and vertices of cubes and cuboids to solve problems in 3D

## Still stuck?

Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.

Find out more about our GCSE maths revision programme.


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# If 1 > 1 - ab > 0, which of the following must be true?

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If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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23 Jun 2015, 23:42
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If 1 > 1 - ab > 0, which of the following must be true?

I. a/b > 0
I. a/b < 1
III. ab < 1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

Kudos for a correct solution.

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Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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24 Jun 2015, 02:36
4
I am going to go with E.

1>1-ab>0; If we split this into two separate equations (1-ab<1 & 1-ab>0), we get the following solutions - ab>0 & ab<1.

I says that a/b>0. Since ab>0, it means that either a,b both are positive or both are negative. In either scenario I holds true.

III says that ab<1. We already got that result be splitting and solving the inequality.

II says that a/b<1. Let's take two cases - Case 1 (a=2, b=0.2) & Case 2 (a=0.2, b=2). II holds true for Case 2 but not Case 1, and therefore cannot always be true.

Which means that only I and III are always true.
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If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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Updated on: 24 Jun 2015, 10:15
1
1
Step 1: Subtracting 1 from all,
0>-ab>-1.
Step 2: Multiplying by -1, signs are reversed. Therefore we get :
0<ab<1
Therefore, we can conclude :
1. both ab have the same sign
2. ab lies between 0 to 1

Hence E !!

Originally posted by NickHalden on 24 Jun 2015, 02:38.
Last edited by NickHalden on 24 Jun 2015, 10:15, edited 1 time in total.
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Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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24 Jun 2015, 07:27
2
If 1 > 1 - ab > 0, which of the following must be true?

I. a/b > 0
I. a/b < 1
III. ab < 1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

Solution -
a. For the part 1 > 1 - ab -> Subtract -1 from both sides gives -ab < 0 -> ab > 0. Both a and b are positive or negative.
b. For the part 1 - ab > 0 -> Add ab on both sides gives -> ab<1.

I. a/b > 0. This will give us both a and b are positive or negative. Meets the condition a above. Sufficient.
II. a/b < 1. This is opposite of condition a above. In Sufficient.
III. ab < 1. This inequality satisfy the condition b above. Sufficient.

Thanks
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Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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24 Jun 2015, 10:09
Rearranging the 1>1-ab>0, equation we get
1>1-ab or ab>0 ---1
1-ab>0 or ab<1 -----2
A and b both has to be same sign from equation 1, so a/b will always be >0 buy in some cases it will be >2 or <1 .Hence choose 1
From equation 2 it’s clear that ab<1, hence choose 3.
[This can also be solved by taking example where we have d>a>l , and cross verify ]

Thanks,
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If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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Updated on: 25 Jun 2015, 03:32
Bunuel wrote:
If 1 > 1 - ab > 0, which of the following must be true?

I. a/b > 0
I. a/b < 1
III. ab < 1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

Kudos for a correct solution.

1 > 1 - ab > 0 has two parts

1 > 1 - ab and 1 - ab > 0
i.e. ab > 0 and 1 > ab

i.e. 0 < ab < 1

I. a/b > 0 will always be true as a and b must have same sign for ab to be between 0 and 1

II. a/b < 1 will not always be true @a=1/3 and b=1/2

III. ab < 1 will always be true as inferred from the given range of ab

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Originally posted by GMATinsight on 25 Jun 2015, 02:59.
Last edited by GMATinsight on 25 Jun 2015, 03:32, edited 1 time in total.
Math Expert
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Posts: 7984
Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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25 Jun 2015, 03:22
1
GMATinsight wrote:
Bunuel wrote:
If 1 > 1 - ab > 0, which of the following must be true?

I. a/b > 0
I. a/b < 1
III. ab < 1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

Kudos for a correct solution.

1 > 1 - ab > 0 has two parts

1 > 1 - ab and 1 - ab > 0
i.e. ab > 0 and 1 > ab

i.e. 0 < ab < 1

I. a/b > 0 will always be true as a and b must have same sign for ab to be between 0 and 1

II. a/b < 1 will not always be true @a=1/3 and b=1/2

III. ab < 1 will not always be true as inferred from the given range of ab

ab<1 for all values and has to be true..
a and b have to have same sign so a/b>0..
ans E
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Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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25 Jun 2015, 03:35
chetan2u wrote:

ab<1 for all values and has to be true..
a and b have to have same sign so a/b>0..
ans E

Yes. A typo error due to copy paste. But thank you! you deserve a Kudos .
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Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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26 Jun 2015, 07:53
Bunuel wrote:
If 1 > 1 - ab > 0, which of the following must be true?

I. a/b > 0
I. a/b < 1
III. ab < 1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

Kudos for a correct solution.

1 > 1 - ab > 0
Hence 0 > -ab > -1
Hence 0< ab < 1

Hence ab is positive and ab is less than 1
if ab is positive, then a and b both are of same sign and a/b is positive.

Hence statement 1 and statement 3 are correct.

Hence option E is correct.
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Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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27 Jun 2015, 10:25
1
E
1<1-ab<0 or 0<ab<1
I. a/b > 0 - true as ab>0 then a/b>0
I. a/b < 1 - could be
III. ab < 1 - true
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Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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27 Jun 2015, 10:46
answer is E.. ab has to be positive, greater than 0 but less than 1
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Posts: 58452
Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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29 Jun 2015, 06:22
Bunuel wrote:
If 1 > 1 - ab > 0, which of the following must be true?

I. a/b > 0
I. a/b < 1
III. ab < 1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

You can manipulate the original compound inequality as follows, making sure to perform each manipulation on every term:

1 > 1 - ab > 0

0 > -a b > -1 Subtract 1 from all three terms.

0 < ab < 1 Multiply all three terms by -1 and flip the inequality signs.

Therefore you know that 0 < ab < 1. This tells you that ab is positive, so a/b must be positive (a and b have the same sign). Therefore, I must be true. However, you do not know whether a/b < 1, so II is not necessarily true. But you do know that ab must be less than 1, so III must be true.

Therefore, the correct answer is (E).
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Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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30 Jun 2015, 08:15
1
If 1 > 1 - ab > 0, which of the following must be true?

I. a/b > 0
I. a/b < 1
III. ab < 1

If you first deal with the right half of the inequality, you can add ab to both sides to get ab<1. Since III is true, you can eliminate A, B, and D.

Now test I. Looking at the left side of the inequality, add ab to both sides and subtract 1 from both sides to yield ab>0. For ab to be positive, a and b must have the same signs. This will also be true if we divide a by b. Statement I is also true.
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
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Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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30 Jun 2015, 09:46
0 < 1-ab < 1, since (1-ab) is both positive and less than 1, 0<ab<1, so III is true. For (ab) to be less than 1 and greater than 0 the fraction a/b must be greater than 0 (must be positive), so I is also true. Ans is E?
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Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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15 Jul 2015, 09:19
Hello Bunuel ,
While solving this problem , i reached up till this inequality ( 0<ab<1) .
However , statement III above says that ab<1 .
This covers a lot many more numbers, which wont be satisfied by the inequality
provided in the stem ( e.g since ab<1 , it will also mean that ab =-3).
Should this not be reason good enough to eliminate statement III (as
this has provided us with values that are unable to satisfy the stem) ?

PS: If the question had asked , "which of the following might be true ?",
then,yes, we could still include statement III.
I am unable to understand this.
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Joined: 02 Sep 2009
Posts: 58452
Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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15 Jul 2015, 09:27
fetchnitin wrote:
Hello Bunuel ,
While solving this problem , i reached up till this inequality ( 0<ab<1) .
However , statement III above says that ab<1 .
This covers a lot many more numbers, which wont be satisfied by the inequality
provided in the stem ( e.g since ab<1 , it will also mean that ab =-3).
Should this not be reason good enough to eliminate statement III (as
this has provided us with values that are unable to satisfy the stem) ?

PS: If the question had asked , "which of the following might be true ?",
then,yes, we could still include statement III.
I am unable to understand this.

