text:
## double angular body label 7

I have a problem, because the angular generates a double label, so being contained in another screen is not complete, no matter what happened?

## 7 – In the URL of the file, a double slash works, but not once

My website contains a URL to a PDF file, as shown in the HTML code below.

``````See the information here
``````

After deploying it on a liver server, I can see the following URL:

``````http://example.com/sites/default/files/media/info.pdf
``````

Recently I have re-downloaded the PDF to the server via FTP. Please note that since I kept the same file name, the old PDF is replaced by the new PDF. Now I can see the new PDF using FTP.

Unfortunately, I am not able to see the changes in PDF. I still see the old PDF.

Surprisingly, when I used the URL below, I can see the new PDF-

``````http://example.com//sites/default/files/media/info.pdf
``````

The URL above contains a double slash character. Please note that I have cleaned the website cache and even tried other recently installed browsers, but the problem persists.

I want to see the new PDF. Any alternative, please?

## Product Identification – Which game uses dice with engraved faces, weapon symbols, double weapon symbols and object symbols?

Note to reviewers, this question is part of several similar questions. It has been broken down so as not to violate any other guidelines. Although they are almost identical, they are not duplicates because they require the identification of different dice. This means that even though the questions are similar, they are not the same. In addition, as they each ask to identify different dice, the answers to each question will be very different, so these are not duplicates.

For readers, if you have already read another part of this series of questions, feel free to skip the text as it is the same, the only section you should focus on is the images themselves.

For reference, here are the rest of the questions in this series:

Which game uses dice with cyborg skulls instead of symbol 1?

Which game uses dice filled with glitter with an arrow-shaped symbol with hook-shaped symbols on each side?

Which game uses six-sided dice with symbols as well as numbers on the 5 and 6 sides?

Which game uses dice with engraved faces, weapon symbols, double weapon symbols and object symbols?

Which game uses dice with directional arrows, forbidden signs, explosions, arrows and reticles?

Some time ago, I bought a few bags of 2nd die from factory. Among these bags, there are a handful of dice that I could not identify immediately. I tried to check dice websites, search Google, and even use reverse image searches, to no avail. For some, I do not even know how to describe them in a concise way to search for them accurately.

What I want to know is what game are these dice? Or, if they do not come from a specific game, how they call. Below pictures of the dice:

## Litecoin is finding resistance and will be double! Enter the free Bitmex Dynamic Bot for the first 500 traders. – Advertising, offers

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.

## For Soulbolt Double Blade Blade skill, is the pair of ranged attacks performed at the highest BAB of the PC?

The skill simply indicates ….

Double shot: In standard action, a soulbolt can use his psychic attack to perform two ranged attacks with his lightning. Attacks do not cause additional damage related to psychic strikes.

Is this pair of ranged attacks at the highest BAB of the PC or are the rules of two-weapon combat applicable?

## linear algebra – Double base in \$ V ^ {**} \$

(All vector spaces mentioned here are finished.)

Suppose we are given isomorphism $$Phi: V to V ^ {**}: v mapsto ( Phi_v: V * to mathbb {K}: phi mapsto phi (v))$$. Consider a base $$( varepsilon_i) _ {i in I}$$ of $$V ^ *$$. Affirm that there is a $$(e_i) _ {i in I}$$ of $$V$$ such as $$( varepsilon_i) _ {i in I}$$ is his double base.

My book says that this statement can be demonstrated by considering the isomorphism $$Phi$$ and looking at the double base of $$( varepsilon_i) _ {i in I}$$ in $$V ^ {**}$$. I'm not really sure what that means. How to find this double base in $$V ^ {**}$$. How can I use it to show the claim?

Thank you!

## How can I solve the double integral \$ iint e ^ {sin (x) cos (y)} dxdy qquad qquad x ^ {2} + y ^ {2} leq \$ 4

How can I solve the double integral $$iint e ^ {sin (x) cos (y)} dxdy qquad qquad and qquad x ^ {2} + y ^ {2} leq 4$$

## linear programming – On optimal double solutions for minimum weight, perfect matching problems in the case of metric weights

Next Lovasz-Plummer (matching theory, North Holland 1986, Theorem 9.2.1),
the problem of perfect matching of minimum weight on a complete graph
$$G$$ with an even number of vertices and weights $$w: E (G) to mathbb {R}$$, $$w 0$$, can be reduced to the following linear
program.

minimize $$w ^ top cdot x$$ (or $$x: E (G) to mathbb {R}$$)

subject to

(1) $$x (e) ge 0$$ for each $$e in E (G)$$

(2) $$x (C) = 1$$ for each trivial odd cup $$C$$ (trivial cut is a cut that
separates a vertex from the rest of the graph)

(3) $$x (C) 1$$ for each non-trivial odd cut $$C$$.

We introduce a variable $$y_C$$ for each odd cut $$C$$.

The double program is:

maximize $$sum_C y_C$$

subject to

(D1) $$y_C 0$$ for each non-trivial odd cut $$C$$

(D2) $$sum_ {C ~ { rm containing ~} e} y_C the w (e)$$ for each $$e in E (G)$$.

Now, we assume that the weight function $$w$$ is a metric on the top
together $$G$$ (C & # 39; s, $$w (e) =$$ the distance between the ends of $$e$$).

Question: Is it possible to show (under this hypothesis) that
among the optimal dual solutions, there is one for which
$$y_C 0$$ for all the strange cuts, including the most trivial.

## Take 1 double 2

We produce high quality undetectable counterfeit banknotes of the following currencies:

US dollars (USD)
-Large British Pounds (GBP)
-Euros
Dirhams -Emirati (AED).
These notes satisfy UV tests and pencil tests with all watermarks and security features, just as you will get real money.

We have a very long experience in the following areas:
-20 years of experience in shipping / shipping illegal products.
-15 years of printing black notes, counterfeit banknotes undetectable premium quality.
– More than 20 employees and 50 agents worldwide with terminals in several countries, enabling us to make shipments all over the world from all angles, with fast delivery.
-partners with workers in 5 major central banks and 3 large transport companies.

These experiences and partnerships, combined with the ever-increasing advanced technology, allow us to present invoices as authentic and authentic as to deliver safe and secure deliveries.

We not only provide excellent invoices, but we guarantee a safe and secure delivery, as we have encrypted everything with us.
In addition to this, we provide our justifications that we are legitimate, and if you are interested in first-year bills, feel free to email us at the following address:


text:
Equations > Physics > Electricity and Magnetism > Magnetic force on a moving charge

### Magnetic force on a moving charge

Latex Code:

MathML Code:

 $F=qv×B$

MathType 5.0:

format: plain


text:
Python implementation of Schoof's algorithm for counting the points on elliptic curves over finite fields

## Project description

Python3 Implementation of Schoof's Algorithm
============================================

This is an implementation of Schoof's algorithm for counting the
points on elliptic curves over finite fields (Schoof, René. Elliptic
curves over finite fields and the computation of square roots mod
p. *Mathematics of Computation*, 44(170):483–494, 1985).

Elliptic curve cryptographic algorithms base their security guarantees
on the number of points on the used elliptic curve. Because naive
point counting is infeasible, having a fast counting algorithm is
important to swiftly decide whether a curve is safe to use in
cryptography or not. René Schoof's algorithm for counting the points
on an elliptic curve over a finite field is the foundation for the
(asymptotically) fastest [Schoof–Elkies–Atkin][sea-algorithm] counting
algorithm.

**Implementation.**
The implementation is written in [Python 3][python] and is kept as
simple as possible. Its goal is to give insight into the mathematics
of the algorithm without the use of (too) high-level concepts. For a
(pretty) gentle introduction into why and how Schoof's algorithm
[*An Elementary Derivation and Implementation of Schoof's Algorithm for Counting Points on Elliptic Curves*][thesis].
In the thesis, I try to explain *how one might get the idea for the
algorithm*. This understanding of *why* things look the way they do
is often neglected in mathematics.

Schoof's algorithm uses arithmetic on elliptic curves, finite fields,
rings of polynomials, and quotient rings. This repository therefore
contains Python modules for all these concepts that can be used on
their own.

System Requirements
-------------------

The algorithm is implemented in version 3.1 of [Python][python], an
open licensed dynamic programming language available on all common
platforms. To find out whether a compatible version of Python is
already installed on your system, execute python --version in a
terminal. The command will return the version number if Python is
available. Note that the version 3 interpreter could also be named
python3. Please see the *Using Python* part of Python's
[documentation][python-using] for system installation instructions;
follow the steps below to set up Python locally in your account.

### Local Installation from Source Code

Installing Python from source code requires a C compiler; on Linux and
Unix systems, one is almost always available. The following steps
install Python on a Linux system:

from the Python website at
* **Compile.** Open a terminal and create a temporary directory, say
${HOME}/tmp/, by executing mkdir${HOME}/tmp/. Change into the
temporary directory and extract the source tar ball: cd
${HOME}/tmp/ and then tar xzvf Python-3.1.2.tgz; adjust the path and file name accordingly. If you downloaded the bzipped source tar ball, use tar xjvf Python-3.1.2.tar.bz2 instead. Next, change into the directory that contains the extracted source code, for instance ${HOME}/tmp/Python-3.1.2/. Configure the build
system by executing ./configure --prefix=${HOME}/python3. The prefix is the path that will be the root of the Python installation, so adjust it to taste. In case required components are missing, the command will exit with an error message. In this case, please install the components and re-execute configure. If everything worked, then the last line of output by configure will be creating Makefile. To start the compilation, execute make. * **Install.** Use make install to install Python after the compilation finished. * **Set Up Environment.** To enable the local Python installation, add its interpreter and modules to the respective search paths: execute export PATH=${HOME}/python3/bin:${PATH} to tell the shell where to find the python3 interpreter; adjust the path to your prefix for configure. Likewise, execute export PYTHONPATH=${HOME}/python3/lib/python3.1 to tell Python where to
find its modules.

Note that the scope of export is the current shell. Thus you have
to issue both commands in every freshly opened terminal you wish to
use for Python 3.1 programs.

Program Execution
-----------------

The implementations work without any installation; they may be
executed directly from the checked out repository. However, they
expect a set up Python 3.1 run-time environment as explained above.

The root directory contains the point counting programs: the file
naive_schoof.py is the implementation discussed in
\autoref{sec:implementation}; the file reduced_computation_schoof.py
is the version with better constants using a smarter computation order
and Hasse's theorem. Curves for counting are specified as
space-separated triples *p*, *A*, and *B*: *p* is the prime size of
the galois field *GF[p]*, and *A* and *B* are the curve parameters.

### Example

Suppose you want to count the number of points on the elliptic curve
over *GF[23]* with parameters *A=4* and *B=2*. If the current
directory in the terminal is the repository root, then
executing python3 naive_schoof.py 23 4 2 yields the output

~~~~~
Counting points on y^2 = x^3 + 4x + 2 over GF<23>: 21
~~~~~

### Command Line Parameters

The program supports the command line options described below, which
for instance allow to read the curves from a file, or to create
execution profiles for performance analysis.

