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# What is atomic mass calculator?
Atomic mass is a key factor in determining the properties of an element. It is the total mass of an atom, including its protons, neutrons, and electrons. It is one of the most important components of an atom, yet it is often misunderstood. That’s why it is so important to understand what atomic mass is, how it is calculated, and how to use an atomic mass calculator to get the most accurate results.
Atomic mass can be determined by summing the number of protons and neutrons in an atom. To make this calculation easier, an atomic mass calculator can be used. This calculator takes the chemical formula of a substance and calculates the atomic mass of each element in the formula. This can help scientists and researchers determine the properties of the substance and how it could be used.
Atomic mass is an important tool for understanding the properties of an element, such as its density, melting point, and boiling point. It can also be used to determine the molecular weight of a compound, as well as the atomic weight of a molecule. With the help of an atomic mass calculator, scientists can also study how different elements interact with each other and how they could be used in various applications.
By understanding what atomic mass is and how it is calculated, scientists can get the most accurate results. This is why it is so important to understand how to use an atomic mass calculator. With this tool, scientists can study the properties of an element and the interaction between different elements in order to further their research.
## What is atomic mass calculator?
Atomic Mass Calculator is a free online tool that displays the atomic mass for the given chemical formula. Atomic Mass can be explained as the overall mass of electrons, neutrons and protons within an atom when the atom is at rest. We can calculate atomic mass by summing number of protons and neutrons in an atom.You will be able to find the atomic mass by using this below formula:
Atomic Mass = Number of Protons + Number of Neutrons
Use our below online atomic mass calculator to calculator to find atomic mass. BYJU’S online atomic mass calculator tool makes the calculation faster and it calculates the atomic mass of a chemical formula in a fraction of seconds.
### How to Use the Atomic Mass Calculator?
The procedure to use the atomic mass calculator is as follows:
Step 1: Enter the chemical formula in the input field
Step 2: Now click the button “Calculate Atomic Mass” to get the result
Step 3: Finally, the atomic mass of the chemical formula will be displayed in the new window
### What is Meant by Atomic Mass?
The atomic mass is defined as the mass/weight of the single atom of a chemical element. It is a unit of measurement for the mass of atoms and molecules and is expressed in unified atomic mass units (u). Atomic mass is also known as atomic weight and is represented by the symbol A. It is determined by the number of protons and neutrons present in the nucleus of the atom.
Atomic mass is an important concept in chemistry and physics. It is used in calculating the mass of a molecule, the mass of a chemical reaction, and the mass of an element. Atomic mass is also used to determine the relative abundance of elements in nature, which is an important factor in understanding the properties of molecules.
Atomic mass is also used to determine the stability of atoms and molecules. The stability of an atom or molecule is determined by the number of protons and neutrons in its nucleus. If the number of protons and neutrons is balanced, then the atom or molecule is said to be stable.
Atomic mass calculator is a useful tool for calculating the atomic mass of a chemical formula. It is a quick and easy way to determine the atomic mass of a chemical element or molecule. Atomic mass is an important concept in chemistry and physics and is used to determine the stability of atoms and molecules. It is also used to determine the relative abundance of elements in nature.
## What is an atomic calculator?
An atomic calculator is a tool that helps to calculate the atomic mass of a chemical formula. It is an important tool in chemistry, as it helps to determine the mass of an atom based on its chemical elements. Atomic calculators are used in a variety of applications, such as in the analysis of organic compounds and in the study of nuclei.
Atomic calculators are often used in research, as they can provide valuable information about the composition of atoms and molecules. They can also be used to determine the mass of a molecule, the number of protons and electrons in an atom, and the atomic charge of a molecule. Atomic calculators can also be used to calculate the mass of a chemical reaction, as well as the energy produced by the reaction.
Using an atomic mass calculator is relatively simple. All you need to do is enter the chemical formula of the substance you’re analyzing. The calculator will then provide the atomic mass and other data associated with the substance.
Atomic calculators can also be used to calculate the mass of a chemical reaction. To do this, you need to enter the number of protons and electrons in the reaction. The calculator will then calculate the energy produced by the reaction and the mass of the reaction.
### Atomic Charge Formula
The formula used to calculate the atomic charge of a molecule is:
AC = P – E
Where AC is the atomic charge (in eV), P is the number of protons, and E is the number of electrons. To calculate an atomic charge, simply subtract the number of electrons from the number of protons.
### How to Calculate Atomic Charge?
Calculating an atomic charge is easy. All you need to do is enter the number of protons and electrons into the atomic calculator. The calculator will then calculate the atomic charge of the molecule.
Atomic calculators can also be used to calculate the mass of a chemical reaction, as well as the energy produced by the reaction. To do this, you need to enter the number of protons and electrons in the reaction. The calculator will then calculate the energy produced by the reaction and the mass of the reaction.
Atomic calculators are an important tool in chemistry and are used in a variety of applications. They can provide valuable information about the composition of atoms and molecules, as well as the energy produced by chemical reactions. They can also be used to calculate the mass of an atom or molecule, the number of protons and electrons in an atom, and the atomic charge of a molecule.
## What is atomic mass and how is it calculated?
Atomic mass is an important concept in physics and chemistry, and knowing how to calculate it can be essential for understanding certain concepts in the fields. Atomic mass is the sum of the masses of the protons, neutrons, and electrons in an atom, or the average mass, in a group of atoms. This article explains what atomic mass is, how it is calculated, and why it is important.
### What Is Atomic Mass?
Atomic mass is the sum of the masses of the protons, neutrons, and electrons in an atom. It is usually expressed in atomic mass units (amu). One atomic mass unit is equal to 1/12 of the mass of a single carbon-12 atom. Atomic mass is also sometimes referred to as atomic weight.
Atomic mass is an important concept in physics and chemistry because it is used to determine the relative masses of different atoms. For example, the atomic mass of hydrogen is 1 amu, while the atomic mass of oxygen is 16 amu. This means that 16 hydrogen atoms have the same mass as 1 oxygen atom.
### How Is Atomic Mass Calculated?
There are two main ways to calculate atomic mass. The first is by adding the masses of the protons and neutrons in the nucleus of the atom. This is known as the “addition of protons and neutrons” method.
The second method is to calculate the average atomic mass. This is done by summing the masses of an element’s isotopes, each multiplied by its natural abundance on Earth. For example, the element carbon has two isotopes, carbon-12 and carbon-13. Carbon-12 has an atomic mass of 12 amu, and carbon-13 has an atomic mass of 13 amu. The average atomic mass of carbon is calculated by adding the contributions of both isotopes:
Average atomic mass of carbon = (12 x 0.9893) + (13 x 0.0107) = 12.011 amu
### Why Is Atomic Mass Important?
Atomic mass is important because it is used to calculate the mass of atoms, molecules, and other particles. Atomic mass is also used to determine the relative masses of different elements, which is important for understanding the structure of molecules and compounds. For example, knowing the atomic masses of hydrogen and oxygen can help us understand why water has a particular formula (H2O).
Atomic mass is also important for understanding the properties of elements. For example, the atomic mass of an element affects its boiling point and melting point. This is because the heavier an element is, the more energy it takes to vaporize the element.
Atomic mass is an important concept in physics and chemistry. It is the sum of the masses of the protons, neutrons, and electrons in an atom, or the average mass, in a group of atoms. Atomic mass can be calculated by adding the masses of the protons and neutrons, or by calculating the average atomic mass of an element’s isotopes. Atomic mass is important for understanding the relative masses of different atoms, molecules, and compounds, as well as the properties of elements.
## What is atomic mass and how do you calculate it?
Atomic mass is a measure of the mass of a single atom, molecule, or group of atoms. It is the sum of the masses of the protons, neutrons, and electrons in an atom, or the average mass in a group of atoms. Calculating atomic mass can help you understand the structure and properties of an atom, or a group of atoms.
Atomic mass is the sum of the masses of the protons, neutrons, and electrons in an atom, or the average mass, in a group of atoms. The atomic mass is the mass of an atomic particle or molecule. With this system, the masses of a proton and a neutron are nearly equal to 1 atomic mass unit, while the electron has a mass close to 0.0005 atomic mass unit. Atomic mass is also called atomic weight.
For atoms, the protons and neutrons of the nucleus account for almost all of the mass, and the atomic mass measured in u has nearly the same value as the mass number. Masses of fundamental atomic particles are often expressed in atomic mass units (u). “One atomic mass unit (1u) is one twelfth of the mass of an atom of carbon with six protons and six neutrons.”
### How to Calculate Atomic Mass
There are several ways to calculate atomic mass. The most common method is to use the mass number (which is the total number of protons and neutrons in the nucleus) and the atomic number (the number of protons in the nucleus) of the atom.
The calculation for atomic mass is:
Atomic Mass = Mass Number – Atomic Number
For example, for an atom with a mass number of 12 and an atomic number of 6, the calculation would be:
Atomic Mass = 12 – 6
Atomic Mass = 6
Another way to calculate atomic mass is to use the average masses of the isotopes. Isotopes are atoms with the same number of protons and electrons, but different numbers of neutrons. The average mass of an element is calculated by taking the weighted average of the masses of the different isotopes.
For example, the average mass of helium is calculated by taking the weighted average of the masses of helium-3 (3.0160293 u) and helium-4 (4.0026022 u).
Average Mass of Helium = (3.0160293 x 0.000137) + (4.0026022 x 0.999863)
Average Mass of Helium = 4.002604
Atomic mass is an important concept in chemistry and physics. It is the sum of the masses of the protons, neutrons, and electrons in an atom, or the average mass in a group of atoms. There are several ways to calculate atomic mass, depending on the information you have. The most common way is to use the mass number and the atomic number of the atom. You can also calculate the average mass of an element by taking the weighted average of the masses of its isotopes.
## How is atomic mass calculated?
Atomic mass is an important concept in chemistry and physics. It is the sum of the masses of the protons, neutrons, and electrons in an atom, or the average mass in a group of atoms. Knowing how to calculate atomic mass is essential for understanding the behavior of elements and compounds.
### What is Atomic Mass?
Atomic mass, also referred to as atomic weight, is the sum of the masses of the protons, neutrons, and electrons in a single atom, or the average mass of a group of atoms. In chemistry, atomic mass is used to define the relative mass of an atom or element. In physics, it is used to calculate the mass of an object, such as a molecule.
The most basic way to calculate atomic mass is by adding the masses of each proton and neutron present in the nucleus of an atom. This is done by using the number of protons and neutrons indicated in the periodic table, which is typically represented by the symbol of an element. For example, the atomic mass of carbon-12 can be calculated by adding the mass of each proton (1 amu) and neutron (1 amu) to get 12 amu, or 12 atomic mass units.
### Calculating Atomic Mass
There are several ways to calculate the atomic mass of an atom. The most common method is to refer to the periodic table and find the atomic number and mass number of the element. The atomic number is the number of protons in the nucleus, and the mass number is the sum of the protons and neutrons. For example, carbon-12 has an atomic number of 6 and a mass number of 12. The atomic mass can then be calculated by subtracting the atomic number from the mass number, giving a result of 6 amu.
### Mass Spectrometry
Mass spectrometry is another method used to calculate atomic mass. This technique uses a device called a mass spectrometer to measure the mass of an atom or molecule. The spectrometer uses an electric field to separate the atoms or molecules into different components, which can then be measured and used to calculate the atomic mass.
### Isotope Abundance
Finally, isotope abundance can be used to calculate the atomic mass of an element. This method uses the relative abundance of different isotopes of an element to calculate the average atomic mass. Isotopes are atoms of the same element that have a different number of neutrons. For example, carbon has three common isotopes: carbon-12, carbon-13, and carbon-14. The average atomic mass of carbon is calculated by taking the relative abundance of each isotope and multiplying it by its mass.
In summary, there are several ways to calculate atomic mass. The method used will depend on the information available. For most calculations, the periodic table and mass spectrometry are the most reliable methods. Isotope abundance can also be used to calculate the average atomic mass of an element. Knowing how to calculate atomic mass is an important part of understanding the behavior of elements and compounds.
## How do you find the mass of an atom?
Atomic mass is a key component of chemistry and physics, and is essential for understanding the structure of atoms and their behavior. But how do you find the mass of an atom?
There are a few different methods you can use, depending on the type of information you’re given. Here, we’ll discuss three ways to find atomic mass, and explain what atomic mass is in the first place.
Atomic mass is the total mass of all the particles in an atom, including protons, neutrons, and electrons. It’s also known as atomic weight and is measured in atomic mass units (amu).
Atomic mass is a fundamental property of an atom that is important for understanding its structure and behavior. It can be used to calculate the mass of a single atom, or a sample of atoms.
### Look Up Atomic Mass on the Periodic Table
The easiest way to find the atomic mass of an element is to look it up on the periodic table. The atomic mass is listed in the bottom right corner of each element’s box.
The atomic mass is the average mass of all the atoms of the same element, taking into account the different isotopes of the element. For example, hydrogen has two isotopes, protium and deuterium, with atomic masses of 1.0078 amu and 2.0141 amu respectively. The average atomic mass of hydrogen is 1.0079 amu, taking into account the relative abundance of the two isotopes.
### Calculate Atomic Mass for a Single Atom
If you know the number of protons and neutrons in an atom, you can calculate its atomic mass. This is done by adding the mass of the protons and neutrons together.
The mass of a proton is 1.0073 amu, and the mass of a neutron is 1.0087 amu. Therefore, if an atom has 6 protons and 7 neutrons, its atomic mass is 6 x 1.0073 amu + 7 x 1.0087 amu = 14.1021 amu.
### Calculate Atomic Mass for a Sample
If you have a sample of atoms, you can calculate the average atomic mass of the sample. This is done by taking into account the different isotopes of the element and their relative abundances.
For example, if a sample of chlorine contains 75.77% of chlorine-35 and 24.23% of chlorine-37, the average atomic mass of the sample can be calculated by multiplying the atomic mass of each isotope by its relative abundance, and then adding the results together.
In this example, the average atomic mass of the sample is (35 x 0.7577) + (37 x 0.2423) = 35.45 amu.
Atomic mass is an important property of atoms and is essential for understanding the structure and behavior of atoms. There are several ways to calculate atomic mass, depending on the type of information you’re given. You can look up the atomic mass on the periodic table, calculate it for a single atom, or calculate it for a sample of atoms.
#### Kernan Sean
As a Science and Chemistry Web Editor at the renowned Chemcafe.net Magazine, I strive to provide readers with insightful content that educates as well as entertains! I'm a renowned author, lecturer, and life coach from Chicago, Illinois. I has been actively writing since she was just 14 years old and has authored multiple novels, short stories, and works of non-fiction. My work has been published in magazines and newspapers across the US and abroad. I'm passionate about inspiring others to reach their full potential by living an intentional life full of purpose. | 3,977 | 18,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-26 | latest | en | 0.922464 |
http://math.stackexchange.com/questions/140654/what-is-the-dimension-of-this-subspace-of-mn-times-n-mathbbr | 1,419,268,862,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802775517.52/warc/CC-MAIN-20141217075255-00076-ip-10-231-17-201.ec2.internal.warc.gz | 176,853,627 | 17,441 | What is the dimension of this subspace of $M(n\times n,\mathbb{R})$
What is the dimension of the space of $\{A\ {^t\!A}: A\in M(n\times n,\mathbb{R})\}$? I think it should be $n(n+1)/2$ if one knows already the dimension of the special orthogonal group, but I would love to derive the latter from the former.
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It's not clear to me that your set is a vector space. It's contained in the $n(n+1)/2$ dimensional space of symmetric matrices, but, for example, $\pmatrix{0&1\cr1&0\cr}$ isn't in it. – Gerry Myerson May 3 '12 at 23:40
Ah, you are right. But you have made me realize what I was missing before, I think. It is sufficient for my purposes to just note that I can define a map from the $\mathbb{R}^{2n}$ to my set considered as a subset of the symmetric matrices. Thanks! – Eric Gregor May 3 '12 at 23:45
The space is the same as the set of (symmetric) positive semi-definite matrices. The dimension could be defined as the dimension of the affine hull of the set, which in this case (since $0$ is in the set) is the span of the set. The span of the set is just the set of symmetric matrices, which has the given dimension. – copper.hat May 4 '12 at 0:52
Talking properly, your set is a cone. – Martin Argerami May 4 '12 at 1:15
Would you explain this intriguing comment, Martin? If you expanded and wrote an answer I would accept it. – Eric Gregor May 4 '12 at 1:18
Typically, a cone can be seen as the "positive part" of your space. Canonical examples are $[0,\infty)\subset\mathbb{R}$, $[0,\infty)\subset\mathbb{C}$, and the positive-semidefinite matrices (either real or complex); this last one is your example.
The main property of the cone is that it is convex (i.e. $x,y$ in the cone, $t\in[0,1]$ imply that $tx+(1-t)y$ is in the cone), and so it can be seen as generated (as a convex set) by its extreme points (that is, every point is a convex combination of extreme points). In your case, the extreme points are the scalar multiplies of the projections, i.e. $\lambda A$ with $\lambda\geq0$, $A=A^t=A^2$.
Many definitions of cone that I have seen just require $\lambda x$ be in the code when $x$ is in the cone and $\lambda>0$. If $0$ is in the cone, it is sometimes called a pointed cone. That is, convexity is not necessarily part of the definition. Not that it matters here. – copper.hat May 4 '12 at 3:09 | 676 | 2,334 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2014-52 | longest | en | 0.925116 |
https://amarineblog.com/2019/10/23/fatigue-on-weld/ | 1,685,769,394,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649105.40/warc/CC-MAIN-20230603032950-20230603062950-00385.warc.gz | 121,138,830 | 33,818 | Fatigue on Weld
Designing for fatigue is very different than designing for static loads. One of the biggest challenges is being able to estimate the loads, timing of the loads and duration of the loads. If you look at a bridge you have to estimate the number of vehicles going on the bridge every day, the weight of these vehicles, and other forces and loads acting on the bridge such as its own weight (steel and concrete), wind loading, and pre-existing stresses such as those caused by welding.
Girders are subjected to fluctuating loads which combined with temperature and fabrication quality will determine fatigue life.
Four of the main factors that will affect the fatigue life of a welded structure are:
3. Temperature
4. Quality of fabrication (workmanship)
Loading magnitudes are very important in estimating fatigue life. However, what is really important is the different between the stress at the peak (maximum stress) and the stress when the load is removed or reversed (minimum stress). For example, the applied load P (see sketch below) translates to 15,000 psi. This load can be a vehicle going over the bridge. When the load is removed (the vehicles is through the bridge) the applied stress is 0 psi (neglecting the weight of the girder, concrete and other static loads for the sake of this example).
The applied load P places the top flange of the girder in compression and the bottom flange in tension.
In this case the stress range is equal to the maximum stress minus the minimum stress…
Stress Range (1) = 15,000 lb/in² – 0 lb/in² = 15,000 lb/in²
If we have another situation in which the applied peak stress is higher, say 25,000 lb/in², but the minimum stress is 18,000 lb/in² (this may be due to the structure’s weight or design) the stress range would be
Stress Range (2) = 25,000 lb/in² – 18,000 lb/in² = 7,000 lb/in²
Although the peak stress is higher in the second scenario, the first scenario would have a shorter fatigue life due to the magnitude of the stress range. So, the larger the stress range the shorter the fatigue life and vice versa.
Temperature is very important for fatigue life. For structural steels, as temperature decreases so does its fracture toughness, or its ability to resist the propagation of a crack. Fracture toughness can be tested in a lab. You may have heard of the Charpy v-notch test. Typically these tests are done at temperature ranging from -80˚F to 0˚F. This is done because structural steels have sharp drop in fracture toughness once the temperature dips below freezing. This is especially important for bridges and other structures that are in service in colder regions.
Since fracture toughness decreases as temperature decreases, the lower the temperature the lower the fatigue life.
Quality of fabrication refers to the presence, or hopefully absence, of fabrication defects. When we weld we can have many different types of defects such as undercut, overlap, porosity, cracks, excessive convexity, etc. All these discontinuities will act as stress risers. This means that a given stress will be magnified to different degrees in the presence of these discontinuities.
This is why welding codes impose more stringent requirements (less allowances) for discontinuities in cyclically loaded members as compared to their statically loaded counterparts.
Conservative design will assume that there are pre-existing flaw, such as those mentioned above. This is a safety net for errors in fabrication and inspection.
You may be wondering why there was no discussion of the strength of the steel when considering fatigue life. This is because after welding, the fatigue life is dominated by the weld discontinuity. This discontinuity is not necessarily a flaw; it could simply be the change in geometry). For structural steels (yield strengths ranging from 36,000 lb/in² to 140,000 lb/in²) fatigue life improves with higher strength, but as mentioned above, as soon as we weld the fatigue life is governed by the quality and geometry of the weld.
If you look at AWS D1.1 you’ll find that the different stress categories (A, B, B’, C, D, E, E’ and F) are not dependent on the yield strength of the steel but strictly dependent on geometry and type of loading as can be seen below.
References: | 940 | 4,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-23 | latest | en | 0.909921 |
https://studdy.ai/shared-solution/a835cfc6-062d-445d-b653-cf6f24382e57 | 1,721,527,186,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517544.48/warc/CC-MAIN-20240720235600-20240721025600-00203.warc.gz | 469,812,259 | 13,445 | # Math Snap
## Question 1: A can of COKE has a base radius of $30 \mathrm{~mm}$ and a height of $130 \mathrm{~mm}$. a) Calculate the volume of the container (answer in $\mathrm{cm}^{3}$ ): convert to \begin{aligned} \mathrm{cm}^{3} & =30 \mathrm{~mm}=3 \mathrm{~cm} \\ & 136 \mathrm{~mm}=13 \mathrm{~cm}\end{aligned} volume $=\pi \times r^{2} \times$ height $3.14 \times(3 \mathrm{~cm})^{2} \times 13$ $3.14 \times 9 \mathrm{~cm}^{2} \times 13=36.9 \mathrm{~cm}^{3}$ b) A conical cover of height $3 \mathrm{~cm}$ is placed on top of the can to form a silo. Calculate the volume of the cone. (HINT: Draw a diagram)
#### STEP 1
Assumptions1. The base radius of the can is30 mm or3 cm. The height of the can is130 mm or13 cm3. The volume of a cylinder is given by the formula $\pi \times r^{} \times h$ 4. The height of the conical cover is3 cm5. The volume of a cone is given by the formula $\frac{1}{3} \times \pi \times r^{} \times h$
#### STEP 2
First, we need to calculate the volume of the can. We can do this by using the formula for the volume of a cylinder. $Volume_{cylinder} = \pi \times r^{2} \times h$
#### STEP 3
Now, plug in the given values for the radius and height of the can to calculate the volume. $Volume_{cylinder} = \pi \times (3 \, cm)^{2} \times13 \, cm$
#### STEP 4
Calculate the volume of the cylinder. $Volume_{cylinder} = \pi \times9 \, cm^{2} \times13 \, cm =3.14 \times9 \, cm^{2} \times13 \, cm =363.51 \, cm^{3}$
#### STEP 5
Now, we need to calculate the volume of the conical cover. We can do this by using the formula for the volume of a cone. $Volume_{cone} = \frac{1}{3} \times \pi \times r^{2} \times h$
#### STEP 6
Now, plug in the given values for the radius and height of the cone to calculate the volume. $Volume_{cone} = \frac{1}{3} \times \pi \times (3 \, cm)^{2} \times3 \, cm$
##### SOLUTION
Calculate the volume of the cone. $Volume_{cone} = \frac{1}{3} \times \pi \times9 \, cm^{2} \times3 \, cm =3.14 \times3 \, cm^{3} =28.26 \, cm^{3}$The volume of the can is363.51 cm³ and the volume of the conical cover is28.26 cm³. | 710 | 2,081 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 16, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2024-30 | latest | en | 0.638163 |
https://aplicaciones.uc3m.es/cpa/generaFicha?est=221&plan=165&asig=14815&idioma=2 | 1,642,731,124,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00615.warc.gz | 176,724,555 | 4,079 | Checking date: 23/06/2021
Course: 2021/2022
Strength of Materials
(14815)
Study: Bachelor in Mechanical Engineering (221)
Coordinating teacher: ZAERA POLO, RAMON EULALIO
Department assigned to the subject: Department of Continuum Mechanics and Structural Analysis
Type: Compulsory
ECTS Credits: 6.0 ECTS
Course:
Semester:
Requirements (Subjects that are assumed to be known)
Mechanics of Structures Elasticity.
Objectives
By the end of this subject, students will be able to have: 1. a systematic understanding of the key aspects and concepts of strength of materials on the behavior of real solids. 2. coherent knowledge of mechanical engineering including some at the forefront of the branch in strength of materials. 3. the ability to apply their knowledge and understanding to identify, formulate and solve problems of strength of materials using established methods 4. the ability to select and apply relevant analytic and modelling methods in strength of materials . 5. the ability to apply their knowledge and understanding to develop and realise designs in strength of materials to meet defined and specified requirements; 6. an understanding of design methodologies in strength of materials , and an ability to use them. 7. the ability to design and conduct appropriate experiments of strength of materials , interpret the data and draw conclusions; 8. workshop and laboratory skills in strength of materials . 9. the ability to select and use appropriate equipment, tools and methods to solve problems of strength of materials ; 10. the ability to combine theory and practice to solve problems of strength of materials ; 11. an understanding of applicable techniques and methods in strength of materials , and of their limitations.
Skills and learning outcomes
Description of contents: programme
1. PART 1: STRESSES IN PRISMATIC ELEMENTS 1.1. Normal stresses: axial force and bending moment 1.2. Shear stresses: shear force and torsion PART 2: MOVEMENTS IN PRISMATIC ELEMENTS 2.1. Navier-Bresse equations 2.2. Mohr theorems 2.3. Equation of the Elastica PART 3: ENERGETIC THEOREMS 3.1. Work of the extrenal forces 3.2. Elastic energy stored in the prismatic element 3.3. Maxwell-Betti theorem 3.4. First Castigliano theorem. Applications PART 4: SOLVING STATICALLY INDETERMINATE STRUCTURES 4.1. Fundamentals of structural analysis 4.2. Beams 4.3. Truss structures 4.4. Frame structures
Learning activities and methodology
In each week one lecture session (master class) and one practical session (in reduced groups) will be taught. The first is geared to the acquisition of theoretical knowledge, and the second to the acquisition of practical skills related to theoretical concepts. In addition to this sessions two laboratory practical sessions in reduced groups (maximum 20 students) will be impart. Students will have the possibility of individual tutorials. Students will have the possibility of individual tutorials. Also, there could be group tutoring session.
Assessment System
• % end-of-term-examination 60
• % of continuous assessment (assigments, laboratory, practicals...) 40
Calendar of Continuous assessment
Basic Bibliography
• F.P. Beer, E.R. Johnston, J.T. DeWolf, D.F. Mazurek. Mechanics of Materials. McGraw-Hill. 2013
• J.M. Gere, S. Timoshenko. Mechanics of Materials (8th Ed.). Cengage Learning. 2009 | 769 | 3,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-05 | latest | en | 0.869912 |
http://math.stackexchange.com/questions/148242/vector-analysis-linear-algebra | 1,469,337,897,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823947.97/warc/CC-MAIN-20160723071023-00324-ip-10-185-27-174.ec2.internal.warc.gz | 165,190,711 | 18,371 | # Vector Analysis & Linear Algebra
I'm given a positive number, a unit vector $u \in \mathbb{R} ^n$ and a sequence of vectors $\{ b_k \} _{ k \geq 1}$ such that $|b_k - ku| \leq d$ for every $k=1,2,...$.
This obviously implies $|b_k| \to \infty$ . But why does this imply $\angle (u,b_k) \to 0$ ? I've tried proving it using some inner-product calculations, but without any success.
In addition, why the given data implies that there must exist $i<j$ such that $|b_i| \leq \frac{\delta}{4} |b_j|$ ?
Thanks a lot !
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You have $|b_k| \leq |b_k -k u| + k=d+k$, and $|b_k| \geq k-|b_k -k u|=k-d$. This gives: $$\lim_{k \to \infty} \frac{\langle u, b_k-ku\rangle}{|b_k|} = 0, \;\; \lim_{k \to \infty} \frac{\langle u, ku\rangle}{|b_k|} = 1,$$ so it follows that $$\lim_{k \to \infty} \frac{\langle u, b_k \rangle}{|b_k|} = \lim_{k \to \infty} \frac{\langle u, b_k -ku\rangle + \langle u, ku\rangle}{|b_k|} = 1.$$
-
Thanks !! have you got any idea about my second question? – joshua May 22 '12 at 15:59
Fix $i$. Since $|b_j| \geq j-d$, you can choose $j$ large enough so that $|b_j| >\frac{4}{\delta} |b_i|$. – copper.hat May 22 '12 at 16:04
Great !!! Thanks !!! – joshua May 22 '12 at 16:14
From $|b_k-k\,u|\le d$ you get that for all sufficiently large $k$ $$k-d\le |b_k|\le k+d$$ and $$k-d\le u\cdot b_k\le k+d.$$ Then $$\frac{k-d}{k+d}\le\cos(\angle(u,b_k))\le\frac{k+d}{k-d}.$$ It follows that $\cos(\angle(u,b_k))\to1$ and $\angle(u,b_k)\to0$
-
Thanks !!!!!!!! – joshua May 22 '12 at 16:43 | 588 | 1,498 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2016-30 | latest | en | 0.680081 |
http://math.stackexchange.com/questions/301264/topology-prerequisites-for-algebraic-topology | 1,466,864,363,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783393332.57/warc/CC-MAIN-20160624154953-00123-ip-10-164-35-72.ec2.internal.warc.gz | 200,185,169 | 20,566 | # Topology Prerequisites for Algebraic Topology
Note: There is another question of the same title, but it is different and asks for group theory prerequisites in algebraic topology, while i want the topology prerequisites.
I am a physics undergrad, and I wish to take up a course on Introduction to Algebraic Topology for the next sem, which basically teaches the first two chapters of Hatcher, on Fundamental Group and Homology. However, I don't have a formal mathematics background in point-set topology, and I don't have enough time to go though whole books such as Munkres. So What part of point set topology from Munkres is actually used in the first two chapters of Hatcher?
More importantly, I wanted to know if the first chapter of the book Topology, Geometry and Gauge Fields by Naber or first 2 chapters of Lee's Topological Manifolds would be sufficient to provide me the necessary background for Hatcher.
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What Sigur wrote in his answer, but also separation axioms, though most spaces you deal with in algebraic topology have all separation properties. Quotient spaces and quotient maps are of particular importance in homology. – Stefan Hamcke Feb 12 '13 at 19:36
You might be interested in A. H. Wallace, An Introduction to Algebraic Topology, Dover, 2007. It is a reprint of the 1957 original, so it is on the older side and a bit "old fashioned" in terms of the exposition. That said, it assumes no prior background in topology and so builds up the necessary prerequisites in the first three chapters (about 60 pages). – cardinal Feb 17 '13 at 3:03
Other than a little "mathematical maturity" there's not very many hard formal prerequisites for studying from Hatcher. On the point-set topology front, you'll want to be familiar with the subspace topology and the quotient topology. You should also be familiar with abelian groups and at least be modestly familiar with abstract (non-abelian) groups up to quotient groups. – Ryan Budney Feb 18 '13 at 5:12
Chapter 1 of Hatcher corresponds to chapter 9 of Munkres. These topology video lectures (syllabus here) do chapters 2, 3 & 4 (topological space in terms of open sets, relating this to neighbourhoods, closed sets, limit points, interior, exterior, closure, boundary, denseness, base, subbase, constructions [subspace, product space, quotient space], continuity, connectedness, compactness, metric spaces, countability & separation) of Munkres before going on to do 9 straight away so you could take this as a guide to what you need to know from Munkres before doing Hatcher, however if you actually look at the subject you'll see chapter 4 of Munkres (questions of countability, separability, regularity & normality of spaces etc...) don't really appear in Hatcher apart from things on Hausdorff spaces which appear only as part of some exercises or in a few concepts tied up with manifolds (in other words, these concepts may be being implicitly assumed). Thus basing our judgement off of this we see that the first chapter of Naber is sufficient on these grounds... However you'd need the first 4 chapters of Lee's book to get this material in, & then skip to chapter 7 (with 5 & 6 of Lee relating to chapter 2 of Hatcher).
There's a crazy amount of abstract algebra involved in this subject (an introduction to which you'll find after lecture 25 in here) so I'd be equally worried about that if I didn't know much algebra.
These video lectures (syllabus here) follow Hatcher & I found the very little I've seen useful mainly for the motivation the guy gives. If you download the files & use a program like IrfanView to view the pictures as you watch the video on vlc player or whatever it's much more bearable since you can freeze the position of the screen on the board as you scroll through 200 + pictures.
I wouldn't recommend you treat point set topology as something one could just rush through, I did & suffered very badly for it...
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Hey, Thanks for the comprehensive answer. I am doing a full masters level course of groups and rings, so I am pretty sure, I will have the algebra prerequisites. I will try to finish Munkres, else I will go through Naber or Lee. – user23238 Feb 18 '13 at 8:52
No worries, just wondering whether you read the introductory chapter of Naber? I found the dirac string stuff fascinating, & the end of the book contains stuff on Donaldson theory that (I think) one of our lecturers contributed to or has some relationship with at any rate. If these books are too brief books like the schaums one or topology without tears are useful for testing definitions out on finite topologies. – sponsoredwalk Feb 18 '13 at 17:12
I did read part of Nabers first chapter including the Dirac monopoles, it is very interesting, I agree! I also want to read the section of Hopf Bundle, but not finding the time to do so. Have you read bout the hopf bundle?P.S: +1 for your answer. I will wait for a few days, before I award you the bounty. – user23238 Feb 19 '13 at 0:16
Yeah, & the mention of how Hopf's work apparently had no relation to Dirac strings reminded me of a moment in this talk by Steven Weinberg where he talks about how he cracked a problem with Lagrangians using abstract algebra :p Good introductory chapter to a book... – sponsoredwalk Feb 19 '13 at 1:37
For sure you'll need continuous functions, homeomorphisms, connectedness, compactness, coverings and many others.
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I prefer Munkres over all topology books.
You might starting with Munkres chapter 2, then read chapters 3, 4, 7 (without " * " sections), but if you have enought time is not bad idea reading all of the first part: Chapters 1-8 (long but fun).
I think that chapter 1 is good for you, is an intuitive approach for set-theory, since you are a physicist probably not like going too deeply into sets, but if you dont have time, skip it.
But my biggest advice is not worry about taking the course as quickly, if you don't feel safe. I was physicist.
- | 1,392 | 5,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2016-26 | longest | en | 0.921659 |
https://www.solutioninn.com/study-help/data-structures-algorithms/suppose-we-are-given-an-array-a-of-n-nonintersecting-835672 | 1,685,898,228,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650201.19/warc/CC-MAIN-20230604161111-20230604191111-00217.warc.gz | 1,078,734,183 | 8,639 | # Suppose we are given an array, A, of n nonintersecting segments, s 1 , s 2 ,...,s n , with endpoints on the lines y = 0 and y = 1, and ordered from left to right. Given a point q with 0 < y(q) < 1, design an algorithm that in O(log n) time computes the segment s i
Chapter 22, Creativity #14
Suppose we are given an array, A, of n nonintersecting segments, s1, s2,...,sn, with endpoints on the lines y = 0 and y = 1, and ordered from left to right. Given a point q with 0 < y(q) < 1, design an algorithm that in O(log n) time computes the segment si of A immediately to right of q, or reports that q is to the right of all the segments. | 190 | 640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-23 | latest | en | 0.917615 |
https://archive.softwareheritage.org/browse/revision/0a03bbb7cf19e479dc77592ed09621eeb8afb470/?path=man/SimGLP.Rd | 1,726,719,838,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00714.warc.gz | 87,519,123 | 8,884 | Revision 0a03bbb7cf19e479dc77592ed09621eeb8afb470 authored by A.I. McLeod on 21 December 2015, 08:55:04 UTC, committed by cran-robot on 21 December 2015, 08:55:04 UTC
1 parent 83deceb
SimGLP.Rd
\name{SimGLP}
\alias{SimGLP}
\title{ Simulate GLP given innovations}
\description{
Simulates a General Linear Time Series that can have nonGaussian innovations.
It uses the FFT so it is O(N log(N)) flops where N=length(a) and N is assumed
to be a power of 2.
The R function \code{convolve} is used which implements the FFT.
}
\usage{
SimGLP(psi, a)
}
\arguments{
\item{psi}{ vector, length Q, of MA coefficients starting with 1. }
\item{a}{ vector, length Q+n, of innovations, where n is the length of time series
to be generated. }
}
\details{
\deqn{ z_t = \sum_{k=0}^Q psi_k a_{t-k} }
where \eqn{t=1,\ldots,n} and the innovations
$a_t, t=1-Q, \ldots, 0, 1, \ldots, n$ are
given in the input vector a.
Since \code{convolve} uses the FFT this is faster than direct computation.
}
\value{
vector of length n, where n=length(a)-length(psi)
}
\author{ A.I. McLeod }
\seealso{
}
\examples{
#Simulate an AR(1) process with parameter phi=0.8 of length n=100 with
# innovations from a t-distribution with 5 df and plot it.
#
phi<-0.8
psi<-phi^(0:127)
n<-100
Q<-length(psi)-1
a<-rt(n+Q,5)
z<-SimGLP(psi,a)
z<-ts(z)
plot(z)
}
\keyword{ ts }
\keyword{ datagen }
Computing file changes ... | 472 | 1,382 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-38 | latest | en | 0.690148 |
http://openwetware.org/wiki/Physics307L_F07:People/Knockel/Lab0 | 1,481,184,517,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542455.45/warc/CC-MAIN-20161202170902-00406-ip-10-31-129-80.ec2.internal.warc.gz | 200,284,076 | 6,012 | # Physics307L F07:People/Knockel/Lab0
## Oscilloscope lab summary
1. Summary of lab: basically, we practiced setting up and using an oscilloscope to get used to the online notebook and the equipment in the lab.
2. Link to my notebook: August 29 lab notebook
3. My values for AC fall time: I received three values for fall time by measuring the time it took for the voltage to drop to 10%: 52ms, 53ms, and 52ms. The average of these is 52.3ms. I'm not too sure what error bars are, but the standard deviation here is $\sigma=\sqrt{\frac{1}{3}\times\left(\left(52-52.3\right)^2+\left(53-52.3\right)^2+\left(52-52.3\right)^2 \right)}ms=0.47ms$
4. What I learned: I learned a lot about how oscilloscopes work and a little about editing a wiki. I don't really get the RC constant...
5. Fun things I explored during lab: I pushed a lot of random buttons which did nothing amazing, but when I didn't plug in a ground cable into the function generator, the screen looked like the the waves moving in the ocean.
6. how to make the lab better: hmmm... maybe say when this summary is due? It was very helpful when you explained the AC Coupling in front of the class. Maybe if you would have done the same for the RC constant things would have gone smoother.
Koch comments (02:15, 30 August 2007 (EDT)):
• 3: That's OK, we will spend a lot of time talking about error bars and the goal is to be an expert in it by the end of the semester (not right away). It's a bit complicated for me to spell out here, but calculating the standard deviation of your measurements is a great start (also nice use of math by the way: I don't know how to do that yet). It's slightly different to calculate the standard error in your estimate of the mean, and we'll talk about it later (probably the next lecture). If you want a preview, you can try this: http://davidmlane.com/hyperstat/A103735.html
• 4: Great that you learned a lot! Yeah, don't worry about RC, you're right I should have explained it more.
• 5: The waves were groovy
• 6: Thank you for the constructive comments. I think it will be more clear when we're doing "real" labs when the summary is due. I was thinking at the end of the second lab session, but maybe that won't be practical. I'll send you an email about your final grade for all of this. | 601 | 2,290 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2016-50 | latest | en | 0.937841 |
http://nrich.maths.org/public/leg.php?code=-68&cl=3&cldcmpid=779 | 1,506,443,450,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818696653.69/warc/CC-MAIN-20170926160416-20170926180416-00540.warc.gz | 244,672,323 | 10,171 | Search by Topic
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Screwed-up
Stage: 3 Challenge Level:
A cylindrical helix is just a spiral on a cylinder, like an ordinary spring or the thread on a bolt. If I turn a left-handed helix over (top to bottom) does it become a right handed helix?
Getting an Angle
Stage: 3 Challenge Level:
How can you make an angle of 60 degrees by folding a sheet of paper twice?
Counting Triangles
Stage: 3 Challenge Level:
Triangles are formed by joining the vertices of a skeletal cube. How many different types of triangle are there? How many triangles altogether?
There and Back Again
Stage: 3 Challenge Level:
Bilbo goes on an adventure, before arriving back home. Using the information given about his journey, can you work out where Bilbo lives?
More Pebbles
Stage: 2 and 3 Challenge Level:
Have a go at this 3D extension to the Pebbles problem.
Squares in Rectangles
Stage: 3 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
Crossing the Atlantic
Stage: 3 Challenge Level:
Every day at noon a boat leaves Le Havre for New York while another boat leaves New York for Le Havre. The ocean crossing takes seven days. How many boats will each boat cross during their journey?
Eight Hidden Squares
Stage: 2 and 3 Challenge Level:
On the graph there are 28 marked points. These points all mark the vertices (corners) of eight hidden squares. Can you find the eight hidden squares? | 2,233 | 9,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-39 | latest | en | 0.904485 |
https://plugincafe.maxon.net/topic/9243/12294_mcommandsubdivide | 1,695,543,371,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506623.27/warc/CC-MAIN-20230924055210-20230924085210-00796.warc.gz | 506,411,513 | 15,039 | # MCOMMAND_SUBDIVIDE
On 05/12/2015 at 21:44, xxxxxxxx wrote:
I have a simple polygon and the MCOMMAND_SUBDIVIDE ** ** divides it in 4.
Now my question:
Is there a way that I can divide it in only 2 instead of 4??
Thanks
On 06/12/2015 at 02:11, xxxxxxxx wrote:
You mean asymmetrically? No, that is topologically not possible.
Imagine the three squares that meet in a corner of a cube. Try subdividing them in only one direction (in any possible combination - there are eight cases, many of which are topologically symmetric). You will see that it is not possible to cut them in a way that will not ALSO generate an additional point on an edge of a poly that has already been cut.
On 07/12/2015 at 01:53, xxxxxxxx wrote:
Hello,
the MCOMMAND_SUBDIVIDE simply applies a subdivision algorithm that creates for polygons for each given quad.
Best wishes,
Sebastian | 225 | 869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-40 | latest | en | 0.922038 |
https://lists.defectivebydesign.org/archive/html/help-octave/2020-03/msg00226.html | 1,686,247,106,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655092.36/warc/CC-MAIN-20230608172023-20230608202023-00101.warc.gz | 404,693,413 | 4,483 | help-octave
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## Re: Interpolation on scattered data
From: Nicklas Karlsson Subject: Re: Interpolation on scattered data Date: Mon, 23 Mar 2020 11:12:04 +0100
Then thinking again. Sometimes point will end up outside the area in between the three nearest neighbours. Or maybe more correctly expressed in mathematical terms, point sometimes end up outside the convex hull spent up by the three nearest neighbours.
Maybe the weighted estimator will work, have to think. Whole data set will most probably work well, there are only about 100 points and guess plot figure use more exuction time. Averaging over points piled up is the correct solution.
For some reason Octave delaunay(...) function return a screwed up mesh, some angles are very small while others are very wide. This might be because most of the points are rather squared with a small scew while in one corner points in the outer corner is squeezed to a line because of limited signal. Then mesh is screwed up interpolation does not work well.
Thanks, Nicklas Karlsson
Den mån 16 mars 2020 kl 10:02 skrev Augustin Lefèvre <address@hidden>:
You can use 3-nearest neighbour interpolation :
y=13j=13yjy = \frac{1}{3} \sum_{j=1}^3 y_j
where j is the index of the three nearest neighbours (based on xx values)
or a weighted estimator :
y=j=1nK(x,xj)yjj=1nK(x,xj)y = \frac{\sum_{j=1}^n K(x,x_j) y_j}{\sum_{j=1}^n K(x,x_j)}
(careful, means the -th point in your data, so x(i,:) or x(:,i) in your code)
If you're trying to use mesh techniques, I guess you will prefer this solution : it's computationally less demanding, and it makes the interpolant function smooth.
In this case, when the density is high, and points "pile up", then the estimated will be averaged over those points : it's up to you to decide whether this corresponds to your model or not.
In the formula above, you can restrict to to the 3 nearest neighbours, or use the whole data set if you can afford the computational load.
For the radial basis function, you can use for instance a gaussian kernel :
if is low, the interpolant will be "blurry", if is high it will be "spiky".
On 12/03/2020 15:53, Nicklas Karlsson wrote:
I have some problem with interpolation on scattered data. griddata(...) or interp2(...) functions do not work well. Delaunay function could triangulate data but maybe not well in all cases.
I have function to find nearest points. Interpolate between the three nearest point should if I think correct be equal to interpolation on delaunay triangulation but me a little bit stupid and can't immidiately figure out the equation. Anyone have it at hand?
I however found two other very interesting functions here http://fourier.eng.hmc.edu/e176/lectures/ch7/node7.html called "Radial Basis Function Method" and "Shepard method". I am however a little bit uncertain what happen then density vary for example then many points are tightly spaced as I expect them to pile up. Anyone used any of these?
Regards Nicklas Karlsson
Den mån 16 mars 2020 kl 10:02 skrev Augustin Lefèvre <address@hidden>:
You can use 3-nearest neighbour interpolation :
y=13j=13yjy = \frac{1}{3} \sum_{j=1}^3 y_j
where j is the index of the three nearest neighbours (based on xx values)
or a weighted estimator :
y=j=1nK(x,xj)yjj=1nK(x,xj)y = \frac{\sum_{j=1}^n K(x,x_j) y_j}{\sum_{j=1}^n K(x,x_j)}
(careful, means the -th point in your data, so x(i,:) or x(:,i) in your code)
If you're trying to use mesh techniques, I guess you will prefer this solution : it's computationally less demanding, and it makes the interpolant function smooth.
In this case, when the density is high, and points "pile up", then the estimated will be averaged over those points : it's up to you to decide whether this corresponds to your model or not.
In the formula above, you can restrict to to the 3 nearest neighbours, or use the whole data set if you can afford the computational load.
For the radial basis function, you can use for instance a gaussian kernel :
if is low, the interpolant will be "blurry", if is high it will be "spiky".
On 12/03/2020 15:53, Nicklas Karlsson wrote:
I have some problem with interpolation on scattered data. griddata(...) or interp2(...) functions do not work well. Delaunay function could triangulate data but maybe not well in all cases.
I have function to find nearest points. Interpolate between the three nearest point should if I think correct be equal to interpolation on delaunay triangulation but me a little bit stupid and can't immidiately figure out the equation. Anyone have it at hand?
I however found two other very interesting functions here http://fourier.eng.hmc.edu/e176/lectures/ch7/node7.html called "Radial Basis Function Method" and "Shepard method". I am however a little bit uncertain what happen then density vary for example then many points are tightly spaced as I expect them to pile up. Anyone used any of these?
Regards Nicklas Karlsson | 1,243 | 4,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-23 | latest | en | 0.874302 |
http://mathhelpforum.com/advanced-statistics/205317-mathematical-expectation-print.html | 1,501,225,923,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549448095.6/warc/CC-MAIN-20170728062501-20170728082501-00557.warc.gz | 204,374,508 | 2,891 | # Mathematical Expectation
Printable View
• Oct 14th 2012, 01:48 PM
Hal2001
Mathematical Expectation
Attachment 25227
I have the above conditions, the book wants me to find the value of c such that E(c^X)=1.
So I tried c^X*p+(1-p)c^X=1, which becomes,
c*p+(1-p)/c=1. From there I got the equation c^2p-c-p=-1. But I'm stuck, and I think may have went off track somewhere.
• Oct 14th 2012, 02:06 PM
Plato
Re: Mathematical Expectation
Quote:
Originally Posted by Hal2001
Attachment 25227
I have the above conditions, the book wants me to find the value of c such that E(c^X)=1.
So I tried c^X*p+(1-p)c^X=1, which becomes,
c*p+(1-p)/c=1. From there I got the equation c^2p-c-p=-1. But I'm stuck, and I think may have went off track somewhere.
No, you are correct. Just solve that quadratic equation for c.
Can you solve $px^2-x+1-p=0$
• Oct 14th 2012, 02:14 PM
Hal2001
Re: Mathematical Expectation
Oh silly me, I didn't think of considering 1-p and setting it equal to zero. Thanks. | 321 | 986 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-30 | longest | en | 0.939073 |
http://exercism.io/exercises/fsharp/rotational-cipher/readme | 1,513,245,488,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948543611.44/warc/CC-MAIN-20171214093947-20171214113947-00217.warc.gz | 95,351,775 | 3,801 | `1` ```exercism fetch fsharp rotational-cipher ```
Create an implementation of the rotational cipher, also sometimes called the Caesar cipher.
The Caesar cipher is a simple shift cipher that relies on transposing all the letters in the alphabet using an integer key between `0` and `26`. Using a key of `0` or `26` will always yield the same output due to modular arithmetic. The letter is shifted for as many values as the value of the key.
The general notation for rotational ciphers is `ROT + <key>`. The most commonly used rotational cipher is `ROT13`.
A `ROT13` on the Latin alphabet would be as follows:
```1 2``` ```Plain: abcdefghijklmnopqrstuvwxyz Cipher: nopqrstuvwxyzabcdefghijklm ```
It is stronger than the Atbash cipher because it has 27 possible keys, and 25 usable keys.
Ciphertext is written out in the same formatting as the input including spaces and punctuation.
## Examples
• ROT5 `omg` gives `trl`
• ROT0 `c` gives `c`
• ROT26 `Cool` gives `Cool`
• ROT13 `The quick brown fox jumps over the lazy dog.` gives `Gur dhvpx oebja sbk whzcf bire gur ynml qbt.`
• ROT13 `Gur dhvpx oebja sbk whzcf bire gur ynml qbt.` gives `The quick brown fox jumps over the lazy dog.` | 311 | 1,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-51 | longest | en | 0.805915 |
https://www.cnblogs.com/JoeFan/p/4322722.html | 1,723,331,413,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640826253.62/warc/CC-MAIN-20240810221853-20240811011853-00243.warc.gz | 570,941,346 | 8,324 | [BZOJ 1098] [POI2007] 办公楼biu 【链表优化BFS】
代码
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
const int MaxN = 100000 + 5, MaxM = 2000000 + 5;
{
char c; c = getchar();
while (c < '0' || c > '9') c = getchar();
Num = c - '0'; c = getchar();
while (c >= '0' && c <= '9')
{
Num = Num * 10 + c - '0';
c = getchar();
}
}
int n, m, Top, R1, R2, Sum;
int Ans[MaxN], Last[MaxN], Next[MaxN];
bool InList[MaxN];
struct Edge
{
int v;
Edge *Next;
} E[MaxM * 2], *P = E, *Point[MaxN];
inline void AddEdge(int x, int y)
{
++P; P -> v = y;
P -> Next = Point[x]; Point[x] = P;
}
queue<int> Q;
inline void Add(int x, int y)
{
Next[y] = Next[x];
if (Next[x]) Last[Next[x]] = y;
Next[x] = y;
Last[y] = x;
}
inline void Delete(int x)
{
if (Last[x]) Next[Last[x]] = Next[x];
if (Next[x]) Last[Next[x]] = Last[x];
}
void BFS()
{
while (!Q.empty()) Q.pop();
Q.push(Next[R1]);
InList[Next[R1]] = false;
Delete(Next[R1]);
int x, y;
++Top;
while (!Q.empty())
{
x = Q.front();
++Sum;
++Ans[Top];
Q.pop();
R2 = n + 2;
Last[R2] = Next[R2] = 0;
for (Edge *j = Point[x]; j; j = j -> Next)
{
y = j -> v;
if (!InList[y]) continue;
Delete(y);
}
for (int i = Next[R1]; i; i = Next[i])
{
Q.push(i);
InList[i] = false;
}
Next[R1] = Next[R2];
Last[Next[R1]] = R1;
}
}
int main()
{
scanf("%d%d", &n, &m);
int a, b;
for (int i = 1; i <= m; ++i)
{
}
Top = 0; Sum = 0;
R1 = n + 1;
for (int i = 1; i <= n; ++i) Add(R1, i);
for (int i = 1; i <= n; ++i) InList[i] = true;
while (Sum < n) BFS();
printf("%d\n", Top);
sort(Ans + 1, Ans + Top + 1);
for (int i = 1; i <= Top; ++i) printf("%d ", Ans[i]);
return 0;
}
posted @ 2015-03-09 09:07 JoeFan 阅读(486) 评论(0编辑 收藏 举报 | 702 | 1,741 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-33 | latest | en | 0.249264 |
http://www.origami-instructions.com/origami-magic-rose-cube.html | 1,656,277,496,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103271864.14/warc/CC-MAIN-20220626192142-20220626222142-00085.warc.gz | 92,605,518 | 7,686 | Origami-Instructions.com spreading joy one fold at a time
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#### Origami Paper
We use standard size 6 inch x 6 inch (15cm x 15cm) square origami paper for this site unless stated otherwise. If you can, use different types of origami paper to change the look of the finished origami and have fun with it!
# Origami Magic Rose Cube Instructions
First we have to thank one of our readers, Imre of Netherlands, for alerting us to this origami. Imre had written to us saying he saw this origami online but couldn't fold it and wanted us to make instructions for it.
It has taken us a while to get around to it but this is a brilliant origami design by Valerie Vann. It transforms from a cube to rose within seconds! It is really a cool design. We love it and we're sure you're going to as well!
Thanks, Imre, for writing to us and we hope you find these instructions helpful!
Made this origami? Comment and Submit your photo using the comment box at the end of this page!
Origami Magic Rose Cube Step 1: We'll start off by making the units for the "petals". We'll need 3 units of these. These will form the rose so you should use a color you want for your flower.
Start with a 6 inch x 6 inch (15cm x 15cm) square origami paper color side down.
Fold paper in half on the vertical axis. Crease well and unfold.
Origami Magic Rose Cube Step 2: Fold both sides to meet in the center. Crease well and unfold.
Origami Magic Rose Cube Step 3: Fold AB to meet AC. Crease well and unfold.
Origami Magic Rose Cube Step 4: Now fold DE to meet DF. Crease well.
Origami Magic Rose Cube Step 5: This is an interesting step that we'll repeat several times.
Make a horizontal fold on the top layer as shown.
Origami Magic Rose Cube Step 6: Now fold the left side back to the right.
Then press paper flat so that the diagonal edges line up on the left.
Origami Magic Rose Cube Step 7: Rotate paper.
Origami Magic Rose Cube Step 8: Now we're going to repeat Steps 5 and 6.
First pry open the top layer on the right. Then make the diagonal fold as shown.
Origami Magic Rose Cube Step 9: Now make the horizontal fold as shown below, folding down the top layer only.
Origami Magic Rose Cube Step 10: Now bring the left side over to the right to close the paper.
Press paper flat so the the diagonal edges line up on the left.
Origami Magic Rose Cube Step 11: Flip paper over. Make the 2 diagonal folds as shown. Crease well.
Origami Magic Rose Cube Step 12: Now make 2 more diagonal folds. Crease well.
Origami Magic Rose Cube Step 13: Unfold paper to Step 10.
Origami Magic Rose Cube Step 14: Flip paper over and your petals unit is complete. Repeat Steps 1 to 13 and make a total of 3 units.
Origami Magic Rose Cube Step 15: We'll now move on to the leaves. We'll be using green paper for these units.
Follow Steps 1 to 6 for the petals as shown above. You should have the following to start with.
Origami Magic Rose Cube Step 16: Make the small diagonal fold on the bottom right as shown.
Origami Magic Rose Cube Step 17: Fold up the bottom tip.
Origami Magic Rose Cube Step 18: Make the diagonal fold. Crease well and unfold.
Origami Magic Rose Cube Step 19: Rotate paper.
Origami Magic Rose Cube Step 20: You should be familiar with the next few steps as they're the same as Step 8 to 10 for the petals.
Pry open the top layer on the right side. Then make the diagonal fold as shown.
Origami Magic Rose Cube Step 21: Make the horizontal fold.
Origami Magic Rose Cube Step 22: Now we're going to bring the left side over to the right to close the paper.
Origami Magic Rose Cube Step 23: Flip paper over. Make the small diagonal fold on the bottom right.
Origami Magic Rose Cube Step 24: One more diagonal fold. Crease well and unfold.
Origami Magic Rose Cube Step 25: Unfold to Step 23.
Origami Magic Rose Cube Step 26: Flip paper over and your leaves unit is complete.
Repeat Steps 15 to 26 and make a total of 3 units.
Origami Magic Rose Cube Step 27: Watch the video for the assembly process.
Origami Magic Rose Cube Assembly Video
Your origami cube will go from this....
...to this like magic!!
Did you make this origami? If so, upload your photo (2MB limit) via the comment box below. You can login with your Facebook, Twitter, Google or Yahoo accounts. | 1,077 | 4,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-27 | latest | en | 0.894223 |
https://mersenneforum.org/showthread.php?s=3a90f619f89b9baa3e60cbcbc61793ee&t=10737 | 1,675,819,204,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500664.85/warc/CC-MAIN-20230207233330-20230208023330-00274.warc.gz | 405,519,893 | 14,313 | mersenneforum.org > Math PRIMALITY PROOF for Wagstaff numbers!
Register FAQ Search Today's Posts Mark Forums Read
2008-10-05, 06:33 #1
AntonVrba
Jun 2005
2×72 Posts
PRIMALITY PROOF for Wagstaff numbers!
At last I have found the proof! Needed some lateral thinking.
This is as per my conjecture 4 in a paralel thread but now hopefully we can upgrade to theorem once checked. I am inviting comment to the attached proof.
$\blue \text{Let } q \text{ be a prime integer } > \ 3 \text{ , } W_q = \frac{1}{3}(2^q+1) \text{.}$
$\blue W_q \text{ is prime } \Longleftrightarrow \ S_{q} \equiv S_2 \ \pmod{W_q} \text{ , where: } S_0=6 \text{ , and } \ S_{i+1}=S_i^2-2 \ \pmod{W_q} \ .$
best regards
Anton Vrba
Attached Files
Waggstaff0.0.pdf (90.3 KB, 572 views)
2008-10-05, 08:02 #2 AntonVrba Jun 2005 2×72 Posts I noticed a small typo, page 3 comment 2 (a+i)^(2k+1)=(a-i) (mod 2k+1) and not (mod (a+i)) as in the paper. so please correct.
2008-10-05, 08:31 #3 Andi47 Oct 2004 Austria 2×17×73 Posts Another typo, page 3, comment 1: "The reasoning presented may not be knew and already documented somewhere and references are kindly requested." I guess this should read: "...may not be new..." Edit: Your link to conjecture 4 does not work - there is one "http" too much. Last fiddled with by Andi47 on 2008-10-05 at 08:35
2008-10-05, 11:53 #5
AntonVrba
Jun 2005
2·72 Posts
AND, after a rest, I noticed a wild and false conclusion regarding the order of the group, this has been corrected as well as the equality (19) regarding the group order.
I also added the alternate cycle S_0 = -3/2
I started of with q but then changewd to p so I must have skipped some. Editing in LaTex is not my strength and a complete new experience.
Attached Files
Waggstaff0.2.pdf (97.3 KB, 332 views)
Last fiddled with by AntonVrba on 2008-10-05 at 12:07
2008-10-05, 12:02 #6
R. Gerbicz
"Robert Gerbicz"
Oct 2005
Hungary
161510 Posts
Quote:
Originally Posted by AntonVrba At last I have found the proof! Needed some lateral thinking. This is as per my conjecture 4 in a paralel thread but now hopefully we can upgrade to theorem once checked. I am inviting comment to the attached proof.
Your proof is wrong. You can not do that step from (15) to (16), because if
a*c==b*c mod d, then you can't say that a==b mod d. You can do it only in that case, if you could prove that gcd(c,d)=1 is true in your ring (so if gcd(delta,W(p))=1 is true).
2008-10-05, 12:07 #7
TimSorbet
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
427910 Posts
Quote:
Originally Posted by AntonVrba $\blue W_q \text{ is prime } \Longleftrightarrow \ S_{q} \equiv S_2 \ \pmod{W_q} \text{ , where: } S_0=6 \text{ , and } \ S_{i+1}=S_i^2-2 \ \pmod{W_q} \ .$
Is the $\pmod{W_q}$ in $\blue \ S_{q} \equiv S_2 \ \pmod{W_q}$ necessary? It seems that since both $S_{q}$ and $S_2$ are already $\pmod{W_q}$ it is completely unnecessary.
2008-10-05, 13:00 #8
AntonVrba
Jun 2005
2×72 Posts
Quote:
Originally Posted by T.Rex Some preliminary comments: - About (9) { S(p-1)=-S(1) (Mod Wp) } , why ? It has been a question for the LLT when the "Penultimate Lucas-Lehmer step" is - or +. No proof. But I remember that I read that, when using LLT in another case, it is always + (can't remember more details...). What is your proof ?
just remember s^2-2 = (-s)^2-2
Quote:
Originally Posted by T.Rex - I do not see in the theorem the extra conditions saying that S2 must be reached at step p and never before step p. Is it no more useful ? With this p-2 iteration, is it impossible for a composite to reach S2 at step i
No, comment two says that S_0 and S_1 will never repeat in the iteration. Further, I believe that S_0=6,-6 and S_0=-3/2 are the only three numbers to guarantee cycles of p-2 and p-1 respectively and that all elemets of S_0 to S_(p-1) are unique for all prime W_p. I do not know what to call these numbers any suggestion.
Quote:
Originally Posted by T.Rex - I'm surprised to see that there is a (p-2) cycle... but that works !
You have not studied the digraphs as you claim you have.
Quote:
Originally Posted by T.Rex - what's inefficient in your test for Wagstaff numbers is your use of (mod Wp). There are technics for that (see LLR test) but they have a cost at each iteration, I think. Using S^2-2 (mod Np) with N=1+2^p enables to use faster FFT with modulo, I think, meaning the Prime95 code can be used with nearly no change. Hey ! you other people experts in this (Crandall, Woltman, Penné, & others), do you agree ?
I am sure that this is correct but I cannot prove it for now - somebody else can. The proof of sufficiency is not valid for 3W_p but the proof of necessity would be valid.
Quote:
Originally Posted by T.Rex - and you say not a word about my idea to use cycles of a Digraph, nor you talk about Shallit&Vasiga work (they gave proofs about the cycles).
Please note I have not named the test but I am presenting the proof. I have not made use of any work of Shallit&Vasiga and thus no need to mention. We are not proofing the form of the diagraph, Shallit&Vasiga have done that, but the proof is making a statement regarding the possible divisors of W_p.
The Lucas-Lehmer test proves that 2^p-1 has no non-trivial factors smaller than 2^(p-1). A consequence of this is that the number is prime.
What this test does assert, is that there are no possible divisors smaller than sqrt(W_p), and as a consequence, W_p is prime - and thus it is very similar to the Lucas-Lehmer test.
Quote:
Originally Posted by T.Rex Again, congratulations !! Tony
Thanks we now need to wait for the experts.
Last fiddled with by AntonVrba on 2008-10-05 at 13:22 Reason: in last sentance change no to now
2008-10-05, 13:03 #9
AntonVrba
Jun 2005
9810 Posts
Quote:
Originally Posted by Mini-Geek Is the $\pmod{W_q}$ in $\blue \ S_{q} \equiv S_2 \ \pmod{W_q}$ necessary? It seems that since both $S_{q}$ and $S_2$ are already $\pmod{W_q}$ it is completely unnecessary.
Thanks for the correction
2008-10-05, 13:18 #10
AntonVrba
Jun 2005
2·72 Posts
Quote:
Originally Posted by R. Gerbicz Your proof is wrong. You can not do that step from (15) to (16), because if a*c==b*c mod d, then you can't say that a==b mod d. You can do it only in that case, if you could prove that gcd(c,d)=1 is true in your ring (so if gcd(delta,W(p))=1 is true).
My old number theory book from 1962 says:
(1) x = y(mod m)
and say that x is congruent to y modulo m. The quantity m is called the
modulus, and all numbers congruent (or equivalent) to x (mod m) are said
to constitute a congruence (or equivalence) class. Congruence classes are
preserved under the rational integral operations, addition, subtraction, and
multiplication.
Mathematica calculates that Mod[7,3x43]=7 and Mod[7*3,3*43]=21 so is it wrong? I think you can withdraw your statement.
2008-10-05, 13:23 #11 R. Gerbicz "Robert Gerbicz" Oct 2005 Hungary 5·17·19 Posts For example: 14==21 mod 7 is true, but after dividing 2==3 mod 7 isn't true. And you've done this. Last fiddled with by R. Gerbicz on 2008-10-05 at 13:23
Similar Threads Thread Thread Starter Forum Replies Last Post Godzilla Miscellaneous Math 40 2018-10-17 00:11 f1pokerspeed FactorDB 14 2014-01-09 21:06 Tony Reix Wagstaff PRP Search 7 2013-10-10 01:23 princeps Math 15 2012-04-02 21:49 ixfd64 Math 12 2010-01-05 16:36
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# The numbers \$1,\$ \$2,\$ \$\dots,\$ \$10\$ are to be entered into the 10 boxes shown below, so that each number is used exactly once: \[P = (\squar
0
33
1
The numbers \$1,\$ \$2,\$ \$\dots,\$ \$10\$ are to be entered into the 10 boxes shown below, so that each number is used exactly once: \[P = (\square + \square + \square + \square + \square)(\square + \square + \square + \square + \square).\]What is the maximum value of \$P\$? What is the minimum value of \$P\$?
May 4, 2020 | 159 | 490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2020-24 | latest | en | 0.716491 |
https://www.techylib.com/el/view/coalitionhihat/dc_series_and_parallel_experiment_worksheet._glacier_high | 1,529,936,706,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867885.75/warc/CC-MAIN-20180625131117-20180625151117-00593.warc.gz | 920,186,997 | 12,357 | # DC - Series and Parallel experiment worksheet. - Glacier High ...
Ηλεκτρονική - Συσκευές
7 Οκτ 2013 (πριν από 4 χρόνια και 9 μήνες)
135 εμφανίσεις
Electricity and Circuits
Learning Goals:
Students will understand basic properties of
circuits
.
Students will understand what is required to complete a circuit.
Students will be able to build series and parallel circuits and understand the
difference bet
ween the different circuits.
Students will learn about conductors and insulators.
CAP
1.1
(RED
)
Ask questions and state hypotheses that lead to different types of scientific investigations (for
example: experimentation, collecting specimens, constructing
models, researching scientific
literature)
2.8 (BLUE)
There are different forms of energy and those forms of energy can be transferred and stored (for
example: kinetic, potential) but total energy is conserved
Open the phET simulator by…
Open the DC on
ly circuit simulator
at
Find a way to
make a single light bulb light up with as FEW parts hooked up as possible
.
When electricity flows through wires and makes something work, like a light bulb, it is
called a circuit.
1.
Sketch your circuit below:
2.
