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Train Humanoid Walker - MATLAB & Simulink - MathWorks Italia
Humanoid Walker Model
Humanoid Walker Training
This example shows how to model a humanoid robot using Simscape Multibody™ and train it using either a genetic algorithm (which requires a Global Optimization Toolbox license) or reinforcement learning (which requires Deep Learning Toolbox™ and Reinforcement Learning Toolbox™ licenses).
This example is based on a humanoid robot model. You can open the model by entering sm_import_humanoid_urdf in the MATLAB® command prompt. Each leg of the robot has torque-actuated revolute joints in the frontal hip, knee, and ankle. Each arm has two passive revolute joints in the frontal and sagittal shoulder. During the simulation, the model senses the contact forces, position and orientation of the torso, joint states, and forward position. The figure shows the Simscape Multibody model on different levels.
The model uses Spatial Contact Force blocks to simulate the contact between the feet and ground. To simplify the contact and speed up the simulation, red spheres are used to represent the bottoms of the robotic feet. For more details, see Use Contact Proxies to Simulate Contact.
The model uses a stiffness-based feedback controller to control each joint [1]. Model the joints as first-order systems with an associated stiffness (K) and damping (B), which you can set to make the joint behavior critically damped. The torque is applied when the setpoint
{\theta }_{0}
differs from the current joint position
\theta
\mathit{T}=\mathit{B}\stackrel{•}{\theta }+\mathit{K}\left({\theta }_{0}-\theta \right)
You can vary the spring set-point
{\theta }_{0}
to elicit a feedback response to move the joint. The figure shows the Simulink model of the controller.
The goal of this example is to train a humanoid robot to walk, and you can use various methods to train the robot. The example shows the genetic algorithm and reinforcement learning methods.
The Walking Objective Function
This example uses an objective function to evaluate different walking styles. The model gives a reward (
{\mathit{r}}_{\mathit{t}}
) at each timestep:
{\mathit{r}}_{\mathit{t}}={\mathit{w}}_{1\text{\hspace{0.17em}}}{\mathit{v}}_{\mathit{y}\text{\hspace{0.17em}}}+{\mathit{w}}_{2}{\mathit{t}}_{\mathit{s}}-{\mathit{w}}_{3\text{\hspace{0.17em}}}\mathit{p}-{\mathit{w}}_{4\text{\hspace{0.17em}}}\Delta \mathit{z}-{\mathit{w}}_{5\text{\hspace{0.17em}}}\Delta \mathit{x}
{\mathit{v}}_{\mathit{y}}
— Forward velocity (rewarded)
\mathit{p}
— Power consumption (penalized)
\Delta \mathit{z}
— Vertical displacement (penalized)
\Delta \mathit{x}
— Lateral displacement (penalized)
{\mathit{w}}_{1,...,5}
: Weights, which represent the relative importance of each term in the reward function
Additionally, not falling over is rewarded.
Consequently, the total reward (
\mathit{R}
) for a walking trial is:
\mathit{R}=\sum _{\mathit{t}=0}^{\mathit{T}}{\mathit{r}}_{\mathit{t}\text{\hspace{0.17em}}}
\mathit{T}
is the time at which the simulation terminates. You can change the reward weights in the sm_humanoid_walker_rl_parameters script. The simulation terminates when the simulation time is reached or the robot falls. Falling is defined as:
The robot drops below 0.5 m.
The robot moves laterally by more than 1 m.
The robot torso rotates by more than 30 degrees.
Train with Genetic Algorithm
To optimize the walking of the robot, you can use a genetic algorithm. A genetic algorithm solves optimization problems based on a natural selection process that mimics biological evolution. Genetic algorithms are especially suited to problems when the objective function is discontinuous, nondifferentiable, stochastic, or highly nonlinear. For more information, see ga (Global Optimization Toolbox).
The model sets the angular demand for each joint to a repeating pattern that is analogous to the central pattern generators seen in nature [2]. The repeating pattern yields an open-loop controller. The periodicity of the signals is the gait period, which is the time taken to complete one full step. During each gait period, the signal switches between different angular demand values. Ideally, the humanoid robot walks symmetrically, and the control pattern for each joint in the right leg is transmitted to the corresponding joint in the left leg, with a delay of half a gait period. The pattern generator aims to determine the optimal control pattern for each joint and to maximize the walking objective function.
To train the robot with a genetic algorithm, open the sm_humanoid_walker_ga_train script. By default, this example uses a pretrained humanoid walker. To train the humanoid walker, set trainWalker to true.
Train with Reinforcement Learning
Alternatively, you can also train the robot using a deep deterministic policy gradient (DDPG) reinforcement learning agent. A DDPG agent is an actor-critic reinforcement learning agent that computes an optimal policy that maximizes the long-term reward. DDPG agents can be used in systems with continuous actions and states. For details about DDPG agents, see rlDDPGAgent (Reinforcement Learning Toolbox).
To train the robot with reinforcement learning, open the sm_humanoid_walker_rl_train script. By default, this example uses a pretrained humanoid walker. To train the humanoid walker, set trainWalker to true.
[1] Kalveram, Karl T., Thomas Schinauer, Steffen Beirle, Stefanie Richter, and Petra Jansen-Osmann. “Threading Neural Feedforward into a Mechanical Spring: How Biology Exploits Physics in Limb Control.” Biological Cybernetics 92, no. 4 (April 2005): 229–40. https://doi.org/10.1007/s00422-005-0542-6.
[2] Jiang Shan, Cheng Junshi, and Chen Jiapin. “Design of Central Pattern Generator for Humanoid Robot Walking Based on Multi-Objective GA.” In Proceedings. 2000 IEEE/RSJ International Conference on Intelligent Robots and Systems (IROS 2000) (Cat. No.00CH37113), 3: 1930–35. Takamatsu, Japan: IEEE, 2000. https://doi.org/10.1109/IROS.2000.895253.
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Class 9 Pearson - Statistics
Class 9 Pearson - StatisticsContact Number: 9667591930 / 8527521718
The mean of first n natural numbers is
\frac{5n}{9}.
Find n.
Mean of a certain number of observations is m. If each observation is divided by x (x ≠ 0) and increased by y, then the mean of new observations is
(1) mx + y
\frac{mx+y}{x}
\frac{m+xy}{x}
(4) m + xy
The mode of the observations 2x + 3, 3x – 2, 4x + 3, x – 1, 3x – 1, 5x + 2 (x is a positive integer) can be
The median of 21 observations is 18. If two observations 15 and 24 are included to the observations, then the median of the new series is
If the quartile deviation of a set of observations is 10 and the third quartile is 35, then the first quartile is
The upper class limit of inclusive type class interval 10–20 is ______.
The semi-inter quartile range of the observations 9, 12, 14, 6, 23, 36, 20, 7, 42 and 32 is
Find the mode of the following discrete series.
The mean deviation of a3 + b3 and a3 – b3 (where a and b > 0) is _________.
(4) 2b3
If the arithmetic mean of the observation x1, x2, x3, ... xn, is 1, then the arithmetic mean of
\frac{{x}_{1}}{k},\frac{{x}_{2}}{k},\frac{{x}_{3}}{k}, ...,\frac{{x}_{n}}{k}\left(k>0\right)
(1) greater than 1.
(3) equal to 1.
Range of 14, 12, 17, 18, 16 and x is 20. Find x (x > 0).
The mean of a set of observation is a. If each observation is multiplied by b and each product is decreased by c. then the mean of new set of observations is ______.
\frac{a}{b}+c
(2) ab – c
\frac{a}{b}-c
(4) ab + c
The mean deviation of first 8 composite numbers is _________.
The the highest score of certain data exceeds its lowest score by 16 and coefficient of range is
\frac{1}{3}.
Find the sum of the highest score and the lowest score.
Find the mean deviation (approximately) about the mode for the following ungrouped data: 20, 25, 30, 18, 15, 40.
\frac{{n}^{2}}{81}.
The arithmetic mean of 12 observations is 15. If two observations 20 and 25 are removed then the arithmetic mean of remaining observations is
The arithmetic mean and mode of a data is 24 and 12 respectively, then the median of the data is _______.
The inter-quartile range of the observations 3, 5, 9, 11, 13, 18, 23, 25, 32 and 39 is
Find the mean deviation from die mode for the following ungrouped data: 2.5, 6.5, 7.3, 12.3, 16.2.
The mean of the following distribution is 5, then find the value of b.
The mean deviation of
\frac{a+b}{2}
\frac{a-b}{2}
(where a and b > 0) is ________.
\frac{b}{2}
\frac{a}{2}
If the mean of x + 2, 2x + 3, 3x + 4 and 4x + 5 is x + 2, then find the value of x.
The range of 15, 14, x, 25, 30, 35 is 23. Find the least possible value of x.
In the following table, pass percentage of three schools from the year 2001 to the year 2006 are given. Which school students' performance is more consistent?
The median of the following discrete series is
Which of the following does not change for the observations 23, 50, 27, 2x, 48, 59, 72, 89, 5x, 100, 120, when x lies between 15 and 20?
(1) Arithmetic mean
If mean of the following distribution is 13, then the value of p is
If the arithmetic mean of the following distribution is 8.2, then find the value of p.