* --version: Show program's version number and exit.
* -h, --help: Show a help message and exit.
* -i *f*, --input-file=*f*: Read the curve parameters from file
*f*. Lines in *f* must contain either a comment, or a
space-separated triple of curve parameters *p*, *A*, and *B*.
Comment lines start with a hash (#); these and empty lines are
skipped.
* -o *f*, --output-file=*f*: Write the results to file *f* instead
of showing them on the terminal. Each line of input generates one
corresponding line of output.
* -t *s*, --timelimit=*s*: Terminate the program if processing an
input triple takes longer than *s* seconds. The program ends if the
time limit expires; no subsequent lines will be read from an input
file. Thus, sort the parameters in input files ascending in length
of the prime *p* to avoid triggering the time limit too early.
* -p, --create-profile: Create an execution profile for each
processed input triple. The profile consists of a call profile
generated with the cProfile Python module, and a file with data
about elapsed time and used memory.
* -d *d*, --profile-directory=*d*: If profiling is enabled with
the -p flag, then the collected data is written to the directory
*d*. The default value is the current directory.

Profiling Tools
---------------

The tools/ directory contains several programs to post-process
profiles resulting from a set -p flag. For example, it includes a
call profile converter that outputs [Callgrind][callgrind] files. Use
[KCacheGrind][kcachegrind] to interactively inspect Callgrind files.

Further Documentation
---------------------

The implementation comes with extensive [API documentation][api] that
explains the purpose and usage of all classes.

Furthermore, the directory _test contains unit tests to verify
correct arithmetic. The unit tests are designed to be easily extended
to further implementations of the same objects. Thus, mistakes in new
designs should be simpler to locate. To run the unit tests, execute
python3 -m _test in the repository root directory.

-------

Copyright (c) 2010--2012 Peter Dinges <pdinges@acm.org>.

The software in this repository is free software: you can redistribute
it and/or modify it under the terms of the GNU General Public License
as published by the Free Software Foundation, either version 3 of the

The software is distributed in the hope that it will be useful, but
WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
General Public License for more details.

You should have received a copy of the [GNU General Public License][gpl3]
along with this program. If not, see <http://www.gnu.org/licenses/>.

[api]: http://pdinges.github.com/python-schoof
[callgrind]: http://valgrind.org/info/tools.html "Callgrind is part of the Valgrind tool suite"
[kcachegrind]: http://kcachegrind.sourceforge.net/html/Home.html "Interactive viewer for Callgrind files."
[python]: http://python.org "Python Programming Language"
[python-using]: http://docs.python.org/py3k/using/index.html "Python Setup and Usage"
[sea-algorithm]: http://en.wikipedia.org/wiki/Schoof%E2%80%93Elkies%E2%80%93Atkin_algorithm "Schoof-Elkies-Atkin algorithm for counting points on elliptic curves over finite fields."

## Project details

### Source Distribution

pyschoof-0.1.tar.gz (93.3 kB view hashes)

Uploaded source


text:
An In-Depth Guide to Infrared Photography: Setup and Capture

This post is part of a series called Infrared Photography.
25 Examples of Infrared Photography
An In-Depth Guide to Infrared Photography: Processing

Infrared photographs are a stunning way to make unique landscape photos, and will add variety to your portfolio. Understanding how to take and process them properly can be very difficult and time consuming due to the steep learning curve involved, but I hope to help.

In this two-part tutorial, you’ll learn about the infrared choices available, how to take infrared photographs using a filter, process your photographs to a high standard and avoid any common mistakes along the way. Let’s begin.

Infrared Photography: A Brief Introduction

Infrared (IR) is a challenging form of photography. When going through this tutorial, you need to be very methodical about the process of setting up your equipment and taking the shot.

It would be a good idea to print out some notes to take with your when shooting as there is a lot to think about, and it can easily go wrong. I have created a handy checklist for you to print and fold up into your bag.

Check to See if Your Camera Can Capture IR Light

Check before you go and buy filter, as certain cameras cannot capture IR. A quick tip to find out if you can capture IR is to put your camera in live view and point a remote control at the lens and press a few buttons. If you notice a flashing red tinted light, your camera can capture IR light well.

If the light is dimmer, this means your camera can still capture IR, but it means your exposure times will skyrocket due to your camera having a strong internal IR blocking filter.

If your camera does not have live view, set a long exposure, start taking the shot, and point and press buttons on the remote towards the lens. The resulting picture should show the red tinted light. If you do not see anything from the remote, this means your camera cannot capture IR, and cannot use this tutorial to effect.

Check for this light.

There are a few offerings available when it comes to infrared filters. There are both screw in types from brands such as Hoya, and square filters from Cokin, for example.

Screw-in filters are a great way to make infrared photographs. They are, however, relatively expensive. My advice is to buy a well-known brand to get the best results. I purchased a Hoya R72 and I'm impressed by the results, although they sell for around $100+ for the larger sizes. Square filters offer the advantage of being able to slot in and out quickly. While these work fine, the risk of the shot being ruined from light leaks is much higher than a screw in type. Prices range from around$60 for normal sizes.

A handy tip is to buy the biggest screw-in filter you can afford, and use a stepping ring so it can fit on all your other lenses without having to buy a new filter for each lens.

A Hoya 720 IR filter.

Wavelength & Other Infrared Options

720nm is considered standard IR. I believe it is best to start off with this wavelength. There are others available such as 900nm models (RM90), but the prices of these are astronomical. I have seen them going for around \$300+. These are reserved for serious IR photographers with deep pockets only!

There is another option for IR photography if you do not want to use a filter. You can have your DSLR converted permanently to capture the IR spectrum. This can be done by sending away your camera and a lens to be calibrated. It is very expensive, and leaves your camera only able to shoot in IR. Check out www.lifepixel.com for a lot more information on this, as there are many options available.

Visible light vs. IR

What/When to Shoot?

One of the most popular IR genres is landscape photography. Because of the effects of IR, foliage can be rendered pure white when processed, making photos look eerie, and have a haunting sense about them. Example subjects to experiment with include trees, flowers and grass.

Ideal conditions for IR photography are a bright sunny days, rendering (when false color processed) skies a deep blue and foliage a bright white. That doesn’t mean that you can’t get brilliant results on overcast days.

With the longer exposure times available to you with the IR filter fitted, results will be similar to a Neutral Density (ND) filter, giving a strong sense of motion to the photograph.

It is up to you to experiment, so don’t be limited by these example subjects and situations!

Example subjects.

Lens Issues

Certain lenses can produce abnormalities when shooting in IR, namely hotspots. When looking at an image, a bright, discolored ring can be seen in the middle of the photograph when it has been affected.

Other issues can occur such as streaks over the image. These can be minimized is post-production, but are a real pain to deal with.

Currently, there is no truly comprehensive list to show what lenses work well, and what lenses produce hotspots, but DPAnswers have produced a respectable guide on a number of popular lenses to help you see what lenses could cause you problems when shooting.

A telephoto lens.

Let's move on to the process of shooting an IR image.

1. Setting Up

Setting up the camera is an important step in the process to getting great IR photos. Do not attach the IR filter until you have completed the necessary steps, including focusing, exposure and white balance.

First, set up your camera on a sturdy tripod. Put your camera bag on the hook of your tripod to add extra weight and minimize movement.

Next are a few settings that are important to help you get the cleanest image possible:

Shoot RAW: Shooting RAW allows you to easily alter the white balance in post-production without any problems clipping colors. Shoot in uncompressed RAW so you get the most bits possible. This will give you greater flexibility when processing later. Never shoot JPEG, otherwise when processing images, noise and clipping will become more apparent.

Turn off long exposure noise reduction: Because of the long exposures involved in IR photography, turning off long exposure noise reduction will mean no in-camera noise processing, resulting in no lag when the shutter closes. This also means you can finely tune the strength of the noise-reduction later.

Turn on exposure delay mode/use mirror lockup: Turning on exposure delay mode or using mirror lock up will make sure vibrations from the mirror are minimized when you release the shutter.

Remote release/self timer: Using a remote release is optional, but will further reduce vibrations as you will not be touching the camera when you release the shutter. Alternatively, you can use a 2 second self-timer if you do not have a remote.

Camera on a tripod.

2. White Balance

White balance is an important part of getting a decent IR image. You can use a pre-set value, or you can use Pre-White Balance to capture the White Balance in the current conditions. Either way, you will need to alter it in post-production to get any desirable results.

Using preset White Balance values is fine. Incandescent is the coolest preset available.

Taking a pre-white balance reading from a subject when on location is even better. Because IR will make foliage under sunlight white, this is what the white balance target will be.

Go into the white balance menu and select ‘PRE’ as the white balance. Then...

• Press OK
• Select ‘Measure’ and press OK.
• Select ‘Yes’ and overwrite any pre-existing data
• Make sure in the viewfinder, the subject is mostly green. A good idea is to aim for a patch of grass.
• Take a shot, and wait for your camera to report back. It will either say ‘Data Acquired’ or ‘Gd’.
• If the camera flashes ‘Unable to acquire’ or ‘No Gd’, then make sure you expose properly.
• The resulting exposure will be heavily tinted red/magenta/orange. This will be corrected later in post-production.

4. Focusing & Stabilization

Focusing is a difficult part of IR photography and can be a bit hit and miss in some respects unless your lens has IR focusing marks (Usually a small red nib, or on older lenses, a longer red line). It is best to use a narrow aperture, such as f/20, to ensure a large depth of field and minimize focusing issues.

If your lens does have IR focusing marks, focus manually, and adjust accordingly to focal length.

If your lens does not have IR focus marks, it is pretty much impossible to focus properly without a bit of a trick. One of the best things you can do is to set a small aperture to ensure you have a large depth of field.

This little trick will make sure your shots are sharp, but it does mean you cannot use wider apertures for shallow DOF effects. Without converting your camera to permanent IR and sending your lens away to be calibrated, there is nothing else that can be done to achieve consistent focus at wider apertures. However, feel free to experiment.

First focus and compose on your subject as normal with Autofocus. Then switch to manual focus. If you have a lens which has a rotating front thread, be very careful not to move it!

Make sure any image stabilization systems are turned off. VR/IS/OS is not required because the camera is on a tripod, and also because the lens will try and make adjustments when not needed, which could result in a blurry photo.

Focusing screen and VR logo.

5. Aperture

One of the most important settings in your exposure setup is to use a small aperture. This is not only great for landscape shots, giving you a larger depth of field, but it also minimizes focusing issues as explained in the focusing section above.

Notice the large depth of field.