What seems to be making the light bulb turn on
in your circuit? (what do you
think
electricity
is
based on the simulator?)
Make a gap in your circuit.
Go to the grab bag and play with the different objects. Find out which objects allow
elect
ricity to flow and fill in the data table:
Objects that allow electricity to flow
(conductors)
Objects that do NOT allow electricity to
flow (insulators)
3.
What do the conductors have in common?
4.
What do the insulators have in common?
For the next
few activities, you need to light up more than 1 bulb at the same time,
using
just
one battery
.
First
circuit:
find a way to hook up your bulbs in a way that if you break the connection at
one bulb, ALL bulbs go out.
5.
Sketch your new circuit:
6.
Why did
the rest of the bulbs go out if you break the connection at one bulb?
7.
This circuit is called a
series circuit
because the bulbs are hooked up in one long
“series” or line. Name somewhere you have seen a string of lights that are also a
series circuit.
Second circuit: find a way to hook up your bulbs in a way that if you break the
co
nnection at one bulb, ONLY that
bulb goes out.
8.
Sketch this circuit:
9.
Why do the rest of the bulbs stay lit if you break the connection at one bulb?
10.
This circuit is call
ed a
parallel circuit,
which has 2 or more single loops
connected to the same battery. When 1 bulb goes out in these circuits, the rest of
the lights stay on! Name somewher
e you have seen many bulbs hooked up to one
power source, where one bulb can go ou
t without affecting the others.
11.
You design toys for a toy company. Your boss wants you to hook up the lights in
the toy car you are working on in the cheapest way possibl
e, without
consideration of the quality of the toy. Which circuit should you use
if you want
to save money by using fewer parts? Why would this circuit be cheaper?
12.
You are an electrician working on a house. What type of circuit should you use
for the house so that the owners don’t call to complain about their wiring? Why
use this
circuit?
Experiment with the simulator, see what you can make it do!!!
13.
What did you do to make light bulbs glow brighter
?
14.
What did you do to make light bulbs glow dimmer
?
15.
How can you c
ause a fire
? (In the simulator… NOT in the real world!)
16.
Can you c
atch
the
puppy on fire
? | 858 | 3,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2018-26 | latest | en | 0.891363 |
https://isiarticles.com/article/21689 | 1,713,244,970,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817043.36/warc/CC-MAIN-20240416031446-20240416061446-00803.warc.gz | 291,986,805 | 7,901 | دانلود مقاله ISI انگلیسی شماره 21689
ترجمه فارسی عنوان مقاله
# استفاده از تئوری لحظه هندسی مربوط به مدیریت پرتفولیو بهینه
عنوان انگلیسی
Applications of geometric moment theory related to optimal portfolio management
کد مقاله سال انتشار تعداد صفحات مقاله انگلیسی
21689 2006 26 صفحه PDF
منبع
Publisher : Elsevier - Science Direct (الزویر - ساینس دایرکت)
Journal : Computers & Mathematics with Applications, Volume 51, Issues 9–10, May 2006, Pages 1405–1430
ترجمه کلمات کلیدی
لحظه ها و تئوری لحظه هندسی - مدیریت پرتفولیو بهینه - مساله مرز بهینه - تقعر
کلمات کلیدی انگلیسی
Moments and geometric moment theory, Optimal portfolio management, Problem of optimal frontier, Concavity
پیش نمایش مقاله
#### چکیده انگلیسی
In this article, we start with the brief description of the essence of geometric moment theory method for optimization of integrals due to Kemperman [1–3]. Then, we solve several new Moment problems with applications to stock market and financial mathematics. That is, we give methods for optimal allocation of funds over stocks and bonds at maximum return. More precisely, we present here the optimal portfolio management under optimal selection of securities so to maximize profit. The above are done within the models of optimal frontier and optimizing concavity.
#### مقدمه انگلیسی
Tile main problem we solve here is the optimal allocation of funds over stocks and bonds and at the same time, given certain level of expectation, best choice of securities on the purpose to maximize return. The results are very general so that they stand by themselves as "formulas" to treat other similar stochastic situations and structures far away from the stock market and financial mathematics. The answers to the above described problem are given under two models of investing, the optimal frontier and optimizing concavity, as being the most natural. There are given many examples all motivated from financial mathematics and of course fitting and working well there. The method of proof derives from the geometric moment theory of Kemperman, see [1-3], and several new moment results of very general nature are presented here. We start the article with basic geometric moment review and we show the proving tool we use next repeatedly. To the best of our knowledge this paper is totally new in l i t e r a t u r e as a whole and nothing similar or prior to it in any form exists there. We hope it is well received by the community of mathematical-economists and that can be useful there, by giving some definite real answers to existing questions in optimal portfolio theory. The continuation of this work will be one to derive algorithms out of this theoretical work and create computer software of implementation and work with actual numerical data of the stock market. | 592 | 2,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-18 | latest | en | 0.54518 |
https://calculator-online.org/equation/e/tg_x_ctg_x_minus_co_sinus_e_x_equal_two_sinus_2x_squared | 1,726,716,041,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00034.warc.gz | 132,145,877 | 124,790 | Mister Exam
tg(x)(ctg(x)-cos(x))=2sin(2x)^2 equation
The teacher will be very surprised to see your correct solution 😉
v
Numerical solution:
Do search numerical solution at [, ]
The solution
You have entered [src]
2
tan(x)*(cot(x) - cos(x)) = 2*sin (2*x)
$$\left(- \cos{\left(x \right)} + \cot{\left(x \right)}\right) \tan{\left(x \right)} = 2 \sin^{2}{\left(2 x \right)}$$
The graph
Sum and product of roots [src]
sum
0
$$0$$
=
0
$$0$$
product
1
$$1$$
=
1
$$1$$
1
x1 = 75.712382951514
x2 = -13.0899693899575
x3 = 68.1725605828985
x4 = 1.57079642323511
x5 = -83.8805238508475
x6 = -31.9395253114962
x7 = -34.0339204138894
x8 = 4.08407044966673
x9 = -61.2610564700665
x10 = -46.6002910282486
x11 = 34.2433599241287
x12 = 84.5088423815654
x13 = 45.553093441597
x14 = -5.96902604182061
x15 = -97.7035315266426
x16 = -54.9778711811089
x17 = 70.6858347066598
x18 = 41.7831822927443
x19 = -100.216805649514
x20 = 60.2138591938044
x21 = -60.0044196835651
x22 = 44.2964564156161
x23 = -73.8274272809227
x24 = -9.73893722612836
x25 = 26.7035376165518
x26 = -86.3937979007636
x27 = 64.402649315709
x28 = 53.9306738866248
x29 = 5.34070751110265
x30 = -63.7743308678728
x31 = 95.8185760408313
x32 = 16.2315620435473
x33 = 78.2256570743859
x34 = -53.7212343763855
x35 = -87.6504350351552
x36 = 92.0486647501809
x37 = -82.2050077689329
x38 = -93.9336203423348
x39 = -13.5088484104361
x40 = 81.9955682586936
x41 = -33.6150413934108
x42 = 62.3082542961976
x43 = -67.5442421525188
x44 = 38.0132711084365
x45 = -29.8451301004081
x46 = 1.57079634265433
x47 = 40.5265452313083
x48 = 89.5353905135251
x49 = 24.1902634326414
x50 = 48.0663675999238
x51 = -39.8982267005904
x52 = 14.1371670741504
x53 = -46.18141200777
x54 = -80.1106125860317
x55 = 74.4557458900781
x56 = 30.473448739821
x57 = -75.9218224617533
x58 = 182.526533173567
x59 = 22.9336263712055
x60 = -2.19911485751286
x61 = -70.0575161750524
x62 = 7.85398172911943
x63 = -12.2522113490002
x64 = 20.4203521589613
x65 = -214.151899219704
x66 = -36.128315428008
x67 = -90.1637091580271
x68 = -17.2787594419908
x69 = -27.7507351067098
x70 = 51.8362788853931
x71 = -38.2227106186758
x72 = -42.4115007735831
x73 = 100.007366139275
x74 = -56.2345084992573
x75 = -47.4380490692059
x76 = -49.9513231920777
x77 = -78.0162175641465
x78 = 49.3230046613598
x79 = -43.6681378848981
x80 = 12.0427718387609
x81 = 58.1194638203606
x82 = -6.80678408277789
x83 = 27.9601746169492
x84 = 85.7654794430014
x85 = -57.4911455606932
x86 = -71.733032256967
x87 = 66.9159235214626
x88 = 9.94837673636768
x89 = 88.2787535658732
x90 = -16.0221225333079
x91 = 564.544199850086
x92 = 31.7300858012569
x93 = 71.9424717672063
x94 = -23.5619449971286
x95 = -19.7920337176157
x96 = 56.025068989018
x96 = 56.025068989018 | 1,323 | 2,763 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.523284 |
http://personalpages.manchester.ac.uk/staff/stefan.guettel/py/01c-python_intro.ipynb | 1,521,365,253,000,000,000 | text/plain | crawl-data/CC-MAIN-2018-13/segments/1521257645604.18/warc/CC-MAIN-20180318091225-20180318111225-00541.warc.gz | 234,531,407 | 25,594 | { "cells": [ { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "# Programming with Python\n", "#### Stefan Güttel, [guettel.com](http://guettel.com)\n", "\n", "\n", "\n", "**Acknowledgement:** These course notes are based on a previous set of notes written by Vedran Šego ([vsego.org](http://vsego.org/)) in 2015.\n", "\n", "\n", "## Contents:\n", "\n", "1. What are algorithms and what are programs?\n", "\n", "2. Basic input and output\n", "\n", "3. Variables, types, and operators\n", "\n", "4. Real number trouble" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## An algorithm\n", "\n", "* _A process or set of rules to be followed in calculations or other problem-solving operations, especially by a computer_
\n", " http://www.oxforddictionaries.com/definition/english/algorithm?q=algorithm\n", "* Informal definition from Wikipedia: *a set of rules that precisely defines a sequence of operations*.\n", "* More precisely (albeit still a bit informal):
\n", " *A **sequence** of actions that are always executed in a **finite** number of steps, used to solve a certain problem.*\n", "* Simplified: a cookbook, often very general\n", "\n", "Important parts of an algorithm:\n", "\n", "1. get the input data,\n", "2. solve the problem,\n", "3. output the result(s)." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "**Example:** preparing a frozen pizza. Steps:\n", " 1. Get temperature *temp* and time *t* (written on the box).\n", " 2. Heat oven to the temperature *temp*.\n", " 3. Discard all packaging (recycle the box).\n", " 4. Cook for a time *t*.\n", " 5. Get pizza out of the oven.
\n", " Use some heat-resistent hands protection or execute the \"go to the ER\" subroutine.\n", "\n", "Example of the input data:\n", "* *temp* = \"conventional oven: 190C; fan oven: 180C\",\n", "* *T* = between 15 and 17 minutes." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## A program\n", "\n", "* _A series of coded software instructions to control the operation of a computer or other machine._
\n", " http://www.oxforddictionaries.com/definition/english/programme\n", "* Simplified: precise instructions written in some programming language" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "## Our focus\n", "\n", "* **Algorithms,** not programs!\n", "* We shall use programs to test algorithms and to see how they work.\n", "* **Aim:** Learn to solve problems using a computer, **regardless of the choice of language** (which may or may not be Python in the courses you take in your future work or studies)." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Types of programming languages (implementations)\n", "\n", "### Compiled\n", "* Require compilation (translation) of the code to the machine code (popular, albeit a bit meaningless, \"zeroes and ones\"),\n", "* Usually strictly typed,\n", "* Usually faster than interpreters.\n", "\n", "### Interpreted\n", "* Translated during the execution,\n", "* Usually untyped or loosly typed,\n", "* Usually slow(ish)." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "## Where is Python?\n", "\n", "* Simplified: an interpreter.\n", "* Programs get semi-translated when run (_pseudocompiler_).\n", "* Untyped, slower than compiled languages (but with various ways to speed up).\n", "\n", "**No details** – too technical and they depend on the specific Python implementation." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "## Python versions\n", "\n", "Many implementations, but two major versions\n", " * Python 2\n", " * Python 3 \n", " \n", "Minor differences in what we need (these will be mentioned)." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## An example: Hello world" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "slideshow": { "slide_type": "-" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Hello, World!\n" ] } ], "source": [ "print(\"Hello, World!\")" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "The output can be nicely formated, but more on that in the near future." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "### What is \"Hello World\"?\n", "\n", "A very simple program, used to show the basic syntax of a programming language.\n", "\n", "See the [The Hello World Collection](http://www.roesler-ac.de/wolfram/hello.htm) for the examples in many other programming languages.\n", "\n", "The need for such an example can be clearly seen from the examples of more complex, but still fairly readable languages (for example, various versions of C++ and Java).
\n", "Beware of the Assembler-Z80-Console and BIT examples. " ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "**Note:** A full Python program should always start with the following:\n", "\n", "* the first line: #!/usr/bin/env python3 \n", " This has no meaning on Windows, but it makes your programs easier to run on Linux and Mac OSX.\n", "* Right after that, a description what the program does, possibly with some other info (system requirements, authors name and contact, etc.), between the triple quotation marks. This is called a *docstring* and it will be further addressed next week.\n", "\n", "So, a full \"Hello World\" program would look like this:" ] }, { "cell_type": "code", "execution_count": 9, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Hello, World!\n" ] } ], "source": [ "#!/usr/bin/env python3\n", "\n", "\"\"\"\n", "A program that prints the \"Hello, World!\" message.\n", "\"\"\"\n", "\n", "print(\"Hello, World!\")" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We typically omit these elements in the lectures and we will mostly present chunks of code (that are not necessarily full programs) to save some space." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "The following line that Spyder adds to new files\n", "python\n", "# -*- coding: utf-8 -*-\n", "\n", "is not necessary in Python 3. However, if you are using international characters in some encoding other than UTF-8 (which you really shouldn't do!), this is a way to specify that encoding.\n", "\n", "In this course we shall not cover encodings, as it is a very technical subject and most of the time it is enough to just use UTF-8 (find it in your editor's settings). However, if you're ever to build an application with the need for international characters, do look up the encodings on the internet (the [Wikipedia page](http://en.wikipedia.org/wiki/Character_encoding) is a good start) and use [UTF-8](http://en.wikipedia.org/wiki/UTF-8) whenever possible, as it is a widely accepted standard and the default in Python 3. You should also make sure that your editor saves files using the UTF-8 encoding (Spyder does that by default).\n", "\n", "In Python 2, the default encoding is [ASCII](http://en.wikipedia.org/wiki/ASCII) and the UTF-8 support has to be enabled manually." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Comments\n", "\n", "It is often useful to add human language notes in the code. These are called *comments* and are ignored by a computer, but they help programmers read the code.\n", "\n", "In Python, comments are made by prepending the hash sign # in front of it. Each comments ends with the end of the line.\n", "\n", "It is a standard to **always write comments in English:**\n", "> Python coders from non-English speaking countries: please write your comments in English, unless you are 120% sure that the code will never be read by people who don't speak your language.\n", "> Source: [PEP 8](https://www.python.org/dev/peps/pep-0008/#comments)\n", "\n", "For example:" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "#!/usr/bin/env python3\n", "\n", "\"\"\"\n", "A program that prints the \"Hello, World!\" message.\n", "\"\"\"\n", "\n", "# A welcome message\n", "print(\"Hello, World!\")\n", "# TODO: ask user their name and save it as a file" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "As you can see, the above code runs just as the previous one, but a programmer reading it can get more information about the code itself.\n", "\n", "Some editors will recognize certain *tags* in comments and highlight them to make them more noticable (like the TODO tag in the previous example). Some of the more common ones are, as listed in the [Wikipedia's comments article](http://en.wikipedia.org/wiki/Comment_%28computer_programming%29#Tags):\n", "\n", "* FIXME to mark potential problematic code that requires special attention and/or review.\n", "\n", "* NOTE to document inner workings of code and indicate potential pitfalls.\n", "\n", "* TODO to indicate planned enhancements.\n", "\n", "* XXX to warn other programmers of problematic or misguiding code." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "We shall often use the comments to denote what certain parts of the code do. These should always be descriptive and not merely rewritten code.\n", "\n", "For example, this is good:\n", "python\n", "# Get the sum of primes in L as prime_sum\n", "for x in L:\n", " if is_prime(x):\n", " prime_sum += x\n", "\n", "as it makes clear what the code is doing, even to someone who doesn't \"speak\" Python.\n", "\n", "This is bad:\n", "python\n", "# For each element x in the list, if x is a prime number,\n", "# add it to prime_sum.\n", "for x in L:\n", " if is_prime(x):\n", " prime_sum += x\n", "\n", "because this comment is just a rewrite of the code that follows it and, as such, it is useless." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "It is advisable to keep all lines (comments and docstrings) wrapped under 80 characters, although it shouldn't be forced when it reduces the code readability." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Input\n", "\n", "How do we **input** some data?\n", "\n", "Not surprisingly, using the function input()." ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "slideshow": { "slide_type": "-" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "17\n", "The value of x is 17\n" ] } ], "source": [ "x = input()\n", "print(\"The value of x is\", x)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "Well, this x looks kind of important here. What could it be?" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Variables\n", "\n", "* Variables are used to store and retrieve data.\n", "* Every variable has a value (which, in Python, *can* be undefined) and a type (partly hidden in Python)." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "### Simple types\n", "\n", "* **Number** -- generally, what you'd consider an integer or a real number in mathematics.
\n", " For example: 17, -19, 17.19, 0, 2e3, ...\n", "* **String** -- a text.
\n", " For example: \"word\", \"This is a mighty deep, philosophical sentence.\", \"ŞƿҿÇïåĿ sɹǝʇɔɐɹɐɥɔ\", \"17\", ...
\n", " The so-called *empty string*, \"\", has no characters (its length is zero).\n", "* **Boolean** -- truth values (True and False).\n", "* **NoneType** -- the type of a special constant None that means \"no value\".
\n", " This is **not** a zero (a number), nor an empty string, nor anything else! None is different from any other constant and any other value that a variable can get.\n", "\n", "**Be careful:** 17 is a number, while \"17\" is a string!" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "### Some not-so-simple types\n", "\n", "* Lists and tuples -- most languages have only lists (usually called *arrays*)\n", "* Dictionaries\n", "* Sets\n", "* Objects\n", "* Functions (yes, functions can be saved in variables as well)
\n", " ...\n", "\n", "More on these in the near future." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## How do variables work?\n", "\n", "Let us analyze this piece of code:" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "slideshow": { "slide_type": "-" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "a mistery\n", "The value of x is a mistery\n" ] } ], "source": [ "x = input()\n", "print(\"The value of x is\", x)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "Whatever is on the right-hand side of the assignment = gets computed first. Then the result is assigned to the variable on the left-hand side. When this is done, the next line of code is executed." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "In our concrete example this means:\n", "\n", "1. The function input() reads a sequence of characters from the standard input (usually the user's keyboard) and returns it as a **string**.\n", "2. That value is then assigned to the variable x (on the left-hand side of the assignment operator =).
\n", " Now x holds - as a string - whatever we have typed up to the first newline, i.e., up to the first Enter key (the newline itself is not part of the string).\n", "3. The function print() now outputs its arguments to the standard output (usually the user's screen), in order in which they were given, separated by a single space character. So,\n", " * First, a string \"The value of x is\" is written out.\n", " * Then a singe space character is written out.\n", " * Then the value of x is written out (**not** the string \"x\" itself, because x is a variable!)." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "In other words, if we type \"17\", our program will write\n", "\n", " The value of x is 17\n", "\n", "And if we type \"a mistery\", our program will write\n", "\n", " The value of x is a mistery\n", "\n", "Don't expect Python to do anything *smart* here. It just writes out the values as they were given." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "**Python 2 remark:** In Python 2, print does not need parentheses (i.e., print \"Hello, World!\" is fine).
\n", "However, do include them even if writing a Python 2 program, to make it easier to port to Python 3, and to avoid problems in some more advanced uses you might encounter in the future." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## On the simple types\n", "\n", "**Be careful!** Even if a value *looks* like a number, it *might* **not be** one!\n", "\n", "Let us try to input two numbers in the following code, also adding the descriptions of what is expected in each of the inputs:" ] }, { "cell_type": "code", "execution_count": 13, "metadata": { "slideshow": { "slide_type": "-" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "x: 17\n", "y: 19\n", "17 + 19 = 1719\n" ] } ], "source": [ "x = input(\"x: \")\n", "y = input(\"y: \")\n", "print(x, \"+\", y, \"=\", x+y)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "#### What did just happen here?\n", "\n", "The user types two numbers, which are saved -- as two **strings** -- in variables x and y. Then the program writes out (among other things) the value of x+y.\n", "\n", "How would we \"add\" one string to another in the real world?\n", "\n", "For example, if x = \"Bruce\" and y = \"Wayne\", what would x + y be?" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "(It may come as a little surprise that \"x + y\" will **not** produce \"Batman\". Python is a well defined language that keeps Bruce's secret identity well hidden.)\n", "\n", "The result of x+y will be \"BruceWayne\". Notice that there is no additional space here: the strings are *glued* (concatenated) one to another, with no extra separators!\n", "\n", "So, what happens if x = \"17\" and y = \"19\"?" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "It would be very bad if Python looked at these and decided that they were numbers only because they have nothing but digits. Maybe we want them concatenated (as opposed to adding them one to another)!\n", "\n", "So, the result is -- by now no longer surprisingly -- \"1719\", because the strings' addition + is a concatenation, regardless of the value of the strings in question.\n" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "How do we tell Python to \"treat these two variables as numbers\"?" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Converting basic types\n", "\n", "We can explicitly tell Python to convert a string to an integer or a real number, and vice versa." ] }, { "cell_type": "code", "execution_count": 14, "metadata": { "slideshow": { "slide_type": "-" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "17\n", "1.23\n", "x = 17\n", "y = 1.23\n", "x+y = 18.23\n", "This is a string: \"18.23\"\n" ] } ], "source": [ "x = int(input())\n", "y = float(input())\n", "print(\"x = \", x)\n", "print(\"y = \", y)\n", "print(\"x+y = \", x+y)\n", "z = 'This is a string: \"' + str(x+y) + '\"'\n", "print(z)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "We see three conversion functions:\n", "* int(), which takes a string and converts it to an integer. If the argument is not a string representation of an integer, an error occurs.\n", "* float(), which takes a string and converts it to a \"real\" number (also called *floating point number*, hence the name of the function). If the argument is not a string representation of a real number, an error occurs.\n", "* str(), which takes a number (among other allowed types) and converts it to a string." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "**Python 2 remark:** In Python 2, input() is similar to float(input()) in Python 3 (actually, eval(input()), but that is well beyond this lecture). This means it loads a number and returns it as a floating point number (causing an error or a strange behaviour if anything else is given as the input). \n", "To load a string in Python 2, one has to call raw_input() (which does not exist in Python 3).\n", "\n", "**A note on the string creation:** There are better ways to form the variable z (using various *formatting* methods), but this will have to wait a few weeks until we cover strings in more depth than here." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## More on the assignments\n", "\n", "What will the following code print?" ] }, { "cell_type": "code", "execution_count": 15, "metadata": { "slideshow": { "slide_type": "-" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The value of x was 17\n", "The value of x is 19\n" ] } ], "source": [ "x = 17\n", "print(\"The value of x was\", x)\n", "x = x + 2\n", "print(\"The value of x is\", x)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "As we said before: whatever is on the right hand side of the assignment =, gets computed **first**. Only **after that**, the result is assigned to the variable on the left hand side.\n", "\n", "So, when Python encounters the command\n", "\n", "python\n", "x = x + 2\n", "\n", "\n", "while the value of x is 17, it first computes x + 2, which is 19. After that, it performs the assignment x = 19, so 19 becomes the **new value** of x (which is then displayed with the second print function)." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "In most of the modern languages, x = x + y can be written as x += y. The same shortcut works for other operators as well, i.e., x = x op y can be written as x op= y.\n", "\n", "For basic numerical operations, this means we have the following shortcuts:\n", "\n", "Expression | Shortcut\n", ":-----------:|:-------:\n", "x = x + y | x += y\n", "x = x - y | x -= y\n", "x = x * y | x *= y\n", "x = x / y | x /= y\n", "x = x // y | x //= y\n", "x = x % y | x %= y\n", "x = x ** y | x **= y\n", "\n", "**A note on other languages:** there are no increment (++) and decrement (--) operators in Python." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "## Some special operators\n", "\n", "Most of the operators in the above table have the same meaning as in mathematics (**for those knowing C:** / means the *usual*, i.e., real division). The three not used in mathematics are defined as follows:\n", "\n", "* x // y means floored quotient of x and y (also called *integer division*), i.e., x // y $:= \\left\\lfloor \\mathsf{x}\\ /\\ \\mathsf{y} \\right\\rfloor$,\n", "* x % y means the remainder of $x / y$, i.e., x % y := x - y * (x // y),\n", "* x ** y means $x^y$ (x to the power y)." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "**Python 2 remark:** In Python 2, the ordinary real division x/y works in a C-like manner, which means that x/y is equivalent to x//y if both x and y are integers.
\n", "In Python 3, x/y always means real division. In other words,\n", "\n", "* Python 2: 3//2 = 3/2 = 1, but 3/2.0 = 3.0 / 2 = 3.0 / 2.0 = 1.5;\n", "* Python 3: 3//2 = 1, but 3/2 = 3/2.0 = 3.0 / 2 = 3.0 / 2.0 = 1.5." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Real number trouble\n", "\n", "When dealing with real numbers, one must be extremely careful!\n", "\n", "### Simple arithmetics\n", "\n", "What happens when we try to compute a + b - a for several different real values of a and b? Fairly often, the result will **not** be b!" ] }, { "cell_type": "code", "execution_count": 16, "metadata": { "slideshow": { "slide_type": "-" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "a = 10 b = 0.1 -> a + b - a = 0.09999999999999964 != 0.1 = b\n", "a = 10000000 b = 1e-07 -> a + b - a = 1.0058283805847168e-07 != 1e-07 = b\n", "a = 100000000000 b = 1e-11 -> a + b - a = 0.0 != 1e-11 = b\n" ] } ], "source": [ "a = 10\n", "b = 0.1\n", "print(\"a =\", a, \" b =\", b, \" -> \", \"a + b - a =\", a + b - a, \"!=\", b, \"= b\")\n", "a = 10**7\n", "b = 10**(-7)\n", "print(\"a =\", a, \" b =\", b, \" -> \", \"a + b - a =\", a + b - a, \"!=\", b, \"= b\")\n", "a = 10**11\n", "b = 10**(-11)\n", "print(\"a =\", a, \" b =\", b, \" -> \", \"a + b - a =\", a + b - a, \"!=\", b, \"= b\")" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "### A division\n", "\n", "There is no such thing as a *real* number in a computer. All numbers are actually (something like) decimals with an upper limit on the number of correctly remembered digits. The rest of the digits is lost, which can produce weird results, like x * (1 / x) ≠ 1." ] }, { "cell_type": "code", "execution_count": 17, "metadata": { "slideshow": { "slide_type": "-" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0.9999999999999999\n" ] } ], "source": [ "x = 474953\n", "y = 1 / x\n", "print(x * y)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "### Use integers whenever possible\n", "\n", "Fibonacci numbers are defined as follows:\n", "$$F_0 := 0, \\quad F_1 := 1, \\quad F_{n+1} := F_n + F_{n-1}, \\quad n \\ge 1.$$\n", "There is also a direct formula for computing $F_n$:\n", "$$F_n = \\frac{\\varphi^n - \\psi^n}{\\sqrt{5}}, \\quad \\varphi := \\frac{1 + \\sqrt{5}}{2}, \\quad \\psi := \\frac{1 - \\sqrt{5}}{2}.$$\n", "Mathematically, both definitions are equivalent. On a computer, however, the second will soon give you wrong results.\n", "\n", "In the following code, fib1(n) returns the n-th Fibonacci number computed by a simple integer-arithmetics algorithm, while fib(2) uses the above formula (never use the recursive definition for computation of Fibonacci numbers!)." ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "slideshow": { "slide_type": "-" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Type n (try to go for 73 or more): 100\n", "|fib1(n) - fib2(n)| = |354224848179261915075 - 354224848179263111168| = 1196093\n" ] } ], "source": [ "def fib1(n):\n", " f0 = 0\n", " f1 = 1\n", " while n > 1:\n", " (f0, f1) = (f1, f0 + f1)\n", " n -= 1\n", " return f1\n", "\n", "def fib2(n):\n", " sqrt5 = 5 ** .5\n", " phi = (1 + sqrt5) / 2\n", " psi = (1 - sqrt5) / 2\n", " return int((phi**n - psi**n) / sqrt5)\n", "\n", "n = int(input(\"Type n (try to go for 73 or more): \"))\n", "fib1n = fib1(n)\n", "fib2n = fib2(n)\n", "print(\"|fib1(n) - fib2(n)| = |\" + str(fib1n), \"-\", str(fib2n) + \"| =\", abs(fib1n - fib2n))" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "### Even a simple addition incurs errors\n", "\n", "The following code computes and prints three sums:\n", "$$\\sum_{i = 0}^{999} 0.1 = 100, \\quad \\sum_{i = 0}^{9999} 0.1 = 1000, \\quad \\text{and} \\quad \\sum_{i = 0}^{9999999} 0.1 = 10^6.$$" ] }, { "cell_type": "code", "execution_count": 21, "metadata": { "slideshow": { "slide_type": "-" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "99.9999999999986\n", "1000.0000000001588\n", "999999.9998389754\n" ] } ], "source": [ "s = 0\n", "for _ in range(1000):\n", " s += 0.1\n", "print(s)\n", "s = 0\n", "for _ in range(10000):\n", " s += 0.1\n", "print(s)\n", "s = 0\n", "for _ in range(10000000):\n", " last = s\n", " s += 0.1\n", "print(s)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "Notice how the result is sometimes smaller and sometimes bigger than the correct result." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "### Associativity of addition\n", "\n", "We all know that for a finite set of real numbers $\\{ a_1, \\dots, a_n \\}$ the following is true:\n", "$$\\sum_{i=1}^n a_i = \\sum_{i=n}^1 a_i = \\sum_{i=1}^n a_{P(i)},$$\n", "for any permutation $P$. However, in a computer, this isn't always so." ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "slideshow": { "slide_type": "-" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "sin1 = -3121.3699495895926\n", "sin2 = -2947.8076865687467\n", "sin3 = -2403.8076865683283\n", "sin4 = -1768.0\n", "|sin1 - sin4| = 1353.3699495895926\n", "the first element: 47.12388980384689\n", "the last element: 5.3966776616824465e-12\n" ] } ], "source": [ "from math import pi\n", "x = 15 * pi\n", "# Create the list of series elements\n", "elts = [ ]\n", "f = 1\n", "for k in range(1, 150, 2):\n", " elts.append(x**k / f)\n", " f *= -(k+1) * (k+2)\n", "# Sum elements in the original order\n", "sin1 = 0\n", "for el in elts:\n", " sin1 += el\n", "print(\"sin1 =\", sin1)\n", "# Sum elements in the reversed order\n", "sin2 = 0\n", "for el in reversed(elts):\n", " sin2 += el\n", "print(\"sin2 =\", sin2)\n", "# Sum elements from the middle one to the ones on the edges\n", "cnt = len(elts)\n", "mid = cnt // 2\n", "sin3 = 0\n", "for i in range(mid + 1):\n", " if mid + i < cnt:\n", " sin3 += elts[mid + i]\n", " if i:\n", " sin3 += elts[mid - i]\n", "print(\"sin3 =\", sin3)\n", "# Sum elements from the ones on the edge to the middle one\n", "sin4 = 0\n", "for i in reversed(range(mid + 1)):\n", " if mid + i < cnt:\n", " sin4 += elts[mid + i]\n", " if i:\n", " sin4 += elts[mid - i]\n", "print(\"sin4 =\", sin4)\n", "print(\"|sin1 - sin4| =\", abs(sin1 - sin4))\n", "print(\"the first element:\", elts[0])\n", "print(\"the last element:\", elts[-1])" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "The above is the computation of $\\sin( 15\\pi )$ via the first $74$ elements of the [Taylor series](http://en.wikipedia.org/wiki/Sine#Series_definition) of the sine function:\n", "* sin1 computation starting from the first element ($a_{1} + a_{2} + a_{3}+ \\cdots$),\n", "* sin2 going from the last to the first element ($a_{74} + a_{73} + a_{72}+ \\cdots$),\n", "* sin3 going from the center out ($a_{37} + a_{36} + a_{38} + a_{35} + a_{39} + \\cdots$),\n", "* sin4 going from the edges in ($a_1 + a_{74} + a_2 + a_{73} + \\cdots$).\n", "\n", "The difference between sin1 and sin4 is roughly $1353$, which may not look like much, but it is far more than the difference between any two sines should be.\n", "\n", "You might also notice that $\\sin(15\\pi)$ shouldn't be anywhere near $-3000$ or $-1768$.\n", "\n", "One might think that we should compute more elements of the sum, but this is not the case: the last element of the sum is only around $5.4 \\cdot 10^{-12}$ (and the following ones would be even smaller)." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "So what happened here?\n", "\n", "A detailed explanation is part of Numerical Analysis, but the key is in the largely varying magnitude and alternating signs of the elements: " ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "slideshow": { "slide_type": "-" } }, "outputs": [ { "data": { "image/png": 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LITqRAIwBO2aDG+L3w3bJM6kIYLraWhUsqww0KwLIrgJK9QJqs4laq4ZFlUZZ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\nEZqWmn1iAAAAAElFTkSuQmCC\n", "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "%matplotlib inline\n", "import matplotlib.pyplot as plt\n", "plt.plot(elts)\n", "plt.show()" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "### Powers\n", "\n", "Let us define\n", "$$f(x,n) := \\underbrace{\\sqrt{\\sqrt{\\dots\\sqrt{x}}}\\,\\hskip-1em}_{n}\\hskip1em, \\quad g(x,n) := \\hskip3.7em\\overbrace{\\hskip-3.7em\\left(\\left(\\dots\\left(x\\right)^2\\dots\\right)^2\\right)^2}^{n}.$$\n", "In other words, $f(x, n)$ is the number that we get by taking the square root of $x$, $n$ times in a row, and $g(x, n)$ is the number we get by computing the second power of $x$, $n$ times in a row.\n", "\n", "Obviously, $x = f(g(x, n), n) = g(f(x, n), n)$ for any $n \\in \\mathbb{N}$ and $x \\in \\mathbb{R}^+_0$. But, let's see what a computer has to say if we input some $x \\ne 1$ and $n = 50, 60, \\dots$:" ] }, { "cell_type": "code", "execution_count": 24, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "x = 1.7\n", "n = 53\n", "g(f(1.7, 53), 53) = 1.0\n", "f(g(1.7, 53), 53) = inf\n" ] } ], "source": [ "from math import sqrt\n", "x = float(input(\"x = \"))\n", "n = int(input(\"n = \"))\n", "t = x\n", "for _ in range(n):\n", " t = sqrt(t)\n", "for _ in range(n):\n", " t *= t\n", "print(\"g(f(\" + str(x) + \", \" + str(n) + \"), \" + str(n) + \") =\", t)\n", "t = x\n", "for _ in range(n):\n", " t *= t\n", "for _ in range(n):\n", " t = sqrt(t)\n", "print(\"f(g(\" + str(x) + \", \" + str(n) + \"), \" + str(n) + \") =\", t)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Do these tiny errors really matter?\n", "\n", "Yes, rounding errors have repeatedly led to catastophic consequences, for example, in engineering, finance, and science. See [http://ta.twi.tudelft.nl/users/vuik/wi211/disasters.html] for an interesting list. Even when solving a linear system of equations, probably the most fundamental problem in scientific computing, rounding errors have to be taken care of.\n", "\n", "Consider the following two systems of linear equations:\n", "\n", "$$\\left\\{\\begin{array}{rcrcr}\n", "1 \\cdot x &+& 1 \\cdot y &=& 2, \\\\\n", "1.000001 \\cdot x &+& 1 \\cdot y &=& 2.000001,\n", "\\end{array}\\right.\n", "\\quad \\text{and} \\quad\n", "\\left\\{\\begin{array}{rcrcr}\n", "1 \\cdot x &+& 1 \\cdot y &=& 2, \\\\\n", "1.000001 \\cdot x &+& 1 \\cdot y &=& 1.999999.\n", "\\end{array}\\right.$$\n", "\n", "What are their solutions?" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "The solution to the first one is $(x, y) = (1, 1)$, but the solution to the second one is $(x, y) = (-1, 3)$. \n", "\n", "Notice that both systems only differ by a tiny change of magnitude $10^{-6}$ in just one element, but their solutions $(x,y)$ are completely different! Such a small change could easily be caused by one of the small errors shown before. Similar results can be achieved with arbitrarily small errors." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Bottom line \n", "\n", "Always be extra careful when working with \"real\" numbers in a computer (or, better, avoid them altogether if possible, like in the Fibonacci example)!\n", "\n", "These errors cannot always be considered insignificant, as they can pile up and/or grow in subsequent computations.\n", "\n", "Without your lecturer being biased at all, anyone intending to do serious computations with computers should take the course \"[Numerical Analysis 1](http://www.maths.manchester.ac.uk/study/undergraduate/courses/mathematics-bsc/course-unit-spec/?unitcode=MATH20602)\" (MATH20602)." ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.6.3" } }, "nbformat": 4, "nbformat_minor": 1 } | 22,407 | 51,394 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-13 | latest | en | 0.718181 |
http://www.markedbyteachers.com/gcse/science/investigating-how-the-amount-of-oxygen-given-off-by-pond-weed-changes-when-the-environment-is-changed.html | 1,526,853,502,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863689.50/warc/CC-MAIN-20180520205455-20180520225455-00579.warc.gz | 414,198,291 | 19,581 | • Join over 1.2 million students every month
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# Investigating how the amount of oxygen given off by Pond weed changes when the environment is changed.
Extracts from this document...
Introduction
Chris Wirt 1st December 2000 Investigating how the amount of oxygen given off by Pond weed, when the environment is changed. Aim: My aim is to do a well planned safe experiment. It should show clearly how the amount of oxygen given off by pond weed can be changed by changing the environment. Photosynthesis Equation: 6CO2 + 6H2O light energy & chlorophyll C6H12O6 + 6O2 Carbon dioxide + water carbon hydroxide + oxygen List of Equipment: When choosing the equipment for my investigation I have to take into account which will be the most appropriate for the task. * beaker 200ml * Thermometer * Stop watch * Funnel * 5cm piece of pond weed (Elodea) * water * lamp * 25w, 40w, 60w, 75w, 100w bulbs * test tube Variable: These are some possible variables I could use: * I could change the intensity of the light shone on the pond weed. I would do this by testing the amount of oxygen given off after a period of time, using different wattage bulbs. * I could change the colour of the light shone on the pond weed. I could easily do this by covering the head of the lamp with coloured translucent material. For the variable in my experiment I will be changing the Intensity of the Light shone apon the pond weed. The different wattage which I will use are 100w, 80w, 60w, 40w, 20w. Once I have allowed for the light to start effecting the weed I will be able to see how much oxygen is produced in a set time. ...read more.
Middle
As a watt is a unit of power, or work done per unit time, equal to 1 joule per second. It is used as a measure of electrical and mechanical power. One watt is the amount of power that is delivered to a component of an electric circuit when a current of 1 ampere flows through the component and a voltage of 1 volt exists across it. So basically the greater the wattage the more power that is applied to the device. So as I increase the wattage of the bulbs there is more power being used, and as more power is used more power is transferred to photosynthesis in the plant. So the rate of photosynthesis will increase as the watts do. This will only continue until the rate is stopped by other limiting factors, such as when all the carbon dioxide in the water is used up the reaction will stop. This can be clearly shown using graph 2 above as an example. Results: Bulb wattage (W) 1st try bubbles per min 2nd try bubbles per min 3rd try bubbles per min average bubbles per min temperature (�C) 25w 3.0 3.0 4.0 3.3 29.0 40w 4.0 6.0 6.0 5.3 29.0 60w 10.0 8.0 9.0 9.0 29.5 75w 17.0 19.0 17.0 17.7 29.5 100w 26.0 30.0 32.0 29.3 30.0 Conclusion: From the results I have gathered I can clearly say that the light intensity does increase the rate of photosynthesis in the plant. This will have happened due to when I increased the intensity of the light the result would be more energy being caught in the chlorophyll. ...read more.
Conclusion
The temperature was also another factor which could affect the rate of photosynthesis. This was a problem as most of the energy from a lamp is heat energy, so this could increase reaction speed. To make sure this was not happening we measured the temperature, which stayed between 29.0�C and 30?C. The method for my experiment could also be improved to generally increase the quality of my results. As I said before a gas syringe could be used to collect all the gas for greater accuracy. I also said that I was finding counting the number of bubbles a problem, this could be solved by only testing the rate of photosynthesis over a shorter period, such as 10seconds. Measuring over a shorter period of time will reduce the chance of human error as there are fewer bubbles to count and they are over a shorter period of time. Because of the way my experiment is set up it is easy for me to adapt to measure another variable of photosynthesis. As it is possible to control the amount of carbon dioxide in the water by using sodium hydrogen carbonate (NaHCO3). This could be easily done by using different volumes of NaHCO3. All other variables will be kept at constants. Another possible adaptation would be to alter the wavelength of the light used for photosynthesis. This would be done my using translucent colour filters in front of the lamps. The only problem is that there is no way to define or measure the wavelength of light. It is because of this point that we only have a basic way to class colours, that the coloured light experiment would not be as affective or interesting as light intensity or CO2 concentration. ...read more.
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# Related GCSE Green Plants as Organisms essays
1. ## How temperature affects the rate of photosynthesis.
Make sure the syringe is squeezed in. (This has to be done because if the syringe is pulled out, there would already be some air trapped inside the syringe affecting the amount of water drawn into the syringe.) 6.
2. ## Investigating the factors, which affect the rate of photosynthesis in a pond plant.
21 27 35 60 529 18 22 20 Analysis: The results that have been retrieved from each experiment and all the trials show good detail and accurate, positive results. The results show that the prediction made was correct. What was conclusive was that the higher the light intensity and the
1. ## Absorption Spectrum of Chlorophyll.
This gives rise to two equations which can be solved simultaneously to give the concentration of each pigment in ?g/ml of extract as shown in equations (2) and (3) below. Ca is the concentration of chlorophyll a and Cb is the concentration of chlorophyll b.
2. ## An Investigation into Species Diversity with distance along a Pingo.
its significance and the strength of the relationship, if any was found. Statistical tests: Statistical tests were conducted on some of the results. All statistical tests have been done separately, by hand on paper. These include Spearman's Rank coefficient on species diversity and distance, species diversity and light intensity and species diversity and soil water content.
1. ## I aim to investigate the effects of temperature on the rate of photosynthesis in ...
Henceforth it is limited, according to Blackman's Law, by the factor in least supply, either H2O or temperature. however there is a physical limitation of the carbon dioxide diffusion and the plant's sunlight absorption. At a lower temperature, the rate of p/s is increasing with increasing LI or CO2 availability, but the LSP or CSP is quickly reached.
2. ## Nuclear Power
Advantages * Solar energy is free - it needs no fuel and produces no waste or pollution. * In sunny countries, solar power can be used where there is no easy way to get electricity to a remote place. * Handy for low-power uses such as solar powered garden lights and battery chargers Disadvantages * Doesn't work at night.
1. ## Find out what factors affect the rate of photosynthesis. Rate being the amount of ...
Therefore if my results are according to distance vs. rate of photosynthesis, I predict the graph will be a negative curve. Without sunlight a plant does not grow and will die. This is because the sunlight is the source of energy that the plant needs.
2. ## Do different coloured wavelengths of light affect the rate of photosynthesis in Canadian Pond ...
Fair Testing; These are things I am going to keep the same; * The amount of sodium carbonate (1 spatula full) * The amount of water (500ml) * The same sized piece of pond weed ( 6cm) * The same amount of light (1 Lamp)
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http://weldnotes.com/?p=55&replytocom=399 | 1,695,603,215,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506669.96/warc/CC-MAIN-20230924223409-20230925013409-00342.warc.gz | 49,899,832 | 14,840 | # Basic Trig – Sine and Inverse Sine
If you have never had a trig class, you can do some pretty useful things with just a few trig functions. Bob Welds discusses the sine of an angle and how to use it. He also tells how to find an angle from two known sides.
Video Transcript:
Hi I’m Bob Welds and this is a first look a trigonometry.
Today we are going to pick up where the Pythagorean theorem left off. If you don’t know how the Pythagorean theorem works then you probably want to go learn about that first. Also, we’ll use a little bit of algebra but i don’t think it will be a problem for you if you already know some basic algebra. To keep it simple, we’re only going to talk about a thing called the sine of an angle today. Later we’ll see other trig functions. But really, if you understand how the sine works, you’ll get the others pretty easily.
Ok. Let’s look at this right triangle. You see that it has a 30 degree angle in one corner. The side opposite of the 30 degree angle is 1 meter long, and the hypotenuse is 2 meters long. (remember that the hypotenuse is the longest side of a right triangle).
let’s compare the opposite side to the hypotenuse by dividing one by two and see what we get….1 divided by two equals .5 — a half. um. so .. that was easy. What’s the big deal? The deal is that we just found the “sine” of thirty degrees. The sine of thirty degrees is one half. we write it like this:
Here, let’s do it again. this right triangle has a height of 4 meters and a hypotenuse of eight meters. The sine of 30 deg is four divided by eight. 4 divided by eight reduces to one half, again, the sine of thirty degrees is one half, or point five. You see it doesn’t matter how big the triangle is; 30 degrees always has a sine of point five. Any 30 degree right triangle has a hypotenuse that is exactly twice as long as the opposite side. Here, you can google it…type sin 30 degrees and see what you get. be sure to type degrees or you’ll get something else.
The sine of an angle tells us the answer to the division problem we get when we put the opposite side (that is, the side across from the angle) over the hypotenuse. So the sine of the angle is “ratio” of the opposite side to the hypotenuse. the opposite over the hypotenuse. In this case, we put one over two and see that it is one half or point five.
You might be asking what good it is to know that ratio is since we already know how to divide one by two or four by eight… well let’s take a closer look. Here is another triangle, but with the same 30 degree angle in the corner. This time, we know the length of the hypotenuse and angle, but we don’t know the opposite side. Look what we could do if we knew the sine of the angle…instead of saying “sine of 30 degrees” we could say point five. Because .5 is the sine of thirty degrees. ….and instead of saying hypotenuse, we could say fourteen, because that’s how long this hypotenuse is. Now instead of using that silly question mark, we could call the length of the opposite side: “a”. And voila! we have an algebra problem instead of a trigonometry problem.
Here is how we could solve the algebra part….multiply both sides by 14… on the left, we see 14 times .5 equals seven. and on the right side .the 14’s cancel each other out…and look! A equals seven. That is something we could not have found if we didn’t know this little bit of trigonometry.
Lets see how useful the sine of an angle can be. We know the sine of thirty is, but with our calculator we can find the sine of ANY angle. Lets try to find the length of the hypotenuse of this triangle. The angle is 25 degrees and the side opposite of the angle is 1.69 meters long. First write down what we know…The sine of 25 degrees is equal to the opposite side–1.690 over the Hypotenuse. Now, we need the sine of 25 degrees…This time, I’ll use my calculator instead of google. First, we’ll turn it to scientific mode…..then I’ll be sure it knows to use degrees….now I’ll type 25 and hit the sin button. Different calculators might have you hit the sine key first, so check your answers to be sure you are doing what your calculator expects….0.422 618 261 and it keeps going. We only need the first few decimal places of that. So instead of saying “sine of 25 degrees” we will say 0.4226. Here, I rewrite our equation.
Now it’s an algebra problem. We need to isolate thy hypotenuse to see what it’s value is. I’m going to call it “C” to keep things neat and clean. I’ll multiply both sides by c….the c’s cancel out on the right…..now i’ll divide both sides by .4226. this leaves C on the left and a little division problem on the right. we use our calculator to type 1.690 divide by .4226 and we get something just a smidgen over 3.999 — a number very close to four meters.
Now we’ve seen that we can use the sine of an angle to find the sides of the triangle, and you can use the opposite side and the hypotenuse to find the sine of the angle. What if you knew the sine of an angle, not the angle itself? There’s a trick for that. Let me show you. Here we have another right triangle. The side opposite of our angle is 2.5 meters long, and the hypotenuse is 3.889 inches long. You may be able to see right away that we could find the sine of the angle like we did a few minutes ago. here, remember? You see we just put 2.5 over 3.889. Since a fraction is just a division problem, we’ll divide 2.5 by 3.889. That gives us .64283. So just like before, we know what the sign of the angle is. Only now we don’t know what the angle itself is. To find the angle when we know the sine of the angle, we just run the sine function in reverse. That’s called taking the ARC Sine or the INVERSE SINE. You will see it on your calculator as a sine with a -1 exponent, or you may see it spelled out as “A”sin or “arc sine” or even inverse sine. A lot of the time it’s the same button as the sine button, but you push another button to enable this feature. I’ll do it with this little calculator so you can see. I’ll type .64284 and push this up arrow button. You’ll see that my sine key turned into an inverse sine key.
Now when i press the inverse sine button i get an angle! Isn’t that teh cats pajamas? My angle is just a tiny bit more than forty degrees. Ill round it off to the nearest tenth of a degree. Here’s the original problem you can remember what we were solving. We knew the side opposite of the angle was 2.5 meters long, and we knew the hypotenuse was 3.889 meters long. We wanted to know the angle. First we found the sine of the angle, opposite over hypotenuse (2.5 over 3.889). We divided and found the sine of the angle was .64283. Then we used our new trick, the inverse sine to find teh angle was very close to forty degrees. I just think that’s great. That’s going to be very useful.
Ok, let’s work some problems. Im going to give you a problem where you find the missing opposite side first. When you see sparky’s paws, Pause teh video and try to fill in the blanks in the equation. You don’t need to solve it yet, just fill the blanks in with what you see on the diagram….[pause]. Ok, here are the blanks filled in. The sine of 30 degrees is equal to “a” over 15. “a” is the opposite side, and 15 is the hypotenuse. teh sine of an angle is the opposite over the hypotenuse. Now lets see if we can turn the trig problem into an ordinary algebra problem. Let’s see instead of saying sine of thirty degrees, what could we say? use a calculator to find the sine of 30 degrees, even if you know what it is. This way you can be sure your calculator is set up correctly.
I’ll pause while you calculate. ok. i know you probably remembered that the sine of thirty degrees was .5, but hopefully you tried it on your calculator. It is really important that you know how to get the sine of an angle. Now we just have a little algebra problem on our hands. See if you can solve for “a.” a is equal to 7.5. If you were able to solve it by looking at it, let me encourage you to go through the algebra steps to get good at them. Here, we multiply both sides by 15 to get rid of that fraction…. Then we see that 15*.5 is equal to 7.5. that’s our answer.
Let’s do one more. Let’s do one where we find the angle by using the opposite side and the hypotenuse…. let’s remember that if we know the opposite side and the hypotenuse, we can find the SINE of the angle. Let’s do that first. Pause the video and see if you can fill in the blanks. Then find the sine of the unknown angle by dividing. That’s right, 1 on top, 4.810 on the bottom. That means the sine of this angle is .2079 . Now for the second part, lets use the arc sine to find what the angle is….remember we know the sine is .2079 Olk, pause the video and be sure you know how to find the angle using the sine of the angle. Fill in the blanks and do the calculation on your calculator. OK, I got an answer that was ridiculously close to 12. I’m going to round it off to the nearest tenth of a degree.
Well that’s our first look at trigonometry. I hope you can see how useful it is to find angle from those lengths and lengths from the angles. There is a LOT more to cover, but if you understand how the sine and inverse sine work you are on you way to using some very useful tools. I’m bob welds, and these are weldnotes!
### 1 Response
1. Robert Vaillancourt says:
I’ve watched four of the video’s and they answered the questions I had. Easy listening Thank you. | 2,429 | 9,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2023-40 | latest | en | 0.912056 |
https://brainmass.com/business/weighted-average-cost-of-capital/pendant-weighted-average-cost-capital-246692 | 1,713,393,704,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817181.55/warc/CC-MAIN-20240417204934-20240417234934-00401.warc.gz | 122,105,324 | 8,078 | Purchase Solution
# Pendant, Inc: Weighted Average Cost of Capital
Not what you're looking for?
The treasurer at Pendant, Inc. is estimating the company's weighted average cost of capital. Everyone in the company evaluating capital investments will use this estimate of WACC. The financial information is as follows:
The company's 6.5% coupon rate bonds pay annual interest, mature eight years from now and sell on the New York Bond Exchange for 92.50. The face value of the bond is 100. Pendant's common and preferred stock are listed on the New York Stock Exchange. The common stock is selling for 58.50 and pays a dividend of 2.20. Analysts expect growth of 6% per year. When Pendant sells common stock, it incurs a flotation cost (investment banking and administrative fees) of 4%, which reduces the amount it receives from selling each share by 4%. The preferred stock sells for 22.63 and carries a dividend of 1.89. Flotation costs are 3% when preferred stock is sold. Since preferred stockholders do not share in the growth of the company, the cost of preferred stock is just the dividend yield calculated based on a price net of flotation cost. The company's beta coefficient is .65. Its optimum financial structure, based on market values, consists of 40% debt, 2% preferred stock, 28% common stock and 30% retained earnings. The company's tax rate is 36%. When the company sells common stock to raise capital, it estimates the cost of common equity with the dividend yield plus percentage growth rate model. It estimates its cost of retained earnings with the Capital Asset Pricing Model (see formula 8.2, p.301 in the Higgins text), and assumes an interest rate on government bond of 5%, and an 8% historical excess return on common stocks. Calculate the company's WACC.
##### Solution Summary
This solution will use financial information to estimate the WACC.
##### Solution Preview
Please see the attachment for the solution.
Given that,
Debt:
Coupon rate=6.5%
Time to maturity=8 years
Price of the bond=\$92.50
Face value of the bond=\$100
We have,
...
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##### Marketing Management Philosophies Quiz
A test on how well a student understands the basic assumptions of marketers on buyers that will form a basis of their marketing strategies.
##### IPOs
This Quiz is compiled of questions that pertain to IPOs (Initial Public Offerings) | 629 | 2,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-18 | latest | en | 0.905247 |
https://www.socomaconstruction.com/appartement/seier-antwert-what-does-nnn-mean-in-commercial-real-estate.html | 1,656,455,573,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103617931.31/warc/CC-MAIN-20220628203615-20220628233615-00770.warc.gz | 1,076,037,896 | 11,678 | # Séier Äntwert: What Does Nnn Mean In Commercial Real Estate?
## What is \$25 NNN?
NNN stands for Triple Net rent. In this type of commercial real estate rent, you pay the amount listed and you also have pay additional costs (usually Operating Expenses) on top of that. For example: say the Office Space listing you’re interested in says the rent is \$24.00 NNN per sqft/year.
## How is NNN calculated?
Lease Rate: \$20.00 /SF NNN (Estimated NNN = \$3.25/SF ), meaning the base rental rate is \$20.00 per square foot per year and the property expenses, which include property taxes and insurance, are estimated to be \$3.25 per square foot per year, though they can fluctuate from year to year.
## Is triple net lease bad?
A triple net lease has risk for both the tenant and landlord (lessor). The Bad: For the tenant, there are some unknown variables that might cause a problem. Take for instance rising costs. A triple net lease might have some sort of cap, but likely, a tenant would be forced to cover rising taxes and insurance rates.
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## Is NNN monthly or yearly?
The estimated operating expenses (aka NNN) are \$10 per square foot per year. The total yearly rent you would pay equals \$40 sf per year. So if you are leasing 3,000 sf then your yearly rent would be \$120,000 or \$10,000 per month.
## Can you negotiate a triple net lease?
Absolutely not! There are many areas where a tenant can negotiate a NNN lease to make it more favorable. If the tenant is taking on all responsibility and risk of the landlord’s overhead, then the tenant may be able to negotiate a more favorable base rental amount.
## What is the average NNN rate?
Now we have to add on the NNN cost which may range from \$1 to \$20 a square foot based on the use and costs. It is typical to see a \$3 a square foot NNN cost in my area, which would add \$15,000 a year or \$1,250 a month to the costs.
## What does NNN mean in a lease agreement?
A triple net lease (triple-Net or NNN) is a lease agreement on a property whereby the tenant or lessee promises to pay all the expenses of the property including real estate taxes, building insurance, and maintenance.
## How is triple n calculated?
Triple net leases are calculated by adding the yearly taxes on the property and the insurance for the space together and dividing that amount by the building total rental square footage. The process of calculating a triple net lease is simplified when an entire building is leased to one tenant.
## Who pays for a new roof in a triple net lease?
As the triple net property owner (unless otherwise specified in the NNN lease), you’ll generally be responsible for maintaining and repairing these 3 main aspects of your building: Roof (repairs, maintenance, upgrades) Exterior Walls. Utility Repairs and Upkeep (for major things such as plumbing and electricity)
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## What are the benefits of a triple net lease?
The most obvious benefit of using a triple net lease for a tenant is a lower price point for the base lease. Since the tenant is absorbing at least some of the taxes, insurance, and maintenance expenses, a triple net lease features a lower monthly rent than a gross lease agreement.
## How do you value a triple net lease?
NNN leases are calculated by adding the estimated common area maintenance expenses, annual property taxes, and the building insurance for the property. This number is then divided by the total square footage of the building and given to the tenants on a per square foot basis.
## What’s NNN on Tiktok?
NNN is an acronym for ” No Nut November,” which is basically a 30-day challenge men partake in where they don’t ejaculate.
## What does sq ft yr mean?
In the commercial leasing industry, \$/SF/year or \$/SF/yr means the rent per square foot per year. This would be calculated as \$20 x 1000 square feet = \$20,000 total (this is the cost for the total year). Now, to get your monthly cost, divide by 12.
## What is triple net lease example?
A triple net lease is an agreement between a property owner and a tenant where the tenant pays property taxes, insurance premiums, and maintenance upkeep and repairs, in addition to a monthly rental fee of the building or space. | 975 | 4,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-27 | latest | en | 0.934134 |
https://www.gatsport3.ru/geometry-problem-solving-15520.html | 1,621,008,357,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991428.43/warc/CC-MAIN-20210514152803-20210514182803-00244.warc.gz | 826,141,898 | 9,714 | # Geometry Problem Solving
Tags: How To Do A Research Paper In Mla FormatCosi Lewis EssayReview Essay Good Faith In Public International LawSolve Math Problem OnlineSae International Research PapersCreative Writing Activities For 4th GradersWebsite That Solves Any Math ProblemResearch Paper On Same-Sex MarriageCommercial Real Estate Business Plan TemplateTerm Papers In Spanish
Solve your probability, combination, permutation problems. Statistics - find median, mean (arithmetic, geometric, quadratic), mode, dispersion, mormal distributions, t-Distribution.
The solver successfully do Statistical hypothesis testing You can online solve chemistry equations.
If the object contains circles, then you may need to think about the circumference of a circle, which is the perimeter of the circle.
Solving geometric problems, such as those found in art and architecture, is an important skill.
If you're seeing this message, it means we're having trouble loading external resources on our website.An extended task designed to exercise students skills with mensuration, including area of triangles, Pythagoras' theorem, area of circles using p and calculations combining the results of each.There are two Power Points - the first setting out the problem in clear stages (ideal for students who need walking through the steps), and the second showing worked answers.From the air, you can see distinct circles from the resulting vegetation.If there are four sprinklers anchored in a field and each sprinkler extends 100 feet from the pivot point, how much area does the farmer irrigate?The JP Morgan Chase Bank Tower in downtown Houston, Texas, is one of the tallest buildings west of the Mississippi River. If 40% of each face is covered with glass windows, what is the amount of surface area covered with glass?.pass_color_to_child_links a.u-inline.u-margin-left--xs.u-margin-right--sm.u-padding-left--xs.u-padding-right--xs.u-relative.u-absolute.u-absolute--center.u-width--100.u-flex-inline.u-flex-align-self--center.u-flex-justify--between.u-serif-font-main--regular.js-wf-loaded .u-serif-font-main--regular.amp-page .u-serif-font-main--regular.u-border-radius--ellipse.u-hover-bg--black-transparent.web_page .u-hover-bg--black-transparent:hover. Some measurements of the building might, of course, be required, but the same problem-solving techniques apply.It behooves us to present a basic approach to solving practical geometry problems.You can simplify and evaluate expressions, factor/multiply polynomials, combine expressions.You can solve all problems from the basic math section plus solving simple equations, inequalities and coordinate plane problems.
## Comments Geometry Problem Solving
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• ###### PROBLEM-SOLVING IN GEOMETRY IN COLLABORATIVE.
This research focused on Geometry problem-solving in collaborative small-group. have seen how learners struggle with the solving of Geometry problems.… | 816 | 4,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-21 | latest | en | 0.849264 |
https://www.reference.com/math/many-edges-pentagonal-pyramid-d00b6cd0bc42efa2 | 1,571,211,089,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666467.20/warc/CC-MAIN-20191016063833-20191016091333-00273.warc.gz | 1,055,544,163 | 17,242 | How Many Edges Does a Pentagonal Pyramid Have?
A pentagonal pyramid is characterized by six faces, six vertices and 10 edges. This three-dimensional polyhedron is a type of pyramid containing five triangular lateral faces and one pentagonal face, which is the base of the pyramid.
Pyramids are solid geometric figures made up of planar surfaces called "faces" that are bounded by line segments referred to as "edges" or "sides." The points where the edges of the faces meet are called "vertices" or "corner points." The faces are all triangular polygons, which are two-dimensional shapes that are completely enclosed. The intersection point of the lateral faces is called the apex of the pyramid, located above the base. Pyramids are named based on the shape of their bases.
The Swiss mathematician Leonhard Euler devised a formula for computing the number of faces, vertices and edges for most three-dimensional polyhedrons. Known as Euler's formula, this mathematical equation is given as F + V - E = 2, where "F" indicates faces, "V" denotes vertices and "E" represents edges. The sum of the faces and vertices minus the number of edges is always equal to 2. The formula for calculating the number of edges can then be expressed by the derived equation F + V - 2 = E. By substituting the correct values for a pentagonal pyramid such that 6 + 6 - 2 = E, the number of edges is equal to 10.
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https://www.studysmarter.co.uk/explanations/engineering/engineering-mathematics/inverse-laplace-transform/ | 1,720,939,967,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00710.warc.gz | 893,423,065 | 77,480 | Inverse Laplace Transform
Explore the fascinating world of inverse Laplace transform, an essential mathematical tool on which much of your engineering coursework will hinge. Gain a solid foundation in the subject through an in-depth explanation of the concept, its historical background, and its importance in engineering mathematics. Delve deeper into its properties, take a look at its practical applications across several engineering disciplines like environmental and mechanical, and utilise the inverse Laplace transform table for quick referencing. Also, familiarise yourself with the key inverse Laplace transform equations, conversion of complex problems using the tool, and tip-off with some useful tips and techniques. Let this be your guide to conquering understanding, applying, and benefiting from the concept of inverse Laplace transform in your engineering journey.
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Understanding the Concept: Inverse Laplace Transform Meaning
The Inverse Laplace Transform is a significant aspect in the field of Engineering Mathematics. Simply put, it is the process of converting a function from the Laplace transform domain back to the time domain.
Historical Background of Inverse Laplace Transform
The Laplace Transform, named after its inventor Pierre-Simon Laplace, is a mathematical procedure widely used in engineering, physics, and many areas of applied mathematics. It is useful in solving linear ordinary differential equations such as those arising in the analysis of electronic circuits.
Laplace initially introduced this transform theory during the late 18th century to solve problems in celestial mechanics. Over the years, the Laplace Transform, and consequently the Inverse Laplace Transform, has found myriad applications not just in engineering, but also in various other branches of science.
Mathematical Explanation: What is Inverse Laplace Transform?
Combined with the Laplace transform, the Inverse Laplace Transform transfigures the complex operations of calculus into simple algebraic procedures. The inverse Laplace transform is symbolically represented as $$L^-1\{F(s)\}=f(t)$$ where:
• $$F(s)$$ is a function in the s domain
• $$f(t)$$ is the equivalent function in the time domain
The process of determining this function $$f(t)$$ from its Laplace transform $$F(s)$$ involves complex integrals and is termed as the Inverse Laplace Transform.
Importance of Inverse Laplace Transform in Engineering Mathematics
In the realm of engineering, particularly in signal processing and control systems, the Inverse Laplace Transform is fundamentally important.
Assume you have an electrical circuit with a particular impedance. You would typically use the Laplace Transform to translate this physical system from the time domain to the s domain and perform intricate mathematical operations. After your calculations, you'd need to interpret these results back into the real world scenario. This interpretation would necessitate the Inverse Laplace Transform.
Fields of Engineering Application of Inverse Laplace Transform Electrical Engineering Solving circuit differential equations, analysing signal problems Mechanical Engineering Studying system dynamics, vibration analysis Civil Engineering Analysing structural systems
So, while it is a complex mathematical tool, the Inverse Laplace Transform holds immense practical importance for engineers, and expertise in this subject can open new avenues in problem-solving and decision making.
Digging Deeper: Inverse Laplace Transform Properties
The concept of the Inverse Laplace Transform, crucial as it is, isn't complete without a detailed look into its underlying properties. These properties, which govern the behaviour of this mathematical tool, are pivotal to its application and understanding.
Linearity of Inverse Laplace Transform
The property of linearity is fundamental in the realm of mathematical transformations. In the context of the Inverse Laplace Transform, it signifies that the inverse transform of the sum of two Laplace transforms is equal to the sum of their respective inverse Laplace transforms. Phrased mathematically, if $$F(s) = L\{f(t)\}$$ and $$G(s) = L\{g(t)\}$$, then $$L^{-1}\{F(s) + G(s)\} = f(t) + g(t)$$. With this property, you can simplify computations involving the Inverse Laplace Transform. The addition or subtraction of functions in the s domain does not stringifyly change the result in the time domain, thereby offering the flexibility to manipulate functions before applying the inverse transform. This makes the handling of complex functions easier.
Time Shift Property
The time shift property is another significant feature of the Inverse Laplace Transform. This property relates to how a shift in the time domain corresponds to a change in the s domain. The formula for the time shift property is as given: if $$L^{-1}\{F(s)\} = f(t)$$, then $$L^{-1}\{e^{-as}F(s)\} = f(t-a)u(t-a)$$. Here, $$u(t)$$ is the step function, and a is the time shift.
The step function, most often represented as $$u(t)$$, is defined to be zero for $$t < 0$$ and one for $$t ≥ 0$$. This is also commonly known as the Heaviside step function after its inventor, Oliver Heaviside.
• If the value of $$a$$ is positive, it depicts a delay in the time region.
• If the value of $$a$$ is negative, it represents an advance in the time region.
Frequency Shift Property
The Frequency Shift property is a major key in understanding how shifts in the frequency domain affect the time domain. The formula representing the frequency property is set as: If $$F(s) = L\{f(t)\}$$, then $$L^{-1}\{F(s+a)\} = e^{at}f(t)$$. Here, $$a$$ is the frequency shift.