We have that 0 < ab < 1. Now, let me asks you is ab < 1 true?
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Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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15 Jul 2015, 09:56
Hello,
yes,this is true .ab<1 .
But my confusion is that it also contains values
that are outside the range of the inequality in the
question stem .
Is it that, while solving these type of questions all
the values of the question stem should be tried and checked
with the options and not the other way round ( i.e all values of the
options should fit the range of stem ). If so ,then could u explain to me
the logic behind this, as i am unable to visualize this .

Thanks
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Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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15 Jul 2015, 10:00
Hello Bunuel,
yes,this is true .ab<1 .
But my confusion is that it also contains values
that are outside the range of the inequality in the
question stem .
Is it that, while solving these type of questions all
the values of the question stem should be tried and checked
with the options and not the other way round ( i.e all values of the
options should fit the range of stem ). If so ,then could u explain to me
the logic behind this, as i am unable to visualize this .

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 58452
Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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15 Jul 2015, 10:03
fetchnitin wrote:
Hello Bunuel,
yes,this is true .ab<1 .
But my confusion is that it also contains values
that are outside the range of the inequality in the
question stem .
Is it that, while solving these type of questions all
the values of the question stem should be tried and checked
with the options and not the other way round ( i.e all values of the
options should fit the range of stem ). If so ,then could u explain to me
the logic behind this, as i am unable to visualize this .

Thanks

I think that you should practice Must or Could be true questions more: search.php?search_id=tag&tag_id=193

Hope it helps.
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Re: If 1 > 1 - ab > 0, which of the following must be true?  [#permalink]

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24 Feb 2018, 08:25
Top Contributor
Bunuel wrote:
If 1 > 1 - ab > 0, which of the following must be true?

I. a/b > 0
I. a/b < 1
III. ab < 1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

Kudos for a correct solution.

GIVEN: 1 > 1 - ab > 0
Multiply all 3 sides by -1 to get: -1 < -1 + ab < 0 [since I multiplied by a NEGATIVE number, I had to REVERSE the inequality symbols]
Add 1 to all 3 sides to get: 0 < ab < 1

First, if 0 < ab, then a and b are the SAME SIGN, which means a/b > 0
So, statement I is TRUE
ELIMINATE B and C

Second, our new inequality clearly tells us that ab < 1
So, statement III is TRUE
ELIMINATE A and D

So, we need not even check whether statement II is true.

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Re: If 1 > 1 - ab > 0, which of the following must be true?   [#permalink] 24 Feb 2018, 08:25

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text:
Evaluate the following integral:
Question:

Evaluate the following integral:

$\int \frac{1}{x\left(x^{n}+1\right)} d x$

Solution:

$\frac{1}{x\left(x^{n}+1\right)}$

Multiply numerator and denominator by $x^{n-1}$, we get

$\int \frac{1}{x\left(x^{n}+1\right)} d x \Rightarrow \int \frac{x^{n-1}}{x\left(x^{n}+1\right) x^{n-1}} d x \Rightarrow \int \frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)} d x$

Let $x^{n}=t \Rightarrow n x^{n-1} d x=d t$

So the above equation becomes,

$\int \frac{\mathrm{x}^{\mathrm{n}-1}}{\mathrm{x}^{\mathrm{n}}\left(\mathrm{x}^{\mathrm{n}}+1\right)} \mathrm{dx} \Rightarrow \frac{1}{\mathrm{n}} \int \frac{1}{\mathrm{t}(\mathrm{t}+1)} \mathrm{dt}$

The denominator is factorized, so let separate the fraction through partial fraction, hence let

$\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{t+1} \ldots \ldots(i)$

$\Rightarrow \frac{1}{t(t+1)}=\frac{A(t+1)+B t}{t(t+1)}$

$\Rightarrow 1=\mathrm{A}(\mathrm{t}+1)+\mathrm{Bt} \ldots \ldots$ (ii)

Put $t=0$ in above equations we get

$1=A(0+1)+B(0)$

$\Rightarrow A=1$

Now put $t=-1$ in equation (ii) we get

$1=A(-1+1)+B(-1)$

$\Rightarrow B=-1$

We put the values of $A$ and $B$ values back into our partial fractions in equation (i) and replace this as the integrand. We get

$\int \frac{\mathrm{x}^{\mathrm{n}-1}}{\mathrm{x}^{\mathrm{n}}\left(\mathrm{x}^{\mathrm{n}}+1\right)} \mathrm{dx} \Rightarrow \frac{1}{\mathrm{n}} \int \frac{1}{\mathrm{t}(\mathrm{t}+1)} \mathrm{dt}$

$\Rightarrow \frac{1}{n} \int\left[\frac{A}{t}+\frac{B}{t+1}\right] d t$

$\Rightarrow \frac{1}{n} \int\left[\frac{1}{t}+\frac{-1}{t+1}\right] d t$

Split up the integral,

$\Rightarrow \frac{1}{n}\left[\int \frac{1}{t} d t-\int \frac{1}{t+1} d t\right]$

Let substitute

$\mathrm{u}=\mathrm{t}+1 \Rightarrow \mathrm{du}=\mathrm{dt}$, so the above equation becomes,

$\Rightarrow \frac{1}{\mathrm{n}}\left[\int \frac{1}{\mathrm{t}} \mathrm{dt}-\int \frac{1}{\mathrm{u}} \mathrm{du}\right]$

Substituting back the values of $u$, we get

$\Rightarrow \frac{1}{\mathrm{n}}[\log |\mathrm{t}|-\log (|\mathrm{t}+1|)]+\mathrm{C}$

Substituting back the values of $t$, we get

$\Rightarrow \frac{1}{n}\left[\log \left|x^{n}\right|-\log \left|x^{n}+1\right|\right]+C$

Applying the logarithm rules, we get

$\Rightarrow \frac{1}{n}\left[\log \left|\frac{x^{n}}{x^{n}+1}\right|\right]+C$

Note: the absolute value signs account for the domain of the natural log function $(x>0)$.

Hence,

$\int \frac{1}{x\left(x^{n}+1\right)} d x=\frac{1}{n}\left[\log \left|\frac{x^{n}}{x^{n}+1}\right|\right]+C$


text:
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text:
全文文献 工具书 数字 学术定义 翻译助手 学术趋势 更多