The median of the series 8, 12,15, 7, x, 19 and 22 lies in the interval.
(2) [7, 15]
(4) [9. 12]
The mode of the following distribution is
The median of the following frequency distribution is
Find the quartile deviation of the following discrete series.
The given figure represents the percentage of marks on X-axis and the number of students on Y-axis.
Find the number of students who scored less than or equal to 50% of marks.
Find the number of students who scored greater than or equal to 90% of marks.
In a class of 15 students, on an average, each student got 12 books. If exactly two students received same number of books, and the average of books received by remaining students be an integer, then which of the following could be the number of books received by each of the two students who received same number of books?
Find the mean deviation (approximately) about the median for the above data.
Find the mean deviation (approximately) about the mean for the following.
If the average mark of 15 students is 60 and the average mark of another 10 students is 70, then find the average mark of 25 students.
(A) Average marks of 25 students
\text{=}\frac{1600}{25}=64
(B) The total marks of 15 students = 15 × 60 = 900
The total marks of 10 students = 10 × 70 = 700
(C) The total marks of 25 students = 900 + 700 = 1600
In a class of 25 boys and 20 girls, the mean weight of the boys is 40 kg and the mean weight of the girls is 35 kg. Find the mean weight of the class.
(A) The total weight of 25 boys = 25 × 40 = 1000 kg
The total weight of 20 girls = 20 × 35 = 700 kg
(B) The mean weight of the class
=\frac{\text{1700}}{\text{45}}=\text{37}\frac{\text{7}}{\text{9}}\text{kg}
(C) The total weight of 45 students = 1000 kg + 700 kg- 1700 kg
If p < q < 2p; the median and mean of p, q and 2p are 36 and 31 respectively, then find the mean of p and q.
If x < y < 2x; the median and the mean of x, y and 2x are 27 and 33 respectively, then find the mean of x and y.
The mean of a set of 12 observations is 10 and another set of 8 observations is 12. The mean of combined set is ________.
A class of 40 students is divided into four groups named as A, B, C and D. Group-wise percentage of marks scored by them are given below in the table.
By using the coefficient of range find which of the group has shown good performance.
Life (in hour) of 10 bulbs from each of four different companies A, B, C and D are given below in the table.
By using the coefficient of range find which company has shown the best consistency in its quality?
If the mode of the observations 5, 4, 4, 3, 5, x, 3, 4, 3, 5, 4, 3 and 5 is 3, then find the median of the observations.
In a colony, the average age of the boys is 14 years and the average age of the girls is 17 years. If the average age of the children in the colony is 15 years, find the ratio of number of boys to that of girls,
In a class of 20 students, 10 boys brought 11 books each and 6 girls brought 13 books each. Remaining students brought atleast one book each and no two students brought the same number of books. If the average number of books brought in the class is a positive integer, then what could be the total number of books brought by the remaining students?
The mean of a set of 20 observations is 8 and another set of 30 observations is 10. The mean of combined set is __________.
Find the approximate value of mean deviation about the mode of the following data.
The mean of the following distribution is 4. Find the value of q.
Find the mean deviation about the median for the following data.
0.\overline{7}
1.\overline{3}
Find the arithmetic mean of the observations x + 5, x + 6, x + 10, x + 11, x + 14, x + 20 (where x is any real number).
Find the mean of the following continuous distribution.
Which of the following is not changed for the observations 31, 48, 50, 60, 25, 8, 3x, 26, 32?
(where x lies between 10 and 15).
Ravi travelled from P to Q at 12 kmph. He then travelled from Q to R at 32 kmph and then from R to S at 10.8 kmph. 2PQ = 3QR = 4RS. Find Ravi’s average speed for his entire journey. (in kmph)
\frac{64}{5}
\frac{72}{5}
\frac{54}{5}
\frac{66}{5}
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Post Video Test - Some Basic Concepts of Chemistry
Post Video Test - Some Basic Concepts of ChemistryContact Number: 9667591930 / 8527521718
{\mathrm{NH}}_{4}{\mathrm{NO}}_{3}
{\left({\mathrm{NH}}_{4}\right)}_{2}HP{O}_{4}
contain 30.40 % mass per cent of nitrogen. What is the mass ratio of the two components in the mixture ?
Calculate the number of millilitres of
{\mathrm{NH}}_{3}\left(\mathrm{aq}\right)
solution (d = 0.986 g/mL) contain 2.5 % by mass NH3, which will be required to precipitate iron as
\mathrm{Fe}{\left(\mathrm{OH}\right)}_{3}
in a 0.8 g sample that contains 50 %
{\mathrm{Fe}}_{2}{\mathrm{O}}_{3}
1. 0.344 mL
3. 17.24 mL
In the preparation of iron from haematite
\left({\mathrm{Fe}}_{2}{\mathrm{O}}_{3}\right)
by the reduction with carbon
{\mathrm{Fe}}_{2}{\mathrm{O}}_{3}+\mathrm{C}\to \mathrm{Fe}+{\mathrm{CO}}_{2}
. How much 80 % pure iron may be produced from 120 kg of 90 % pure
\left({\mathrm{Fe}}_{2}{\mathrm{O}}_{3}\right)
A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of 12.5 % by mass. 2.8 g of the mineral on heating lost 1.32 g of
{\mathrm{CO}}_{2}
. What is the % by mass of the other metal ?
6.2 g of a sample containing
{\mathrm{Na}}_{2}{\mathrm{CO}}_{3}, {\mathrm{NaHCO}}_{3}
and non-volatile inert impurity on gentle heating loses 5 % of its mass due to reaction
2{\mathrm{NaHCO}}_{3}\to {\mathrm{Na}}_{2}{\mathrm{CO}}_{3}+{\mathrm{H}}_{2}\mathrm{O}+{\mathrm{CO}}_{2}
. Residue if dissolved in water and formed 100 mL solution and its 10 mL portion requires 7.5 mL of 0.2 M aqueous solution of
{\mathrm{BaCl}}_{2}
for complete precipitation of carbonates.
Determine mass (in gram) of
{\mathrm{Na}}_{2}{\mathrm{CO}}_{3}
in the original sample
1 M NaOH solution was slowly added into 1000 mL of 183.75 g impure
{\mathrm{H}}_{2}{\mathrm{SO}}_{4}
solution and the following plot was obtained. The percentage purity of
{\mathrm{H}}_{2}{\mathrm{SO}}_{4}
sample and slope of the curve respectively are :
1. 75 %,
-\frac{1}{3}
2. 80%,
-\frac{1}{2}
3. 80 %, -1
{\mathrm{MnO}}_{2}
on ignition converts into
{\mathrm{Mn}}_{3}{\mathrm{O}}_{4}
. A sample of pyrolusite having 75 %
{\mathrm{MnO}}_{2}
, 20 % inert impurities and rest water is ignited in air to constant mass. What is the percentage of Mn in the ignited sample ?
A 1.0 g sample of a pure organic compund containing chlorine is fused with
{\mathrm{Na}}_{2}{\mathrm{O}}_{2}
to convert chlorine to NaCl. The sample is then dissolved in water, and the chlorine precipitated with
{\mathrm{AgNO}}_{3}
giving 1.96 g of AgCl. If the molecular mass of organic compound is 147, how many chlorine atoms does each molecule contain ?
A 0.60 g sample consisting of only
{\mathrm{CaC}}_{2}{\mathrm{O}}_{4}
{\mathrm{MgC}}_{2}{\mathrm{O}}_{4}
is heated at 500
°
C, converting the two salts of
{\mathrm{CaCO}}_{3} \mathrm{and} {\mathrm{MgCO}}_{3}
. The sample then weighs 0.465 g. If the sample had been heating to 900
°
C, where the products are CaO and MgO. What would the mixtures of oxides have weighed ?
A metal M forms the sulphate
{\mathrm{M}}_{2}{\left({\mathrm{SO}}_{4}\right)}_{3}
. A 0.596 gram sample of the sulphate reacts with excess
{\mathrm{BaCl}}_{2}
to give 1.220 g
{\mathrm{BaSO}}_{4}
. What is the atomic mass of M
(Atomic mass : S = 32, Ba = 137.3)
{\mathrm{SO}}_{2}{\mathrm{Cl}}_{2}
(sulphuryl chloride) reacts with water to given a mixture of
{\mathrm{H}}_{2}{\mathrm{SO}}_{4}
and HCl. What volume of 0.2 M
\mathrm{Ba}{\left(\mathrm{OH}\right)}_{2}
is need to completely neutralize 25 mL of 0.2 M
{\mathrm{SO}}_{2}{\mathrm{Cl}}_{2}
5 g sample contain only
{\mathrm{Na}}_{2}{\mathrm{CO}}_{3}
{\mathrm{Na}}_{2}{\mathrm{SO}}_{4}
. This sample is dissolved and the volume made up to 250 mL. 25 mL of this solution neutralizes 20 mL of 0.1 M
{\mathrm{H}}_{2}{\mathrm{SO}}_{4}
. Calculate the % of
{\mathrm{Na}}_{2}{\mathrm{SO}}_{4}
in the sample.