The exponential function, denoted by $$e^{at}$$, describes exponential growth or decay depending on whether $$a$$ is positive or negative respectively.
• A positive $$a$$ shows a growth in frequency, leading to increasingly rapid oscillations in the time domain.
• A negative $$a$$ symbolises a decay in frequency, resulting in diminishing oscillations in the time domain.
Interplay of Different Inverse Laplace Transform Properties
It's not uncommon in engineering scenarios to deal with signals that require multiple property applications. These could involve combinations of linearity, time shift, frequency shift, and even other properties of Inverse Laplace Transform that we haven't covered here. Consider, for instance, a function in the s domain that involves both a shift in frequency and an addition of functions. With a firm grasp of the linearity and frequency shift properties, such signals would not pose a significant challenge.
$$L^{-1} \{ F(s+a) + G(s+a) + H(s) \} = e^{at}f(t) + e^{at}g(t) + h(t)$$
Where:
• $$F(s) = L\{f(t)\}$$
• $$G(s) = L\{g(t)\}$$
• $$H(s) = L\{h(t)\}$$
It's worth noting that these properties not only permit a preliminary simplification of the problem but also put forth a step-by-step methodology to handle complicated signals. Overall, the apt understanding and usage of these properties can immensely facilitate an engineer's problem-solving abilities with Inverse Laplace Transforms. The goal is to cultivate a mindset towards problem-solving that is as much about the journey as it is about the destination, so keep exploring these properties to maximise your potential.
Practical Utilisation: Inverse Laplace Transform Applications
The learning and understanding of the Inverse Laplace Transform reaches its pinnacle when you start seeing how it is applied in real-world contexts. In the field of engineering, the practical applications of this mathematical tool are vast and span multiple disciplines. By examining some of these applications, you can get a comprehensive sense of why this concept is so integral to engineering.
Use of Inverse Laplace Transform in Environmental Engineering
In environmental engineering, the Inverse Laplace Transform manifests its presence in a number of critical areas. A commendable testimony of this fact would be the way environmental engineers leverage this transform for modelling contaminant transport in groundwater. In this domain, groundwater quality models are of paramount importance. These models work on quantifying the concentration of contaminants over time and space to orchestrate beneficial remediation measures.
The term contaminant refers to any physical, chemical, biological, or radiological substance or matter in water, which is generally unwanted or harmful.
Inverse Laplace Transform is used to solve associated differential equations that represent the variation of contaminant concentration with time. By applying the Laplace Transform on the given equations, the problem is simplified into an algebraic one. After solving for the unknowns in the s domain, engineers utilise the Inverse Laplace Transform to ascertain the solutions in the time domain—the actual physical scenario. Additionally, this mathematical tool sees extensive usage in delineating uncertainty in environmental models. Uncertainty analysis, dealing with the quantification of errors and their potential impact on decision-making, relies heavily on probability and statistics. The Inverse Laplace Transform here serves to transform probability density functions between the time and frequency domains, thereby allowing more detailed and algebraically simplified studies. Therefore, from contamination modelling to decision-making procedures, the Inverse Laplace Transform stands as an indispensable tool in the field of environmental engineering.
How is the Inverse Laplace Transform used in Mechanical Engineering?
Mechanical Engineering is known for its broad applicability and diversity in the problems it endeavours to solve. One such area is the analysis of vibrational systems which is predominant in this field. Structures like beams, plates, and shells are subject to vibrations and oscillations of various frequencies. Accurately determining these frequencies and the behaviour of these vibrational systems are crucial to maintain structural integrity and prevent failures. A mathematical representation of such systems often culminates in the form of differential equations, which are easier to manipulate using transforms. The Inverse Laplace Transform finds its application here as it enables the transformation of these complex differential equations about vibrational systems into simple algebraic ones. Once solved, the results are converted back into the time domain using the Inverse Laplace Transform providing meaningful interpretations about the system’s behaviour. Furthermore, the design and analysis of feedback control systems, a fundamental aspect of mechanical engineering, is heavily dependent on Inverse Laplace transforms. Control systems, dealing with the management and regulation of different systems, use feedback to compare desired and actual outputs. Transforms, particularly the Inverse Laplace Transform, aid in the interpretation of system responses from transfer functions, helping engineers improve system performance and stability. So, whether it is about ensuring the structural safety of buildings or the optimal performance of mechanical systems, the Inverse Laplace Transform has branched its reach deep into the roots of mechanical engineering.
Inclusive Applications through Various Fields of Engineering
It’s evident that the Inverse Laplace Transform is universal in its utility, straddling many different branches of engineering. While we have focused on its applications in environmental and mechanical engineering, it's pertinent to point out that its usage isn't just restricted to these. Here are a few more examples:
• In Electrical Engineering, circuit analysis heavily relies on Inverse Laplace Transforms. Analysing transient responses, AC and DC circuit behaviour, filter designs, signal analysis, etc., are all domains where this tool is indispensable.
• In Civil Engineering, studying vibrational modes and frequencies of buildings and bridges, evaluating structural responses, deducing intensities of stress and strain within structures—all of these tasks find Inverse Laplace Transforms at their core.
• Even in the realm of Biomedical Engineering, Inverse Laplace Transforms find their place. They are used for filtering out noise from ECG signals or MRI images, understanding the dynamic behaviour of various physiological systems, and so on.
All these examples testify to the importance of the Inverse Laplace Transform in engineering. They collectively underline the fact that this mathematical concept is truly universal in its application across different fields.
Quick Reference: Inverse Laplace Transform Table
The utility of comprehending the Inverse Laplace Transform extends to a variety of domains within the field of engineering. To aid students and professionals alike, building a handy quick-reference table can serve as a beneficial resource while solving problems concerning this mathematical transformation. This table would encompass the most commonly used functions and their corresponding Inverse Laplace Transforms.
Overview of Key Components in the Inverse Laplace Transform Table
The Inverse Laplace Transform table typically comprises three key components: Laplace Transform pairs, equations in the s-domain, and the corresponding equations in the time-domain. The comprehension of these components is of utmost importance to effectively employ this resource.
Laplace Transform pairs
The first major component of the table is the Laplace Transform pairs. This refers to a function in the time domain and its corresponding Laplace Transform in the s domain. The concept underlying Laplace Transform pairs is the direct relation between a function $$f(t)$$ (where $$t ≥ 0$$) and its Laplace Transform $$F(s)$$. In other words, whenever you come across the function $$F(s)$$ in the s domain, you know it corresponds to the function $$f(t)$$ in the time domain.
As an example, if we have the exponential function $$f(t) = e^{at}$$, the Laplace Transform of this function would be $$F(s) = 1/(s-a)$$. Thus, $$e^{at}$$ and $$1/(s-a)$$ are a pair, with the former in the time domain and the latter in the s domain.
Equations in the s-domain
The next important component present in the table is the equations represented in the s domain. As discussed earlier, the critical advantage of the s domain is the simplification of counting equations. By applying the Laplace Transform to differential equations, you can convert them into algebraic equations, easing the operations involved in solving equations.
$$L\{f(t)\} = F(s)$$
The equations provided in the s domain therefore correspond to the Laplace Transforms of different functions and serve as the base for calculating inverse transforms.
Corresponding Equations in the time-domain
The third main component of the table is the corresponding equations in the time domain. These equations denote the original functions from which the Laplace Transforms were derived.
$$L^{-1}\{F(s)\} = f(t)$$
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The time-domain function holds substantial context in real-world scenarios and is often the primary concern in engineering problem-solving. Providing proper handles over an extensive variety of functions in the time domain, thereby, can offer a significant advantage towards solving complex problems.
Remember, when you're working in the s domain, always maintain a clear sense of what the original function (in the time domain) was. Maintaining this link to real-world situations is crucial as it helps maintain the physical significance of the problem, something that might otherwise get lost in mathematical abstraction.
This overview of the key components in the Inverse Laplace Transform table provides you a glimpse into understanding how this quick reference tool works. It's now time to dynamically engage with this knowledge and cruise through your mathematical journey with a greater understanding of the Inverse Laplace Transform!
Working with Formulas: Inverse Laplace Transform Equations
Understanding the role and application of formulas in the realm of the Inverse Laplace Transform becomes a prerequisite for establishing a firm grounding in this mathematical concept. From basic equation structures to converting complex equations, and honing practical problem-solving skills, each dimension throws light on the intrinsic relationship between the Inverse Laplace Transform and various fields of engineering.
Basic Equation Structures in Inverse Laplace Transform
Take the typical format of an Inverse Laplace Transform equation: it would usually be represented as the inverse transform of a laplace-expressed function. Symbolically, this can be characterised as $$L^{-1}\{F(s)\} = f(t)$$. If the Laplace of a function $$f(t)$$ is $$F(s)$$, the Inverse Laplace of $$F(s)$$ will revert the function back to the time or spatial domain. A good understanding of how to proceed with the inverse Laplace transformation fundamentally rests on recognising standard transformed forms. For instance, it is essential to recognise that $$F(s) = 1/s$$ denotes the transform of a unit step function and $$F(s) = 1/(s^2 + a^2)$$ corresponds to a sine function. Familiarity with these common transform pairs can aid in expediting the inverse transformation process. Another significant feature of Inverse Laplace Transform equations is their reliance on the region of convergence, known as the ROC. The ROC denotes the range of values of the complex variable $$s$$ for which the Inverse Laplace Transform exists. This becomes crucial when determining the inverse for more complex functions as the ROC can sometimes influence which inverse is chosen to avoid violations of causality or stability.
Converting Complex Equations using Inverse Laplace Transform
In (electrical) engineering, complex systems are often characterized by differential equations. Using the Laplace Transform, these equations can be transferred to an algebraic form making them easier to manipulate, solve, and understand. However, often, you will need to convert these 'solves for' algebraic equations back into the timely and applicable real-world forms, that's where the Inverse Laplace Transform comes in handy. Consider solving a second order differential equation. Through Laplace transformation, this becomes a simple quadratic equation which one might typically solve by factoring, completing the square or using the quadratic formula. At this stage, you are essentially working with two parts: the homogeneous solution derived from auxiliary roots, and the particular solution. It's next expected that you transform this solution in the 's' domain back into a time-domain equation. Here you apply the Inverse Laplace Transform to 're-humanize' each algebraic term or expression. Typically, the complexity in this stage stems from dealing with quadratic or higher order polynomials in the denominator of an expression. Remember, you are not looking at an arbitrary equation in 's', rather an Inverse Laplace Transform. The goal is to bring the expression into a form that is identifiable as the Laplace Transform of a standard function. It often involves breaking down the complex expression into simpler fractions (partial fraction decomposition) or manipulating it into a standard form to make it recognisable. This process might require substantial algebraic skills, but it lays the groundwork to finally invert these expressions to time-domain functions.
Useful Techniques in Solving Inverse Laplace Transform Equations
Application of Inverse Laplace Transform can potentially involve confident manipulation of complex mathematical expressions. While the basic approach remains converting algebraic equations in 's' domain into simple, recognisable forms, different techniques can be leveraged to simplify the process. For instance, one might encounter complex numbers in equations due to undamped or underdamped systems in circuits or mechanical vibrations. Here you can benefit from Euler's formula, $$e^{ix} = cos x + i sin x$$, that links exponential functions with trigonometric functions. This formula helps convert such complex expressions into forms corresponding to damped sinusoidal functions. Partial fraction decomposition remains another pivotal technique when you are confronted with a larger or more complex function trying to return to the time domain. By breaking the complex expression down into simpler terms, the resulting fractions usually correspond to the Laplace of standard functions, expediting the inverse transformation process. Tables of transforms provide an indispensable tool, especially when dealing with transforms of derivatives, integrals, or more complex functions. Knowledge of standard transforms and their inverses can often offer a quick solution to the problem at hand.
Continual practice remains a prerequisite to mastering the art of Inverse Laplace Transforms. It not only tests the theoretical acumen gained but also enhances the understanding about its application in engineering scenarios. Here are a few problems which you can practice on: 1. $$L^{-1}\{s/(s^2 + 4)\} 2. \(L^{-1}\{(1 + 3s)/(s^2 + s + 1)\} 3. \(L^{-1}\{s^2/(s^3 + 2s^2 + 2s + 1)\} The aim should be to apply the mathematical skills discussed and try identifying standard patterns within the 's' domain expressions. It is through such exercises and practical application that the concept of Inverse Laplace Transform ingrains within the problem-solving sphere and serves its purpose as a significant tool within the engineering domain. Remember, continuous practice increases familiarity and understanding of this mathematical tool. Happy problem-solving! Inverse Laplace Transform - Key takeaways • Inverse Laplace Transform properties include the Linearity Property: a mathematical principle stating that the sum of the inverse laplace transformation of multiple functions in the 's' domain equates to the sum of the original functions in the time domain. This can be represented by the equation \(L^{-1}\{F(s) + G(s)\} = f(t) + g(t)$$.
• Time Shift Property pertains to modifications in the time domain resulting in corresponding changes in the 's' domain, represented by the equation $$L^{-1}\{e^{-as}F(s)\} = f(t-a)u(t-a)$$. Here, $$u(t)$$ is the step function, and a is the time shift which can either indicate a delay (if positive) or an advance (if negative) in the time region.
• Frequency Shift Property relates to how shift in frequency domain affects the time domain, portrayed by the equation $$L^{-1}\{F(s+a)\} = e^{at}f(t)$$. Here, 'a' is the frequency shift: A positive 'a' shows a growth in frequency leading to rapid oscillations, while a negative 'a' shows a decay leading to diminishing oscillations.
• Applications of Inverse Laplace Transforms are prominent in Engineering fields such as Environmental, Mechanical, Electrical, Civil, and Biomedical. For instance, in Environmental Engineering, it is essential for modeling contaminant transportation in groundwater, and in Mechanical Engineering for analysis of vibrational systems and feedback control systems.
• An Inverse Laplace Transform table is a critical reference tool including three primary components: Laplace Transform pairs (function in time domain and corresponding Laplace Transform in 's' domain), equations in 's' domain (Laplace Transforms of different functions), and corresponding equations in the time domain (original function from which Laplace transform is derived).
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How can I find the inverse Laplace transform?
To find the inverse Laplace transform, you need to match the given Laplace function with the standard Laplace transforms. If it doesn't match any standard form, break it down into simpler parts. Then, use the corresponding inverse transforms of the standard forms to find the original function.
What is the inverse Laplace transform?
The inverse Laplace transform is a mathematical operation used in engineering to recover a function from its Laplace transform. It transforms a function of a real variable "s" back to a function of a time variable "t".
Is the inverse Laplace transform linear?
Yes, the inverse Laplace transform is linear. This means it maintains the properties of superposition and homogeneity. This linearity allows the separate transformation of individual terms in an equation.
What is the inverse Laplace transform of a constant?
The inverse Laplace transform of a constant, say 'a', is a unit step function scaled by the constant. It is given as a*u(t), where u(t) is the unit step function and 't' is time.
What is an example of an inverse Laplace transform?
The inverse Laplace Transform of F(s) = 1/s is f(t) = 1 for t ≥ 0. This is because the Laplace Transform of a constant function '1' is 1/s. Therefore, by definition, the inverse Laplace Transform of 1/s is 1.
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What is the significance of the Region of Convergence (ROC) in the Inverse Laplace Transform?
Why is Inverse Laplace Transform important in electrical engineering?
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• Checked by StudySmarter Editorial Team | 5,217 | 26,321 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.90461 |
https://www.physicsforums.com/threads/describe-square-root-of-zero.259299/ | 1,511,242,217,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806316.80/warc/CC-MAIN-20171121040104-20171121060104-00662.warc.gz | 866,263,158 | 13,850 | # Describe Square Root of Zero
1. Sep 25, 2008
### eric7521
1. describe the square root of 0 from the following perspectives:
A. Algebraic
B. Complex Numbers
C. Limits
2. Sep 26, 2008
### Gib Z
A) Zero.
b) Zero.
c) $$\lim_{x\to 0^+} \sqrt{x} = 0$$, $$\lim_{x\to 0^-} \sqrt{x}$$ does not exist. | 117 | 299 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-47 | longest | en | 0.514528 |
https://dfi-geophysics-tool.org/bssDeckStability.htm | 1,723,455,655,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036895.73/warc/CC-MAIN-20240812092946-20240812122946-00829.warc.gz | 157,325,187 | 5,850 | Deck Stability Analysis
This section discusses periodic and permanent monitoring of bridge decks to assess their stability.
The stability of bridge decks has been of concern to designers and owners ever since the infamous destruction of the Tacoma Narrows Bridge in 1940 (figure 64) under moderate winds. Geophysical instruments, including accelerometers and tilt meters, as well as strain gages, are increasingly used for short and long term monitoring of bridge decks. Issues regarding stability monitoring and methods used will be discussed here. There is additional information on vibration monitoring provided in this web manual under Vibration Measurements.
Figure 64. Collapse of Tacoma Narrows Bridge. (Ed Elliott, The Camera Shop, Tacoma, WA)
The basic problems considered here are summarized briefly as follows:
1. The aerodynamic stability of bridge decks when subject to high winds.
2. The long-term stability of bridge decks due to fatigue caused by a variety of cyclic stresses including temperature, increased loading from traffic, and heavily loaded trucks. (See http://www.pubs.asce.org/ceonline/1098feat.html and http://www.kinemetrics.com/newprojects.html#newprojects.bangkok)
3. The stability of bridge decks during earthquakes. (See http://www.kinemetrics.com/halkis.html)
The primary method used to evaluate stability is to measure the baseline performance characteristics of a structure and then monitor the change in these characteristics with time. This can be done periodically, such as during an inspection cycle, or continuously using a permanent monitoring system.
The geophysical techniques applied to this problem are principally vibration monitoring, including strains, displacements, rotations, and accelerations, but also some environmental monitoring including wind and temperature, since these affect the response.
Initial Monitoring or Structural Verification
The purpose of initial monitoring is first to measure the as-built structural modal characteristics and use these data to evaluate the accuracy of the designer's model, and through this, the adequacy of the design. The second purpose is to measure the baseline performance to monitor stability over time. These data can also be used to assist in modeling older structures for the purpose of better retrofitting and upgrading.
Advantages: Better model accuracy means better predictions of performance under extreme loads and lower cost because structural upgrades can be planned more accurately.
Baseline performance is vital to long-term monitoring.
Limitations: None. This step is highly recommended for significant bridges.
Features: Two important questions are, "What characteristics should be monitored, and at how many locations?" For each bridge, this will depend on which structural elements the design engineer feels are most critical. Typical measurements include the following:
• Temperature and wind conditions during the study. Bridges are non-linear structures, and response characteristics can change under different conditions.
• Acceleration, both global movement, and individual elements. Sufficient channels are needed to measure simultaneous movements, with preferably at least 64 channels.
• Strains in critical structural elements under specific loads.
• Precise baseline locations (position or tilt) should be measured at critical locations.
The recording system should have the following minimum characteristics:
• Data should be recorded digitally with at least 16 bits of dynamic range.
• Sensors must be highly sensitive and capable of recording very small motions (0.005m/sec2 or less).
• Bandwidth will depend on the structure. For example, the fundamental mode of the Golden Gate Bridge has an 18-second period; therefore, sensors with DC (zero Hertz) response are required. Typically, DC to 50 Hz is recommended.
The recording system must have sufficient capacity to record all 64 channels for many minutes. This will provide statistical redundancy when analyzing the data.
Periodic Monitoring
The purpose of periodic monitoring is to evaluate long-term behavioral and performance changes through intermittent measurements, separated by months or years.
Advantages: The primary advantage is cost. A temporary monitoring system can be mobilized for the survey at any time. Long-term monitoring systems are expensive and require maintenance to stay in optimal condition.
Another advantage is flexibility. New parameters can be added to or removed from the system as needed during each survey.
The third advantage is the potential for improvement. As new instruments and methods are developed, these can be deployed for better monitoring.
Limitations: The primary limitation to periodic monitoring is lack of availability. Usually, the system is not available during peak motions, such as an earthquake or windstorm.
Periodic monitoring can lead to inconsistency. It might be difficult to repeat the same measurements using the same instruments.
Features>: Again, two important questions are, "What characteristics should be monitored, and at how many locations?" As with initial monitoring or structural verification, this will depend on which structural elements the design engineer feels are most critical. Typical measurements include the following:
• Temperature and wind conditions during each study. Bridges are non-linear structures, and response characteristics can change under different conditions.
• Strains in critical structural elements under specific loads. Even though measurements are temporary, permanent mounting locations should be established for periodic mounting of strain gages.
• Permanent displacements or rotations of piers and decks. These should be measured at critical locations.
Permanent Monitoring
The purpose of a permanent monitoring system is to monitor long-term behavioral and performance changes through continuous measurements. High-speed acquisition can be triggered by important events, such as an earthquake or high winds. Low-speed acquisition can be programmed for daily, weekly, or monthly measurements for long-term studies.
Advantages: The availability of the system is a primary advantage. The system is always available during peak motions, such as an earthquake or windstorm.
The consistency of the system is a second advantage. Measurements recorded today can be directly compared with measurements obtained 10 years ago.
Limitations: A critical limitation is cost. Long-term monitoring systems are expensive and require maintenance to stay in optimal condition.
As new instruments and methods are developed, the system may need to be upgraded to provide maximum utility.
Features: The same two important questions remain: "What characteristics should be monitored, and at how many locations?" Again, for each bridge this will depend on which structural elements the design engineer feels are most critical. Typical measurements include:
• Temperature and wind conditions. Bridges are non-linear structures, and response characteristics can change under different conditions.
• Strains in critical structural elements under specific loads.
• Permanent displacements or rotations of piers and decks. These should be measured at critical locations
• Triggered event recording is an important benefit of a permanent monitoring system. This is particularly true if the bridge is located in a seismic zone. The system must be able to turn on automatically and record the event. The sample rate for these channels must be adequate to record full response, to at least 100Hz.
Recommendations
Table 6 summarizes the recommendations discussed above.
Table 6. Bridge deck monitoring. | 1,370 | 7,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.925787 |
https://www.codespeedy.com/data-type-objects-in-numpy-python/ | 1,675,701,183,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500356.92/warc/CC-MAIN-20230206145603-20230206175603-00028.warc.gz | 718,853,857 | 15,493 | # Data Type objects in NumPy Python
Data Type Objects describe the interpretation of the bytes in the fixed-size block corresponding to an array. These objects are instances of Python’s numpy.dtype classIt mainly focuses on :
• Data Type (int, float, python object, etc.)
• Data size (number of bytes)
• The byte order of the data ()
• If data is sub-array, then the shape and datatype of it.
Remember that the data type objects and scalar types are not the same. Although scalar types can be used when there is a requirement of data type specification in NumPy.
### 1. Construct a data type object:
Data type object is an instance and can be constructed using NumPy library of Python. Syntax of data type object:
`numpy.dtype(object, align, copy)`
Significance of the parameter are:
• Object is the object to convert as the data-type object.
• Align (boolean): adds padding to the fields to make them comparative to C-struct if align is True.
• Copy (boolean): creates a new copy of a data type object if True. Otherwise, the output returns a built-in data type object reference on having a False value for the copy field.
In the example below, dtype function gives the data type of the object passed to it.
```# Demonstration
import numpy as np
# to convert np.int32 into a dtype object.
data_type = (np.dtype(np.int32))
print(data_type)```
```Output:
int32```
```# Program to construct a data type object
import numpy as np
# integer of size 8 bit represented as i8.
data_type = np.dtype('i8')
# Byte order of data type
print(data_type.byteorder)
# size of data type
print(data_type.itemsize)
#data type
print(data_type.name)```
```Output:
=
8
int64```
`1`
The functions byteorder and itemsize give the byte order and size of the data type respectively. In the above example, the type specifier is ‘i8’ that is equivalent to int64. Type specifiers can be of different forms such as:
```b1 : byte
i1, i2, i4, i8, etc : ints
u1, u2, u4, u8, etc : unsigned ints
f1, f2, f4, f8, etc : floats
c8, c16 : complex```
`1`
for example: i1 is int8, i2 is int16 and i4 is int 32, etc.
### 2. Create a structured array using data type object:
data-type object is used for creating a structured array. The structured array is also known as “Record Array”. This provides the ability to have each column with different data types.
```# Program to create a structured array using data type object
import numpy as np
employee = np.dtype([('name','S20'), ('age', 'i1'), ('salary', '>i4')])
print(employee)```
```Output:
[('name', 'S20'), ('age', 'i1'), ('salary', '>i4')]```
```# Program to create a structured array using data type object
import numpy as np
employee = np.dtype([('name','S20'), ('age', 'i1'), ('salary', '>i4')])
# struc_array is a structure array
struc_array = np.array([('Jim', 32, 1200000),('Jam', 39, 2000000)], dtype = employee)
print(struc_array)
print(struc_array[1])```
```Output:
[(b'Jim', 32, 1200000) (b'Jam', 39, 2000000)]
(b'Jam', 39, 2000000)``` | 795 | 2,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-06 | latest | en | 0.692464 |
https://www.physicsforums.com/threads/fermi-level-as-a-function-of-temperature.331710/ | 1,527,303,497,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867277.64/warc/CC-MAIN-20180526014543-20180526034543-00031.warc.gz | 821,385,544 | 18,412 | # Fermi level as a function of temperature
1. Aug 20, 2009
### TheDestroyer
Hello guys,
very short question :), why does Fermi level go down with temperature? is there some physical explanation or discussion for this? apart from mathematics (well, some mathematics would be cool if necessary
Thanks guys :)
#### Attached Files:
• ###### Fermi.jpg
File size:
6.4 KB
Views:
860
2. Aug 21, 2009
### TheDestroyer
Oh! come on guys! is it that mysterious? at least tell me why the question is that difficult?!?!?
3. Aug 21, 2009
### Mapes
The slope of the curve $(\partial \mu/\partial T)_P$ at any point is $-S$, the negative value of the entropy, and entropy is always positive.
4. Aug 21, 2009
### sokrates
What is your y-axis in that plot?
As far as I know, Fermi energy (Ef) does not move with respect to temperature. It stays at a constant value but for increasing temperature fermi function is broadened because fermions are thermally excited and therefore the probability of occupying higher energy states is increased (thus the broadening)
See this:
http://hyperphysics.phy-astr.gsu.edu/HBASE/solids/fermi.html
5. Aug 21, 2009
### TheDestroyer
This is not the answer pal, I understand the graph very well. I'm asking physically, I've just put the graph to illustrate the situation.
I got the answer, for who's interested, I thought it over and got something, if you think the answer is wrong tell me.
We're going to use the definition of chemical potential, that it's the energy required to add a particle to the system.
The Fermi distribution at T=0 is a step function, meaning if we want to add any other electron to the system we need to put it after those levels, meaning we need to insert it after the step function, and so the energy required to add this electron to the system is precisely EF(T=0) (Fermi energy at T=0).
http://www.doitpoms.ac.uk/tlplib/semiconductors/images/fermiDirac.jpg
At higher temperature, the distribution loses the step function profile (as we can see in the link), and therefore adding another electron is possible before EF(T=0) since there are vacancies.
At very high temperatures, kT becomes very large that it dominates over EF(T=0), and therefore without consuming any energy, we can find vacancies for new electrons as in the following figure
and so, continuing to increase temperature makes the system extract heat upon adding electron, in other words we get negative Fermi level or chemical potential. The heat extracted is because the system tries to sustain the Fermi distribution function.
I hope this is right, tell me what you think guys.
6. Aug 22, 2009
### TheDestroyer
7. Aug 22, 2009
### TheDestroyer
Hey guys! come on! is this physics forum or wheat farm!??!?!
8. Aug 22, 2009
### Mapes
I gave you the quantitative value and interpretation of the slope in my post #3 above. If your physical picture makes more sense to you, use it!
Note, though, that your description of the chemical potential is a little less precise than the standard definition, which is $(\partial E/\partial N)_{P,T}$, the infinitesimal increase in system energy when matter is added (usually at constant pressure and temperature). The differential definition is a bit more precise because only energy differences are meaningful (energy can only be measured relative to a reference value).
9. Aug 22, 2009
### TheDestroyer
But in the thermodynamic limit we can consider adding 1 particle a continuous event, right?
10. Aug 22, 2009
### Mapes
For individual particles, we necessarily have to replace $(\partial E/\partial N)_{P,T}$ by $(\Delta E/\Delta N)_{P,T}$, it being exceedingly difficult to add a fraction of an electron. But my point isn't with $\partial N$ vs. $\Delta N$, it's with making sure your description of the chemical potential specifies an energy change of a system. "Energy required" is vague because it's not clear who supplies this energy: the particle? the system? an external entity? Know what I mean?
11. Aug 22, 2009
### TheDestroyer
Well, is it that big problem where the energy will come from? we just give it, let's say as heat! isn't that good enough?
12. Aug 22, 2009
### olgranpappy
Hi TheDestroyer,
Maybe you would have got a better/quicker answer if you posted this in the Condensed Matter forum... then again, maybe not.
The standard explanation of the *decrease* of the chemical potential with temperature is that because the Fermi function becomes broader, and because there is typically a higher density of states at higher energy (think of the DOS for a free particle), the chemical potential must decrease in order to keep the total number of particles constant.
In general the chemical potential will not *always* decrease, but rather it will generally "be repelled" by the region of higher density of states--typically, as stated above, the DOS is higher at higher energies, but not always.
See, Callen's books for an explanation of this effect. Cheers,
13. Aug 22, 2009
### TheDestroyer
14. Aug 22, 2009
### olgranpappy
I don't think so... but your explanation is a little hard to follow.
Check out the reference I gave. Another good one which works out how the chemical potential depends on temperature in great detail is Ashcroft and Mermin's book "Solid State Physics"
15. Aug 23, 2009
Thanks :) | 1,274 | 5,329 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-22 | latest | en | 0.907048 |
https://studysoup.com/tsg/calculus/293/algebra-and-trigonometry/chapter/12766/3-5 | 1,603,401,252,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880038.27/warc/CC-MAIN-20201022195658-20201022225658-00432.warc.gz | 553,224,046 | 11,032 | ×
×
# Solutions for Chapter 3.5: One-to-One Functions and Inverse Functions
## Full solutions for Algebra and Trigonometry, | 3rd Edition
ISBN: 9780840068132
Solutions for Chapter 3.5: One-to-One Functions and Inverse Functions
Solutions for Chapter 3.5
4 5 0 335 Reviews
16
4
##### ISBN: 9780840068132
Algebra and Trigonometry, was written by and is associated to the ISBN: 9780840068132. Since 96 problems in chapter 3.5: One-to-One Functions and Inverse Functions have been answered, more than 45878 students have viewed full step-by-step solutions from this chapter. This textbook survival guide was created for the textbook: Algebra and Trigonometry,, edition: 3. Chapter 3.5: One-to-One Functions and Inverse Functions includes 96 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions.
Key Calculus Terms and definitions covered in this textbook
• Algebraic model
An equation that relates variable quantities associated with phenomena being studied
• Basic logistic function
The function ƒ(x) = 1 / 1 + e-x
• Coefficient of determination
The number r2 or R2 that measures how well a regression curve fits the data
• Common ratio
See Geometric sequence.
• Dependent event
An event whose probability depends on another event already occurring
• Equation
A statement of equality between two expressions.
• Fitting a line or curve to data
Finding a line or curve that comes close to passing through all the points in a scatter plot.
• Horizontal component
See Component form of a vector.
• Horizontal shrink or stretch
See Shrink, stretch.
• Irrational zeros
Zeros of a function that are irrational numbers.
• n factorial
For any positive integer n, n factorial is n! = n.(n - 1) . (n - 2) .... .3.2.1; zero factorial is 0! = 1
• Negative angle
Angle generated by clockwise rotation.
• Position vector of the point (a, b)
The vector <a,b>.
• Reference triangle
For an angle ? in standard position, a reference triangle is a triangle formed by the terminal side of angle ?, the x-axis, and a perpendicular dropped from a point on the terminal side to the x-axis. The angle in a reference triangle at the origin is the reference angle
• Sample space
Set of all possible outcomes of an experiment.
• Series
A finite or infinite sum of terms.
• Standard deviation
A measure of how a data set is spread
• Standard form of a complex number
a + bi, where a and b are real numbers
A graph in which (-x, -y) is on the the graph whenever (x, y) is; or a graph in which (-r, ?) or (r, ? + ?) is on the graph whenever (r, ?) is
• Unit vector in the direction of a vector
A unit vector that has the same direction as the given vector.
× | 664 | 2,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-45 | latest | en | 0.879385 |
https://www.aqua-calc.com/one-to-one/density/gram-per-metric-tablespoon/tonne-per-metric-teaspoon/1 | 1,570,998,407,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986647517.11/warc/CC-MAIN-20191013195541-20191013222541-00262.warc.gz | 911,936,385 | 8,560 | # 1 gram per metric tablespoon in tonnes per metric teaspoon
## g/metric tbsp to t/metric tsp unit converter of density
1 gram per metric tablespoon [g/metric tbsp] = 3 × 10-7 tonne per metric teaspoon [t/metric tsp]
### grams per metric tablespoon to tonnes per metric teaspoon density conversion cards
• 1
through
25
grams per metric tablespoon
• 1 g/metric tbsp to t/metric tsp = 3 × 10-7 t/metric tsp
• 2 g/metric tbsp to t/metric tsp = 7 × 10-7 t/metric tsp
• 3 g/metric tbsp to t/metric tsp = 1 × 10-6 t/metric tsp
• 4 g/metric tbsp to t/metric tsp = 1.3 × 10-6 t/metric tsp
• 5 g/metric tbsp to t/metric tsp = 1.7 × 10-6 t/metric tsp
• 6 g/metric tbsp to t/metric tsp = 2 × 10-6 t/metric tsp
• 7 g/metric tbsp to t/metric tsp = 2.3 × 10-6 t/metric tsp
• 8 g/metric tbsp to t/metric tsp = 2.7 × 10-6 t/metric tsp
• 9 g/metric tbsp to t/metric tsp = 3 × 10-6 t/metric tsp
• 10 g/metric tbsp to t/metric tsp = 3.3 × 10-6 t/metric tsp
• 11 g/metric tbsp to t/metric tsp = 3.7 × 10-6 t/metric tsp
• 12 g/metric tbsp to t/metric tsp = 4 × 10-6 t/metric tsp
• 13 g/metric tbsp to t/metric tsp = 4.3 × 10-6 t/metric tsp
• 14 g/metric tbsp to t/metric tsp = 4.7 × 10-6 t/metric tsp
• 15 g/metric tbsp to t/metric tsp = 5 × 10-6 t/metric tsp
• 16 g/metric tbsp to t/metric tsp = 5.3 × 10-6 t/metric tsp
• 17 g/metric tbsp to t/metric tsp = 5.7 × 10-6 t/metric tsp
• 18 g/metric tbsp to t/metric tsp = 6 × 10-6 t/metric tsp
• 19 g/metric tbsp to t/metric tsp = 6.3 × 10-6 t/metric tsp
• 20 g/metric tbsp to t/metric tsp = 6.7 × 10-6 t/metric tsp
• 21 g/metric tbsp to t/metric tsp = 7 × 10-6 t/metric tsp
• 22 g/metric tbsp to t/metric tsp = 7.3 × 10-6 t/metric tsp
• 23 g/metric tbsp to t/metric tsp = 7.7 × 10-6 t/metric tsp
• 24 g/metric tbsp to t/metric tsp = 8 × 10-6 t/metric tsp
• 25 g/metric tbsp to t/metric tsp = 8.3 × 10-6 t/metric tsp
• 26
through
50
grams per metric tablespoon
• 26 g/metric tbsp to t/metric tsp = 8.7 × 10-6 t/metric tsp
• 27 g/metric tbsp to t/metric tsp = 9 × 10-6 t/metric tsp
• 28 g/metric tbsp to t/metric tsp = 9.3 × 10-6 t/metric tsp
• 29 g/metric tbsp to t/metric tsp = 9.7 × 10-6 t/metric tsp
• 30 g/metric tbsp to t/metric tsp = 1 × 10-5 t/metric tsp
• 31 g/metric tbsp to t/metric tsp = 1.03 × 10-5 t/metric tsp
• 32 g/metric tbsp to t/metric tsp = 1.07 × 10-5 t/metric tsp
• 33 g/metric tbsp to t/metric tsp = 1.1 × 10-5 t/metric tsp
• 34 g/metric tbsp to t/metric tsp = 1.13 × 10-5 t/metric tsp
• 35 g/metric tbsp to t/metric tsp = 1.17 × 10-5 t/metric tsp
• 36 g/metric tbsp to t/metric tsp = 1.2 × 10-5 t/metric tsp
• 37 g/metric tbsp to t/metric tsp = 1.23 × 10-5 t/metric tsp
• 38 g/metric tbsp to t/metric tsp = 1.27 × 10-5 t/metric tsp
• 39 g/metric tbsp to t/metric tsp = 1.3 × 10-5 t/metric tsp
• 40 g/metric tbsp to t/metric tsp = 1.33 × 10-5 t/metric tsp
• 41 g/metric tbsp to t/metric tsp = 1.37 × 10-5 t/metric tsp
• 42 g/metric tbsp to t/metric tsp = 1.4 × 10-5 t/metric tsp
• 43 g/metric tbsp to t/metric tsp = 1.43 × 10-5 t/metric tsp
• 44 g/metric tbsp to t/metric tsp = 1.47 × 10-5 t/metric tsp
• 45 g/metric tbsp to t/metric tsp = 1.5 × 10-5 t/metric tsp
• 46 g/metric tbsp to t/metric tsp = 1.53 × 10-5 t/metric tsp
• 47 g/metric tbsp to t/metric tsp = 1.57 × 10-5 t/metric tsp
• 48 g/metric tbsp to t/metric tsp = 1.6 × 10-5 t/metric tsp
• 49 g/metric tbsp to t/metric tsp = 1.63 × 10-5 t/metric tsp
• 50 g/metric tbsp to t/metric tsp = 1.67 × 10-5 t/metric tsp
• 51
through
75
grams per metric tablespoon
• 51 g/metric tbsp to t/metric tsp = 1.7 × 10-5 t/metric tsp
• 52 g/metric tbsp to t/metric tsp = 1.73 × 10-5 t/metric tsp
• 53 g/metric tbsp to t/metric tsp = 1.77 × 10-5 t/metric tsp
• 54 g/metric tbsp to t/metric tsp = 1.8 × 10-5 t/metric tsp
• 55 g/metric tbsp to t/metric tsp = 1.83 × 10-5 t/metric tsp
• 56 g/metric tbsp to t/metric tsp = 1.87 × 10-5 t/metric tsp
• 57 g/metric tbsp to t/metric tsp = 1.9 × 10-5 t/metric tsp
• 58 g/metric tbsp to t/metric tsp = 1.93 × 10-5 t/metric tsp
• 59 g/metric tbsp to t/metric tsp = 1.97 × 10-5 t/metric tsp
• 60 g/metric tbsp to t/metric tsp = 2 × 10-5 t/metric tsp
• 61 g/metric tbsp to t/metric tsp = 2.03 × 10-5 t/metric tsp
• 62 g/metric tbsp to t/metric tsp = 2.07 × 10-5 t/metric tsp
• 63 g/metric tbsp to t/metric tsp = 2.1 × 10-5 t/metric tsp
• 64 g/metric tbsp to t/metric tsp = 2.13 × 10-5 t/metric tsp
• 65 g/metric tbsp to t/metric tsp = 2.17 × 10-5 t/metric tsp
• 66 g/metric tbsp to t/metric tsp = 2.2 × 10-5 t/metric tsp
• 67 g/metric tbsp to t/metric tsp = 2.23 × 10-5 t/metric tsp
• 68 g/metric tbsp to t/metric tsp = 2.27 × 10-5 t/metric tsp
• 69 g/metric tbsp to t/metric tsp = 2.3 × 10-5 t/metric tsp
• 70 g/metric tbsp to t/metric tsp = 2.33 × 10-5 t/metric tsp
• 71 g/metric tbsp to t/metric tsp = 2.37 × 10-5 t/metric tsp
• 72 g/metric tbsp to t/metric tsp = 2.4 × 10-5 t/metric tsp
• 73 g/metric tbsp to t/metric tsp = 2.43 × 10-5 t/metric tsp
• 74 g/metric tbsp to t/metric tsp = 2.47 × 10-5 t/metric tsp
• 75 g/metric tbsp to t/metric tsp = 2.5 × 10-5 t/metric tsp
• 76
through
100
grams per metric tablespoon
• 76 g/metric tbsp to t/metric tsp = 2.53 × 10-5 t/metric tsp
• 77 g/metric tbsp to t/metric tsp = 2.57 × 10-5 t/metric tsp
• 78 g/metric tbsp to t/metric tsp = 2.6 × 10-5 t/metric tsp
• 79 g/metric tbsp to t/metric tsp = 2.63 × 10-5 t/metric tsp
• 80 g/metric tbsp to t/metric tsp = 2.67 × 10-5 t/metric tsp
• 81 g/metric tbsp to t/metric tsp = 2.7 × 10-5 t/metric tsp
• 82 g/metric tbsp to t/metric tsp = 2.73 × 10-5 t/metric tsp
• 83 g/metric tbsp to t/metric tsp = 2.77 × 10-5 t/metric tsp
• 84 g/metric tbsp to t/metric tsp = 2.8 × 10-5 t/metric tsp
• 85 g/metric tbsp to t/metric tsp = 2.83 × 10-5 t/metric tsp
• 86 g/metric tbsp to t/metric tsp = 2.87 × 10-5 t/metric tsp
• 87 g/metric tbsp to t/metric tsp = 2.9 × 10-5 t/metric tsp
• 88 g/metric tbsp to t/metric tsp = 2.93 × 10-5 t/metric tsp
• 89 g/metric tbsp to t/metric tsp = 2.97 × 10-5 t/metric tsp
• 90 g/metric tbsp to t/metric tsp = 3 × 10-5 t/metric tsp
• 91 g/metric tbsp to t/metric tsp = 3.03 × 10-5 t/metric tsp
• 92 g/metric tbsp to t/metric tsp = 3.07 × 10-5 t/metric tsp
• 93 g/metric tbsp to t/metric tsp = 3.1 × 10-5 t/metric tsp
• 94 g/metric tbsp to t/metric tsp = 3.13 × 10-5 t/metric tsp
• 95 g/metric tbsp to t/metric tsp = 3.17 × 10-5 t/metric tsp
• 96 g/metric tbsp to t/metric tsp = 3.2 × 10-5 t/metric tsp
• 97 g/metric tbsp to t/metric tsp = 3.23 × 10-5 t/metric tsp
• 98 g/metric tbsp to t/metric tsp = 3.27 × 10-5 t/metric tsp
• 99 g/metric tbsp to t/metric tsp = 3.3 × 10-5 t/metric tsp
• 100 g/metric tbsp to t/metric tsp = 3.33 × 10-5 t/metric tsp
#### Foods, Nutrients and Calories
BACON JERKY, UPC: 049022805309 contain(s) 393 calories per 100 grams or ≈3.527 ounces [ price ]
GREEK FROZEN YOGURT ICE CREAM, UPC: 079893093215 weigh(s) 213.45 gram per (metric cup) or 7.13 ounce per (US cup), and contain(s) 168 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
#### Gravels, Substances and Oils
Substrate, Flourite weighs 1 005 kg/m³ (62.7401 lb/ft³) with specific gravity of 1.005 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Chromium ore weighs 2 162 kg/m³ (134.96925 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-422A, liquid (R422A) with temperature in the range of -40°C (-40°F) to 60°C (140°F)
#### Weights and Measurements
A kilowatt hour per mile (kWh/mi) measures how many kilowatt hours of electricity a car consumes to travel distance of 1 mile;
A force that acts upon an object can cause the acceleration of the object.
kg/mm² to lb/in² conversion table, kg/mm² to lb/in² unit converter or convert between all units of surface density measurement.
#### Calculators
Volume to Weight conversions for sands, gravels and substrates | 3,280 | 8,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-43 | latest | en | 0.328766 |
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# please review my gmat study approach and answer a wild q
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18 Sep 2007, 20:37
Hey everybody, Im new here and wanted to consult with you GMAT gurus regarding my approach. there are two things I wantd to ask
1) My appointment is on the 29th Oct. I took the GMAT once before, but I went in w/o any study at all, scored 540. 34q 31v
I finished Princeton Review Cracking the GMAT, I also finished a Manhattan Prep Book for geometry (formerly a week area).
Until the test I plan to complete the OG as well as the GMAT 800 book, I will aslo be finishing 5-6 online sample GMATs and the PowerPrep SOftware GMATS.
How does this approach sound? I am aiming for 700 +.
2)I scored 95% or better on all sections of the OG diagnostic test but got 560 on powerprep when I tried it two days ago, what the heck does this mean? how could this happen ?
thank you very much!
Viktor
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20 Sep 2007, 12:10
The diagnostic test in the OG, like all other paper tests, is basically worthless when it comes to assessing your probable score on the real thing.
Paper tests aren't adaptive and they don't put you on the clock. It may not be what you want to hear, but your 560 is the best indicator you have of what you're likely to score on the GMAT if you took it tomorrow.
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21 Sep 2007, 07:16
jeffsackmann wrote:
The diagnostic test in the OG, like all other paper tests, is basically worthless when it comes to assessing your probable score on the real thing.
Paper tests aren't adaptive and they don't put you on the clock. It may not be what you want to hear, but your 560 is the best indicator you have of what you're likely to score on the GMAT if you took it tomorrow.
--
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i tried to simulate the adaptive format by flipping and doing random OG book excercises, from all area tough and easy, and got above 95% right...but the thing is these were untimed, so I know how to do them but it takes too long...
i understand 560 is the score i would get under the same conditions that i took it under, but maybe I can i mprove by adjusting to these conditions during the next month= learn to do them in less time ??? or gain some skills/knowledge in a month
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22 Sep 2007, 13:30
You should be taking CATs, not trying to simulate CATs by flipping through the OG. It just isn't the same. Did you take the tests in the PR book? How did you score? Have you taken the GMATPrep tests yet?
I would plan on taking those 6 tests over the next few weeks. Carefully review your answers and then work on weak areas.
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22 Sep 2007, 18:38
Calamity Jane wrote:
You should be taking CATs, not trying to simulate CATs by flipping through the OG. It just isn't the same. Did you take the tests in the PR book? How did you score? Have you taken the GMATPrep tests yet?
I would plan on taking those 6 tests over the next few weeks. Carefully review your answers and then work on weak areas.
I took PR tests and got over 90% right, I took gmat prep as stated above and got 560 mysteriously
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22 Sep 2007, 20:45
viktorv wrote:
I took PR tests and got over 90% right, I took gmat prep as stated above and got 560 mysteriously
Its ALL in the timing Viktor. There are a bunch of posters on these forums (including me) who have 80% - 90% accuracy when doing untimed practice. We had some threads discussing these issues also.
22 Sep 2007, 20:45
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,487 | 5,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-22 | latest | en | 0.915106 |
https://www.physicsforums.com/threads/complexification-of-subspace.424973/ | 1,521,771,298,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648113.87/warc/CC-MAIN-20180323004957-20180323024957-00303.warc.gz | 827,456,195 | 16,012 | # Complexification of subspace
1. Aug 28, 2010
### julian
Let F be a subspace of a real vector space V and let
G \subset V_C
i.e. a subspace of its complexification. Define the real subspace of G by
G_R := G \cap V.
There is a symplectic form w[u,v]. The annihilator subspace F^perp of V is defined by
F^perp = {v \in V : w[u,v] = 0 for all u \in F}
The annihilator subspace G^perp of V_C is defined by
G^perp = {v \in V_C : w[u,v] = 0 for all u \in G}.
Then the following results hold:
(F^perp)_C = (F_C)^perp,
It is easy to see that (F^perp)_C subset of (F_C)^perp. Could somone prove it the other way? I'm thinking F_C is larger than F so the annihilator condition is more restrictive?
thanks
2. Aug 29, 2010
### lanedance
had a quick look & think it should work both ways
$$\Rightarrow$$
$$p \in F^{\perp}$$
$$p_c \in (F^{\perp})_c, \ \ p_c = p_1 + ip_2, \ \ p_1, p_2 \in F^{\perp}$$
now take
$$f_c \in F_c, \ \ f_c = f_1 + i f_2, \ \ f_1, f_2 \in F$$
$$w[p_c,f_c] = w[ f_1 + if_2, p_1 + ip_2 ] = 0$$
so
$$p_c \in (F_c)^{\perp}$$
3. Aug 29, 2010
### lanedance
and the other
$$\Leftarrow$$
$$f_c \in F_c, \ \ f_c = f_1 + i f_2, \ \ f_1, f_2 \in F$$
$$q \in (F_c)^{\perp}, \ \ w[f_c,q] = 0 , \ \ \forall f_c \in F_c$$
$$q \in V_c, \ \ q = q_1 + iq_2, \ \ q_1,q_2 \in V$$
$$w[q,f_c]=0 \ \implies \ w[q_1, f_1] = 0 \ \implies \ q_1 \in F^{\perp}$$
Last edited: Aug 29, 2010
4. Aug 29, 2010
### julian
Thank you.
to prove the very last part do you put f_1=f_2 which implies from real and imaginary parts of w [q,f_c] are
w [q_1 + q_2 , f_1] =0
w[q_1 - q_2, f_1] = 0
w[q_1,f_1] = 0 ?
5. Aug 29, 2010
### lanedance
no i'ev assumed w is a linear operator
w[q_1 + i q_2, f_1] = w[q_1, f_1] + w[i q_2, f_1] = 0
so it follows that each part must be zero (though you may need to be careful with conjugates)
6. Aug 30, 2010
### julian
I'm not sure you can put f_2 = 0 - it's not necessarily an element of F
7. Aug 30, 2010
### lanedance
F is a subspace, so must contain the zero vector by def'n
8. Aug 31, 2010
### julian
Are you assuming the subspace is a Vector subspace?
thanks
9. Aug 31, 2010
### lanedance
yeah thats generally what's meant by a subspace of a vector space...
Last edited: Aug 31, 2010
10. Sep 15, 2010
### julian
A subspace will be a vector space if it is closed under addition and multiplication by the field, but what if these conditions are not imposed?
I have another question (I dont know if it has a relation to the above). I mentioned a G at the begining. In the book he also says that if G=G^* then (G_R)_C. I know that we cannot have \$z w = w^* in general and maybe that has something to do with it.
11. Sep 15, 2010
### julian
sorry I meant
(G_R)_C = G | 1,049 | 2,735 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-13 | longest | en | 0.815165 |
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7
1.
Gr (Ts T )
4 1
(c) A = 1 4
Nu (Ts T )
1/ 4
Now A l I = 0, where l = eigen value
4 l
1
N u2
1
=0
4 l
N u1
N u2 =
(4 l ) 2 1 = 0
(4 l ) 2 (1) 2 = 0
2.
3.
4.
(4 l 1)(4 l + 1) = 0
\ l = 3, l = 5
(a) A line integral and a surface integral is connected by
stokes theorem.
(d) At the time of intake, intake valve operates only when
the pressure inside the cylinder is lower than the
atmospheric pressure. And appreciable large pressure
difference at the time of suction starting, causing
fluttering at initial.
At the time of compression, process followed is
polytorpic and continues till pressure exceeds the
delivery line pressure for poening the delivery valve.
(b) Potential energy at A = mg ( l d )
8.
9.
10.
5.
(c)
6.
(c)
s = ut +
1 2
gt
2
h1=
1
1 2
gt1 , and h2 = gt 22
2
2
1/2
t1 h1
Hence =
t2 h2
Apparently ratio of times taken is independent of the
mass of bodies.
1
7.
(c)
Gr 4
Nu =
( Pr )
4
(c)
(c)
(12735.48 N)
Free expansion, D = aa t La + as t Ls
= 11 106 20 1000 + 12 106 20 3000 = 0.94
\
or
11.
PL s
PL a
+
A a E a As E s
(a)
1000
3000
P
+
= 0.94
600 70 1000 300 200 1000
P = 12735.48 N.
x < 1, hence it is underdamped vibration case.
\ Frequency of the system,
wd =
1 - x 2 . wn
= 1 - 0.64 100 = 60
12.
13.
14.
15.
16.
17.
18.
and u = 0
48
= 24
2
Here, Ea = 70 1000 N/mm2
and
Es = 200 1000 N/mm2
\ Change in energy
1
= kx 2 mgx.d
2
1/ 4
If P is support reaction, then D =
1 2
Total energy at B = mg l (d + x ) + kx
2
1
= mgl mg (d + x ) + kx 2 mgl + mgd .
2
30 20
=
180 20
19.
(a)
(c)
(a)
(a)
(b)
(c)
(b) Here,
tDBT = 40, tWBT = 20
Water sprayed at temperature = 42
Since,
twater spray > tDBT so heating and humidification.
(c) Degrees of freedom, m = 3 (n 1) 2j1 j2
where n = number of links
j1 = number of single degree of freedom
and j2 = number of two degree of freedom in the
problem
n = 5, j1 = 5, j2 = 0
\ m=2
8
20.
(b)
26.
27.
28.
21.
22.
23.
\ dpt < wbt
(c) Order : 2; Degree = 2
Only 2 rows in the determinant is interchanged. By
property, the determinant remain same with opposite sign.
1.214 (m)
Q = A1V1 = A2V2 (Continuity equation)
(V1 - V2 )2
(6.508 - 1.627)
=
= 1.214 m
2 9.81
2g
A 2nd order partial differential equation of form
2 y
+B
2 y
2 y
+C
+
xy
y 2
x
F(x, y, u, p, q) = 0
is parabolic if B2 4AC = 0
Given equation can be written as
2
Now, if S = S and m = 2m., then
2
Q
0.115
=
= 6.508 m /s
p
A1
0.32
4
24. (a)
1 S
2p m
where, S = stiffness of spring
m = mass suspended
N=
0.115
Q
=
= 1.627 m /s
p
A2
0.32
4
V2 =
V1 =
29.
30.
3 1 1
0 0 2
0 0 0
Rank (A) = 2
(d)
(b)
(c) dQ = du + dw
2000 = du + ( 5000 kJ)
\ du = 3000 kJ.
(a)
(a) Frequency of natural vibration,
N =
31.
(c)
32.
(d) C =
33.
34.
35.
(d)
(a)
1 S / 2 = 1 1 S = N
2p 2 m 2
2 p 2m
rRT = 1.4 .4210 1000 500
(4/8 lit/s)
Apply Bernoulies equation at (1) and (2), we get
p1 v12
p v2
+ z1 = 2 + 2 + z2 + h f
+
lg 2 g
lg 2 g
Neglecting the minor losses, we get
f
f f
+0
+ - = 0
t
x
x
t 2 t
B2 4AC = 02 4 1 0
=0
\ It is parabolic
Equation is elliptical if B2 4AC < 0
hyperbolic if B2 4AC > 0
2
+0
0 0 -3
25. A = 9 3 5
3 1 1
R1 R3
R2 R2 3R1
3 1 1
0 0 2
0 0 -3
R3 2R3 3R3
0 + 0 + 30 = 0 +
4 flv2 2
v2 2
+0+
d 2g
2g
v2 2 4 .04 1000 v2 2
+
2g
0.6 2 g
or
30 =
or
30 2 g = v2 2 [267.67]
or
v2 2 =
30 2 9.81
267.67
\ v2 = 1/ 48m / sec
Now discharge in pipe,
9
Q = Area velocity
p
p
2
= d 2 v2 = ( 0.6 ) 1.48
4
4
= 0.418 m3/sec or 418 lit/sec
36.
(c) Given: Pressure ratio,
1
1
+
2pr1hi 2pK st
Rt =
p2
=4
p1
ln
r2
1
+
r1 2pKG / am
ln
r3
1
+
-m
r2 2pr3 h0
T1 = 27C = 300K
g = 1.4
T3 = 1000 C = 1273.K
Power output, W = m c p (T3 - T4 ) - c p (T2 - T1 )
T p
Now, 2 = 2
T1 p1
p
= 2
p1
g-1
g
1.4 -1
1.4
= ( 4)
T2 = 446K
and T3 = p3
T4 p4
37.
g-1
g
r -1
r
p
= 2
p1
r -1
r
In
1
1
+
2 p 9 10 -3 150 2p 400
10
1
+
9 2p 0.75
20
1
+
10 2 p 20 10 -3 5
= 3.18 deg/watt
Hence heat loss per unit length of the pipe
In
ti - to
= R
t
1.4 -1
1.4
= ( 4)
150 - 27
= 38.679 W/m
3.18
Hence total heat loss in one hour per metre length of
the pipe
= 38.679 3600 J/sec-m
= 139.245 kJ/hour-m
Hence rate of steam condensation
=
\ T4 = 856.6 K
Now, 100 = m[1(1273 856.6) 1 (446 300)]
\ m = 0.36kJ/kg
(0.06244) Given, D1 = 20 2 1 = 18 mm
r1 = 9 mm, r2 = 10 mm
D2 = 20 mm
139.245
kg/hr
2230
= 0.06244 kg/hr-m
38. (d)
39. (b)
40. (b) X2 X + I = 0
Multiply by X1
X I + X1 = 0
\ X1 = I X
a
1
1 0
= 0 1 - 2
-a + a - 1 1 - a
hi = 150 W/m2 K
h0 = 5 W/m2 K
-1
1- a
= 2
a - a + 1 a
K st = 400 W/m K
K G / am = 0.075 W/mK
Critical thickness of insulation is given by,
rc =
41. (b)
K
h0
0.075
= .015m = 15 cm
=
5
Hence, r3 = rc + 5 = 20 mm
Total thermal resistance per unit length of the pipe
L{u(t a)}
=
e
0
42. (b)
-st
- as
u ( t - a ) dt = e
s
Newton Raphson formula
x n +1 = x n -
f ( xn )
f ( xn )
f(x) = x3 x2 + 4x 4 = 0
10
f' (x) = 3x2 2x + 4
x0 = 2
x1 = x 0 -
f ( x0 )
f (x0 )
50.
( 23 - 22 + 4 ( 2 ) - 4 )
= 2= 2-
48.
49.
(b)
(a) W0 = h0CV 150 = 35.54 kJ/kg
WB = h0CP 100 = 33.165 kJ/kg
(c) Total kinetic energy
= K.E. due to translation
+ K.E. due to rotation
1
1 mr 2 x
3 2
= mx& 2 +
= mx&
2
2 2 r
4
2
3( 2) - 2 (2) + 4
2
8
12
\ x1 =
4
3
43. (d) n ( s ) = 5c1 - x5c1 = 25
Let E be the went of picking one chip from each box such
that product of numbers on chips is even
\ n (E) = (2 5) + (3 3)
= 19
19
25
44. (a) Given (D2 4D + 4) y = 0
f(D) y = 0 where,
f(D) = D2 4D + 4
f(D) = 0
\ D2 4D + 4 = 0
(D 2)2 = 0
\ D = 2, 2
\ yc = (c1 + c2x) e2x
45. Solution : 3.5 m
\ Prob =
s10
51.
52.
Q
(B+ 2) 2
16.57 =
46.
(B+ 2) 2
\ B = 3.5 m
(c) Total discharge = 0.32 + 0.29 = 0.61 m3/s
Design discharge =
47. (c)
0 to 1
Q
0.61
= 0.76 m3/s
=
TF
0.8
3
1
&&& + k .2 xx& = 0
m.2 xx
4
2
3
mx&& + kx = 0
2
wn =
2k
3m
D gh f
2 L
0.30 848.3 9.81 10
2
3000
= 4.16 Pa
53.
(a) At r = 10 cm,
shear stress, t =
=
54.
55.
t0
r
R
4.16 0.10
= 2.773 Pa
0.15
(d)
(b)
(2 H : 1V)
500
(c) Wall shear stress, t0 =
s1 + Ds
C
Sf = H 0 c log w 0 1
1 + e0
s0
0.36
37.2 + Ds
log w
0.120 = 4
1 + 0.92
37.2
\ Ds = 16.57 kPa
d
d 3
1
(T + 0 ) = mx& 2 + kx2 = 0
dt
dt 4
2
= [2 + 2]r
= 4 (19.3 10) = 37.2 kPa
Final settlement,
k
x+
x=0
(b) \ &&
( 3m / 2 )
Ds =
1 2
kx
2
U=
GENERAL APTITUDE
56.
57.
58.
59.
60.
61.
62.
63.
64.
65.
(d)
(a)
(a)
(a)
(b)
(b)
(d)
(d)
(b)
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2011-01-06, 20:57 #155 bchaffin Sep 2010 Portland, OR 37110 Posts Done with all the 137Ks: 137106: added 51 terms (min 104 digits): i820 (size 113) = 2^4 * 29 * 31 * 2141 * c105 137120: added 51 terms (min 103 digits): i837 (size 110) = 2^2 * 3 * 7 * 2213 * 3797 * c101 137184: added 80 terms (min 101 digits): i1295 (size 110) = 2^2 * 3^2 * 7 * 337 * 9811 * c101 137232: added 71 terms (min 105 digits): i1008 (size 114) = 2^4 * 3 * 7^2 * 53 * 9391 * c105 137394: added 49 terms (min 101 digits): i2409 (size 118) = 2^3 * 3 * 5 * 13 * 401 * 1873493579 * c103 137478: added 30 terms (min 105 digits): i1213 (size 112) = 2^4 * 3 * 7 * 31 * 37 * 13441 * c103 137820: added 116 terms (min 100 digits): i695 (size 112) = 2 * 3 * 5 * c111 137940: added 38 terms (min 104 digits): i2397 (size 116) = 2^3 * 3 * 5^4 * 251 * c110 Taking 138K and 139K: 138144, 138426, 139056, 139386, 139650, 139854, 139992.
2011-01-07, 11:04 #156 RobertS Aug 2009 somewhere 197 Posts Done with (all currently reserved ones): 125514: i1403, sz110, 2^2 ! (C107, fully ECMed) 125928: i1254, sz110, 2^2*7 128136: i346, sz114, 2^3*3*5^2 128928: i1291, sz111, 2^3*3 128952: i933, sz115, 2^5*3*7 129168: i2357, sz110, 2*3 129540: i1161, sz110, 2^4*31 129696: i633, sz111, 2^4**3*5 129924: i2598, sz112, 2*3^4 133938: i2962, sz110, 2^5*7^2*19 Reserving rest of 13xxxx: 133248, 133560, 133728, 133992, 134832,
2011-01-07, 15:48 #157 EdH "Ed Hall" Dec 2009 Adirondack Mtns 5·11·61 Posts reserving 140742, 141540
2011-01-07, 17:28 #158 bchaffin Sep 2010 Portland, OR 7×53 Posts Done with all of 136K: 136020: added 38 terms (min 103 digits): i1489 (size 111) = 2^2 * 7 * c109 136062: added 51 terms (min 104 digits): i1141 (size 110) = 2^2 * 3 * 7 * c108 136356: added 54 terms (min 100 digits): i1042 (size 113) = 2^6 * 3 * 59 * c109 136416: added 107 terms (min 100 digits): i1322 (size 110) = 2 * 3 * 5 * 7 * 11 * 41 * 185993 * c100 136464: added 137 terms (min 101 digits): i714 (size 113) = 2 * 3 * 89 * c111 136500: added 237 terms (min 103 digits): i1361 (size 118) = 2^5 * 3^2 * 7 * 35917067 * c107 136584: added 59 terms (min 103 digits): i1491 (size 110) = 2^4 * 5 * 31 * 359 * c104 136776: added 112 terms (min 102 digits): i668 (size 112) = 2^3 * 5 * c110 136780: added 17 terms (min 105 digits): i540 (size 112) = 2^3 * 3^3 * 5 * 707191 * c103 Next I'll take all of 141K: 141008, 141036, 141102, 141408, 141444, 141480, 141486, 141720, 141816, 141822, 141864, 141930, 141966.
2011-01-08, 05:01 #159 EdH "Ed Hall" Dec 2009 Adirondack Mtns 5×11×61 Posts 138738 is finished: size 113, i1426, 24 * 32 * 52 * 31 * 881 * 229954753780051 * c90
2011-01-08, 21:21 #160 EdH "Ed Hall" Dec 2009 Adirondack Mtns D1B16 Posts 134310 is finished: size 110, i1439, 24 * 31 * 173 * c105 138960 is finished: size 114, i2009, 25 * 3 * 72 * 271 * 281 * 3539 * 1578223427747 * c90 139830 is finished: size 110, i3409, 22 * 3 * 7 * 61 * 14975717 * c99 The db is again refusing to accept factors reported via Aliqueit! reserving 142764
2011-01-09, 00:53 #161 bchaffin Sep 2010 Portland, OR 7·53 Posts Done with 138K and 139K: 138144: added 564 terms (min 65 digits): i1943 (size 111) = 2^2 * 3 * 5 * 7 * 23 * 109 * 463 * c102 138426: added 37 terms (min 103 digits): i2074 (size 115) = 2^3 * 3 * 5 * 19 * 83 * 227 * c108 139056: added 484 terms (min 99 digits): i2248 (size 116) = 2^2 * 3 * 5 * 23 * 6941749 * c106 139386: added 25 terms (min 105 digits): i932 (size 113) = 2^3 * 3 * 5 * 37 * 307 * 8737 * c103 139650: added 48 terms (min 104 digits): i883 (size 111) = 2^2 * 3 * 7 * 13 * 43797703 * c100 139854: added 130 terms (min 100 digits): i1419 (size 110) = 2 * 3 * 7 * 11 * c108 139992: added 92 terms (min 100 digits): i927 (size 110) = 2^3 * 3 * 7 * 43 * c106 I'll take all the 142Ks and 143Ks: 142050, 142080, 142092, 142140, 142248, 142302, 142362, 142446, 142452, 142584 143160, 143268, 143310, 143332, 143676, 143850 Last fiddled with by bchaffin on 2011-01-09 at 01:05
2011-01-09, 04:10 #162 EdH "Ed Hall" Dec 2009 Adirondack Mtns 335510 Posts 138372 is finished: size 111, i899, 2 * 32 * 5 * 509 * c106 reserving 144078
2011-01-10, 04:24 #163 EdH "Ed Hall" Dec 2009 Adirondack Mtns 5·11·61 Posts 140742 is finished: size 110, i596, 26 * 32 * 17 * 19 * 2399974411 * c95 141540 is finished: size 110, i1468, 23 * 32 * 52 * 331 * 773 * c101 reserving 140832 and 144522 Last fiddled with by EdH on 2011-01-10 at 04:35
2011-01-10, 17:04 #164 EdH "Ed Hall" Dec 2009 Adirondack Mtns 5·11·61 Posts 125376 is "finally" finished: size 110, i3437, 23 * 3 * 86209 * c103 I had to add over 1350 lines to get it from 100 to 110 digits. It had several downdriver runs; the most recent, as low as 28 digits. But, it finally gave in to the goal of this subproject... reserving 140688 Last fiddled with by EdH on 2011-01-10 at 17:13 Reason: added reservation
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All times are UTC. The time now is 08:37.