 sum 在 物理学 分类中 的翻译结果: 查询用时：0.188秒
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 sum 求和(410)
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 sum
 We show that ${\mathcal M}(G,R)$ is a symmetric tensor category, i.e., the motive of the product of two projective homogeneous G-varieties is a direct sum of twisted motives of projective homogeneous G-varieties. We also study the problem of uniqueness of a direct sum decomposition of objects in ${\mathcal M}(G,R).$ We prove that the Krull--Schmidt theorem holds in many cases. For a split reductive algebraic group, this paper observes a homological interpretation for Weyl module multiplicities in Jantzen's sum formula. The new interpretation makes transparent for GLn (and conceivable for other classical groups) a certain invariance of Jantzen's sum formula under "Howe duality" in the sense of Adamovich and Rybnikov. From elements of the invariant algebra C[V]G we obtain, by polarization, elements of C[kV]G, where k ≥ 1 and kV denotes the direct sum of k copies of V. 更多
 The assignment of aberration tolerances for optical systems suffering from wave-front aberrations greater than λ/4 is worth further studying. In this paper it is shown that "low" contrast resolving power of optical systems can be taken as an appropriate criterion for assessing image quality, where the optical transfer function can be evaluated approximately in terms of the sums of Legendre polynomials. Accordingly, the best program of aberration correction and tolerences in the case where certain aberration... The assignment of aberration tolerances for optical systems suffering from wave-front aberrations greater than λ/4 is worth further studying. In this paper it is shown that "low" contrast resolving power of optical systems can be taken as an appropriate criterion for assessing image quality, where the optical transfer function can be evaluated approximately in terms of the sums of Legendre polynomials. Accordingly, the best program of aberration correction and tolerences in the case where certain aberration constants are inevitably present and others are adjustable in optical design can be obtained. The results so obtained as applied to aberrations of small magnitude comparable to λ/4 conform well to the Strehl Deffinitionshelligkeit method. 具有大象差的光学系统象差公差问题一直还未得到很好解决。本文中提出以低对比分辨本领作为光学系统的质量指标,并将光学系统的传递函数近似表示为正交多项式之和,象差任意大小的光学系统的最佳校正方案和公差之值就可以求出。在小象差情况下,所得的结果与Strehl判断的结果基本一致。 This paper suggests an approximation method for solving the problems of diffraction due to perfectly conducting cylinder, the section of which is a smooth curve C of arbitrary form. The principle of the method is similar to that of H. Bremmer[8]: The field of diffraction due to a cylinder with a polygonal section (which is an inscribed polygon of the curve C) is expanded into a series. The first term of the series is the geometrical field. The second term of the series is the sum of the elementary diffraction... This paper suggests an approximation method for solving the problems of diffraction due to perfectly conducting cylinder, the section of which is a smooth curve C of arbitrary form. The principle of the method is similar to that of H. Bremmer[8]: The field of diffraction due to a cylinder with a polygonal section (which is an inscribed polygon of the curve C) is expanded into a series. The first term of the series is the geometrical field. The second term of the series is the sum of the elementary diffraction fields due to the wedges of the polygonal cylinder. These fields are taken as those of Sommerfeld's problem, i.e., both sides of each wedge are infinitely extended. Each of these elementary fields falls on the neighbour wedge and is diffracted by the latter, and this diffracted field in turn falls on the next neighbour wedge and is again diffracted by the latter, etc. The field diffracted by the wedges one after another in such a way is called the main tangential elementary field. The third term of the series is the sum of these main tangential elementary fields. The field diffracted by wedge A, being diffracted again by the neighbour wedge B, reflects back on wedge A again, and then propagates in this direction progressively in a manner mentioned above. Such a field is called once-reflected elementary field. The fourth term of the series is the sum of these once-reflected elementary fields, etc. In general, the m-th term of the series is the sum of the (m-3) times-reflected elementary fields. Every elementary diffracted field due to any wedge is taken as the solution of Sommerfeld's problem for this wedge in the manner mentioned above. As the sides of the inscribed polygon approach to zero, the inscribed polygon approaches to the curve C, and each term of the series becomes an integral, the limit of the summation of the series approaching to the rigorous solution of the initial problem. 本文对于任意形状的光滑柱状理想导体的衍射提出一种级数解法。方法的原理与层变媒质的Bremmer级数相似:先以内接多面稜柱代替上述光滑柱体;将此稜柱产生的衍射场展为一个级数。级数之首项为几何光学场;级数之第二项为稜柱的所有各稜产生的元衍射场之和,其中每个元衍射场皆取Sommerfeld问题的解,即将该稜之两侧面视为半无限大的平面。上述每一元衍射场皆投射在其相邻稜上,并在相邻稜上发生衍射;这一衍射场随之又投射在下一个相邻稜上而发生衍射;依此类推。按此方式依次被各稜所衍射的场称为“主掠射元场”。级数之第三项即为这些主掠射元场之和。被某一稜A衍射而后又在相邻的稜B上衍射的某一元场,同样会回射到A上;然后以上述“主掠射”方式传递下去,这样的场称为“一次反射元场”。级数的第四项即为这些一次反射元场之和。依此类推。一般说来,级数之第m项(m>3)为m-3次反射元场之和。元场在任何一稜上的衍射皆取Sommerfeld解。当内接多面稜柱之面数趋向无穷,且每面之宽度趋向零时,多面稜柱即趋于光滑柱体,且级数每一项的求和变为一个积分。这时该级数总和之极限即为原问题之解。 对级数之前三项单独进行了推导。对于一般的第m项(m>3),导出...本文对于任意形状的光滑柱状理想导体的衍射提出一种级数解法。方法的原理与层变媒质的Bremmer级数相似:先以内接多面稜柱代替上述光滑柱体;将此稜柱产生的衍射场展为一个级数。级数之首项为几何光学场;级数之第二项为稜柱的所有各稜产生的元衍射场之和,其中每个元衍射场皆取Sommerfeld问题的解,即将该稜之两侧面视为半无限大的平面。上述每一元衍射场皆投射在其相邻稜上,并在相邻稜上发生衍射;这一衍射场随之又投射在下一个相邻稜上而发生衍射;依此类推。按此方式依次被各稜所衍射的场称为“主掠射元场”。级数之第三项即为这些主掠射元场之和。被某一稜A衍射而后又在相邻的稜B上衍射的某一元场,同样会回射到A上;然后以上述“主掠射”方式传递下去,这样的场称为“一次反射元场”。级数的第四项即为这些一次反射元场之和。依此类推。一般说来,级数之第m项(m>3)为m-3次反射元场之和。元场在任何一稜上的衍射皆取Sommerfeld解。当内接多面稜柱之面数趋向无穷,且每面之宽度趋向零时,多面稜柱即趋于光滑柱体,且级数每一项的求和变为一个积分。这时该级数总和之极限即为原问题之解。 对级数之前三项单独进行了推导。对于一般的第m项(m>3),导出了一个递推公式。最后,对该级数之收敛条件进行了探讨。 The general formulas for computing actual shielding factors of high-conductivity overhead ground wires hung on power and tele-communication lines are given. These formulas are derived from a system of finite difference equations representing the relation between the currents in successive sections of the multiply-grounded shielding conductors, and the subsequent substitution of terminal conditions corresponding to the particular circuit arrangement under consideration. It is found that the actual shielding factor... The general formulas for computing actual shielding factors of high-conductivity overhead ground wires hung on power and tele-communication lines are given. These formulas are derived from a system of finite difference equations representing the relation between the currents in successive sections of the multiply-grounded shielding conductors, and the subsequent substitution of terminal conditions corresponding to the particular circuit arrangement under consideration. It is found that the actual shielding factor can be written as the sum of the intrinsic shielding factor t0 and a correction term evaluating the effect of finite conductance of the ground connections. It is also shown that the shielding factor formulas given in Engineering Report No. 26 of the joint subcommitte on development and research of the Edison electric institute and the Bell telephone system are only special cases of the general formulas given in this article. 本文分别就良导体架空地线架设在电力线路上和电信线路上的两种情况,给出了架空地线对于电信线路的实际屏蔽系数的普遍公式。这些公式的推导,是用有限差分方程组代表多点接地的屏蔽导线的相邻两档之间电流的关系,再代入相应的边界条件而解出。所得实际屏蔽系数的公式可以写成架空地线的固有屏蔽系数t_0和一个考虑了接地电阻影响的校正项之和。同时并证明了爱迪生电气协会和贝尔电话系统共同发展和研究小组的工程报告第26号中所列出的屏蔽系数公式是本文普遍公式的一个特殊情况。 << 更多相关文摘
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text:
# Why do isomeric bis-ethylenediamine cobalt complexes react differently?

A compound $\ce{Co(en)2(NO2)2Cl}$ has been prepared in a number of isomeric forms.One form undergoes no reaction with $\ce{AgNO3}$ or $\ce{(en)}$ and is optically inactive.

A second form reacts with $\ce{AgNO3}$ but not with $\ce{(en)}$ and is optically inactive.

A third from is optically active and reacts with both $\ce{AgNO3}$ and $\ce{en}$. Identify each of these isomeric forms. Name and sketch each of the structures.

My attempt:

I understood how to use the information of reaction with $\ce{AgNO3}$ and optical activity but couldn't understand what does reaction with $\ce{en}$ represent (I know $\ce{en }$ means ethylene diamine).

In solution, it is given that, if the compound reacts with en it means monodentate ligands are cis to each other,if not it is trans to each other. I couldn't understand why it is so. please help.