{\mathrm{cm}}^{3}
of a solution of an acid (Molar mass = 98) containing 29.4 g of the acid per litre were completely neutralized by 90.0
{\mathrm{cm}}^{3}
of aq. NaOH containing 20 g of NaOH per 500
{\mathrm{cm}}^{3}
. The basicity of the acid is :
4. data sufficient
A 150 mL of solution of
{\mathrm{I}}_{2}
is divided into two unequal parts. I part reacts with hypo solution in acidic medium. 15 mL of 0.4 M hypo was consumed. II part was added with 100 mL of 0.3 M NaOH solution. Residual base required 10 mL of 0.3 M
{\mathrm{H}}_{2}{\mathrm{SO}}_{4}
solution for complete neutralization. What was the initial concentration of
{\mathrm{I}}_{2}
The concentration of an oxalic acid solution is 'x' mol litre-1, 40 mL of this solution reacts with 16 mL of 0.05 M acidified
{\mathrm{KMnO}}_{4}
. What iss the pH of 'x' M oxalic acid solution ? (Assume that oxalic acid dissociates completely)
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Calculate power gain from two-port S-parameters - MATLAB powergain - MathWorks France
{G}_{t}=\frac{{P}_{L}}{{P}_{\text{avs}}}=\frac{\left(1-{|{\Gamma }_{S}|}^{2}\right){|{S}_{21}|}^{2}\left(1-{|{\Gamma }_{L}|}^{2}\right)}{{|\left(1-{S}_{11}{\Gamma }_{S}\right)\left(1-{S}_{22}{\Gamma }_{L}\right)-{S}_{12}{S}_{21}{\Gamma }_{S}{\Gamma }_{L}|}^{2}}
\begin{array}{l}{\Gamma }_{S}=\frac{{Z}_{S}-{Z}_{0}}{{Z}_{S}+{Z}_{0}}\\ {\Gamma }_{L}=\frac{{Z}_{L}-{Z}_{0}}{{Z}_{L}+{Z}_{0}}\end{array}
{G}_{a}=\frac{{P}_{\text{avn}}}{{P}_{\text{avs}}}=\frac{\left(1-{|{\Gamma }_{S}|}^{2}\right){|{S}_{21}|}^{2}}{{|1-{S}_{11}{\Gamma }_{S}|}^{2}\left(1-{|{\Gamma }_{out}|}^{2}\right)}
{\Gamma }_{\text{out}}={S}_{22}+\frac{{S}_{12}{S}_{21}{\Gamma }_{S}}{1-{S}_{11}{\Gamma }_{S}}
{G}_{p}=\frac{{P}_{L}}{{P}_{\text{in}}}=\frac{{|{S}_{21}|}^{2}\left(1-{|{\Gamma }_{L}|}^{2}\right)}{\left(1-{|{\Gamma }_{\text{in}}|}^{2}\right){|1-{S}_{22}{\Gamma }_{L}|}^{2}}
{\Gamma }_{\text{in}}={S}_{11}+\frac{{S}_{12}{S}_{21}{\Gamma }_{L}}{1-{S}_{22}{\Gamma }_{L}}
{G}_{\text{mag}}=\frac{|{S}_{21}|}{|{S}_{12}|}\left(K-\sqrt{{K}^{2}-1}\right)
{G}_{msg}=\frac{|{S}_{21}|}{|{S}_{12}|}
{\mathit{G}}_{\mathrm{msg}}
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Research | Flashbots
Research from the Flashbots team.
Speeding up the EVM (part 1)
Thanks to Alejo Salles, Hongbo Zhang, Alex Obadia, and Kushal Babel for feedback and review of this post.
With a more performant Ethereum Virtual Machine (EVM), we can achieve better network scalability and more efficient maximal-extractable-value (MEV) extraction. This series of posts analyzes several approaches to speed up the EVM with a focus on parallelization and shared data conflict analysis.
MEV-Boost -- Merge ready Flashbots Architecture
\textrm{MEV}
\textrm{MEV}
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Class 10 - Science - Our Environment
Class 10 - Science - Our EnvironmentContact Number: 9667591930 / 8527521718
Why are some substances biodegradable and and some non-biodegradable?
What are tropic levels? Give an example of a food chain and state the different tropic level in it.
1. Will the impact of removing all the organisms in a trophic level be different for different trophic levels?
2. Can the organisms of any trophic level be removed without causing any damage to the eco system?
What is biological magnification ? will the levels of this magnification be different at different levels of the ecosystem?
If all the waste we generate is biodegradable, will this have no imoact on the environment?
Why is damage to the ozone layer a cause fo concern? What steps are being taken to limit this damage?
Now a days, our government is stressing upon the use of jute or paper bags instead of plastic bag. What purpose is supposed to be acheived by the government?
Which of the following materials are non-biodegradable in each of the following?
1. Paper, leather, nylon, egg shell, glass
2. Tea leaves, ,glass, glucose, cotton cloth, silver foil
3. Glass, ,glucose, leather, silver foil, nylon cloth
State one diffrence between autotrophs and heterotrophs.
Write the appropriates names of the trophic levels z and x in the figure given below:
Draw a food cahin most likely to be a part of a forest ecosystem.
Rearrange the following according to their ascending trophic levels in a food chain. Hawk, grass, snake, frog, grasshopper.
Rearrange the following according to their trophic levels in a food chain.
Fish, zooplankton, seal, phytoplankton.
Which of the following belongs to the first trophic level?
grasshopper, rose plant, cockroach, vulture, neem plant.
Which one of the following is always at the third trophic level in a food chain?
In a food chain comprising lion, grass and deer, which will
1. transfer the maximum amount of energy.
2. receive the minimum amount of energy.
Among all four types of animals, i.e. carnivores, decomposers, herbivores and producers, how does energy flow in an ecosystem, occur through these organisms.
In the following food chain, 20 J of energy was available to the hawks. How much would have been present in the plants?
\to
\to
\to
If a harmful chemical enetrs in food chain comprising snakes, peacock, mice and plants, which of these organisms is likely to have the maximum concentration of this chemical in its body?
Mention the amount of solar energy captured by the green plants out of the total energy that reaches on the earth from the sun and mentioned the percentage of energy, which is transferred from one trophic level to next higher one.
"The level of energy reaching one to next higher trophic level decreases". Suggest a reason for this occurence.
If the ozone layer disappears completely, what will be the consequences?
Name the group of chemical compounds which adversely affects the ozone layer.
Among the substance given below, select the ones which have posed a threat to the environment.
Aerosol, consumers, bacteria, CFCs
Why did Unites Nations act to control the production of CFCs used in refrigerators?
To keep the environment clean, garbage disposal programme is an essential prerequisite. What method constitute the garbage disposal programme?
List two methods of safe disposal of the non-biodegradable waste.
List two items which can be easily recycled, but we generally throw them in the dustbins.
A lake has been polluted by sewage. On comparison with the sample of unpolluted water the water in the lake is found to have increased contents of some components. Identify these components.
Why should biodegradable and non-biodegradable wastes be discarded in two seperate dustbins?
Write one negative effect, of affluent life style of few persons of a society on the environment.
What si meant by non-biodegradable waste? Identify biodegradable waste from the following.
Empty packet of chips, empty plastic bottle of mineral water, empty paper box of sweets, empty tin of cold drink.
Write the harmful effect of using plastic bags, on the environment. Suggest alternatives to plastic bags.
List two reasons to show that the existence of decomposers is essential in an ecosystem.
What would happen, if all the microorganisms are removed from the environment?
Microorganisms are often reffered to as the 'scavengers of the environment'. Explain.
In an ecosystem, how the biotic and abiotic components are dependent on each other?
Write the food chain operating in a freshwater pond. Mention the food habit of each trophic level in this food chain.
Give an example of a food chain consisting of four organisms at different trophic levels. Give the scientific terms used to indicate the first and third trophic level.
We already know that a food chain contains different organisms at different trophic level in a typical ecosystem.
In the diagram (of energy flow in an ecosystem) given below identify the secondary consumers and explain your choice.
Which of the following belong to the first trophic level?
1. Grasshopper, mango tree, hawk, snake
2. Sunflower plant, grasshopper, cockroach, banyan tree.
Which of the following belong to the second trophic level?
1. frog, butterfly, spider, rice weevil
2. parrot, frog, butterfly, spider
State the 10% law of energy transfer.
In the given figure, the various trophic levels are shown in a pyramid. At which trophic level is maximum energy available?
Mention the form of energy transfer, if a grasshopper is eaten by a frog.
A good web is given below, observe the figure and answer the questions given below:
1. Identify the primary consumer in the food web.
2. If all the foxes are killed due to a disease, what will your observations about food web be?
Explain an agriculture practice that has a harmful effect on ecosystem.
Give some methods that could be applied to reduce our intake of pesticides.
Give some examples along with reasons of insecticides that are banned due to their fatal effects.
Describe how ozone present in the atmosphere is important for sustaining life on earth.
Mention the harmful effects of UV radiations. Name the gas which protects us from it.
Why is improper disposal of waste a curse to environment.
Carnivores cannot be self dependent and have to depend on herbivores. Explain
Mention the ecologically amplified elements that lead to the decline in population of predator birds.
We often observe domestic waste decomposing in the bylanes of residential colonies. Suggest ways to make people realise that the improper disposal of waste is harmful to the environment.