Thu Oct 1 08:37:12 UTC 2020 up 21 days, 5:48, 0 users, load averages: 1.61, 1.68, 1.63 | 2,385 | 5,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-40 | latest | en | 0.548288 |
https://ch.mathworks.com/matlabcentral/cody/problems/961-robust-alignment-of-coordinate-system/solutions/720562 | 1,604,111,125,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107912593.62/warc/CC-MAIN-20201031002758-20201031032758-00484.warc.gz | 223,715,841 | 17,194 | Cody
# Problem 961. Robust alignment of coordinate system
Solution 720562
Submitted on 24 Aug 2015
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
%% nocheat = isempty(regexp(evalc('type align_frame'),'(eval|regexprep|inline|str2func)')); x = {[1 0 0],'zyx'}; y_correct = [0;0;1]; assert(isequal(align_frame(x{:}),y_correct) && nocheat)
2 Fail
%% nocheat = isempty(regexp(evalc('type align_frame'),'(eval|regexprep|inline|str2func)')); x = {1:3,'XYZ'}; y_correct = [0.963624111659432 -0.148249863332220 -0.222374794998330].'; assert(isequal(align_frame(x{:}),y_correct) && nocheat)
3 Fail
%% nocheat = isempty(regexp(evalc('type align_frame'),'(eval|regexprep|inline|str2func)')); x = {[1;1;0],'xy'}; y_correct = null([1 0;1 0;0 1].'); assert(isequal(align_frame(x{:}),y_correct) && nocheat)
4 Fail
%% nocheat = isempty(regexp(evalc('type align_frame'),'(eval|regexprep|inline|str2func)')); x = {[0 0 -2],'Z'}; y_correct = []; assert(isequal(align_frame(x{:}),y_correct) && nocheat) | 364 | 1,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-45 | latest | en | 0.17199 |
http://mathhelpforum.com/geometry/83310-geo-trig-problem.html | 1,481,457,413,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544678.42/warc/CC-MAIN-20161202170904-00049-ip-10-31-129-80.ec2.internal.warc.gz | 172,064,150 | 10,121 | 1. ## Geo/trig Problem.
The ‘London Eye’ can be considered to be a circular frame of radius 67.5m on the
circumference of which are ‘capsules’ carrying a number of people round the circle. Take
a co-ordinate system where O is the base of the circle and Oy is a diameter. At any time
after starting off round the frame, the capsule will be at height h metres when it has
rotated .°θ
Thanks
2. Originally Posted by p4pri
The ‘London Eye’ can be considered to be a circular frame of radius 67.5m on the
circumference of which are ‘capsules’ carrying a number of people round the circle. Take
a co-ordinate system where O is the base of the circle and Oy is a diameter. At any time
after starting off round the frame, the capsule will be at height h metres when it has
rotated .°θ | 203 | 779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2016-50 | longest | en | 0.916068 |
https://open.kattis.com/problems/captainobvious | 1,656,221,074,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037089.4/warc/CC-MAIN-20220626040948-20220626070948-00502.warc.gz | 509,534,644 | 7,526 | Hide
# Captain Obvious and the Rabbit-Man
“It’s you, Captain Obvious!”—cried the evil Rabbit-Man—“you came here to foil my evil plans!”
“Yes, it’s me.”—said Captain Obvious.
“But…how did you know that I would be here, on 625 Sunflower Street?! Did you crack my evil code?”
“I did. Three days ago, you robbed a bank on 5 Sunflower Street, the next day you blew up 25 Sunflower Street, and yesterday you left quite a mess under number 125. These are all powers of 5. And last year you pulled a similar stunt with powers of 13. You seem to have a knack for Fibonacci numbers, Rabbit-Man.”
“That’s not over! I will learn…
arithmetics!”—Rabbit-Man screamed as he was dragged into custody—“You will never know what to expect... Owww! Not my ears, you morons!”
“Maybe, but right now you are being arrested.”—Captain added proudly.
Unfortunately, Rabbit-Man has now indeed learned some more advanced arithmetics. To understand it, let us define the sequence $F_ n$ (being not completely unlike the Fibonacci sequence):
\begin{align*} F_1 & = 1\\ F_2 & = 2\\ F_ n & = F_{n-1} + F_{n-2} \text { for $n \geq 3$.} \end{align*}
Rabbit-Man has combined all his previous evil ideas into one master plan. On the $i$-th day, he does a malicious act on the spot number $p(i)$, defined as follows:
\begin{equation*} p(i) = a_1 \cdot F_1^ i + a_2 \cdot F_2^ i + \cdots + a_ k \cdot F_ k^ i\; . \end{equation*}
The number $k$ and the integer coefficients $a_1, \ldots , a_ k$ are fixed. Captain Obvious learned $k$, but does not know the coefficients. Given $p(1), p(2), \ldots , p(k)$, help him to determine $p(k + 1)$. To avoid overwhelmingly large numbers, do all the calculations modulo a fixed prime number $M$. You may assume that $F_1 , F_2 , \ldots , F_ n$ are pairwise distinct modulo $M$. You may also assume that there always exists a unique solution for the given input.
## Input
The first line of input contains the number of test cases $T$. The descriptions of the test cases follow:
The first line of each test case contains two integers $k$ and $M$, $1 \leq k \leq 4\, 000$, $3 \leq M \leq 10^9$. The second line contains $k$ space-separated integers—the values of $p(1), p(2), \ldots , p(k)$ modulo $M$.
## Output
Print the answers to the test cases in the order in which they appear in the input. For each test case print a single line containing one integer: the value of $p(k + 1)$ modulo $M$.
Sample Input 1 Sample Output 1
2
4 619
5 25 125 6
3 101
5 11 29
30
83
CPU Time limit 9 seconds
Memory limit 1024 MB
Statistics Show | 750 | 2,539 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2022-27 | latest | en | 0.876192 |
http://www.caida.org/projects/ark/statistics/oua2-bf/dest_rtt_ccdf.html | 1,539,812,160,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511216.45/warc/CC-MAIN-20181017195553-20181017221053-00269.warc.gz | 422,065,314 | 3,772 | Center for Applied Internet Data Analysis
projects : ark : statistics
Archipelago (Ark): CAIDA's active measurement infrastructure serving the network research community since 2007.
| Ark Monitors: Hosting an Ark Monitor Locations Data Coverage Statistics |
# oua2-bf
Airtel Mobile
### CCDF of destination RTTs
• percentile 10th 25th 50th 75th 90th
RTT (ms) 144.744 184.710 267.207 390.396 440.200
Use the following link to download the data used to render this graph in ASCII, comma-separated values format here: (CSV output)
### Description
This graph shows the complementary cumulative distribution function (CCDF) of round-trip times (RTTs) to the destination host.
### Motivation
By showing the distribution of RTT values to all responding destinations, we can get a sense of how varied the speeds are for connecting to different points in the Internet.
### Background
The complementary cumulative distribution function shows the fraction of collected data points that are greater than a given value. This is backwards from how percentiles are given, as those show the percentage lower than a given value. On this graph, you would find the 80th percentile at the 0.2 Y value. The round trip time of a probe is the time (in milliseconds) that it takes for a packet to be sent from an Ark monitor to a destination and for that destination's response to be received by the monitor. Therefore, no RTT values are recorded when a probe does not reach a destination.
### Analysis
When the CCDF graph has a nearly vertical dropoff point, that indicates that RTT values fall within a narrow range. This tends to mean that a bottleneck exists within the monitor's connectivity that dominates over individual destination path variation. A more gradual curve, on the other hand, indicates greater variability in the response times of destinations, which tends to scale directly with the path length distribution. | 426 | 1,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-43 | latest | en | 0.822388 |
https://sampleprograms.io/projects/longest-palindromic-substring/php/ | 1,726,698,524,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00468.warc.gz | 458,377,058 | 4,582 | # Longest Palindromic Substring in Php
Published on 27 March 2023 (Updated: 27 March 2023)
Welcome to the Longest Palindromic Substring in Php page! Here, you'll find the source code for this program as well as a description of how the program works.
## Current Solution
``````<?php
// Find longest palindromic string using matching array
// Source: https://www.geeksforgeeks.org/longest-palindromic-substring-using-dynamic-programming/
function longest_palindromic_substring(\$str)
{
// Initialize array indicating whether there is a character match
// between two characters to indicate that nothing matches
\$n = strlen(\$str);
\$matches = array_fill(0, \$n, array_fill(0, \$n, FALSE));
// Indicate all length 1 strings match
for (\$i = 0; \$i < \$n; \$i++)
{
\$matches[\$i][\$i] = TRUE;
}
// Convert string to lowercase
\$temp_str = strtolower(\$str);
// Find all length 2 matches
\$start = 0;
\$max_len = 1;
for (\$i = 0; \$i < \$n - 1; \$i++)
{
if (\$temp_str[\$i] == \$temp_str[\$i + 1])
{
\$matches[\$i][\$i + 1] = TRUE;
if (\$max_len < 2)
{
\$start = \$i;
\$max_len = 2;
}
}
}
// Find all length 3 or higher matches
for (\$len = 3; \$len <= \$n; \$len++)
{
// Loop through each starting character
for (\$i = 0; \$i < \$n - \$len + 1; \$i++)
{
// If match for one character in from start and end characters
// and start and end characters match, set match for start and
// end characters, and update max length
\$j = \$i + \$len - 1;
if (\$matches[\$i + 1][\$j - 1] && \$temp_str[\$i] == \$temp_str[\$j])
{
\$matches[\$i][\$j] = TRUE;
if (\$len > \$max_len)
{
\$start = \$i;
\$max_len = \$len;
}
}
}
}
return substr(\$str, \$start, \$max_len);
}
function usage()
{
exit("Usage: please provide a string that contains at least one palindrome");
}
// Exit if 1st command-line argument is missing is empty
if (count(\$argv) < 2 || empty(\$argv[1]))
{
usage();
}
// Get longest palindromic substring. Exit if none found
\$str = \$argv[1];
\$longest = longest_palindromic_substring(\$str);
if (strlen(\$longest) < 2)
{
usage();
}
// Show longest palindromic substring
echo "\${longest}\n";
``````
Longest Palindromic Substring in Php was written by:
• rzuckerm
If you see anything you'd like to change or update, please consider contributing.
## How to Implement the Solution
No 'How to Implement the Solution' section available. Please consider contributing.
## How to Run the Solution
No 'How to Run the Solution' section available. Please consider contributing. | 735 | 2,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.395919 |
https://www.schooltube.com/createdby/eyJpdiI6InhLVmUzOWRPTGRzSHVFVmxJUGUrdmc9PSIsInZhbHVlIjoiVVBVVWd4ckxNMUJsU0JyWnJEQWdjZz09IiwibWFjIjoiNDI5YTNhYTQyYzcwYTFkNTViNzE3NDY2MWM2NWVkZDlkZWM4MTlhZjg4ZGNkYmU3ODU0YWRkYzA2YmJhMmM5YiJ9/sort/-name/keyword/Michael+Perry | 1,642,921,594,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304134.13/warc/CC-MAIN-20220123045449-20220123075449-00715.warc.gz | 1,020,392,679 | 24,707 | # Search for media from: "Michael Perry"
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From Michael Perry 0 likes 0 plays | 383 | 1,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-05 | latest | en | 0.791366 |
https://www.includehelp.com/cpp-programs/print-Lucas-series-upto-n-terms.aspx | 1,726,590,603,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00190.warc.gz | 736,952,176 | 36,493 | # C++ program to print Lucas series upto N terms
Here, we are going to implement a C++ program to print Lucas series upto N terms.
Submitted by Aditi S., on May 30, 2019
Given N and we have to print lucas series upto N terms.
## Lucas series
The Lucas series is an integer series very similar to the Fibonacci series, named after the French mathematician François Édouard Anatole Lucas. Each term of the Lucas series is defined as the sum of the previous two terms of the series with the first two terms being 2 and 1 respectively. The Lucas series and Fibonacci series are complementary to each other. The terms of the series are integer powers of the golden ratio rounded to the closest whole number. Given below is the code to find the Terms of the Lucas series up to n iterations.
Code
```/*Program to print the Lucas series for n terms.*/
#include <iostream>
using namespace std;
int main()
{
int n, i, t1 = 2, t2 = 1, tn;
cout << "Enter the number of terms desired in the lucas series: ";
cin >> n;
if (n == 1)
cout << endl << 2 << endl;
else if (n == 2)
cout << endl << 2 << endl << 1 << endl;
else if (n > 2)
{
cout <<endl<<"Lucas series for "<< n<< " terms is:"<<endl<< t1 << endl << t2 << endl;
for (i = 0; i < n-2; i++)
{
tn = t1 + t2;
cout << tn << endl;
t1 = t2;
t2 = tn;
}
}
return 0;
}
```
Output
```First run:
Enter the number of terms desired in the lucas series: 5
Lucas series for 5 terms is:
2
1
3
4
7
Second run:
Enter the number of terms desired in the lucas series: 10
Lucas series for 10 terms is:
2
1
3
4
7
11
18
29
47
76
``` | 452 | 1,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-38 | latest | en | 0.859544 |
http://www.enotes.com/homework-help/prove-that-iidentity-true-any-triangle-b-cosc-c-173331 | 1,477,378,267,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719960.60/warc/CC-MAIN-20161020183839-00100-ip-10-171-6-4.ec2.internal.warc.gz | 424,657,098 | 10,549 | Prove that the iidentity is true in any triangle: b*cosC+c*cosB=a
hala718 | High School Teacher | (Level 1) Educator Emeritus
Posted on
bcosC +c cosb = a
Since abc is a triangle, then we know that:
b^2 = a^2+c^2-2ac cos(b)
==> cosb=-(b^2-a^2-c^2)/2ac
and c^2 = a^2+b^2 -2ab cos(c)
==> cos (c)= -(c^2-a^2-b^2)ab
Now we need to prove that b cos(c)+ C cos(b) = a
then,
-b(c^2-a^2-b^2)/2ab -C (b^2-a^2-c^2)/2ac
= -(c^2 -a^2-b^2)/2a - (b^2-a^2-c^2)/2a
= (-c^2+a^2+b^2-b^2+a^2+c^2)/2a
= 2a^2/2a = a
neela | High School Teacher | (Level 3) Valedictorian
Posted on
Let ABC be the triangle. Draw a perpendicular from A to BC to meet BC at D. Then,
BC = a = BD+DC.
But BD = AB cos B from the right angled triangleABD . And DC = AC cos C from the right angled triangle ACD. Therefore,
BC = a = BD+DC = c*cos B+b*CosC. So the given equality is true for any triangle and is therefore an identity.
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
We'll prove this with the help from the cosine theorem:
b^2=c^2+a^2-2ac*cosB, cos B = (c^2+a^2-b^2)/2ac
c^2=a^2+b^2-2ab*cosC, cos C = (a^2+b^2-c^2)/2ab
Now, we'll substitute cos B and cos C, by the found expressions, so that:
b*cosC+c*cosB=a
b*(a^2+b^2-c^2)/2ab + c*(c^2+a^2-b^2)/2ac = a
After reducing the terms, we'll obtain the irreducible quotients:
(a^2+b^2-c^2)/2a + (c^2+a^2-b^2)/2a = a
We'll multiply the term from the right side, by 2a:
a^2+b^2-c^2+c^2+a^2-b^2 = 2a^2
After reducing similar terms, we'll obtain:
a^2+a^2 = 2a^2
2a^2 = 2a^2
The last result tells us that the identity, b*cosC+c*cosB=a, is valid for any triangle. | 659 | 1,626 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2016-44 | latest | en | 0.696778 |
https://www.numbersaplenty.com/5164 | 1,713,891,569,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818732.46/warc/CC-MAIN-20240423162023-20240423192023-00314.warc.gz | 831,007,755 | 3,061 | Search a number
5164 = 221291
BaseRepresentation
bin1010000101100
321002021
41100230
5131124
635524
721025
oct12054
97067
105164
113975
122ba4
132473
141c4c
1517e4
hex142c
• 51642 = 26666896 is the smallest square that contains exactly five digits '6'.
5164 has 6 divisors (see below), whose sum is σ = 9044. Its totient is φ = 2580.
The previous prime is 5153. The next prime is 5167. The reversal of 5164 is 4615.
It is an Ulam number.
It is not an unprimeable number, because it can be changed into a prime (5167) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (5) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 642 + ... + 649.
25164 is an apocalyptic number.
It is an amenable number.
5164 is a deficient number, since it is larger than the sum of its proper divisors (3880).
5164 is a wasteful number, since it uses less digits than its factorization.
5164 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1295 (or 1293 counting only the distinct ones).
The product of its digits is 120, while the sum is 16.
The square root of 5164 is about 71.8609768929. The cubic root of 5164 is about 17.2847092218.
Adding to 5164 its reverse (4615), we get a palindrome (9779).
It can be divided in two parts, 5 and 164, that multiplied together give a triangular number (820 = T40).
The spelling of 5164 in words is "five thousand, one hundred sixty-four".
Divisors: 1 2 4 1291 2582 5164 | 463 | 1,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-18 | latest | en | 0.881014 |
https://www.geeksforgeeks.org/system-linear-equations-three-variables-using-cramers-rule/?ref=leftbar-rightbar | 1,627,645,775,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.52/warc/CC-MAIN-20210730091645-20210730121645-00713.warc.gz | 786,168,742 | 29,780 | Related Articles
# System of Linear Equations in three variables using Cramer’s Rule
• Difficulty Level : Easy
• Last Updated : 11 May, 2021
Cramer’s rule: In linear algebra, Cramer’s rule is an explicit formula for the solution of a system of linear equations with as many equations as unknown variables. It expresses the solution in terms of the determinants of the coefficient matrix and of matrices obtained from it by replacing one column by the column vector of the right-hand-sides of the equations. Cramer’s rule is computationally inefficient for systems of more than two or three equations.
Suppose we have to solve these equations:
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
Following the Cramer’s Rule, first find the determinant values of all four matrices.
[Tex]D_1 = \begin{vmatrix} d_1 & b_1 & c_1\\ d_2 & b_2 & c_2\\ d_3 & b_3 & c_3\\ \end{vmatrix} [/Tex][Tex]D_3 = \begin{vmatrix} a_1 & b_1 & d_1\\ a_2 & b_2 & d_2\\ a_3 & b_3 & d_3\\ \end{vmatrix} [/Tex]
There are 2 cases:
Case I : When D ≠ 0 In this case we have,
x = D1/D
y = D2/D
z = D3/D
Hence unique value of x, y, z will be obtained.
Case II : When D = 0
(a) When at least one of D1, D2 and D3 is non zero: Then no solution is possible and hence system of equations will be inconsistent.
(b) When D = 0 and D1 = D2 = D3 = 0: Then the system of equations will be consistent and it will have infinitely many solutions.
Example
Consider the following system of linear equations.
[2x – y + 3z = 9], [x + y + z = 6], [x – y + z = 2]
[Tex]D_1 = \begin{vmatrix} 9 & -1 & 3\\ 6 & 1 & 1\\ 2 & -1 & 1\\ \end{vmatrix} [/Tex][Tex]D_3 = \begin{vmatrix} 2 & -1 & 9\\ 1 & 1 & 6\\ 1 & -1 & 2\\ \end{vmatrix} [/Tex]
[x = D1/D = 1], [y = D2/D = 2], [z = D3/D = 3]
Below is the implementation.
## C++
// CPP program to calculate solutions of linear// equations using cramer's rule#include using namespace std; // This functions finds the determinant of Matrixdouble determinantOfMatrix(double mat[3][3]){ double ans; ans = mat[0][0] * (mat[1][1] * mat[2][2] - mat[2][1] * mat[1][2]) - mat[0][1] * (mat[1][0] * mat[2][2] - mat[1][2] * mat[2][0]) + mat[0][2] * (mat[1][0] * mat[2][1] - mat[1][1] * mat[2][0]); return ans;} // This function finds the solution of system of// linear equations using cramer's rulevoid findSolution(double coeff[3][4]){ // Matrix d using coeff as given in cramer's rule double d[3][3] = { { coeff[0][0], coeff[0][1], coeff[0][2] }, { coeff[1][0], coeff[1][1], coeff[1][2] }, { coeff[2][0], coeff[2][1], coeff[2][2] }, }; // Matrix d1 using coeff as given in cramer's rule double d1[3][3] = { { coeff[0][3], coeff[0][1], coeff[0][2] }, { coeff[1][3], coeff[1][1], coeff[1][2] }, { coeff[2][3], coeff[2][1], coeff[2][2] }, }; // Matrix d2 using coeff as given in cramer's rule double d2[3][3] = { { coeff[0][0], coeff[0][3], coeff[0][2] }, { coeff[1][0], coeff[1][3], coeff[1][2] }, { coeff[2][0], coeff[2][3], coeff[2][2] }, }; // Matrix d3 using coeff as given in cramer's rule double d3[3][3] = { { coeff[0][0], coeff[0][1], coeff[0][3] }, { coeff[1][0], coeff[1][1], coeff[1][3] }, { coeff[2][0], coeff[2][1], coeff[2][3] }, }; // Calculating Determinant of Matrices d, d1, d2, d3 double D = determinantOfMatrix(d); double D1 = determinantOfMatrix(d1); double D2 = determinantOfMatrix(d2); double D3 = determinantOfMatrix(d3); printf("D is : %lf \n", D); printf("D1 is : %lf \n", D1); printf("D2 is : %lf \n", D2); printf("D3 is : %lf \n", D3); // Case 1 if (D != 0) { // Coeff have a unique solution. Apply Cramer's Rule double x = D1 / D; double y = D2 / D; double z = D3 / D; // calculating z using cramer's rule printf("Value of x is : %lf\n", x); printf("Value of y is : %lf\n", y); printf("Value of z is : %lf\n", z); } // Case 2 else { if (D1 == 0 && D2 == 0 && D3 == 0) printf("Infinite solutions\n"); else if (D1 != 0 || D2 != 0 || D3 != 0) printf("No solutions\n"); }} // Driver Codeint main(){ // storing coefficients of linear equations in coeff matrix double coeff[3][4] = { { 2, -1, 3, 9 }, { 1, 1, 1, 6 }, { 1, -1, 1, 2 }, }; findSolution(coeff); return 0;}
## Java
// Java program to calculate solutions of linear// equations using cramer's ruleclass GFG{ // This functions finds the determinant of Matrixstatic double determinantOfMatrix(double mat[][]){ double ans; ans = mat[0][0] * (mat[1][1] * mat[2][2] - mat[2][1] * mat[1][2]) - mat[0][1] * (mat[1][0] * mat[2][2] - mat[1][2] * mat[2][0]) + mat[0][2] * (mat[1][0] * mat[2][1] - mat[1][1] * mat[2][0]); return ans;} // This function finds the solution of system of// linear equations using cramer's rulestatic void findSolution(double coeff[][]){ // Matrix d using coeff as given in cramer's rule double d[][] = { { coeff[0][0], coeff[0][1], coeff[0][2] }, { coeff[1][0], coeff[1][1], coeff[1][2] }, { coeff[2][0], coeff[2][1], coeff[2][2] }, }; // Matrix d1 using coeff as given in cramer's rule double d1[][] = { { coeff[0][3], coeff[0][1], coeff[0][2] }, { coeff[1][3], coeff[1][1], coeff[1][2] }, { coeff[2][3], coeff[2][1], coeff[2][2] }, }; // Matrix d2 using coeff as given in cramer's rule double d2[][] = { { coeff[0][0], coeff[0][3], coeff[0][2] }, { coeff[1][0], coeff[1][3], coeff[1][2] }, { coeff[2][0], coeff[2][3], coeff[2][2] }, }; // Matrix d3 using coeff as given in cramer's rule double d3[][] = { { coeff[0][0], coeff[0][1], coeff[0][3] }, { coeff[1][0], coeff[1][1], coeff[1][3] }, { coeff[2][0], coeff[2][1], coeff[2][3] }, }; // Calculating Determinant of Matrices d, d1, d2, d3 double D = determinantOfMatrix(d); double D1 = determinantOfMatrix(d1); double D2 = determinantOfMatrix(d2); double D3 = determinantOfMatrix(d3); System.out.printf("D is : %.6f \n", D); System.out.printf("D1 is : %.6f \n", D1); System.out.printf("D2 is : %.6f \n", D2); System.out.printf("D3 is : %.6f \n", D3); // Case 1 if (D != 0) { // Coeff have a unique solution. Apply Cramer's Rule double x = D1 / D; double y = D2 / D; double z = D3 / D; // calculating z using cramer's rule System.out.printf("Value of x is : %.6f\n", x); System.out.printf("Value of y is : %.6f\n", y); System.out.printf("Value of z is : %.6f\n", z); } // Case 2 else { if (D1 == 0 && D2 == 0 && D3 == 0) System.out.printf("Infinite solutions\n"); else if (D1 != 0 || D2 != 0 || D3 != 0) System.out.printf("No solutions\n"); }} // Driver Codepublic static void main(String[] args){ // storing coefficients of linear // equations in coeff matrix double coeff[][] = {{ 2, -1, 3, 9 }, { 1, 1, 1, 6 }, { 1, -1, 1, 2 }}; findSolution(coeff); }} // This code is contributed by PrinciRaj1992
## Python3
# Python3 program to calculate# solutions of linear equations# using cramer's rule # This functions finds the# determinant of Matrixdef determinantOfMatrix(mat): ans = (mat[0][0] * (mat[1][1] * mat[2][2] - mat[2][1] * mat[1][2]) - mat[0][1] * (mat[1][0] * mat[2][2] - mat[1][2] * mat[2][0]) + mat[0][2] * (mat[1][0] * mat[2][1] - mat[1][1] * mat[2][0])) return ans # This function finds the solution of system of# linear equations using cramer's ruledef findSolution(coeff): # Matrix d using coeff as given in # cramer's rule d = [[coeff[0][0], coeff[0][1], coeff[0][2]], [coeff[1][0], coeff[1][1], coeff[1][2]], [coeff[2][0], coeff[2][1], coeff[2][2]]] # Matrix d1 using coeff as given in # cramer's rule d1 = [[coeff[0][3], coeff[0][1], coeff[0][2]], [coeff[1][3], coeff[1][1], coeff[1][2]], [coeff[2][3], coeff[2][1], coeff[2][2]]] # Matrix d2 using coeff as given in # cramer's rule d2 = [[coeff[0][0], coeff[0][3], coeff[0][2]], [coeff[1][0], coeff[1][3], coeff[1][2]], [coeff[2][0], coeff[2][3], coeff[2][2]]] # Matrix d3 using coeff as given in # cramer's rule d3 = [[coeff[0][0], coeff[0][1], coeff[0][3]], [coeff[1][0], coeff[1][1], coeff[1][3]], [coeff[2][0], coeff[2][1], coeff[2][3]]] # Calculating Determinant of Matrices # d, d1, d2, d3 D = determinantOfMatrix(d) D1 = determinantOfMatrix(d1) D2 = determinantOfMatrix(d2) D3 = determinantOfMatrix(d3) print("D is : ", D) print("D1 is : ", D1) print("D2 is : ", D2) print("D3 is : ", D3) # Case 1 if (D != 0): # Coeff have a unique solution. # Apply Cramer's Rule x = D1 / D y = D2 / D # calculating z using cramer's rule z = D3 / D print("Value of x is : ", x) print("Value of y is : ", y) print("Value of z is : ", z) # Case 2 else: if (D1 == 0 and D2 == 0 and D3 == 0): print("Infinite solutions") elif (D1 != 0 or D2 != 0 or D3 != 0): print("No solutions") # Driver Codeif __name__ == "__main__": # storing coefficients of linear # equations in coeff matrix coeff = [[2, -1, 3, 9], [1, 1, 1, 6], [1, -1, 1, 2]] findSolution(coeff) # This code is contributed by Chitranayal
## C#
// C# program to calculate solutions of linear// equations using cramer's ruleusing System; class GFG{ // This functions finds the determinant of Matrixstatic double determinantOfMatrix(double [,]mat){ double ans; ans = mat[0,0] * (mat[1,1] * mat[2,2] - mat[2,1] * mat[1,2]) - mat[0,1] * (mat[1,0] * mat[2,2] - mat[1,2] * mat[2,0]) + mat[0,2] * (mat[1,0] * mat[2,1] - mat[1,1] * mat[2,0]); return ans;} // This function finds the solution of system of// linear equations using cramer's rulestatic void findSolution(double [,]coeff){ // Matrix d using coeff as given in cramer's rule double [,]d = { { coeff[0,0], coeff[0,1], coeff[0,2] }, { coeff[1,0], coeff[1,1], coeff[1,2] }, { coeff[2,0], coeff[2,1], coeff[2,2] }, }; // Matrix d1 using coeff as given in cramer's rule double [,]d1 = { { coeff[0,3], coeff[0,1], coeff[0,2] }, { coeff[1,3], coeff[1,1], coeff[1,2] }, { coeff[2,3], coeff[2,1], coeff[2,2] }, }; // Matrix d2 using coeff as given in cramer's rule double [,]d2 = { { coeff[0,0], coeff[0,3], coeff[0,2] }, { coeff[1,0], coeff[1,3], coeff[1,2] }, { coeff[2,0], coeff[2,3], coeff[2,2] }, }; // Matrix d3 using coeff as given in cramer's rule double [,]d3 = { { coeff[0,0], coeff[0,1], coeff[0,3] }, { coeff[1,0], coeff[1,1], coeff[1,3] }, { coeff[2,0], coeff[2,1], coeff[2,3] }, }; // Calculating Determinant of Matrices d, d1, d2, d3 double D = determinantOfMatrix(d); double D1 = determinantOfMatrix(d1); double D2 = determinantOfMatrix(d2); double D3 = determinantOfMatrix(d3); Console.Write("D is : {0:F6} \n", D); Console.Write("D1 is : {0:F6} \n", D1); Console.Write("D2 is : {0:F6} \n", D2); Console.Write("D3 is : {0:F6} \n", D3); // Case 1 if (D != 0) { // Coeff have a unique solution. Apply Cramer's Rule double x = D1 / D; double y = D2 / D; double z = D3 / D; // calculating z using cramer's rule Console.Write("Value of x is : {0:F6}\n", x); Console.Write("Value of y is : {0:F6}\n", y); Console.Write("Value of z is : {0:F6}\n", z); } // Case 2 else { if (D1 == 0 && D2 == 0 && D3 == 0) Console.Write("Infinite solutions\n"); else if (D1 != 0 || D2 != 0 || D3 != 0) Console.Write("No solutions\n"); }} // Driver Codepublic static void Main(){ // storing coefficients of linear // equations in coeff matrix double [,]coeff = {{ 2, -1, 3, 9 }, { 1, 1, 1, 6 }, { 1, -1, 1, 2 }}; findSolution(coeff); }} // This code is contributed by 29AjayKumar
## Javascript
Output:
D is : -2.000000
D1 is : -2.000000
D2 is : -4.000000
D3 is : -6.000000
Value of x is : 1.000000
Value of y is : 2.000000
Value of z is : 3.000000
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My Personal Notes arrow_drop_up | 4,695 | 13,033 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2021-31 | latest | en | 0.759636 |
https://www.physicsforums.com/threads/how-does-entanglement-manifest-itself.84056/ | 1,714,012,319,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00107.warc.gz | 828,943,105 | 25,036 | # How does entanglement manifest itself?
• Ron Smith
In summary: A knows the state of particle X, then B MUST know the state of particle Y. There is no way for information to be 'switched' between particles without violating this constraint.There is a lot of debate surrounding the concept of 'superluminal' information transfer, but to summarize, there is no such thing. In fact, the reduced density matrix of one system is unaffected by whatever measurement is done on the other system (or even whether a measurement is done). All the rest depends on how you interpret quantum theory. There are many different interpretations, but the majority of them would lead to the same conclusion- that there is no superluminal information transfer.
Ron Smith
I am confused by the descriptions of entanglement I have read. Some seem to imply there is a superluminal transfer of information going on, while other descriptions do not. Can anyone explain this simply? I have a B.A. in Math, but am self taught in physics.
Ron Smith said:
I am confused by the descriptions of entanglement I have read. Some seem to imply there is a superluminal transfer of information going on, while other descriptions do not. Can anyone explain this simply? I have a B.A. in Math, but am self taught in physics.
One thing is sure: there is NO superluminal transfer of information. This can be shown mathematically, in that the reduced density matrix of one system is unaffected by whatever measurement is done on the other system (or even whether a measurement is done).
All the rest depends on how you interpret quantum theory.
What is true, is that Bell's inequalities can be violated in quantum theory. Bell's inequalities put a condition upon correlation functions under very "reasonable" hypotheses, which come down to claiming that (1) the apparent randomness of quantum events is only apparent, and that there is a deterministic underlying theory giving their outcomes, but which uses extra variables we don't know about and which are statistically distributed according to some probability law satisfying Kolmogorov's axioms and (2) that the dynamics of this underlying deterministic theory respects locality.
Theories respecting (1) and (2) are called "local realistic hidden variable" theories, and you can show then that correlations between outcomes of such theories satisfy certain inequalities (Bell's inequalities).
Quantum theory violates these inequalities in its statistical predictions (simply because it doesn't satisfy (1) and (2)). Experiments seem to be in agreement with the quantum predictions, although there are still people debating the validity of these experiments.
This kills Einstein's view that quantum theory was in fact a kind of statistical mechanics and we'd find soon a theory satisfying (1) and (2) which explains the apparent randomness of quantum theory.
Mind you that there EXISTS a theory which satisfies (1) (but not 2) and which is equivalent to QM: it is Bohmian mechanics. It contains non-local dynamics (action at a distance, which poses problems with special relativity).
BTW, I've written up quite some stuff concerning EPR in my journal (to the left of this post).
cheers,
Patrick.
Entanglement can be understood also as a way of conecting the states of two or more subsystems. If you cut a coin in two equal parts through a plane perpendicular to its simetry axis, and send one of this parts to person A e the other to the person B inside envelopes, when person A finds a head, he knows B had enconuntered a tail and vice versa. We know, according to classical machanics, that the states of this two parts were already well defined in the instant you sent the envelopes. There were only a statistical uncertainty, a classical statistical uncertainty. And the fact that makes this uncertainty classical is that there exists a well difined experimental method which could have been used to let us know which side of the coin you have sent to me at the moment you sent. You had a well defined intention, and ultimately you are choosing which half coin I will get. One could ask you before I receive the letter.
In quantum mechanics there exists no material experimentally accessible source of information on which side of coin was sent to A and B. It is not just a question of not being interested or not being technically capable of pursuing this information. Quantum mechanics assumes this information does not exists until A and/or B open their letters. So, the physical reality of "the state of the piece of metal inside the letter going to A" is a bad defined issue within quantum mechanics. Although if A find tail, necessarily B will find head, i.e. although this correlation still exists, there is no well defined state of the half coin inside the letter : the two pieces of metal are ENTANGLED.
Last edited:
Ron Smith said:
I am confused by the descriptions of entanglement I have read. Some seem to imply there is a superluminal transfer of information going on, while other descriptions do not. Can anyone explain this simply? I have a B.A. in Math, but am self taught in physics.
Welcome to PhysicsForums!
Entanglement is many things to different people. If you look at the 1935 EPR paper and Bell's Theorem, you will get a pretty good idea of some of the base issues. You can treat these as source documents for most of the debate. There are lot of different ways to describe entanglement and you will want to find a description that is comfortable to you.
While particles are entangled, there is a conservation or symmetry constraint present. The constraint usually is stated in terms of spin, but also applies to other observables as well. Usually there are 2 particles, and it is possible to entangle more. The following are some pretty good things to consider:
1. There is no superluminal information transfer (as Vanesch pointed out).
2. The HUP is not violated.
3. The entangled relationship extends over space and time beyond the limits implied by Special Relativity.
4. Experiments support all of the above, and also that the separate and independent existence of the individual particle observables is not supported (per Bell's Theorem). I.e. attributes such as spin are "contextual" (are not independent of the act of observation).
Interpreting the finer meaning of the above leads to a lot of philosophical issues. Good luck!
Ron Smith said:
I am confused by the descriptions of entanglement I have read. Some seem to imply there is a superluminal transfer of information going on, while other descriptions do not. Can anyone explain this simply? I have a B.A. in Math, but am self taught in physics.
The analog of the entanglement is two coins. Every of the coins gives a random results - head I1> or tail I0> with the probability 1/2.
The coefficient of correlation K of this coins is zero because we have two different random process.
In quantum world for the entanglement we have non-zero coefficient of correlation. In the ideal case is K=1.
You are remember but I employ the definition of coefficient of correlation K=<xy>-<x><y>. Here x and y is two random process between 1 and 0. <> is average. If the process is independent the K=0. For quantum case we are think that the processes is independent but K non-zero. It is very strange for the classical viewpoint.
cartuz said:
The analog of the entanglement is two coins. Every of the coins gives a random results - head I1> or tail I0> with the probability 1/2.
The coefficient of correlation K of this coins is zero because we have two different random process.
In quantum world for the entanglement we have non-zero coefficient of correlation. In the ideal case is K=1.
You are remember but I employ the definition of coefficient of correlation K=<xy>-<x><y>. Here x and y is two random process between 1 and 0. <> is average. If the process is independent the K=0. For quantum case we are think that the processes is independent but K non-zero. It is very strange for the classical viewpoint.
I must say that it seems to me that we may have entaglement with less that 1 correlation. Supose you have two DICES. One entangled state of the upper side numbers of these dices may be:
| 2 > X N {| 2 > + | 3 > + | 5 > - | 6 >} +
| 5 > X N {| 1 > + | 4 > + | 2 > - | 3 >} +
N {| 1 > + | 2 > + | 3 > + | 6 >} X | 1 >
where N is a normalization constant asd the first bra (or set of bras) always is related to dice 1 and the expression after the direct product X always has to do with the state of dice 2.
If the first dice shows | 2 > we can expect the second dice to show 2,3,5,6 and 1.
But if the second dice shows | 2 > we may expect the first to be 2 or 5 only.
The concept of entanglement in nuclear physics is often based on a false logical premise, that is, that the "object" [O] and the "act of perception" [P] are independent. It is nonsense to say that neither A nor B as [O] exist in the envelop until [P] event occurs, likewise that [P] must exist before A or B can be realized. Both positions are false. As discussed above, experiments in quantum reality show that [O] and [P] are not separate nor independent.
Entanglement is the "product" of the interaction between [O] and [P], thus entanglement [E] = { [O] * [P] }. Entanglement cannot be exclusively identified with either [O] nor [P]--which thus forms the third alternative of quantum reality, quantum entanglement is the integration operation process of "object [O] as perceived [P]".
Finally, entanglement as { [O] * [P] } is preformed within [O]. Thus [E] = {([O]*[P]) * [P]}, and we see then that [E] = [O] * [P]2. Thus the concept of entanglement is the source of the metaphysical nature of reality itself. We also know this fundamental reality of entanglement by its more specific notation: E = MC2.
Thus the concept of entanglement is the source of the metaphysical nature of reality itself. We also know this fundamental reality of entanglement by its more specific notation: E = MC2.
This is not an accurate usage of the terms. Entanglement is a specialized quantum state involving 2 or more particles. Entanglement itself has nothing to do with the observer.
On the other hand: all quantum state statistics will obey the Heisenberg Uncertainty relations, and these are affected by the observer. This element of QM is often subjected to metaphysical analysis.
Entanglement is NOT the source of this, and it is certainly not described by E=mc^2.
DrChinese said:
This is not an accurate usage of the terms. Entanglement is a specialized quantum state involving 2 or more particles. Entanglement itself has nothing to do with the observer.
With respect, I disagree. Entanglement in quantum reality makes no sense without observer. To support my position I offer the following paper summary (see this link for entire paper:http://citebase.eprints.org/cgi-bin/fulltext?format=application/pdf&identifier=oai%3AarXiv.org%3Aquant-ph%2F0106003 )
Is entanglement observer-dependent?
Italo Vecchi
Vicolo del Leoncorno 5 - 44100 Ferrara - Italy
email: vecchi@isthar.com
Abstract: The properties of quantum entanglement are examined and the role of the observer is pointed out.
...We can now go back to entanglement and ask the question: ”What is entanglement?” .The answer may be: ”Entanglement is the observer’s blueprint for state-vector reduction”. It should be clear that entanglement can be defined only in terms of the observer-dependent basis. Prior to observation all bases are equivalent so that speaking about entanglement is meaningless. It is only when state-vector reduction takes place that the system’s state-vector is cast according to an observer-dependent set of rules . Entanglement has an observer-independent support, since the observer’s perceptions are based on the information it extracts from its interaction with the system’s state vector, which is determined by the system’s evolution. However for state-vector reduction the physical features of the system, as encoded in the system’s state-vector, must be interpreted through a blueprint that depends on the observer. Loosely speaking we may say that physical interaction, as described by the relevant Schroedinger equation, may leave ”marks” on 4 the system‘s state-vecto affecting the measurement outcome, e.g. the scrambling/vanishing of superpositions, but such ”marks” are read according to an observer-dependent blueprint only when state-vector reduction takes place. Without an observer the ”marks” are meaningless ripples on the system’s wave-function...
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DaTario said:
...Supose you have two DICES...
FYI, the singular of this word is "die" and the plural is "dice". I know no other English word that follows this rule ... what a crazy language!
With respect, I disagree. Entanglement in quantum reality makes no sense without observer...
The reason this is not correct is simple. The observer plays a similarly important role an all quantum interactions, including systems where entanglement is not a feature. There is nothing wrong with your quotes per se but they don't tell the entire story.
The reason entanglement is so associated with the observer is because it disproves the idea of local hidden variables via Bell's Theorem. That elevates the role of the observer. But it is not a feature solely of entangled systems, it is a general feature of QM. Your quote would be more consistent if you had said: "quantum reality makes no sense without observer".
Nicky said:
FYI, the singular of this word is "die" and the plural is "dice". I know no other English word that follows this rule ... what a crazy language!
Mouse/mice?
With respect, I disagree. Entanglement in quantum reality makes no sense without observer. To support my position I offer the following paper summary (see this link for entire paper:http://citebase.eprints.org/cgi-bin...nt-ph%2F0106003 )
DrChinese said:
The reason this is not correct is simple. The observer plays a similarly important role an all quantum interactions, including systems where entanglement is not a feature. There is nothing wrong with your quotes per se but they don't tell the entire story.
The reason entanglement is so associated with the observer is because it disproves the idea of local hidden variables via Bell's Theorem. That elevates the role of the observer. But it is not a feature solely of entangled systems, it is a general feature of QM. Your quote would be more consistent if you had said: "quantum reality makes no sense without observer".
I agree with DrChinese. I don't understand why this "quantum reality" bit is only applied to entanglement and NOT to the rest of QM. If one has a problem with "observer dependent reality", then why pick only on the entanglement phenomenon? The Schrodinger-Cat type observation is a HUGE part of this and can't be separated out of any QM measurement, entangled or NOT. The fact that the system of entangled states are in a superposition of a number of possible states IS the whole reason why it is different than a simple "conservation of angular momentum" system in classical mechanics.
Zz.
Last edited by a moderator:
To summarize what I think I am hearing about concept of entanglement from the above comments: (?):
1. Reality-[R] exists [An axiomatic statement]
2. Quantum-reality is a subset of Reality-[R], thus quantum-reality exists
3. Quantum-reality makes no sense without observer [DrChinese]
4. Entanglement is a sub-set of quantum reality [DrChinese, ZapperZ]
5. Thus, by logic, entanglement makes no sense without observer [Vecchi paper]
however,
6. Non-quantum reality = Reality-[R] that is not Quantum-reality
7. Non-quantum reality makes sense without observer
8. Is entanglement also a state of Non-quantum reality ?
Entanglement IMO is related to making ones capable of performing an interference experiment. Otherwise it has only the aspect of statistical uncertainty.
I have always been confused by entanglement but recently got this gist after reading the fabric of the cosmos (brian green).
It goes like this:
Consider first just a single particle that can be in state A or B. Now this is the important point - when you perform a measurement it can either be in A or B, then if you perform the measurement again a little later - it can AGAIN be in A or B. Just because you measured it A say the first time it doesn't mean it will be in A the second time.
Once you grasp that then entanglement is easy to understand.
So, now consider 2 entangled particles.
1 has a 50:50 chance or being in state A or B.
2 has a 50:50 chance of being in state A or B.
Now, release both particles in opposite directions say. The point is that when you perform the measurement they should both give RANDOM answers, but they end up correlating exactly. i.e particle 1 really is in both A and B until the measurement and then when you finish measuring it goes back to being in some superposition of A and B.
The very fact that 1 and 2 correlate is amazing because without an observer they really are never in state A or B - just a superposition, but when an observer comes in and has a look it collapses both wavefunctions simultaneously!
But this part of measuring in sequence has to do with the specific Hamiltonian which will ultimately rule the time evolution of your system. I am not sure that you can count on the fact the 50:50 pattern will come back after the first measurement.
Best Regards
DaTario
To summarize what I think I am hearing about concept of entanglement from the above comments: (?):
1. Reality-[R] exists [An axiomatic statement]
2. Quantum-reality is a subset of Reality-[R], thus quantum-reality exists
3. Quantum-reality makes no sense without observer [DrChinese]
4. Entanglement is a sub-set of quantum reality [DrChinese, ZapperZ]
5. Thus, by logic, entanglement makes no sense without observer [Vecchi paper]
however,
6. Non-quantum reality = Reality-[R] that is not Quantum-reality
7. Non-quantum reality makes sense without observer
8. Is entanglement also a state of Non-quantum reality ?
Is this physics or philosophy?
What is "a state of non-quantum reality"? This is odd because entanglement came out of QM's own formulation. How can it be a "non-quantum" anything?
You do know what mathematics represented by "entanglement", don't you? Now see if it is separable. If it isn't, you have just arrived at its definition.
Zz.
robousy said:
Consider first just a single particle that can be in state A or B. Now this is the important point - when you perform a measurement it can either be in A or B, then if you perform the measurement again a little later - it can AGAIN be in A or B. Just because you measured it A say the first time it doesn't mean it will be in A the second time.
This portion of the post is not fully accurate. If you make a measurement of a particle, subsequent measurements will be consistent with the Heisenberg Uncertainty Principle (HUP). A photon with a known polarity will continue to have that polarity as many times as you care to check it. If you make a different measurement, you may then get different results, still obeying the HUP.
ZapperZ said:
Is this physics or philosophy?
You are too kind.
DrChinese said:
This portion of the post is not fully accurate. If you make a measurement of a particle, subsequent measurements will be consistent with the Heisenberg Uncertainty Principle (HUP). A photon with a known polarity will continue to have that polarity as many times as you care to check it. If you make a different measurement, you may then get different results, still obeying the HUP.
Ok, thanks DrC. I wasn't 100% sure I was saying things right.
To summarize what I think I am hearing about concept of entanglement from the above comments: (?):
1. Reality-[R] exists [An axiomatic statement]
2. Quantum-reality is a subset of Reality-[R], thus quantum-reality exists
3. Quantum-reality makes no sense without observer [DrChinese]
4. Entanglement is a sub-set of quantum reality [DrChinese, ZapperZ]
5. Thus, by logic, entanglement makes no sense without observer [Vecchi paper]
however,
6. Non-quantum reality = Reality-[R] that is not Quantum-reality
7. Non-quantum reality makes sense without observer
8. Is entanglement also a state of Non-quantum reality ?
I think I know what you're asking. What is it in *nature* that's
essential to producing the entangled measurements that qm describes?
Qm isn't designed to answer that, but we can make some
guesses. Whatever is being measured to produce the
predictable (entangled) results would seem to have to be
related in some way. Maybe the spatially separated, correlated
results are due to a prior interaction, or a common origin,
or maybe you're looking at separate parts of a single,
encompassing submicroscopic system.
## 1. What is entanglement?
Entanglement is a phenomenon in quantum physics where two or more particles become connected in such a way that the state of one particle cannot be described without considering the state of the other particle, even if they are separated by large distances.
## 2. How does entanglement occur?
Entanglement occurs when two or more particles interact with each other and become entangled. This can happen through various processes such as collisions, interactions with other particles, or through the measurement of one particle affecting the state of another.
## 3. What is the significance of entanglement?
The significance of entanglement lies in its implications for our understanding of quantum mechanics and its potential applications in technology. Entangled particles can be used for secure communication, quantum computing, and teleportation.
## 4. Can entanglement be observed in everyday life?
No, entanglement is a phenomenon that occurs at the quantum level and cannot be observed in everyday life. It requires precise experimental conditions and equipment to be observed.
## 5. How does entanglement challenge our classical understanding of physics?
Entanglement challenges our classical understanding of physics because it violates the principle of local realism, which states that an object can only be influenced by its immediate surroundings. Entangled particles can influence each other regardless of the distance between them, which goes against our classical understanding of how the world works.
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3K | 5,106 | 22,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-18 | latest | en | 0.950673 |
https://ablconnect.harvard.edu/book/roman-numeral-analysis-and-class-game | 1,674,827,211,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494986.94/warc/CC-MAIN-20230127132641-20230127162641-00649.warc.gz | 94,906,229 | 21,449 | # Roman Numeral Analysis and Class Game
Students practice and review the fundamental skills of Roman numeral analysis through a competitive question/answer game.
Goal/s:
For students to become comfortable with the fundamental skills of Roman numeral analysis: chord rotation, building chords on scale degrees, functionality within a key and a few voice-leading techniques.
Class: Mus 51a: Tonal Harmony I
Introduction/Background:
Teaching Fellow, Olivia Lucas, designed a game for students to learn and review the fundamental skills of Roman numeral analysis. Roman numeral analysis forms the backbone of much of what is taught in the undergraduate music theory curriculum, and thus, familiarity and agility with its tools and techniques will serve the students well not only in the class, but also in future semesters.
Prior to the activity, students will have learned the fundamentals of a Roman numeral analysis, a technique for analyzing music. This analytical technique relates chords to the scale degree on which they are based, and uses additional symbols to express inversion and applied functionality. They will already understand how to a) generate a chord when given a roman numeral and key b) identify the correct roman numeral to label a chord in a piece of music, though they will have not necessarily had much practice applying these skills yet.
In this activity, students are put into teams of three and have to work together to answer questions on Roman Numeral Analysis posed by the instructor.
Procedure:
1. For the first half of the section, the class reviews the relevant terms and techniques that were just covered in lecture.
2. Then, the instructor breaks the students into two teams, of 3 students each.
3. These three students must take turns being the team's "scribe" - the person who writes the answer to the problem on the board. The scribe is not allowed to write anything on the board until her/his teammates say what to write. (This ensures that if one student is significantly quicker at solving the problems than the others, there will be turns when s/he must stay quiet and let the others think through the problem.)
4. The instructor, then, reads out a problem for the teams to solve. The problems will start simple, for example, "what is IV in F major?" When instructed by their teammates, the scribe must write the correct key signature, and the correct chord.
5. The first team to have the correct answer gets a point. A tally can be kept on the board.
6. The game proceeds with the scribes rotating and the questions getting more complex, asking them to not only use what they've learned, but put more and more pieces of it together. Later questions would include asking them to write inverted chords, correctly voice-led cadences, or short chord progressions, all from the starting point of the roman numerals. The competitive aspect encourages them to work quickly without losing accuracy, which will in turn help them in completing their assignments and taking quizzes, which are timed.
7. The game continues for about twenty minutes, and the team with the most points could get a small prize, but this is optional. At the end, five minutes are left for questions. Students can ask questions about anything that came up during the game, or about the upcoming assignment and quiz on the material. The class also reflects, briefly, as a group on the value of using this kind of abstract analysis to examine pieces of tonal music. | 686 | 3,480 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-06 | latest | en | 0.951158 |
https://astronomy.stackexchange.com/questions/44703/if-we-watched-extremely-red-shifted-galaxies-near-the-edge-of-the-observable-uni/44741 | 1,716,623,920,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058789.0/warc/CC-MAIN-20240525065824-20240525095824-00219.warc.gz | 85,446,790 | 40,239 | # If we watched extremely red-shifted galaxies near the edge of the observable universe for a very long time, how would they change? Would more appear?
I have understanding sphere eversion as #1 on my bucket list (if I ever get a round tuit) but understanding metric expansion seems to be a rapidly receding possibility :-)
Question: Suppose it takes me 100 million years to evert a sphere and I manage to live that long. If I watched extremely red-shifted galaxies near the edge of the observable universe for a very long time, how would their appearance change (aside from their own natural evolution)? Would their red shift z remain constant or increase/decrease? Would more of them appear?
If the radius observable universe is growing and by the rate and mechanisms described in the answer linked below, I'm thinking that at a given observing red-shifted wavelength new galaxies would appear farther away, and so the ones that were at the limit before must have somehow become less red-shifted.
This question was inspired by this answer to Does the mass of the observable universe ever change? but here I'd like to ask for supporting sources as that will allow me to read further. Thanks!
Related in Physics SE, my previous efforts to try to understand metric expansion:
Somewhat related:
• The redshift of galaxies entering the horizon decreases from infinity to a minimum value, after which is increase to infinity again. I started writing an answer, but then I found this one and this one on physics.SE, which I think answer your question.
– pela
Jul 4, 2021 at 11:16
• Your question about observing at a given redshift is also answered there: If you follow the dashed lines (e.g. the one called "z = 50") in answer #2's spacetime diagram, you'll see that in the beginning more and more distant galaxies will have that redshift, but at some point in the future, galaxies with that redshift will be progressively nearby.
– pela
Jul 4, 2021 at 11:18
• @pela I checked briefly and I can see I will need to look at them again in the morning here, and I'm not confident I'll be able to understand them sufficiently to know the answer to my question here with any confidence. If those answers tell us just how the current horizon galaxies' appearance will be different in 100 million years, if their red shift z will go up or down and if we'll see more, and that can be summarized briefly here, that would be an ideal answer; a few sentences linking to those answers as authoritative sources.
– uhoh
Jul 4, 2021 at 12:47
• Okay, I tried to have a go at it.
– pela
Jul 5, 2021 at 11:13
• I made my own spacetime diagram, omitting some information that wasn't necessary.
– pela
Jul 6, 2021 at 13:56
tl;dr Their redshift would first decrease from $$\infty$$ to $$\sim60$$, then increase to $$\infty$$ again. And more eventually appear.
The answer to this question is somewhat non-trivial, and will depend on the cosmology of the universe you're considering. But in our Universe, in which dark energy supposedly accelerates expansion, what happens can be summarized as this:
### Qualitative description
As time goes, light from more and more distant regions will have had the time to reach us, so new matter enters our (particle) horizon all the time. For an infinitesimally small period of time, the redshift $$z$$ of that matter is infinite, but will then decrease with time, as the matter moves further into our observable Universe. However, at some point later on, the accelerated expansion will speed up the matter, increasing its redshift.
### Quantitative description
The behavior is best understood using comoving coordinates, i.e. the coordinates that expand along with the Universe. In these coordinates, galaxies and other matter stay fixed (except for a peculiar velocity which is of no importance to the principle). The relation between real, physical distances and comoving distances is $$d_\mathrm{phys} = a\,d_\mathrm{com},$$ where $$a$$ is the scale factor (the "size") of the Universe, normalized to be $$a=1$$ today.
Consider an observer at a time when the scale factor of the Universe was $$a_1$$, observing light emitted earlier when the scale factor was $$a_2 < a_1$$. The redshift $$z_{12}$$ observed by this observer is $$z_{12} = \frac{a_1}{a_2} - 1.$$ Now this is simple enough; the non-trivial part enters when calculating how $$a$$ evolves with time, which is done by integrating the Friedmann equation. This can only be solved analytically in simple cases; in general it must be solved numerically.
Numerically solving the Friedmann equation can also yield the distance to an observed object. In this way you can convert a redshift, or a corresponding scale factor, to a distance. But you can also do the opposite; that is, convert a given, observed redshift at a time $$t_1=t(a_1)$$ of light emitted at time $$t_2=t(a_2)$$ to a (comoving) distance $$d_{12}$$.
I can add the full set of equations, if you like, or you can have a look at this and this answers on physics.SE. For now, I've just implemented them in Python and plotted the (hopefully) illuminating diagram below.
### Example
Below is a spacetime diagram, showing the Universe at all times (along the $$y$$ axis) and, correspondingly, at all scale factors, as a function of comoving distances (along the $$x$$ axis). A given epoch is a horizontal line (e.g. "now"), and a given position — also called a worldline — is along a vertical line (e.g. "here").
Horizons in the spacetime diagram
Everything we observe lie on our part light cone (red) which, as time goes, converges toward our event horizon (orange); the part of the Universe we will ever see. The observable Universe is the part of the "Now" line inside the green lines marking the particle horizon.
Curves of constant redshift
The dashed, cyan lines show the curves of constant redshift, calculated as described above. That is, at any time (given by a horizontal slice of your choice), a redshift (given by one of the dashed lines of your choice) will be observed for objects lying at a comoving distance given by the crossing of those two chosen lines.
As an example, let's consider the highest-redshift observed galaxy, GN-z11, which is currently observed at $$z=11.1$$.
The worldline of GN-z11 is show as a black, dashed line, and our view of GN-z11 today is marked by a star lying on our past light cone. GN-z11 was not inside our observable Universe at the time we see it today; it only entered our cosmic horizon (the green line) when the Universe was just over $$t = 4\,\mathrm{Gyr}$$ old. As time went, its redshift decreased from $$\infty$$, to $$100$$ at $$t\sim6\,\mathrm{Gyr}$$, to $$30$$ at $$t\sim8\,\mathrm{Gyr}$$, to $$11.1$$ today.
In the future, GN-z11's redshift will continue to decrease to around $$t\sim20\,\mathrm{Gyr}$$, at which time we will observe it to have $$z\sim9$$. After this, it will increase without bounds.
Example #2
Now you ask specifically to "extremely redshifted galaxies", i.e. galaxies appearing now (which we would see "at the Big Bang" and hence not yet as a galaxy). The comoving distance to such a (proto-)galaxy — let's call it "$$\textsf{uhoh}$$" — would be (almost) equal to the distance to the particle horizon. Its worldline is marked in the spacetime diagram, and as indicated, we will see its redshift decreasing from $$\infty$$ to $$z\simeq60$$, and back to $$\infty$$.
When we observe $$z_\mathsf{uhoh}\simeq60$$, the age of the Universe will be just over $$t=30\,\mathrm{Gyr}$$, but we will see $$\textsf{uhoh}$$ as it was when the Universe was $$t\simeq\mathrm{200}\,\mathrm{Myr}$$ which, coincidentally, was just around the time when the first galaxies formed.
This can be seen from noticing that the worldline of $$\textsf{uhoh}$$ just grazes the $$z_\mathrm{obs}=60$$ line at $$t\sim30\,\mathrm{Gyr}$$, then following the $$45\circ$$ line from the point $$\{t,d\}=\{30,0\}$$ (our past light cone in the future) back to $$\textsf{uhoh}$$'s worldline, seeing that these lines cross at $$t\simeq200\,\mathrm{Gyr}$$.
In the far future, as $$z_\mathsf{uhoh}\rightarrow\infty$$, we will see $$\textsf{uhoh}$$ converge toward an age of $$t\simeq500\,\mathrm{Myr}$$, because this is where $$\textsf{uhoh}$$ crosses our event horizon.
• @uhoh What "dz/dt = 0" means is that, if at a given time you observe a bunch of galaxies with different redshifts, then all those with a redshift z_obs that is smaller than the value, z', that satisfies dz/dt = 0 will — if you wait a little — acquire a larger redshift. Conversely, all those with z_obs > z' will after a little while have a smaller z_obs. This "threshold redshift" slowly increases with time, and is currently z_obs ≈ 2. I'm not sure there's anything meaningful about this particular value, though. Anyway, thanks a lot, I'll consider writing it up somewhere :)
– pela
Jul 9, 2021 at 9:41
• @uhoh Okay, I’ll do the same then :)
– pela
Jul 13, 2021 at 7:29
• Too late :D I’ll leave the first one, which elaborates on dz/dt.
– pela
Jul 13, 2021 at 7:32 | 2,290 | 9,036 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 46, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-22 | latest | en | 0.967333 |
https://faafl.org/t-account-worksheet-examples/ | 1,566,798,029,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330968.54/warc/CC-MAIN-20190826042816-20190826064816-00225.warc.gz | 453,258,299 | 31,482 | # T Account Worksheet Examples
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T Account Worksheet Examples. The worksheet is an assortment of 4 intriguing pursuits that will enhance your kid's knowledge and abilities. The worksheets are offered in developmentally appropriate versions for kids of different ages. Adding and subtracting integers worksheets in many ranges including a number of choices for parentheses use.
You can begin with the uppercase cursives and after that move forward with the lowercase cursives. Handwriting for kids will also be rather simple to develop in such a fashion. If you're an adult and wish to increase your handwriting, it can be accomplished. As a result, in the event that you really wish to enhance handwriting of your kid, hurry to explore the advantages of an intelligent learning tool now!
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T Account Worksheet Examples. Try to remember, you always have to care for your child with amazing care, compassion and affection to be able to help him learn. You may also ask your kid's teacher for extra worksheets. Your son or daughter is not going to just learn a different sort of font but in addition learn how to write elegantly because cursive writing is quite beautiful to check out. As a result, if a kid is already suffering from ADHD his handwriting will definitely be affected. Accordingly, to be able to accomplish this, if children are taught to form different shapes in a suitable fashion, it is going to enable them to compose the letters in a really smooth and easy method. Although it can be cute every time a youngster says he runned on the playground, students want to understand how to use past tense so as to speak and write correctly. Let say, you would like to boost your son's or daughter's handwriting, it is but obvious that you want to give your son or daughter plenty of practice, as they say, practice makes perfect.
Without phonics skills, it's almost impossible, especially for kids, to learn how to read new words. Techniques to Handle Attention Issues It is extremely essential that should you discover your kid is inattentive to his learning especially when it has to do with reading and writing issues you must begin working on various ways and to improve it. Use a student's name in every sentence so there's a single sentence for each kid. Because he or she learns at his own rate, there is some variability in the age when a child is ready to learn to read. Teaching your kid to form the alphabets is quite a complicated practice.
Author: Molly Nilsson
Have faith. But just because it's possible, doesn't mean it will be easy. Know that whatever life you want, the grades you want, the job you want, the reputation you want, friends you want, that it's possible.
Top | 1,424 | 7,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-35 | latest | en | 0.923742 |
https://nrich.maths.org/1406 | 1,485,028,006,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281202.94/warc/CC-MAIN-20170116095121-00545-ip-10-171-10-70.ec2.internal.warc.gz | 856,229,497 | 5,743 | ### Algebra Match
A task which depends on members of the group noticing the needs of others and responding.
### Simplifying Doughnut
An algebra task which depends on members of the group noticing the needs of others and responding.
# Algebra from Geometry
##### Stage: 3 and 4 Challenge Level:
Michael Brooker tackled the problem Bull's Eye and noticed something interesting when finding the area of a ring, which he has followed up. The result he found will be familiar to many older mathematicians.
Michael could see that the formula for the area of a ring must be pi(R 2 - r 2 ). What he noticed was that in the examples he worked out, the area also appeared to be pi(R+r). This interested him, and he realised that you can find the difference between numbers with consecutive square roots by adding the roots:
144 - 121 = 23 = 12 + 11
Michael realised that this result is the reason why the second formula works for rings where R = r + 1, and tried to prove why.
He used a geometrical method to show that (r + 1) 2 = r 2 + 2r + 1:
So, if R = r + 1,
R 2 - r 2 = (r 2 + 2r + 1) - r 2 = 2r + 1 = R + r
We asked Michael what happened if the ring was 2 wide, or 3 wide. Could he find an alternative formula for these rings, similar to the simple one he found above? He started by trying some examples, to see if he could spot anything interesting.
r R area width R+r
2 4 12pi 2 6
2 5 21pi 3 7
2 6 32pi 4 8
Have you spotted what Michael did?
He then set out to prove that the formula for the area of a ring is pi(R - r)(R + r)
Michael ignored the pi, and tried to show that
R 2 - r 2 = (R - r)(R + r).
This result is known as "the difference of two squares". It is a very useful result, and Michael did well to find out about it for himself.
He introduced the letter n for the width, so R = r + n, and then rewrote the formulas above using r and n instead of r and R.
(R - r)(R + r) = n(r + n + r) = n(2r + n)
R 2 - r 2 = (r + n) 2 - r 2 = (r + n)(r + n) - r 2
Even if you don't know how to do (r + n)(r + n), you can probably see what it is from the diagram below.
So (r + n)(r + n) - r 2 = r 2 + rn + rn + n 2 - r 2 = 2rn + n 2 = n(2r + n)
So R 2 - r 2 = (R - r)(R + r), and Michael has proved the result. Well done!
You may like to look at a couple of problems which are based on the the "Difference Of Two Squares" result: What's Possible? and Plus Minus | 704 | 2,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2017-04 | longest | en | 0.959209 |
https://www.studypool.com/discuss/342717/if-f-x-x-2-1-find-f-a-1?free | 1,481,464,242,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544679.86/warc/CC-MAIN-20161202170904-00159-ip-10-31-129-80.ec2.internal.warc.gz | 1,001,008,612 | 13,911 | ##### If f(x) = x 2 + 1, find f(a + 1).
Algebra Tutor: None Selected Time limit: 1 Day
If f(x) = x 2 + 1, find f(a + 1).
Jan 14th, 2015
f(x) =x^2 +1
f(a+1) = (a+1)^2 +1 = a^2+2a+1+1 = a^2+2 +1
Jan 14th, 2015
...
Jan 14th, 2015
...
Jan 14th, 2015
Dec 11th, 2016
check_circle | 148 | 281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2016-50 | longest | en | 0.57418 |
https://cryptlabs.com/what-is-linear-algebra/ | 1,713,057,206,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816863.40/warc/CC-MAIN-20240414002233-20240414032233-00037.warc.gz | 169,674,317 | 15,248 | # What is Linear Algebra?
Linear algebra is a branch of mathematics that deals with linear equations, linear transformations, and vector spaces. It plays an important role in various fields, including physics, engineering, computer science, and economics.
At its core, linear algebra studies the properties of linear equations and linear systems of equations, which are equations that can be expressed in the form Ax = b, where A is a matrix, x is a vector, and b is a constant vector. Linear algebra also deals with the concept of linear transformations, which are functions that map vectors to other vectors in a linear fashion.
Some of the key topics in linear algebra include:
• Vector spaces and subspaces
• Linear independence and basis vectors
• Linear transformations and matrices
• Eigenvalues and eigenvectors
• Inner products and orthogonality
• Determinants
Linear algebra has many applications, including:
• Solving systems of linear equations
• Image and signal processing
• Computer graphics
• Machine learning and data analysis
• Optimization problems
Some of the most widely used textbooks on linear algebra include:
• “Linear Algebra and Its Applications” by Gilbert Strang
• “Introduction to Linear Algebra” by Serge Lang
• “Matrix Analysis and Applied Linear Algebra” by Carl Meyer
• “Linear Algebra Done Right” by Sheldon Axler
• “Linear Algebra: A Modern Introduction” by David Poole
Overall, linear algebra is a foundational area of mathematics that has many practical applications across a wide range of fields.
Reference:
• Strang, G. (2006). Linear algebra and its applications. Cengage Learning.
• Lang, S. (1987). Introduction to linear algebra. Springer.