• If the monodentate ligands are cis to each other, then one en may substitute them both, which it does quite willingly. Otherwise it can't. – Ivan Neretin Jan 23 '16 at 7:24
• why is it so? how does being cis or trans affect chemical properties?after all stereoisomers have smae chemical properties. – Ajay Sabarish Jan 23 '16 at 8:39
• Simple geometry. Ethylenediamine occupies two positions in the coordination sphere, and these two positions must be cis. Trans positions won't do, they are just too far apart. – Ivan Neretin Jan 23 '16 at 10:15
• sir,please explain in detail,what does configuration has to do with reaction with ethylene diammine?and and why should ethylene diammine in molecule be trans to each other for external ethylene diammine to react with the compound.? – Ajay Sabarish Jan 23 '16 at 11:53
• I never said that ethylenediamine should (or should not) be trans to each other. This does not even make sense. Likewise, my words don't make sense to you. Sorry, I don't think I can help. – Ivan Neretin Jan 23 '16 at 20:08


text:
# Nilpotent Lie Algebras and 2-dimensional Lie Subalgebras

Let be $\mathcal{L}$ a finite-dimensional Lie algebra. How I can prove that if every $2-$dimensional Lie subalgebra of $\mathcal{L}$ is abelian, then $\mathcal{L}$ is nilpotent?

• You mean complex Lie algebra. This result is false, say, for real Lie algebras (e.g., the Lie algebra of the group of motions of the Euclidean plane has only abelian 2-dimensional subalgebras, but is not nilpotent). – YCor Jun 2 '16 at 21:31

Let $x\in L$ and consider the adjoint operator $ad(x)$ for $x\in L$. Its eigenvectors $y\neq 0$ are given by $[x,y]=\lambda y$. Suppose that $\lambda\neq 0$. Then $\langle x,y\rangle$ is a $2$-dimensional non-abelian subalgebra, a contradiction to the assumption. Hence $\lambda=0$ and all adjoint operators have only $\lambda=0$ as eigenvalue. Hence they are all nilpotent. By Engel's theorem, $L$ is nilpotent.
• You are welcome. The other direction is also true. If $L$ is nilpotent, then all $2$-dimensional subalgebras are abelian. – Dietrich Burde May 24 '16 at 19:34


text:
## Posts

Showing posts from July, 2014

### Embarrassing Questions, German Tanks and Estimations

Q: You are conducting a survey and want to ask an embarrassing yes/no question to subjects. The subjects wouldn't answer that embarrassing question honestly unless they are guaranteed complete anonymity. How would you conduct the survey?

Machine Learning: The Art and Science of Algorithms that Make Sense of Data

A: One way to do this is to assign a fair coin to the subject and ask them to toss it in private. If it came out heads then answer the question truthfully else toss the coin a second time and record the result (heads = yes, tails = no). With some simple algebra you can estimate the proportion of users who have answered the question with a yes.

Assume total population surveyed is $$X$$. Let $$Y$$ subjects have answered with a "yes". Let $$p$$ be the sort after proportion. The tree diagram below shows the user flow.

The total expected number of "yes" responses can be estimated as
$$\frac{pX}{2} + \frac{X}{4} = Y$$
which …


text:
# Modified Coupon Collector’s Problem in Java

Posted on

Problem

What is Coupon Collector’s Problem?

In probability theory, the coupon collector’s problem describes the “collect all coupons and win” contests. It asks the following question: Suppose that there is an urn of n different coupons, from which coupons are being collected, equally likely, with replacement. What is the probability that more than t sample trials are needed to collect all n coupons? An alternative statement is: Given n coupons, how many coupons do you expect you need to draw with replacement before having drawn each coupon at least once? from here.

What is my Modified Coupon Collector’s Problem?

Instead of “at least once”, I check for “at least 2 times”.

I loop for a crore times. I should get 14.7,24.134$14.7,24.134$$14.7,24.134$. My code returns 14.760683$14.760683$$14.760683$ and 23.83069$23.83069$$23.83069$. Fairly correct. Maybe Java is not so randomly accurate?

Here’s is my code:

package test;

import java.util.ArrayList;
import java.util.Random;

public class CouponCollector {
private static final int mMaxCouponNumber = 6;
private static final int mMinCouponNumber = 1;
private static final int EXPERIMENT_REPETITIONS = 10000000;
private static final int EACH_COUPONS_ATLEAST_OCCURENCE = 2;

public static void main(String args[]) {
float mSumOfDraws = 0;
for (int i = 1; i <= EXPERIMENT_REPETITIONS; i++) {
ArrayList<Integer> mCouponList = new ArrayList<Integer>();
int mDrawsRequired = 0;
do {
.nextInt((mMaxCouponNumber - mMinCouponNumber) + 1)
+ mMinCouponNumber);
// mCouponList.get(mDrawsRequired));
++mDrawsRequired;
if (allCouponsCollected(mCouponList)) {
break;
}
} while (true);
if(i%10000==0){
// Enable this if you are too impatient! :D
//DEBUG System.out.println("Process: "+(((float)i)*100/EXPERIMENT_REPETITIONS)+"% complete.");
}
mSumOfDraws += mDrawsRequired;
// DEBUG System.out.println("Draws required in " + i
// + " st/nd/rd/th trial is " + mDrawsRequired);
}
System.out.println("Average value of draws required is "
+ (mSumOfDraws / EXPERIMENT_REPETITIONS));
}

private static boolean allCouponsCollected(ArrayList<Integer> pCouponList) {
int[] mCouponCount = new int[mMaxCouponNumber - mMinCouponNumber + 1];
for (int i = 0; i < pCouponList.size(); i++) {
++mCouponCount[pCouponList.get(i) - 1];
// DEBUG System.out.println("Count of " + pCouponList.get(i) +
// " is " + mCouponCount[pCouponList.get(i) - 1]);
}
for (int i = 0; i < mCouponCount.length; i++) {
if (mCouponCount[i] < EACH_COUPONS_ATLEAST_OCCURENCE) {
return false;
}
}
return true;
}
}

Solution

The deviation you see from the expected results is probably in large part due to the poor use of Random. When I time your code, I get:

Average value of draws required is 23.832188
Complete in 23473.917s

If I use Random the way it should be (re)used, and convert this code:

        mCouponList.add(new Random()
.nextInt((mMaxCouponNumber - mMinCouponNumber) + 1)
+ mMinCouponNumber);

to be:

private static final Random RANDOM = new Random();

....

.nextInt((mMaxCouponNumber - mMinCouponNumber) + 1)
+ mMinCouponNumber);

it completes in half the time:

Average value of draws required is 23.827381
Complete in 11552.228s

Hmm.. I was expecting this to correct the errors in the ‘randomness’ too, but, now I see the problem there…. you are using float, and not double. Float has relatively low accuracy. If I change this line here:

float mSumOfDraws = 0;

to be:

double mSumOfDraws = 0;

My results become:

Average value of draws required is 24.1312123

That’s much closer to the mathematical expectation.

Now, about performance… your code is really slow. There’s four prime reasons:

1. there’s no need to record every draw you make in an ArrayList – it will grow large, and require garbage collection. You only need to count how many times you ‘pull’ each coupon.
2. there is no need to number the coupons from minimum to maximum – you just need to know how many there are.
3. You use Integer values instead of int primitives.
4. You have too much code in your main method – the main method is hard for the compiler to optimize – it is only ever called once.

Additionally, apart from the performance, the code style is… unconventional. You should parameterize a lot of the static values so that you can reuse the code better too.

Here is the results of my simplified code:

Average for 2 sets of 6 coupons is 24.131
Complete in 3.983s

(compared to your results with the improved Random, and double):

Average value of draws required is 24.1291875
Complete in 11.109s

The code I use is:

import java.util.Arrays;
import java.util.Random;

public class CouponClipper {

private final int[] counts;
private int remaining;

public CouponClipper(int couponCount, int sets) {
this.counts = new int[couponCount];
Arrays.fill(counts, sets);
this.remaining = couponCount;
}

public boolean pull(int coupon) {
if (--counts[coupon] == 0) {
remaining--;
}
return remaining == 0;
}

private static final Random RANDOM = new Random();

public static int countRequired(int coupons, int sets) {
CouponClipper clipper = new CouponClipper(coupons, sets);
int count = 0;
do {
count++;
} while (!clipper.pull(RANDOM.nextInt(coupons)));
return count;
}

public static double averageCounts(int coupons, int sets, int iterations) {
long sum = 0;
for (int i = 0; i < iterations; i++) {
sum += countRequired(coupons, sets);
}
return sum / (double)iterations;
}

public static void main(String[] args) {
long nanos = System.nanoTime();
System.out.printf("Average for 2 sets of 6 coupons is %.3fn", averageCounts(6, 2, 10000000));
nanos = System.nanoTime() - nanos;
System.out.printf("Complete in %.3fsn", nanos / 1000000000.0);

}

}


text:
# Intuition for $\lim\sup$ and $\lim\inf$

After reading several alternative definitions of $$\lim\sup$$ and $$\lim\inf$$, such as $$\lim\sup$$ being the supremum of the set of all subsequential limits, I'm still having trouble building the intuition for $$\lim\sup$$.