Insecticides are used in crop fields to kill any current or recurring infections. In a crop field an insecticide is sprayed to kill the caterpillars. A food web involving caterpillar is given below.
Identify the organism from the food web which will have the highest level of insecticides accumulated in it after some time.
News paper reports about the alarming increase in pesticides level in packed food items have appeared. As a result some of the states have banned these food items.
1. What are the sources of pesticides in the food items?
2. Name the biological phenomenon associated with accumulation of pesticides in the food chain.
Why is government of India imposing a ban on the use of polythene bags? Suggest two alternatives to these bags and explain how this ban is likely to improve the environment.
Why are crop fields known as artificial ecosystem?
We do not clean ponds or lakes, but an aquarium needs to be cleaned, why?
State 10% law. Explain with an example how energy flows through different trophic levels.
Indicate the flow of energy in an ecosystem.Why is it unidirectional?
With the help of an example, involving four organisms, describe how energy flows through different trophic levels.
'Vegetarian food habits can sustain a larger number of people'. Justify the statement in terms of food chain.
Give reason to justify the following
1. The existence of decomposers is essential in a biosphere.
2. Flow of energy in a food chain is unidirectional.
Correct the explanation given for the terms if wrong.
1. Biomagnification - Decrease of chemical at the successive trophic levels of a food chain.
2. Ecosystem- Biotic components of environment.
3. Aquarium- Natural ecosystem.
Select the mis matched pair in the following and correct it.
1. Biomagnification Accumulation of chemicals at the successive trophic levels of a food chain.
2. Ecosystem Biotic components of environment.
3. Aquarium A man-made ecosystem.
4. Parasites Oganisms which obtain from other living organisms.
A modern insecticides has been introduced with certain new properties like, accumulation in the bodies of predators, broken down by soil bacteria, easily washed into lakes and rivers and taken up by plant roots. Among all these properties which one will help in reducing or keeping the level of environment pollution to lowest.
Explain how pesticides get accumulated in the environment.
Our food grains, ,such as wheat and rice, the vegetables and fruits and even meat are found to contain varying amounts of pesticides residue. State the reason to explain how and why it happens.
Number of vultures is decreasing remarkably in recent years which has become a matter of concern.
1. Vultures belong to which category of animal?
2. What is their role in nature regarding the maintenance of ecological balance?
3. Give the position the vultures occupy in a food chain.
Name the radiations absorbed by ozone layer. Give any one cause of the depletion of the ozone layer. Name a disease likely to be caused due to depletion.
1. What harm is caused to skin by ultraviolet rays?
2. At which level, pesticides enter the food chain?
3. How are most of the solid wastes in urban areas disposed of?
What is garbage? What does garbage consists of? What is disposal of waste?
How would you dispose the following waste
1. Domestic wastes like vegetables peel?
2. Industrial wastes like metallic cans?
3. Plastic material?
Draw a sequence of suitable methods of disposal of waste produced at your home to minimize environmental pollution?
It is the responsibilty of the government to arrange for the management and disposal of waste. As an individual you have no role to play. Do you agree? Support your answers with two reasons.
1. How do food chains get shortened? How does the shortening of food chain affect the biosphere?
2. How will you justify that vegetarian food habits give us more calories?
1. Energy flow in a food chain is unidirectional. Justify this statement.
2. Explain how the pesticides enter a food chain and subsequently get into our body.
Suggest suitable mechanism (s) for waste management in fertilisers industries.
What are byproducts of fertiliser industries? How do they affect the environment?
Write in detail about garbage management.
Suggest any five activities in daily life, which are eco-friendly.
One day Mohan found his neighbours burning plastic wastes in open space near to his house. He explained three method to save the environment from plastic waste to them. Imagine yourself in the place of Mohan and suggest the methods that Mohan might have told to his neighbours. What value was exhibited by Mohan in this situation?
Naman got into a quarrel with some farmers who were spraying DDT in their fields. Many people gathered at the spot to see and enjoy the incident. The angry mob demanded that Naman should not interfere in the farmers' job. Naman tried to explain his point and finally succeeded,the farmers gave up spraying DDT.
The activities of a man had adverse effects on all forms of living organisms in the biosohere. Unlimited exploitation of nature by man disturbed the delicate ecological balance between the living and non-living componenets of the biosphere. The unfavourable conditions created by man himself threatend the survival not only of himself but also of the entire living organisms on the mother earth. One of your classmates is an active member of 'Eco Club' of your school, which is creating environmental awareness amongst the school students, spreading the same in the society and also working hard for preventing environmental degradation of the surroundings.
1. Why is it necessary to conserve our environment?
2. State the importance of green and blue dustbins in the safe disposal of the household waste?
3. List two values exhibited by your classmates who is an active member of Eco-club of your school.
World environment day is celebrated on
{5}^{\mathrm{th}}
June to raise global awareness to protect nature and the planet Earth. It is run by the United Nations Environment Programme (UNEP).
The 'Paper of India' uses recycled paper to commemorate this day.
1. How does the use of recycled paper help in protecting the environment?
2. Mention any two values portrayed by the 'Paper of India' group?
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Undulator - Simple English Wikipedia, the free encyclopedia
Working of the undulator. 1: magnets, 2: electron beam, 3: synchrotron radiation
A multipole wiggler, as used in the storage ring at the Australian Synchrotron to generate synchrotron radiation
An undulator is an insertion device from high-energy physics and usually part of a larger installation, a synchrotron storage ring. It consists of a periodic structure of dipole magnets. A static magnetic field is alternating along the length of the undulator with a wavelength
{\displaystyle \lambda _{u}}
. Electrons traversing the periodic magnet structure are forced to undergo oscillations. So the electrons give off energy as electronmagnetic radiation. The radiation produced in an undulator is very intense and concentrated in narrow energy bands in the spectrum. The light beam is also collimated on the orbit plane of the electrons. This radiation is guided through beamlines for experiments in various scientific areas.
The important dimensionless parameter
{\displaystyle K={\frac {eB\lambda _{u}}{2\pi \beta m_{e}c}}}
where e is the particle charge, B is the magnetic field,
{\displaystyle \beta =v/c}
{\displaystyle m_{e}}
is the electron rest mass and c is the speed of light, characterizes the nature of the electron motion. For
{\displaystyle K\ll 1}
the oscillation amplitude of the motion is small and the radiation displays interference patterns which lead to narrow energy bands. If
{\displaystyle K\gg 1}
the oscillation amplitude is bigger and the radiation contributions from each field period sum up independently, leading to a broad energy spectrum. When K is much bigger than 1, the device is no longer called an undulator; it is called a wiggler.
Physicist think about undulators both using classical physics and relativity. This means that though the precision calculation is tedious the undulator can be seen as a black box. An electron enters this box and an electromagnetic pulse exits through a small exit slit. The slit should be small enough so that only the main cone passes, so that the side lobes may be ignored.
Undulators can provide hundreds of time more magnetic flux than a simple bending magnet and as such are in high demand at synchrotron radiation facilities. For an undulator that repeat N times (N periods), the brightness can be up to
{\displaystyle N^{2}}
more than a bending magnet. The intensity is enhanced up to a factor of N at harmonic wavelengths due to the constructive interference of the fields emitted during the N radiation periods. The usual pulse is a sine wave with some envelope. The second factor of N comes from the reduction of the emission angle associated with these harmonics, which is reduced in proportion to 1/N. When the electrons come with half the period, they interfere destructively. So, the undulator stays dark. The same is true if the electrons come as a bead chain. Because the bunch of electron spreads out the more times that they travel around the synchrotron, physicists want to design new machines that throw out the electron bunches before they have a chance to spread out. This change will produced more useful synchrotron radiation.[1]
The polarization of the emitted radiation can be controlled by using permanent magnets to induce different periodic electron trajectories through the undulator. If the oscillations are confined to a plane the radiation will be linearly polarized. If the oscillation trajectory is helical, the radiation will be circularly polarized, with the handedness determined by the helix.
If the electrons follow the Poisson distribution, a partial interference leads to a linear increase in intensity. In the free-electron laser the intensity increases exponentially with the number of electrons.
Physicists measure an undulator's effectiveness in terms of spectral radiance.
The first undulator was built by Hans Motz and his coworkers at Stanford in 1953. One of their undulators produced the first ever coherent infrared radiation. Their total frequency range was from visible light down to millimeter waves. The Russian physicist V.L. Ginzburg showed that undulators could be made in principle in a 1947 paper.