6. ISO

Next set the ISO. Generally, it is best to use as low an ISO as possible to minimize noise. Taking into account the length of the exposure, I would recommend a maximum of ISO 800 for shots between 10 seconds and 1 minute, and a maximum of ISO 400 for shots over 1 minute.

Any greater than these values and you will risk of having copious amounts of noise and hot pixels in post-production. Sometimes however, this is unavoidable.

Also remember that going from ISO 100 to 200, for example, will half the time you are waiting for an IR exposure. So an 8 minute exposure at ISO 100, can be a 4 minute exposure at ISO 200. Only a slight increase in noise will be noticeable and it’s a good idea if you are pushed for time!

Hot pixels when zoomed in.

7. Shutter Speed

The last setting to choose is shutter speed. At first it will be an experimental process to determine exposure time. Most often you will need bulb mode, so be sure to bring a stopwatch!

IR filters are dense and force slower shutter speeds, in the same way as ND filters, so you can calculate the amount of stops you need to compensate for by using an exposure calculator/chart, or you can work it out for yourself.

So for example, if your visible light exposure is 1/30th, ISO 100, f/11, and your best resulting exposure in IR after experimentation is 1 second, then you know your filter is equivalent to a 5-stop reduction in light.

An exposure calculator.

8.Taking the Shot, Finally!

Finally, you can screw the IR filter to the lens/stepping ring. Be careful not to adjust any settings or accidentally nudge the focus ring. You are ready to take the shot, so go ahead. Press the button, and await the results.

Be sure to read the second part of this tutorial tomorrow to learn how to process your images!

An unprocessed IR image.


text:
## Why do We Use the Portland Cement?

In ancient times people used the lime mortar or mud for construction works. Both clay and lime are the binding material. Lime has been used as a strong and durable mortar for thousands of years. However, cement has a following special properties over lime, it is abundantly used in construction works.

• The setting time of cement is comparatively faster than lime.
• Mortar containing lime harden slowly as compared to cement mortar.Also, cement has a property to set under water.
• The curing period of lime mortar is also more than cement.
• A cement gains strength earlier than lime.

Portland cement is an excellent building material generally used for its excellent binding properties giving strength to structural elements. Cement is stronger under compressive load.Portland cement is used either in the form of cement mortar or cement concrete.

##### Must Read: 15+ Types of Cement Used in Modern Construction
Courtesy - 123rf

When Portland cement is mixed with required quantity of water and fine aggregate (sand) ’cement mortar’ is obtained and when cement is mixed with required quantity of water, coarse aggregates and fine aggregates (sand)‘cement concrete’ is obtained.

After adding water in cement, it acts like a gel type glue. it is therefore also used to stick various building materials. (i.e. brick masonry works, stone masonry work etc.)

Cement mortar is used for constructing wall, roofs, also for finishing (i.e. plaster to walls & ceilings). Cement concrete is used almost for all types of construction including dams & ports, hospitals, roads, schools, roads, housing and alsofor decorative applications (i.e. patios, floors, staircases, driveways, pool decks)

Portland cement can be partly replaced by various pozzolanic materials such as fly-ash, rice-husk ash, slags with a required amount which otherwise are harmful to the environment. We can change proportions of pozzolanas depending upon on the ultimate use and importance ofstructure. In this way, we can use the cement as a green building material and also a resolve the disposal problem of by-products and save the environment.

There are some very important compounds in cement responsible for its strength. It is therefore mostly used in construction for various applications.Furthermore, using of cement avoid the frequent repairs. Cement is also a cheaper material than other modern construction chemicals, if you view from the binding properties. The raw materials required for cement are along available globally to a larger or smaller extent.


text:
Now showing items 1-20 of 55

• #### Accessibility and selective stabilization of the principal spin states of iron by pyridyl versus phenolic ketimines: modulation of the (6)a(1) <-> t-2(2) ground-state transformation of the [fen4o2](+) chromophore

(AMER CHEMICAL SOC, 2012-08-06)
Several potentially tridentate pyridyl and phenolic Schiff bases (apRen and HhapRen, respectively) were derived from the condensation reactions of 2-acetylpyridine (ap) and 2'-hydroxyacetophenone (Hhap), respectively, with ...
• #### Biologically motivated quantum neural networks

(IEEE, 2017)
This paper presents one step toward creating the building blocks for machine intelligence that is inspired by its biological equivalent. The authors' quantum learning methods (deep quantum learning) are applied to quantum ...
• #### Cation distribution and dielectric properties of nanocrystalline gallium substituted nickel ferrite

(Elsevier B.V., 2014-02-16)
Nano crystalline NiFe2−xGaxO4, 0.0 ⩽ x ⩽ 1 ferrite was prepared by citrate method. The structural and micro structural evolutions of the nanophase have been studied using X-ray powder diffraction and the Rietveld method. ...
• #### Chapter 1 -- Quantum neural computation of entanglement is robust to noise and decoherence

(Elsevier, 2017)
Measurement and witnesses of entanglement remain an important issue in quantum computing. Most witnesses will work for only a very restricted class of states, while measurements commonly require lengthy procedures. Quantum ...
• #### Constraints on oscillation parameters from $\nu_e$ appearance and $\nu_\mu$ disappearance in NOvA

(American Physical Society, 2017-06-05)
Results are reported from an improved measurement of $\nu_\mu \rightarrow \nu_e$ transitions by the NOvA experiment. Using an exposure equivalent to $6.05\times10^{20}$ protons-on-target 33 $\nu_e$ candidates were observed ...
• #### Constraints on the distance moduli, helium, and metal abundances, and ages of globular clusters from their RR Lyrae and non-variable horizontal branch stars. II. multiple stellar populations in 47 Tuc, M3, and M13

(IOP Publishing, 2017-11-10)
We present a new set of horizontal branch (HB) models computed with the MESA stellar evolution code. The models adopt alpha-enhanced Asplund et al. metal mixtures and include the gravitational settling of He. They are used ...
• #### Effect of gallium additions on reduction, carburization and Fischer-Tropsch activity of iron catalysts

(Springer US, 2018-07)
Undoped and gallium-doped iron oxide catalysts (100Fe, 100Fe:2Ga, and 100Fe:5Ga) were prepared by following a continuous co-precipitation technique using ammonium hydroxide as precipitant. The catalysts were characterized ...
• #### Effect of H2S in syngas on the Fischer-Tropsch synthesis performance of a precipitated iron catalyst

(Elsevier B.V., 2016-03-05)
The sulfur limit, the relationship between the sulfur added and the surface Fe atoms lost (Fe/S), and mechanism of sulfur poisoning were studied using an iron Fischer-Tropsch synthesis (FTS) catalyst (100 Fe/5.1 Si/2.0Cu/3.0K). ...
• #### Effect of support pretreatment temperature on the performance of an iron Fischer-Tropsch catalyst supported on silica-stabilized alumina

(MDPI, 2018-02-12)
The effect of support material pretreatment temperature, prior to adding the active phase and promoters, on Fischer-Tropsch activity and selectivity was explored. Four iron catalysts were prepared on silica-stabilized ...
• #### The effects of individual metal contents on isochrones for C, N, O, Na, Mg, Al, Si, and Fe

(IOP Publishing, 2016-07-28)
The individual characteristics of C, N, O, Na, Mg, Al, Si, and Fe on isochrones have been investigated in this study. Stellar models have been constructed for various mixtures in which the content of each element is changed ...
• #### Enhancement of the critical current density in FeO-coated MgB2 thin films at high magnetic fields

(BEILSTEIN-INSTITUT, 2011-12-14)
The effect of depositing FeO nanoparticles with a diameter of 10 nm onto the surface of MgB2 thin films on the critical current density was studied in comparison with the case of uncoated MgB2 thin films. We calculated the ...
• #### Experimental pairwise entanglement estimation for an N-qubit system: a machine learning approach for programming quantum hardware

(Springer, 2020-11-03)
Designing and implementing algorithms for medium- and large-scale quantum computers is not easy. In the previous work, we have suggested, and developed, the idea of using machine learning techniques to train a quantum ...
• #### First measurement of electron neutrino appearance in NOvA

(American Physical Society, 2016-04-13)
We report results from the first search for $\nu_\mu\to\nu_e$ transitions by the NOvA experiment. In an exposure equivalent to $2.74\times10^{20}$ protons-on-target in the upgraded NuMI beam at Fermilab, we observe 6 events ...
• #### First measurement of muon-neutrino disappearance in NOvA

(American Physical Society, 2016-03-25)
This paper reports the first measurement using the NOvA detectors of $\nu_\mu$ disappearance in a $\nu_\mu$ beam. The analysis uses a 14 kton-equivalent exposure of $2.74 \times 10^{20}$ protons-on-target from the Fermilab ...
• #### Fischer-Tropsch synthesis: Activity and selectivity of chi-Fe5C2 and circle minus-Fe3C carbides

(American Chemical Society, 2014-08-10)
Two different iron carbides, Hägg (χ-Fe5C2) and cementite (ϴ-Fe3C) for Fischer-Tropsch synthesis (FTS) was investigated. Both iron carbides were oxidized to Fe3O4 during FTS. Hägg carbide (χ-Fe5C2), initially exhibits ...
• #### Fischer-Tropsch synthesis: Deactivation as a function of potassium promoter loading for precipitated iron catalyst

(Springer, 2014-10)
The effect of potassium promoter loading (0, 0.5, 1.0 and 2.0 atomic ratio) on the performance of precipitated iron catalysts was investigated during Fischer-Tropsch synthesis using a continuously stirred tank reactor. ...


text:
# blog/Why is My Laptop CPU So Slow

Recently, my laptop, with an Intel i7-4700MQ CPU running Debian Linux began running very slowly (very is largely relative, I didn’t notice for months…) when I changed from the current stable Debian Linux kernel 3.16 to the 4.4 kernel available in Debian backports. I investigated and noticed that the CPU was often running at only 600-700Mhz! A simple command to show the speed of all of your cores is watch -n 1 'for i in /sys/devices/system/cpu/cpu?/cpufreq/cpuinfo_cur_freq ; do cat \$i ; done'. Initially I assumed there was some misconfiguration of cpufreq, the Linux system which manages CPU speed. However, changing cpufreq settings, such as changing the governor to performance, changing the minimum speed, etc., had no effect at all. There are several other systems which may slow down the CPU:

• Thermal overload - in my system there are several different methods that the machine may detect that it is over-temperature and reduce the clock speed of the CPU to compensate. These include:
• Thermal Monitor 1 - Introduced in the Pentium 4 and only supports being enabled or disabled. Do not disable this. It seems to be completely automatic within the CPU itself.
• Thermal Monitor 2 - Introduced in later Penium 4 and Penitium M CPUs, which also should not be disabled. In addition to supporting being disabled, it supports setting a target maximum temperature for the CPU. Increasing this target is probably a very bad idea.
• Adaptive Thermal Monitor - Introduced in Core 2 CPUs, it will slow the CPU down more and more over time if the CPU fails to cool down. One can check whether the CPU is running slowly because of Adaptive TM, TM1, or TM2 activation by running the following command: rdmsr 0x19c -f 1:0: in the output, bit 0 indicates if the these systems are currently throttling the CPU and bit 1 indicates if it has ever done so since the CPU was reset. Bit 1 can also be reset by writing a 0 to it.
• Voltage regulator thermal overload: if the voltage regulators on the mainboard become too hot.
• The OS can use the same system used by the thermal monitors systems to slow down the CPU via the clock modulation interface.
• Autonomous Utilization-based Frequency Control - Used when the CPU has turbo enabled to determine which core(s) should be turbo boosted.
• Hardware-Controlled Performance States (HWP), though this is not available on my CPU. I haven’t looked into it further.
• C1E - Reduces frequency, but only supposed to activate when the core is idle.
• Power throttling - The system can specify either a power or a current limit (or both) which the CPU will try not to exceed by limiting its speed. On my system there are limits configured at both 47W and 58.75W, it is unclear which limit is active when. This is configured by the BIOS and may not be changed.
• Graphics driver - The graphics driver can request that the CPU be throttled. It is unclear why this happens or by how much.
• External PROCHOT or FORCEPR lines - external chips can be wired to the CPU pins to force the CPU to throttle, such as power or temperature sensing chips on the mainboard.
• External STPCLK lines - external chips can stop the CPU clock by connecting to a pin on the CPU itself.
• ACPI - there are probably more methods for the BIOS/ACPI firmware system to throttle the CPU than I am aware of.
• At the very least the thermal_cooling sysfs interface can throttle the CPU in a way that isn’t seemingly detectible by probing the CPU’s model-specific registers. For example, a thermal_cooling sysfs interface is created for each ACPI CPU which can be activated by writing to e.g. /sys/devices/virtual/thermal/cooling_device0/cur_state. Since this is capable of slowing my CPU down below its P-state minimum of 800Mhz, I assume it’s using clock modulation or STPCLK.

In the end, I discovered that on Linux 4.4 the /sys/devices/virtual/thermal/cooling_device0/cur_state would end up being set by something unknown to 4 when plugging or unplugging the power cable to my laptop resulting in 600-700Mhz operation. At the same time, the CPU reports seeing the PROCHOT or FORCEPR lines asserted as the power cable is plugged/unplugged. It’s my guess that this is the result of some interaction between the kernel, ACPI, and an external chip on the system. This hasn’t occurred since upgrading to Linux 4.5, but the behavior can be half-reproduced by writing 4 to /sys/devices/virtual/thermal/cooling_device0/cur_state.

I have developed a script to detect some possible reasons why my CPU might be running slowly, it’s available at: https://github.com/orezpraw/scripts/blob/master/whyslow.pl.


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# Equation $wxyz(w+x+y+z)=1$ in $\mathbb{Q}_+^4$

In this thread on Math.SE, Noam D. Elkies give the following parametric family of solutions in $$\mathbb{Q_+}^3$$ of the equation $$xyz(x+y+z)=1$$ (found by Euler) :

$$x = \frac{6 t^3 (t^4-2)^2} {(4 t^4 + 1) (2 t^8 + 10 t^4 - 1)},$$ $$y = \frac{ 3 (4 t^4 + 1)^2} {2t (t^4-2) (2 t^8 + 10 t^4 - 1)},$$ $$z = \frac{ 2 (2 t^8 + 10 t^4 - 1)} {3t (4 t^4 + 1)}.$$

Can we do the same for the equation $$wxyz(w+x+y+z)=1$$ ?

Thank you.

Remarks :

1) I crossposted to Math.SE.

2) A computer investigation leads to the following solutions $$(w\,;x\,;y\,;z)$$ :

$$\left(\dfrac{1}{6}\,;\dfrac{1}{3}\,;2\,;2\right)$$ , $$\left(\dfrac{1}{6}\,;\dfrac{2}{3}\,;\dfrac{2}{3}\,;3\right)$$ , $$\left(\dfrac{1}{12}\,;\dfrac{1}{4}\,;\dfrac{2}{3}\,;8\right)$$ , $$\left(\dfrac{1}{12}\,;\dfrac{1}{4}\,;\dfrac{8}{3}\,;3\right)$$ , $$\left(\dfrac{1}{12}\,;\dfrac{1}{3}\,;1\,;\dfrac{16}{3}\right)$$ , $$\left(\dfrac{1}{10}\,;\dfrac{2}{5}\,;2\,;\dfrac{5}{2}\right)$$ , $$\left(\dfrac{1}{14}\,;1\,;\dfrac{7}{4}\,;\dfrac{7}{4}\right)$$ , $$\left(\dfrac{1}{30}\,;\dfrac{5}{6}\,;\dfrac{9}{5}\,;\dfrac{10}{3}\right)$$ , $$\left(\dfrac{1}{24}\,;\dfrac{9}{8}\,;\dfrac{3}{2}\,;\dfrac{8}{3}\right)$$ , $$\left(\dfrac{1}{4}\,;\dfrac{1}{4}\,;\dfrac{4}{3}\,;\dfrac{8}{3}\right)$$ , $$\left(\dfrac{1}{10}\,;\dfrac{2}{3}\,;\dfrac{9}{10}\,;\dfrac{10}{3}\right)$$ , $$\left(\dfrac{1}{10}\,;\dfrac{5}{6}\,;\dfrac{16}{15}\,;\dfrac{5}{2}\right)$$ , $$\left(\dfrac{1}{10}\,;\dfrac{16}{15}\,;\dfrac{3}{2}\,;\dfrac{3}{2}\right)$$ , $$\left(\dfrac{1}{10}\,;\dfrac{6}{5}\,;\dfrac{6}{5}\,;\dfrac{5}{3}\right)$$ , $$\left(\dfrac{1}{6}\,;\dfrac{3}{10}\,;\dfrac{6}{5}\,;\dfrac{10}{3}\right)$$ , $$\left(\dfrac{1}{6}\,;\dfrac{3}{5}\,;1\,;\dfrac{12}{5}\right)$$ , $$\left(\dfrac{2}{9}\,;\dfrac{1}{4}\,;\dfrac{16}{9}\,;\dfrac{9}{4}\right)$$ , $$\left(\dfrac{3}{10}\,;\dfrac{5}{6}\,;1\,;\dfrac{6}{5}\right)$$ , $$\left(\dfrac{2}{5}\,;\dfrac{3}{5}\,;\dfrac{5}{6}\,;\dfrac{3}{2}\right)$$ , $$\left(\dfrac{1}{3}\,;\dfrac{5}{12}\,;\dfrac{5}{4}\,;\dfrac{8}{5}\right)$$ , $$\left(\dfrac{5}{12}\,;\dfrac{1}{2}\,;\dfrac{16}{15}\,;\dfrac{27}{20}\right)$$ and $$\left(\dfrac{9}{20}\,;\dfrac{2}{3}\,;\dfrac{5}{6}\,;\dfrac{5}{4}\right)$$.

3) I found the following parametrization but $$w$$, $$x$$, $$y$$, and $$z$$ are never simultaneously positive :

$$w=\frac{296352\,t^4}{(2401\,t^5-16)(2401t^5-4)(2401\,t^5+8)}$$ $$x=-\frac{(2401\,t^5-4)^2}{21\,t\,(2401\,t^5-16)}$$ $$y=\frac{(2401\,t^5-16)^2}{168\,t\,(2401\,t^5+8)}$$ $$z=\frac{2(2401\,t^5+8)}{7\,t\,(2401\,t^5-4)}$$

• Aug 26, 2022 at 15:24

Let
\begin{align*} w &= \frac{c_1t^{t_1}}{ABC}\\ x &= \frac{c_2B^2t^{t_2}}{A}\\ y &= \frac{c_3A^2t^{t_3}}{C}\\ z &= \frac{c_4Ct^{t_4}}{B} \end{align*} Then $$wxyz(w+x+y+z) = \frac{c_1t^{t_1}c_2t^{t_2}c_3t^{t_3}c_4t^{t_4}(c_1t^{t_1}+c_2B^3t^{t_2}C+c_3A^3t^{t_3}B+c_4C^2t^{t_4}A)}{C^2AB}$$

Hence
\begin{align*} &(c_1t^{t_1}c_2t^{t_2}c_3t^{t_3}c_4^2(t^{t_4})^2A-AB)C^2\\ &+c_1t^{t_1}c_2^2(t^{t_2})^2c_3t^{t_3}c_4t^{t_4}B^3C\\ &+c_1t^{t_1}c_2t^{t_2}c_3t^{t_3}c_4t^{t_4}(c_1t^{t_1}+c_3A^3t^{t_3}B)=0 \end{align*}

Since $$C$$ must be rational number then discriminant must be square number.
\begin{align*} v^2 &= (4c_3t^{t_3}B^2-4Bc_3^2(t^{t_3})^2c_1t^{t_1}c_2t^{t_2}c_4^2(t^{t_4})^2)c_1t^{t_1}c_2t^{t_2}c_3t^{t_3}c_4t^{t_4}A^4\\ &+(4Bc_1t^{t_1}-4c_1^2(t^{t_1})^2c_2t^{t_2}c_3t^{t_3}c_4^2(t^{t_4})^2)c_1t^{t_1}c_2t^{t_2}c_3t^{t_3}c_4t^{t_4}A\\ &+c_1^2(t^{t_1})^2c_2^4(t^{t_2})^4c_3^2(t^{t_3})^2c_4^2(t^{t_4})^2B^6 \end{align*}

Obviously, quartic is birationally equivalent to an elliptic curve.

For instance, take $$(c_1,c_2,c_3)=(1,1,1),(t1,t2,t3)=(1,1,1),B=1$$, then

$$v^2 = (4t^5-4t^{10})A^4+(4t^5-4t^{10})A+t^{10}$$

Quartic is transformed to an elliptic curve below.

$$E: Y^2+(4-4t^5)YX = X^3+(-4+8t^5-4t^{10})X^2+(-16t^{15}+16t^{20})X+64t^{15}-192t^{20}+192t^{25}-64t^{30}$$

$$E$$ has a point $$P(X,Y)=(4-8t^5+4t^{10}, -16+48t^5-48t^{10}+16t^{15})$$.
We obtain $$2P(X)=\frac{1-26t^{15}+33t^{20}-24t^{25}+8t^{30}-6t^5+15t^{10}}{(t^{10}-2t^5+1)^2}$$
According to Nagell-Lutz theorem, the point $$P(X,Y)$$ is not a point of finite order, hence we can obtain infinitely many parametric solutions.