One thing that I feel is true, but not sure, is that $$\lim\sup$$ represents the greatest real number that infinitely many $$a_n$$ gets close to, and $$\lim\inf$$ represents the smallest value that infinitely many $$a_n$$ gets close to. Are these correct statements? If so, how would one go about showing it? Thanks

• Well both lim sup and lim inf can be infinite, so perhaps you should say "extended real number." Nov 7, 2018 at 5:10
• yes, that's what I meant. Thanks for pointing out.
– gws
Nov 7, 2018 at 5:14
• Which is the definition of $\limsup$ that you are using? Nov 7, 2018 at 5:14
• Yes, $\limsup$ is the largest possible limiting value achievable over a subsequence of times that go to infinity, while $\liminf$ is the smallest possible. Intuition comes by considering $\limsup_{t\rightarrow\infty} \cos(t) = 1$ and $\liminf_{t\rightarrow\infty} \cos(t) = -1$. We have $\limsup = \liminf = \lim$ if and only if the regular limit $\lim$ exists. Nov 7, 2018 at 5:27
• You may have a look at this answer math.stackexchange.com/a/1893725/72031 Nov 8, 2018 at 13:31

One thing that helped me: You could try to prove that if $$s$$ is the $$\limsup$$ then for any number $$s' < s$$, the sequence exceeds $$s'$$ infinitely often. And for any number $$s'' > s$$, the sequence exceeds $$s''$$ only finitely often.

• I tried and I think I proved this. But I was just wondering would s be the unique real number that satisfies these two properties?
– gws
Nov 7, 2018 at 6:38
• Yes, it is. Any $t< s$ fails to satisfy the property: "for any number $u>t$, the sequence exceeds $u$ only finitely often" (to show that, just pick $u\in(t,s)$, since $u<s$, the sequence must exceed $u$ infinitely often). Analogously, we prove that any $s<t$ fails to satisfy the property: "for any number $u<t$, the sequence exceeds $u$ infinitely often". So $s$ is the only one that satisfies both those properties. Nov 7, 2018 at 10:25

One intuition I certainly recommend is that the $$\limsup$$ is the infimum of the numbers that are passed by only finitely many $$a_n$$'s. More precisely, $$\limsup a_n=\inf\{\alpha\in\mathbb R:\mbox{the set \{a_n: a_n>\alpha\} is finite}\}$$

Pick $$\alpha$$ satisfying this. Then, there exists $$n\in\mathbb N$$ such that for every $$k\geq n$$, $$a_n\leq \alpha$$. Therefore, $$\sup_{k\geq n} a_n \leq \alpha \Rightarrow \limsup a_n = \inf_{n\in\mathbb N}\left(\sup_{k\geq n} a_n\right) \leq \alpha,$$ so $$\limsup a_n \leq \inf\{\alpha\in\mathbb R:\mbox{the set \{a_n: a_n>\alpha\} is finite}\}.$$

To prove the remaining inequality, note that if $$\beta<\inf\{\alpha\in\mathbb R:\mbox{the set \{a_n: a_n>\alpha\} is finite}\},$$ then certainly $$\beta\notin \{\alpha\in\mathbb R:\mbox{the set \{a_n: a_n>\alpha\} is finite}\},$$ so there are infinite many $$a_n$$'s greater than $$\beta$$, so for any $$n\in\mathbb N$$, $$\sup_{k\geq n} a_n > \beta,$$ then taking the infimum on the $$n$$'s, we get $$\limsup a_n = \inf_{n\in\mathbb N}\left(\sup_{k\geq n} a_n\right) \geq \beta.$$ Since this happens for any $$\beta<\inf\{\alpha\in\mathbb R:\mbox{the set \{a_n: a_n>\alpha\} is finite}\}$$, we get that $$\limsup a_n = \inf_{n\in\mathbb N}\left(\sup_{k\geq n} a_n\right) \geq \inf\{\alpha\in\mathbb R:\mbox{the set \{a_n: a_n>\alpha\} is finite}\}.$$

Yes it can be proved that if we consider the set $$S\subseteq \mathbb{\bar R}$$ of all the limits of the subsequences of $$a_n$$ we have that

$$\max\{S\}=\limsup a_n \in \mathbb{\bar R}$$ $$\min\{S\}=\liminf a_n\in \mathbb{\bar R}$$

extending the notation/definition also to the infinity cases.

That property with bounding evauation is used to prove what $$\limsup$$ and $$\liminf$$ are.

Refer also to the related


text:
# How to include a latex document as layout example?

I would like to include a latex document in an other latex document as figure to give an example for an layout, that means, the result of a latex document should be shown as, so to say a mini page in an other document:

This is a Title of my Main Document

This is some text in the document that talks about a layout and so on an
refers to a figure that should illustrate this talking (fig. 1).
+-------------------------+
|                         |
|   This is a latex       |
|   document, embedded    |
|   to the main document  |
|                         |
|                         |
|                         |
|                         |
|                         |
|                         |
|                         |
+-------------------------+
fig. 1

Is this possible?

-
do you want to show the source or the result? (both are possible) – David Carlisle Jul 16 '14 at 15:31
Do you want to include the output (for example, a single page) or the code? If you compile the example code in a separate file you can insert the pdf page as figure. – Sigur Jul 16 '14 at 15:32
@DavidCarlisle The result should be shown. – user5950 Jul 16 '14 at 15:34
@Sigur I updated my question. Is it possible to directly insert a latex file without compile the example code in a separate file? – user5950 Jul 16 '14 at 15:44
@user5950 Either you display the LaTeX code or you insert the result of compiling the LaTeX code, the result is (usually, nowadays) a PDF. So what -- PDF or *.tex file? – Keks Dose Jul 16 '14 at 15:58

You ask in comments if a separate file can be avoided, but it is much more accurate to just typeset the document to a separate file and include the resulting page with \includegraphics from graphicx package.

Obviously for small sections such as a single \textbf{hello} or even an enumerate environment it is possible just to put the code being demonstrated in a minipage, but in general it's very hard to isolate the style settings for the outer document from affecting the example.

For the LaTeX companion book which has dozens of these kinds of examples, the source code of the example document is in the main file, and is written verbatim to a separate file, a makefile controlling the process ensures that these generate files are run through latex to produce the result images to re-include back into the original document.

-

Simulating the more general case (just for fun). You can adapt it to suit your need.

\documentclass{article}

\usepackage{filecontents}
\begin{filecontents*}{layout.tex}
\documentclass[preview,border=12pt,12pt]{standalone}
\usepackage[a6paper,margin=1cm,showframe]{geometry}
\usepackage{lipsum}
\begin{document}
\section*{Minipage}
\noindent
\fbox{%
\begin{minipage}{.45\linewidth}
Left Minipage
\end{minipage}}
\hfill
\fbox{
\begin{minipage}{.45\linewidth}
Right Minipage
\end{minipage}}

\section*{Hello World}
{\tiny\lipsum[1]}
\end{document}
\end{filecontents*}

\immediate\write18{pdflatex layout.tex}

\usepackage[a5paper,vmargin=1.5cm,hmargin=1cm,landscape]{geometry}
\usepackage{xcolor,showexpl,lipsum}
\lstset
{
language={[LaTeX]TeX},
basicstyle=\scriptsize\tt,
backgroundcolor=\color{yellow!50},
numbers=none,
frame=single,
breaklines,
explpreset={},
pos=r,
varwidth,
}

\begin{document}
\section*{How to write a report with \LaTeX}
\LTXinputExample[graphic=layout.pdf,width=.5\textwidth]{layout.tex}
\lipsum[1-2]
\begin{figure}
\centering
\includegraphics[width=.5\linewidth]{layout}
\caption{The layout of our project report.}
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## FANDOM

38,121 Pages

Earth element attack

Description

Quake2 is an attack magic spell in Final Fantasy VII. It is granted by the Earth Materia at level two, and is the second-tier Earth-elemental spell above Quake and below Quake3.