↑ Donald H Bilderback, Pascal Elleaume and Edgar Weckert (2005). "Review of third and next generation synchrotron light sources" (PDF). J. Phys. B: At. Mol. Opt. Phys. 38 (9): S773–S797. doi:10.1088/0953-4075/38/9/022. Retrieved December 22, 2011. {{cite journal}}: CS1 maint: uses authors parameter (link)
D. T. Attwood's page at Berkeley: Soft X-Rays and Extreme Ultraviolet Radiation Archived 2009-05-03 at the Wayback Machine. His lecture and viewgraphs are available online.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Undulator&oldid=8068650"
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Magnetism and Matter- Live session - NEET 2020
Magnetism and Matter- Live session - NEET 2020Contact Number: 9667591930 / 8527521718
An iron rod of length L and magnetic moment M is bent in the form of a semicircle. Now its magnetic moment will be
\frac{2\mathrm{M}}{\mathrm{\pi }}
\frac{\mathrm{M}}{\mathrm{\pi }}
\mathrm{M\pi }
A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It
1. Will become rigid showing no experiment
2. Will stay in any position
3. Will stay in north-south direction only
4. Will stay in east-west direction only
The figure shows the various positions (labelled by subscripts ) of small magnetised needles P and Q. The arrows show the direction of their magnetic moment. Which configuration corresponds to the lowest potential energy among all the configurations shown
{\mathrm{PQ}}_{3}
{\mathrm{PQ}}_{4}
{\mathrm{PQ}}_{5}
{\mathrm{PQ}}_{6}
A current carrying coil is placed with its axis perpendicular to N-S direction. Let horizontal component of earth's magnetic field be
{\mathrm{H}}_{\mathrm{o}}
and magnetic field inside the loop be H. If a magnet is suspended inside the loop, it makes angle
\mathrm{\theta }
with H. Then
\mathrm{\theta }
{\mathrm{tan}}^{-1}\left(\frac{{H}_{o}}{H}\right)
{\mathrm{tan}}^{-1}\left(\frac{H}{{H}_{\mathit{o}}}\right)
\mathrm{cos}e{c}^{-1}\left(\frac{H}{{H}_{\mathit{o}}}\right)
co{t}^{-1}\left(\frac{{H}_{o}}{H}\right)
Two identical bar magnets are placed one above the other such that they are mutually perpendicular and bisect each other. The time period of this combination in a horizontal magnetic field is T. The time period of each magnet in the same field is
\sqrt{2}T
{2}^{\frac{1}{4}}\mathrm{T}
{2}^{-\frac{1}{4}}\mathrm{T}
{2}^{-\frac{1}{2}}\mathrm{T}
A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is T' , then ratio T'/T is
\frac{1}{4}
\frac{1}{2\sqrt{2}}
\frac{1}{2}
Two tangent galvanometers having coils of the same radius are connected in series. A current flowing in them produces deflections of 60
°
°
respectively. The ratio of the number of turns in the coils is
\left(\sqrt{3}+1\right)/1
\left(\sqrt{3}+1\right)/\left(\sqrt{3}-1\right)
\sqrt{3}/1
Core of electromagnets are made of ferromagnetic material which has
1. High permeability and high retentivity
2. Low permeability and low retentivity
3. High permeability and low retentivity
4. Low permeability and high retentivity
If a bar magnet of length l and cross-sectional area A is cut into two equal parts as shown in figure, then the pole strength of each pole becomes
Two identical short bar magnets, each having magnetic moment M, are placed a distance of 2d apart axes perpendicular to each in a horizontal plane. The magnetic induction at a point midway between them is
\frac{{\mathrm{\mu }}_{0}}{4\mathrm{\pi }}\left(\sqrt{2}\right)\frac{M}{{d}^{3}}
\frac{{\mathrm{\mu }}_{0}}{4\mathrm{\pi }}\left(\sqrt{3}\right)\frac{M}{{d}^{3}}
\left(\frac{2{\mathrm{\mu }}_{0}}{\mathrm{\pi }}\right)\frac{M}{{d}^{3}}
\frac{{\mathrm{\mu }}_{0}}{4\mathrm{\pi }}\left(\sqrt{5}\right)\frac{M}{{d}^{3}}
A dip circle is adjusted so that its needle moves freely in the magnetic meridian. In this position, the angle of dip is 40
°
. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of 30
°
with the magnetic meridian. In this position the needle will dip by an angle
°
°
3. More than 40
°
4. Less than 40
°
The true value of angle of dip at a place is 60
°
, the apparent dip in a plane inclined at an angle of 30
°
with magnetic meridian is
{\mathrm{tan}}^{-1}\left(\frac{1}{2}\right)
{\mathrm{tan}}^{-1}\left(2\right)
{\mathrm{tan}}^{-1}\left(\frac{2}{3}\right)
The variation of magnetic susceptibility
\left(\mathrm{Χ}\right)
with absolute temperature T for a ferromagnetic material is
The figure illustrates how B, the flux density inside a sample of unmagnetised ferromagnetic material, varies with
{\mathrm{B}}_{0}
. the magnetic flux density in which the sample is kept. For the sample to be suitable for making a permanent magnet
1. OQ should be large, OR should be small
2. OQ and OR should both be large
3. OQ should be small and OR should be large
4. OQ and OR should both be small
A magnet of length 14 cm and magnetic moment M is broken into two parts of lengths 6 cm and 8 cm. They are put at right angle to each other with opposite poles together. The magnetic moment of the combination is
3. M/1.4
A current carrying loop is placed in a uniform magnetic field in four different orientations I,II,III and IV, arrange them in the decreasing order of potential energy
1. I>III>II>IV
2. I>II>III>IV
3. I>IV>II>III
4. III>IV>I>II
A wire of length L metre carrying current i ampere is bent in the form of a circle. What is the magnitude of magnetic dipole moment ?
{\mathrm{iL}}^{2}/4\mathrm{\pi }
{\mathrm{i}}^{2}{\mathrm{L}}^{2}/4\mathrm{\pi }
{\mathrm{i}}^{2}\mathrm{L}/8\mathrm{\pi }
{\mathrm{iL}}^{2}/8\mathrm{\pi }
The magnetic susceptibility of a paramagnetic substance at -73
°
C is 0.0060, then its value of -173
°
C will be
If the B-H curves of two samples of P and Q of iron are as shown below, then which one of the following statements is correct ?
1. Both P and Q are suitable for making permanent magnet
2. P is suitable for making permanent magnet and Q for making electromagnet
3. P is suitable for making electromagnet and Q is suitable for permanent magnet
4. Both P and Q are suitable for making electromagnets
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Help:Preferences - Wikinews, the free news source
This is a copy of m:Help:Preferences using Wikinews-specific templates. Don't edit this page, except to replace the contents with the current version of that master page. To make changes applicable on all projects, edit the master page; to change Wikinews-specific content, edit the templates; to make demos work, copy templates to this project, and images to Wikimedia Commons. Please see the history of the master page for author details. Please note that all meta content is licensed under the GFDL. As a result this page is licensed under the GFDL.- Edit this template
Check the rules of your project (for the English Wikipedia, see Wikipedia:Username policy) and note that:
"Beobachtungsliste" - German Wikipedia - long - large font is hardly usable
Warning: Broken links!
{\displaystyle a^{b^{c}}}
{\displaystyle a^{b^{\,\!c}}}
Format broken links like this. An internal link to a non-existing pages is automatically a link to the edit page. By default the link label of [[b]] and [[a|b]] is "b", just like for links to existing pages. Alternatively the link label is a question mark inserted after "b", like this: The weather in London?. This helps if you have trouble seeing colors or if the colors are overridden in your browser settings. The appearance of the link is further determined by the style specified for css selectors "a.new" and "a.new:hover" (the example on the preferences page wrongly uses class="internal" for the question mark). Internal links to pages which do not yet exist currently appear on your browser like this: the weather in London. Normally, this is underlined and in red. With the trailing question mark link one can then not distinguish between a single word being linked or a phrase of more than one word, without pointing at the question mark with the cursor. Also, remember that the question mark does not mean that the information is uncertain.
Retrieved from "https://en.wikinews.org/w/index.php?title=Help:Preferences&oldid=4385137"
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Human Health & Disease - Live Session - NEET & AIIMS 2019Contact Number: 9667591930 / 8527521718
The sustained of 39 -40 C, weakness, stomach pain, Constipation, headache and loss
of apetite are the common symptoms of:
3. Amobiasis
1. was a cook by profession in United states
2. Was identified as asymptomatic carrier of pathogen associated with typhoid fever
3. Continued to spread Typhoid for several years through the food she prepared
4. is related to all above
Arrange the stages of lifecycle of Plasmodium starting from human host and ending in
l. RBC release the toxic haemozoin
ll. Parasite reproduces asexually in liver cells
lll. Parasutes attack RBC
lV. Sporozoites enter human body through the mosquito bite
V. Mature sporozoites escape from intestine and migrate to the mosquito's salivary
Vl. Fertilization in gametocytes occur in mosquito's sallvary gland
Vlll. Mosquite takes up gametocytes while sucking blood of an infected person
1. lV
\to
\to
\to
\to
\to
\to
\to
\to
\to
\to
\to
\to
\to
\to
\to
\to
\to
\to
4. lll
\to
\to
\to
\to
\to
\to
The primary and secondary host of Plasmodium is ___ and ___ respectively :
1. Human; female Anopheles
2. Female Anopheles; Human
3. male Anopheles; Human
4. Human; male Anopheles
A person is showing the following symptoms.
constipation, abdominal pain and cramps, stools with excees mucous, blood suffering
clots and without fever.