Thus we can obtain a quartic point $$2Q(A) = \frac{-2t^5(t-1)^2(t^4+t^3+t^2+t+1)^2}{(2t^5-1)(t^2-t+1)(t^8+t^7-t^5-t^4-t^3+t+1))}$$

Then we obtain $$(A,C)$$

$$A = \frac{-2t^5(t-1)^2(t^4+t^3+t^2+t+1)^2}{(2t^5-1)(t^2-t+1)(t^8+t^7-t^5-t^4-t^3+t+1)}$$ $$C = \frac{-(3t^{10}-3t^5+1)}{-1+3t^5-3t^{10}+2t^{15}}$$

Finally we obtain $$(w,x,y,z)$$ \begin{align*} w &= \frac{1}{2}\frac{(2t^5-1)^2(t^2-t+1)^2(t^8+t^7-t^5-t^4-t^3+t+1)^2}{t^4(t-1)^2(t^4+t^3+t^2+t+1)^2(3t^{10}-3t^5+1)}\\ x &= \frac{-1}{2}\frac{(2t^5-1)(t^2-t+1)(t^8+t^7-t^5-t^4-t^3+t+1)}{t^4(t-1)^2(t^4+t^3+t^2+t+1)^2}\\ y &= \frac{-4(t^4+t^3+t^2+t+1)^4(t-1)^4t^{11}}{(3t^{10}-3t^5+1)(2t^5-1)(t^2-t+1)(t^8+t^7-t^5-t^4-t^3+t+1)}\\ z &= \frac{-(3t^{10}-3t^5+1)t}{(2t^5-1)(t^2-t+1)(t^8+t^7-t^5-t^4-t^3+t+1)} \end{align*}

This parametrization gives positive solution where $$0\lt t \lt 1/2^{1/5}$$.

Needless to say, this method can be applied to $$xyz(x+y+z)=a$$.

• Do you think your method can be applied more generally to the equation $\left(\prod\limits_{k=1}^{n}x_k\right)\left(\sum\limits_{k=1}^{n}x_k\right)=a$ with $n\geqslant4$ ? Aug 11, 2022 at 7:25
• It may be possible with the suitable combination of A,B,C,…. Aug 11, 2022 at 9:07

I finally found the following parametrization :

$$w=\frac{32916734851\,t^4}{(219896613895728\,t^5+1)(370352191824384\,t^5+1)(479774430317952\,t^5+1)}$$ $$x=\frac{32\,(219896613895728\,t^5+1)(479774430317952\,t^5+1)}{5187\,t\,(370352191824384\,t^5+1)}$$ $$y=\frac{3\,(370352191824384\,t^5+1)}{1232\,t\,(479774430317952\,t^5+1)}$$ $$z=\frac{(370352191824384\,t^5+1)(479774430317952\,t^5+1)}{4368\,t\,(219896613895728\,t^5+1)}$$

In computer-readable format :

w:=32916734851*t^4/((219896613895728*t^5+1)*(370352191824384*t^5+1)*(479774430317952*t^5+1))

x:=32*(219896613895728*t^5+1)*(479774430317952*t^5+1)/(5187*t*(370352191824384*t^5+1))

y:=3*(370352191824384*t^5+1)/(1232*t*(479774430317952*t^5+1))

z:=(370352191824384*t^5+1)*(479774430317952*t^5+1)/(4368*t*(219896613895728*t^5+1))

• I think a method is more important here than any particular parametrization. So it would be more informative if you describe how you found this parametrization. Aug 8, 2022 at 0:48
• @MaxAlekseyev You're right. I will give some explanation soon. Aug 8, 2022 at 5:56

My idea is to try to find a polynomial equality of the form $$X+\lambda P^2Q^2+\mu PR^2+\varepsilon Q^2R^2=\theta P^2QR \;\;\;(1)$$ with $$\lambda$$, $$\mu$$, $$\varepsilon$$, $$\theta$$ in $$\mathbb{Q}$$ and $$P$$, $$Q$$, $$R$$ in $$\mathbb{Q}[X]$$.

Let $$A=X$$, $$B=\lambda P^2Q^2$$, $$C=\mu PR^2$$ and $$D=\varepsilon Q^2R^2$$, then $$ABCD(A+B+C+D)=\lambda\mu\varepsilon\theta X(PQR)^5 \;\;\;(E_a)\;.$$ Taking $$X=kat^5$$ with $$k\in\mathbb{Q}$$ such that $$k\lambda\mu\varepsilon\theta=K^5$$ with $$K\in\mathbb{Q }$$, we have $$wxyz(w+x+y+z)=a \;\;\;(E_a)\;,$$ with $$w=\dfrac{A}{KtPQR}$$, $$x=\dfrac{B}{KtPQR}$$, $$w=\dfrac{C}{KtPQR}$$ and $$w=\dfrac{D}{KtPQR }$$.

In fact, we will see that we can find an equality of the form $$(1)$$ with $$P$$, $$Q$$ and $$R$$ of degree 1.

Take $$P=X-\alpha$$, $$Q=X-\beta$$ and $$R=X-\gamma$$ with $$\alpha$$, $$\beta$$ and $$\gamma$$ in $$\mathbb{Q}$$ (such that $$\gamma\neq\beta$$).

Suppose $$(1)$$ holds ; by replacing $$X$$ by $$\alpha$$, $$\beta$$ then $$\gamma$$, by derivating then by replacing $$X$$ by $$\alpha$$, and by considering the dominant terms, we obtain

$$\left\{\begin{array}{l}\alpha+\varepsilon(\alpha-\beta)^2(\alpha-\gamma)^2=0\\\beta+\mu(\beta-\alpha)(\beta-\gamma)^2=0\\\gamma+\lambda(\gamma-\alpha)^2(\gamma-\beta)^2=0\\1+\mu(\alpha-\gamma)^2+2\varepsilon(\alpha-\beta)(\alpha-\gamma)(2\alpha-\gamma-\beta)=0\\\theta=\lambda+\varepsilon\\\end{array}\right.$$ which is equivalent to $$\left\{\begin{array}{l} \varepsilon=-\dfrac{\alpha}{(\alpha-\beta)^2(\alpha-\gamma)^2}\\ \mu=-\dfrac{\beta}{(\beta-\alpha)(\beta-\gamma)^2}\\ \lambda=-\dfrac{\gamma}{(\gamma-\alpha)^2(\gamma-\beta)^2}\\ (\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)^2+\beta(\alpha-\gamma)^3-2\alpha(\beta-\gamma)^2(2\alpha-\gamma-\beta) \;\;\;(2)\\ \theta=\lambda+\varepsilon\\ \end{array}\right.$$ Let $$u=\dfrac{\alpha-\gamma}{\gamma-\beta}$$; the identity $$(2)$$ gives $$(\alpha-\beta)u(\gamma-\beta)^3+\beta u^3(\gamma-\beta)^3-2\alpha(\gamma-\beta)^2(2( \alpha-\gamma)+\gamma-\beta) \;,$$ which is equivalent to $$(\alpha-\beta)u+\beta u^3-2\alpha(2u+1)=0 \;,$$ which leads to $$\left\{\begin{array}{l} \beta=\dfrac{3u+2}{u^3-u}\alpha\\ \gamma=\dfrac{u^2+3u+1}{(1+u)(u^2-1)}\alpha\\ \varepsilon=-\dfrac{u^2(u+1)^6(u-1)^4}{\alpha^3(u^3-4u-2)^4}\\ \mu=\dfrac{u^2(u+1)^4(u-1)^2(3u+2)}{\alpha^2(u^3-4u-2)^3}\\ \lambda=-\dfrac{u^2(u+1)^6(u-1)^3(u^2+3u+1)}{\alpha^3(u^3-4u-2)^4}\\ \theta=-\dfrac{u^3(u+1)^6(u-1)^3(u+4)}{\alpha^3(u^3-4u-2)^4}\\ \end{array}\right.$$ Conversely, one can verify that if $$\beta$$, $$\gamma$$, $$\varepsilon$$, $$\mu$$, $$\lambda$$ and $$\theta$$ are defined in terms of $$u$$ and $$\alpha$$ by the previous relations, then the polynomial relation $$(1)$$ is satisfied.

Taking $$k=-\dfrac{\alpha u}{(u+1)^2(u-1)^2(u^2+3u+1)(u+4)(3u+2)}$$, we have $$k\lambda\mu\varepsilon\theta=K^5$$ with $$K=\dfrac{u^2(u+1)^4(u-1)^2}{\alpha^2(u^ 4-4u-2)^3}$$; we then choose $$X=kat^5$$. Let's set $$\left\{\begin{array}{l} F(u)=(u+1)^2(u-1)^2(u^2+3u+1)(u+4)(3u+2)\\ G(u)=(u+1)(u-1)(u^2+3u+1)(u+4)(3u+2)^2\\ H(u)=(u-1)(u^2+3u+1)^2(u+4)(3u+2)\\ J(u)=(u+4)(3u+2)(u^3-4u-2)\\ \end{array}\right.$$ we obtain (after simplifications) the following $$2-$$parameterization of an infinite subset of solutions of $$(E_a)$$: $$\left\{\begin{array}{l} w=\dfrac{a(u-1)^2(u^2+3u+1)^2(u^3-4u-2)J(u)^2t^4}{(aut^5+F(u))(au^2t^5+G(u))(aut^5+H(u))}\\ x=\dfrac{(aut^5+F(u))(au^2t^5+G(u))}{u(u-1)J(u)t(aut^5+H(u))}\\ y=\dfrac{u(3u+2)(aut^5+H(u))}{t(au^2t^5+G(u))}\\ z=\dfrac{(au^2t^5+G(u))(aut^5+H(u))}{u(u^2+3u+1)J(u)t(aut^5+F(u))}\\ \end{array}\right.$$ By choosing for example $$a=1$$ and $$u=3$$, we obtain the following $$1-$$parametrization of an infinite subset of solutions of $$(E_1)$$ $$\left\{\begin{array}{l} w=\dfrac{18809562772t^4}{(3t^5+55594)(3t^5+93632)(9t^5+128744)}\\ x=\dfrac{(3t^5+93632)(9t^5+128744)}{6006t(3t^5+55594)}\\ y=\dfrac{33(3t^5+55594)}{t(9t^5+128744)}\\ z=\dfrac{(3t^5+55594)(9t^5+128744)}{57057t(3t^5+93632)}\\ \end{array}\right.$$

• I made some simplifications between the first answer and the second. So we find the first parametrization from the second by setting $a=1$, $u=3$ and substituting $4\times7\times11\times19\times t$ for $t$. Aug 9, 2022 at 19:07


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# Why didn't the space shuttle SRBs have wings and tires?

Why didn't the space shuttle solid rocket boosters have wings and tires to land the same way the orbiter lands? I don't think they haven't thought of that so there must be something that led them not to choose that design, right?