• Meyer, C. D. (2000). Matrix analysis and applied linear algebra (Vol. 1). Society for Industrial and Applied Mathematics.
• Axler, S. (2015). Linear algebra done right. Springer.
• Poole, D. (2010). Linear algebra: a modern introduction. Cengage Learning. | 413 | 1,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-18 | longest | en | 0.901111 |
https://billsschedulechallenge.com/w86bk6s3/ | 1,606,602,933,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195929.39/warc/CC-MAIN-20201128214643-20201129004643-00699.warc.gz | 204,660,729 | 8,952 | Free Printable Fifth Grade Math Worksheets
Zurie Abby November 17, 2020 Worksheet
The same is true for mastery of working with fractions. So, in general, kids who don’t learn their multiplication tables and arithmetic with fractions usually don’t do well in math. It’s been proven that success in science depends to a great extent on success in math. So, these same kids usually don’t do well with science, either. In other words, they’re locked out of most of the growth that our economy is going to experience in the future. According to the federal government, eight out of ten jobs in the next twenty years are going to be computer related. If you want your child to have access to the major portion of jobs in the future, your child must master math. It’s that simple.
Teaching needs to be more than passing out worksheets. Whether you are the classroom teacher, instructional specialist, or parent, the methods you use greatly impact the level of understanding achieved by your students. Here are five reasons why math worksheets don’t work if you want students to understand math, enjoy math, and think mathematically.
No matter what materials you choose, it is most important that you supervise your child constantly so that mistakes get caught rather than practiced. I learned this particular lesson the hard way. When my daughter was young, she did something that needed ”attention.” I no longer remember what it was that she did, but I told her to write the sentence ”I will not disobey my parents again” 50 times. I should have known better, but I didn’t check on her at the beginning and then I got busy. So, sometime later, she brought me 50 sentences of ”I will not disobey my parents agen.” She had just practiced misspelling ”again” as ”agen” — 50 times! I’m not certain that we ever really got that fixed.
The present generation seems to be blessed immensely with intellect and the benefits of mastering math are something worth considering. It is a well-known fact that math is not a subject that one learns by simply reading the problems and its solutions. In order to master the subject, earnest practice on multiple problems is the best way to go. However, not every person is bestowed with required materials like math worksheets to receive adequate amount of practice.
it does need is a change in attitude, and a solid foundation of basic skills on which to build. Mathematics worksheets can help you provide your preschooler with a solid grounding that will help them conquer math.
Math worksheets rarely ask students to think critically or creatively. They usually present multiple examples of the same problem type with the hope of reinforcing a skill or procedure. They do not challenge students to use higher order thinking skills such as comparing, analyzing, deducing, and synthesizing. These skills are built through activities in which students discover concepts, explore ideas, test a hypothesis, solve a problem, and discuss their thinking with their peers. Exploring concepts and problems in many different ways builds interest and promotes critical thinking.
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Latest News | 842 | 3,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-50 | latest | en | 0.962556 |
http://mathhelpforum.com/discrete-math/152373-reflection-principle-binary-operator.html | 1,527,233,418,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867046.34/warc/CC-MAIN-20180525063346-20180525083346-00504.warc.gz | 185,065,663 | 10,329 | # Thread: reflection principle and binary operator
1. ## reflection principle and binary operator
Consider a computation like
a1 + a2 + a3 + ...... + an
Now to perform this you need to fully paranthesize the sum. Question is - how many ways can you fully paranthesize this. Let this be called Pn (let me call this problem 1)
For e.g. for n = 3 we can do it in 2 ways
((a1 + a2) + a3)
(a1 + (a2 + a3))
Now, I have been able to derive the following recurrence
$\displaystyle P_n = \sum_{k=1}^{n-1} P_k.P_{n-k}$
I read that this problem can be solved using reflection principle. I have used reflection princile to solve the problems like - how many ways you can permute a sequence of n '(' and n ')' so that it forms a valid matched pattern. (Which is nothing but catalan number) (let me call this problem 2)
I know how to solve problem 2 using reflection principle.
For problem 2 also I have been able to derive a recurrence relation which is similar to the one delrived for Problem 1 above (with an adjustment to sub-script).
So yes - I can interpret the recurrence relation for the problem 1 - convert it into a problem of problem type 2 and then solve this using reflection principle. And hence I have solved problem 1 using reflection principle.
But apart from the similarity of the recurrence relations of the two problems is there any other combinatorial argument as to why these problem are similar? More specifically can problem 1 be solved without deriving its recurrence relation and rather using reflection principle directly.
Hope I was clear in my explanation !
2. If one takes a fully parenthesized sum of $\displaystyle n$ terms and removes actual terms and plus signs, then one gets a well-matched string of $\displaystyle n-1$ pairs of ('s and )'s. Moreover, different parethetizations yield different strings and all strings can be obtained in this way. Thus, $\displaystyle P_n=C_{n-1}$ ($\displaystyle C_n$ is the $\displaystyle n$th Catalan number).
To show the first fact, assume that $\displaystyle S_n$ is a fully parenthesized sum of $\displaystyle n$ terms and proceed by (strong) induction on $\displaystyle n$ starting with $\displaystyle n=1$.
3. Thanks for your induction idea ! | 530 | 2,222 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-22 | latest | en | 0.899619 |
https://georgewoodbury.wordpress.com/2009/11/24/mymathlab-%E2%80%93-homework-grades-monday-112309/ | 1,555,629,821,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526904.20/warc/CC-MAIN-20190418221425-20190419003425-00292.warc.gz | 438,022,508 | 16,915 | ## MyMathLab – Homework Grades (Monday 11/23/09)
This week I’ve decided to focus on MyMathLab homework and the grades we assign to it. (Sorry for the delay – exams, exams, exams …)
How I’ve Always Done It
For the last few years I have made MyMathLab 25% of my students’ grade. I divide the 25% equally among the following 3 categories: homework, chapter quizzes, SLO quizzes. I have always felt comfortable assigning a little over 8% of my total grade to homework. I know some instructors get a little worried about students getting high homework grades without actually learning, and therefore setting themselves up for failure on exams.
I spend a significant amount of time at the beginning of the semester talking about how to do the homework to maximize understanding. I explain each of the learning aids and how to incorporate them in an effective plan to learn, while also explaining some of the pitfalls associated with the learning aids. (See my article on “Student Pointers” on my web site for more information.) With this approach I have a very small number of students who do well on the homework but do not learn.
Here are a few other strategies you can use if you worry about the homework.
Option 1 – Limit the number of attempts per question.
MyMathLab now allows the instructor to limit the number of attempts per question. I have a colleague who currently uses this strategy, limiting his students to 3 attempts, and has found it to be successful. First, he says, this is actually 9 attempts – 3 chances to answer each of the 3 questions. Second, he feels that his students take each question more seriously due to the limited number of attempts, rather than just going through the motions.
Another benefit to this approach is that the use of the “Help Me Solve This” learning aid uses up one of their attempts. This makes students less reliant on the learning aids, which should improve their exam scores.
(Keep up the good work, JB.)
Option 2- Turn off the learning aids.
MyMathLab now allows you to turn of some or all of the learning aids for an assignment. So, if you don’t want your students to abuse the learning aids, turn them off. You could create one assignment that leaves them on, and another that leaves them off. You could turn them off in homework assignments, and direct students to Study Plan problems for help.
Option 3 – Make the homework worth 0%.
What?!?!?! If the homework is worth 0%, the students won’t do it. Not so fast, my friend.
At Pioneer VI in Kentucky, I got a great idea from two instructors (one from Hazzard County) that I plan to implement next semester. For each section, create a homework assignment and a quiz. Leave the learning aids on for the homework, but not for the quiz. Make the homework a prerequisite for the quiz; for example, set the prerequisite to 90% for access to the quiz.
Students still need to do the homework to learn, because they will need to be able to solve the problems on the quiz. This way, they use the learning aids as they were intended – to increase understanding.
This is a win-win situation. We can get students to do the homework and demonstrate proficiency on the quizzes, without inflating their grades.
Summary
MyMathLab homework is not the “magic bullet”. It needs to be one part of a comprehensive plan to help students understand. Using quizzes, without learning aids, is one way to make sure students understand and are not simply clicking through problems until they get one they have seen before. Also, if you want to make sure your students are understanding the material, find a way to check out their written work. I give quick quizzes at the beginning of class and group assignments at the end of class so I can take a look at their work.
If you have any questions, leave a comment or drop me a line through the contact page on my web site. – George
Note – MyMathLab related articles appear every Monday.
I am a math instructor at College of the Sequoias in Visalia, CA. If there are topics you’d like me to address in future MyMathLab articles, send in your requests through the contact page on my web site. – George | 889 | 4,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-18 | latest | en | 0.960007 |
https://root.cern.ch/root/html526/src/TGeoArb8.cxx.html | 1,685,932,767,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650620.66/warc/CC-MAIN-20230605021141-20230605051141-00541.warc.gz | 536,292,640 | 37,807 | ```// @(#)root/geom:\$Id: TGeoArb8.cxx 27731 2009-03-09 17:40:56Z brun \$
// Author: Andrei Gheata 31/01/02
/*************************************************************************
* *
* For the licensing terms see \$ROOTSYS/LICENSE. *
* For the list of contributors see \$ROOTSYS/README/CREDITS. *
*************************************************************************/
#include "Riostream.h"
#include "TGeoManager.h"
#include "TGeoVolume.h"
#include "TGeoArb8.h"
#include "TGeoMatrix.h"
#include "TMath.h"
ClassImp(TGeoArb8)
//________________________________________________________________________
// TGeoArb8 - a arbitrary trapezoid with less than 8 vertices standing on
// two paralel planes perpendicular to Z axis. Parameters :
// - dz - half length in Z;
// - xy[8][2] - vector of (x,y) coordinates of vertices
// - first four points (xy[i][j], i<4, j<2) are the (x,y)
// coordinates of the vertices sitting on the -dz plane;
// - last four points (xy[i][j], i>=4, j<2) are the (x,y)
// coordinates of the vertices sitting on the +dz plane;
// The order of defining the vertices of an arb8 is the following :
// - point 0 is connected with points 1,3,4
// - point 1 is connected with points 0,2,5
// - point 2 is connected with points 1,3,6
// - point 3 is connected with points 0,2,7
// - point 4 is connected with points 0,5,7
// - point 5 is connected with points 1,4,6
// - point 6 is connected with points 2,5,7
// - point 7 is connected with points 3,4,6
// Points can be identical in order to create shapes with less than
// 8 vertices.
//
//Begin_Html
/*
<img src="gif/t_arb8.gif">
*/
//End_Html
////////////////////////////////////////////////////////////////////////////
// //
// TGeoTrap //
// //
// TRAP is a general trapezoid, i.e. one for which the faces perpendicular//
// to z are trapezia and their centres are not the same x, y. It has 11 //
// parameters: the half length in z, the polar angles from the centre of //
// the face at low z to that at high z, H1 the half length in y at low z, //
// LB1 the half length in x at low z and y low edge, LB2 the half length //
// in x at low z and y high edge, TH1 the angle w.r.t. the y axis from the//
// centre of low y edge to the centre of the high y edge, and H2, LB2, //
// LH2, TH2, the corresponding quantities at high z. //
// //
////////////////////////////////////////////////////////////////////////////
//Begin_Html
/*
<img src="gif/t_trap.gif">
*/
//End_Html
//
//Begin_Html
/*
<img src="gif/t_trapdivZ.gif">
*/
//End_Html
////////////////////////////////////////////////////////////////////////////
// //
// TGeoGtra //
// //
// Gtra is a twisted trapezoid, i.e. one for which the faces perpendicular//
// to z are trapezia and their centres are not the same x, y. It has 12 //
// parameters: the half length in z, the polar angles from the centre of //
// the face at low z to that at high z, twist, H1 the half length in y at low z, //
// LB1 the half length in x at low z and y low edge, LB2 the half length //
// in x at low z and y high edge, TH1 the angle w.r.t. the y axis from the//
// centre of low y edge to the centre of the high y edge, and H2, LB2, //
// LH2, TH2, the corresponding quantities at high z. //
// //
////////////////////////////////////////////////////////////////////////////
//Begin_Html
/*
<img src="gif/t_gtra.gif">
*/
//End_Html
//
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//_____________________________________________________________________________
TGeoArb8::TGeoArb8()
{
// Default ctor.
fDz = 0;
fTwist = 0;
for (Int_t i=0; i<8; i++) {
fXY[i][0] = 0.0;
fXY[i][1] = 0.0;
}
SetShapeBit(kGeoArb8);
}
//_____________________________________________________________________________
TGeoArb8::TGeoArb8(Double_t dz, Double_t *vertices)
:TGeoBBox(0,0,0)
{
// Constructor. If the array of vertices is not null, this should be
// in the format : (x0, y0, x1, y1, ... , x7, y7)
fDz = dz;
fTwist = 0;
SetShapeBit(kGeoArb8);
if (vertices) {
for (Int_t i=0; i<8; i++) {
fXY[i][0] = vertices[2*i];
fXY[i][1] = vertices[2*i+1];
}
ComputeTwist();
ComputeBBox();
} else {
for (Int_t i=0; i<8; i++) {
fXY[i][0] = 0.0;
fXY[i][1] = 0.0;
}
}
}
//_____________________________________________________________________________
TGeoArb8::TGeoArb8(const char *name, Double_t dz, Double_t *vertices)
:TGeoBBox(name, 0,0,0)
{
// Named constructor. If the array of vertices is not null, this should be
// in the format : (x0, y0, x1, y1, ... , x7, y7)
fDz = dz;
fTwist = 0;
SetShapeBit(kGeoArb8);
if (vertices) {
for (Int_t i=0; i<8; i++) {
fXY[i][0] = vertices[2*i];
fXY[i][1] = vertices[2*i+1];
}
ComputeTwist();
ComputeBBox();
} else {
for (Int_t i=0; i<8; i++) {
fXY[i][0] = 0.0;
fXY[i][1] = 0.0;
}
}
}
//_____________________________________________________________________________
TGeoArb8::TGeoArb8(const TGeoArb8& ga8) :
TGeoBBox(ga8),
fDz(ga8.fDz),
fTwist(ga8.fTwist)
{
//copy constructor
for(Int_t i=0; i<8; i++) {
fXY[i][0]=ga8.fXY[i][0];
fXY[i][1]=ga8.fXY[i][1];
}
}
//_____________________________________________________________________________
TGeoArb8& TGeoArb8::operator=(const TGeoArb8& ga8)
{
//assignment operator
if(this!=&ga8) {
TGeoBBox::operator=(ga8);
fDz=ga8.fDz;
fTwist=ga8.fTwist;
for(Int_t i=0; i<8; i++) {
fXY[i][0]=ga8.fXY[i][0];
fXY[i][1]=ga8.fXY[i][1];
}
}
return *this;
}
//_____________________________________________________________________________
TGeoArb8::~TGeoArb8()
{
// Destructor.
if (fTwist) delete [] fTwist;
}
//_____________________________________________________________________________
Double_t TGeoArb8::Capacity() const
{
// Computes capacity of the shape in [length^3].
Int_t i,j;
Double_t capacity = 0;
for (i=0; i<4; i++) {
j = (i+1)%4;
capacity += 0.25*fDz*((fXY[i][0]+fXY[i+4][0])*(fXY[j][1]+fXY[j+4][1]) -
(fXY[j][0]+fXY[j+4][0])*(fXY[i][1]+fXY[i+4][1]) +
(1./3)*((fXY[i+4][0]-fXY[i][0])*(fXY[j+4][1]-fXY[j][1]) -
(fXY[j][0]-fXY[j+4][0])*(fXY[i][1]-fXY[i+4][1])));
}
return TMath::Abs(capacity);
}
//_____________________________________________________________________________
void TGeoArb8::ComputeBBox()
{
// Computes bounding box for an Arb8 shape.
Double_t xmin, xmax, ymin, ymax;
xmin = xmax = fXY[0][0];
ymin = ymax = fXY[0][1];
for (Int_t i=1; i<8; i++) {
if (xmin>fXY[i][0]) xmin=fXY[i][0];
if (xmax<fXY[i][0]) xmax=fXY[i][0];
if (ymin>fXY[i][1]) ymin=fXY[i][1];
if (ymax<fXY[i][1]) ymax=fXY[i][1];
}
fDX = 0.5*(xmax-xmin);
fDY = 0.5*(ymax-ymin);
fDZ = fDz;
fOrigin[0] = 0.5*(xmax+xmin);
fOrigin[1] = 0.5*(ymax+ymin);
fOrigin[2] = 0;
SetShapeBit(kGeoClosedShape);
}
//_____________________________________________________________________________
void TGeoArb8::ComputeTwist()
{
// Computes tangents of twist angles (angles between projections on XY plane
// of corresponding -dz +dz edges). Computes also if the vertices are defined
// clockwise or anti-clockwise.
Double_t twist[4];
Bool_t twisted = kFALSE;
Double_t dx1, dy1, dx2, dy2;
for (Int_t i=0; i<4; i++) {
dx1 = fXY[(i+1)%4][0]-fXY[i][0];
dy1 = fXY[(i+1)%4][1]-fXY[i][1];
if (TMath::Abs(dx1)<TGeoShape::Tolerance() && TMath::Abs(dy1)<TGeoShape::Tolerance()) {
twist[i] = 0;
continue;
}
dx2 = fXY[4+(i+1)%4][0]-fXY[4+i][0];
dy2 = fXY[4+(i+1)%4][1]-fXY[4+i][1];
if (TMath::Abs(dx2)<TGeoShape::Tolerance() && TMath::Abs(dy2)<TGeoShape::Tolerance()) {
twist[i] = 0;
continue;
}
twist[i] = dy1*dx2 - dx1*dy2;
if (TMath::Abs(twist[i])<TGeoShape::Tolerance()) {
twist[i] = 0;
continue;
}
twist[i] = TMath::Sign(1.,twist[i]);
twisted = kTRUE;
}
if (twisted) {
if (fTwist) delete [] fTwist;
fTwist = new Double_t[4];
memcpy(fTwist, &twist[0], 4*sizeof(Double_t));
}
Double_t sum1 = 0.;
Double_t sum2 = 0.;
Int_t j;
for (Int_t i=0; i<4; i++) {
j = (i+1)%4;
sum1 += fXY[i][0]*fXY[j][1]-fXY[j][0]*fXY[i][1];
sum2 += fXY[i+4][0]*fXY[j+4][1]-fXY[j+4][0]*fXY[i+4][1];
}
if (sum1*sum2 < -TGeoShape::Tolerance()) {
Fatal("ComputeTwist", "Shape %s type Arb8: Lower/upper faces defined with opposite clockwise", GetName());
return;
}
if (sum1>0.) {
Error("ComputeTwist", "Shape %s type Arb8: Vertices must be defined clockwise in XY planes. Re-ordering...", GetName());
Double_t xtemp, ytemp;
xtemp = fXY[1][0];
ytemp = fXY[1][1];
fXY[1][0] = fXY[3][0];
fXY[1][1] = fXY[3][1];
fXY[3][0] = xtemp;
fXY[3][1] = ytemp;
xtemp = fXY[5][0];
ytemp = fXY[5][1];
fXY[5][0] = fXY[7][0];
fXY[5][1] = fXY[7][1];
fXY[7][0] = xtemp;
fXY[7][1] = ytemp;
}
// Check for illegal crossings.
Bool_t illegal_cross = kFALSE;
illegal_cross = TGeoShape::IsSegCrossing(fXY[0][0],fXY[0][1],fXY[1][0],fXY[1][1],
fXY[2][0],fXY[2][1],fXY[3][0],fXY[3][1]);
if (!illegal_cross)
illegal_cross = TGeoShape::IsSegCrossing(fXY[4][0],fXY[4][1],fXY[5][0],fXY[5][1],
fXY[6][0],fXY[6][1],fXY[7][0],fXY[7][1]);
if (illegal_cross) {
Error("ComputeTwist", "Shape %s type Arb8: Malformed polygon with crossing opposite segments", GetName());
InspectShape();
}
}
//_____________________________________________________________________________
Double_t TGeoArb8::GetTwist(Int_t iseg) const
{
// Get twist for segment I in range [0,3]
if (!fTwist) return 0.;
if (iseg<0 || iseg>3) return 0.;
return fTwist[iseg];
}
//_____________________________________________________________________________
void TGeoArb8::ComputeNormal(Double_t *point, Double_t *dir, Double_t *norm)
{
// Compute normal to closest surface from POINT.
Double_t safe = TGeoShape::Big();
Double_t safc;
Int_t i; // current facette index
Double_t x0, y0, z0, x1, y1, z1, x2, y2;
Double_t ax, ay, az, bx, by;
Double_t fn;
safc = fDz-TMath::Abs(point[2]);
if (safc<1E-4) {
memset(norm,0,3*sizeof(Double_t));
norm[2] = (dir[2]>0)?1:(-1);
return;
}
Double_t vert[8], lnorm[3];
SetPlaneVertices(point[2], vert);
//---> compute safety for lateral planes
for (i=0; i<4; i++) {
x0 = vert[2*i];
y0 = vert[2*i+1];
z0 = point[2];
x1 = fXY[i+4][0];
y1 = fXY[i+4][1];
z1 = fDz;
ax = x1-x0;
ay = y1-y0;
az = z1-z0;
x2 = vert[2*((i+1)%4)];
y2 = vert[2*((i+1)%4)+1];
bx = x2-x0;
by = y2-y0;
lnorm[0] = -az*by;
lnorm[1] = az*bx;
lnorm[2] = ax*by-ay*bx;
fn = TMath::Sqrt(lnorm[0]*lnorm[0]+lnorm[1]*lnorm[1]+lnorm[2]*lnorm[2]);
if (fn<1E-10) continue;
lnorm[0] /= fn;
lnorm[1] /= fn;
lnorm[2] /= fn;
safc = (x0-point[0])*lnorm[0]+(y0-point[1])*lnorm[1]+(z0-point[2])*lnorm[2];
safc = TMath::Abs(safc);
// printf("plane %i : (%g, %g, %g) safe=%g\n", i, lnorm[0],lnorm[1],lnorm[2],safc);
if (safc<safe) {
safe = safc;
memcpy(norm,lnorm,3*sizeof(Double_t));
}
}
if (dir[0]*norm[0]+dir[1]*norm[1]+dir[2]*norm[2] < 0) {
norm[0] = -norm[0];
norm[1] = -norm[1];
norm[2] = -norm[2];
}
}
//_____________________________________________________________________________
Bool_t TGeoArb8::Contains(Double_t *point) const
{
// Test if point is inside this shape.
// first check Z range
if (TMath::Abs(point[2]) > fDz) return kFALSE;
// compute intersection between Z plane containing point and the arb8
Double_t poly[8];
// memset(&poly[0], 0, 8*sizeof(Double_t));
//SetPlaneVertices(point[2], &poly[0]);
Double_t cf = 0.5*(fDz-point[2])/fDz;
Int_t i;
for (i=0; i<4; i++) {
poly[2*i] = fXY[i+4][0]+cf*(fXY[i][0]-fXY[i+4][0]);
poly[2*i+1] = fXY[i+4][1]+cf*(fXY[i][1]-fXY[i+4][1]);
}
return InsidePolygon(point[0],point[1],poly);
}
//_____________________________________________________________________________
Double_t TGeoArb8::DistToPlane(Double_t *point, Double_t *dir, Int_t ipl, Bool_t in) const
{
// Computes distance to plane ipl :
// ipl=0 : points 0,4,1,5
// ipl=1 : points 1,5,2,6
// ipl=2 : points 2,6,3,7
// ipl=3 : points 3,7,0,4
Double_t xa,xb,xc,xd;
Double_t ya,yb,yc,yd;
Double_t eps = 100.*TGeoShape::Tolerance();
Int_t j = (ipl+1)%4;
xa=fXY[ipl][0];
ya=fXY[ipl][1];
xb=fXY[ipl+4][0];
yb=fXY[ipl+4][1];
xc=fXY[j][0];
yc=fXY[j][1];
xd=fXY[4+j][0];
yd=fXY[4+j][1];
Double_t dz2 =0.5/fDz;
Double_t tx1 =dz2*(xb-xa);
Double_t ty1 =dz2*(yb-ya);
Double_t tx2 =dz2*(xd-xc);
Double_t ty2 =dz2*(yd-yc);
Double_t dzp =fDz+point[2];
Double_t xs1 =xa+tx1*dzp;
Double_t ys1 =ya+ty1*dzp;
Double_t xs2 =xc+tx2*dzp;
Double_t ys2 =yc+ty2*dzp;
Double_t dxs =xs2-xs1;
Double_t dys =ys2-ys1;
Double_t dtx =tx2-tx1;
Double_t dty =ty2-ty1;
Double_t a=(dtx*dir[1]-dty*dir[0]+(tx1*ty2-tx2*ty1)*dir[2])*dir[2];
Double_t b=dxs*dir[1]-dys*dir[0]+(dtx*point[1]-dty*point[0]+ty2*xs1-ty1*xs2
+tx1*ys2-tx2*ys1)*dir[2];
Double_t c=dxs*point[1]-dys*point[0]+xs1*ys2-xs2*ys1;
Double_t s=TGeoShape::Big();
Double_t x1,x2,y1,y2,xp,yp,zi;
if (TMath::Abs(a)<eps) {
if (TMath::Abs(b)<eps) return TGeoShape::Big();
s=-c/b;
if (s>eps) {
if (in) return s;
zi=point[2]+s*dir[2];
if (TMath::Abs(zi)<fDz) {
x1=xs1+tx1*dir[2]*s;
x2=xs2+tx2*dir[2]*s;
xp=point[0]+s*dir[0];
y1=ys1+ty1*dir[2]*s;
y2=ys2+ty2*dir[2]*s;
yp=point[1]+s*dir[1];
zi = (xp-x1)*(xp-x2)+(yp-y1)*(yp-y2);
if (zi<=0) return s;
}
}
return TGeoShape::Big();
}
Double_t d=b*b-4*a*c;
if (d>=0) {
if (a>0) s=0.5*(-b-TMath::Sqrt(d))/a;
else s=0.5*(-b+TMath::Sqrt(d))/a;
if (s>eps) {
if (in) return s;
zi=point[2]+s*dir[2];
if (TMath::Abs(zi)<fDz) {
x1=xs1+tx1*dir[2]*s;
x2=xs2+tx2*dir[2]*s;
xp=point[0]+s*dir[0];
y1=ys1+ty1*dir[2]*s;
y2=ys2+ty2*dir[2]*s;
yp=point[1]+s*dir[1];
zi = (xp-x1)*(xp-x2)+(yp-y1)*(yp-y2);
if (zi<=0) return s;
}
}
if (a>0) s=0.5*(-b+TMath::Sqrt(d))/a;
else s=0.5*(-b-TMath::Sqrt(d))/a;
if (s>eps) {
if (in) return s;
zi=point[2]+s*dir[2];
if (TMath::Abs(zi)<fDz) {
x1=xs1+tx1*dir[2]*s;
x2=xs2+tx2*dir[2]*s;
xp=point[0]+s*dir[0];
y1=ys1+ty1*dir[2]*s;
y2=ys2+ty2*dir[2]*s;
yp=point[1]+s*dir[1];
zi = (xp-x1)*(xp-x2)+(yp-y1)*(yp-y2);
if (zi<=0) return s;
}
}
}
return TGeoShape::Big();
}
//_____________________________________________________________________________
Double_t TGeoArb8::DistFromOutside(Double_t *point, Double_t *dir, Int_t /*iact*/, Double_t step, Double_t * /*safe*/) const
{
// Computes distance from outside point to surface of the shape.
Double_t sdist = TGeoBBox::DistFromOutside(point,dir, fDX, fDY, fDZ, fOrigin, step);
if (sdist>=step) return TGeoShape::Big();
Double_t dist[5];
// check lateral faces
Int_t i;
for (i=0; i<4; i++) {
dist[i]=DistToPlane(point, dir, i, kFALSE);
}
// check Z planes
dist[4]=TGeoShape::Big();
if (TMath::Abs(point[2])>fDz) {
if (!TGeoShape::IsSameWithinTolerance(dir[2],0)) {
Double_t pt[3];
if (point[2]>0) {
dist[4] = (fDz-point[2])/dir[2];
pt[2]=fDz;
} else {
dist[4] = (-fDz-point[2])/dir[2];
pt[2]=-fDz;
}
if (dist[4]<0) {
dist[4]=TGeoShape::Big();
} else {
for (Int_t j=0; j<2; j++) pt[j]=point[j]+dist[4]*dir[j];
if (!Contains(&pt[0])) dist[4]=TGeoShape::Big();
}
}
}
Double_t distmin = dist[0];
for (i=1;i<5;i++) if (dist[i] < distmin) distmin = dist[i];
return distmin;
}
//_____________________________________________________________________________
Double_t TGeoArb8::DistFromInside(Double_t *point, Double_t *dir, Int_t /*iact*/, Double_t /*step*/, Double_t * /*safe*/) const
{
// Compute distance from inside point to surface of the shape.
#ifdef OLDALGORITHM
Int_t i;
Double_t dist[6];
dist[0]=dist[1]=TGeoShape::Big();
if (dir[2]<0) {
dist[0]=(-fDz-point[2])/dir[2];
} else {
if (dir[2]>0) dist[1]=(fDz-point[2])/dir[2];
}
for (i=0; i<4; i++) {
dist[i+2]=DistToPlane(point, dir, i, kTRUE);
}
Double_t distmin = dist[0];
for (i=1;i<6;i++) if (dist[i] < distmin) distmin = dist[i];
return distmin;
#else
// compute distance to plane ipl :
// ipl=0 : points 0,4,1,5
// ipl=1 : points 1,5,2,6
// ipl=2 : points 2,6,3,7
// ipl=3 : points 3,7,0,4
Double_t distmin;
Bool_t lateral_cross = kFALSE;
if (dir[2]<0) {
distmin=(-fDz-point[2])/dir[2];
} else {
if (dir[2]>0) distmin =(fDz-point[2])/dir[2];
else distmin = TGeoShape::Big();
}
Double_t dz2 =0.5/fDz;
Double_t xa,xb,xc,xd;
Double_t ya,yb,yc,yd;
Double_t eps = 100.*TGeoShape::Tolerance();
for (Int_t ipl=0;ipl<4;ipl++) {
Int_t j = (ipl+1)%4;
xa=fXY[ipl][0];
ya=fXY[ipl][1];
xb=fXY[ipl+4][0];
yb=fXY[ipl+4][1];
xc=fXY[j][0];
yc=fXY[j][1];
xd=fXY[4+j][0];
yd=fXY[4+j][1];
Double_t tx1 =dz2*(xb-xa);
Double_t ty1 =dz2*(yb-ya);
Double_t tx2 =dz2*(xd-xc);
Double_t ty2 =dz2*(yd-yc);
Double_t dzp =fDz+point[2];
Double_t xs1 =xa+tx1*dzp;
Double_t ys1 =ya+ty1*dzp;
Double_t xs2 =xc+tx2*dzp;
Double_t ys2 =yc+ty2*dzp;
Double_t dxs =xs2-xs1;
Double_t dys =ys2-ys1;
Double_t dtx =tx2-tx1;
Double_t dty =ty2-ty1;
Double_t a=(dtx*dir[1]-dty*dir[0]+(tx1*ty2-tx2*ty1)*dir[2])*dir[2];
Double_t b=dxs*dir[1]-dys*dir[0]+(dtx*point[1]-dty*point[0]+ty2*xs1-ty1*xs2
+tx1*ys2-tx2*ys1)*dir[2];
Double_t c=dxs*point[1]-dys*point[0]+xs1*ys2-xs2*ys1;
Double_t s=TGeoShape::Big();
if (TMath::Abs(a)<eps) {
if (TMath::Abs(b)<eps) continue;
s=-c/b;
if (s>eps && s < distmin) {
distmin =s;
lateral_cross=kTRUE;
}
continue;
}
Double_t d=b*b-4*a*c;
if (d>=0.) {
if (a>0) s=0.5*(-b-TMath::Sqrt(d))/a;
else s=0.5*(-b+TMath::Sqrt(d))/a;
if (s>eps) {
if (s < distmin) {
distmin = s;
lateral_cross = kTRUE;
}
} else {
if (a>0) s=0.5*(-b+TMath::Sqrt(d))/a;
else s=0.5*(-b-TMath::Sqrt(d))/a;
if (s>eps && s < distmin) {
distmin =s;
lateral_cross = kTRUE;
}
}
}
}
if (!lateral_cross) {
// We have to make sure that track crosses the top or bottom.
if (distmin > 1.E10) return TGeoShape::Tolerance();
Double_t pt[2];
pt[0] = point[0]+distmin*dir[0];
pt[1] = point[1]+distmin*dir[1];
// Check if propagated point is in the polygon
Double_t poly[8];
Int_t i = 0;
if (dir[2]>0.) i=4;
for (Int_t j=0; j<4; j++) {
poly[2*j] = fXY[j+i][0];
poly[2*j+1] = fXY[j+i][1];
}
if (!InsidePolygon(pt[0],pt[1],poly)) return TGeoShape::Tolerance();
}
return distmin;
#endif
}
//_____________________________________________________________________________
TGeoVolume *TGeoArb8::Divide(TGeoVolume *voldiv, const char * /*divname*/, Int_t /*iaxis*/, Int_t /*ndiv*/,
Double_t /*start*/, Double_t /*step*/)
{
// Divide this shape along one axis.
Error("Divide", "Division of an arbitrary trapezoid not implemented");
return voldiv;
}
//_____________________________________________________________________________
Double_t TGeoArb8::GetAxisRange(Int_t iaxis, Double_t &xlo, Double_t &xhi) const
{
// Get shape range on a given axis.
xlo = 0;
xhi = 0;
Double_t dx = 0;
if (iaxis==3) {
xlo = -fDz;
xhi = fDz;
dx = xhi-xlo;
return dx;
}
return dx;
}
//_____________________________________________________________________________
void TGeoArb8::GetBoundingCylinder(Double_t *param) const
{
//--- Fill vector param[4] with the bounding cylinder parameters. The order
// is the following : Rmin, Rmax, Phi1, Phi2
//--- first compute rmin/rmax
Double_t rmaxsq = 0;
Double_t rsq;
Int_t i;
for (i=0; i<8; i++) {
rsq = fXY[i][0]*fXY[i][0] + fXY[i][1]*fXY[i][1];
rmaxsq = TMath::Max(rsq, rmaxsq);
}
param[0] = 0.; // Rmin
param[1] = rmaxsq; // Rmax
param[2] = 0.; // Phi1
param[3] = 360.; // Phi2
}
//_____________________________________________________________________________
Int_t TGeoArb8::GetFittingBox(const TGeoBBox *parambox, TGeoMatrix *mat, Double_t &dx, Double_t &dy, Double_t &dz) const
{
// Fills real parameters of a positioned box inside this arb8. Returns 0 if successfull.
dx=dy=dz=0;
if (mat->IsRotation()) {
Error("GetFittingBox", "cannot handle parametrized rotated volumes");
return 1; // ### rotation not accepted ###
}
//--> translate the origin of the parametrized box to the frame of this box.
Double_t origin[3];
mat->LocalToMaster(parambox->GetOrigin(), origin);
if (!Contains(origin)) {
Error("GetFittingBox", "wrong matrix - parametrized box is outside this");
return 1; // ### wrong matrix ###
}
//--> now we have to get the valid range for all parametrized axis
Double_t dd[3];
dd[0] = parambox->GetDX();
dd[1] = parambox->GetDY();
dd[2] = parambox->GetDZ();
//-> check if Z range is fixed
if (dd[2]<0) {
dd[2] = TMath::Min(origin[2]+fDz, fDz-origin[2]);
if (dd[2]<0) {
Error("GetFittingBox", "wrong matrix");
return 1;
}
}
if (dd[0]>=0 && dd[1]>=0) {
dx = dd[0];
dy = dd[1];
dz = dd[2];
return 0;
}
//-> check now vertices at Z = origin[2] +/- dd[2]
Double_t upper[8];
Double_t lower[8];
SetPlaneVertices(origin[2]-dd[2], lower);
SetPlaneVertices(origin[2]+dd[2], upper);
Double_t ddmin=TGeoShape::Big();
for (Int_t iaxis=0; iaxis<2; iaxis++) {
if (dd[iaxis]>=0) continue;
ddmin=TGeoShape::Big();
for (Int_t ivert=0; ivert<4; ivert++) {
ddmin = TMath::Min(ddmin, TMath::Abs(origin[iaxis]-lower[2*ivert+iaxis]));
ddmin = TMath::Min(ddmin, TMath::Abs(origin[iaxis]-upper[2*ivert+iaxis]));
}
dd[iaxis] = ddmin;
}
dx = dd[0];
dy = dd[1];
dz = dd[2];
return 0;
}
//_____________________________________________________________________________
void TGeoArb8::GetPlaneNormal(Double_t *p1, Double_t *p2, Double_t *p3, Double_t *norm)
{
// Computes normal to plane defined by P1, P2 and P3
Double_t cross = 0.;
Double_t v1[3], v2[3];
Int_t i;
for (i=0; i<3; i++) {
v1[i] = p2[i] - p1[i];
v2[i] = p3[i] - p1[i];
}
norm[0] = v1[1]*v2[2]-v1[2]*v2[1];
cross += norm[0]*norm[0];
norm[1] = v1[2]*v2[0]-v1[0]*v2[2];
cross += norm[1]*norm[1];
norm[2] = v1[0]*v2[1]-v1[1]*v2[0];
cross += norm[2]*norm[2];
if (TMath::Abs(cross) < TGeoShape::Tolerance()) return;
cross = 1./TMath::Sqrt(cross);
for (i=0; i<3; i++) norm[i] *= cross;
}
//_____________________________________________________________________________
Bool_t TGeoArb8::GetPointsOnFacet(Int_t /*index*/, Int_t /*npoints*/, Double_t * /* array */) const
{
// Fills array with n random points located on the surface of indexed facet.
// The output array must be provided with a length of minimum 3*npoints. Returns
// true if operation succeeded.
// Possible index values:
// 0 - all facets togeather
// 1 to 6 - facet index from bottom to top Z
return kFALSE;
/*
if (index<0 || index>6) return kFALSE;
if (index==0) {
// Just generate same number of points on each facet
Int_t npts = npoints/6.;
Int_t count = 0;
for (Int_t ifacet=0; ifacet<6; ifacet++) {
if (GetPointsOnFacet(ifacet+1, npts, &array[3*count])) count += npts;
if (ifacet<5) npts = (npoints-count)/(5.-ifacet);
}
if (count>0) return kTRUE;
return kFALSE;
}
Double_t z, cf;
Double_t xmin=TGeoShape::Big();
Double_t xmax=-xmin;
Double_t ymin=TGeoShape::Big();
Double_t ymax=-ymin;
Double_t dy=0.;
Double_t poly[8];
Double_t point[2];
Int_t i;
if (index==1 || index==6) {
z = (index==1)?-fDz:fDz;
cf = 0.5*(fDz-z)/fDz;
for (i=0; i<4; i++) {
poly[2*i] = fXY[i+4][0]+cf*(fXY[i][0]-fXY[i+4][0]);
poly[2*i+1] = fXY[i+4][1]+cf*(fXY[i][1]-fXY[i+4][1]);
xmin = TMath::Min(xmin, poly[2*i]);
xmax = TMath::Max(xmax, poly[2*i]);
ymin = TMath::Min(ymin, poly[2*i]);
ymax = TMath::Max(ymax, poly[2*i]);
}
}
Int_t nshoot = 0;
Int_t nmiss = 0;
for (i=0; i<npoints; i++) {
Double_t *point = &array[3*i];
switch (surfindex) {
case 1:
case 6:
while (nmiss<1000) {
point[0] = xmin + (xmax-xmin)*gRandom->Rndm();
point[1] = ymin + (ymax-ymin)*gRandom->Rndm();
}
return InsidePolygon(point[0],point[1],poly);
*/
}
//_____________________________________________________________________________
Bool_t TGeoArb8::InsidePolygon(Double_t x, Double_t y, Double_t *pts)
{
// Finds if a point in XY plane is inside the polygon defines by PTS.
Int_t i,j;
Double_t x1,y1,x2,y2;
Double_t cross;
for (i=0; i<4; i++) {
j = (i+1)%4;
x1 = pts[i<<1];
y1 = pts[(i<<1)+1];
x2 = pts[j<<1];
y2 = pts[(j<<1)+1];
cross = (x-x1)*(y2-y1)-(y-y1)*(x2-x1);
if (cross<0) return kFALSE;
}
return kTRUE;
}
//_____________________________________________________________________________
void TGeoArb8::InspectShape() const
{
// Prints shape parameters
printf("*** Shape %s: TGeoArb8 ***\n", GetName());
if (IsTwisted()) printf(" = TWISTED\n");
for (Int_t ip=0; ip<8; ip++) {
printf(" point #%i : x=%11.5f y=%11.5f z=%11.5f\n",
ip, fXY[ip][0], fXY[ip][1], fDz*((ip<4)?-1:1));
}
printf(" Bounding box:\n");
TGeoBBox::InspectShape();
}
//_____________________________________________________________________________
Double_t TGeoArb8::Safety(Double_t *point, Bool_t in) const
{
// Computes the closest distance from given point to this shape.
Double_t safz = fDz-TMath::Abs(point[2]);
if (!in) safz = -safz;
Int_t iseg;
Double_t safe = TGeoShape::Big();
Double_t lsq, ssq, dx, dy, dpx, dpy, u;
if (IsTwisted()) {
if (!in) {
if (!TGeoBBox::Contains(point)) return TGeoBBox::Safety(point,kFALSE);
}
// Point is also in the bounding box ;-(
// Compute closest distance to any segment
Double_t vert[8];
Double_t *p1, *p2;
Int_t isegmin=0;
Double_t umin = 0.;
SetPlaneVertices (point[2], vert);
for (iseg=0; iseg<4; iseg++) {
if (safe<TGeoShape::Tolerance()) return 0.;
p1 = &vert[2*iseg];
p2 = &vert[2*((iseg+1)%4)];
dx = p2[0] - p1[0];
dy = p2[1] - p1[1];
dpx = point[0] - p1[0];
dpy = point[1] - p1[1];
lsq = dx*dx + dy*dy;
u = (dpx*dx + dpy*dy)/lsq;
if (u>1) {
dpx = point[0]-p2[0];
dpy = point[1]-p2[1];
} else {
if (u>=0) {
dpx -= u*dx;
dpy -= u*dy;
}
}
ssq = dpx*dpx + dpy*dpy;
if (ssq < safe) {
isegmin = iseg;
umin = u;
safe = ssq;
}
}
if (umin<0) umin = 0.;
if (umin>1) {
isegmin = (isegmin+1)%4;
umin = 0.;
}
Int_t i1 = isegmin;
Int_t i2 = (isegmin+1)%4;
Double_t dx1 = fXY[i2][0]-fXY[i1][0];
Double_t dx2 = fXY[i2+4][0]-fXY[i1+4][0];
Double_t dy1 = fXY[i2][1]-fXY[i1][1];
Double_t dy2 = fXY[i2+4][1]-fXY[i1+4][1];
dx = dx1 + umin*(dx2-dx1);
dy = dy1 + umin*(dy2-dy1);
safe *= 1.- 4.*fDz*fDz/(dx*dx+dy*dy+4.*fDz*fDz);
safe = TMath::Sqrt(safe);
return safe;
}
Double_t saf[5];
saf[0] = safz;
for (iseg=0; iseg<4; iseg++) saf[iseg+1] = SafetyToFace(point,iseg,in);
if (in) safe = saf[TMath::LocMin(5, saf)];
else safe = saf[TMath::LocMax(5, saf)];
if (safe<0) return 0.;
return safe;
}
//_____________________________________________________________________________
Double_t TGeoArb8::SafetyToFace(Double_t *point, Int_t iseg, Bool_t in) const
{
// Estimate safety to lateral plane defined by segment iseg in range [0,3]
// Might be negative: plane seen only from inside.
Double_t vertices[12];
Int_t ipln = (iseg+1)%4;
// point 1
vertices[0] = fXY[iseg][0];
vertices[1] = fXY[iseg][1];
vertices[2] = -fDz;
// point 2
vertices[3] = fXY[ipln][0];
vertices[4] = fXY[ipln][1];
vertices[5] = -fDz;
// point 3
vertices[6] = fXY[ipln+4][0];
vertices[7] = fXY[ipln+4][1];
vertices[8] = fDz;
// point 4
vertices[9] = fXY[iseg+4][0];
vertices[10] = fXY[iseg+4][1];
vertices[11] = fDz;
Double_t safe;
Double_t norm[3];
Double_t *p1, *p2, *p3;
p1 = &vertices[0];
p2 = &vertices[9];
p3 = &vertices[6];
if (IsSamePoint(p2,p3)) {
p3 = &vertices[3];
if (IsSamePoint(p1,p3)) return -TGeoShape::Big(); // skip single segment
}
GetPlaneNormal(p1,p2,p3,norm);
safe = (point[0]-p1[0])*norm[0]+(point[1]-p1[1])*norm[1]+(point[2]-p1[2])*norm[2];
if (in) return (-safe);
return safe;
}
//_____________________________________________________________________________
void TGeoArb8::SavePrimitive(ostream &out, Option_t * /*option*/ /*= ""*/)
{
// Save a primitive as a C++ statement(s) on output stream "out".
if (TObject::TestBit(kGeoSavePrimitive)) return;
out << " // Shape: " << GetName() << " type: " << ClassName() << endl;
out << " dz = " << fDz << ";" << endl;
out << " vert[0] = " << fXY[0][0] << ";" << endl;
out << " vert[1] = " << fXY[0][1] << ";" << endl;
out << " vert[2] = " << fXY[1][0] << ";" << endl;
out << " vert[3] = " << fXY[1][1] << ";" << endl;
out << " vert[4] = " << fXY[2][0] << ";" << endl;
out << " vert[5] = " << fXY[2][1] << ";" << endl;
out << " vert[6] = " << fXY[3][0] << ";" << endl;
out << " vert[7] = " << fXY[3][1] << ";" << endl;
out << " vert[8] = " << fXY[4][0] << ";" << endl;
out << " vert[9] = " << fXY[4][1] << ";" << endl;
out << " vert[10] = " << fXY[5][0] << ";" << endl;
out << " vert[11] = " << fXY[5][1] << ";" << endl;
out << " vert[12] = " << fXY[6][0] << ";" << endl;
out << " vert[13] = " << fXY[6][1] << ";" << endl;
out << " vert[14] = " << fXY[7][0] << ";" << endl;
out << " vert[15] = " << fXY[7][1] << ";" << endl;
out << " TGeoShape *" << GetPointerName() << " = new TGeoArb8(\"" << GetName() << "\", dz,vert);" << endl;
TObject::SetBit(TGeoShape::kGeoSavePrimitive);
}
//_____________________________________________________________________________
void TGeoArb8::SetPlaneVertices(Double_t zpl, Double_t *vertices) const
{
// Computes intersection points between plane at zpl and non-horizontal edges.
Double_t cf = 0.5*(fDz-zpl)/fDz;
for (Int_t i=0; i<4; i++) {
vertices[2*i] = fXY[i+4][0]+cf*(fXY[i][0]-fXY[i+4][0]);
vertices[2*i+1] = fXY[i+4][1]+cf*(fXY[i][1]-fXY[i+4][1]);
}
}
//_____________________________________________________________________________
void TGeoArb8::SetDimensions(Double_t *param)
{
// Set all arb8 params in one step.
// param[0] = dz
// param[1] = x0
// param[2] = y0
// ...
fDz = param[0];
for (Int_t i=0; i<8; i++) {
fXY[i][0] = param[2*i+1];
fXY[i][1] = param[2*i+2];
}
ComputeTwist();
ComputeBBox();
}
//_____________________________________________________________________________
void TGeoArb8::SetPoints(Double_t *points) const
{
// Creates arb8 mesh points
for (Int_t i=0; i<8; i++) {
points[3*i] = fXY[i][0];
points[3*i+1] = fXY[i][1];
points[3*i+2] = (i<4)?-fDz:fDz;
}
}
//_____________________________________________________________________________
void TGeoArb8::SetPoints(Float_t *points) const
{
// Creates arb8 mesh points
for (Int_t i=0; i<8; i++) {
points[3*i] = fXY[i][0];
points[3*i+1] = fXY[i][1];
points[3*i+2] = (i<4)?-fDz:fDz;
}
}
//_____________________________________________________________________________
void TGeoArb8::SetVertex(Int_t vnum, Double_t x, Double_t y)
{
// Set values for a given vertex.
if (vnum<0 || vnum >7) {
Error("SetVertex", "Invalid vertex number");
return;
}
fXY[vnum][0] = x;
fXY[vnum][1] = y;
if (vnum == 7) {
ComputeTwist();
ComputeBBox();
}
}
//_____________________________________________________________________________
void TGeoArb8::Sizeof3D() const
{
// Fill size of this 3-D object
TGeoBBox::Sizeof3D();
}
//_____________________________________________________________________________
void TGeoArb8::Streamer(TBuffer &R__b)
{
// Stream an object of class TGeoManager.
ComputeTwist();
} else {
R__b.WriteClassBuffer(TGeoArb8::Class(), this);
}
}
ClassImp(TGeoTrap)
//_____________________________________________________________________________
TGeoTrap::TGeoTrap()
{
// Default ctor
fDz = 0;
fTheta = 0;
fPhi = 0;
fH1 = fH2 = fBl1 = fBl2 = fTl1 = fTl2 = fAlpha1 = fAlpha2 = 0;
}
//_____________________________________________________________________________
TGeoTrap::TGeoTrap(Double_t dz, Double_t theta, Double_t phi)
:TGeoArb8("", 0, 0)
{
// Constructor providing just a range in Z, theta and phi.
fDz = dz;
fTheta = theta;
fPhi = phi;
fH1 = fH2 = fBl1 = fBl2 = fTl1 = fTl2 = fAlpha1 = fAlpha2 = 0;
}
//_____________________________________________________________________________
TGeoTrap::TGeoTrap(Double_t dz, Double_t theta, Double_t phi, Double_t h1,
Double_t bl1, Double_t tl1, Double_t alpha1, Double_t h2, Double_t bl2,
Double_t tl2, Double_t alpha2)
:TGeoArb8("", 0, 0)
{
// Normal constructor.
fDz = dz;
fTheta = theta;
fPhi = phi;
fH1 = h1;
fH2 = h2;
fBl1 = bl1;
fBl2 = bl2;
fTl1 = tl1;
fTl2 = tl2;
fAlpha1 = alpha1;
fAlpha2 = alpha2;
fXY[0][0] = -dz*tx-h1*ta1-bl1; fXY[0][1] = -dz*ty-h1;
fXY[1][0] = -dz*tx+h1*ta1-tl1; fXY[1][1] = -dz*ty+h1;
fXY[2][0] = -dz*tx+h1*ta1+tl1; fXY[2][1] = -dz*ty+h1;
fXY[3][0] = -dz*tx-h1*ta1+bl1; fXY[3][1] = -dz*ty-h1;
fXY[4][0] = dz*tx-h2*ta2-bl2; fXY[4][1] = dz*ty-h2;
fXY[5][0] = dz*tx+h2*ta2-tl2; fXY[5][1] = dz*ty+h2;
fXY[6][0] = dz*tx+h2*ta2+tl2; fXY[6][1] = dz*ty+h2;
fXY[7][0] = dz*tx-h2*ta2+bl2; fXY[7][1] = dz*ty-h2;
ComputeTwist();
if ((dz<0) || (h1<0) || (bl1<0) || (tl1<0) ||
(h2<0) || (bl2<0) || (tl2<0)) {
SetShapeBit(kGeoRunTimeShape);
}
else TGeoArb8::ComputeBBox();
}
//_____________________________________________________________________________
TGeoTrap::TGeoTrap(const char *name, Double_t dz, Double_t theta, Double_t phi, Double_t h1,
Double_t bl1, Double_t tl1, Double_t alpha1, Double_t h2, Double_t bl2,
Double_t tl2, Double_t alpha2)
:TGeoArb8(name, 0, 0)
{
// Constructor with name.
SetName(name);
fDz = dz;
fTheta = theta;
fPhi = phi;
fH1 = h1;
fH2 = h2;
fBl1 = bl1;
fBl2 = bl2;
fTl1 = tl1;
fTl2 = tl2;
fAlpha1 = alpha1;
fAlpha2 = alpha2;
for (Int_t i=0; i<8; i++) {
fXY[i][0] = 0.0;
fXY[i][1] = 0.0;
}
fXY[0][0] = -dz*tx-h1*ta1-bl1; fXY[0][1] = -dz*ty-h1;
fXY[1][0] = -dz*tx+h1*ta1-tl1; fXY[1][1] = -dz*ty+h1;
fXY[2][0] = -dz*tx+h1*ta1+tl1; fXY[2][1] = -dz*ty+h1;
fXY[3][0] = -dz*tx-h1*ta1+bl1; fXY[3][1] = -dz*ty-h1;
fXY[4][0] = dz*tx-h2*ta2-bl2; fXY[4][1] = dz*ty-h2;
fXY[5][0] = dz*tx+h2*ta2-tl2; fXY[5][1] = dz*ty+h2;
fXY[6][0] = dz*tx+h2*ta2+tl2; fXY[6][1] = dz*ty+h2;
fXY[7][0] = dz*tx-h2*ta2+bl2; fXY[7][1] = dz*ty-h2;
ComputeTwist();
if ((dz<0) || (h1<0) || (bl1<0) || (tl1<0) ||
(h2<0) || (bl2<0) || (tl2<0)) {
SetShapeBit(kGeoRunTimeShape);
}
else TGeoArb8::ComputeBBox();
}
//_____________________________________________________________________________
TGeoTrap::~TGeoTrap()
{
// destructor
}
//_____________________________________________________________________________
Double_t TGeoTrap::DistFromInside(Double_t *point, Double_t *dir, Int_t iact, Double_t step, Double_t *safe) const
{
// Compute distance from inside point to surface of the trapezoid
if (iact<3 && safe) {
// compute safe distance
*safe = Safety(point, kTRUE);
if (iact==0) return TGeoShape::Big();
if (iact==1 && step<*safe) return TGeoShape::Big();
}
// compute distance to get ouside this shape
// return TGeoArb8::DistFromInside(point, dir, iact, step, safe);
// compute distance to plane ipl :
// ipl=0 : points 0,4,1,5
// ipl=1 : points 1,5,2,6
// ipl=2 : points 2,6,3,7
// ipl=3 : points 3,7,0,4
Double_t distmin;
if (dir[2]<0) {
distmin=(-fDz-point[2])/dir[2];
} else {
if (dir[2]>0) distmin =(fDz-point[2])/dir[2];
else distmin = TGeoShape::Big();
}
Double_t xa,xb,xc;
Double_t ya,yb,yc;
for (Int_t ipl=0;ipl<4;ipl++) {
Int_t j = (ipl+1)%4;
xa=fXY[ipl][0];
ya=fXY[ipl][1];
xb=fXY[ipl+4][0];
yb=fXY[ipl+4][1];
xc=fXY[j][0];
yc=fXY[j][1];
Double_t ax,ay,az;
ax = xb-xa;
ay = yb-ya;
az = 2.*fDz;
Double_t bx,by;
bx = xc-xa;
by = yc-ya;
Double_t ddotn = -dir[0]*az*by + dir[1]*az*bx+dir[2]*(ax*by-ay*bx);
if (ddotn<=0) continue; // entering
Double_t saf = -(point[0]-xa)*az*by + (point[1]-ya)*az*bx + (point[2]+fDz)*(ax*by-ay*bx);
if (saf>=0.0) return 0.0;
Double_t s = -saf/ddotn;
if (s<distmin) distmin=s;
}
return distmin;
}
//_____________________________________________________________________________
Double_t TGeoTrap::DistFromOutside(Double_t *point, Double_t *dir, Int_t iact, Double_t step, Double_t *safe) const
{
// Compute distance from outside point to surface of the trapezoid
if (iact<3 && safe) {
// compute safe distance
*safe = Safety(point, kFALSE);
if (iact==0) return TGeoShape::Big();
if (iact==1 && step<*safe) return TGeoShape::Big();
}
// Check if the bounding box is crossed within the requested distance
Double_t sdist = TGeoBBox::DistFromOutside(point,dir, fDX, fDY, fDZ, fOrigin, step);
if (sdist>=step) return TGeoShape::Big();
// compute distance to get ouside this shape
Bool_t in = kTRUE;
Double_t pts[8];
Double_t snxt;
Double_t xnew, ynew, znew;
Int_t i,j;
if (point[2]<-fDz+TGeoShape::Tolerance()) {
if (dir[2]<=0) return TGeoShape::Big();
in = kFALSE;
snxt = -(fDz+point[2])/dir[2];
xnew = point[0] + snxt*dir[0];
ynew = point[1] + snxt*dir[1];
for (i=0;i<4;i++) {
j = i<<1;
pts[j] = fXY[i][0];
pts[j+1] = fXY[i][1];
}
if (InsidePolygon(xnew,ynew,pts)) return snxt;
} else if (point[2]>fDz-TGeoShape::Tolerance()) {
if (dir[2]>=0) return TGeoShape::Big();
in = kFALSE;
snxt = (fDz-point[2])/dir[2];
xnew = point[0] + snxt*dir[0];
ynew = point[1] + snxt*dir[1];
for (i=0;i<4;i++) {
j = i<<1;
pts[j] = fXY[i+4][0];
pts[j+1] = fXY[i+4][1];
}
if (InsidePolygon(xnew,ynew,pts)) return snxt;
}
snxt = TGeoShape::Big();
// check lateral faces
Double_t dz2 =0.5/fDz;
Double_t xa,xb,xc,xd;
Double_t ya,yb,yc,yd;
Double_t ax,ay,az;
Double_t bx,by;
Double_t ddotn, saf;
Double_t safmin = TGeoShape::Big();
Bool_t exiting = kFALSE;
for (i=0; i<4; i++) {
j = (i+1)%4;
xa=fXY[i][0];
ya=fXY[i][1];
xb=fXY[i+4][0];
yb=fXY[i+4][1];
xc=fXY[j][0];
yc=fXY[j][1];
xd=fXY[4+j][0];
yd=fXY[4+j][1];
ax = xb-xa;
ay = yb-ya;
az = 2.*fDz;
bx = xc-xa;
by = yc-ya;
ddotn = -dir[0]*az*by + dir[1]*az*bx+dir[2]*(ax*by-ay*bx);
saf = (point[0]-xa)*az*by - (point[1]-ya)*az*bx - (point[2]+fDz)*(ax*by-ay*bx);
if (saf<=0) {
// face visible from point outside
in = kFALSE;
if (ddotn>=0) return TGeoShape::Big();
snxt = saf/ddotn;
znew = point[2]+snxt*dir[2];
if (TMath::Abs(znew)<=fDz) {
xnew = point[0]+snxt*dir[0];
ynew = point[1]+snxt*dir[1];
Double_t tx1 =dz2*(xb-xa);
Double_t ty1 =dz2*(yb-ya);
Double_t tx2 =dz2*(xd-xc);
Double_t ty2 =dz2*(yd-yc);
Double_t dzp =fDz+znew;
Double_t xs1 =xa+tx1*dzp;
Double_t ys1 =ya+ty1*dzp;
Double_t xs2 =xc+tx2*dzp;
Double_t ys2 =yc+ty2*dzp;
if (TMath::Abs(xs1-xs2)>TMath::Abs(ys1-ys2)) {
if ((xnew-xs1)*(xs2-xnew)>=0) return snxt;
} else {
if ((ynew-ys1)*(ys2-ynew)>=0) return snxt;
}
}
} else {
if (saf<safmin) {
safmin = saf;
if (ddotn>=0) exiting = kTRUE;
else exiting = kFALSE;
}
}
}
if (!in) return TGeoShape::Big();
if (exiting) return TGeoShape::Big();
return 0.0;
}
//_____________________________________________________________________________
TGeoVolume *TGeoTrap::Divide(TGeoVolume *voldiv, const char *divname, Int_t iaxis, Int_t ndiv,
Double_t start, Double_t step)
{
//--- Divide this trapezoid shape belonging to volume "voldiv" into ndiv volumes
// called divname, from start position with the given step. Only Z divisions
// are supported. For Z divisions just return the pointer to the volume to be
// divided. In case a wrong division axis is supplied, returns pointer to
// volume that was divided.
TGeoShape *shape; //--- shape to be created
TGeoVolume *vol; //--- division volume to be created
TGeoVolumeMulti *vmulti; //--- generic divided volume
TGeoPatternFinder *finder; //--- finder to be attached
TString opt = ""; //--- option to be attached
if (iaxis!=3) {
Error("Divide", "cannot divide trapezoids on other axis than Z");
return 0;
}
Double_t end = start+ndiv*step;
Double_t points_lo[8];
Double_t points_hi[8];
finder = new TGeoPatternTrapZ(voldiv, ndiv, start, end);
voldiv->SetFinder(finder);
finder->SetDivIndex(voldiv->GetNdaughters());
opt = "Z";
vmulti = gGeoManager->MakeVolumeMulti(divname, voldiv->GetMedium());
Double_t txz = ((TGeoPatternTrapZ*)finder)->GetTxz();
Double_t tyz = ((TGeoPatternTrapZ*)finder)->GetTyz();
Double_t zmin, zmax, ox,oy,oz;
for (Int_t idiv=0; idiv<ndiv; idiv++) {
zmin = start+idiv*step;
zmax = start+(idiv+1)*step;
oz = start+idiv*step+step/2;
ox = oz*txz;
oy = oz*tyz;
SetPlaneVertices(zmin, &points_lo[0]);
SetPlaneVertices(zmax, &points_hi[0]);
shape = new TGeoTrap(step/2, fTheta, fPhi);
for (Int_t vert1=0; vert1<4; vert1++)
((TGeoArb8*)shape)->SetVertex(vert1, points_lo[2*vert1]-ox, points_lo[2*vert1+1]-oy);
for (Int_t vert2=0; vert2<4; vert2++)
((TGeoArb8*)shape)->SetVertex(vert2+4, points_hi[2*vert2]-ox, points_hi[2*vert2+1]-oy);
vol = new TGeoVolume(divname, shape, voldiv->GetMedium());
((TGeoNodeOffset*)voldiv->GetNodes()->At(voldiv->GetNdaughters()-1))->SetFinder(finder);
}
return vmulti;
}
//_____________________________________________________________________________
TGeoShape *TGeoTrap::GetMakeRuntimeShape(TGeoShape *mother, TGeoMatrix * /*mat*/) const
{
// In case shape has some negative parameters, these have to be computed
// in order to fit the mother.
if (!TestShapeBit(kGeoRunTimeShape)) return 0;
if (mother->IsRunTimeShape()) {
Error("GetMakeRuntimeShape", "invalid mother");
return 0;
}
Double_t dz, h1, bl1, tl1, h2, bl2, tl2;
if (fDz<0) dz=((TGeoTrap*)mother)->GetDz();
else dz=fDz;
if (fH1<0) h1 = ((TGeoTrap*)mother)->GetH1();
else h1 = fH1;
if (fH2<0) h2 = ((TGeoTrap*)mother)->GetH2();
else h2 = fH2;
if (fBl1<0) bl1 = ((TGeoTrap*)mother)->GetBl1();
else bl1 = fBl1;
if (fBl2<0) bl2 = ((TGeoTrap*)mother)->GetBl2();
else bl2 = fBl2;
if (fTl1<0) tl1 = ((TGeoTrap*)mother)->GetTl1();
else tl1 = fTl1;
if (fTl2<0) tl2 = ((TGeoTrap*)mother)->GetTl2();
else tl2 = fTl2;
return (new TGeoTrap(dz, fTheta, fPhi, h1, bl1, tl1, fAlpha1, h2, bl2, tl2, fAlpha2));
}
//_____________________________________________________________________________
Double_t TGeoTrap::Safety(Double_t *point, Bool_t in) const
{
// Computes the closest distance from given point to this shape.
Double_t safe = TGeoShape::Big();
Double_t saf[5];
Double_t norm[3]; // normal to current facette
Int_t i,j; // current facette index
Double_t x0, y0, z0=-fDz, x1, y1, z1=fDz, x2, y2;
Double_t ax, ay, az=z1-z0, bx, by;
Double_t fn;
//---> compute safety for lateral planes
for (i=0; i<4; i++) {
if (in) saf[i] = TGeoShape::Big();
else saf[i] = 0.;
x0 = fXY[i][0];
y0 = fXY[i][1];
x1 = fXY[i+4][0];
y1 = fXY[i+4][1];
ax = x1-x0;
ay = y1-y0;
az = z1-z0;
j = (i+1)%4;
x2 = fXY[j][0];
y2 = fXY[j][1];
bx = x2-x0;
by = y2-y0;
if (TMath::Abs(bx)<TGeoShape::Tolerance() && TMath::Abs(by)<TGeoShape::Tolerance()) {
x2 = fXY[4+j][0];
y2 = fXY[4+j][1];
bx = x2-x1;
by = y2-y1;
if (TMath::Abs(bx)<TGeoShape::Tolerance() && TMath::Abs(by)<TGeoShape::Tolerance()) continue;
}
norm[0] = -az*by;
norm[1] = az*bx;
norm[2] = ax*by-ay*bx;
fn = TMath::Sqrt(norm[0]*norm[0]+norm[1]*norm[1]+norm[2]*norm[2]);
if (fn<1E-10) continue;
saf[i] = (x0-point[0])*norm[0]+(y0-point[1])*norm[1]+(-fDz-point[2])*norm[2];
if (in) {
saf[i]=TMath::Abs(saf[i])/fn; // they should be all positive anyway
} else {
saf[i] = -saf[i]/fn; // only negative values are interesting
}
}
saf[4] = fDz-TMath::Abs(point[2]);
if (in) {
safe = saf[0];
for (j=1;j<5;j++) if (saf[j] <safe) safe = saf[j];
} else {
saf[4]=-saf[4];
safe = saf[0];
for (j=1;j<5;j++) if (saf[j] >safe) safe = saf[j];
}
return safe;
}
//_____________________________________________________________________________
void TGeoTrap::SavePrimitive(ostream &out, Option_t * /*option*/ /*= ""*/)
{
// Save a primitive as a C++ statement(s) on output stream "out".
if (TObject::TestBit(kGeoSavePrimitive)) return;
out << " // Shape: " << GetName() << " type: " << ClassName() << endl;
out << " dz = " << fDz << ";" << endl;
out << " theta = " << fTheta << ";" << endl;
out << " phi = " << fPhi << ";" << endl;
out << " h1 = " << fH1<< ";" << endl;
out << " bl1 = " << fBl1<< ";" << endl;
out << " tl1 = " << fTl1<< ";" << endl;
out << " alpha1 = " << fAlpha1 << ";" << endl;
out << " h2 = " << fH2 << ";" << endl;
out << " bl2 = " << fBl2<< ";" << endl;
out << " tl2 = " << fTl2<< ";" << endl;
out << " alpha2 = " << fAlpha2 << ";" << endl;
out << " TGeoShape *" << GetPointerName() << " = new TGeoTrap(\"" << GetName() << "\", dz,theta,phi,h1,bl1,tl1,alpha1,h2,bl2,tl2,alpha2);" << endl;
TObject::SetBit(TGeoShape::kGeoSavePrimitive);
}
//_____________________________________________________________________________
void TGeoTrap::SetDimensions(Double_t *param)
{
// Set all arb8 params in one step.