Column-ll Column-ll
A. Ascariasis (i) inflammation of Lymphatic vessels of lower limbs and genital organs
B. Filariasis (ii) lnternal bleeding muscular pain, fever, Anaemia, blockage of intestinal passage
C. Amoebiasis (iii) Dry, scaly lesions on varous body parts like skin, nail and scalp with intense itching
C. Ringworm disease (iv) Abdominal pain, stools with excess muccous and blood clots
1. A= (ii), B=(iv), C=(iii), D=(i)
2. A=(ii), B=(i), C=(iv), D=(iii)
3. A=(i), B=(ii), C=(iii), D=(iv)
4. A= (iv), B=(ii), C=(i), D=(iii)
The helminth that cause elephantiasis is:
1. Wuchereria
Aedes is the vector of viral disease (s):
Trypanosoma cause :
4. Dysentry
The cellular barrier(s) in innate immunity involves:
1. PMNL-neutrophilis
An antibody is represented as:
{}_{2}
{}_{4}
{}_{3}
{}_{3}
{}_{4}
{}_{4}
{}_{2}
{}_{2}
Major histocompatibility complex(MHC):
1. ls a set of cell surface proteins essential for acqire immune system to recognize
foreign molecules
2. Binds to antigens derived from pathogens and display on the cell surface for
recognition by the appropriate T-cells
3. Determines compatibility of donous for organ transplant as well as one's
susceptibility to an autoimmune disease via cross-reacting immunization
4. ls related to all of the above
lnterferons are :
HV/AlDS :
1. Symptoms always appear within a month of infectin
2. Can spread by mere touch or physical contact
3. Spreads when a healthy person's body fluid cones in contact with the fluid fo an infected person
4. ls related ot all of the above options
lonizing radiations like ___ (A) ___ and ___ (B) ____ and non-ionizing radiations like ____ and (c) ___ cause DNA damage leading to neoplastic transformation :
1. A= X-rays, B=Gamma-rays, C=uv-rays
2. A=X=rays, B=UV-rays, C=Gamma-rays
3. A=X-rays, B=Gamma-rays, C=lnfra-red rays
4. A= lnfra-red, B=X-rays, B=X-rays, C=UV-rays
MRI uses strong _____ and ____ to accurately to detect pathological and physiological changes in the living tissue:
1. Magnetic fields ; non-ionizing radiations
2. Magnetic fields ; ionizing radiations
3. Beam of electrons; lonizing radiations
4. Beam of electrons; non-ionizing radiations
'Smack' is :
1. Chemically Fromaline
2. Sweet to taste
3. Obtained by acetylation of morphine
4. Extracted from the latex of cannabis sativa
Marijuana, hashish, charas and ganja are obtained from :
2. Aveng sativa
3. Papaaver sominiferum
4. Phalaris
Read the following statements about a drug and identify the drug.
i. These are known for their effects on cardiovascular system of the body
ii. These interact with the receptors present principally in brain
iii. Generally taken by inhalation and oral ingestion
iv. These are obtained from a variety of hemp plants
1. Papaver sominiferum
2. Erythroxylon coca
Column-i Column-ll
A. lSD and cannabionids ` (i) Stimulant
B. Amphetamines (ii) Hallucinogen
C.Barbiturates and benzodiazepines (iii) lnterferes with fransport of dopamine
D.Cocaine (iv) Sedative
1. A=(i),(iii), B=(iv), c=(ii), D=(ii)
2. A= (i), B=(ii), c=(i),(iv), D=(iv)
3. A=(ii), B=(i), C=(i), D=(i), (iii)
4. A=(iii), B=(ii),(iii), C=(iv), D=(i)
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Laws of Motion (in English)- Live Session - NEET 2020
Laws of Motion (in English)- Live Session - NEET 2020Contact Number: 9667591930 / 8527521718
1. Rolling friction is smaller than sliding friction.
2. Limiting value of static friction is directly proportional to normal reaction.
3. Coefficient of sliding friction has dimensions of length.
4. Frictional force opposes the relative motion.
A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration a towards the right. The relation between a and θ for the block to remain stationary on the wedge is
\mathrm{a}=\frac{\mathrm{g}}{\mathrm{cosec} \mathrm{\theta }}
\mathrm{a}=\frac{\mathrm{g}}{\mathrm{sin} \mathrm{\theta }}
\mathrm{a}=\mathrm{g} \mathrm{tan\theta }
\mathrm{a}=\mathrm{g} \mathrm{cos\theta }
One end of string of length l is connected to a particle of mass ‘m’ and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed ‘v’, the net force on the particle (directed towards center) will be (T represents the tension in the string)
\mathrm{T}+\frac{{\mathrm{mv}}^{2}}{\mathrm{l}}
\mathrm{T}-\frac{{\mathrm{mv}}^{2}}{\mathrm{l}}
Suppose the charge of a proton and an electron differ slightly. One of them is –e, the other is (e + Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of [Given mass of hydrogen mh = 1.67 × 10–27 kg]
A rigid ball of mass m strikes a rigid wall at 60° and gets reflected without loss of speed as shown in the figure below. The value of impulse imparted by the wall on the ball will be
\frac{\mathrm{mV}}{2}
\frac{\mathrm{mV}}{3}
A car is negotiating a curved road of radius R. The road is banked at an angle θ. The coefficient of friction between the tyres of the car and the road is μs. The maximum safe velocity on this road is
\sqrt{\frac{\mathrm{g}}{{\mathrm{R}}^{2}}\frac{{\mathrm{\mu }}_{\mathrm{s}}+\mathrm{tan\theta }}{1-{\mathrm{\mu }}_{\mathrm{s}}\mathrm{tan\theta }}}
\sqrt{g{R}^{2}\frac{{\mathrm{\mu }}_{\mathrm{s}}+\mathrm{tan\theta }}{1-{\mathrm{\mu }}_{\mathrm{s}}\mathrm{tan\theta }}}
\sqrt{gR\frac{{\mathrm{\mu }}_{\mathrm{s}}+\mathrm{tan\theta }}{1-{\mathrm{\mu }}_{\mathrm{s}}\mathrm{tan\theta }}}
\sqrt{\frac{\mathrm{g}}{\mathrm{R}}\frac{{\mathrm{\mu }}_{\mathrm{s}}+\mathrm{tan\theta }}{1-{\mathrm{\mu }}_{\mathrm{s}}\mathrm{tan\theta }}}
Two stones of masses m and 2 m are whirled in horizontal circles, the heavier one in a radius
\frac{\mathrm{r}}{2}
and lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is
A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and table is μk. When the block A is sliding on the table, the tension in the string is
\frac{{\mathrm{m}}_{1}{\mathrm{m}}_{2}\left(1-{\mathrm{\mu }}_{\mathrm{k}}\right)\mathrm{g}}{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)}
\frac{\left({\mathrm{m}}_{1}+{\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{2}\right)\mathrm{g}}{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)}
\frac{\left({\mathrm{m}}_{1}-{\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{2}\right)\mathrm{g}}{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)}
\frac{{\mathrm{m}}_{1}{\mathrm{m}}_{2}\left(1+{\mathrm{\mu }}_{\mathrm{k}}\right)\mathrm{g}}{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)}
A system consists of three masses m1, m2 and m3 connected by a string passig over a pulley P. The mass m1 hangs freely and m2 and m3 are on a rough horizontal table (The coefficient of friction = μ). The pulley is frictionless and of negligible mass. The downward acceleration of mass m1 is: (Assume m1 = m2 = m3 = m)
\frac{\mathrm{g}\left(1-\mathrm{g\mu }\right)}{9}
\frac{2\mathrm{g\mu }}{3}
\frac{\mathrm{g}\left(1-2\mathrm{\mu }\right)}{3}
\frac{\mathrm{g}\left(1-2\mathrm{\mu }\right)}{2}
The force F acting on a particle of mass m is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from 0 s to 8 s is
1. 24 Ns
4. 6 Ns
A balloon with m is descending down with an acceleration a (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration a?
\frac{2\mathrm{ma}}{\mathrm{g}+\mathrm{a}}
\frac{2\mathrm{ma}}{\mathrm{g}-\mathrm{a}}
\frac{\mathrm{ma}}{\mathrm{g}+\mathrm{a}}
\frac{\mathrm{ma}}{\mathrm{g}-\mathrm{a}}
The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by
\mathrm{\mu }=\frac{2}{\mathrm{tan} \mathrm{\theta }}
\mathrm{\mu }=2 \mathrm{tan\theta }
\mathrm{\mu }=\mathrm{tan\theta }
\mathrm{\mu }=\frac{1}{\mathrm{tan\theta }}
An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. the first part of mass 1 kg moves with a speed of 12 ms–1 and the second part of mass 2 kg moves with 8 ms–1 speed. If the third part flies off with 4 ms–1 speed, then its mass is
1. 5 ms–1
2. 10 ms–1
A car of mass m is moving on a level circular track of radius R. If μs represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by
\sqrt{{\mathrm{\mu }}_{\mathrm{s}}\mathrm{mRg}}
\sqrt{\frac{\mathrm{Rg}}{{\mathrm{\mu }}_{\mathrm{s}}}}
\sqrt{\frac{\mathrm{mRg}}{{\mathrm{\mu }}_{\mathrm{s}}}}
\sqrt{{\mathrm{\mu }}_{\mathrm{s}}\mathrm{Rg}}
A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on control panel. The lift starts moving upwards with an acceleration 1 m/s2. If g = 10 ms–2, the tension in the supporting cable is
3. 1.5Mv
A radioactive nucleus of mass M emits a photon of frequency ν and the nucleus recoils. The recoil energy will be:
1. hν
2. Mc2 – hν
\frac{{\mathrm{h}}^{2}{\mathrm{v}}^{2}}{2{\mathrm{Mc}}^{2}}
A block of mass m is in contact with the cart C as shown in the figure.