• There is no runway in the ocean. – Erik May 1 '17 at 19:01
• It'd be interesting to try to fly them back like remote control airplanes! – Cort Ammon May 1 '17 at 19:59
• @CortAmmon The Russians have had such plans for their Baikal booster for the Angara modular launcher family. – LocalFluff May 2 '17 at 0:19

Wings, engines to fly back to the launch site, and landing gear are heavier than parachutes. Using this design on the SRBs could have resulted in a significant payload to orbit reduction (every 10 pounds added to the SRBs resulted in 1 pound less payload that could be carried to orbit).

However, this was studied for the case of liquid fueled flyback boosters. Another interesting shuttle upgrade that never went through. And, resulting in a crew member comment "I guess we would be 3rd in line for the runway if we had to do an RTLS."

• See Russell Borogove's answer but yes, the original shuttle proposal was for a fully reusable system where the shuttle orbiter launched on a giant reusable manned flyback booster. When NASA was told they had only \$4 billion for development, they changed the design to the expendable tank and refurbishable SRBs (increased operations cost, lower development cost). The flyback LRBs were a later proposal to make the existing system more reusable, only the tank would be expended. – Organic Marble May 2 '17 at 0:14
• Agreed on the third in line quote, that's priceless :) and @ALz, to answer your question about to orbit reduction, that's just a sentence fragment - the meaning can be understood better if you read it as a whole: could have resulted in a significant payload-to-orbit reduction, rephrasing for clarity would give us it could have resulted in a significant reduction in what payload could be lifted into orbit. – flith May 2 '17 at 10:45
• With all new liquid fueled boosters there would be no reduction in payload capability; the new boosters would provide more total impulse than the original SRBs, canceling the drag losses. – Russell Borogove May 2 '17 at 13:54
• It's, let us say, interesting? ...that the flyback booster proposal handwaves away the development of a large RP-LOX staged combustion engine properly sized for the booster. – Russell Borogove May 2 '17 at 14:00
• You'd need to scale up the Merlin about 2x. The Boeing proposal calls for 4x 900klbf engines per booster at 340s Isp which might as well be a flashing sign reading "RD-180". – Russell Borogove May 2 '17 at 15:39

In addition to the booster upgrade proposals that Organic Marble mentions, some early shuttle proposals considered using a single large liquid-fueled winged booster that would fly back to the launch site. The development budget for the shuttle didn't allow that strategy to be pursued.

http://www.nss.org/resources/library/shuttledecision/chapter08.htm

• What an awesome article! – Jason C May 2 '17 at 4:11

It's all about trade-offs. Parachutes are much simpler, lighter, and safer for recovering a "dumb" booster. Wings, landing gear, and control systems add an awful lot of weight (therefore reducing payload) and complexity for pretty much zero return on investment. I also imagine there would've been all kinds of stability problems from adding two extra sets of side-mounted wings to the stack.


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# AddForce not moving imported model in Unity

I've got a Unity project with an imported .fbx model I'm attempting to add movement to.

As a test, I've added a sphere primitive with a sphere collider, a rigid body, and the following script:

public class ball_test : MonoBehaviour {

private Rigidbody rb;

void Start ()
{
rb = GetComponent<Rigidbody> ();
}

void FixedUpdate ()
{
Vector3 movement = new Vector3 (1f, 0.0f, 1f);

}
}

This is partly copied from the Roll A Ball tutorial, and this works fine. The ball slowly rolls with no problem.

I've imported a custom model, added a box collider, a rigid body, and assigned to it the same script. It doesn't move.

Here's how they're set up:

• This seems like a problem with friction. Try to make a Physic material with Zero friction and set it in the Collider component. – Aditya Yadav Nov 24 '17 at 5:50
• @AdityaYadav, No change. – Nightmare Games Nov 24 '17 at 5:58
• Have you tried just cranking up the force? It's harder to slide a crate than roll a wheel. ;) – DMGregory Nov 24 '17 at 6:06
• @DMGregory, yep, that was it. – Nightmare Games Nov 24 '17 at 6:12
• Cool. Maybe share the force values you needed to use as an answer so future users can get a sense of how much they need to crank it up in similar situations. – DMGregory Nov 24 '17 at 6:12


text:
Volume 17, Issue 5
Modifying the Split-Step $θ$-Method with Harmonic-Mean Term for Stochastic Differential Equations

Int. J. Numer. Anal. Mod., 17 (2020), pp. 662-678.

Published online: 2020-08

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• Abstract

In this paper, we design a class of general split-step methods for solving Itô stochastic differential systems, in which the drift or deterministic increment function can be taken from special ordinary differential equations solver, based on the harmonic-mean. This method is justified to have a strong convergence order of $\frac{1}{2}$. Further, we investigate mean-square stability of the proposed method for linear scalar stochastic differential equation. Finally, some examples are included to demonstrate the validity and efficiency of the introduced scheme.

• Keywords

Itô stochastic differential system, split-step method, ODE solver, harmonic-mean, strong convergence, mean-square stability.

• AMS Subject Headings

• BibTex
• RIS
• TXT
@Article{IJNAM-17-662, author = {Nouri , Kazem and Ranjbar , Hassan and Carlos Cortés López , Juan}, title = {Modifying the Split-Step $θ$-Method with Harmonic-Mean Term for Stochastic Differential Equations}, journal = {International Journal of Numerical Analysis and Modeling}, year = {2020}, volume = {17}, number = {5}, pages = {662--678}, abstract = {

In this paper, we design a class of general split-step methods for solving Itô stochastic differential systems, in which the drift or deterministic increment function can be taken from special ordinary differential equations solver, based on the harmonic-mean. This method is justified to have a strong convergence order of $\frac{1}{2}$. Further, we investigate mean-square stability of the proposed method for linear scalar stochastic differential equation. Finally, some examples are included to demonstrate the validity and efficiency of the introduced scheme.

}, issn = {2617-8710}, doi = {https://doi.org/}, url = {http://global-sci.org/intro/article_detail/ijnam/17874.html} }
TY - JOUR T1 - Modifying the Split-Step $θ$-Method with Harmonic-Mean Term for Stochastic Differential Equations AU - Nouri , Kazem AU - Ranjbar , Hassan AU - Carlos Cortés López , Juan JO - International Journal of Numerical Analysis and Modeling VL - 5 SP - 662 EP - 678 PY - 2020 DA - 2020/08 SN - 17 DO - http://doi.org/ UR - https://global-sci.org/intro/article_detail/ijnam/17874.html KW - Itô stochastic differential system, split-step method, ODE solver, harmonic-mean, strong convergence, mean-square stability. AB -

In this paper, we design a class of general split-step methods for solving Itô stochastic differential systems, in which the drift or deterministic increment function can be taken from special ordinary differential equations solver, based on the harmonic-mean. This method is justified to have a strong convergence order of $\frac{1}{2}$. Further, we investigate mean-square stability of the proposed method for linear scalar stochastic differential equation. Finally, some examples are included to demonstrate the validity and efficiency of the introduced scheme.

Kazem Nouri, Hassan Ranjbar & Juan Carlos Cortés López. (2020). Modifying the Split-Step $θ$-Method with Harmonic-Mean Term for Stochastic Differential Equations. International Journal of Numerical Analysis and Modeling. 17 (5). 662-678. doi:
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# on the adjointness of the global section functor and the Spec functor

In fact, it is an exercise on Hartshorne, Ex 2.4 to the second chapter of it (P79): let $A$ be a ring and $(X,\mathcal{O}_X)$ be a scheme. Given a morphism $f:X\longrightarrow \operatorname{Spec} A$,we have an associated map on sheaves $f^\sharp :\mathcal{O}_{\operatorname{Spec} A}\longrightarrow f_*\mathcal{O}_X$.Taking global sections we obtain a homomorphism $A\longrightarrow \Gamma (X,\mathcal{O}_X)$.Thus there is a natural map $\alpha:\operatorname{Hom}_{\mathcal{Sch}}(X,\operatorname{Spec} A)\longrightarrow \operatorname{Hom}_{\mathcal{Rings}}(A,\Gamma (X,\mathcal{O}_X))$.How to show that $\alpha$ is bijective?My approach is to construct an inverse for $\alpha$.Starting from a ring hom $\phi :A\longrightarrow \Gamma (X,\mathcal{O}_X)$,and an affine covering $X=\bigcup_i \operatorname{Spec} B_i$,restricting $\Gamma (X,\mathcal{O}_X)$ to $\mathcal{O}_X (\operatorname{Spec}B_i)$,we get maps $\phi_i:A\longrightarrow B_i$,thus inducing $(f_i,f_{i}^{\sharp}):\operatorname{Spec}B_i\longrightarrow \operatorname{Spec} A$.Then I want to glue these $\operatorname{Spec}B_i$ together to get a morphism $(f,f^\sharp):(X,\mathcal{O}_X)\longrightarrow (\operatorname{Spec}A,\mathcal{O}_{\operatorname{Spec}A})$.I get stuck here:I don't know how to glue them together and verify that $(f,f^\sharp)$ is independent of the covering $\{ \operatorname{Spec}B_i\}_i$.Is my idea right? Will someone be kind enough to give me some hints on this problem?Thank you very much!

If you want to prove the statement in question using a gluing argument, then you need to prove the result in the special case that $X=\mathrm{Spec}(B)$ is also affine. This will allow you to conclude that your maps on intersections of affine opens agree on overlaps (because $\mathrm{Hom}(\mathrm{Spec}(B),\mathrm{Spec}(A))\rightarrow\mathrm{Hom}(A,B)$ is a bijection).

The result for $X=\mathrm{Spec}(B)$ is proved in Hartshorne if I remember correctly. In any case, the idea is that a ring map $\varphi:A\rightarrow B$ induces a continuous map $\alpha:\mathrm{Spec}(B)\rightarrow\mathrm{Spec}(A)$ under which the inverse image of a standard open $D(f)\subseteq\mathrm{Spec}(A)$ is $D(\varphi(f))$ ($f\in A$). The homomorphism $\varphi$ naturally induces, for each $f\in A$, a ring map $A_f\rightarrow B_{\varphi(f)}$, i.e., a map $\mathcal{O}_{\mathrm{Spec}(A)}(D(f))\rightarrow\mathcal{O}_{\mathrm{Spec}(B)}(D(\varphi(f))$, compatible with restriction, and since the sets $D(f)$ give a base for the topology of $\mathrm{Spec}(A)$, this is extends uniquely to a sheaf map $\mathcal{O}_{\mathrm{Spec}(A)}\rightarrow\alpha_*\mathcal{O}_{\mathrm{Spec}(B)}$. For $\mathfrak{q}\in\mathrm{Spec}(B)$, $\mathfrak{p}=\alpha(\mathfrak{q})=\varphi^{-1}(\mathfrak{q})$, the stalk map $A_\mathfrak{p}\rightarrow B_\mathfrak{q}$ is also induced (via the universal property of localization) by $\varphi$, and is easily seen to be local (by construction). This morphism $\alpha$ also recovers $\varphi$ on global sections.