## StatsEdit

Magic Materia Earth, Master Magic Moderate earth-elemental damage. 22 Added Cut, All, Final Attack, HP Absorb, Magic Counter, MP Absorb, MP Turbo, Quadra Magic, Sneak Attack, Steal as well Yes

Quake2 deals damage in the following formula:

$(24 / 16) * [ (Level + Magical Attack) * 6 ]$

where "Level" is the caster's current level and "Magical Attack" is their Magic atk stat. This is effectively 1.5x the base magic damage.

## UseEdit

Quake2 can be used by characters with the Earth Materia at level 2. It can be a useful spell that deals strong damage, particularly against enemies weak to Earth. Its main drawback is that it misses against flying opponents (see below) and takes 6000 AP to learn, compared to the similar spells Fire2, Ice2, and Bolt2, which only require 2000 AP to learn and have comparable damage. It also has a high MP cost early on, meaning it is best to save it against tougher enemies or bosses.

Final Fantasy VII does not have a "flying" state like some other entries to the series, but most flying enemies have their Earth affinity set to Immune. Usually when an element is used on an enemy that is immune it does 0 damage, but in the case of Earth it will miss. Earth will miss Earth immune enemies even if they are grounded (this also applies to when player's own characters are immune to Earth).

Quake2 can be amplified by pairing the Earth Materia with Support Materia. All Materia allows it to hit all enemies, while MP Turbo enhances its damage but also increases its MP cost. Quadra Magic can also make it useful against groups of enemies.

Quake2 becomes outclassed by Quake3 at later levels. However, as characters' max MP increases, it can be used much more frequently, especially if paired with MP Absorb. This allows it to overtake the role of Quake as the basic Earth spell. However, as characters' max MP increase further, it itself is overtaken by Quake3. The spell is later outclassed by Break, gained from the Contain Materia.

Quake2 can be used by the enemies Bagrisk, Ultimate Weapon (on land and Mideel), and Lost Number (Magic form).

Community content is available under CC-BY-SA unless otherwise noted.


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## Marquina, Antonio

Compute Distance To:
 Author ID: marquina.antonio Published as: Marquina, Antonio; Marquina, A. External Links: ORCID
 Documents Indexed: 52 Publications since 1973 Co-Authors: 23 Co-Authors with 42 Joint Publications 668 Co-Co-Authors
all top 5

### Co-Authors

 10 single-authored 7 Castro, Manuel J. 7 Gallardo, José M. 7 Serna, Susana 6 Donat, Rosa 5 Carreras, P. Perez 4 Candela, Vicente F. 3 Martínez, Vicente 2 Ibáñez, José María 2 Mulet, Pep 2 Osher, Stanley Joel 2 Peris, Rosa M. 2 Sanz-Serna, Jesús María 1 Chan, Tony Fan-Cheong 1 Chiavassa, Guillaume 1 Dussoulliez, Ph. 1 Fedkiw, Ronald P. 1 Font, José A. 1 Fuster, Robert 1 Jung, Miyoun 1 Merriman, Barry 1 Schmets, Jean 1 Stiriba, Youssef 1 Vese, Luminita A.
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### Serials

 7 Journal of Computational Physics 4 Journal of Scientific Computing 4 Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales de Madrid 3 Archiv der Mathematik 3 SIAM Journal on Scientific Computing 2 Computers and Fluids 2 Applied Mathematics and Computation 2 Computing 2 Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales de Madrid 1 American Mathematical Monthly 1 Computers & Mathematics with Applications 1 Computer Physics Communications 1 International Journal for Numerical Methods in Fluids 1 Bulletin de la Société Royale des Sciences de Liège 1 Collectanea Mathematica 1 Journal of Computational and Applied Mathematics 1 Manuscripta Mathematica 1 SIAM Journal on Numerical Analysis 1 Applied Numerical Mathematics 1 Physics of Fluids 1 SIAM Journal on Imaging Sciences
all top 5

### Fields

 24 Numerical analysis (65-XX) 19 Fluid mechanics (76-XX) 15 Partial differential equations (35-XX) 15 Functional analysis (46-XX) 2 Operator theory (47-XX) 2 Computer science (68-XX) 2 Information and communication theory, circuits (94-XX) 1 Ordinary differential equations (34-XX) 1 Integral transforms, operational calculus (44-XX) 1 Integral equations (45-XX) 1 Calculus of variations and optimal control; optimization (49-XX) 1 Mechanics of deformable solids (74-XX) 1 Classical thermodynamics, heat transfer (80-XX) 1 Relativity and gravitational theory (83-XX)