// param[0] = dz
// param[1] = theta
// param[2] = phi
// param[3] = h1
// param[4] = bl1
// param[5] = tl1
// param[6] = alpha1
// param[7] = h2
// param[8] = bl2
// param[9] = tl2
// param[10] = alpha2
fDz = param[0];
fTheta = param[1];
fPhi = param[2];
fH1 = param[3];
fH2 = param[7];
fBl1 = param[4];
fBl2 = param[8];
fTl1 = param[5];
fTl2 = param[9];
fAlpha1 = param[6];
fAlpha2 = param[10];
fXY[0][0] = -fDz*tx-fH1*ta1-fBl1; fXY[0][1] = -fDz*ty-fH1;
fXY[1][0] = -fDz*tx+fH1*ta1-fTl1; fXY[1][1] = -fDz*ty+fH1;
fXY[2][0] = -fDz*tx+fH1*ta1+fTl1; fXY[2][1] = -fDz*ty+fH1;
fXY[3][0] = -fDz*tx-fH1*ta1+fBl1; fXY[3][1] = -fDz*ty-fH1;
fXY[4][0] = fDz*tx-fH2*ta2-fBl2; fXY[4][1] = fDz*ty-fH2;
fXY[5][0] = fDz*tx+fH2*ta2-fTl2; fXY[5][1] = fDz*ty+fH2;
fXY[6][0] = fDz*tx+fH2*ta2+fTl2; fXY[6][1] = fDz*ty+fH2;
fXY[7][0] = fDz*tx-fH2*ta2+fBl2; fXY[7][1] = fDz*ty-fH2;
ComputeTwist();
if ((fDz<0) || (fH1<0) || (fBl1<0) || (fTl1<0) ||
(fH2<0) || (fBl2<0) || (fTl2<0)) {
SetShapeBit(kGeoRunTimeShape);
}
else TGeoArb8::ComputeBBox();
}
ClassImp(TGeoGtra)
//_____________________________________________________________________________
TGeoGtra::TGeoGtra()
{
// Default ctor
fTwistAngle = 0;
}
//_____________________________________________________________________________
TGeoGtra::TGeoGtra(Double_t dz, Double_t theta, Double_t phi, Double_t twist, Double_t h1,
Double_t bl1, Double_t tl1, Double_t alpha1, Double_t h2, Double_t bl2,
Double_t tl2, Double_t alpha2)
:TGeoTrap(dz, theta, phi, h1, bl1, tl1, alpha1, h2, bl2, tl2, alpha2)
{
// Constructor.
fTheta = theta;
fPhi = phi;
fH1 = h1;
fH2 = h2;
fBl1 = bl1;
fBl2 = bl2;
fTl1 = tl1;
fTl2 = tl2;
fAlpha1 = alpha1;
fAlpha2 = alpha2;
Double_t x, y, dx, dy, dx1, dx2, th, ph, al1, al2;
dx = 2*dz*TMath::Sin(th)*TMath::Cos(ph);
dy = 2*dz*TMath::Sin(th)*TMath::Sin(ph);
fDz = dz;
dx1 = 2*h1*TMath::Tan(al1);
dx2 = 2*h2*TMath::Tan(al2);
fTwistAngle = twist;
Int_t i;
for (i=0; i<8; i++) {
fXY[i][0] = 0.0;
fXY[i][1] = 0.0;
}
fXY[0][0] = -bl1; fXY[0][1] = -h1;
fXY[1][0] = -tl1+dx1; fXY[1][1] = h1;
fXY[2][0] = tl1+dx1; fXY[2][1] = h1;
fXY[3][0] = bl1; fXY[3][1] = -h1;
fXY[4][0] = -bl2+dx; fXY[4][1] = -h2+dy;
fXY[5][0] = -tl2+dx+dx2; fXY[5][1] = h2+dy;
fXY[6][0] = tl2+dx+dx2; fXY[6][1] = h2+dy;
fXY[7][0] = bl2+dx; fXY[7][1] = -h2+dy;
for (i=4; i<8; i++) {
x = fXY[i][0];
y = fXY[i][1];
}
ComputeTwist();
if ((dz<0) || (h1<0) || (bl1<0) || (tl1<0) ||
(h2<0) || (bl2<0) || (tl2<0)) SetShapeBit(kGeoRunTimeShape);
else TGeoArb8::ComputeBBox();
}
//_____________________________________________________________________________
TGeoGtra::TGeoGtra(const char *name, Double_t dz, Double_t theta, Double_t phi, Double_t twist, Double_t h1,
Double_t bl1, Double_t tl1, Double_t alpha1, Double_t h2, Double_t bl2,
Double_t tl2, Double_t alpha2)
:TGeoTrap(name, dz, theta, phi, h1, bl1, tl1, alpha1, h2, bl2, tl2, alpha2)
{
// Constructor providing the name of the shape.
SetName(name);
fTheta = theta;
fPhi = phi;
fH1 = h1;
fH2 = h2;
fBl1 = bl1;
fBl2 = bl2;
fTl1 = tl1;
fTl2 = tl2;
fAlpha1 = alpha1;
fAlpha2 = alpha2;
Double_t x, y, dx, dy, dx1, dx2, th, ph, al1, al2;
dx = 2*dz*TMath::Sin(th)*TMath::Cos(ph);
dy = 2*dz*TMath::Sin(th)*TMath::Sin(ph);
fDz = dz;
dx1 = 2*h1*TMath::Tan(al1);
dx2 = 2*h2*TMath::Tan(al2);
fTwistAngle = twist;
Int_t i;
for (i=0; i<8; i++) {
fXY[i][0] = 0.0;
fXY[i][1] = 0.0;
}
fXY[0][0] = -bl1; fXY[0][1] = -h1;
fXY[1][0] = -tl1+dx1; fXY[1][1] = h1;
fXY[2][0] = tl1+dx1; fXY[2][1] = h1;
fXY[3][0] = bl1; fXY[3][1] = -h1;
fXY[4][0] = -bl2+dx; fXY[4][1] = -h2+dy;
fXY[5][0] = -tl2+dx+dx2; fXY[5][1] = h2+dy;
fXY[6][0] = tl2+dx+dx2; fXY[6][1] = h2+dy;
fXY[7][0] = bl2+dx; fXY[7][1] = -h2+dy;
for (i=4; i<8; i++) {
x = fXY[i][0];
y = fXY[i][1];
}
ComputeTwist();
if ((dz<0) || (h1<0) || (bl1<0) || (tl1<0) ||
(h2<0) || (bl2<0) || (tl2<0)) SetShapeBit(kGeoRunTimeShape);
else TGeoArb8::ComputeBBox();
}
//_____________________________________________________________________________
TGeoGtra::~TGeoGtra()
{
// Destructor.
}
//_____________________________________________________________________________
Double_t TGeoGtra::DistFromInside(Double_t *point, Double_t *dir, Int_t iact, Double_t step, Double_t *safe) const
{
// Compute distance from inside point to surface of the shape.
if (iact<3 && safe) {
// compute safe distance
*safe = Safety(point, kTRUE);
if (iact==0) return TGeoShape::Big();
if (iact==1 && step<*safe) return TGeoShape::Big();
}
// compute distance to get ouside this shape
return TGeoArb8::DistFromInside(point, dir, iact, step, safe);
}
//_____________________________________________________________________________
Double_t TGeoGtra::DistFromOutside(Double_t *point, Double_t *dir, Int_t iact, Double_t step, Double_t *safe) const
{
// Compute distance from inside point to surface of the shape.
if (iact<3 && safe) {
// compute safe distance
*safe = Safety(point, kTRUE);
if (iact==0) return TGeoShape::Big();
if (iact==1 && step<*safe) return TGeoShape::Big();
}
// compute distance to get ouside this shape
return TGeoArb8::DistFromOutside(point, dir, iact, step, safe);
}
//_____________________________________________________________________________
TGeoShape *TGeoGtra::GetMakeRuntimeShape(TGeoShape *mother, TGeoMatrix * /*mat*/) const
{
// In case shape has some negative parameters, these has to be computed
// in order to fit the mother
if (!TestShapeBit(kGeoRunTimeShape)) return 0;
if (mother->IsRunTimeShape()) {
Error("GetMakeRuntimeShape", "invalid mother");
return 0;
}
Double_t dz, h1, bl1, tl1, h2, bl2, tl2;
if (fDz<0) dz=((TGeoTrap*)mother)->GetDz();
else dz=fDz;
if (fH1<0)
h1 = ((TGeoTrap*)mother)->GetH1();
else
h1 = fH1;
if (fH2<0)
h2 = ((TGeoTrap*)mother)->GetH2();
else
h2 = fH2;
if (fBl1<0)
bl1 = ((TGeoTrap*)mother)->GetBl1();
else
bl1 = fBl1;
if (fBl2<0)
bl2 = ((TGeoTrap*)mother)->GetBl2();
else
bl2 = fBl2;
if (fTl1<0)
tl1 = ((TGeoTrap*)mother)->GetTl1();
else
tl1 = fTl1;
if (fTl2<0)
tl2 = ((TGeoTrap*)mother)->GetTl2();
else
tl2 = fTl2;
return (new TGeoGtra(dz, fTheta, fPhi, fTwistAngle ,h1, bl1, tl1, fAlpha1, h2, bl2, tl2, fAlpha2));
}
//_____________________________________________________________________________
void TGeoGtra::SavePrimitive(ostream &out, Option_t * /*option*/ /*= ""*/)
{
// Save a primitive as a C++ statement(s) on output stream "out".
if (TObject::TestBit(kGeoSavePrimitive)) return;
out << " // Shape: " << GetName() << " type: " << ClassName() << endl;
out << " dz = " << fDz << ";" << endl;
out << " theta = " << fTheta << ";" << endl;
out << " phi = " << fPhi << ";" << endl;
out << " twist = " << fTwistAngle << ";" << endl;
out << " h1 = " << fH1<< ";" << endl;
out << " bl1 = " << fBl1<< ";" << endl;
out << " tl1 = " << fTl1<< ";" << endl;
out << " alpha1 = " << fAlpha1 << ";" << endl;
out << " h2 = " << fH2 << ";" << endl;
out << " bl2 = " << fBl2<< ";" << endl;
out << " tl2 = " << fTl2<< ";" << endl;
out << " alpha2 = " << fAlpha2 << ";" << endl;
out << " TGeoShape *" << GetPointerName() << " = new TGeoGtra(\"" << GetName() << "\", dz,theta,phi,twist,h1,bl1,tl1,alpha1,h2,bl2,tl2,alpha2);" << endl;
TObject::SetBit(TGeoShape::kGeoSavePrimitive);
}
//_____________________________________________________________________________
void TGeoGtra::SetDimensions(Double_t *param)
{
// Set all arb8 params in one step.
// param[0] = dz
// param[1] = theta
// param[2] = phi
// param[3] = h1
// param[4] = bl1
// param[5] = tl1
// param[6] = alpha1
// param[7] = h2
// param[8] = bl2
// param[9] = tl2
// param[10] = alpha2
// param[11] = twist
fDz = param[0];
fTheta = param[1];
fPhi = param[2];
fH1 = param[3];
fH2 = param[7];
fBl1 = param[4];
fBl2 = param[8];
fTl1 = param[5];
fTl2 = param[9];
fAlpha1 = param[6];
fAlpha2 = param[10];
fTwistAngle = param[11];
Double_t x, y, dx, dy, dx1, dx2, th, ph, al1, al2;
dx = 2*fDz*TMath::Sin(th)*TMath::Cos(ph);
dy = 2*fDz*TMath::Sin(th)*TMath::Sin(ph);
dx1 = 2*fH1*TMath::Tan(al1);
dx2 = 2*fH2*TMath::Tan(al2);
Int_t i;
for (i=0; i<8; i++) {
fXY[i][0] = 0.0;
fXY[i][1] = 0.0;
}
fXY[0][0] = -fBl1; fXY[0][1] = -fH1;
fXY[1][0] = -fTl1+dx1; fXY[1][1] = fH1;
fXY[2][0] = fTl1+dx1; fXY[2][1] = fH1;
fXY[3][0] = fBl1; fXY[3][1] = -fH1;
fXY[4][0] = -fBl2+dx; fXY[4][1] = -fH2+dy;
fXY[5][0] = -fTl2+dx+dx2; fXY[5][1] = fH2+dy;
fXY[6][0] = fTl2+dx+dx2; fXY[6][1] = fH2+dy;
fXY[7][0] = fBl2+dx; fXY[7][1] = -fH2+dy;
for (i=4; i<8; i++) {
x = fXY[i][0];
y = fXY[i][1];
}
ComputeTwist();
if ((fDz<0) || (fH1<0) || (fBl1<0) || (fTl1<0) ||
(fH2<0) || (fBl2<0) || (fTl2<0)) SetShapeBit(kGeoRunTimeShape);
else TGeoArb8::ComputeBBox();
}
```
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TGeoArb8.cxx:1853 | 39,165 | 84,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-23 | latest | en | 0.468149 |
https://www.numbersaplenty.com/2253342391 | 1,611,616,666,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704792131.69/warc/CC-MAIN-20210125220722-20210126010722-00172.warc.gz | 909,048,577 | 3,246 | Cookie Consent by FreePrivacyPolicy.com
Search a number
2253342391 is a prime number
BaseRepresentation
bin1000011001001111…
…0100011010110111
312211001001122122121
42012103310122313
514103323424031
61011332540411
7106561044226
oct20623643267
95731048577
102253342391
11a56a53543
1252a781707
1329baba1c1
141753a4cbd
15d2c57011
hex864f46b7
2253342391 has 2 divisors, whose sum is σ = 2253342392. Its totient is φ = 2253342390.
The previous prime is 2253342389. The next prime is 2253342431. The reversal of 2253342391 is 1932433522.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 2253342391 - 21 = 2253342389 is a prime.
Together with 2253342389, it forms a pair of twin primes.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (2253342341) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1126671195 + 1126671196.
It is an arithmetic number, because the mean of its divisors is an integer number (1126671196).
Almost surely, 22253342391 is an apocalyptic number.
2253342391 is a deficient number, since it is larger than the sum of its proper divisors (1).
2253342391 is an equidigital number, since it uses as much as digits as its factorization.
2253342391 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 38880, while the sum is 34.
The square root of 2253342391 is about 47469.3837225637. The cubic root of 2253342391 is about 1311.0192310727.
The spelling of 2253342391 in words is "two billion, two hundred fifty-three million, three hundred forty-two thousand, three hundred ninety-one". | 538 | 1,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-04 | latest | en | 0.850487 |
https://www.shaalaa.com/question-bank-solutions/what-number-should-be-subtracted-x3-3x2-8x-14-so-that-dividing-it-x-2-remainder-10-remainder-theorem_28820 | 1,596,509,594,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735851.15/warc/CC-MAIN-20200804014340-20200804044340-00450.warc.gz | 784,717,344 | 8,974 | Share
# What Number Should Be Subtracted from X3 + 3x2 – 8x + 14 So that on Dividing It with X – 2, the Remainder is 10. - Mathematics
Course
#### Question
What number should be subtracted from x3 + 3x2 – 8x + 14 so that on dividing it with x – 2, the remainder is 10.
#### Solution
Let the number to be subtracted be k and the resulting polynomial be f(x).
So, f(x)=x^3+3x^2-8x+14-k
it is given that when f(x) is divided by (x-2), the remainder is 10.
f(2)=10
(2)^3+3(2)^2-8(2)+14-k=10
8+12-16+14-k=10
18-k=10
k=8
Thus, the requied number is 8.
Is there an error in this question or solution? | 220 | 608 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2020-34 | latest | en | 0.826058 |
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Depdendent Variable
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• "lcm calculator using variables" | 979 | 4,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-09 | latest | en | 0.868053 |
http://www.math.chalmers.se/Stat/Research/particle.html | 1,369,189,209,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701153213/warc/CC-MAIN-20130516104553-00068-ip-10-60-113-184.ec2.internal.warc.gz | 591,434,694 | 2,740 | ## Particle systems and statistical physics
For more than a hundred years, physicists have put great effort into trying to understand the connection between the microscopic and macroscopic laws of nature. How can the seemingly ordered structure of matter on the macroscopic level be consistent with the highly disordered structure on the atomic and molecular level? This subdiscipline of physics is known as statistical mechanics.
During the last few decades, probability theory has played an important role in statistical mechanics, through the construction and analysis of large (sometimes infinite) spatial stochastic particle system models that capture various dichotomies between the microscopic and macroscopic behaviors observed in nature.
Much of our work is concerned with such models. A famous example is the Ising model for spontaneous magnetization. Each integer point in 2- or 3-dimensional space is assigned the value +1 or -1 randomly, according to rules which favor agreement between nearest neighbors. How strongly favored, depends on the interaction parameter J. When J=0, different spins are independent, while as J increases, the tendency to take the same values becomes stronger and stronger. It turns out that there is a critical value for J, below which the macroscopic behavior is disordered and above which it is ordered; this is an example of an important phenomenon known as phase transition, which in this case explains the occurrence of spontaneous magnetization.
The Ising model and many of its relatives are so called Markov random fields, meaning that they exhibit a spatial variant of the property that characterizes Markov chains (see the Markov theory group). Such models are used extensively in many fields outside of physics, such as spatial statistics and image analysis.
We also study so called percolation models, which are models for the connectivity behavior of random media. These have a variety of applications, ranging from the spread of liquid through porous media and electrical conductivity, to forest fires and epidemics. An example is given in the figure below: disks of unit radius are spread out randomly in 2-dimensional space (according to a so called Poisson process) with an average of L disks per unit area. Again we see a phase transition phenomenon: there is a critical value for L, below which the set of disks exhibits short-range connectivity only, and above which there are long- (in fact, infinite-) range connections. The figure shows realizations below, near, and above the critical value.
A poetic name for this model is the lilypad model. Imagine a pond with lilypads, and a snail trying to cross the pond but being limited to going on the lilypads. How large must the "lilypad density" L be for this to be possible?
We also study clumping phenomena in such models. This is relevant e.g. for understanding strength of inhomogeneous materials (see the material fatigue group) and can also be viewed as an extreme value problem.
Rigorous mathmematical solutions to a large number of difficult problems in this area have been obtained in recent years, by us and others. However, many of the problems appear to be, at present, too difficult to admit such solutions. When that happens, it is common practice to revert to simulation studies, in which case the so called Markov chain Monte Carlo algorithms (developed e.g. in the Markov theory group) are of great use.
Senior researchers: Olle Häggström, Johan Jonasson, Marianne Månsson, Jeff Steif.
Ph.D. students: Erik Broman, Marcus Isaksson, Fredrik Lundin, Oskar Sandberg, Johan Tykesson, Marcus Warfheimer.
Back to the list of research groups. | 744 | 3,671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2013-20 | latest | en | 0.955547 |
https://www.jiskha.com/display.cgi?id=1349474637 | 1,516,475,840,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889681.68/warc/CC-MAIN-20180120182041-20180120202041-00135.warc.gz | 959,562,202 | 4,133 | posted by .
I'm having trouble with this word problem. Two friends Jodi and Abe go horseback riding on the same trail in the same direction. Jodi's horse travels at 5 miles per hour while Abes horse travels at a slower pace. After 2 hours they are 2 miles apart. Find Abes horse speed.
I set this up as a table and did 2x+2(x+5)=2 solved it but I guess I got it wrong can anyone help me please
Jodi: 5 * 2 = 10 miles in 2 hours
Abe: 10 - 2 = 8 miles
8/2 = 4 mph
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http://www.gurufocus.com/term/grossmargin/FFH/Gross%252BMargin/Fairfax%2BFinancial%2BHoldings%2BLimited | 1,417,247,455,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931014049.81/warc/CC-MAIN-20141125155654-00036-ip-10-235-23-156.ec2.internal.warc.gz | 581,649,671 | 22,111 | Switch to:
(:)
Gross Margin
0.00% (As of . 20)
Gross Margin is calculated as gross profit divided by its revenue. 's gross profit for the six months ended in . 20 was \$0.00 Mil. 's revenue for the six months ended in . 20 was \$0.00 Mil. Therefore, 's Gross Margin for the quarter that ended in . 20 was 0.00%.
FFH' s 10-Year Gross Margin Range
Min: 0 Max: 0
Current: 0
During the past 0 years, the highest Gross Margin of was %. The lowest was %. And the median was %.
FFH's Gross Marginis ranked lower than
100% of the Companies
in the Global industry.
( Industry Median: vs. FFH: )
had a gross margin of % for the quarter that ended in . 20 => No sustainable competitive advantage
The 3-Year average Growth Rate of Gross Margin for was % per year.
Definition
Gross Margin is the percentage of Gross Profit out of sales or Revenue.
's Gross Margin for the fiscal year that ended in . 20 is calculated as
Gross Margin (A: . 20 ) = Gross Profit (A: . 20 ) / Revenue (A: . 20 ) = 0 / = (Revenue - Cost of Goods Sold) / Revenue = ( - ) / = %
's Gross Margin for the quarter that ended in . 20 is calculated as
Gross Margin (Q: . 20 ) = Gross Profit (Q: . 20 ) / Revenue (Q: . 20 ) = 0 / = (Revenue - Cost of Goods Sold) / Revenue = ( - ) / = %
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
A positive Gross Profit is only the first step for a company to make a net profit. The gross profit needs to be big enough to also cover related labor, equipment, rental, marketing/advertising, research and development and a lot of other costs in selling the products.
Explanation
Warren Buffett believes that firms with excellent long term economics tend to have consistently higher margins.
Durable competitive advantage creates a high Gross Margin because of the freedom to price in excess of cost. Companies can be categorized by their Gross Margin
1. Greater than 40% = Durable competitive advantage
2. Less than 40% = Competition eroding margins
3. Less than 20% = no sustainable competitive advantage
Consistency of Gross Margin is key
had a gross margin of % for the quarter that ended in . 20 => No sustainable competitive advantage
Be Aware
If a company loses its competitive advantages, usually its gross margin declines well before its sales declines. Watching Gross Margin and Operating Margin closely helps avoid value trap situations.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Annual Data
Gross Margin 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
Semi-Annual Data
Gross Margin 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
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Calculating Common Area Maintenance ( C A M)
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Calculating the Common Area Maintenance cost for retail properties
Calculating the Common Area Maintenance cost for retail properties
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• 1. How Common Area Maintenance (CAM) is Calculated? Marcus Bourn Triplenetmarcus.wordpress.com [email_address]
• 2. Definitions: <ul><li>Common Area Maintenance (CAM) </li></ul><ul><li>The amount of money charged to tenants for their shares of maintaining a center’s common areas or shared services and utilities. </li></ul>
• 3. Definitions: <ul><li>Gross Leasable Area (GLA) </li></ul><ul><li>The square footage of a shopping center that can generate income by being leased to tenants. </li></ul>
• 4. Explanation <ul><li>The formula for calculating the CAM is based upon taking the square footage of the space being occupied by the Tenant (Tenant’s Square Footage) and dividing that number by the GLA of the shopping center. The resulting number will be the prorated percentage of the common area maintenance the Tenant is expected to pay to the Landlord for maintaining the premises. </li></ul>
• 5. The Formula <ul><li>Tenant’s Square Footage </li></ul><ul><li>Gross Leasable Area (GLA) </li></ul><ul><li>Common Area Maintenance (CAM) </li></ul>=
• 6. Example <ul><li>Tenant’s Space: 3,000 Square Feet (SF) </li></ul><ul><li>GLA: 50,000 Square Feet (SF) </li></ul><ul><li>3,000 SF </li></ul><ul><li>50,000 SF </li></ul><ul><li>You must multiply the 0.06 by 100 to get the percentage of the shopping center represented by the Tenant which is 6% </li></ul>= 0.06
• 7. Resources <ul><li>ICSC’s Dictionary of Shopping Center Terms </li></ul><ul><li>Dealmaker’s Guide to Commercial Real Estate by Ray Alcorn </li></ul> | 630 | 2,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2015-27 | longest | en | 0.743942 |
https://www.binary-code.org/binary/9bit/110010001/ | 1,597,512,499,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740929.65/warc/CC-MAIN-20200815154632-20200815184632-00033.warc.gz | 572,928,327 | 4,479 | ### Binary calculator
=
Binary 110010001 401 191 9 256 + 128 + 16 + 1
Decimal Binary Spelt
+256100000000Two hundred fifty-six
+12810000000One hundred twenty-eight
+1610000Sixteen
+11One
= 401 110010001 Four hundred one | 75 | 219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-34 | latest | en | 0.574805 |
http://mathoverflow.net/questions/44737/invertible-matrices-satisfying-x-y-yx/44762 | 1,369,176,475,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700842908/warc/CC-MAIN-20130516104042-00089-ip-10-60-113-184.ec2.internal.warc.gz | 166,829,123 | 13,135 | ## Invertible matrices satisfying $[x,y,y]=x$.
I have been thinking about this question for quite some time but now this question by Denis Serre revived some hope.
Question. Let $x,y$ be invertible matrices (say, over $\mathbb C$) and $[x,y,y]=x$ where $[a,b]=a^{-1}b^{-1}ab$, $[a,b,c]=[[a,b],c]$. Does it follow that some power of $x$ is unipotent?
The motivation is this. Consider the one-relator group $\langle x,y \mid [x,y,y]=x\rangle$. It is hyperbolic (proved by A. Minasyan) and residually finite (that is proved in my paper with A. Borisov). If the answer to the above question is "yes", then that group would be non-linear which would provide an explicit example of non-linear hyperbolic group.
Update 1. Can $x$ in the above be a diagonal matrix and not a root of 1?
Update 2. The group is residually finite, so it has many representations by matrices such that $x, y$ have finite orders (hence their powers are unipotents).
Update 3. The group has presentation as an ascending HNN extension of the free group: $\langle a,b,t \mid a^t=ab, b^t=ba\rangle$. So it is related to the Morse-Thue map. Properties of that map may have something to do with the question. See two quasi-motivations of the question as my comments below.
-
Are there any one-relator groups known not to be linear? – Ćukasz Grabowski Nov 3 2010 at 23:06
@Lukasz: Yes, there are even non-residually finite ones: $BS(2,3)=\langle x,y \mid y^{-1}x^2y=x^3\rangle$. There are also residually finite 1-related groups which are not linear. Those were constructed in our paper with Cornelia Drutu (in J. Algebra). The point is that this group is hyperbolic. There is an example of a non-linear hyperbolic group due to M. Kapovich (which easily follows from the super-rigidity of certain rank 1 lattices and a Gromov-Olshanskii theorem). But that example has no explicit presentation. This one would be the first explicit example. – Mark Sapir Nov 3 2010 at 23:12
Here is one of the quasi-reasons why I think the answer is positive. If $G=\langle x,y \mid [x,y,y]=x\rangle$ is linear, then it has a representation over a number field, hence over $\mathbb{Q}$. Therefore the sequence of indexes of subgroups of finite index of $G$ must grow polynomially (take congruence subgroups). This would imply that certain polynomial maps over finite fields have many quasi-fixed points with long orbits (see our paper with Borisov). The latter seems to be impossible. – Mark Sapir Mar 18 2012 at 14:04
A trivial observation: setting $z := [x,y]$, the condition $[x,y,y]=x$ is equivalent to the assertion that the pair $(x,z)$ is conjugate to $(xz,zx)$ after conjugation by $y$. So the question is equivalent to the question of whether a pair of matrices $(x,z)$ which has the property of being conjugate to $(xz,zx)$ is such that all the eigenvalues of $x$ (or equivalently, $z$, which is necessarily conjugate to $x$) are roots of unity. Unfortunately, I got stuck after this observation: the conjugacy does give a number of trace identities involving various words in z,x, but not enough of them... – Terry Tao Mar 18 2012 at 19:40
@Terry: Yes, this was another quasi-reason. Consider $G=\langle x,y\mid x^y=x^2\rangle$. Then in every linear representation of $G$, conjugating $x$ by powers of $y^{-1}$ will produce matrices that are closer and closer to 1. So if $x,y$ are matrices $\lim_{n\to\infty} x^{y^{-n}}=1$. This means that $x$ is a unipotent element "of $y$" in Margulis' terminology, hence $x$ is unipotent. Now we have a similar presentation $\langle x,z,y\mid x^y=xz, z^y=zx\rangle$, so the idea was to show that some power of $x$ satisfies the limit property above. – Mark Sapir Mar 18 2012 at 21:02
Here's a quick test which might disprove your hopes very quickly:
Take $n$ to be small: Try $2$ first, and $5$ is probably near the limit of a computer algebra system. Choose $x$ to be a random $n \times n$ diagonal matrix with determinant $1$, for example, $\mathrm{diag}(17, 1/17)$. Write out your relation, leaving all the elements of $y$ as variables. After clearing denominators, you have $n^2$ simultaneuous homogenous equations in $n^2$ variables. (If I haven't made any dumb errors, they have degree $3n$.) Ask your favorite computer algebra system to solve them for you. If any of the roots are not on the hypersurface $\det y=0$, then you have a counterexample!
-
I did it for $n=2$, of course. The conjecture is true in that case. For $n=2$ you can use the trace identities. That reduces dimension to 3 (every pair of matrices is determined by three traces, if I remember correctly). It is written in the paper with Drutu which I mentioned above. For other $n$'s I did not check. There are no trace identities and the computation is too large. – Mark Sapir Nov 4 2010 at 1:37
@David: Just to clarify my previous comment. Every pair of $2\times 2$- invertible matrices of det 1 is determined by the traces $tr(a), tr(b), tr(ab)$ up to conjugacy. There are polynomial identities allowing to compute the trace of the word $tr(w(a,b))$ if you know $tr(a), tr(b), tr(ab)$. Then the relation $[x,y,y]x^{-1}=1$ gives that certain trace is equal to 2, etc. – Mark Sapir Nov 4 2010 at 2:23
OK, got it. Yeah, trace identities would be the way to do this for $n=2$, and maybe for $n=3$. I think just writing out the equations should win for $n=4$, though I haven't tried it. But my point was just that you should be doing these basic low dimensional checks, and it sounds like you are. – David Speyer Nov 4 2010 at 2:59
@David: My favorite CAS (Maple) refuses to deal even with the 3-dim case. What is your favorite CAS that can do it? – Mark Sapir Nov 4 2010 at 20:11
@David: You don't need to clear denominators, as you can suppose that y is in SL_n. The degree of the polynomials will be n^2+n, though, not 3n. @Mark: Playing around, I found that there is a matrix x in SL_2(C) of order 6 and y of order 8 such that [x,y,y]=x - this is of course not an answer to your question as x^6 is unipotent. Do you have an explanation for this example, though? – Guntram Nov 5 2010 at 18:41
Ignore this, it is wrong
I migth miss something simple, but
$[a,b]^n=[a,b]$ for all $n$ hence $x^2=[x,y,y]^2=[x,y,y]=x$.
Since $x$ is invertible and $x^2=x$ it follows $x=I$.
using this it is easy to show that $[x,y,y]=x$ for x,y invertible if and only if $x=I$.
-
Why $\left[a,b\right]^n=\left[a,b\right]$ ???? – darij grinberg Nov 3 2010 at 22:48 | 1,855 | 6,427 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2013-20 | latest | en | 0.923359 |
https://www.nag.com/numeric/nl/nagdoc_28.5/flhtml/g02/g02ajf.html | 1,675,583,591,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500250.51/warc/CC-MAIN-20230205063441-20230205093441-00232.warc.gz | 919,680,280 | 7,804 | # NAG FL Interfaceg02ajf (corrmat_h_weight)
## ▸▿ Contents
Settings help
FL Name Style:
FL Specification Language:
## 1Purpose
g02ajf computes the nearest correlation matrix, using element-wise weighting in the Frobenius norm and optionally with bounds on the eigenvalues, to a given square, input matrix.
## 2Specification
Fortran Interface
Subroutine g02ajf ( g, ldg, n, h, ldh, x, ldx, iter, norm,
Integer, Intent (In) :: ldg, n, ldh, maxit, ldx Integer, Intent (Inout) :: ifail Integer, Intent (Out) :: iter Real (Kind=nag_wp), Intent (In) :: alpha, errtol Real (Kind=nag_wp), Intent (Inout) :: g(ldg,*), h(ldh,*), x(ldx,*) Real (Kind=nag_wp), Intent (Out) :: norm
C Header Interface
#include <nag.h>
void g02ajf_ (double g[], const Integer *ldg, const Integer *n, const double *alpha, double h[], const Integer *ldh, const double *errtol, const Integer *maxit, double x[], const Integer *ldx, Integer *iter, double *norm, Integer *ifail)
The routine may be called by the names g02ajf or nagf_correg_corrmat_h_weight.
## 3Description
g02ajf finds the nearest correlation matrix, $X$, to an approximate correlation matrix, $G$, using element-wise weighting, this minimizes ${‖H\circ \left(G-X\right)‖}_{F}$, where $C=A\circ B$ denotes the matrix $C$ with elements ${C}_{ij}={A}_{ij}×{B}_{ij}$.
You can optionally specify a lower bound on the eigenvalues, $\alpha$, of the computed correlation matrix, forcing the matrix to be strictly positive definite, if $0<\alpha <1$.
Zero elements in $H$ should be used when you wish to put no emphasis on the corresponding element of $G$. The algorithm scales $H$ so that the maximum element is $1$. It is this scaled matrix that is used in computing the norm above and for the stopping criteria described in Section 7.
Note that if the elements in $H$ vary by several orders of magnitude from one another the algorithm may fail to converge.
## 4References
Borsdorf R and Higham N J (2010) A preconditioned (Newton) algorithm for the nearest correlation matrix IMA Journal of Numerical Analysis 30(1) 94–107
Jiang K, Sun D and Toh K-C (2012) An inexact accelerated proximal gradient method for large scale linearly constrained convex SDP SIAM J. Optim. 22(3) 1042–1064
Qi H and Sun D (2006) A quadratically convergent Newton method for computing the nearest correlation matrix SIAM J. Matrix AnalAppl 29(2) 360–385
## 5Arguments
1: $\mathbf{g}\left({\mathbf{ldg}},*\right)$Real (Kind=nag_wp) array Input/Output
Note: the second dimension of the array g must be at least ${\mathbf{n}}$.
On entry: $G$, the initial matrix.
On exit: $G$ is overwritten.
2: $\mathbf{ldg}$Integer Input
On entry: the first dimension of the array g as declared in the (sub)program from which g02ajf is called.
Constraint: ${\mathbf{ldg}}\ge {\mathbf{n}}$.
3: $\mathbf{n}$Integer Input
On entry: the order of the matrix $G$.
Constraint: ${\mathbf{n}}>0$.
4: $\mathbf{alpha}$Real (Kind=nag_wp) Input
On entry: the value of $\alpha$.
If ${\mathbf{alpha}}<0.0$, $0.0$ is used.
Constraint: ${\mathbf{alpha}}<1.0$.
5: $\mathbf{h}\left({\mathbf{ldh}},*\right)$Real (Kind=nag_wp) array Input/Output
Note: the second dimension of the array h must be at least ${\mathbf{n}}$.
On entry: the matrix of weights $H$.
On exit: a symmetric matrix $\frac{1}{2}\left(H+{H}^{\mathrm{T}}\right)$ with its diagonal elements set to zero and the remaining elements scaled so that the maximum element is $1.0$.
Constraint: $\mathit{H}\left(\mathit{i},\mathit{j}\right)\ge 0.0$, for all $i$ and $j=1,2,\dots ,n$, $i\ne j$.
6: $\mathbf{ldh}$Integer Input
On entry: the first dimension of the array h as declared in the (sub)program from which g02ajf is called.
Constraint: ${\mathbf{ldh}}\ge {\mathbf{n}}$.
7: $\mathbf{errtol}$Real (Kind=nag_wp) Input
On entry: the termination tolerance for the iteration. If ${\mathbf{errtol}}\le 0.0$, is used. See Section 7 for further details.
8: $\mathbf{maxit}$Integer Input
On entry: specifies the maximum number of iterations to be used.
If ${\mathbf{maxit}}\le 0$, $200$ is used.
9: $\mathbf{x}\left({\mathbf{ldx}},*\right)$Real (Kind=nag_wp) array Output
Note: the second dimension of the array x must be at least ${\mathbf{n}}$.
On exit: contains the nearest correlation matrix.
10: $\mathbf{ldx}$Integer Input
On entry: the first dimension of the array x as declared in the (sub)program from which g02ajf is called.
Constraint: ${\mathbf{ldx}}\ge {\mathbf{n}}$.
11: $\mathbf{iter}$Integer Output
On exit: the number of iterations taken.
12: $\mathbf{norm}$Real (Kind=nag_wp) Output
On exit: the value of ${‖H\circ \left(G-X\right)‖}_{F}$ after the final iteration.
13: $\mathbf{ifail}$Integer Input/Output
On entry: ifail must be set to $0$, $-1$ or $1$ to set behaviour on detection of an error; these values have no effect when no error is detected.
A value of $0$ causes the printing of an error message and program execution will be halted; otherwise program execution continues. A value of $-1$ means that an error message is printed while a value of $1$ means that it is not.
If halting is not appropriate, the value $-1$ or $1$ is recommended. If message printing is undesirable, then the value $1$ is recommended. Otherwise, the value $0$ is recommended. When the value $-\mathbf{1}$ or $\mathbf{1}$ is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6Error Indicators and Warnings
If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
${\mathbf{ifail}}=1$
On entry, ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{n}}>0$.
${\mathbf{ifail}}=2$
On entry, ${\mathbf{ldg}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{ldg}}\ge {\mathbf{n}}$.
${\mathbf{ifail}}=3$
On entry, ${\mathbf{ldh}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{ldh}}\ge {\mathbf{n}}$.
${\mathbf{ifail}}=4$
On entry, ${\mathbf{ldx}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{ldx}}\ge {\mathbf{n}}$.
${\mathbf{ifail}}=5$
On entry, ${\mathbf{alpha}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{alpha}}<1.0$.
${\mathbf{ifail}}=6$
On entry, one or more of the off-diagonal elements of $H$ were negative.
${\mathbf{ifail}}=7$
Routine failed to converge in $⟨\mathit{\text{value}}⟩$ iterations.
Increase maxit or check the call to the routine.
${\mathbf{ifail}}=8$
Failure to solve intermediate eigenproblem. This should not occur. Please contact NAG with details of your call.
${\mathbf{ifail}}=-99$
An unexpected error has been triggered by this routine. Please contact NAG.
See Section 7 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 9 in the Introduction to the NAG Library FL Interface for further information.
## 7Accuracy
The returned accuracy is controlled by errtol and limited by machine precision. If ${e}_{i}$ is the value of norm at the $i$th iteration, that is
$ei = ‖H∘(G-X)‖F ,$
where $H$ has been scaled as described above, then the algorithm terminates when:
$|ei-ei-1| 1+ max(ei,ei-1) ≤ errtol .$
## 8Parallelism and Performance
Background information to multithreading can be found in the Multithreading documentation.
g02ajf is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.
g02ajf makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
## 9Further Comments
Arrays are internally allocated by g02ajf. The total size of these arrays is $15×{\mathbf{n}}+5×{\mathbf{n}}×{\mathbf{n}}+\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(2×{\mathbf{n}}×{\mathbf{n}}+6×{\mathbf{n}}+1,120+9×{\mathbf{n}}\right)$ real elements and $5×{\mathbf{n}}+3$ integer elements. All allocated memory is freed before return of g02ajf.
## 10Example
This example finds the nearest correlation matrix to:
$G = ( 2 -1 0 0 -1 2 -1 0 0 -1 2 -1 0 0 -1 2 )$
weighted by:
$H = ( 0.0 10.0 0.0 0.0 10.0 0.0 1.5 1.5 0.0 1.5 0.0 0.0 0.0 1.5 0.0 0.0 )$
with minimum eigenvalue $0.04$.
### 10.1Program Text
Program Text (g02ajfe.f90)
### 10.2Program Data
Program Data (g02ajfe.d)
### 10.3Program Results
Program Results (g02ajfe.r) | 2,787 | 9,019 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 102, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-06 | latest | en | 0.610068 |
https://datascience.stackexchange.com/questions/67089/anomaly-detection-novelty-detection/69509 | 1,638,980,739,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363515.28/warc/CC-MAIN-20211208144647-20211208174647-00157.warc.gz | 248,976,886 | 34,886 | # Anomaly Detection/Novelty detection
I have a data-set that has over 6 million normal data and around 50 anomaly data. Those anomaly data is identified manually (by monitoring the user`s activity over camera and identify). I need to develop a model to detect these anomalies.
My problem is that the anomaly data looks like normal data, which means they are not outliers or has a certain pattern. If I plot the normal data over anomaly data they are in the same distribution.
I tried several anomaly detection approaches:
1. Multivariate Gaussian Distribution Approach to identify anomalies
• I tried to create new features that anomaly data will be outliers and then I can use Multivariate Gaussian Distribution Approach, but could not able to find any combination to isolate the anomalies.
2. I guess there is no point of using a classification algorithm since dataset is highly imbalanced.
• I tried OneClassSVM, DecisionTree, RandomForest but AUC is 0.5 (as good as random).
How to implement a model for this kind of scenario?
Other methods which I can think about:
• Develop a NN with AutoEncoders
• Try Generate Synthetic Samples and resample the dataset
• How many features does your data have? How about using Lasso Regression to select features (using 50 anomaly data and another random 50/100 normal data, assuming that normal data come from the same distribution) and then see when plotting normal vs. abnormal data points, are abnormal points separated out? Feb 10 '20 at 7:43
• How can you tell they are anomalous data? If they are not outliers, statistically speaking, where can you tell they are anomalies? There must be some difference that lets your mind separate them from the rest. Let's follow this hint to find the right features for your model Mar 11 '20 at 8:04
• that specific records are identified as anomalous by looking at the user behaviour at that specific time..Issue is they are not outliers,they are exactly looks like normal data. Mar 11 '20 at 10:37
• ok, but how can you as a human say it's an anomaly? In plain words, not mathematically. Mar 11 '20 at 10:43
• dataset is related daily usage of pachinko and slot (pinball ) game users.for example customer has identified some particular machine,at specific time some user has done a fraud or illegal game play(monitoring user behaviors).so he recorded that entry as a fraudulent or illegal input to the machine at that specific time.dataset is already separated to normal and anomaly data. Mar 11 '20 at 10:56 | 547 | 2,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-49 | latest | en | 0.930626 |
https://www.ferienwohnung-colditz.de/ball-mill/s3g9m41s/feed.html | 1,603,338,681,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878879.33/warc/CC-MAIN-20201022024236-20201022054236-00260.warc.gz | 724,009,699 | 7,081 | feed size ball mill wikipedia ukraine
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# feed size ball mill wikipedia ukraine
A ball mill is a type of grinder used to grind, blend and sometimes for mixing of materials for use in mineral dressing processes, paints, pyrotechnics, ceramics and selective laser works on the principle of impact and attrition size reduction is done by impact as the balls drop from near the top of the shell. A ball mill consists of a hollow cylindrical shell rotati
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Based on his work, this formula can be derived for ball diameter sizing and selection Dm lt 6 log dk d0.5 where D m the diameter of the singlesized balls in mm.d the diameter of the largest chunks of ore in the mill feed in mm. dk the P90 or fineness of the finished product in microns umwith this the finished product is
• Cement Mill Wikipedia
Ball mills are normally operated at around 75 of critical speed, so a mill with diameter 5 metres will turn at around 14 rpm. The mill is usually divided into at least two chambers although this depends upon feed input size mills including a roller press are mostly singlechambered, allowing the use of different sizes of grinding media.
• Vsi Mill Wikipedia
A VSI mill vertical shaft impactor mill is a mill that comminutes particles of material into smaller finer particles by throwing them against a hard surface inside the mill called the wear plate. Any hard or friable materials can be ground with low value of metal waste. This type of mill is combined with a classifier for fine tuning of a product size.
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A mill is a device that breaks solid materials into smaller pieces by grinding, crushing, or cutting. Such comminution is an important unit operation in many are many different types of mills and many types of materials processed in them. Historically mills were powered by hand e.g., via a hand crank, working animal e.g., horse mill, wind or water .
• Grinding In Ball Mills Modeling And Process Control
Keywords Ball mills, grinding circuit, process control. I. Introduction Grinding in ball mills is an important technological process applied to reduce the size of particles which may have different nature and a wide diversity of physical, mechanical and chemical characteristics. Typical examples are the various ores, minerals, limestone, etc.
• Feed Size Vs Ball Size Grinding Amp Classification
One can argue the f50 is a more realistic value since balls wear and the average ball size might be closer to half the maximum. Then the multiple would be closer to 13 times the maximum particle size fed to the ball mill. A f80 6 mm would need a ball size of 75 mm. An empirical model from 34Emerging Trends in Mineral Processing34, 2005, is
• Ball Mill Wikipedia
A ball mill is a type of grinder used to grind, blend and sometimes for mixing of materials for use in mineral dressing processes, paints, pyrotechnics, ceramics and selective laser works on the principle of impact and attrition size reduction is done by impact as the balls drop from near the top of the shell. A ball mill consists of a hollow cylindrical shell rotati
• Isamill Wikipedia
IsaMill Operating Principles. The IsaMill is a stirredmedium grinding mill, in which the grinding medium and the ore being ground are stirred rather than being subjected to the tumbling action of older highthroughput mills such as ball mills and rod mills.Stirred mills often consist of stirrers mounted on a rotating shaft located along the central axis of the mill.
• Pdf Design And Fabrication Of Mini Ball Mill Methodology
DESIGN AND FABRICATION OF MINI BALL MILL. Variations of t he ore hardness and feed partic le size. and P.T. Luckie, The effect of ball size on mill performance. Powder Technology, 1976. 14
• Ball Mill Cermica Wiki Fandom
Espaol Molino de bolas A ball mill is a type of grinder used to grind materials into extremely fine powder for use in mineral dressing processes, paints, pyrotechnics, and ceramics. Indicemostrar Description A ball mill, a type of grinder, is a cylindrical device used in grinding or mixing materials like ores, chemicals, ceramic raw materials and paints . Ball mills rotate around a
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milling circuit consists of a rod mill followed by a ball mill in series. Crusher product 9 mm is fed to the rod mill, and the water is fed in ratio to the ore feed mass. The rod mill discharge is pumped, without any further water addition, to the first ball mill. The ball mill discharges to a sump where water is
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Ball mills are among the most variable and effective tools when it comes to size reduction of hard, brittle or fibrous materials. The variety of grinding modes, usable volumes and available grinding tool materials make ball mills the perfect match for a vast range of applications. Feed material mediumhard, hard, brittle, fibrous dry or wet.
• Top 5 Productspecific Milling Technologies Used In
Impact and attrition sizereduction methods include air classifying mills, pin mills, hammer mills and jet mills. Shear, impact and compression methods are used in media or ball mills. The five types of milling technologies discussed in this article cover more than 90 percent of sizereduction applications in major chemical, food
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First, a Jar Mill grindability test requires a 5 lb. 2 kg sample and produces a direct measured specific energy net Hphrt to grind from the design feed size to the required product size. The second test, a Bond Work Index determination, results in a specific energy value net Hphrt from an empirical formula.
• The Operating Principle Of The Ball Mill Primo Pizza
The operating principle of the ball mill consists of following steps. In a continuously operating ball mill, feed material fed through the central hole one of the caps into the drum and moves therealong, being exposed by grinding media. The material grinding occurs during impact falling grinding balls and abrasion the particles between the balls.
• How Can One Select Ball Size In Ball Milling And How Much
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• Practical 1 Ball Milling Tf Lab 1
Practical 1 Title Ball Milling Objective To grind the coarse salt to a smaller size by using a ball mill and to obtain the particle size distribution of the initial and the sieved final mixture. Introduction 39Ball milling is a method used to break down the solids to smaller sizes or into a powder. A | 1,500 | 7,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-45 | latest | en | 0.928925 |
https://jp.mathworks.com/matlabcentral/cody/problems/2958-finding-perimeter-of-a-rectangle/solutions/577820 | 1,606,811,948,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141672314.55/warc/CC-MAIN-20201201074047-20201201104047-00079.warc.gz | 349,050,746 | 16,631 | Cody
# Problem 2958. Finding perimeter of a rectangle
Solution 577820
Submitted on 9 Feb 2015 by Zikobrelli
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 8; w = 3; y_correct = 22; assert(isequal(perimeter(x, w),y_correct))
ans = 22
2 Pass
%% x = 10; w = 7; y_correct = 34; assert(isequal(perimeter(x, w),y_correct))
ans = 34
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http://www.coderanch.com/t/375794/java/java/BigDecimal-rounding | 1,462,365,449,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860123023.37/warc/CC-MAIN-20160428161523-00074-ip-10-239-7-51.ec2.internal.warc.gz | 431,762,977 | 10,972 | This week's book giveaways are in the Refactoring and Agile forums.We're giving away four copies each of Re-engineering Legacy Software and Docker in Action and have the authors on-line!See this thread and this one for details.
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# BigDecimal rounding problem
Vijay Dharap
Ranch Hand
Posts: 32
referrinf to following code,
but my belief is if i use Rounding by BigDecimal.ROUND_HALF_EVEN then it should do round up for 4.225 to 4.23.
even the documentation suggests that way.. then y it doesnt do it??
any clues
Vijay
Vijay Dharap
Ranch Hand
Posts: 32
Sorry correct snippet is as following. sorry for all the trouble..
Robert Hayes
Ranch Hand
Posts: 116
Replace your round method with this, and you'll see why it does this:
[ February 04, 2005: Message edited by: Robert Hayes ]
Vijay Dharap
Ranch Hand
Posts: 32
thanks robert reason stood in front of my eyes,
But is there any other way to achieve what i want to achieve??
I want 4.225 to be rounded to 2 decimal places giving output as 4.23.
any help will be appriciated.
marc weber
Sheriff
Posts: 11343
There are two things going on here.
First, as Robert's code illustrates, the BigDecimal values are probably not what you expect them to be. The reason is that they are initially doubles, and so they are stored according to IEEE 754 standards, which are not precise. Indeed, this is actually why we use BigDecimal. But in the code above, by the time these values are passed to the BigDecimal constructor, the precision is already gone.
The trick is in avoiding that floating-point representation. One way to do this is by using Strings instead of doubles. This is critical in using BigDecimal, and is explained in the API under the descriptions for BigDecimal constructors -- especially BigDecimal(double)...
http://java.sun.com/j2se/1.5.0/docs/api/java/math/BigDecimal.html
So if you use the BigDecimal(String) constructor with Robert's addition to your code, you'll see that the BigDecimal values are exactly what you expect. But there's still a problem, and that brings us to the second issue...
I believe you want ROUND_HALF_UP instead of ROUND_HALF_EVEN.
[ February 04, 2005: Message edited by: marc weber ]
Vijay Dharap
Ranch Hand
Posts: 32
Hi Marc and Robert,
I tried your change by trying to pass the String to the BigDecimal's constructor. this is the code that i changed
Still output now shows exact 4.225 for bigdecimals value. but still rounding was not a success.
my output was
About the rounding way to use, i dont want to use ROUND_UP as the value for 4.224 must round to 4.22 while 4.225 and 4.226 should round to 4.23. Thats is what is most logical i believe.
Vijay Dharap
Ranch Hand
Posts: 32
hey marc,
it worked man.. with ROUND_HALF_UP, it worked.
thanks for pointing out both the problems in my code
thanks again.
marc weber
Sheriff
Posts: 11343
Yes, BigDecimal has a lot of different ways to round.
But as the API says about ROUND_HALF_UP, "Note that this is the rounding mode that most of us were taught in grade school."
[ February 04, 2005: Message edited by: marc weber ] | 797 | 3,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-18 | longest | en | 0.904623 |
https://herebeanswers.com/things-you-need-to-consider-before-playing-online-casino-games.html | 1,643,066,430,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304686.15/warc/CC-MAIN-20220124220008-20220125010008-00566.warc.gz | 347,322,582 | 14,320 | Things you need to consider before playing online casino games
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• Terms and conditions | 666 | 3,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-05 | latest | en | 0.975171 |
https://math.stackexchange.com/questions/388532/weird-3n-in-an-identity-to-be-combinatorially-proved | 1,561,433,133,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999787.0/warc/CC-MAIN-20190625031825-20190625053825-00530.warc.gz | 508,269,991 | 36,602 | # Weird $3^n$ in an identity to be combinatorially proved
Give a combinatorial proof of the following identity:
$$3^n=\sum_{i=0}^{n}\binom{n}{i}2^{n-i}$$
I can't see any counting argument that would yield $3^n$, and the right hand side is also pretty opaque. For some reason I really really suck at doing these proofs - I just started my first combinatorics course.
• This is the Newton binomial for $(2+1)^n$ – Ewan Delanoy May 11 '13 at 14:48
• just take out $2^n$ and use Binomial theorem – Alex May 11 '13 at 14:52
• I want a combinatorics proof which means no algebraic rearranging of the stuff. – ithisa May 11 '13 at 14:59
• @EricDong It might help to modify your question as "I want a proof by a combinatorial argument" to get better responses. – user75930 May 11 '13 at 15:13
## 3 Answers
Suppose you have $n$ balls to be put into 3 boxes. You may keep any or all boxes empty.Then the number of ways of putting $n$ balls into 3 boxes is $3^n$. For the right hand side of the equation, you may fix a box $C$. The box $C$ may contain $0,1,2,\dots n$ balls. If the box $C$ has $i$ balls, you may choose the $i-\text{balls}$ to be put into $C$ in $\binom{n}{i}$ ways and the remaining $n-i$ balls may be put into boxes A or B in $2^{n-i}$ ways. Hence the number of ways of putting $n$ balls into 3 boxes is $\displaystyle \sum_{i=0}^n \binom{n}{i}2^{n-i}$ ways and hence $$3^n=\displaystyle \sum_{i=0}^n \binom{n}{i}2^{n-i}.$$ Please let me know if my solution is incorrect.
Hints:
$$(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k$$
Hint: Color $n$ balls with 3 colors. | 515 | 1,579 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-26 | latest | en | 0.865127 |
https://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Writing_Equations_for_Redox_Reactions | 1,500,676,435,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423812.87/warc/CC-MAIN-20170721222447-20170722002447-00441.warc.gz | 640,031,909 | 17,331 | # Writing Equations for Redox Reactions
This page explains how to work out electron-half-reactions for oxidation and reduction processes, and then how to combine them to give the overall ionic equation for a redox reaction. This is an important skill in inorganic chemistry.
### Electron-half-equations
The ionic equation for the magnesium-aided reduction of hot copper(II) oxide to elemental copper is given below :
$Cu^{2+} + Mg \rightarrow Cu + Mg^{2+}$
The equation can be split into two parts and considered from the separate perspectives of the elemental magnesium and of the copper(II) ions. This arrangement clearly indicates that the magnesium has lost two electrons, and the copper(II) ion has gained them.
$Mg \rightarrow Mg^{2+} + 2e^-$
$Cu^{2+} + 2e^- \rightarrow Cu$
These two equations are described as "electron-half-equations," "half-equations," or "ionic-half-equations," or "half-reactions." Every redox reaction is made up of two half-reactions: in one, electrons are lost (an oxidation process); in the other, those electrons are gained (a reduction process).
### Working out electron-half-equations and using them to build ionic equations
In the example above, the electron-half-equations were obtained by extracting them from the overall ionic equation. In practice, the reverse process is often more useful: starting with the electron-half-equations and using them to build the overall ionic equation.
Example 1: The reaction between Chlorine and Iron (III) Ions
Chlorine gas oxidizes iron(II) ions to iron(III) ions. In the process, the chlorine is reduced to chloride ions. From this information, the overall reaction can be obtained. The chlorine reaction, in which chlorine gas is reduced to chloride ions, is considered first:
$Cl_2 \rightarrow Cl^-$
The atoms in the equation must be balanced:
$Cl_2 \rightarrow 2Cl^-$
This step is crucial. If any atoms are unbalanced, problems will arise later.
To completely balance a half-equation, all charges and extra atoms must be equal on the reactant and product sides. In order to accomplish this, the following can be added to the equation:
• electrons
• water
• hydrogen ions (unless the reaction is being done under alkaline conditions, in which case, hydroxide ions must be added and balanced with water)
In the chlorine case, the only problem is a charge imbalance. The left-hand side of the equation has no charge, but the right-hand side carries 2 negative charges. This is easily resolved by adding two electrons to the left-hand side. The fully balanced half-reaction is:
$Cl_2 +2 e^- \rightarrow 2Cl^-$
Next the iron half-reaction is considered. Iron(II) ions are oxidized to iron(III) ions as shown:
$Fe^{2+} \rightarrow Fe^{3+}$
The atoms balance, but the charges do not. There are 3 positive charges on the right-hand side, but only 2 on the left. To reduce the number of positive charges on the right-hand side, an electron is added to that side:
$Fe^{2+} \rightarrow Fe^{3+} + e-$
The next step is combining the two balanced half-equations to form the overall equation. The two half-equations are shown below:
It is obvious that the iron reaction will have to happen twice for every chlorine reaction. This is accounted for in the following way: each equation is multiplied by the value that will give equal numbers of electrons, and the two resulting equations are added together such that the electrons cancel out:
At this point, it is important to check once more for atom and charge balance. In this case, no further work is required.
Example 2: The reaction between Hydrogen Peroxide and Magnanate Ions
The first example concerned a very simple and familiar chemical equation, but the technique works just as well for more complicated (and perhaps unfamiliar) chemistry.
Manganate(VII) ions, MnO4-, oxidize hydrogen peroxide, H2O2, to oxygen gas. The reaction is carried out with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulfuric acid. As the oxidizing agent, Manganate(VII) is reduced to manganese(II).
The hydrogen peroxide reaction is written first according to the information given:
$H_2O_2 \rightarrow O_2$
The oxygen is already balanced, but the right-hand side has no hydrogen.
All you are allowed to add to this equation are water, hydrogen ions and electrons. Adding water is obviously unhelpful: if water is added to the right-hand side to supply extra hydrogen atoms, an additional oxygen atom is needed on the left. Hydrogen ions are a better choice.
Adding two hydrogen ions to the right-hand side gives:
$H_2O_2 \rightarrow O_2 + 2H^+$
Next the charges are balanced by adding two electrons to the right, making the overall charge on both sides zero:
$H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-$
Next the manganate(VII) half-equation is considered:
$MnO_4^- \rightarrow Mn^{2+}$
The manganese atoms are balanced, but the right needs four extra oxygen atoms. These can only come from water, so four water molecules are added to the right:
$MnO_4^- \rightarrow Mn^{2+} + 4H_2O$
The water introduces eight hydrogen atoms on the right. To balance these, eight hydrogen ions are added to the left:
$MnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O$
Now that all the atoms are balanced, only the charges are left. There is a net +7 charge on the left-hand side (1- and 8+), but only a charge of +2 on the right. 5 electrons are added to the left-hand side to reduce the +7 to +2:
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} _ 4H_2O$
This illustrates the strategy for balancing half-equations, summarized as followed:
• Balance the atoms apart from oxygen and hydrogen.
• Balance the oxygens by adding water molecules.
• Balance the hydrogens by adding hydrogen ions.
• Balance the charges by adding electrons.
Now the half-equations are combined to make the ionic equation for the reaction.
As before, the equations are multiplied such that both have the same number of electrons. In this case, the least common multiple of electrons is ten:
The equation is not fully balanced at this point. There are hydrogen ions on both sides which need to be simplified:
This often occurs with hydrogen ions and water molecules in more complicated redox reactions. Subtracting 10 hydrogen ions from both sides leaves the simplified ionic equation.
$2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + 5O_2$
Example 3: Oxidation of Ethanol of Acidic Potassium Dichromate (IV)
This technique can be used just as well in examples involving organic chemicals. Potassium dichromate(VI) solution acidified with dilute sulfuric acid is used to oxidize ethanol, CH3CH2OH, to ethanoic acid, CH3COOH.
The oxidizing agent is the dichromate(VI) ion, Cr2O72-, which is reduced to chromium(III) ions, Cr3+. The ethanol to ethanoic acid half-equation is considered first:
$CH_3CH_2OH \rightarrow CH_3COOH$
The oxygen atoms are balanced by adding a water molecule to the left-hand side:
$CH_3CH_2OH + H_2O \rightarrow CH_3COOH$
Four hydrogen ions to the right-hand side to balance the hydrogen atoms:
$CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4H^+$
The charges are balanced by adding 4 electrons to the right-hand side to give an overall zero charge on each side:
$CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4H^+ + 4e^-$
The unbalanced dichromate (VI) half reaction is written as given:
$Cr_2O_7^{2-} \rightarrow Cr^{3+}$
At this stage, students often forget to balance the chromium atoms, making it impossible to obtain the overall equation. To avoid this, the chromium ion on the right is multiplied by two:
$Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$
The oxygen atoms are balanced by adding seven water molecules to the right:
$Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O$
The resulting hydrogen atoms are balanced by adding fourteen hydrogen ions to the left:
$Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O$
Six electrons are added to the left to give a net +6 charge on each side.
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
The two balanced half reactions are summarized:
$CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4H^+ + 4e^-$
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
The least common multiple of 4 and 6 is 12. Therefore, the first equation is multiplied by 3 and the second by 2, giving 12 electrons in each equation:
Simplifying the water molecules and hydrogen ions gives final equation:
### Balancing reactions under alkaline conditions
Working out half-equations for reactions in alkaline solution is decidedly more tricky than the examples above. As some curricula do not include this type of problem, the process for balancing alkaline redox reactions is covered on a separate page.
### Contributors
Jim Clark (Chemguide.co.uk) | 2,282 | 8,761 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-30 | longest | en | 0.916688 |
https://justhindi.in/the-ordinary-differential-equations-and-partial-differential-equation-differ-by/ | 1,675,632,995,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00153.warc.gz | 351,842,691 | 18,836 | Question
1. The Ordinary Differential Equations And Partial Differential Equation Differ By
Differential equations are a mathematical tool that can be used to understand a variety of different phenomena. In this blog post, we will explore the ordinary differential equation and how it differs from the partial differential equation. Both equations deal with changes in quantities over time, but they have some key differences that you should be aware of if you want to use them correctly. By understanding these differences, you will be better equipped to solve differential equations and understand the models that underlie various physical systems.
What are Ordinary Differential Equations?
Ordinary differential equations are a subset of differential equations that are solved using the Method Of Ordinary Differentiation. Partial differential equations are a subset of differential equations that are solved by integrating over certain portions of the domain. These two subsets of differential equations differ in several ways, most notably in how they are solved.
Ordinary differential equation can be solved by the Method Of Ordinary Differentiation, which involves taking derivatives of both the function and the vector field with respect to one another. This process is repeated until all derivatives have been taken and the function has at least one solution in its domain. Partial differential equation can be solved by integrating over certain portions of the domain, but this process can be more complicated than solving an ordinary differential equation because partial derivatives need to be taken with respect to multiple variables at once.
What is Partial Differential Equation?
There are two types of differential equations: ordinary and partial. Ordinary differential equations describe the change in a function over time, while partial differential equations allow for some functions to vary in place. Partial differential equations differ by how variables vary with one another.
Partial differential equations can be broken down into two categories: homogeneous and heterogeneous. Homogeneous partial differential equations have the same variable everywhere, while heterogeneous partial differential equations involve different variables at different locations.
Both types of partial differential equations can be solved using either linear or nonlinear methods. Linear methods use algebra to solve for solutions, while nonlinear methods break down the equation into several smaller ones that can be solved more easily.
Difference between Ordinary and Partial Differential Equations
Ordinary differential equations and partial differential equations differ by their properties. Ordinary differential equations are linear and can be solved for the unknown function in terms of derivatives at a few points. Partial differential equations are nonlinear and can only be solved for the unknown function in terms of general expressions. In addition, ordinary differential equations are independent of initial conditions, while partial differential equations need an initial condition to be valid. Additionally, ordinary differential equations always have the same shape, while partial differential equations may have more complicated shapes.
Difference Between ODE and PDE
ODEs and PDEs are two very different types of equations. ODEs are used to model physical systems, while PDEs are used to model chemical and biological systems. Here is a quick summary of the key differences between ODEs and PDEs:
1) ODEs can be solved for analytical solutions, while PDEs cannot.
2) ODEs can be linear or nonlinear, while PDEs always require a nonlinear equation in order to be solved.
3) ODEs can be treated as first-order equations, while PDEs always require a higher order equation in order to solve them.
4) ODE solutions typically only vary slowly over time, while PDE solutions can vary rapidly over time.
Applications of ODE and PDE in Engineering
There is a huge range of applications that Ordinary Differential Equations (ODEs) and Partial Differential Equations (PDEs) can be put to in engineering. They are used for a wide variety of problems, from mathematical modeling to physical simulations to design analysis.
One of the most common uses of ODEs and PDEs is in mathematical modeling. Modeling is the process of creating a model or simulation of some aspect of reality in order to understand it better. Mathematical models are often used to simulate complex systems, such as weather patterns or financial markets. ODEs and PDEs are extremely versatile tools for modeling because they can be used to describe a wide range of behavior.
Another common use for ODEs and PDEs is in physical simulations. Physical simulations are used to study how different objects or systems behave under various conditions. They can help engineers better understand how a particular system will behave under different conditions and help them make more accurate predictions about how the system will behave in future scenarios.
Finally, ODEs and PDEs are also frequently used for design analysis. Design analysis is the process of analyzing how different designs would impact the performance and reliability of a system. By understanding how different designs affect performance and reliability, engineers can make better decisions when designing systems
Summary
The ordinary differential equations and partial differential equation differ by their properties. The ordinary differential equation is simpler to solve, but the partial differential equation can give more accurate results.
2. The ordinary differential equation (ODE) and partial differential equation (PDE) are powerful mathematical tools used by scientists and engineers in many disciplines. The two equations have much in common, yet they differ in a few important ways.
The ODE is used when the dependent variable is a function of only one independent variable, such as time or distance. These equations can be solved using analytical techniques such as separation of variables, Laplace transforms and series solutions. They are also relatively simple to solve numerically using well-established methods like the Runge-Kutta algorithm.
In contrast, PDEs model relationships that involve multiple independent variables such as space and time, or other physical parameters like temperature or pressure.
3. 🤔 What is the difference between Ordinary Differential Equations (ODEs) and Partial Differential Equations (PDEs)?
The main difference between ODEs and PDEs is that ODEs involve functions of one or more variables, while PDEs involve multiple functions of multiple variables. In other words, ODEs involve one or more independent variables and a single dependent variable, while PDEs involve multiple independent and dependent variables.
An ODE is a mathematical equation that relates the rate of change of one or more variables to the values of the variables themselves. In other words, an ODE describes how a certain variable changes over time. An example of an ODE is Newton’s second law of motion, which states that the acceleration of an object is proportional to the net force acting on it.