Uniqueness follows ultimately because of the requirement that the stalk maps of a morphism of locally ringed spaces are local and because morphisms of sheaves are determined by the morphisms on stalks. The requirement that the stalk maps be local forces $\alpha(\mathfrak{q})=\varphi^{-1}(\mathfrak{q})$, and then, again using the universal property of localization, there is a unique map of stalks $A_{\varphi^{-1}(\mathfrak{q})}\rightarrow B_\mathfrak{q}$ compatible with $\varphi$. In summary, if you want to recover $\varphi$ on global sections and you want your stalk maps to be local, you only have one choice (everything is induced by the universal property of localization).

I want to point out that this argument actually extends to prove the result you're after with $X$ replaced not just by an arbitrary scheme, but by an arbitrary locally ringed space. This means the result actually has nothing to do with gluing (since an arbitrary locally ringed space needn't be built from affine schemes). The crucial thing is (as mentioned above) that for a morphism $\alpha:X\rightarrow\mathrm{Spec}(A)$ the stalk map $\alpha_x^\sharp:A_{f(x)}\rightarrow\mathcal{O}_{X,x}$ is local for any $x\in X$. Note that here $f(x)$ is a prime ideal of $A$ and $A_{f(x)}$ is the localization at that prime (the stalk of the structure sheaf of $\mathrm{Spec}(A)$ at $f(x)$). If $\alpha^\sharp:A\rightarrow\mathcal{O}_X(X)$ is the map on global sections of the morphism $\alpha$, then its compatibility with the stalk map $\alpha_x^\sharp$ and the fact that $\alpha_x^\sharp$ is local actually implies that $f(x)$ is the inverse image of the maximal ideal $\mathfrak{m}_x\subseteq\mathcal{O}_{X,x}$ under the ring map $A\rightarrow\mathcal{O}_X\rightarrow\mathcal{O}_{X,x}$ (the first arrow is $\alpha^\sharp$ and the second is taking the stalk at $x$). So the map on global sections determines the morphism $\alpha$ on the underlying topological spaces. Once you know this, it also follows that the stalk maps are uniquely determined.

The point of all this is that the map you want to prove is a bijection is injective. To prove that it is surjective, you basically run the above argument backwards. Given a ring map $\varphi:A\rightarrow\mathcal{O}_X(X)$, you can define $\alpha:X\rightarrow\mathrm{Spec}(A)$ on topological spaces by taking $f(x)$ to the prime ideal that is the inverse image of $\mathfrak{m}_x\subseteq\mathcal{O}_{X,x}$ under the map mentioned in the previous paragraph. You can prove then that for any $f\in A$, $\alpha^{-1}(D(f))$ is $X_{\varphi(f)}$, defined as the set of all $x\in X$ such that $\varphi(f)_x$ is not in the maximal ideal $\mathfrak{m}_x$ of $\mathcal{O}_{X,x}$. This is an open set of $X$ (the analogue of a standard open in an affine scheme), so $\alpha$ is continuous. The universal property of localization together with the fact that $\varphi(f)\vert_{X_{\varphi(f)}}\in\mathcal{O}_X(X_{\varphi(f)})$ is a unit (this follows from the definition of $X_{\varphi(f)}$) shows that the ring map $A\rightarrow\mathcal{O}_X(X)\rightarrow\mathcal{O}_X(X_{\varphi(f)})$ ($\varphi$ followed by restriction from $X$ to $X_{\varphi(f)})$ induces a unique map $A_f\rightarrow\mathcal{O}_X(X_f)$ compatible with restriction. This data is what you need for a map of sheaves $\mathcal{O}_{\mathrm{Spec}(A)}\rightarrow\alpha_*\mathcal{O}_X$. This $\alpha$ recovers $\varphi$ on global sections and proves surjectivity.

Just to explain, the reason I went through all this was to illustrate that the adjunction between the global sections functor and $\mathrm{Spec}$ actually works on the entire category of locally ringed spaces (not just the full subcategory of schemes) and so really has nothing to do with gluing maps on affine patches. I personally (when learning algebraic geometry) found this fact really helped me and shaped the way I think of affine schemes among locally ringed spaces (or just schemes). For example, any locally ringed space $X$ admits a unique morphism to $\mathrm{Spec}(\mathbb{Z})$, and the method of proof I've described shows that it sends a point $x\in X$ to the prime ideal generated by the characteristic of the residue field $\kappa(x)$.

• Can you please elaborate your statement so really has nothing to do with gluing maps on affine patches We are definitely not attaching as maps $X\rightarrow \text{Spec}(A)$. We are defining for each $x\in X$ some $\mathfrak{p}\in \text{Spec}(A)$. It would be useful if you can say more about this – user312648 Jul 22 '17 at 16:04

I think it's unfortunate that Hartshorne doesn't say, "you can glue morphisms". Here's a general lemma that might help:

Let $(X, \mathscr{O}_X)$ and $(Y, \mathscr{O}_Y)$ be ringed spaces. Suppose I have an open cover $\{U_i\}$ of $X$ and morphisms $f_i\colon (U_i, \mathscr{O}_X|_{U_i}) \to Y$ such that for each $i, j$ we have $f_i|_{U_i \cap U_j} = f_j|_{U_i \cap U_j}$. Then there is a unique morphism $f\colon X \to Y$ such that each $f|_{U_i} = f_i$.

It seems to me that one way to avoid worrying about canonicity is to take the family of all affine open sets in $X$ as your cover. There are some remaining technical points in order to even apply the lemma; hopefully I can comment on those later.

• But Hartshorne does say that! He glues schemes in Exercise 2.12, page 80, and he glues morphisms in page 88, Step 3. It's even in the index under "Glueing". I am always careful before I say Hartshorne doesn't say something, because Hartshorne says a lot of things in its exercises. – Patrick Da Silva Nov 23 '14 at 23:05


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# Question regarding Sobolev embeddings

1. Jan 1, 2012

### phibonacci

Hi,

I'm having a bit of trouble understanding certain Sobolev embedding theorems. For instance, if 1 ≤p<∞, and m,j are non-negative integers and mp>n then Wj+m,p(ℝn) embeds into CjB(ℝn). (See for instance Adams and Fournier, Sobolev Spaces, 2nd edition, Theorem 4.12 p. 85). But does this result mean that every function in Wj+m,p(ℝn) is continuously differentiable in the classical sense up to order j (with the function and its classical derivatives up to order j being bounded and continuous), or does it simply mean that the function has weak derivatives up to order j, and that the function and these weak derivatives are continuous and bounded?


text:
# Circles on a plane

$n$ circles with total area A have been drawn on the plane (overlapping circles are not counted multiple times). Prove that we can select a disjoint union of circles that has area greater than $\frac{A}{9}$.

I'm thinking that if there is a circle that has an area that is greater than 1/9 then we're basically done. But if not AND the statement is false then we must be full of overlaps. More specifically a point has been covered by at least 9 circles.

• Funny problem; I like it. Where does it come from? Why do you think it's true? What have you tried? – user228113 Feb 1 '16 at 3:25
• @G.Sassatelli This is the same as being able to cover a table with 10 dots on it by 10 circles. See here very instructive. – Shailesh Feb 1 '16 at 3:54
• Thank you for the reference, @Shailesh. – user228113 Feb 1 '16 at 3:57
• @Shailesh Why is that the same? – Michael Biro Feb 1 '16 at 4:08
• Updated the question – The Math Penguin Feb 1 '16 at 4:08

Now, if we accept a disk of radius $r$, we will disallow any disk that intersects it (which must have radius $\leq r$), all of which are completely covered by a disk of radius $3r$.
Therefore, we cover $\pi r^2$ with our chosen disk, while disallowing at most $\pi (3r)^2$ in area from $A$. In other words, we cumulatively cover a total area of at least $\frac{A}{9}$.


text:
×

# Expected value

I don't know anything about "expected value". I am moderately good in combinatorics, but I never heard this term before today. Like I have got 6 coins numbered a, b, c, d, e & f; and I pick up a coin randomly, what is the expected value (I mean a to f) of the coin?. Can anyone help me?

Note by Joseph Gomes
3 years, 6 months ago

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Comment deleted Apr 11, 2013

Thanks a lot Now I can solve Expected Raffle Value Problem · 3 years, 6 months ago

Thanks a lot. I had a misconception that expected value referred to the outcome having highest probability. · 3 years, 6 months ago

X=46; Y=91. I would choose the 5-envelope version, because there is more chance of getting the envelope with more money when the number of envelopes is smaller. My explanation is probably wrong, you are requested to solve the last part of your problem. · 3 years, 6 months ago

Comment deleted Apr 11, 2013

I would preface your post with a statement on risk-neutrality. Given 2 games with the same expected outcome, a risk-adverse person will prefer a lower variance, and a risk-loving person will prefer a higher variance. It is not immediately obvious that a person must always want to choose a game with lower variance. Even in the financial sector, not everyone picks the option with lower variance due to lack of information (or can't be bothered to figure out variance), agency problem, focus on kurtosis, etc.

Your statement "larger chance of seeing a net gain or net loss" is not true, and doesn't follow from simply knowing about the variance. $$P(X \neq 0 )$$ is not influence by the variance. It is easily to construct games with a higher variance but a low probability of change in outcome (e.g. the game that with probability $$\frac{1}{N}$$, pays out $$M$$, with probability $$\frac {1}{N}$$ pays out $$- M$$ and with probability $$1 - \frac {2}{N}$$ pays out $$0$$. has a high variance as M increases, but a low chance of seeing a net gain or net loss, especially as we increase $$N$$.)

Your statement on 'ruin probability' is also not necessarily true. Variance is independent of direction of movement. As a counter example, consider the 10-envelope game where 7 of them give you 1, 2 of them give you 10 and 1 of them gives you 73. Then, this has a higher variance then your 5-envelope game (with $$X=46$$) (453.6 against 324). However, the probability of ruin in bounded below by the probability of ruin in the 5-envelope game.

It is a peeve of mine that people have overly simplified all risk to just a few numbers (expected value and variance), and they do not understand the actual / valid implications, or even the real implications. Another example of a misleading statistic is the "Value at Risk" measure, which failed to account for compounding events, leading to the current financial crisis. Staff · 3 years, 6 months ago

Thank you, with the help of your explanation I have solved the Expected Raffle Value problem of this week. · 3 years, 6 months ago


text:
### hars123's blog

By hars123, history, 22 months ago,

I am finding it difficult to find the relation for transition from one state to another. i.e. if dp[i][j] represents no of ways of distributing j candies among positions [1,i]. Can someone explain what should i think to find that transition. Or i am just unable to observe the pattern.