The coefficient of static friction between the block and the cart is μ. The acceleration
\mathrm{\alpha }
of the cart that will prevent the block from falling satisfies :
\mathrm{\alpha }>\frac{\mathrm{mg}}{\mathrm{\mu }}
\mathrm{\alpha }>\frac{\mathrm{g}}{\mathrm{\mu m}}
\mathrm{\alpha }\ge \frac{\mathrm{g}}{\mathrm{\mu }}
\alpha <\frac{\mathrm{g}}{\mathrm{\mu }}
A gramophone record is revolving with an angular velocity ω. A coin is placed at a distance r from the centre of the record. The static coefficient of friction is μ. The coin will revolve with the record if:
\mathrm{r}={\mathrm{\mu g\omega }}^{2}
\mathrm{r}<\frac{{\mathrm{\omega }}^{2}}{\mathrm{\mu g}}
\mathrm{r}<\frac{\mathrm{\mu g}}{{\mathrm{\omega }}^{2}}
\mathrm{r}\ge \frac{\mathrm{\mu g}}{{\mathrm{\omega }}^{2}}
A body, under the action of a force
\stackrel{\to }{\mathrm{F}}=6\stackrel{^}{i}-8\stackrel{^}{j}+10\stackrel{^}{k}
, acquires an acceleration of 1 m/s2. The mass of this
body must be:
3. 10 2 kg
\sqrt{10}
1. 4 ms–2 upwards
2. 4 ms–2 downwards
3. 14 ms–2 upwards
4. 30 ms–2 downwards
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EUDML | Approximation in by solutions of quasi-elliptic equations. EuDML | Approximation in by solutions of quasi-elliptic equations.
Approximation in
{L}_{p}
by solutions of quasi-elliptic equations.
Alborova, M.S.. "Approximation in by solutions of quasi-elliptic equations.." Vladikavkazskiĭ Matematicheskiĭ Zhurnal 5.1 (2003): 10-13. <http://eudml.org/doc/123982>.
keywords = {homogeneous metric},
title = {Approximation in by solutions of quasi-elliptic equations.},
TI - Approximation in by solutions of quasi-elliptic equations.
KW - homogeneous metric
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EUDML | On the -adic spectral analysis and multiwavelet on . EuDML | On the -adic spectral analysis and multiwavelet on .
p
-adic spectral analysis and multiwavelet on
{L}^{2}\left({ℚ}_{p}^{n}\right)
Chong, Wonyong; Kim, Min-Soo; Kim, Taekyun; Son, Jin-Woo
Chong, Wonyong, et al. "On the -adic spectral analysis and multiwavelet on .." International Journal of Mathematics and Mathematical Sciences 2003.41 (2003): 2619-2633. <http://eudml.org/doc/50644>.
@article{Chong2003,
author = {Chong, Wonyong, Kim, Min-Soo, Kim, Taekyun, Son, Jin-Woo},
keywords = {fractional differentiation; Vladimirov operator; wavelet},
title = {On the -adic spectral analysis and multiwavelet on .},
AU - Chong, Wonyong
AU - Son, Jin-Woo
TI - On the -adic spectral analysis and multiwavelet on .
KW - fractional differentiation; Vladimirov operator; wavelet
fractional differentiation, Vladimirov operator, wavelet
𝐑
𝐂
Articles by Chong
Articles by Son
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A geometric characterization of arithmetic Fuchsian groups
15 March 2008 A geometric characterization of arithmetic Fuchsian groups
Slavyana Geninska, Enrico Leuzinger
Slavyana Geninska,1 Enrico Leuzinger1
1Department of Mathematics, Universität Karlsruhe, Germany
The trace set of a Fuchsian group
\Gamma
encodes the set of lengths of closed geodesics in the surface
\Gamma \\mathbb{H}
. Luo and Sarnak [3] showed that the trace set of a cofinite arithmetic Fuchsian group satisfies the bounded clustering (BC) property. Sarnak [5] then conjectured that the BC property actually characterizes arithmetic Fuchsian groups. Schmutz [6] stated the even stronger conjecture that a cofinite Fuchsian group is arithmetic if its trace set has linear growth. He proposed a proof of this conjecture in the case when the group
\Gamma
contains at least one parabolic element, but unfortunately, this proof contains a gap. In this article, we point out this gap, and we prove Sarnak's conjecture under the assumption that the Fuchsian group
\Gamma
contains parabolic elements.
Slavyana Geninska. Enrico Leuzinger. "A geometric characterization of arithmetic Fuchsian groups." Duke Math. J. 142 (1) 111 - 125, 15 March 2008. https://doi.org/10.1215/00127094-2008-002
Slavyana Geninska, Enrico Leuzinger "A geometric characterization of arithmetic Fuchsian groups," Duke Mathematical Journal, Duke Math. J. 142(1), 111-125, (15 March 2008)
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(Redirected from Time/space complexity)
2 Run-time analysis
2.1 Shortcomings of empirical metrics
2.2 Orders of growth
2.3 Empirical orders of growth
2.4 Evaluating run-time complexity
2.5 Growth rate analysis of other resources
4 Constant factors
Cost models[edit]
Time efficiency estimates depend on what we define to be a step. For the analysis to correspond usefully to the actual run-time, the time required to perform a step must be guaranteed to be bounded above by a constant. One must be careful here; for instance, some analyses count an addition of two numbers as one step. This assumption may not be warranted in certain contexts. For example, if the numbers involved in a computation may be arbitrarily large, the time required by a single addition can no longer be assumed to be constant.
Two cost models are generally used:[2][3][4][5][6]
the logarithmic cost model, also called logarithmic-cost measurement (and similar variations), assigns a cost to every machine operation proportional to the number of bits involved
The latter is more cumbersome to use, so it's only employed when necessary, for example in the analysis of arbitrary-precision arithmetic algorithms, like those used in cryptography.
A key point which is often overlooked is that published lower bounds for problems are often given for a model of computation that is more restricted than the set of operations that you could use in practice and therefore there are algorithms that are faster than what would naively be thought possible.[7]
Run-time analysis[edit]
Run-time analysis is a theoretical classification that estimates and anticipates the increase in running time (or run-time or execution time) of an algorithm as its input size (usually denoted as n) increases. Run-time efficiency is a topic of great interest in computer science: A program can take seconds, hours, or even years to finish executing, depending on which algorithm it implements. While software profiling techniques can be used to measure an algorithm's run-time in practice, they cannot provide timing data for all infinitely many possible inputs; the latter can only be achieved by the theoretical methods of run-time analysis.
Shortcomings of empirical metrics[edit]
Since algorithms are platform-independent (i.e. a given algorithm can be implemented in an arbitrary programming language on an arbitrary computer running an arbitrary operating system), there are additional significant drawbacks to using an empirical approach to gauge the comparative performance of a given set of algorithms.
Take as an example a program that looks up a specific entry in a sorted list of size n. Suppose this program were implemented on Computer A, a state-of-the-art machine, using a linear search algorithm, and on Computer B, a much slower machine, using a binary search algorithm. Benchmark testing on the two computers running their respective programs might look something like the following:
n (list size)
Computer A run-time
Computer B run-time
Based on these metrics, it would be easy to jump to the conclusion that Computer A is running an algorithm that is far superior in efficiency to that of Computer B. However, if the size of the input-list is increased to a sufficient number, that conclusion is dramatically demonstrated to be in error:
63,072 × 1012 31,536 × 1012 ns,
or 1 year 1,375,000 ns,
or 1.375 milliseconds
Computer A, running the linear search program, exhibits a linear growth rate. The program's run-time is directly proportional to its input size. Doubling the input size doubles the run-time, quadrupling the input size quadruples the run-time, and so forth. On the other hand, Computer B, running the binary search program, exhibits a logarithmic growth rate. Quadrupling the input size only increases the run-time by a constant amount (in this example, 50,000 ns). Even though Computer A is ostensibly a faster machine, Computer B will inevitably surpass Computer A in run-time because it's running an algorithm with a much slower growth rate.
Orders of growth[edit]
Main article: Big O notation
Informally, an algorithm can be said to exhibit a growth rate on the order of a mathematical function if beyond a certain input size n, the function f(n) times a positive constant provides an upper bound or limit for the run-time of that algorithm. In other words, for a given input size n greater than some n0 and a constant c, the run-time of that algorithm will never be larger than c × f(n). This concept is frequently expressed using Big O notation. For example, since the run-time of insertion sort grows quadratically as its input size increases, insertion sort can be said to be of order O(n2).
Big O notation is a convenient way to express the worst-case scenario for a given algorithm, although it can also be used to express the average-case — for example, the worst-case scenario for quicksort is O(n2), but the average-case run-time is O(n log n).