PDEs, on the other hand, describe how a quantity changes over a region of space rather than a single point. PDEs are much more complicated than ODEs and involve multiple independent and dependent variables. An example of a PDE is the heat equation, which describes how heat flows through a material.
In general, ODEs are used to model physical systems that involve one or more variables, while PDEs are used to model physical systems that involve multiple variables. ODEs are often used to solve problems in physics, engineering, and other sciences, while PDEs are often used to solve problems in mathematics and other fields.
So, while ODEs and PDEs are both important mathematical models, they differ in the way they are used and the scope of the problems they are used to solve. 🤓 | 1,471 | 7,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-06 | longest | en | 0.95447 |
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To solve a factorial, simply multiply the given number by every number below it until you reach one. So, to solve 5!, you would multiply 5 by 4, then 3, then 2, and then 1. The answer would be 120. It is important to start with the given number and work your way down, rather than starting with one and working your way up. This is because the factorial operation is not commutative - that is, 5! is not the same as 1 x 2 x 3 x 4 x 5. When solving factorials, always start with the given number and work your way down to one.
Solving a system of equations by graphing is a visual way to find the point of intersection for two linear equations. To do this, first plot the two equations on a coordinate plane. Then, use a straightedge to draw a line through the points of intersection. The point where the line intersects the x-axis is the solution to the system of equations. This method can be used to solve systems of two or more equations. However, it is important to note that not all systems of equations will have a unique solution. In some cases, the lines may be parallel and will not intersect. In other cases, the lines may intersect at more than one point. When this happens, the system of equations is said to be inconsistent and has no solution. | 603 | 2,886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-49 | latest | en | 0.965772 |
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Picking cards (Posted on 2006-08-17)
A player picks random cards from an ordinary card deck, without returning them to the deck. How many cards should he pick so as to get at least one ace with 70% probability?
No Solution Yet Submitted by atheron Rating: 3.0000 (1 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
answer Comment 7 of 7 |
14 cards;
if we pick 14 cards then prob. of having not a single Ace is
48C14/52C14=.2726
so having at least one ace = 1-.2726=.7274 >.7
Posted by saurabh on 2006-08-30 16:50:46
Search: Search body:
Forums (0) | 202 | 694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-51 | latest | en | 0.88987 |
https://tw.forumosa.com/t/is-the-progressive-difference-cumulative/26238 | 1,607,083,292,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141735600.89/warc/CC-MAIN-20201204101314-20201204131314-00054.warc.gz | 512,371,823 | 5,582 | # Is the progressive difference cumulative?
hello everyone,
Could anyone help me understand how the tax system works?
If my taxable income is 2.000.000 NTD, how will I be taxed?
Will it be
0 to 370000 x 6% - 0 = 22000
370001 to 990000 x 13% - 25900 = 54700
990001 to 1980000 x 21% - 105,100 = 102800
1980001 to 2000000 x 30% - (pro rata 283300) = let’s say 4000
total tax = 184000
–> 9% average tax rate
OR
2)
2000000 x 30 % = 600000 - 283300 = 316700
–> 16% average tax rate
What I need to know is whether the progressive difference is cumulative or whether it is taken as a lump sum when you reach a particular bracket.
This is confusing as the way the tax scale is presented is that you remove the progressive difference in each scale
You’re in finance, aren’t you?
Sorry, so am I, and I can’t help either.
HG
Option No. 2.
This is the way it’s calculated:
First 370000 at 6% = 22200
990000 - 370000 at 13% = 80600
1980000 - 990000 at 21% = 207900
2000000 - 1980000 at 30% = 6000
22200 + 80600 + 207900 + 6000 = 316700
Which is the same as 2000000 at 30% - 283300
However, there is a complication:
When you start work, they take a flat rate 20% (at least) tax for the first 6 months (i.e. they overtax you). This is because you’re a dirty foreigner who’s almost certainly about to flee the country without paying the tax you owe. If they overtax you, you’ll have to claim it back at the end of the year - so they don’t care if you disappear.
thanks for the bad news… looks like i will get hit by 16% … versus the 9 % i am paying today in Singapore.
But the way they present it is very stange:
Net Income (NT\$) Rate (%) Less Progressive Difference (NT\$)
1 - 370,000 6 0
370,001 - 990,000 13 25,900
990,001 - 1,980,000 21 105,100
1,980,001 - 3,700,000 30 283,300
as is supposes you are have to remove the progressive difference at each scale.
So is 16% a reasonable percentage i should consider when going to Taiwan assuming i am getting 2M NTD?
BTW what kind of lifestyle should i expect with 2MNTD a year +/- 160K a month?
thanks
I believe you choose the appropriate entry point, not work your way up the scale, to find your tax calculation.
With 2M NTD a year, you can live VERY well. More money than I ever had in Taiwan, and I lived very comfortably and saved lots of money too.
ok clear.
from what i can read from the other posts, tax matters seem to be a very sensitive topic.
thank you for the feedbacks. | 734 | 2,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-50 | latest | en | 0.899555 |
https://web2.0calc.com/questions/what-is-the-five-number-summary-of-the-data-set-12 | 1,586,386,295,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371824409.86/warc/CC-MAIN-20200408202012-20200408232512-00124.warc.gz | 755,819,217 | 6,087 | +0
# What is the five-number summary of the data set? {12, 14, 15, 14, 16, 18, 19, 18}
0
183
1
What is the five-number summary of the data set?
{12, 14, 15, 14, 16, 18, 19, 18}
minimum value = 12, Q1 = 13, median = 16, Q3 = 17, maximum value = 19
minimum value = 12, Q1 = 13, median = 15.5, Q3 = 18, maximum value = 19
minimum value = 12, Q1 = 14, median = 15, Q3 = 17, maximum value = 19
minimum value = 12, Q1 = 14, median = 15.5, Q3 = 18, maximum value = 19
Sep 25, 2019
#1
+109740
+1
Order the data from low to high
(12, 14, 14, 15, 16, 18, 18, 19)
The median = [15 + 16] / 2 = 31/2 = 15.5
Q1 = [ 14 + 14] / 2 = 28/2 = 14
Q3 = [18 + 18]/2 = 36/2 = 18
Min value = 12
Max value = 19
Sep 25, 2019
edited by CPhill Sep 25, 2019 | 374 | 764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-16 | latest | en | 0.582315 |
https://en.wikipedia.org/wiki/Region-growing | 1,495,496,912,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607242.32/warc/CC-MAIN-20170522230356-20170523010356-00320.warc.gz | 775,134,417 | 14,109 | # Region growing
(Redirected from Region-growing)
Region growing is a simple region-based image segmentation method. It is also classified as a pixel-based image segmentation method since it involves the selection of initial seed points.
This approach to segmentation examines neighboring pixels of initial seed points and determines whether the pixel neighbors should be added to the region. The process is iterated on, in the same manner as general data clustering algorithms. A general discussion of the region growing algorithm is described below.
## Region-based segmentation
The main goal of segmentation is to partition an image into regions. Some segmentation methods such as thresholding achieve this goal by looking for the boundaries between regions based on discontinuities in grayscale or color properties. Region-based segmentation is a technique for determining the region directly. The basic formulation is:
${\displaystyle (a){\text{ }}\bigcup \nolimits _{i=1}^{n}{R_{i}=R.}}$
${\displaystyle (b){\text{ }}R_{i}{\text{ is a connected region}},{\text{ i}}={\text{1}},{\text{ 2}},{\text{ }}...,{\text{n}}}$
${\displaystyle (c){\text{ }}R_{i}\bigcap R_{j}=\varnothing {\text{ for all }}i=1,2,...,n.}$
${\displaystyle (d){\text{ }}P(R_{i})=TRUE{\text{ for }}i=1,2,...,n.}$
${\displaystyle (e){\text{ }}P(R_{i}\bigcup R_{j})=FALSE{\text{ for any adjacent region }}R_{i}{\text{ and }}R_{j}.}$
${\displaystyle P(R_{i})}$ is a logical predicate defined over the points in set ${\displaystyle R_{i}}$ and ${\displaystyle \varnothing }$ is the null set.
(a) means that the segmentation must be complete; that is, every pixel must be in a region.
(b) requires that points in a region must be connected in some predefined sense.
(c) indicates that the regions must be disjoint.
(d) deals with the properties that must be satisfied by the pixels in a segmented region. For example, ${\displaystyle P(R_{i})={\text{TRUE}}}$ if all pixels in ${\displaystyle R_{i}}$ have the same grayscale.
(e) indicates that region ${\displaystyle R_{i}}$ and ${\displaystyle R_{j}}$ are different in the sense of predicate ${\displaystyle P}$.
## Basic concept of seed points
The first step in region growing is to select a set of seed points. Seed point selection is based on some user criterion (for example, pixels in a certain grayscale range, pixels evenly spaced on a grid, etc.). The initial region begins as the exact location of these seeds.
The regions are then grown from these seed points to adjacent points depending on a region membership criterion. The criterion could be, for example, pixel intensity, grayscale texture, or color.
Since the regions are grown on the basis of the criterion, the image information itself is important. For example, if the criterion were a pixel intensity threshold value, knowledge of the histogram of the image would be of use, as one could use it to determine a suitable threshold value for the region membership criterion.
There is a very simple example followed below. Here we use 4-connected neighborhood to grow from the seed points. We can also choose 8-connected neighborhood for our pixels adjacent relationship. And the criteria we make here is the same pixel value. That is, we keep examining the adjacent pixels of seed points. If they have the same intensity value with the seed points, we classify them into the seed points. It is an iterated process until there are no change in two successive iterative stages. Of course, we can make other criteria, but the main goal is to classify the similarity of the image into regions.
## Some important issues
Figure 0. The histogram of Figure 1
Figure 1. The original figure
Figure 2. The seed points: 255~255
Figure 3. Threshold: 225~255
Figure 4. Threshold: 190~255
Figure 5. Threshold: 155~255
Then we can conclude several important issues about region growing：
1.The suitable selection of seed points is important.
The selection of seed points is depending on the users. For example, in a grayscale lightning image, we may want to segment the lightning from the background. Then probably, we can examine the histogram and choose the seed points from the highest range of it.
Obviously, the connectivity or pixel adjacent information is helpful for us to determine the threshold and seed points.
3.The value, “minimum area threshold”.
No region in region growing method result will be smaller than this threshold in the segmented image.
4.The value, “Similarity threshold value“.
If the difference of pixel-value or the difference value of average grayscale of a set of pixels less than “Similarity threshold value”, the regions will be considered as a same region.
The criteria of similarities or so called homogeneity we choose are also important. It usually depends on the original image and the segmentation result we want.
Some criteria often used are grayscale (average intensity or variance), color, and texture or shape.
## Simulation examples
Here we show a simple example for region growing.
Figure 1 is the original image which is a grayscale lightning image. The grayscale value of this image is from 0 to 255. The reason we apply region growing on this image is that we want to mark the strongest lightning part of the image and we also want the result to be connected without being split apart. Therefore, we choose the points having the highest grayscale value which is 255 as the seed points shown in the Figure 2.
After determining the seed points, we have to determine the range of threshold. Always keep in mind that the objective is to mark the strongest light in the image. The third figure is the region growing result from choosing the threshold between 225 and the value of seed points (which is 255). Hence we only mark out the points whose grayscale values are above 225.
If we make the range of threshold wider, we will get a result having a bigger area of the lightning region shown as the Figure 4 and the Figure 5.
We can observe the difference between the last two figures which have different threshold values. Region growing provides the ability for us to separate the part we want connected.
As we can see in Figure 3 to Figure 5, the segmented results in this example are seed-oriented connected. That means the result grew from the same seed points are the same regions. And the points will not be grown without being connected with the seed points.
Therefore, there are still lots of points in the original image having the grayscale values above 155 which are not marked in Figure 5.
This characteristic ensures the reliability of the segmentation and provides the ability to resist noise. For this example, this characteristic prevents us marking out the non-lightning part in the image because the lightning is always connected as one part.
1. Region growing methods can correctly separate the regions that have the same properties we define.
2. Region growing methods can provide the original images which have clear edges with good segmentation results.
3. The concept is simple. We only need a small number of seed points to represent the property we want, then grow the region.
4. We can determine the seed points and the criteria we want to make.
5. We can choose the multiple criteria at the same time.
1. Computationally expensive
2. It is a local method with no global view of the problem.
3. Sensitive to noise.
4. Unless the image has had a threshold function applied to it, a continuous path of points related to colour may exist which connects any two points in the image.
## References
• Jian-Jiun Ding, The class of "Time-Frequency Analysis and Wavelet Transform", the Department of Electrical Engineering, National Taiwan University (NTU), Taipei, Taiwan, 2007.
• Jian-Jiun Ding, The class of "Advanced Digital Signal Processing", the Department of Electrical Engineering, National Taiwan University (NTU), Taipei, Taiwan, 2008.
• W. K. Pratt, Digital Image Processing 4th Edition, John Wiley & Sons, Inc., Los Altos, California, 2007
• M. Petrou and P. Bosdogianni, Image Processing the Fundamentals, Wiley, UK, 2004.
• R. C. Gonzalez and R.E. Woods, Digital Image Processing 2nd Edition, Prentice Hall, New Jersey, 2002. | 1,818 | 8,220 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 13, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2017-22 | latest | en | 0.861637 |
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# Notre Dame MBA Candidate - Taking Questions
new topic post reply Update application status
Author Message
Current Student
Status: Student
Affiliations: University of Notre Dame
Joined: 05 Sep 2011
Posts: 6
Kudos [?]: 6 [1], given: 0
Location: United States (IN)
Schools: University of Notre Dame (Mendoza) - Class of 2013
Notre Dame MBA Candidate - Taking Questions [#permalink]
### Show Tags
05 Sep 2011, 09:54
1
KUDOS
Hey guys - having recently gone through the process of applying to B-school, I wanted to offer to field any questions you may have about the process, The University of Notre Dame and the Mendoza College of Business. Feel free to ask away - academics, life at Notre Dame, career development, etc. - and I'd be happy to provide you with answers. As a current student, I can probably give you some feedback on a lot of questions running through your mind.
Go Irish!
Joe
_________________
MBA Candidate - Class of 2013
University of Notre Dame
Kudos [?]: 6 [1], given: 0
Current Student
Status: Student
Affiliations: University of Notre Dame
Joined: 05 Sep 2011
Posts: 6
Kudos [?]: 6 [0], given: 0
Location: United States (IN)
Schools: University of Notre Dame (Mendoza) - Class of 2013
Re: Notre Dame MBA Candidate - Taking Questions [#permalink]
### Show Tags
19 Sep 2011, 13:35
All - there have been a few questions about the GMAT Range and Average Work Experience for Notre Dame's Class of 2013, which are posted below:
GMAT Range (Middle 80%): 620-760
Average Work Experience: 5 Years
You can check out the recently released statistics and class profile for the Class of 2013 at business.nd.edu.
Good luck with your applications, and Go Irish!
Joe
_________________
MBA Candidate - Class of 2013
University of Notre Dame
Kudos [?]: 6 [0], given: 0
Current Student
Status: Student
Affiliations: University of Notre Dame
Joined: 05 Sep 2011
Posts: 6
Kudos [?]: 6 [0], given: 0
Location: United States (IN)
Schools: University of Notre Dame (Mendoza) - Class of 2013
Re: Notre Dame MBA Candidate - Taking Questions [#permalink]
### Show Tags
25 Sep 2011, 15:50
Just a quick note this Sunday: if you're interested in staying up to date on the latest news and updates from the MBA admissions team at Notre Dame, I'd encourage you to follow them on Twitter and Facebook:
As always, feel free to get in touch if I can answer any question!
Joe
_________________
MBA Candidate - Class of 2013
University of Notre Dame
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Re: Notre Dame MBA Candidate - Taking Questions [#permalink]
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27 Sep 2011, 23:42
Hi Joe,
I have a few questions.
1) ND has 5 rounds of application, more than most of the other schools. how different are the chances of a student applying in R2 or R3 compared to that of one applying in R1? ( I wanted to know about scholarships awarded as well)
2) I want to do an MBA to change my career. I have 4yrs of IT exp (India). I want to get into the media and entertainment industry and
I have read that ND has an active media club. Could you shed some light on the activities of the club and how strong is the school in placing students within this industry?
Thanks
Ganesh
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Current Student
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Affiliations: University of Notre Dame
Joined: 05 Sep 2011
Posts: 6
Kudos [?]: 6 [0], given: 0
Location: United States (IN)
Schools: University of Notre Dame (Mendoza) - Class of 2013
Re: Notre Dame MBA Candidate - Taking Questions [#permalink]
### Show Tags
02 Oct 2011, 09:24
Ganesh -
Thanks! The past two months at Notre Dame have been absolutely great. The professors, MBA office and administration have worked really hard to make the program rigorous but rewarding, and it definitely shows. I've learned a lot this first module, have had some great experiences, and am really looking forward to more.
1) If you're serious about Notre Dame, I would encourage you to apply in the earlier rounds if your application is where you want it be. In the same beat, I would also say don't rush the application and be sure that it reflects the most complete picture of yourself. As for scholarships, there are absolutely opportunities, but they are handled on an individual applicant basis by the Admissions office.
2) I'm a career switcher as well, and Notre Dame does a great job at laying the groundwork of business fundamentals while keeping you engaged and letting you pursue your interests. One of the nice things about being in a smaller class size (my class has 130 MBA candidates) is that there are always opportunities for students to pursue their passions. We've had two new clubs founded already this year because students were driven by their interests, and the administration is very encouraging. If media and entertainment is something you're passionate about, then you'll find support for your efforts. I believe in recent years Notre Dame has had students placed in either internships or jobs at Disney, Fox News, ABC, NBC, MTV and EA.
If you're interested in Notre Dame, I would definitely encourage coming up for a visit. We host prospective applicants every week, and it's a great opportunity to get a feel for the campus and program, and you'll get to spend time with students both in class and one-on-one. You can find information here: business.nd.edu/MBA/Get_in_Touch/Visit_the_Campus/
Hope this helps, and let me know if I answer any other questions.
Go Irish!
Joe
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03 Oct 2011, 06:08
Joe - thanks for taking questions.
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03 Oct 2011, 08:44
Although I haven't had the course yet, Notre Dame's Business on the Frontlines is a great class and was a major draw for me to enroll in the program. It's an application only course that accepts 16 MBAs each year. Offered in the spring, it covers the entire semester and interterm break for a total of 6 credit hours and focuses on post-war torn countries (a minimum of three years since the end of conflict).
Students spend the first 7 weeks of the semester doing research and analysis on what caused the conflicts, why fighting occurred, and how business could have either positively or negatively impacted the situation. During the interterm break, the class will spend 12 days in country where they'll observe conditions since the end of the conflict, talk with local citizens and hear from business leaders on their efforts and needs.
The second half of the semester is spent processing your findings and understanding the business problems the country is facing. You work with your class and professor to come up with recommendations and create a presentation that three of your classmates (1 MBA, 1 JD, and 1 Masters in Peace Studies) and Professor Bartkus will present to Catholic Relief Services in Baltimore (which helps coordinate efforts while students are on the ground in country).
Past countries visited through this course included Lebanon, Kenya, and Uganda. I know that more country options are under review for the next class.
Two-year MBA Candidates are eligible to take the class spring semester of their second year.
If you have any other specific questions, I'm happy to answer them.
Go Irish!
Joe
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27 Oct 2011, 05:47
Thanks for taking questions Joe. Appreciate you taking time.
I'm considering applying and just wondering what the best part of the experience has been for you so far?
They talk a lot about small class size, camaraderie, etc. Is that really true?
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27 Oct 2011, 13:36
Since I started at Notre Dame in August, there have been a lot of great experiences. Students here are kept pretty busy with core classes and career development, but also strike a good balance. Football Saturdays are a blast, while the course work is really interesting and challenging at the same time.
One of the best experiences so far has been interterm. Interterm takes place over the two weeks between modules in the fall and again in the spring. As a first year, you're required to participate in a case competition during the first week. During these competitions, major companies will present a challenge they're facing to MBA students and provide vital information pertaining to the issue. Your team then works for a few days to create your own solution to the problem, and will present to company executives at the end. There are usually prizes to the top teams, and often times the winning plan, or some variation of it, is actually implemented by the company.
There's a little more flexibility with your second week. Here you're either able to go on fall break, or take advantage of a career trek. As a finance major, I took the week long trip to New York City that was set up by the MBA office. Taking advantage of the extensive Notre Dame alumni network, the office worked to set up meetings with 16 financial services firms/offices. Several of my classmates and I were invited to the New York Fed and banks such as Morgan Stanley and Citi for presentations and networking opportunities. Similar trips took place in San Francisco and Chicago. These are great ways to boost career development.
One you're out of your first semester, you have the opportunity to take advantage of international trips during interterm. For instance, during spring interterm, I'll be heading to China for two weeks, while several classmates will be heading to Chile and Argentina. Both are organized through the school.
As for class size, it is a smaller, personalized program. We have about 135 students in my class, and the overall program (first years, second years, and one years), is just over 300. During your first two modules, when you're taking core classes, you're in one of two sections with 65 or so students in each class. That's the largest class you'll be in while at Notre Dame. When you start taking concentration-specific classes, the class size gets much smaller. I personally like it because the smaller class sizes and better access to professors.
The smaller class size also means we have a very close class. I know the name and background of nearly everyone in my year and a good chunk of the second years and one years. Everyone gets along really well, but that isn't to say we aren't still competitive. It's a great environment because while you are competing with your classmates, its done in a respectful way where we're not undercutting or going after each other. Rather, you're able to build off the experience of others.
Happy to answer any other questions.
Go Irish!
Joe
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07 Feb 2012, 05:31
Dear js9657a,
I am an international applicant looking at Mendoza School of Business as a possible option. I have an interest in pursuing opportunities in the Commodities Supply Chain and Trading of Energy and Agricultural Commodities. Would like to know the following things:
1. How good is the placement scenario for International Candidates
2. Is it difficult to get a work Visa for those opting to work for sectors other than consulting
3. Companies like Exxon Mobil and GE come to the campus do they recruit international candidates as well.
4. How many international candidates return back to their home country upon graduation.
5. Does the school support international candidates who want to start an Entrepreneurial venture.
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Re: Notre Dame MBA Candidate - Taking Questions [#permalink] 07 Feb 2012, 05:31
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# Notre Dame MBA Candidate - Taking Questions
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# Abdula sold goods to Tahir on Jan 17, 2017 for Rs 18,000. He drew a bill of exchange for the same amount on Tahir for 45 days. On the same date Tahir accepted the bill and returned it to Abdulla. On the due date Abdulla presented the bill to Tahir which was dishonoured. Abdulla paid Rs 40 as noting charges. Five days after the dishonour of his acceptance Tahir settled his debt by making a payment of Rs 18,700 including interest and noting charges. Record the necessary journal entries in the books of Abdulla and Tahir. Also prepare Tahir's account in the books of Abdulla and Abdulla's account in the books of Tahir.
Solution
## Books of Abdula Journal Date Particulars L.F. Debit Amount Rs Credit Amount Rs 2017 Jan.17 Tahir Dr. 18,000 To Sales A/c 18,000 (Goods sold to Tahir) Jan.17 Bills Receivable A/c Dr. 18,000 To Tahir 18,000 (Tahir's acceptance received) Mar.06 Tahir Dr. 18,040 To Bills Receivable A/c 18,000 To Cash 40 (Tahir's acceptance dishonoured and Rs 40 paid as noting charges) Mar.06 Tahir Dr. 660 To Interest A/c 660 (Interest charged from Tahir on account of bill dishonoured) Mar.12 Cash A/c Dr. 18,700 To Tahir 18,700 (Tahir cleared his account by paying cash) Ledger Tahir’s Account Dr. Cr. Date Particulars J.F. Amount Rs Date Particulars J.F. Amount Rs 2017 2017 Jan.17 Sales 18,000 Jan.17 Bills Receivable 18,000 Mar.06 Bills Receivable 18,000 Mar.11 Cash 18,700 Mar.06 Cash 40 Mar.06 Interest 660 36,700 36,700 Books of Tahir Journal Date Particulars L.F. Debit Amount Rs Credit Amount Rs 2017 Jan.17 Purchases A/c Dr. 18,000 To Abdula 18,000 (Goods bought from Abdula) Jan.17 Abdula Dr. 18,000 To Bills Payable A/c 18,000 (Bill drawn by Abdula accepted, payable after 15 days) Mar.06 Bills Payable A/c Dr. 18,000 Noting Charges A/c Dr. 40 To Abdula 18,040 (Abula’s bill dishonoured) Mar.07 Interest A/c Dr. 660 To Abdula 660 (Interest charged on account of bill dishonoured) Mar.11 Abdula Dr. 18,700 To Cash A/c 18,700 (Cash paid to Abdula) Ledger Abdula’sAccount Dr. Cr. Date Particulars J.F. Amount Rs Date Particulars J.F. Amount Rs 2017 2017 Jan.17 Bills Payable 18,000 Jan.17 Purchases 18,000 Mar.11 Cash 18,700 Mar.06 Bills Payable 18,000 Mar.06 Noting Charges 40 Mar.06 Interest 660 36,700 36,700 AccountancyFinancial Accounting- I - NCERT Solutions (2019)All
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# The Three Formulas of the Circuit (Chap. 2.4)
## The Three Formulas of the Circuit (Chap. 2.4) Lyrics
Chapter 4: The Three Formulas of the Circuit
The three formulas may be set down in the following manner, using Tc for “total circulation process”:
I. M — C ... P ... C' — M'
II. P ... Tc ... P
III. Tc ... P (C')
If we combine all three forms, all premises of the process appear as its result, as a premise produced by it itself. Every element appears as a point of departure, of transit, and of return. The total process presents itself as the unity of the processes of production and circulation. The process of production becomes the mediator of the process of circulation and vice versa.
All three circuits have the following in common: The self-expansion of value as the determining purpose, as the compelling motive. In I this is expressed in its form. Formula II begins with P, the very process of creating surplus-value. In III the circuit begins with the self-expanded value, even if the movement is repeated on the same scale.
As C — M means M — C for the buyer, and M — C means C — M for the seller, the circulation of capital presents only the ordinary metamorphosis of commodities, and the laws evolved with regard to it (Buch I, Kap. III, 2) [English edition: Ch. III, 2. — Ed.] on the mass of money in circulation are valid here. However, if we do not cling to this formal aspect but rather consider the actual connection between the metamorphoses of the various individual capitals, in other words, if we study the connection between the circuits of individual capitals as partial movements of the process of reproduction of the total social capital, then the mere change of form of money and commodities cannot explain the connection.
In a constantly revolving circle every point is simultaneously a point of departure and a point of return. If we interrupt the rotation, not every point of departure is a point of return. Thus we have seen that not only does every individual circuit presuppose (implicite) the others, but also that the repetition of the circuit in one form comprises the performance of the circuit in the other forms. The entire difference thus appears to be a merely formal one, or as a merely subjective distinction existing solely for the observer.
Since every one of these circuits is considered a special form of this movement in which various individual industrial capitals are engaged, this difference exists only as an individual one. But in reality every individual industrial capital is present simultaneously in all three circuits. These three circuits, the forms of reproduction assumed by the three forms of capital, are made continuously side by side. For instance, one part of the capital-value, which now performs the function of commodity-capital, is transformed into money-capital, but at the same time another part leaves the process of production and enters the circulation as a new commodity-capital. The circuit form C' ... C' is thus continuously described; and so are the other two forms. The reproduction of capital in each one of its forms and stages is just as continuous as the metamorphosis of these forms and the successive passage through the three stages. The entire circuit is thus a unity of its three forms.
We assumed in our analysis that capital-value in its entire magnitude acts as money-capital, productive-capital or commodity-capital. For instance, we had those £422 first entirely as money-capital, then we transformed them wholly into productive capital, and finally into commodity-capital, into yarn of the value of £500 (containing £78 worth of surplus-value). Here the various stages are just so many interruptions. So long as, e.g., those £422 retain their money-form, that is to say, until the purchases M — C (L plus MP) are made, the entire capital exists and functions only as money-capital. As soon as it is transformed into productive capital, it performs neither the function of money-capital nor of commodity-capital. Its entire process of circulation is interrupted, as soon as it functions in one of its two circulation stages, either as M or as C'. Consequently, the circuit P ... P would represent not only a periodical renewal of the productive capital but also the interruption of its function, the process of production, up to the time when the process of circulation is completed. Instead of proceeding continuously, production would take place in jerks and would be renewed only in periods of accidental duration, according to whether the two stages of the process of circulation are got through with quickly or slowly. This would apply for instance to a Chinese artisan who works only for private customers and whose process of production ceases until he receives a new order.
This is indeed true of every single part of capital that is in motion, and all parts of capital go through this motion in succession. Suppose that the 10,000 lbs. of yarn are the weekly product of some spinner. These 10,000 lbs. of yarn leave the sphere of production entirely and enter the sphere of circulation; the capital-value contained in it must all be converted into money-capital, and so long as this value continues in the form of money-capital it cannot enter anew into the process of production. It must first go into circulation and be reconverted into the elements of productive capital, L plus MP. The circuit-describing process of capital means constant interruption, the leaving of one stage and the entering into the next, the discarding of one form and the assuming of another. Each one of these stages not only presupposes the next but also excludes it.
But continuity is the characteristic mark of capitalist production, necessitated by its technical basis, although not always absolutely attainable. Let us see then what happens in reality. While, e.g., the 10,000 lbs. of yarn appear in the market as commodity-capital and are transformed into money (regardless of whether it is a paying or purchasing medium or only money of account), new cotton, coal, etc., take the place of the yarn in the process of production, have therefore already been reconverted from the money-form and commodity-form into that of productive capital, and begin to function as such. At the same time that these 10,000 lbs. of yarn are being reconverted into money, the preceding 10,000 lbs. of yarn are going through the second stage of their circulation and are being reconverted from money into the elements of productive capital. All parts of capital successively describe circuits, are simultaneously at its different stages. The industrial capital, continuously progressing along its orbit, thus exists simultaneously at all its stages and in the diverse functional forms corresponding to these stages. That part of industrial capital which is converted for the first time from commodity-capital into money begins the circuit C' ... C', while industrial capital as a moving whole has already passed through that circuit. One hand advances money, the other receives it. The inauguration of the circuit M ... M' at one place coincides with the return of the money at another place. The same is true of productive capital.
The actual circuit of industrial capital in its continuity is therefore not alone the unity of the processes of circulation and production but also the unity of all its three circuits. But it can be such a unity only if all the different parts of capital can go through the successive stages of the circuit, can pass from one phase, from one functional form to another, so that the industrial capital, being the whole of all these parts, exists simultaneously in its various phases and functions and thus describes all three circuits at the same time. The succession ]das Nacheinander] of these parts is here governed by their co-existence [das Nebeneinander], that is to say, by the division of capital. In a ramified factory system the product is constantly in the various stages of its process of formation and constantly passes from one phase of production to another. As the individual industrial capital has a definite size which depends on the means of the capitalist and which has a definite minimum magnitude for every branch of industry, it follows that its division must proceed according to definite proportions. The magnitude of the available capital determines the dimensions of the process of production, and this again determines the dimensions of the commodity-capital and money-capital in so far as they perform their functions parallel with the process of production. However co-existence, by which continuity of production is determined, is only due to the movement of those parts of capital in which they successively pass through their different stages. Co-existence is itself merely the result of succession. If for instance C' — M' as far as one part is concerned, if the commodity cannot be sold, then the circuit of this part is interrupted and no replacement by its means of reproduction takes place; the succeeding parts, which emerge from the process of production in the shape of C', find the change of their functions blocked by their predecessors. If this lasts for some time, production is restricted and the entire process brought to a halt. Every stagnation in succession carries disorder into co-existence, every stagnation in one stage causes more or less stagnation in the entire circuit of not only the stagnant part of capital but also of the total individual capital.
The next form in which the process presents itself is that of a succession of phases, so that the transition of capital into a new phase is made necessary by its departure from another. Every separate circuit has therefore one of the functional forms of capital for its point of departure and point of return. On the other hand the aggregate process is in fact the unity of the three circuits, which are the different forms in which the continuity of the process expresses itself. The aggregate circuit presents itself to every functional form of capital as its specific circuit and every one of these circuits is a condition of the continuity of the total process. The cycle of each functional form is dependent upon the others. It is a necessary prerequisite of the aggregate process of production, especially for the social capital, that it is at the same time a process of reproduction and hence a circuit of each one of its elements. Various fractional parts of capital pass successively through the various stages and functional forms. Thanks to this every functional form passes simultaneously with the others through its own circuit, although always a different part of capital finds its expression in it. One part of capital, continually changing, continually reproduced, exists as a commodity-capital which is converted into money; another as money-capital which is converted into productive capital; and a third as productive capital which is transformed into commodity-capital. The continuous existence of all three forms is brought about by the circuit the aggregate capital describes in passing through precisely these three phases.
Capital as a whole, then, exists simultaneously, spatially side by side, in its different phases. But every part passes constantly and successively from one phase, from one functional form, into the next and thus functions in all of them in turn. Its forms are hence fluid and their simultaneousness is brought about by their succession. Every form follows another and precedes it, so that the return of one capital part to a certain form is necessitated by the return of the other part to some other form. Every part describes continuously its own cycle, but it is always another part of capital which exists in this form, and these special cycles form only simultaneous and successive elements of the aggregate process.
The continuity — instead of the above-described interruption — of the aggregate process is achieved only in the unity of the three circuits. The aggregate social capital always has this continuity and its process always exhibits the unity of the three circuits.
The continuity of the reproduction is at times more or less interrupted so far as individual capitals are concerned. In the first place the masses of value are frequently distributed at various periods in unequal portions over the various stages and functional forms. In the second place these portions may be differently distributed, according to the character of the commodity to be produced, hence according to the particular sphere of production in which the capital is invested. In the third place the continuity may be more or less broken in those branches of production which are dependent on the seasons, either on account of natural conditions (agriculture, herring catch, etc.) or on account of conventional circumstances, as for instance in so-called seasonal work. The process goes on most regularly and uniformly in the factories and mines. But this difference in the various branches of production does not cause any difference in the general forms of the circular process.
Capital as self-expanding value embraces not only class relations, a society of a definite character resting on the existence of labour in the form of wage-labour. It is a movement, a circuit-describing process going through various stages, which itself comprises three different forms of circuit-describing process. Therefore it can be understood only as a motion, not as a thing at rest. Those who regard the gaining by value of independent existence as a mere abstraction forget that the movement of industrial capital is this abstraction in actu. Value here passes through various forms, various movements in which it maintains itself and at the same time expands, augments. As we are here concerned primarily with the mere form of this movement, we shall not take into consideration the revolutions which capital-value may undergo during its circuit. But it is clear that in spite of all the revolutions of value, capitalist production exists and can endure only so long as capital-value is made to create surplus-value, that is, so long as it describes its circuit as a value that has gained independence, so long therefore as the revolutions in value are overcome and equilibrated in some way. The movements of capital appear as the action of some individual industrial capitalist who performs the functions of a buyer of commodities and labour, a seller of commodities, and an owner of productive capital, who therefore promotes the circuit by this activity. If social capital experiences a revolution in value, it may happen that the capital of the individual capitalist succumbs to it and fails, because it cannot adapt itself to the conditions of this movement of values. The more acute and frequent such revolutions in value become, the more does the automatic movement of the now independent value operate with the elemental force of a natural process, against the foresight and calculation of the individual capitalist, the more does the course of normal production become subservient to abnormal speculation, and the greater is the danger that threatens the existence of the individual capitals. These periodical revolutions in value therefore corroborate what they are supposed to refute, namely, that value as capital acquires independent existence, which it maintains and accentuates through its movement.
This succession of the metamorphoses of capital in process includes continuous comparison of the change in the magnitude of value of the capital brought about in the circuit with the original value. If value’s acquisition of independence of the value-creating power, labour-power, is inaugurated by the act M — L (purchase of labour-power) and is effected during the process of production as exploitation of labour-power, this acquisition of independence on the part of value does not re-appear in that circuit, in which money, commodities, and elements of production are merely alternating forms of capital-value in process, and the former magnitude of value is compared with capital’s present changed magnitude of value.
“Value,” argues Bailey against the acquisition of independence by value, an independence which is characteristic of the capitalist mode of production and which he treats as an illusion of certain economists; “value is a relation between contemporary commodities, because such only admit of being exchanged for each other.” [See Bailey, Samuel, A Critical Dissertation on the Nature, Measures, and Causes of Value; Chiefly in Reference to the Writings of Mr. Ricardo and His Followers By the Author of Essays on the Formation and Publication of Opinions, London, 1825, p. 72. — Ed.]
This he says against the comparison of commodity-values of different epochs, a comparison which amounts only to comparing the expenditure of labour required in various periods for the production of the same sort of commodities, once the value of money has been fixed for every period. This comes from his general misunderstanding, for he thinks that exchange-value is equal to value, that the form if value is value itself; consequently commodity-value can no longer be compared, if they do not function actively as exchange-values and thus cannot actually be exchanged for one another. He has not the least inkling of the fact that value functions as capital-value or capital only in so far as it remains identical with itself and is compared with itself in the different phases of its circuit, which are not at all “contemporary” but succeed one another.
In order to study the formula of the circuit in its purity it is not sufficient to postulate that commodities are sold at their value; it must also be assumed that this takes place with other things being equal. Take for instance the form P ... P, disregarding all technical revolutions within the process of production by which the productive capital of a certain capitalist might be depreciated; disregarding furthermore all reactions which a change in the elements of value of the productive capital might have on the value of the existing commodity-capital, which might appreciate or depreciate if a stock of it is on hand. Suppose the 10,000 lbs. of yarn, C', have been sold at their value of £500; 8,440 lbs. equal to £422, replace the capital-value contained in C'. But if the value of cotton, coal, etc., has increased (we do not consider mere fluctuations in price), these £422 may not suffice for the full replacement of the elements of productive capital; additional money-capital is required, money-capital is tied up. The opposite takes place when those prices fall. Money-capital is set free. The process takes a wholly normal course only when the value-relations remain constant; its course is practically normal so long as the disturbances during the repetitions of the circuit balance one another. But the greater these disturbances the greater the money-capital which the industrial capitalist must possess to tide over the period of readjustment; and as the scale of each individual process of production and with it the minimum size of the capital to be advanced increases in the process of capitalist production, we have here another circumstance to be added to those others which transform the function of the industrial capitalist more and more into a monopoly of big money-capitalists, who may operate singly or in association.
We remark incidentally that if a change in the value of the elements of production occurs a difference appears between the form M...M' on one side and of P ... P and C' ... C' on the other.
In M ... M', the formula of newly-invested capital, which first appears as money-capital, a fall in the value of the means of production, such as raw material, auxiliary material, etc., will permit of a smaller expenditure of money-capital than before this fall for the purpose of starting a business of a definite size, because the scale of the process of production (productive power development remaining the same) depends on the mass and volume of the means of production which a given quantity of labour-power can cope with; but it does not depend on the value of these means of production nor on that of the labour-power (the latter value affects only the magnitude of self-expansion). Take the reverse case. If there is a rise in the value of the elements of production of the commodities which constitute the elements of the productive capital, then more money-capital is needed for the establishment of a business of definite proportions. In both cases it is only the amount of the money-capital required for new investment that is affected. In the former case money-capital becomes surplus, in the latter it is tied up, provided the accession of new individual industrial capital proceeds in the usual way in a given branch of production.
The circuits P ... P and C' ... C' present themselves as M ... M' only to the extent that the movement of P and C' is at the same time accumulation, hence to the extent that additional m, money, is converted into money-capital; here, too, we do not take into consideration the reaction of such changes in value on those constituent parts of capital which are engaged in the process of production. It is not the original expenditure which is directly affected here, but an industrial capital engaged in its process of reproduction and not in its first circuit; i.e., C' ... Clmp, the reconversion of commodity-capital into its elements of production, so far as they are composed of commodities. When value (prices) fall three cases are possible: The process of reproduction is continued on the same scale; in that event a part of the money-capital existing hitherto is set free and money-capital is accumulated, although no real accumulation (production on an extended scale) or transformation of m (surplus-value) into an accumulation-fund initiating and accompanying such accumulation has previously taken place. Or the process of reproduction is carried on a more extensive scale than ordinarily would have been the case, provided the technical proportions admit it. Or, finally, a larger stock of raw materials, etc., is laid in.
The opposite occurs if the value of the elements of replacement of a commodity-capital increases. In that case reproduction no longer takes place on its normal scale (e.g., the working-day gets shorter); or additional money-capital must be employed in order to maintain the old volume of work (money-capital is tied up); or the money-fund for accumulation, when one exists, is employed entirely or partially for the operation of the process of reproduction on its old scale instead of for the enlargement of this process. This is also tying up money-capital, except that here the additional money-capital does not come from the outside, from the money-market, but from the means of the industrial capitalist himself.
However, there may be modifying circumstances in P ... P and C' ... C'. If our spinning-mill proprietor for example has a large stock of cotton (a large proportion of his productive capital in the form of a stock of cotton), a part of his productive capital is depreciated by a fall in the prices of cotton; but if on the contrary these prices rise, this part of his productive capital appreciates. On the other hand, if he has tied up huge quantities in the form of commodity-capital, for instance of cotton yarn, a part of his commodity-capital, hence of his circuit describing capital in general, is depreciated by a fall of cotton, or appreciated by a rise in its prices. Finally take the process C' — M — Clmp. If C' — M, the realisation of the commodity-capital, has taken place before a change in the value of the elements of C, then capital is affected only in the way indicated in the first case, namely in the second act of circulation, M — Clmp; but if such a change has occurred before C' — M has been effected, then, other conditions remaining equal, a fall in the price of cotton causes a corresponding fall in the price of yarn, and a rise in the price of cotton means conversely a rise in the price of yarn. The effect on the various individual capitals invested in the same branch of production may differ widely, according to the circumstances in which they find themselves.
Money-capital may also be set free or tied up on account of differences in the duration of the process of circulation, hence also in the speed of circulation. But this belongs in the discussion on turnover. At this point we are only interested in the real difference that becomes evident, with regard to changes of values of the elements of productive capital, between M ... M' and the other two circuit forms.
In the circulation section M — Clmp, in the epoch of the already developed and hence prevailing capitalist mode of production, a large portion of the commodities composing MP, the means of production, is itself functioning as the commodity-capital of someone else. From the standpoint of the seller, therefore, C' — M', the transformation of commodity-capital into money-capital, takes place. But this is not an absolute rule. On the contrary. Within its process of circulation, in which industrial capital functions either as money or as commodities, the circuit of industrial capital, whether as money-capital or as commodity-capital, crosses the commodity circulation of the most diverse modes of social production, so far as they produce commodities. No matter whether commodities are the output of production based on slavery, of peasants (Chinese, Indian ryots). of communes (Dutch East Indies), of state enterprise (such as existed in former epochs of Russian history on the basis of serfdom) or of half-savage hunting tribes, etc. — as commodities and money they come face to face with the money and commodities in which the industrial capital presents itself and enter as much into its circuit as into that of the surplus-value borne in the commodity-capital, provided the surplus-value is spent as revenue; hence they enter in both branches of circulation of commodity-capital. The character of the process of production from which they originate is immaterial. They function as commodities in the market, and as commodities they enter into the circuit of industrial capital as well as into the circulation of the surplus-value incorporated in it. It is therefore the universal character of the origin of the commodities, the existence of the market as world-market, which distinguishes the process of circulation of industrial capital. What is true of the commodities of others is also true of the money of others. Just as commodity-capital faces money only as commodities, so this money functions vis-à-vis commodity-capital only as money. Money here performs the functions of world-money.
However two points must be noted here.
First: as soon as act M — MP is completed, the commodities (MP) cease to be such and become one of the modes of existence of industrial capital in its functional form of P, productive capital. Thereby however their origin is obliterated. They exist henceforth only as forms of existence of industrial capital, are embodied in it. However it still remains true that to replace them they must be reproduced, and to this extent the capitalist mode of production is conditional on modes of production lying outside of its own stage of development. But it is the tendency of the capitalist mode of production to transform all production as much as possible into commodity production. The mainspring by which this is accomplished is precisely the involvement of all production into the capitalist circulation process. And developed commodity production itself is capitalist commodity production. The intervention of industrial capital promotes this transformation everywhere, but with it also the transformation of all direct producers into wage-labourers.
Secondly: the commodities entering into the circulation of industrial capital (including the requisite means of subsistence into which variable capital, after being paid to the labourers, is transformed for the purpose of reproducing their labour-power), regardless of their origin and of the social form of the productive process by which they were brought into existence, come face to face with industrial capital itself already in the form of commodity-capital, in the form of commodity-dealer’s or merchant’s capital. And merchant’s capital, by its very nature comprises commodities of all modes of production.
The capitalist mode of production presupposes not only large-scale production but also, and necessarily so, sales on a large scale, hence sale to the merchant, not to the individual consumer. If this consumer is himself a productive consumer, hence an industrial capitalist, i.e., if the industrial capital of one branch of production supplies some other branch of industry with means of production, direct sale by one industrial capitalist to many others take place (in the form of orders, etc.). To this extent every industrial capitalist is a direct seller and his own merchant, which by the way is when he sells to a merchant.
Trading in commodities as the function of merchant’s capital is a premise of capitalist production and develops more and more in the course of development of such production. Therefore we occasionally take its existence for granted to illustrate particular aspects of the process of capitalist circulation; but in the general analysis of this process we assume direct sale, without the intervention of a merchant, because this intervention obscures various facets of the movement.
Cf. Sismondi, who presents the matter somewhat naively:
“Commerce employs considerable capital, which at first sight does not seem to be a part of that capital whose movement we have described. The value of the cloth accumulated in the stores of the cloth-merchant seems at first to be entirely foreign to that part of the annual production which the rich gives to the poor as wages in order to make him work. However this capital has simply replaced the other of which we have spoken. For the purpose of clearly understanding the progress of wealth, we have begun with its creation and followed it to its consumption. Then the capital employed in cloth manufacturing, for instance, always seemed the same to us; it was exchanged for the revenue of the consumer, it was divided into only two parts, one of them serving as revenue of the manufacturer in the form of the profit, the other serving as revenue of the labourers in the form of wages for the time they were manufacturing new cloth.
“But it was soon found that it would be to the advantage of all if the different parts of this capital were to replace one another and that, if 100,000 ècus were sufficient for the entire circulation between the manufacturer and the consumer, they should be divided equally between the manufacturer, the wholesale merchant, and the retail merchant. The first then did with only one-third of this capital the same work as he had done with the entire capital, because as soon as his work of manufacturing was completed he found out that a merchant would rather buy from him than a consumer would. On the other hand the capital of the wholesaler was much sooner replaced by that of the retailer... The difference between the sums advanced for wages and the purchase price paid by the ultimate consumer was considered the profit of those capitals. It was divided between the manufacturer, the merchant and the retailer, from the moment that they had divided their functions among themselves, and the work performed was the same, although it had required three persons and three parts of capital instead of one.” (Nouveaux Principes, I, pages 139-140.)
“All of them [the merchants] contributed indirectly to the production; for having consumption for its object, production cannot be regarded as completed until the thing produced is placed within the reach of the consumer.” (Ibid., p. 137)
In the discussion of the general forms of the circuit and in the entire second book in general, we take money to mean metallic money, with the exception of symbolic money, mere tokens of value, which are designed for specific use in certain states, and of credit-money, which is not yet developed. In the first place, this is the historical order; credit-production plays only a very minor role, or none at all, during the first epoch of capitalist production. In the second place, the necessity of this order is demonstrated theoretically by the fact that everything of a critical nature which Tooke and others hitherto expounded in regard to the circulation of credit-money compelled them to hark back again and again to the question of what would be the aspect of the matter if nothing but metal-money were in circulation. But it must not be forgotten that metal-money may serve as a purchasing medium and also as a paying medium. For the sake of simplicity, we consider it in this second book generally only in its first functional form.
The process of circulation of industrial capital, which is only a part of its individual circuit, is determined by the general laws previously set forth (Buch I, Kap. III), [English edition: Ch. III. — Ed.] in so far as it is only a series of acts within the general circulation of commodities. The greater the velocity of the currency of money, the more rapidly therefore every individual capital passes through the series of its commodity or money metamorphoses, the more numerous are the industrial capitals (or individual capitals in the form of commodity-capitals) started circulating successively by a given mass of money, for example £500. The more the money functions as a paying medium, the more therefore — for instance in the replacement of some commodity-capital by its means of production — nothing but balances have to be squared, and the shorter the periods of time when payments fall due, as for instance in paying wages, the less money a given mass of capital-value therefore requires for its circulation. On the other hand, assuming that the velocity of the circulation and all other conditions remain the same, the amount of money required to circulate as money-capital is determined by the sum of the prices of the commodities (price multiplied by the volume of commodities), or, if the quantity and value of the commodities are fixed, by the value of the money itself.
But the laws of the general circulation of commodities are valid only when capital’s circulation process consists of a series of simple acts of circulation; they do not apply when the latter constitute functionally determined sections of the circuit of individual industrial capitals.
In order to make this plain, it is best to study the process of circulation in its uninterrupted interconnection, such as it appears in the following two forms:
As series of acts of circulation in general, the process of circulation (whether in the form of C — M — C or of M — C — M) represents the two antithetical series of commodity metamorphoses, every single one of which in its turn implies an opposite metamorphosis on the part of the alien commodity or alien money confronting the commodity.
C — M on the part of the owner of a commodity means M — C on the part of its buyer; the first metamorphosis of the commodity appearing in the form of M; the opposite applies to M — C. What has been shown concerning the intertwining of the metamorphosis of a certain commodity in one stage with that of another in another stage applies to the circulation of capital so far as the capitalist functions as a buyer and seller of commodities, and his capital on that account functions in the form of money opposed to the commodities of another. But this intertwining is not to be identified with the intertwining of the metamorphoses of capitals.
In the first place M — C (MP), as we have seen, may represent an intermingling of the metamorphoses of different individual capitals. For instance the commodity-capital of the spinning-mill owner, yarn is partly replaced by coal. One part of his capital exists in the form of money and is converted into the form of commodities, while the capital of the capitalist producer of coal is in the form of commodities and is therefore converted into the form of money; the same act of circulation represents in this case opposite metamorphoses of two industrial capitals (in different branches of production), hence an intertwining of the series of metamorphoses of these capitals. But as we have seen the MP into which M is transformed need not be commodity-capital in the categorical sense, i.e., need not be a functional form of industrial capital, need not be produced by a capitalist. It is always M — C on one side and C — M on the other, but not always an intermingling of metamorphoses of capitals. Furthermore M — L, the purchase of labour-power, is never an intermingling of metamorphoses of capitals, for labour-power, though the commodity of the labourer, does not become capital until it is sold to the capitalist. On the other hand in the process C' — M', it is not necessary that M' should represent converted commodity-capital; it may be the realisation in money of the commodity labour-power (wages), or of the product of some independent labourer, slave, serf, or community.
In the second place however it is not at all required for the discharge of the functionally determined role played by every metamorphosis occurring within the process of circulation of some individual capital that this metamorphosis should represent the corresponding opposite metamorphosis in the circuit of the other capital, provided we assume that the entire production of the world-market is carried on capitalistically. For instance in the circuit P ... P, the M which converts C' into money may be to the buyer only the realisation in money of his surplus-value (if the commodity is an article of consumption); or in M' — C'lmp (where therefore already accumulated capital enters) M' may, as far as the vendor of MP is concerned, enter into the circulation of his capital only to replace his advanced capital or it may not re-enter at all by being diverted into revenue expenditure.
Therefore the manner in which the various component parts of the aggregate social capital, of which the individual capitals are but constituents functioning independently, mutually replace one another in the process of circulation — in regard to capital as well as surplus-value — is not ascertained from the simple intertwinings of the metamorphoses in the circulation of commodities — intertwinings which the acts of capital circulation have in common with all other circulation of commodities. That requires a different method of investigation. Hitherto one has been satisfied with uttering phrases which upon closer analysis are found to contain nothing but indefinite ideas borrowed from the intertwining of metamorphoses common to all commodity circulation.
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Natural Money and Credit Economy
One of the most obvious peculiarities of the movement in circuits of industrial capital, and therefore also of capitalist production, is the fact that on one hand the component elements of productive capital are derived from the commodity-market and must be continually renewed out of it, bought as commodities; and that on the other hand the product of the labour-process emerges from it as a commodity and must be continually sold anew as a commodity. Compare for instance a modern farmer of the Scotch lowlands with an old-fashioned small peasant on the Continent. The former sells his entire product and has therefore to replace all its elements, even his seed, in the market; the latter consumes the greater part of his product directly, buys and sells as little as possible, fashions tools, makes clothing, etc., so far as possible himself.
Natural economy, money-economy, and credit-economy have therefore been placed in opposition to one another as being the three characteristic economic forms of movement in social production.
In the first place these three forms do not represent equivalent phases of development. The so-called credit-economy is merely a form of the money-economy, since both terms express functions or modes of exchange among the producers themselves. In developed capitalist production, the money-economy appears only as the basis of the credit-economy. The money-economy and credit-economy thus correspond only to different stages in the development of capitalist production, but they are by no means independent forms of exchange vis-à-vis natural economy. With the same justification one might contrapose as equivalents the very different forms of natural economy to those two economies.
In the second place, since it is not the economy, i.e., the process of production itself that is emphasised as the distinguishing mark of the two categories, money-economy and credit-economy, but rather the mode of exchange — corresponding to that economy — between the various agents of production, or producers, the same should apply to the first category. Hence exchange economy instead of natural economy. A completely isolated natural economy, such as the Inca state of Peru, would not come under any of these categories.
In the third place the money-economy is common to all commodity production and the product appears as a commodity in the most varied organisms of social production. Consequently what characterises capitalist production would then be only the extent to which the product is created as an article of commerce, as a commodity, and hence the extent also to which its own constituent elements must enter again as articles of commerce, as commodities, into the economy from which it emerges.
As a matter of fact capitalist production is commodity production as the general form of production. But it is so and becomes so more and more in the course of its development only because labour itself appears here as a commodity, because the labourer sells his labour, that is, the function of his labour-power, and our assumption is that he sells it at its value, determined by its cost of reproduction. To the extent that labour becomes wage-labour, the producer becomes an industrial capitalist. For this reason capitalist production (and hence also commodity production) does not reach its full scope until the direct agricultural producer becomes a wage-labourer. In the relation of capitalist and wage-labourer, the money-relation, the relation between the buyer and the seller, becomes a relation inherent in production. But this relation has its foundation in the social character of production, not in the mode of exchange. The latter conversely emanates from the former. It is, however, quite in keeping with the bourgeois horizon, everyone being engrossed in the transaction of shady business, not to see in the character of the mode of production the basis of the mode of exchange corresponding to it, but vice versa.[7]
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The Meeting of Demand and Supply
The capitalist throws less value in the form of money into the circulation than he draws out of it, because he throws into it more value in the form of commodities than he withdrew from it in the form of commodities. Since he functions simply as a personification of capital, as an industrial capitalist, his supply of commodity-value is always greater than his demand for it. If his supply and demand in this respect covered each other it would mean that his capital had not produced any surplus-value, that it had not functioned as productive capital, that the productive capital had been converted into commodity-capital not big with surplus-value; that it had not drawn any surplus-value in commodity form out of labour-power during the process of production, had not functioned at all as capital. The capitalist must indeed “sell dearer than he has bought, ” but he succeeds in doing so only because the capitalist process of production enables him to transform the cheaper commodity he bought — cheaper because it contains less value — into a commodity of greater value, hence a dearer one. He sells dearer, not because he sells above the value of his commodity, but because his commodity contains value in excess of that contained in the ingredients of its production.
The rate at which the capitalist makes the value of his capital expand is the greater, the greater the difference between his supply and his demand, i.e., the greater the excess of the commodity-value he supplies over the commodity-value he demands. His aim is not to equalize his supply and demand, but to make the inequality between them, the excess of his supply over his demand, as great as possible.
What is true of the individual capitalist applies to the capitalist class.
In so far as the capitalist merely personifies industrial capital, his own demand is confined to means of production and labour-power. In point of value, his demand for MP is smaller than his advanced capital; he buys means of production of a smaller value than that of his capital, and therefore of a still smaller value than that of the commodity-capital which he supplies.
As regards his demand for labour-power, it is determined in point of value by the relation of his variable capital to his total capital, hence equals v : C. In capitalist production this demand thereby grows smaller than his demand for means of production. His purchases of MP steadily rise above his purchases of L.
Since the labourer generally converts his wages into means of subsistence, and for the overwhelmingly larger part into absolute necessities, the demand of the capitalist for labour-power is indirectly also a demand for the articles of consumption essential to the working-class. But this demand is equal to v and not one iota greater (if the labourer saves a part of his wages — we necessarily discard here all credit relations — he converts part of his wages into a hoard and to that extent does not act as a bidder, a purchaser). The upper limit of a capitalist’s demand is C, equal to c + v, but his supply is equal to c + v + s. Consequently if the composition of his commodity-capital is 80c + 20v + 20s, his demand is equal to 80c + 20v, hence, considered from the angle of the value it contains, one-fifth smaller than his supply. The greater the percentage of the mass of surplus-value produced by him (his rate of profit) the smaller becomes his demand in relation to his supply. Although with the further development of production the demand of the capitalist for labour-power, and thus indirectly for necessary means of subsistence, steadily decreases compared with his demand for means of production, it must not be forgotten on the other hand that his demand for MP is always smaller than his capital. His demand for means of production must therefore always be smaller in value than the commodity-product of the capitalist who, working with a capital of equal value and under equal conditions, furnishes him with those means of production. That many capitalists and not only one do the furnishing does not alter the case. Take it that his capital is £1,000, and its constant part £800; then his demand on all these capitalists is equal to £800. Together they supply means of production worth £1,200 for each £1,000 (regardless of what share in each £1,000 may fall to each one of them and of the fraction of his total capital which the share of each may represent), assuming that the rate of profit is the same. Consequently his demand covers only two-thirds of their supply, while his own total demand amounts to only four-fifths of his own supply, measured in value.
It still remains for us, incidentally, to investigate the problem of turnover. Let the total capital of the capitalist be £5,000, of which £4,000 is fixed and £1,000 circulating capital; let this 1,000 be composed of 800c plus 200v, as assumed above. His circulating capital must be turned over five times a year for his total capital to turn over once. His commodity-product is then equal to £6,000, i.e., £1,000 more than his advanced capital, which results in the same ratio of surplus-value as above:
5,000 C : 1,000(c + v) : 20s
This turnover therefore does not change anything in the ratio of his total demand to his total supply. The former remains one-fifth smaller than the latter.
Suppose his fixed capital has to be renewed in 10 years. So the capitalist pays every year one-tenth, or £400, into a sinking fund and thus has only a value of £3,600 of fixed capital left plus £400 in money. If the repairs are necessary and do not exceed the average, they represent nothing but capital invested later. We may look at the matter the same as if he had allowed for the cost of repairs beforehand, when calculating the value of his investment capital, so far as this enters into the annual commodity-product, so that it is included in the one-tenth sinking fund payment. (If his need for repairs is below average he is so much money to the good, and the reverse if above. But this evens out for the entire class of capitalists engaged in the same branch of industry.) At any rate, although his annual demand still remains £5,000, equal to the original capital-value he advanced (assuming his total capital is turned over once a year), this demand increases with regard to the circulating part of the capital, while it steadily decreases with regard to its fixed part.
We now come to reproduction. Let us assume that the capitalist consumes the entire surplus-value m and reconverts only capital C of the original magnitude into productive capital. Then the demand of the capitalist is equal in value to his supply; but this does not refer to the movement of his capital. As a capitalist he exercises a demand for only four-fifths of his supply (in terms of value). He consumes one-fifth as a non-capitalist, not in his function as capitalist but for his private requirements or pleasures.
His calculation, expressed in percentages, is then as follows:
Demand as capitalist . . . . . . . . . . . 100, supply 120
Demand as man about town . . . . . . . 20, supply —
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Total demand . . . . . . . . . . . . . . . . 120, supply 120
This assumption is tantamount to assuming that capitalist production does not exist, and therefore that the industrial capitalist himself does not exist. For capitalism is abolished root and branch by the bare assumption that it is personal consumption and not enrichment that works as the compelling motive.
But such an assumption is impossible also technically. The capitalist must not only form a reserve capital to cushion price fluctuations and enable him to wait for favorable buying and selling conditions. He must accumulate capital in order to extend his production and build technical progress into his productive organism.
In order to accumulate capital he must first withdraw in money-form from circulation a part of the surplus-value which he obtained from that circulation, and must hoard it until it has increased sufficiently for the extension of his old business or the opening of a side-line. So long as the formation of the hoard continues, it does not increase the demand of the capitalist. The money is immobilised. It does not withdraw from the commodity-market any equivalent in commodities for the money equivalent withdrawn from it for commodities supplied.
Credit is not considered here. And credit includes for example deposits by the capitalist of accumulating money in a bank on current account paying interest.
Notes
7. End of Manuscript V. What follows to the end of the chapter, is a note contained in a notebook of 1877 or 1878 amid extracts from various books. — F.E.
More on Genius
4. The Three Formulas of the Circuit (Chap. 2.4) | 10,194 | 52,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-50 | latest | en | 0.919853 |
http://www.recordingblogs.com/sa/Articles/Post/3892/Impulse-based-reverb-through-deconvolution-Part-2-using-the-impulse-response-to-produce-reverberations | 1,519,545,722,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816178.71/warc/CC-MAIN-20180225070925-20180225090925-00322.warc.gz | 517,556,627 | 8,564 | Impulse based reverb through deconvolution – Part 2 – using the impulse response to produce reverberations
By mic on 8/31/2013
This is the second post in a series of articles about deconvolving a reverberated sound to get the impulse response of natural reverberations and then using this impulse response as the artificial reverb. In the previous post we created the impulse response of an example reverb. In this post we will take a short sound – a drum hit – and we will reverberate it with the impulse response developed before. Later, we will take the reverberated drum and we will pretend that we do not know the impulse response. We will try to recreate the impulse response and see how well we do. As a reminder, it should be possible to record the natural reverb of a short sound, such as a drum hit, in a room or a hall and to deconvolve this natural reverb to create an artificial reverb.
A short drum sound may look something like this.
This is, of course, a fake drum hit. A real drum hit may be longer than 50 ms. In the real world, a longer wave recorded at the 44.1 KHz standard sampling rate will also have too many samples, making the drawing of nice intuitive graphs difficult. This, this is simply a sine wave that decays, sampled at 2000 Hz. We will, however, pretend that this is the recording of a drum hit.
We can take this drum hit and run it through the same operations as last time – a two-tap delay, two Shroeder all pass filters, and a feedforward comb filter. Alternatively, we can simply apply the reverb impulse as a filter and convolve the two signals.
I am used to working with finite impulse response filters that are symmetric around the middle and I often forget that the impulse response has to be flipped as in the picture above, to be applied properly. The initial impulse notches (with the least delay from the original signal), have to hit the signal first. With symmetric filters this is not important – they are symmetric – but here this is needed.
The end result will be as in the following figure (the original signal included; note the change in the scale of the vertical access). This is probably not as expected. The peaks around 100 ms are perhaps too large. This is correct however. With this specific signal and with the specific parameters of the reverb, we get some overlapping of the drum hit repetitions around 100 ms that boost the signal peaks.
We can break this result down by following the steps of the previous post. First, here is the result of the early reflections – a tapped delay line with two taps, one with a delay of 43 ms and decay of 0.7 and the second one with delay of 67 ms and decay of 0.5 (at the 2000 Hz sampling rate, the delays are 86 samples and 134 samples). The original signal is omitted from the graph below.
The following is the result of running the original signal through a Shroeder all pass filter with a delay of 3.7 ms (7 samples) and a decay of 0.7. A delay of 83 ms (166 samples) was introduced before the all pass filter.
Taking the output of the first all pass filter and putting it through the second all pass filter produces the graph below. The second all pass filter has a delay of 1.23 ms (3 samples) and a decay of 0.65. It does not look that there is much change in the signal, but this is to be expected with this signal and the small amount of delay.
Finally, we take the output of the second all pass filter and put it through a comb. The comb filter has a delay of 11 ms (23 samples) and a decay of 0.7. Again not much difference from the previous picture, but we will obviously not see as much as we can see when working with impulses. It is, however, this comb that introduces the peak in the final signal below.
The final signal itself is the sum of the original signal, the two taps of the tapped delay line, and the output of this comb. Our next task is to take the final signal and, pretending that we do not know the exact specifications of the filters, create the corresponding impulse response.
authors: mic
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Copyright 2006 by Kaliopa Publishing, LLC | 935 | 4,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-09 | latest | en | 0.947259 |
https://web2.0calc.com/questions/help-1_18 | 1,524,491,100,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946011.29/warc/CC-MAIN-20180423125457-20180423145457-00520.warc.gz | 710,282,596 | 6,004 | +0
# Help #1
0
24
1
+2206
Help #1
NotSoSmart Apr 16, 2018
Sort:
#1
+85726
+1
Taking two marbles from a box without replacement
P(Q) = .41 P(R) = .44
P(Q ∩ R ) = .41 * .44 = 0.1804
Mutually exclusive means that it impossible for both things to happen at the same time
Obviously...it is impossible to roll an even number and an odd number at the same time
CPhill Apr 16, 2018
### 28 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 203 | 659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-17 | longest | en | 0.850657 |
https://mathexamination.com/class/yin-and-yang.php | 1,618,952,192,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039490226.78/warc/CC-MAIN-20210420183658-20210420213658-00138.warc.gz | 503,029,480 | 6,974 | ## Do My Yin And Yang Class
A "Yin And Yang Class" QE" is a standard mathematical term for a generalized consistent expression which is utilized to fix differential formulas and has options which are routine. In differential Class solving, a Yin And Yang function, or "quad" is utilized.
The Yin And Yang Class in Class kind can be expressed as: Q( x) = -kx2, where Q( x) are the Yin And Yang Class and it is a crucial term. The q part of the Class is the Yin And Yang continuous, whereas the x part is the Yin And Yang function.
There are 4 Yin And Yang functions with correct option: K4, K7, K3, and L4. We will now take a look at these Yin And Yang functions and how they are fixed.
K4 - The K part of a Yin And Yang Class is the Yin And Yang function. This Yin And Yang function can likewise be written in partial portions such as: (x2 - y2)/( x+ y). To fix for K4 we increase it by the correct Yin And Yang function: k( x) = x2, y2, or x-y.
K7 - The K7 Yin And Yang Class has an option of the type: x4y2 - y4x3 = 0. The Yin And Yang function is then increased by x to get: x2 + y2 = 0. We then have to increase the Yin And Yang function with k to get: k( x) = x2 and y2.
K3 - The Yin And Yang function Class is K3 + K2 = 0. We then multiply by k for K3.
K3( t) - The Yin And Yang function equationis K3( t) + K2( t). We multiply by k for K3( t). Now we multiply by the Yin And Yang function which gives: K2( t) = K( t) times k.
The Yin And Yang function is likewise known as "K4" because of the initials of the letters K and 4. K means Yin And Yang, and the word "quad" is noticable as "kah-rab".