• +5

 » 22 months ago, # |   0 Your dp state is correct, to do the transition in O(1) time you need to do the dp calculation iteratively and after you finish for some i, you need to make prefix sums in order to calculate the sum in range [l,r] for i-1 in constant time (O(1) ).Here is how to implement it: First set the base states, dp[0][0]=1 and dp[0][1...k]=0, Iterate over i from 1 to n and do the following: Make an array of size k and calculate prefix sums of dp[i-1][0...k], Now for every j from 0 to k, you can calculate in O(1) the number of ways to get to state dp[i][j], This is done by finding the sum of dp[i-1][max(0,j-a[i])...j] using the prefix sums. Here is my implementation of the problem if you need it. (I added comments to the code so you can understand it better :) )
 » 22 months ago, # |   0 In general, DP problems can often be thought of as a decision making process, where you make some "local" decision and your state after that local decision becomes some state you have previously calculated the answer for. So, when you are on a given state, you try all the local decisions you can, and take the one that results in the best answer (or in this case, sum up the number for each local decision to get the total count for the current state).Your dp state characterization is correct: dp[i][j] represents the number of ways of distributing j candies among the first i people. Now, say you are trying to calculate dp[i][j]. What is the "local decision" that you must make at this state? It will just be how many candies you want to give to the i'th person (in fact, deciding what to do with the i'th person is often the decision that must be made when your dp state deals with "something among the first i people"). So, say you want to give the i'th person k candies, where k is somewhere between 0 and min(ai, j) (we can't give them more than ai candies because of the problem constraint, and can't give them more than j candies because we only have j candies). Then, we see that if we give k candies to person i, then the number of ways to do that is dp[i - 1][j - k] (because we precisely want to find the number of ways to distribute j - k candies to the first i - 1 people). If we sum up these numbers over all the k from 0 to min(ai, j), then we should get the count for dp[i][j].You'll notice, though, that this results in O(NK) states and O(K) operations per state, or O(NK2) runtime, which is too slow. There is something about the state transitions that allows you to calculate the sum over all the different local decisions much faster, in fact in O(1). Can you think of what to do? Hint: all of the local decisions for a given dp[i][j] recurse into a contiguous block in the dp[i - 1] row.
•  » » 22 months ago, # ^ |   0 Is it possible to implement it in a topdown manner ?
•  » » » 22 months ago, # ^ |   0 I think it technically is, if you keep track of which states you've computed and calculate the prefix sum once you've gotten all of them... but it seems super complicated. Just do bottom up. If you don't usually do bottom up, good; this is a great opportunity to learn, since both top-down and bottom-up are useful to know.
•  » » 22 months ago, # ^ |   +6 Thank you very much for that detailed description. I would give 10 upvotes if I could. Using words instead of math makes my brain function better :)
•  » » 22 months ago, # ^ |   0 Just a quick question. When you see a problem and know its dp. What are your steps to coming up with the solution? I have a lot of trouble with bottom up but am very good at top down dp.For me, 1. Establish the state(pretty easy) 2. Establish transition(very difficult for me.. any tips?) 3. Establish base case(easy as well)I usually have to draw out the table on paper and try to see if I can find any patterns for the transition. How would you recommend going about finding transitions?
•  » » » 22 months ago, # ^ | ← Rev. 2 →   0 So I think those three steps are pretty good; as for the transition, this is where the "local decisions" come into play. If, at a given state, you can break the problem down into making one of multiple choices (often times it's something as simple as "use this element or don't use this element", like in knapsack), then you look at what states you'll end up with in the case of each one of your choices.Also, as you get to more difficult DP problems, establishing the state starts becoming much harder. For example, for this problem, once you get the state down, the transition is pretty simple, but the state is pretty hard to come up with. In my mind, this is the toughest part :)
•  » » 6 months ago, # ^ |   0 Really great explanation! Thank you!
•  » » 6 months ago, # ^ |   0 Here, how do we decide for the base case?as in what should be the values for dp[i][j] when either of i or j are zero ?i figured, that dp[0][0] is 1, coz there is one way of giving 0 coins to 0 children. There will be 0 ways of giving 1 coin to 0 children. Right? I am new to dp too. so I am thinking my way through.
•  » » » 6 months ago, # ^ |   0 Yup, so there should be 1 way of giving 0 coins to 0 children as you said, and 0 ways of giving $i > 0$ coins to 0 children. You don't need a separate base case for the case of $j = 0$, since that state should have a transition from dp[i-1][0] anyway.
 » 6 months ago, # |   +1 I have a non-dp solution to the problem. Here it is:Imagine that you have no constraints at first. What would that mean? The problem can be viewed as a variation of statement "There are k distinguishable urns and n in-distinguishable candies. Find the number of ways to put candies in urns." The answer is $(n+k-1)C(n)$.Ofc, we have over-counted.Now, consider the constraints. Subtract all the possibilities that gives $a_i + 1$ candies to $ith$ child.My intermediate solution is bit time consuming but I think, it can be easily optimized.
•  » » 6 months ago, # ^ | ← Rev. 3 →   0 The problem with this is that you will be double subtracting as well. For example, suppose $n=2$ and $a_1=3$, $a_2=2$, and say the number of candies to distribute is $7$, you will be subtracting the case of distributing candies [4, 3] twice, among others. This would require some inclusion exclusion to fix, and its not clear how that would be implemented here.
•  » » » 6 months ago, # ^ | ← Rev. 2 →   -8 No that would not happen.I have submitted solution with this logic at atcoder, it gave me 8 test cases correct and in rest of them (total 16 test cases) it gave me TLE.The example you provided is an invalid test case (The answer would be 0).However with a valid example, $sum(a_i) >= k$. We can do some simple algebraic maipulations and convince ourselves that multiple subtraction would not happen. Your code here... #include #include #include #include #include #include #include #include #include #include #include #define PI 3.14159265 #define mod 998244353 #define mod1 1000000007 using namespace std; // A Dynamic Programming based solution to compute nCr % p // Returns nCr % p int nCrModp(long long n, long long r, long long p) { // Optimization for the cases when r is large if (r > n - r) r = n - r; // The array C is going to store last row of // pascal triangle at the end. And last entry // of last row is nCr long long C[r + 1] = {0}; C[0] = 1; // Top row of Pascal Triangle // One by constructs remaining rows of Pascal // Triangle from top to bottom for (int i = 1; i <= n; i++) { // Fill entries of current row using previous // row values for (int j = min((long long)(i), r); j > 0; j--) // nCj = (n-1)Cj + (n-1)C(j-1); C[j] = (C[j] + C[j - 1]) % p; } return C[r]; } void func(){ long long n , k ; cin >> n >> k ; //k is the total chocos and n is total no. of student. long long ans = nCrModp((n + k - 1) , k , mod1) , arr[n]; //cout << ans << " "; for(int i = 0 ; i < n ; ++i){ cin >> arr[i]; } long long sum = 0; for(int i = 0 ; i < n; ++i){ // cout << (n + k - 1 - arr[i]) << " " << k - arr[i] << "\n"; if(arr[i] == k){ continue; } else sum += nCrModp((n + k - 2 - arr[i]) , k - 1 - arr[i] , mod1); } cout << (ans - sum) % mod1 << "\n"; } int main(){ int t = 1 , cnt = 0 ; //cin >> t; while(t--){ func(); } }
•  » » » » 6 months ago, # ^ | ← Rev. 3 →   0 My initial example was bad, but your logic is still incorrect and does double subtract; I'm not sure what algebraic manipulations you used to convince yourself that this wouldn't happen. I ran your code on the case $n=3, k=25, a_1=a_2=a_3=10$. My code output $21$, and yours output $-9$.Word of note: just because your code TLE'd doesn't mean it wouldn't also WA.
•  » » » » » 6 months ago, # ^ |   +8 My bad! You are absolutely right. I made a shitty assumption while going through proof.
 » 5 months ago, # | ← Rev. 3 →   0 My solution Done via bottom-up approach by calculating prefix sums for 1 less number of persons present.
 » 5 weeks ago, # |   0 hey, does anyone have recursive solution for this problem ?


text:
Math Help - Polar Coordinate 03

1. Polar Coordinate 03

Find the area inside r= 3 Cos[x] and outside of r= 1 + Cos[x]

2. Hello, camherokid!

Find the area inside $r \,= \,3\cos x$ and outside $r \:= \:1 + \cos x$
I hope you made a sketch . . .

$r\:=\:3\cos x$ is a circle with center $\left(\frac{3}{2},\,0\right)$ and radius $r = \frac{3}{2}$

$r\:=\:1 + \cos x$ is a cardioid with intercepts: $(2,\,0),\: \left(1,\,\frac{\pi}{2}\right),\:\left(1,\,\frac{3 \pi}{2}\right)$
. . and "dimples in" to the origin from the left.

The polar formula for the area between two curves is: . $A \;=\;\frac{1}{2}\int^{\beta}_{\alpha}\left(r_{_2}^ 2 - r_{_1}^2\right)\,d\theta$

The curves intersect when: . $3\cos x \;=\;1 + \cos x\quad\Rightarrow\quad 2\cos x \:=\:1\quad\Rightarrow\quad \cos x \:=\:\frac{1}{2}$
. . Hence, they intersect at: . $\theta \:=\:\pm\frac{\pi}{3}$

Therefore: . $A \;=\;\frac{1}{2}\int^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\bigg[(3\cos\theta)^2 - (1 + \cos x)^2\bigg]\,d\theta$

3. Originally Posted by camherokid
Find the area inside r= 3 Cos[x] and outside of r= 1 + Cos[x]
We will use the formula:

$A = \int_{ \alpha}^{ \beta} \frac {1}{2} (r_o^2 - r_i^2)~dx$

where $A$ is the area between the curves $r_o$ and $r_i$, $\alpha$ and $\beta$ are the limits of integration (the points of intersection), $r_o$ is the outer curve, and $r_i$ is the inner curve.

First find the points of intersection:

this is where $3 \cos x = 1 + \cos x$

$\Rightarrow \cos x = \frac {1}{2}$

$\Rightarrow x = \frac {\pi}{3}, \frac {5 \pi}{3}$

we want to go from $\frac {5 \pi}{3}$ to $\frac {\pi}{3}$, but we must go from a smaller angle to a bigger angle. changing $\frac {5 \pi}{3}$ to $- \frac {\pi}{3}$ fixes this problem. so our area is given by:

$A = \int_{- \pi / 3}^{ \pi / 3} \frac {1}{2} \left[ (3 \cos x)^2 - (1 + \cos x )^2 \right]~dx$

EDIT: Beaten by Soroban! any way, that's ok. I'm a bit rusty on polar areas, so it's good to have a confirmation that I did the right thing. Soroban, can you check the other posts camherokid put up today and make sure I didn't make any mistakes

4. Originally Posted by camherokid
Find the area inside r= 3 Cos[x] and outside of r= 1 + Cos[x]
the graph