### Citations contained in zbMATH Open

34 Publications have been cited 918 times in 722 Documents Cited by Year
High-order total variation-based image restoration. Zbl 0968.68175
Chan, Tony; Marquina, Antonio; Mulet, Pep
2000
Recurrence relations for rational cubic methods. II: The Chebyshev method. Zbl 0714.65061
Candela, V.; Marquina, A.
1990
Recurrence relations for rational cubic methods. I: The Halley method. Zbl 0701.65043
Candela, V.; Marquina, A.
1990
Capturing shock reflections: An improved flux formula. Zbl 0847.76049
Donat, Rosa; Marquina, Antonio
1996
Local piecewise hyperbolic reconstruction of numerical fluxes for nonlinear scalar conservation laws. Zbl 0805.65088
Marquina, Antonio
1994
Power ENO methods: A fifth-order accurate weighted power ENO method. Zbl 1044.65071
Serna, Susana; Marquina, Antonio
2004
Explicit algorithms for a new time dependent model based on level set motion for nonlinear deblurring and noise removal. Zbl 0969.65081
Marquina, Antonio; Osher, Stanley
2000
A isobaric fix for the overheating problem in multimaterial compressible flows. Zbl 0933.76075
Fedkiw, Ronald P.; Marquina, Antonio; Merriman, Barry
1999
A flux-split algorithm applied to conservative models for multicomponent compressible flows. Zbl 1064.76078
Marquina, Antonio; Mulet, Pep
2003
Image super-resolution by TV-regularization and Bregman iteration. Zbl 1203.94014
Marquina, Antonio; Osher, Stanley J.
2008
A flux-split algorithm applied to relativistic flows. Zbl 0930.76054
Donat, R.; Font, J. A.; Ibáñez, J. M.; Marquina, A.
1998
A class of incomplete Riemann solvers based on uniform rational approximations to the absolute value function. Zbl 1299.76181
Castro, Manuel J.; Gallardo, José M.; Marquina, Antonio
2014
Capturing shock waves in inelastic granular gases. Zbl 1138.76378
Serna, Susana; Marquina, Antonio
2005
Diffusion front capturing schemes for a class of Fokker-Planck equations: application to the relativistic heat equation. Zbl 1423.35365
Marquina, Antonio
2010
Nonlinear inverse scale space methods for total variation blind deconvolution. Zbl 1177.65193
Marquina, Antonio
2009
Barrelledness conditions on $$c_0(E)$$. Zbl 0396.46005
Marquina, A.; Sanz Serna, J. M.
1979
The convergence of the perturbed Newton method and its application for ill-conditioned problems. Zbl 1255.65095
Peris, R.; Marquina, A.; Candela, V.
2011
Approximate Osher-Solomon schemes for hyperbolic systems. Zbl 1410.76325
Castro, Manuel J.; Gallardo, José M.; Marquina, Antonio
2016
Equilibrium real gas computations using Marquina’s scheme. Zbl 1047.76084
Stiriba, Youssef; Marquina, Antonio; Donat, Rosa
2003
Shooting methods for one-dimensional diffusion-absorption problems. Zbl 0821.65062
Martínez, V.; Marquina, A.; Donat, R.
1994
A note on the closed graph theorem. Zbl 0351.46003
Marquina, Antonio
1977
On bornological c//0(E) spaces. Zbl 0514.46025
Marquina, Antonio; Schmets, Jean
1982
Geometric series in incomplete normed algebras. Zbl 0553.46032
Fuster, R.; Marquina, A.
1984
Capturing blast waves in granular flow. Zbl 1194.76100
Serna, Susana; Marquina, Antonio
2007
Approximate Osher-Solomon schemes for hyperbolic systems. Zbl 1402.35019
Castro, M. J.; Gallardo, J. M.; Marquina, A.
2016
Anomalous wave structure in magnetized materials described by non-convex equations of state. Zbl 1321.76079
Serna, Susana; Marquina, Antonio
2014
Barrelledness conditions on $$c_0(E)$$. Zbl 0387.46004
Marquina, A.; Sanz Serna, J. M.
1979
Jacobian-free approximate solvers for hyperbolic systems: application to relativistic magnetohydrodynamics. Zbl 1411.65119
Castro, Manuel J.; Gallardo, José M.; Marquina, Antonio
2017
Shooting methods for 1D steady-state free boundary problems. Zbl 0769.65057
Marquina, A.; Martínez, V.
1993
Computation of travelling wave solutions of scalar conservation laws with a stiff source term. Zbl 1048.76041
Martínez, Vicente; Marquina, Antonio
2003
On quasibarrelled spaces. Zbl 0281.46004
Marquina, A.; Perez Carreras, P.
1974
Afternotes on PHM: Harmonic ENO methods. Zbl 1065.65107
Marquina, Antonio; Serna, Susana
2003
Fronts propagating with signal dependent speed in limited diffusion and related Hamilton-Jacobi formulations. Zbl 1394.35132
Serna, Susana; Marquina, Antonio
2013
On the numerical approximation of the length of (implicit) level curves. Zbl 1203.65040
Candela, Vicente F.; Marquina, Antonio
2008
Jacobian-free approximate solvers for hyperbolic systems: application to relativistic magnetohydrodynamics. Zbl 1411.65119
Castro, Manuel J.; Gallardo, José M.; Marquina, Antonio
2017
Approximate Osher-Solomon schemes for hyperbolic systems. Zbl 1410.76325
Castro, Manuel J.; Gallardo, José M.; Marquina, Antonio
2016
Approximate Osher-Solomon schemes for hyperbolic systems. Zbl 1402.35019
Castro, M. J.; Gallardo, J. M.; Marquina, A.
2016
A class of incomplete Riemann solvers based on uniform rational approximations to the absolute value function. Zbl 1299.76181
Castro, Manuel J.; Gallardo, José M.; Marquina, Antonio
2014
Anomalous wave structure in magnetized materials described by non-convex equations of state. Zbl 1321.76079
Serna, Susana; Marquina, Antonio
2014
Fronts propagating with signal dependent speed in limited diffusion and related Hamilton-Jacobi formulations. Zbl 1394.35132
Serna, Susana; Marquina, Antonio
2013
The convergence of the perturbed Newton method and its application for ill-conditioned problems. Zbl 1255.65095
Peris, R.; Marquina, A.; Candela, V.
2011
Diffusion front capturing schemes for a class of Fokker-Planck equations: application to the relativistic heat equation. Zbl 1423.35365
Marquina, Antonio
2010
Nonlinear inverse scale space methods for total variation blind deconvolution. Zbl 1177.65193
Marquina, Antonio
2009
Image super-resolution by TV-regularization and Bregman iteration. Zbl 1203.94014
Marquina, Antonio; Osher, Stanley J.
2008
On the numerical approximation of the length of (implicit) level curves. Zbl 1203.65040
Candela, Vicente F.; Marquina, Antonio
2008
Capturing blast waves in granular flow. Zbl 1194.76100
Serna, Susana; Marquina, Antonio
2007
Capturing shock waves in inelastic granular gases. Zbl 1138.76378
Serna, Susana; Marquina, Antonio
2005
Power ENO methods: A fifth-order accurate weighted power ENO method. Zbl 1044.65071
Serna, Susana; Marquina, Antonio
2004
A flux-split algorithm applied to conservative models for multicomponent compressible flows. Zbl 1064.76078
Marquina, Antonio; Mulet, Pep
2003
Equilibrium real gas computations using Marquina’s scheme. Zbl 1047.76084
Stiriba, Youssef; Marquina, Antonio; Donat, Rosa
2003
Computation of travelling wave solutions of scalar conservation laws with a stiff source term. Zbl 1048.76041
Martínez, Vicente; Marquina, Antonio
2003
Afternotes on PHM: Harmonic ENO methods. Zbl 1065.65107
Marquina, Antonio; Serna, Susana
2003
High-order total variation-based image restoration. Zbl 0968.68175
Chan, Tony; Marquina, Antonio; Mulet, Pep
2000
Explicit algorithms for a new time dependent model based on level set motion for nonlinear deblurring and noise removal. Zbl 0969.65081
Marquina, Antonio; Osher, Stanley
2000
A isobaric fix for the overheating problem in multimaterial compressible flows. Zbl 0933.76075
Fedkiw, Ronald P.; Marquina, Antonio; Merriman, Barry
1999
A flux-split algorithm applied to relativistic flows. Zbl 0930.76054
Donat, R.; Font, J. A.; Ibáñez, J. M.; Marquina, A.
1998
Capturing shock reflections: An improved flux formula. Zbl 0847.76049
Donat, Rosa; Marquina, Antonio
1996
Local piecewise hyperbolic reconstruction of numerical fluxes for nonlinear scalar conservation laws. Zbl 0805.65088
Marquina, Antonio
1994
Shooting methods for one-dimensional diffusion-absorption problems. Zbl 0821.65062
Martínez, V.; Marquina, A.; Donat, R.
1994
Shooting methods for 1D steady-state free boundary problems. Zbl 0769.65057
Marquina, A.; Martínez, V.
1993
Recurrence relations for rational cubic methods. II: The Chebyshev method. Zbl 0714.65061
Candela, V.; Marquina, A.
1990
Recurrence relations for rational cubic methods. I: The Halley method. Zbl 0701.65043
Candela, V.; Marquina, A.
1990
Geometric series in incomplete normed algebras. Zbl 0553.46032
Fuster, R.; Marquina, A.
1984
On bornological c//0(E) spaces. Zbl 0514.46025
Marquina, Antonio; Schmets, Jean
1982
Barrelledness conditions on $$c_0(E)$$. Zbl 0396.46005
Marquina, A.; Sanz Serna, J. M.
1979
Barrelledness conditions on $$c_0(E)$$. Zbl 0387.46004
Marquina, A.; Sanz Serna, J. M.
1979
A note on the closed graph theorem. Zbl 0351.46003
Marquina, Antonio
1977
On quasibarrelled spaces. Zbl 0281.46004
Marquina, A.; Perez Carreras, P.
1974
all top 5