The Yin And Yang Class is among the primary methods of fixing differential equations. In the Yin And Yang function Class, the Yin And Yang function is first increased by the appropriate Yin And Yang function, which will provide the Yin And Yang function.
The Yin And Yang function is then divided by the Yin And Yang function which will divide the Yin And Yang function into a genuine part and a fictional part. This offers the Yin And Yang term.
Finally, the Yin And Yang term will be divided by the numerator and the denominator to get the quotient. We are entrusted to the right-hand man side and the term "q".
The Yin And Yang Class is an essential idea to comprehend when fixing a differential Class. The Yin And Yang function is simply one technique to resolve a Yin And Yang Class. The techniques for resolving Yin And Yang equations consist of: particular worth decay, factorization, optimum algorithm, numerical option or the Yin And Yang function approximation.
## Pay Me To Do Your Yin And Yang Class
If you want to end up being acquainted with the Quartic Class, then you require to first start by checking out the online Quartic page. This page will reveal you how to utilize the Class by utilizing your keyboard. The explanation will likewise show you how to create your own algebra formulas to help you study for your classes.
Before you can comprehend how to study for a Yin And Yang Class, you should first comprehend using your keyboard. You will discover how to click on the function keys on your keyboard, along with how to type the letters. There are three rows of function keys on your keyboard. Each row has 4 functions: Alt, F1, F2, and F3.
By pushing Alt and F2, you can multiply and divide the worth by another number, such as the number 6. By pushing Alt and F3, you can utilize the 3rd power.
When you press Alt and F3, you will type in the number you are trying to multiply and divide. To increase a number by itself, you will push Alt and X, where X is the number you wish to increase. When you push Alt and F3, you will enter the number you are attempting to divide.
This works the same with the number 6, except you will only type in the two digits that are six apart. Finally, when you push Alt and F3, you will utilize the 4th power. However, when you push Alt and F4, you will use the real power that you have found to be the most appropriate for your problem.
By utilizing the Alt and F function keys, you can increase, divide, and then use the formula for the third power. If you need to increase an odd number of x's, then you will need to go into an even number.
This is not the case if you are attempting to do something complex, such as multiplying two even numbers. For instance, if you want to multiply an odd number of x's, then you will need to enter odd numbers. This is especially true if you are trying to figure out the response of a Yin And Yang Class.
If you want to convert an odd number into an even number, then you will require to press Alt and F4. If you do not know how to multiply by numbers by themselves, then you will need to use the letters x, a b, c, and d.
While you can multiply and divide by utilize of the numbers, they are much easier to use when you can look at the power tables for the numbers. You will need to do some research study when you first start to utilize the numbers, however after a while, it will be second nature. After you have developed your own algebra formulas, you will be able to develop your own reproduction tables.
The Yin And Yang Formula is not the only method to resolve Yin And Yang formulas. It is essential to discover trigonometry, which utilizes the Pythagorean theorem, and after that use Yin And Yang solutions to resolve issues. With this approach, you can learn about angles and how to fix problems without having to take another algebra class.
It is essential to attempt and type as quickly as possible, because typing will assist you learn about the speed you are typing. This will assist you write your answers faster.
## Hire Someone To Take My Yin And Yang Class
A Yin And Yang Class is a generalization of a linear Class. For instance, when you plug in x=a+b for a given Class, you get the worth of x. When you plug in x=a for the Class y=c, you obtain the worths of x and y, which offer you a result of c. By using this fundamental idea to all the equations that we have attempted, we can now solve Yin And Yang equations for all the worths of x, and we can do it rapidly and effectively.
There are numerous online resources readily available that offer free or affordable Yin And Yang equations to solve for all the worths of x, including the expense of time for you to be able to make the most of their Yin And Yang Class assignment help service. These resources typically do not need a membership fee or any kind of investment.
The answers supplied are the outcome of complex-variable Yin And Yang formulas that have actually been fixed. This is also the case when the variable used is an unidentified number.
The Yin And Yang Class is a term that is an extension of a linear Class. One benefit of using Yin And Yang equations is that they are more general than the direct formulas. They are easier to resolve for all the values of x.
When the variable used in the Yin And Yang Class is of the kind x=a+b, it is simpler to fix the Yin And Yang Class because there are no unknowns. As a result, there are fewer points on the line defined by x and a constant variable.
For a right-angle triangle whose base points to the right and whose hypotenuse points to the left, the right-angle tangent and curve graph will form a Yin And Yang Class. This Class has one unknown that can be found with the Yin And Yang formula. For a Yin And Yang Class, the point on the line specified by the x variable and a continuous term are called the axis.
The presence of such an axis is called the vertex. Given that the axis, vertex, and tangent, in a Yin And Yang Class, are a provided, we can discover all the values of x and they will sum to the given values. This is attained when we use the Yin And Yang formula.
The element of being a consistent factor is called the system of equations in Yin And Yang equations. This is in some cases called the main Class.
Yin And Yang equations can be solved for other values of x. One way to resolve Yin And Yang equations for other worths of x is to divide the x variable into its aspect part.
If the variable is offered as a positive number, it can be divided into its factor parts to get the typical part of the variable. This variable has a magnitude that amounts to the part of the x variable that is a constant. In such a case, the formula is a third-order Yin And Yang Class.
If the variable x is unfavorable, it can be divided into the same part of the x variable to get the part of the x variable that is increased by the denominator. In such a case, the formula is a second-order Yin And Yang Class.
Solution help service in solving Yin And Yang formulas. When utilizing an online service for fixing Yin And Yang formulas, the Class will be fixed quickly. | 1,984 | 8,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-17 | latest | en | 0.893027 |
http://comments.gmane.org/gmane.politics.election-methods/22312 | 1,369,428,028,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705043997/warc/CC-MAIN-20130516115043-00008-ip-10-60-113-184.ec2.internal.warc.gz | 58,682,132 | 3,051 | 27 Jun 2012 18:45
## Typo. Convenient terms.
In my most recent posting, I said:
"I didn't ask to minimize the max s/p. I asked to minimize, over all of the pairs of states, the amount by which one state's s/p differs from that of the other."
The meaning is probably clear, but I should add "...maximum...", before the word "amount".
So it should say:
"I didn't ask to minimize the maximum s/p. I asked to minimize, over all of the pairs of states, the maximum amount by which the s/p of one of those two states differs from
that of the other."
Also, p, a state's population, is really an unwieldly and impractically large number for these purposes.
I've therefore often been speaking of Hare quotas as a measure of population.
For the purposes of discussing Sainte-Lague/Webster, or other divisor methods such as d'Hondt, I'd rather use "q" to stand for the quotient resulting from division of the state's population by the final divisor, the divisor that results in the desired number of seats.
When not discussing divisor methods, the Hare quota might be the most convenient population measure to speak of. If I'm going to use "q" to refer to a state's number of Hare quotas, I'll say so in advance. Otherwise, in my usage, "q" means the result of dividing the state's population by the final divisor, as described above. That's the number that I've previously called the state's "cps".
So, I'll speak of "s/q" instead of s/p. And, instead of the unwieldly "s/q_i", I'll just use "M" standing for "middle", to represent the ideal s/q value for all of the states.
I don't know if there will be any more discussion about this, but, if there is, I wanted to say what terms I'd use. Anyway,even if there isn't, I like clarifying these things.
Mike Ossipoff
```----
Election-Methods mailing list - see http://electorama.com/em for list info
```
Gmane | 457 | 1,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2013-20 | latest | en | 0.952959 |
http://betterlesson.com/lesson/resource/2081642/3-4-interceptions-docx | 1,487,631,243,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170613.8/warc/CC-MAIN-20170219104610-00354-ip-10-171-10-108.ec2.internal.warc.gz | 27,352,887 | 24,617 | ## 3.4 Interceptions.docx - Section 4: Interceptions
3.4 Interceptions.docx
3.4 Interceptions.docx
# Adding and Subtracting Integers on a Number Line
Unit 3: Integers and Rational Numbers
Lesson 5 of 17
## Big Idea: What happens when you add positive 5 and negative 3? What happens when you subtract positive 2 from negative 3? Students work with a number line and football to model adding and subtracting integers.
Print Lesson
30 teachers like this lesson
Standards:
Subject(s):
Math, Integers, Number Sense and Operations, number line, 6th grade, master teacher project, adding integers, subtracting integers
60 minutes
### Andrea Palmer
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Environment: Suburban | 332 | 1,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-09 | latest | en | 0.776709 |
https://www.instructables.com/How-To-Solve-The-Rubiks-Cube-1/ | 1,708,647,636,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473871.23/warc/CC-MAIN-20240222225655-20240223015655-00570.warc.gz | 849,197,079 | 32,246 | Introduction: How to Solve the Rubik's Cube
The Rubik's cube is the best selling toy in the U.S.
some may think it is impossible to solve, even Erno Rubik didn't think there was a combination to
there are 43,000,000,000,000,000,000 (that is 43 quintillion) possible combinations.
So how can you solve the cube without memorizing 43 quintillion different combinations?
there are things called algorithms that are based on mathematical equations that get pieces to places without messing up the whole cube.
Some of you may be wondering why i wanted to learn to solve the Rubik's cube, well I am very big into math and I am Very good at memorizing things in fact I have an IQ of 120 but you don't need a high iq to learn how to solve the Rubik's cube. I learn this method last Christmas when I was bored on Christmas break and wanted something to do. It only took me a week to learn to solve it, but it wasn't like I was doing it for a week straight 24 hours a day, in fact i probably spent an hour a day max that's 7 hours max it took me to solve the Rubik's cube. So in this tutorial i will show you a beginners method that I used on how to solve the Rubik's cube and maby later i might teach you an advance method on how to solve it, but you will need to know how to solve it with the beginners method first before you move onto the advanced method.
What You Need
- A Rubik's cube
- an Attentive mindset
Step 1:
To get how to solve the Rubik's cube i will explain to you how this method solves it.
Most people think you solve the cube face by face but that is wrong you are supposed to solve one face then layer by layer.
the blue picture below shows the notation of moves that make up an algorithm
first things first you need to get the cross it doesn't matter what color it is as long as you have a cross where the colors on the sides match
Note: what I mean by the colors on the side matching is if you are solving the red faces cross the color of the edge piece should match up with it's side
Step 2:
Now you will need to get the rest of the first side
to get the white piece up to where it needs to be
the algorithm for it with the green side facing you is
R, Fi, Ri, F
the algorithm for the bottom picture is
F, D, Fi, Di, R, Fi, Ri, F
these are algorithms that I invented
Step 3:
Congratulations you have solved the first side but that isn't very impressive so lets go on.
Now you need to flip the solved side so its on the bottom so you can get the edge pieces aligned also known as the second layer
you need to find the edge pieces that belong in the middle row
you need to find an edge piece and line it up with its color
if it has to the right the algorithm you need to use is
U, R, Ui, Ri, Ui, Fi, U, F
if it has to go to the left use this algorithm
Ui, Li, U, L, U, F, Ui, Fi,
now do that to all the edge pieces
you are 2/3rd's of the way done
Step 4:
Now you have the first 2 layers finished
you now need the top cross
keep doing this algorithm until the top cross is solved
F, U, R, Ui, Ri, Fi
now you should have the top cross
to solve the top face of the Rubik's cube you will need to do this algorithm
until the top of you Rubik's cube looks like the picture then turn the whole cube to the right and do the again. if the top of you cube is solved your good but if it isn't get turn it so it looks like the picture then turn it to the right then do the algorithm again
R, U, Ri, U, R, U, U, Ri
now all you need to do is get the 3rd layer done
Step 5:
Now you have the first 2 layers and the top your so close
if you have a combination like the orange side where there is only one one odd piece turn the top layer so it is on the back side then do this algorithm
Ri, F, Ri, B, B, R, Fi, Ri, B, B, R, R, Ui
Now your cube should look like the 2nd and 3rd picture with 3 faces with only 1 piece off
go to the opposite face off the one fully solve face and if the odd piece has to go to the right do this algorithm
F, F, Ui, L, Ri, F, F, Li, R, Ui, F, F
If it needs to go left do this algorithm
F, F, U, L, Ri, F, F, Li, R, U, F, F
Step 6: Finished
Congratulations You have now solved the Modern day mystery; The Rubik's Cube
but if all this went right over your head try checking out this link it does a very good job explaining how to solve it http://www.rubiks.com/solving-center/3x3_guide/
Thank you for viewing my instructable please like, comment, and share it
if you liked this instructable please follow me and check out some of my other projects | 1,170 | 4,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-10 | latest | en | 0.96999 |
https://icsehelp.com/algebra-class-6-ml-aggarwal-icse-maths-apc-solutions/ | 1,621,081,421,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991801.49/warc/CC-MAIN-20210515100825-20210515130825-00466.warc.gz | 326,088,488 | 32,673 | # Algebra Class-6 ML Aggarwal ICSE Maths APC Solutions
Algebra Class-6 ML Aggarwal ICSE Maths APC Solutions Chapter-9. We provide step by step Solutions of Exercise / lesson-9 Algebra ICSE Class-6th ML Aggarwal Mathematics .
Our Solutions contain all type Questions with Exe- 9.1, Exe-9.2, Exe-9.3, Exe-9.4, Exe- 9.5, Objective Type Questions (includes : Mental Maths , MCQs ) and Check Your Progress to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-6 Maths.
## Algebra Class-6 ML Aggarwal ICSE Maths APC Solutions Chapter-9
–: Select Topic :–
Exercise 9.1 ,
Exercise-9.2,
Exercise-9.3,
Exercise 9.4 ,
Exercise 9.5 ,
Objective Type Questions,
Mental Maths,
Multiple Choice Questions ,(MCQ)
### Exercise – 9.1, Algebra Class-6 ML Aggarwal ICSE Maths APC Solutions
#### Question 1
Find the rule which gives the number of matchsticks required to make the following matchsticks patterns. Use a variable to write the rule.
(i) A pattern of letter T as
(ii) A pattern of letter V as
(iii) A pattern of letter Z as
(iv) A pattern of letter U as
(v) A pattern of letter F as
(vi) A pattern of letters S as
(i) Number of matchsticks requried = 2n
(ii) Number of matchsticks requried = 2n
(iii) Number of matchsticks requried = 3n
(iv) Number of matchsticks requried = 3n
(v) Number of matchsticks requried = An
(vi) Number of matchsticks requried = 5n
#### Question 2
If there are 24 mangoes in a box, how will you write the number of mangoes in terms of the number of boxes? (use b for the number of boxes.)
Total number of mangoes = 24b
#### Question 3.
Anuradha is drawing a dot Rangoli (a beautiful pattern of lines joining dots). She has 8 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 12 rows?
∵ Numbers of dots in 1 row = 8
∴ Number of dots in r rows = 8 × r = 8
Number of dots in 12 rows = 12 × 8 = 96
#### Question 4
Anu and Meenu are sisters. Anu is 5 years younger than Meenu. Can you write Anu’s age in terms of Meenu’s age? Take Meenu’s age as x years.
Yes! we can write Anu’s age in terms of Meenu’s age.
Age of Meenu = x
∵ Anu is 5 years younger than Meenu
∴ Age of Anu = (x – 5) years
#### Question 5
Oranges are to be transferred from larger boxes to smaller boxes. When a larger box is empited, the oranges from it fill 3 samller boxes and still 7 oranges are left. If the number of oranges in a small box are taken to be x, then what is the number of oranges in the larger box? .
Let the number of oranges in a smaller box be x.
∴ Number of oranges in three smaller boxes = 3x
Number of oranges remained outside = 7
∴ Number of oranges in the larger box = 3x + 7
#### Question 6
Harsha’s score in Mathematics is 15 more than three-fourth of her score in Science. If she scores x marks in Science, find her score in Mathemstics?
Let the score of science be x
Harsha score’s in Mathematics = $\frac{3}{4} \text { th of } x+15$
∴ Score of Harsha’s in Mathematics = $\frac{3}{4} x+15$
#### Question 7
Look at the following matchstick pattern of equilateral triangles. The triangles are not separate. Two neighbouring triangles have a common matchstick. Observe the pattern and find the rule that gives the number of matchsticks.
Number of matchsticks required = 2x + 1, where x is the number of triangles.
#### Question 8
Look at the following matchstick pattern of letter A. The A’s are not separate. Two neighbouring A’s have two common matchsticks. Observe the pattern and find the rule that gives the number of matchsticks.
Solution:
Number of matchsticks required = 4x + 2, where x is the number of letter ‘A’ formed.
### Algebra Class-6 ML Aggarwal ICSE Maths APC Solutions Exercise – 9.2
#### Question 1.
If the side of an equilateral triangle is l, then express the perimeter of the triangle in terms of l.
Solution:
Perimeter (P) of the equilateral triangle with side l = Sum of the lengths of sides of the equilateral triangle = l + l + l = 3l
#### Question 2.
The side of a regular hexagon is l. Express its perimeter in terms of l.
Perimeter (P) of the regular hexagon with side l = Sum of the lengths of all sides of the regular hexagon
= l + l + l + l + l + l = 6l
#### Question 3
The length of an edge of a cube isl. Find the formula for the sum of lengths of all the edges of the cube.
Total length (L) of the edges of a cube = Sum of the lengths of all (12) edges of the cube.
= l + l + l + l + l+ l + l + l + l + l + l + l = 12l
#### Question 4
If the radius of a circle is r units, then express the length of a diameter of the circle in terms of r.
### ML Aggarwal Solutions for ICSE Class-6 Maths Algebra Exercise – 9.3
#### Question 1
Form four expressions with numbers 7, 5 and 8 (no variables) using operations of addition, subtraction or multiplication with the condition that every number should be used but not more than once.
The possible expressions are:
5 × 7 + 8, 5 × 8 – 7
(5 + 8) – 7, 8 × (5 + 7)
#### Question 2
Which out of following are expressions with numbers only?
(i) 2y + 3
(ii) (7 × 20) – 82
(iii) 5 × (21 – 7) + 9 x 2
(iv) 5 -11 n
(v) (5 × 4) – 45 + p
(vi) 3 × (11 + 7) – 24 + 3
(iii) 5 × (21 – 7) + 9 × 2,
(vi) 3 × (11 + 7) – 24 ÷ 3 are expressions with numbers only.
#### Question 3
Identify the operations (addition, subtraction, multiplication, division) in forming the following expressions and tell how the expressions have been formed:
(i) x + 5
(ii) y – 7
(iii) 3z
(iv) $\frac{p}{5}$
(v) 2x + 17
(vi) 3y – 5
(vii) $-7 m+\frac{2}{3}$
(viii) $\frac{x}{3}-15$
(i) x + 5
(ii) y – 7
Subtraction → 7 subtracted from y.
(iii) 3z
Multiplication → z multiplied by 3.
(iv) $\frac{p}{5}$
Division → p divided by 5.
(v) 2x + 17
Multiplication and addition → First x mutliplied by 2 and then 17 added to the product.
(vi) 3y – 5
Multiplication and subtraction → First y multiplied by 3 and then 5 subtracted from the product.
(vii) $-7 m+\frac{2}{3}$
Multiplication and addition → First m multiplied by -7 and then $\frac{2}{3}$ added to the product.
(viii) $\frac{x}{3}-15$
Division and subtraction → First x divided by 3 and then 15 subtracted from the quotient.
#### Question 4
Write expression for the following:
(ii) p subtracted from 7
(iii) p multiplied by 7
(iv) p divided by 7
(v) 7 divided by p
(vi) 7 subtracted from -m
(vii) p multiplied by -5
(viii) -p divided by 5
(i) p + 7
(ii) 7 – p
(iii) 7p
(iv) $\frac{p}{7}$
(v) $\frac{7}{p}$
(vi) -m – 7
(vii) -5p
(viii) $\frac{-p}{5}$
#### Question 5
Write expression for the following:
(i) 11 added to 2 m
(ii) 11 subtracted from 2 m
(iii) 3 added to 5 times y
(iv) 3 subtracted from 5 times y
(v) y is multiplied by -8 and then 5 is added to the result
(vi) y is multiplied by 5 and then the result is subtracted from 16.
(i) 2m + 11
(ii) 2m – 11
(iii) 5y + 3
(iv) 5y – 3
(v) -8y + 5
(vi) 16 – 5y
#### Question 6
Write the following in mathematical form using signs and symbols:
(i) 6 more than thrice a number x.
(ii) 7 taken away from y.
(iii) 3 less than quotient of x by y.
(i) 3x + 6
(ii) $\frac{x}{y}-3$
(iii) y – 7
#### Question 7
Form six expressions using t and 4. Use not more than one number operation and every expression must have t in it.
t + 4, t – 4, 4 – t, 4t, $\frac{t}{4}, \frac{4}{t}$
#### Question 8
Form expressions using y, 2 and 7. Use only two different number operations and every expression must have y in it.
Solution:
2y + 7, 2y- 1, 7y + 2, 7y – 2, $\frac{y}{2}+7, \frac{y}{2}$ – 7, …..
#### Question 9
A student scored x marks in English but the teacher deducted 5 marks for bad handwriting. What was the student’s final score in English?
Marks in English = x
Deducted = 5
Final score = x – 5
#### Question 10
Raju’s father’s age is 2 years more than 3 times Raju’s age. If Raju’s present age is y years, then what is his father’s age?
(3y + 2) years
#### Question 11
Mohini is x years old. Express the following in algebraic form:
(i) three times Mohini’s age next year.
(ii) four times Mohini’s age 3 years ago.
(iii) the present age of Mohini’s uncle, if his uncle is 5 times as old as Mohini will be two years from now.
(iv) the present age of Mohini’s cousin, if her cousin is two years less than one-third of Mohini’s age five years ago
#### Question 12
A cuboidal box has height h cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.
Length of the box = 5h cm
Breadth of the box = (5h – 10) cm
#### Question 13
A bus travels at v km per hour. It is going from Delhi to Jaipur. After the bus has travelled 5 hours, Jaipur is still 20 km away. What is the distance from Delhi to Jaipur?
Speed of the bus = v km/hr
Distance travelled in 5 hours = 5v km
∴ Total distance = (5v + 20) km
#### Question 14
Change the following statements using expressions into statements in ordinary language:
(i) A notebook cost ₹ p. A book costs ₹ 3p.
(ii) The cost of rice per kg is ₹ p. The cost of oil per litre is ₹5p.
(iii) The speed of a truck is v km per hour. The speed of a bus is (v + 10) km per hour.
(iv) Tony’s box contains 8 times the marbles he puts on the table.
(v) The total number of students in the school is 20 times that of our class.
(vi) Raju is x years old. His uncle is 4x years old and his aunt is (4x – 3) years old.
(vii) In arrangement of dots there are r rows. Each row contains 5 dots.
(i) The cost of a book is 3 times the cost of a note book.
(ii) The cost of oil per litre is 5 times the cost of rice per kg.
(iii) The speed of a bus is 10 km per hour more than the speed of a truck.
(iv) Tony puts q marbles on the table. He has 8q marbles in his box.
(v) Our class has n students. The school has 20 n students.
(vi) Raju’s uncle is 4 times older than Raju and his aunt is 3 years younger than his uncle.
(vii) The total number of dots is 5 times the number of rows.
### Algebra Exercise – 9.4 for ICSE Class-6 ML Aggarwal Solutions
#### Question 1
Find the value of the following:
(i) 43
(ii) (-6)4
(iii) $\left(\frac{2}{3}\right)^{4}$
(iv) (-2)3 × 52
#### Question 2
Find the value of:
(i) 3x + 2y when x = 3 and y – 2
(ii) 5x – 3y when x = 2 and y = -5
(iii) a + 2b – 5c when a = 2, b = -3 and c = 1
(iv) 2p + 3q + 4r + pqr when p = -1, q = 2 and r = 3
(v) 3ab + 4bc – 5ca when a = 4, 6 = 5 and c = -2.
(i) 3x + 2y, x = 3,y = 2
(3 × 3) + (2 × 2) = 9 + 4 = 13
(ii) 5x – 3y, x = 2, y = -5
(5 × 2) – (3 × -5) = 10 + 15 = 25
(iii) a + 2b – 5c, a =2,b = -3, c = 1
2 + (2 × -3) -5 × (1)
= 2 – 6 – 5 = -9
(iv) 2p + 3q + 4r + pqr, p = -1, q = 2, r = 3
= (2 × -1) + (3 × 2) + (4 × 3) + (-1) × 2 × 3
= -2 + 6 + 12 – 6= 10
(v) 3ab + 4bc – 5ca, a = 4, b = 5, c = -2 (3 × 4 × 5) + (4 × 5 × -2) -5 × -2 × 4
= 60 – 40 + 40 = 60
#### Question 3
Find the value of:
(i) 2x2 – 3x + 4 when × = 2
(ii) 4x3 – 5x2 – 6x + 7 when x = 3
(iii) 3x3 + 9x2 – x + 8 when x = -2
(iv) 2x4 – 5x3 + 7x – 3 when x = -3
(i) 2x2 – 3x + 4, x = 2
= 2 × (2)2 -3x2 + 4
= 8 – 6 + 4 = 6
(ii) 4x3 – 5x2 – 6x + 7, x = 3
= 4(3)3 – 5(3)2 – 6(3) + 7
= 108 – 45 – 18 + 7 = 52
(iii) 3x3 + 9x2 – x + 8, x = -2
= 3(-2)3 + 9(-2)2 – (-2) + 8
= -24 + 36 + 2 + 8 = 22
(iv) 2x4 – 5x3 + 7x – 3, x = -3
= 2(-3)4 – 5(-3)3 + 7(-3) – 3
= 162 + 135 – 21 – 3 = 273
#### Question 4
If x = 5, find the value of:
(i) 6 – 7x2
(ii) 3x2 + 8x – 10
(iii) 2x3 – 4x2 – 6x + 25
(i) 6 – 7x2 = 6 – 7(5)2 = 6 – 7(25)
= 6 – 175 = -169
(ii) 3(5)2 + 8(5) – 10
= 3(25) + 40 – 10
= 75 + 40 – 10 = 75 + 30 = 105
(iii) 2(5)3 – 4(5)2 – 6(5) + 25
= 2(125) – 4(25) – 30 + 25
= 250 – 100 – 30 + 25= 145
#### Question 5
If x = 2, y = 3 and z = -1, find the values of:
(i) x + y
(ii) $\frac{x y}{z}$
(iii) $\frac{2 x+3 y-4 z}{3 x-z}$
#### Question 6
If a = 2, b = 3 and c = -2, find the value of a2 + b2 + c2 – 2ab – 2bc – 2ca + 3abc.
a = 2,b = 3,c = -2
a2 + b2 + c2 – 2ab – 2bc – 2ca + 3abc
= (2)2 + (3)2 + (-2)2 – 2 × 2 × 3 – 2 × 3 × – 2 – 2 × -2 × 2 + 3 × 2 × 3 × -2
= 4 + 9 + 4 – 12 + 12 + 8 – 36
= 25 – 36 = -11
#### Question 7
If p = 4, q = -3 and r = 2, find the value of: p3 + q3 – r3 – 3pqr.
p = 4, q = -3, r = 2
p3 + q3 – r3 – 3pqr
= (4)3 + (-3)3 – (2)3 – 3 × 4 × -3 × 2
= 64 – 27 – 8 + 72
= 136 – 35 = 101
#### Question 8
If m = 1, n = 2 and p = -3, find the value of 2mn4 – 15m2n + p.
m = 1, n = 2, p = -3
2mn4 – 15m2n + p
= 2(1 )(2)4 – 15(1 )2(2) + (-3)
= 32 – 30 – 3 = —1
#### Question 9
State true or false:
(i) The value of 3x – 2 is 1 when x = 0.
(ii) The value of 2x2 – x – 3 is 0 when x = -1.
(iii) p2 + q2 – r2 when p = 5, q = 12 and r = 13.
(iv) 16 – 3x = 5x when x = 2.
(i) The value of 3x – 2 is 1 when x = 0. False
Correct :
∵ 3 × 0 – 2 = -2
(ii) The value of 2x2 – x – 3 is 0 when x = -1. True
2(-1 )2 – (-1) – 3
= 2 + 1 – 3 = 0
(iii) p2 + q2 = r2 when p = 5, q = 12 and r = 13. True
(5)2 + (12)2 = (13)2
= 25 + 144 = 169
⇒ 169= 169
(iv) 16 – 3x = 5x when x = 2. True
16 – 3x2 = 5x2
16 – 6 = 10
⇒ 10 = 10
#### Question 10
For x = 2 and y = -3, verify the following:
(i) (x + y)2 = x2 + 2xy + y2
(ii) (x – y)2 = x2 – 2xy + y2
(iii) x2 – y2 = (x + y) (x – y)
(iv) (x + y)2 = (x – y)2 + 4xy
(v) (x + y)3 = x3 + y3 + 3x2y + 3xy2
x = 2, y = -3
(i) (x + y)2 = x2 + 2xy + y2
L.H.S. = (x + y)2 = (2 – 3)2 = (-1)2 = 1
R.H.S. = x2 + 2xy + y2
= (2)2 + 2 × 2 (-3) + (-3)2
= 4 – 12 + 9 = 13 – 12 = 1
L.H.S. = R.H.S.
(ii) (x – y)2 = x2 – 2xy + y2
L.H.S. = (x – y)2 = [2 – (-3)]2
= (2 + 3)2 = (5)2 = 25
R.H.S. = x2 – 2xy + y2
= (2)2 – 2 × 2 × (-3) + (-3)2
= 4 + 12 + 9 = 25
∴ L.H.S. = R.H.S.
(iii) x2 – y2 = (x + y)(x – y)
L.H.S. = (x)2 – (y)2 = (2)2 – (-3)2
= 4 – 9 = -5
R.H.S. = (x + y)(x – y)
= (2 – 3) [2 – (-3)]
= (2 – 3) (2 + 3) = -1 × 5 = -5
∴ L.H.S. = R.H.S.
(iv) (x + y)2 = (x – y)2 + 4xy
L.H.S. = (x + y)2 = [2 + (-3)]2
= (2 – 3)2 = (-1)2 = 1
R.H.S. = (x – y)2 + 4xy
= [2 – (-3)]2 + 4 × 2 × (-3)
= (2 + 3)2 + 4 × 2 × (-3)
= (5)2 – 24 = 25 – 24 = 1
∴ L.H.S. = R.H.S.
(v) (x + y)3 = x3 + y2 + 3x2y + 3xy2
L.H.S. = (x + y)3 = [2 + (-3)]3 = (2 – 3)3
= (-1)3 = (-1) × (-1) × (-1) = -1
R.H.S. = x3 + y3 + 3x2y + 3xy2
= (2)3 + (-3)3 + 3 (2)2 (-3) + 3 × 2 (-3)2
=8 – 27 + 3 × 4 × (-3) + 6 (9)
= 8 – 27 – 36 + 54 = 62 – 63 = -1
∴ L.H.S. = R.H.S.
### ML Aggarwal Solutions of Algebra Exercise – 9.5 for ICSE Class-6
#### Question 1
State which of the following are equations with a variable. In case of an equation with a variable, identify the variable.
(i) 17 + x = 5
(ii) 2b – 3 = 7
(iii) (y – 7) > 5
(iv) $\frac{9}{3}=3$
(v) 7 × 3 – 19 = 2
(vi) 5 × 4 – 8 = 31
(vii) 2p < 15
(viii) 7 = 11 × 5 – 12 × 4
(ix) $\frac{3}{2} q=5$
(i) 17 + x= 5 Is an equation → L.H.S. = R.H.S. → Related variable x.
(ii) 2b – 3 = 7 Is an equation → L.H.S. = R.H.S. → Related variable b.
(iii) (y- 7) >5
Is not an equation → L.H.S. ≠ R.H.S.
It has no sign of equality (=).
(iv) $\frac{9}{3}=3$
Is an equation = L.H.S. = R.H.S.
It has no variable.
(v) 7 × 3 – 19 = 2
Is an numerical equation = L.H.S. = R.H.S. It has no variable.
(vi) 5 × 4 – 8 = 31
Is an equation = L.H.S. = R.H.S. → Related variable t.
(vii) 2p < 15
Is not an equation = L.H.S. ≠ R.H.S.
It has no sign of equality.
(viii) 7 = 11 × 5 – 12 × 4
Is an numerical equation = L.H.S. = R.H.S. It has no variable.
(ix) $\frac{3}{2} q=5$
Is an equation → L.H.S. = R.H.S. → Related variable q.
#### Question 2
Solve each of the following equations :
(i) x + 6 = 8
(ii) 2 – x = 5
(iii) 4x = -6
(iv) $\frac{x}{2}=5$
(v) 2y – 3 = 2
(vi) 4 – 5y = 2
(i) x + 6 = 8
= x = 8 – 6 ⇒ x = 2
(ii) 2 – x = 5
= -x = 5 – 2 ⇒ -x = 3 ⇒ x = – 3
(iii) 4x = -6
$=x=\frac{-6}{4}=\frac{-3}{2}$
(iv) $\frac{x}{2}=5$
= x = 5 × 2 ⇒ x = 10
(v) 2y – 3 = 2
= 2y = 2 + 3 ⇒ 2y = 5 ⇒ y = $\frac{5}{2}$
(vi) 4 – 5y = 2
= 4 – 2 = 5y ⇒ 5y = 2
⇒ y = $\frac{2}{5}$
#### Question 3
Solve the following linear equations:
(i) 5(x + 1) = 25
(ii)2(3x – 1) = 10
(iii) $\frac{3 x-1}{4}=11$
(i) Givrn 5(x + 1) = 25
⇒ $\frac{5(x+1)}{5}=\frac{25}{5}$ (dividing both sides by 5)
⇒ x + 1 = 5
⇒ x + 1 – 1 = 5 – 1 (Subtracting 1 from both sides)
⇒ x = 4
(ii) 2(3x – 1) = 10
⇒ $\left(\frac{2(3 x-1)}{2}\right)=\frac{10}{2}$ (dividing both sides by 2)
⇒ 3x – 1 = 5
⇒ 3x – 1 + 1 = 5 + 1 (adding 1 to both sides)
⇒ 3x = 6
⇒ $\frac{3 x}{3}=\frac{6}{3}$ (dividing both sides by 3)
⇒ x = 2
(iii) Given $\frac{3 x-1}{4}=11$
⇒ $4 \times \frac{3 x-1}{4}=4 \times 11$ (multiplying both sides by 4)
⇒ 3x – 1 = 44
⇒ 3x – 1 + 1 = 44 + 1 (adding 1 to both sides)
⇒ 3x = 45
⇒ $\frac{3 x}{3}=\frac{45}{3}$ (dividing both sides by 3)
⇒ x = 15
#### Question 4
Solve the following linear equations:
(i) 5x – 6 = 12 – x
⇒ 5x + x = 12 + 6
#### Question 5
(i) 3(x + 7) = 18
(ii) 2(x- 1) = x + 2
(iii) $3 x-\frac{1}{3}=2\left(x-\frac{1}{2}\right)+5$
(iv) 4(2x – 1) -2(x – 5) = 5(x + 1) + 3
⇒ 4(4 – 1) -2 (-3) = 5(3) + 3
⇒ 4 × 3 + 6 = 15 + 3
⇒12 + 6 = 18
⇒ 18 = 18
### Objective Type Questions, Algebra Class-6 ML Aggarwal
#### Question 1
Fill in the blanks:
(i) In algebra, we use …………… to represent variables (generalized numbers).
(ii) A symbol or letter which can be given various numerical values is called a ……………
(iii) If Jaggu’s present age is x years, then his age 7 years from now is ……………
(iv) If one pen costs ₹X x, then the cost of 9 pens is ……………
(v) An equation is a statement that the two expressions are ……………
(vi) Trial an error is one of methods to obtain …………… of an equation.
(vii) 7 less than thrice a number y is ……………
(viii) If 3x + 4 = 19, then the value of x is ……………
(ix) The number of pencils bought for ₹ x at the rate of ₹2 per pencil is ……………
(x) In the expression (-7)5, base = …………… and exponent = ……………
(xi) If base = 6 and exponent = 5, then the exponential form = …………… .
(i) In algebra, we use letters to represent variables (generalized numbers).
(ii) A symbol or letter which can be given various numerical values is called a variable.
(iii) If Jaggu’s present age is x years, then his age 7 years from now is (x + 7) years.
(iv) If one pen costs ₹ x, then the cost of 9 pens is ₹9x.
(v) An equation is a statement that the two expressions are equal.
(vi) Trial an error is one of methods to obtain the solution of an equation.
(vii) 7 less than thrice a num bery is 3y – 7.
(viii) If 3x + 4 = 19, then the value of x is 5.
(ix) The number of pencils bought for ₹ x at the rate of ₹2 per pencil is $\frac{x}{2}$.
(x) In the expression (-7)5, base = -7 and exponent = 5.
(xi) If base = 6 and exponent = 5, then the exponential form = 65.
#### Question 2
State whether the following statements are ture (T) or false (F):
(i) If x is variable then 5x is also variable.
(ii) If y is variable then y – 5 is also variable.
(iii) The number of angles in a triangle is a variable.
(iv) The value of an algebraic expression changes with the change in the value of the variable.
(v) If the length of a rectangle is twice its breadth, then its area is a constant.
(vi) An equation is satisfied only for a definite value of the variable.
(vii) If x toffees are distributed equally among 5 children, then each child gets 5x toffees.
(viii) t minutes are equal to 60 t seconds.
(ix) If x is a negative integer, then -x is a positive integer.
(x) x = 5 is a solution of the equation 3x + 2 = 13.
(xi) 2y- 7 > 13 is an equation.
(xii) ‘One third of a number x added to itself gives 8’ can be expressed as $\frac{x}{3}$ + 8 = x.
(xiii)The difference between the ages of two sisters Lata and Asha is a variable.
(i) If x is variable then 5x is also variable. True
(ii) If y is variable then y – 5 is also variable. True
(iii) The number of angles in a triangle is a variable. False
(iv) The value of an algebraic expression changes with the change in the value of the variable. True
(v) If the length of a rectangle is twice its breadth, then its area is a constant. False
(vi) An equation is satisfied only for a definite value of the variable. True
(vii) If x toffees are distributed equally among 5 children, then each child gets 5x toffees. False
(viii) t minutes are equal to 60 t seconds. True
(ix) If x is a negative integer, then -x is a positive integer. True
(x) x = 5 is a solution of the equation 3x + 2 = 13. False
(xi) 2y – 1 > 13 is an equation. False
(xii) ‘One third of a number x added to itself gives 8’ can be expressed as $\frac{x}{3}$ + 8 = x. False
(xiii)The difference between the ages of two sisters Lata and Asha is a variable. False
#### Question 3
Choose the correct answer from the given four options (3 to 19):
Question 3.
I think of a number x, add 5 to it. The result is then multiplied by 2 and the final result is 24. The correct algebraic statement is
(a) x + 5 × 2 = 24
(b) (x + 5) × 2 = 24
(c) 2 × x + 5 = 24
(d) x + 5 = 2 × 24
Let number = x
⇒ i.e. x + 5
Now multiply result with 2
i.e. (x + 5) × 2
Now, final result is 24
i.e. (x + 5) × 2 = 24 (b)
#### Question 4
Which of the following is an equation?
(a) x + 5
(b) 7x
(c) 2y + 3 = 11
(d) 2p < 1
2y + 3 = 11 (c)
#### Question 5
If each matchbox contains 48 matchsticks, then the number of matchsticks required to fill n such boxes is
(i) 48 + n
(b) 48 – n
(c) 48 ÷ n
(d) 48n
Matchstick required to fill 1 matchbox
= 48 × 1 = 48
Matchstick required to fill 2 matchbox
= 48 × 2 = 96
Matchstick required to fill 3 matchbox
= 48 × 3 – 144
∴ Matchstick required to fill n matchbox
= 48 n (d)
#### Question 6
If the perimeter of a regular hexagon is x metres, then the length of each of its sides is
(a) (x + 6) metres
(b) (x – 6) metres
(c) (x ÷ 6) metres
(d) (6 ÷ x) metres
Perimeter of hexagon = x metres
6(side) = x metres
Side = (x ÷ 6) metres
∴ Side = (x ÷ 6) metres (c)
#### Question 7
x = 3 is the solution of the equation
(a) x + 7 = 4
(b) x + 10 = 7
(c) x + 7 = 10
(d) x + 3 = 7
When put the value of x = 3
3 + 7=10 (c)
#### Question 8
The solution of the equation 3x – 2 = 10 is
(a) x = 1
(b) x = 2
(c) x = 3
(d) x = 4
3x – 2 = 10
3x = 10 + 2
$x=\frac{12}{3}=x=4$ (d)
#### Question 9
The operation not involved in forming the expression 5x + $\frac{5}{x}$ from the variable x and number 5 is
(b) subtraction
(c) multiplication
(d) division
Subtraction (b)
#### Question 10
The quotient of x by 3 added to 7 is written as
$\frac{x}{3}+7$ (a)
#### Question 11
If there are x chairs in a row, then the number of persons that can be seated in 8 rows are
(a) 64
(b) x + 8
(c) 8x
(d) none of these
Let the no. of chairs in a row = x
⇒ Number of persons that can be seated in a row = x
Hence, number of persons that can be seoted in 8 row = 8x (c)
#### Question 12
If Arshad earns ₹ x per day and spends ₹ y per day, then his saving for the month of March is
(a) ₹(31x – y)
(b) ₹31(x – y)
(c) ₹31 (x + y)
(d) ₹31 (y – x)
Earning of Arshad for 1 day = ₹ x
Spending of Arshad for 1 day = ₹ y
Saving for 1 day = ₹(x – y)
Saving for 31 days = ₹31 (x – y) (b)
#### Question 13
If the length of a rectangle is 3 times its breadth and the breadth is x units, then its perimeter is
(a) 4x units
(b) 6x units
(c) 8x units
(d) 10x units
Breadth of rectangle = x units
Length of rectangle = 3(Breadth) = 3x
Perimeter of rectangle = 2(l + b)
= 2(3x + x)
= 2(4x) = 8x units (c)
#### Question 14
Rashmi has a sum of ₹ x. She spend ₹800 on grocery, ₹600 on cloths and ₹500 on education and received as ₹200 as a gift. How much money (in ₹) is left with her?
(a) x – 1700
(b) x – 1900
(c) x + 200
(d) x – 2100
Total money = ₹ x
Money spent = ₹800 on grocery
Money spent = ₹600 on cloths .
Money spent = ₹500 on education
Money left with Rashmi
= x – ₹800 + ₹600 + ₹500
= x – 1900
∴ Money left = x – 1900 + 200
= x – 1700 (a)
#### Question 15
For any two integers a and b, which of the following suggests that the operation of addition is commutative?
(a) a × b = b × a
(b) a + b = b + a
(c) a – b = b – a
(d) a + b > a
a + b = b + a
#### Question 16
In $\left(\frac{3}{4}\right)^{5}$, the base is
(a) 3
(b) 4
(c) 5
(d) $\frac{3}{4}$
$\frac{3}{4}$
#### Question 17
a × a × b × b × b can be written as
(a) a2b3
(b) a3b2
(c) a3b3
(d) a5b5
a × a × b × b × b
= a2 × b3 = a2b3 (a)
#### Question 18
(-5)2 × (-1)3 is equal to
(a) 25
(b) -25
(c) 10
(d) -10
(-5)2 × (-1)3
⇒ (-5) × (-5) x (-1) × (-1) × (-1)
⇒ 25 × (-1) = -25 (b)
#### Question 19
(-2)3 × (-3)2 is equal to
(a) 65
(b) (-6)5
(c) 72
(d) -72
(-2)3 × (-3)2
⇒ (-2) × (-2) × (-2) × (-3) × (-3)
⇒ -8 × 9 = -72 (d)
#### Question 1
Look at the following matchstick pattern of polygons. Complete the table. Also write the general rule that gives the number of matchsticks.
#### Question 2
Write an algebraic expression for each of the following:
(i) If 1 metre cloth costs ₹ x, then what is cost of 6 metre cloth?
(ii) If the cost of a notebook is ₹ x and the cost of a book is ₹ y, then what is the cost of 5 notebooks and 2 books?
(iii) The score of Ragni in Mathematics is 23 more than two-third of her score in English. If she scores x marks in English, what is her score in Mathematics?
(iv) If the length of a side of a regular pentagon is x cm, then what is the perimeter of the pentagon?
(i) Cost of 1 metre cloth = ₹ x
Cost of 6 metre cloth = ₹ x × 6 = ₹6x
(ii) Cost of 1 notebook = ₹ x
Cost of 1 book = ₹y
Cost of 5 notebooks = 5 × (₹x) = ₹5x
Cost of 2 books = (₹y) × 2 = ₹2y
Total cost of 5 notebooks and 2 books = ₹(5x + 2y)
(iii) Score of Ragni in English = x marks
Score of Ragni in Mathematics = 23 more than two-third of her score in English
i.e. = $23+\frac{2}{3}(x)$
$23+\frac{2}{3} x$
(iv) Side of regular pentagon = x cm
Perimeter of pentagon = 5 × Side = 5x cm
#### Question 3
When x = 4 and y = 2, find the value of:
(i) x + y
(ii) x – y
(iii) x2 + 2
(iv) x2 – 2xy + y2
(i) x + y
Put x = 4 and y = 2
We get,
⇒ 4 + 2 = 6
(ii) x – y
Put x = 4, y = 2
We get,
⇒ 4 – 2 = 2
(iii) x2 + 2
Put x = 4 We get,
⇒ (4)2 + 2
= 16 + 2= 18
(iv) x2 – 2xy + y2
Put x = 4 and y = 2
#### Question 4
When a = 3 and b = -1, find the value of 2a3 – b4 + 3a2b3.
2a3 – b4 + 3a2b3
Put the value of a = 3, b = -1
= 2(3)3 – (-1)4 + 3(3)2 (-1)3
= 2 × 3 × 3 × 3 – (-1) × (-1) × (-1) × (-1) + 3 × 3 × 3 × (-1) × (-1) × (-1)
= 54 – 1 – 27 = 26
#### Question 5
When a = 3, b = 0, c = -2, find the values of:
(i) ab + 2bc + 3ca + 4abc
(ii) a3 + b3 + c3 – 3abc
(i) ab + 2bc + 3ca + 4abc
Put the values of a = 3, b = 0, c = -2
3 × 0 + 2 × 0 × -2 + 3 × -2 × 3 + 4 × 3 × 0 × -2
= 0 + 0 – 18 + 0 = -18
(ii) a3 + b3 + c3 – 3abc
Put the values of a = 3, b = 0, c
= -2 (3)3 + (0)3 + (-2)3 – 3 × 3 × 0 × -2
= 27 + 0 – 8 – 0 = 19
#### Question 6
Solve the following linear equations:
(i) 2x – $1 \frac{1}{2}=4 \frac{1}{2}$
(ii) 3(y – 1) = 2(y + 1)
(ii) n – 3 = 5n + 21
(iv) $\frac{1}{3}(7 x-1)=\frac{1}{4}$
-: End of Algebra Class-6 ML Aggarwal Solutions :– | 10,812 | 26,590 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 48, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-21 | longest | en | 0.812145 |
https://www.hometalk.com/diy/outdoor/how-to-build-natural-stone-steps-44414183 | 1,643,445,353,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300573.3/warc/CC-MAIN-20220129062503-20220129092503-00019.warc.gz | 842,935,668 | 39,028 | # How to Build Natural Stone Steps
\$350
5 Hours
We recently laid solid stone steps connecting our new pool area to our existing sidewalk. At first glance, a project like this looks pretty simple. At least, that’s what we thought! That is until we tried to move one of the stones. These things weigh 450 lbs each, which made this project one of the most challenging things I’ve ever done. But it is doable! In this article, I’m going to show you how we did it and what you need to know going into a project like this.
Materials & Tools
• solid stone steps (number and size will be determined by your space)
• shovel
• tape measure
• level
• small gravel
• dolly
Step 1: Measure Your Space
Before you purchase your stone steps, you need to do some measuring to determine the size and quantity of stones you need. Our measurements were as follows:
• 5′ 7″ from the sidewalk to the pool decking
• 4′ 7″ from the house to the edge of the pool decking
• 20″ in elevation from the pool decking to the sidewalk
Typically these types of solid stone steps come in 2 lengths: 3ft and 4ft. We quickly ruled out the 4′ stones because we knew that we would be placing these by hand and the 3′ ones would be much more manageable. They would also fit better in the space and allow room against the house for some small plants.
Because the stone steps are only 1′ 6″ in length, we needed 4 of them. And the height of each step is 7″, but we only needed all of the steps to equal 20″. This required a little figuring. But in the end we were able to find one stone that was only 5 1/2″ thick. This meant that the first step would need to be in the ground and flush with the pool decking and the last step would hit just below the top of the sidewalk, which we could live with. We definitely didn’t want it to come up above the sidewalk.
Keep in mind, for solid stone steps to work out perfectly, you would need to do a lot planning before anything is built, or have custom steps cut, which will mean a very long wait time.
Step 2: Dig Area for Stone
Before you place your first stone step, you’ll want to make sure the ground is level. I used a shovel to dig out the area where I would be placing the stone. I dug it a little deeper than the height of the stone and got it as close to level as I could.
Once I had it pretty close, I used a tape measure to make sure the stone would fit. It’s a good idea to make the space several inches bigger than the stone to ensure that you have some wiggle room.
Then I shoveled some gravel onto the area and spread it out evenly, and checked it with a level.
Step 3: Place the Stone Step
Now here’s the hard part! I did this by myself with a little help from Brooke. I don’t recommend this. Instead, call up 2 or 3 of your strongest friends to help you. I used a dolly to move the stones from the trailer to the area where I was placing them.
I positioned the dolly so that the stone was lined up with the area. Then I flipped it over into place.
It landed pretty close. But to get it into the exact position I wanted, I used a large pry bar to lift and scoot it.
Step 4: Repeat Steps 2 & 3
For the rest of the stone steps, repeat steps 2 and 3. Dig out the dirt so that it is slightly below the previous stone. This will allow some room for the gravel. When placing each of the remaining stones, let them overlap the previous stone by one or two inches. Or you may need to adjust the amount of overlap to allow the steps to end where they need to.
We couldn’t be happier with the end results of the stone steps. We finished the area off with some English dwarf boxwoods and a limelight hydrangea tree.
## Enjoyed the project?
Want more details about this and other DIY projects? Check out my blog post!
## Frequently asked questions
1 question
• Mary Coakley on Jun 23, 2021
What colour is Limelight Hydrangea please?
## Join the conversation
2 of 5 comments
• Carolyn on Jun 23, 2021
Excellent job for sure! Looks wonderful!!
• Ginger Johnson Lambert on Jul 10, 2021
Great job and great instructions. It's one thing to see a beautiful job... But quite another to really get the detailed instruments to do it. Thank you | 1,002 | 4,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-05 | latest | en | 0.960311 |
https://www.scribd.com/document/3381253/dalamberian | 1,566,458,330,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316785.68/warc/CC-MAIN-20190822064205-20190822090205-00130.warc.gz | 976,118,405 | 60,152 | You are on page 1of 3
# Please purchase PDFcamp Printer on http://www.verypdf.com/ to remove this watermark.
Examples
> with(Physics):
Set the default differentiation variables for dAlembertian and d_, define a spacetime
tensor function A and use the enhanced display scheme of the differential equation
packages
> Setup(differentiationvariables = X);
Systems of spacetime Coordinates are: {X}
Default differentiation variables for d_ and dAlembertian are: [X]
[ differentiationvariables = [ X ] ]
> Define(A);
Defined objects with tensor properties
{ A, εµ, α , β, ν, g_ µ, ν, d_ µ, δµ, ν, γµ, σµ, Xµ }
## > PDEtools[declare]((f, A, Phi)(X));
f( x1 , x2 , x3 , x4 ) will now be displayed as f
A( x1 , x2 , x3 , x4 ) will now be displayed as A
Φ ( x1 , x2 , x3 , x4 ) will now be displayed as Φ
The dAlembertian operator is the double application of d_[mu] with the index
contracted:
> d_[mu](d_[mu](f(X)));
(f)
In the default 4 = 3+1 dimensional Minkowski spacetime, the form of dAlembertian as a
sum of diff constructions is
> convert(%, diff);
−fx1 , x1 − fx2 , x2 − fx3 , x3 + fx4 , x4
The dAlembertian deals normally with derivatives expressed using any of the Maple
differential operators including D, diff and d_; it also distributes over sums and products
> d_[mu](f(X) + A[mu](X));
ð µ( f ) + ðµ( Aµ )
> dAlembertian(%);
ð µ( ( f ) ) + ð µ( ( Aµ ) )
> d_[mu](f[mu](X))*A[nu](X);
ð µ( fµ ) Aν
> dAlembertian(%);
ð µ( ( fµ ) ) Aν + 2 ðα ( ðµ( fµ ) ) ð α ( Aν ) + ðµ( fµ ) ( Aν )
## Because dAlembertian is a second order differential operator (dAlembertian(f) =
ð[ µ ](ð[ µ ](f))), the differentiation rule, when applied to a product as in above, requires
Please purchase PDFcamp Printer on http://www.verypdf.com/ to remove this watermark.
## decomposing the dAlembertian operation into the sequence of ð[ µ ] operations. Note
also the introduction of a dummy index α which could be any spacetime index - these
indices are chosen after checking that they are not assigned and not already present in the
given expression at the time of introducing them.
Regardless of having set the default differentiation variables to X you can call
dAlembertian or d_ with other coordinates as differentiation variables; for instance, set
one more coordinate system
> Coordinates(Y);
Systems of spacetime Coordinates are: {X, Y}
{ X, Y }
> dAlembertian(f(X));
(f)
> dAlembertian(f(Y), [Y]);
( f( Y ), [ Y ] )
Note the output above displays the differentiation variables [Y]; this is so because they
are not the "default differentiation variables"; if you set them to be [Y]
> Setup(differentiationvariables = Y);
Default differentiation variables for d_ and dAlembertian are: [Y]
[ differentiationvariables = [ Y ] ]
> dAlembertian(f(X)); # now the omitted differentiation
variables are [Y]
0
> dAlembertian(f(X), [X]); # [X] are now displayed
( f, [ X ] )
> dAlembertian(f(Y), [Y]); # [Y] are omitted
( f( Y ) )
The dAlembertian enters the classical field equations in various models; this is the
electromagnetic field tensor
> Setup(differentiationvariables = X);
Default differentiation variables for d_ and dAlembertian are: [X]
[ differentiationvariables = [ X ] ]
> F[mu, nu] := d_[mu](A[nu](X)) - d_[nu](A[mu](X));
Fµ, ν := ð µ( Aν ) − ðν( Aµ )
Maxwell equations result from taking the functional derivative of the Action
> 'Fundiff'( Intc(F[mu,nu]^2, X), A[rho](Y)); # <- delay
evaluation quotes: peformed in the next line
δ ⌠ ∞ ⌠ ∞ ⌠ ∞ ⌠ ∞
( ð ( A ) − ð ( A ) ) d x1 d x2 dx3 d x4
2
δ Aρ( Y )
µ ν ν µ
⌡ ⌡ ⌡ ⌡
−∞ −∞ −∞ −∞
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## > subs(Y = X, %);
( 2 ðµ( ð ν( Aν ) ) − 2 ( Aµ ) ) g_ µ, ρ + ( −2 ( Aν ) + 2 ð µ( ðν( Aµ ) ) ) g_ ν, ρ
## To simplify the contracted spacetime indices use Simplify
> Simplify(%);
−4 ( Aρ ) + 4 ð µ( ð ρ( Aµ ) )
The Lagrangean of the λ Φ ^4 model, the corresponding Action and the field equations
> L := 1/2 * d_[mu](Phi(X)) * d_[mu](Phi(X)) - m^2/2 *
Phi(X)^2 + lambda/4*Phi(X)^4;
1 2 m2 Φ 2 λ Φ 4
L := ðµ( Φ ) − +
2 2 4
> 'Fundiff'(Intc(L,X), Phi(Y)); # <- delay evaluation
quotes: peformed in the next line
⌠ ∞ ⌠ ∞ ⌠ ∞ ⌠ ∞
δ 1 m2 Φ 2 λ Φ 4
Φ
2
− +
δ Φ ( Y ) 2 µ ð ( ) d x1 d x2 d x3 d x4
2 4
⌡−∞ ⌡−∞ ⌡−∞ ⌡−∞
> subs(Y = X, %);
− ( Φ ) + Φ ( −m 2 + λ Φ 2 )
>
> | 1,715 | 4,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-35 | latest | en | 0.798224 |
https://www.physicsforums.com/threads/gravitational-acceleration.113370/ | 1,508,819,920,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828134.97/warc/CC-MAIN-20171024033919-20171024053919-00640.warc.gz | 970,811,758 | 16,730 | # Gravitational Acceleration
1. Mar 7, 2006
### Milind_shyani
hello
The gravitational acceleration at the equator is the least and at the pole is the maximum due to the rotation of the earth on its axis.Why???
2. Mar 7, 2006
### Galileo
The earth is not precisely spherical, it is flattened at the poles because of the earth's rotation. This is simply because of the centrifugal force, objects at the equator have the greatest distance from the axis of rotation so are 'flung outwards' more than at the poles. So at the poles you are closer to the center of the earth and experience a greater gravitational acceleration.
3. Mar 7, 2006
### arildno
It should be borne in mind that "g" is generally meant to mean "effective gravitational acceleration" and includes the effects of a rotating Earth (which predicts a lower g at equator than by the poles), the deviatioric correction due to the non-spherical form of the Earth, as the two main correctional effects.
4. Mar 7, 2006
### BobG
Galileo is right - the oblateness of the Earth is the biggest factor in the difference in gravitational attraction.
However, the way the question is worded, I think it's addressing centripetal acceleration. Which point will have a larger linear velocity due to the rotation of the Earth: a point on the equator or a point on one of the poles?
Both the oblateness of the Earth and centripetal force contribute to reducing the net force, and the resulting acceleration, at the equator (with oblateness having nearly twice as much affect).
5. Mar 7, 2006
### arildno
I never said that Galileo was wrong..
6. Mar 7, 2006
### BobG
I didn't comment on your answer because I didn't understand it. :rofl:
7. Mar 7, 2006
### Staff: Mentor
To the OP -- just think about what happens as you spin the earth faster and faster....what force is acting on the person at the equator that is different from the person at the pole? Quiz question -- how fast to you have to spin the earth to spit off the person at the equator?
8. Mar 8, 2006
### andrevdh
Think of a girl on a merry-go-round. She will move at a larger speed when she is further away from the center, since the outer regions cover a larger distance in the same time. If the merry-go-round would spin faster and faster the girl would have to hold on for dear life or she would be flung off. She would conclude that some force is pulling her outwards from the center. This force we call the centrifugal force, it is not a real force. It results from her inertia - she wants to keep on moving in a straight line and have to hold on or press up against something in order to keep on the merry-go-round. The same happens with an object on the earth. The further it is away from the rotation axis of the earth the larger the centrifugal force it experiences. This leads to a decrease in the weight of the object - it wants to fly off the earth. Since the weight is related to the gravitational acceleration and mass (which stays the same under all circumstances) we conclude that gravity was somehow reduced by this effect.
9. Mar 8, 2006
### Hootenanny
Staff Emeritus
The effect of 'centrifugal' acceleration can be quanitfied by;
$$g_{a} = g - r\omega^2\cos^2\theta$$
where $g_{a}$ is the apparent force of gravity and $\theta$ is latitude. This assumes the earth is spherical, but is a good approximation for a non-spherical earth. | 819 | 3,388 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-43 | longest | en | 0.935363 |
https://math.stackexchange.com/questions/4067216/irreducible-characters-of-a-group-extension | 1,709,342,744,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475711.57/warc/CC-MAIN-20240301225031-20240302015031-00619.warc.gz | 370,752,794 | 35,226 | # irreducible characters of a group extension
Let $$G$$ be a group and $$M$$ is a minimal normal subgroup of $$G$$. Suppose that $$M$$ is an elementary abelian $$2$$-group and $$M=C_G(M)$$, so that $$G/M$$ acts on $$M$$. The action is faithful and assume that is also irreducible. Suppose that furthermore $$G/M$$ is a non abelian simple group (for my purposes, $$G/M$$ can be $$A_5$$ or $$PSL_2(8)$$, by the way).
I want to know some information about the irreducible complex characters of $$G$$, for example the degree of real characters.
My approach is check all the irreducible representations of $$G/M$$ over $$GF(2)$$ (that are known), build the semidirect product of $$G/M$$ by $$M$$ on GAP and print the character table. Unfortunately, this may not be the case since the group $$M$$ may not have a complement in $$G$$. It would be possible to calculate the second cohomology group $$H^2(G/M,M)$$ with GAP, check that is trivial (if it is true) and conclude that $$M$$ has a complement in $$G$$ (the extension splits, namely). There is a package in GAP, cohomolo, maintained by D. Holt, that does this kind of calculation, but I'm struggling to load it (I'm having some technical problems) and I wandering if there a more theoretical approach.
Well, if you really only interested in $$G/M = A_5$$ or $${\rm PSL}(2,8)$$, then I can tell you what you want to know.
$$A_5$$ has just two nontrivial irreducible modules over $${\rm GF}(2)$$, both of dimension $$4$$, and they both have trivial $$2$$-cohomology.
$${\rm PSL}(2,8)$$ has four such modules, of dimensions $$6,8$$ and $$12$$. Of these, only the one of dimension $$6$$ (which is the natural module) has a nonsplit extension, and that one is unique.
You can access all of these groups except for the split extension $$2^{12}:L_2(8)$$ from the perfect groups database in GAP.
For example, the nonsplit extension $$2^6\!\cdot\! L_2(8)$$ can be accessed by:
> gap> G := PerfectGroup( IsPermGroup,32256,2);
L2(8) N 2^6
and then you can just compute its character table directly.
• Thank you! Seems that $2\cdot L_2(8)$ and $2:L_2(8)$ have the same character table (if I used gap correctly ). Is it a coincidence? May I ask you where I can find some information about the cohomology groups $H^2(G,V)$ that you kindly wrote? The representation of degree $12$ can be built if the matrix representations of generetors are given: brauer.maths.qmul.ac.uk/Atlas/lin/L28 Mar 19, 2021 at 10:00
• There is no nonsplit extension $2 \cdot L_2(8)$. Mar 19, 2021 at 10:55
• sorry I meant $2^6\cdot L_2(8)$ and $2^6: L_2(8)$, it was a mistake Mar 19, 2021 at 10:57
• Yes, there are $8$ equivalence classes of extensions, but they give rise to only two isomorphism classes of groups, one of which is the split extension, corresponding to the zero element of the cohomology group. The groups corresponding to the other seven extensions are all isomorphic. In this example that is easy to see, because the endomorphism ring of the module is cyclic of order $7$. Mar 19, 2021 at 16:35
• I was also going to say that in GAP 4.11 there is new function $\mathtt{TwoCohomologyGeneric}$, which you could use to calcualte $H^2(G,M)$, but it might be slower than the cohomolo functions. Mar 19, 2021 at 16:37 | 932 | 3,252 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.903631 |
https://community.smartsheet.com/discussion/comment/332542 | 1,726,511,572,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00299.warc.gz | 161,237,061 | 109,407 | # Taking Weekly Average Every Friday.
✭✭✭
edited 06/09/22
Hi all!
I have a spreadsheet, on this sheet I have some values I would like the average of as well as dates assigned to these values. Taking the average is not the only thing I need to do, I need to take the average of these values, grouped by week, from Friday to Thursday. Meaning I have to find a way to group the values by week given just their dates, then I must further modify it so the 7 days recorded are from Friday to Thursday, I THEN need to return the average of those values. If anyone could suggest to me any kind of way I can do this, I would really appreciate it, any help is useful. I will provide some mock data below, as well as some mock results that would be ideal.
*Note: The average is the days to complete over the amount of days recorded in that week, shown as the count.
Tags:
• ✭✭✭✭✭✭
I see the issue. It looks like the structure has changed.
You now need to use this:
=AVG(COLLECT([Days to Complete]:[Days to Complete], [Week Of]:[Week Of], @cell = [Week Of]@row))
• ✭✭✭✭✭✭
Try an AVG/COLLECT combo.
=AVG(COLLECT({Days To Complete}, {Dates}, AND(@cell >= [Week of]@row - 7, @cell <= [Week of]@row)))
NOTE: The above is assuming the "Week of" column dates are the ending Friday.
• ✭✭✭
edited 06/10/22
Hi @Paul Newcome Thank you much for your response! I am currently trying this formula on one sheet (both the dates and the days to complete are on the same sheet), so the formula you wrote is returning and "unparseable" error for me. I have tried to modify it it to the following:
=AVG(COLLECT([Days to Complete]:[Days to Complete], [Date Completed]:[Date Completed], AND(@cell >= [Week Of]@row - 7, @cell <= [Week Of]@row)))
but now I am getting a "divide by 0" error.
Here is a screen shot of the dummy data I am using, please ignore the irrelevant columns, I have been trying out other solutions in the meantime.
Thank you again, this is a lot of help!!!
***Note: In the end the solution will be implemented cross referencing sheets, as in the final solution will grab the "dates" and "days to complete" from another sheet and display the averages on a separate sheet, however before implementing the actual solution I am just testing it on the same sheet.
• ✭✭✭✭✭✭
Make sure the Date Completed column is set as a date type column and make sure that the entries are actually dates and not just text values that look like dates.
• ✭✭✭
edited 06/10/22
@Paul Newcome I have checked and the "Date Completed" column is a date column and all values are indeed dates. To give an update on my progress thus far, I have been able to now assign a "Week Of" to every date and value, now I am currently trying to compute the average based on the "Week Of" value. Basically I want to take the average of all values with corresponding "Week Of" values, but I'm in a little bit of a stump as to how to do this w/o the use of variables. Let me show you the updated progress as well as the formula I made to assign a "Week Of" value to rows.
So now, I am stuck with this formula, I have the span to take the average, as well as the span for the criterion, where I'm stuck is what the syntax of the criterion, I don't know how to group them by identical "Week Of" values. Thank you again for all the help Paul, what you sent gave me a lot of insights and ideas as to how to approach this problem!
***Edit: Here is my Week Of formula as well as my average formula thus far:
Week Of:
=IF((IF(WEEKDAY([Date Completed]@row) = 6, WEEKNUMBER([Date Completed]@row) + 1, WEEKNUMBER([Date Completed]@row))) > 19, DATE(2020, 12, 30) + ((IF(WEEKDAY([Date Completed]@row) = 6, WEEKNUMBER([Date Completed]@row) + 1, WEEKNUMBER([Date Completed]@row))) * 7 - 4), DATE(2022, 1, 4) + (((IF(WEEKDAY([Date Completed]@row) = 6, WEEKNUMBER([Date Completed]@row) + 1, WEEKNUMBER([Date Completed]@row))) - 1) * 7 - 4))
Average:
=AVG(COLLECT([Days to Complete]:[Days to Complete], [Week Of]:[Week Of], AND()))
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I see the issue. It looks like the structure has changed.
You now need to use this:
=AVG(COLLECT([Days to Complete]:[Days to Complete], [Week Of]:[Week Of], @cell = [Week Of]@row))
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@Paul Newcome Thank you so much, this solution worked perfectly!
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Happy to help. 👍️
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