### Cited by 1,051 Authors

 47 Argyros, Ioannis Konstantinos 26 Amat, Sergio P. 25 George, Santhosh 23 Hernández, Miguel Ángel 18 Kou, Jisheng 18 Marquina, Antonio 17 Wang, Xiuhua 14 Mulet, Pep 13 Castro, Manuel J. 13 Ezquerro, José Antonio 12 Gutiérrez, José Manuel 12 Shu, Chi-Wang 11 Busquier, Sonia 11 Chen, Ke 11 Magreñán, Ángel Alberto 10 Parés Madroñal, Carlos 10 Tang, Huazhong 9 Baeza, Antonio 9 Candela, Vicente F. 9 Donat, Rosa 9 Gallardo, José M. 9 Lv, Xiaoguang 9 Zorío, David 8 Fedkiw, Ronald P. 8 Huang, Ting-Zhu 8 Laghrib, Amine 8 Liandrat, Jacques 8 Serna, Susana 8 Tai, Xuecheng 7 Guo, Zhichang 7 Liu, Jun 7 Pang, Zhifeng 7 Wu, Boying 6 Gu, Chuanqing 6 Gupta, Dharmendra Kumar 6 Hadri, Aissam 6 Hakim, Abdelilah 6 Osher, Stanley Joel 6 Qamar, Shamsul 6 Schönlieb, Carola-Bibiane 6 Trillo, Juan Carlos 5 Behl, Ramandeep 5 Burger, Martin 5 Bürger, Raimund 5 Kang, Myungjoo 5 Ling, Yonghui 5 Liu, Tiegang 5 Neta, Beny 5 Prashanth, Maroju 5 Romero, Natalia 5 Shen, Zuowei 5 Shyue, Keh-Ming 5 Steidl, Gabriele 5 Wei, Guowei 5 Yi, Dokkyun 5 Zhao, Xile 5 Zheng, Lin 4 Bhuruth, Muddun 4 Caselles Costa, Vicent 4 Dadourian, Karine 4 Ji, Hui 4 Jiang, Le 4 Jung, Miyoun 4 Khoo, Boo Cheong 4 Liu, Xinwu 4 Lou, Yifei 4 Peer, Arshad Ahmud Iqbal 4 Peris, Rosa M. 4 Salanova, M. Amparo 4 Shi, Baoli 4 Sicilia, Juan Antonio 4 Sun, Jiebao 4 Udaykumar, Holavanahalli S. 4 Xu, Xiubin 4 Yang, Xiaoping 3 Abedian, Rooholah 3 Aslam, Tariq D. 3 Bergmann, Ronny 3 Calvo, Juan G. 3 Chan, Raymond Hon-Fu 3 Chan, Tony Fan-Cheong 3 Chang, Qianshun 3 Chun, Changbum 3 Dauhoo, Muhammad Zaid 3 Deng, Xi 3 Dong, Fangfang 3 Dumbser, Michael 3 Farhat, Charbel H. 3 Font, José A. 3 Huang, Jin 3 Iollo, Angelo 3 Jaiswal, Jai Prakash 3 Liu, Pengfei 3 Liu, Yuan 3 Ma, Tian-Hui 3 Motsa, Sandile Sydney 3 Orcos, Lara 3 Ortega, Sergio 3 Papafitsoros, Konstantinos 3 Parhi, Sanjaya Kumar ...and 951 more Authors
all top 5

### Cited in 164 Serials

 99 Journal of Computational Physics 52 Journal of Scientific Computing 41 Applied Mathematics and Computation 29 Numerical Algorithms 28 Computers & Mathematics with Applications 25 Journal of Computational and Applied Mathematics 23 Journal of Mathematical Imaging and Vision 21 Applied Numerical Mathematics 18 Computers and Fluids 14 International Journal of Computer Mathematics 12 SIAM Journal on Imaging Sciences 11 Mathematical Problems in Engineering 10 Inverse Problems and Imaging 9 Communications in Computational Physics 9 S$$\vec{\text{e}}$$MA Journal 8 Nonlinear Analysis. Theory, Methods & Applications. Series A: Theory and Methods 8 Multidimensional Systems and Signal Processing 8 SIAM Journal on Scientific Computing 8 Abstract and Applied Analysis 7 Journal of Mathematical Analysis and Applications 7 Advances in Computational Mathematics 6 International Journal of Computer Vision 5 Journal of Fluid Mechanics 5 Mathematical Methods in the Applied Sciences 5 Archiv der Mathematik 5 Journal of Optimization Theory and Applications 5 Applied Mathematical Modelling 5 Pattern Recognition 5 Applicationes Mathematicae 5 Journal of Mathematical Chemistry 5 Journal of Inequalities and Applications 5 Journal of Applied Mathematics and Computing 5 International Journal of Applied and Computational Mathematics 4 Inverse Problems 4 Information Sciences 4 Numerical Methods for Partial Differential Equations 4 Applied Mathematics Letters 4 Applied and Computational Harmonic Analysis 4 Nonlinear Analysis. Real World Applications 4 International Journal of Computational Methods 3 Computer Methods in Applied Mechanics and Engineering 3 Computer Physics Communications 3 Journal of the Franklin Institute 3 BIT 3 Computing 3 Mathematische Annalen 3 Mathematical and Computer Modelling 3 Computational Optimization and Applications 3 Physics of Fluids 3 International Journal of Computational Fluid Dynamics 3 Living Reviews in Relativity 3 International Journal for Numerical Methods in Biomedical Engineering 2 Bulletin of the Australian Mathematical Society 2 International Journal for Numerical Methods in Fluids 2 Mathematics of Computation 2 International Journal for Numerical Methods in Engineering 2 Numerical Functional Analysis and Optimization 2 Results in Mathematics 2 SIAM Journal on Numerical Analysis 2 Physica D 2 Acta Mathematicae Applicatae Sinica. English Series 2 Signal Processing 2 M$$^3$$AS. Mathematical Models & Methods in Applied Sciences 2 The Journal of Analysis 2 Applied Mathematics. Series B (English Edition) 2 Computational and Applied Mathematics 2 Journal of Inverse and Ill-Posed Problems 2 Novi Sad Journal of Mathematics 2 European Journal of Mechanics. B. Fluids 2 The ANZIAM Journal 2 Journal of Applied Mathematics 2 Milan Journal of Mathematics 2 Acta Numerica 2 International Journal of Numerical Analysis and Modeling 2 Boundary Value Problems 2 Journal of Fixed Point Theory and Applications 2 Advances in Mathematical Physics 2 Advances in Applied Mathematics and Mechanics 2 Numerical Algebra, Control and Optimization 2 East Asian Journal on Applied Mathematics 2 Cogent Mathematics 1 Applicable Analysis 1 Classical and Quantum Gravity 1 Indian Journal of Pure & Applied Mathematics 1 Jahresbericht der Deutschen Mathematiker-Vereinigung (DMV) 1 Journal of Mathematical Biology 1 Journal of Mathematical Physics 1 Rocky Mountain Journal of Mathematics 1 Shock Waves 1 Bulletin of the Polish Academy of Sciences. Technical Sciences 1 Calcolo 1 International Journal of Mathematics and Mathematical Sciences 1 Inventiones Mathematicae 1 Journal of Differential Equations 1 Journal of the Korean Mathematical Society 1 Kodai Mathematical Journal 1 Manuscripta Mathematica 1 Mathematics and Computers in Simulation 1 Numerische Mathematik 1 Proceedings of the American Mathematical Society ...and 64 more Serials
all top 5

### Cited in 38 Fields

 475 Numerical analysis (65-XX) 197 Information and communication theory, circuits (94-XX) 176 Fluid mechanics (76-XX) 143 Partial differential equations (35-XX) 110 Computer science (68-XX) 109 Operator theory (47-XX) 69 Calculus of variations and optimal control; optimization (49-XX) 38 Operations research, mathematical programming (90-XX) 20 Biology and other natural sciences (92-XX) 19 Mechanics of deformable solids (74-XX) 18 Integral equations (45-XX) 16 Functional analysis (46-XX) 15 Approximations and expansions (41-XX) 14 Real functions (26-XX) 12 Statistics (62-XX) 12 Classical thermodynamics, heat transfer (80-XX) 11 Ordinary differential equations (34-XX) 9 Harmonic analysis on Euclidean spaces (42-XX) 8 Relativity and gravitational theory (83-XX) 6 Astronomy and astrophysics (85-XX) 6 Geophysics (86-XX) 5 Differential geometry (53-XX) 5 Statistical mechanics, structure of matter (82-XX) 5 Systems theory; control (93-XX) 4 Global analysis, analysis on manifolds (58-XX) 3 Functions of a complex variable (30-XX) 3 Dynamical systems and ergodic theory (37-XX) 3 Probability theory and stochastic processes (60-XX) 3 Mechanics of particles and systems (70-XX) 2 Linear and multilinear algebra; matrix theory (15-XX) 2 Integral transforms, operational calculus (44-XX) 2 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 1 General and overarching topics; collections (00-XX) 1 Measure and integration (28-XX) 1 Difference and functional equations (39-XX) 1 Sequences, series, summability (40-XX) 1 General topology (54-XX) 1 Optics, electromagnetic theory (78-XX)