Empirical orders of growth[edit]
Assuming the run-time follows power rule, t ≈ kna, the coefficient a can be found [8] by taking empirical measurements of run-time {t1, t2} at some problem-size points {n1, n2}, and calculating t2/t1 = (n2/n1)a so that a = log(t2/t1)/log(n2/n1). In other words, this measures the slope of the empirical line on the log–log plot of run-time vs. input size, at some size point. If the order of growth indeed follows the power rule (and so the line on log–log plot is indeed a straight line), the empirical value of will stay constant at different ranges, and if not, it will change (and the line is a curved line)—but still could serve for comparison of any two given algorithms as to their empirical local orders of growth behaviour. Applied to the above table:
Local order of growth
(n^_)
65 32 1.04 150,000 0.28
250 125 1.01 200,000 0.21
1,000 500 1.00 250,000 0.16
1,000,000 500,000 1.00 500,000 0.10
4,000,000 2,000,000 1.00 550,000 0.07
16,000,000 8,000,000 1.00 600,000 0.06
It is clearly seen that the first algorithm exhibits a linear order of growth indeed following the power rule. The empirical values for the second one are diminishing rapidly, suggesting it follows another rule of growth and in any case has much lower local orders of growth (and improving further still), empirically, than the first one.
Evaluating run-time complexity[edit]
The run-time complexity for the worst-case scenario of a given algorithm can sometimes be evaluated by examining the structure of the algorithm and making some simplifying assumptions. Consider the following pseudocode:
1 get a positive integer n from input
3 print "This might take a while..."
7 print "Done!"
A given computer will take a discrete amount of time to execute each of the instructions involved with carrying out this algorithm. The specific amount of time to carry out a given instruction will vary depending on which instruction is being executed and which computer is executing it, but on a conventional computer, this amount will be deterministic.[9] Say that the actions carried out in step 1 are considered to consume time T1, step 2 uses time T2, and so forth.
In the algorithm above, steps 1, 2 and 7 will only be run once. For a worst-case evaluation, it should be assumed that step 3 will be run as well. Thus the total amount of time to run steps 1-3 and step 7 is:
{\displaystyle T_{1}+T_{2}+T_{3}+T_{7}.\,}
The loops in steps 4, 5 and 6 are trickier to evaluate. The outer loop test in step 4 will execute ( n + 1 ) times (note that an extra step is required to terminate the for loop, hence n + 1 and not n executions), which will consume T4( n + 1 ) time. The inner loop, on the other hand, is governed by the value of j, which iterates from 1 to i. On the first pass through the outer loop, j iterates from 1 to 1: The inner loop makes one pass, so running the inner loop body (step 6) consumes T6 time, and the inner loop test (step 5) consumes 2T5 time. During the next pass through the outer loop, j iterates from 1 to 2: the inner loop makes two passes, so running the inner loop body (step 6) consumes 2T6 time, and the inner loop test (step 5) consumes 3T5 time.
Altogether, the total time required to run the inner loop body can be expressed as an arithmetic progression:
{\displaystyle T_{6}+2T_{6}+3T_{6}+\cdots +(n-1)T_{6}+nT_{6}}
which can be factored[10] as
{\displaystyle \left[1+2+3+\cdots +(n-1)+n\right]T_{6}=\left[{\frac {1}{2}}(n^{2}+n)\right]T_{6}}
The total time required to run the outer loop test can be evaluated similarly:
{\displaystyle {\begin{aligned}&2T_{5}+3T_{5}+4T_{5}+\cdots +(n-1)T_{5}+nT_{5}+(n+1)T_{5}\\=\ &T_{5}+2T_{5}+3T_{5}+4T_{5}+\cdots +(n-1)T_{5}+nT_{5}+(n+1)T_{5}-T_{5}\end{aligned}}}
{\displaystyle {\begin{aligned}&T_{5}\left[1+2+3+\cdots +(n-1)+n+(n+1)\right]-T_{5}\\=&\left[{\frac {1}{2}}(n^{2}+n)\right]T_{5}+(n+1)T_{5}-T_{5}\\=&\left[{\frac {1}{2}}(n^{2}+n)\right]T_{5}+nT_{5}\\=&\left[{\frac {1}{2}}(n^{2}+3n)\right]T_{5}\end{aligned}}}
Therefore, the total run-time for this algorithm is:
{\displaystyle f(n)=T_{1}+T_{2}+T_{3}+T_{7}+(n+1)T_{4}+\left[{\frac {1}{2}}(n^{2}+n)\right]T_{6}+\left[{\frac {1}{2}}(n^{2}+3n)\right]T_{5}}
{\displaystyle f(n)=\left[{\frac {1}{2}}(n^{2}+n)\right]T_{6}+\left[{\frac {1}{2}}(n^{2}+3n)\right]T_{5}+(n+1)T_{4}+T_{1}+T_{2}+T_{3}+T_{7}}
As a rule-of-thumb, one can assume that the highest-order term in any given function dominates its rate of growth and thus defines its run-time order. In this example, n2 is the highest-order term, so one can conclude that f(n) = O(n2). Formally this can be proven as follows:
{\displaystyle \left[{\frac {1}{2}}(n^{2}+n)\right]T_{6}+\left[{\frac {1}{2}}(n^{2}+3n)\right]T_{5}+(n+1)T_{4}+T_{1}+T_{2}+T_{3}+T_{7}\leq cn^{2},\ n\geq n_{0}}
{\displaystyle {\begin{aligned}&\left[{\frac {1}{2}}(n^{2}+n)\right]T_{6}+\left[{\frac {1}{2}}(n^{2}+3n)\right]T_{5}+(n+1)T_{4}+T_{1}+T_{2}+T_{3}+T_{7}\\\leq &(n^{2}+n)T_{6}+(n^{2}+3n)T_{5}+(n+1)T_{4}+T_{1}+T_{2}+T_{3}+T_{7}\ ({\text{for }}n\geq 0)\end{aligned}}}
{\displaystyle {\begin{aligned}&T_{6}(n^{2}+n)+T_{5}(n^{2}+3n)+(n+1)T_{4}+T_{1}+T_{2}+T_{3}+T_{7}\leq k(n^{2}+n)+k(n^{2}+3n)+kn+5k\\=&2kn^{2}+5kn+5k\leq 2kn^{2}+5kn^{2}+5kn^{2}\ ({\text{for }}n\geq 1)=12kn^{2}\end{aligned}}}
{\displaystyle \left[{\frac {1}{2}}(n^{2}+n)\right]T_{6}+\left[{\frac {1}{2}}(n^{2}+3n)\right]T_{5}+(n+1)T_{4}+T_{1}+T_{2}+T_{3}+T_{7}\leq cn^{2},n\geq n_{0}{\text{ for }}c=12k,n_{0}=1}
A more elegant approach to analyzing this algorithm would be to declare that [T1..T7] are all equal to one unit of time, in a system of units chosen so that one unit is greater than or equal to the actual times for these steps. This would mean that the algorithm's run-time breaks down as follows:[11]
{\displaystyle 4+\sum _{i=1}^{n}i\leq 4+\sum _{i=1}^{n}n=4+n^{2}\leq 5n^{2}\ ({\text{for }}n\geq 1)=O(n^{2}).}
Growth rate analysis of other resources[edit]
The methodology of run-time analysis can also be utilized for predicting other growth rates, such as consumption of memory space. As an example, consider the following pseudocode which manages and reallocates memory usage by a program based on the size of a file which that program manages:
while file is still open:
let n = size of file
for every 100,000 kilobytes of increase in file size
double the amount of memory reserved
In this instance, as the file size n increases, memory will be consumed at an exponential growth rate, which is order O(2n). This is an extremely rapid and most likely unmanageable growth rate for consumption of memory resources.
Algorithm analysis is important in practice because the accidental or unintentional use of an inefficient algorithm can significantly impact system performance. In time-sensitive applications, an algorithm taking too long to run can render its results outdated or useless. An inefficient algorithm can also end up requiring an uneconomical amount of computing power or storage in order to run, again rendering it practically useless.
Constant factors[edit]
Analysis of algorithms typically focuses on the asymptotic performance, particularly at the elementary level, but in practical applications constant factors are important, and real-world data is in practice always limited in size. The limit is typically the size of addressable memory, so on 32-bit machines 232 = 4 GiB (greater if segmented memory is used) and on 64-bit machines 264 = 16 EiB. Thus given a limited size, an order of growth (time or space) can be replaced by a constant factor, and in this sense all practical algorithms are O(1) for a large enough constant, or for small enough data.
This interpretation is primarily useful for functions that grow extremely slowly: (binary) iterated logarithm (log*) is less than 5 for all practical data (265536 bits); (binary) log-log (log log n) is less than 6 for virtually all practical data (264 bits); and binary log (log n) is less than 64 for virtually all practical data (264 bits). An algorithm with non-constant complexity may nonetheless be more efficient than an algorithm with constant complexity on practical data if the overhead of the constant time algorithm results in a larger constant factor, e.g., one may have
{\displaystyle K>k\log \log n}
{\displaystyle K/k>6}
{\displaystyle n<2^{2^{6}}=2^{64}}
For large data linear or quadratic factors cannot be ignored, but for small data an asymptotically inefficient algorithm may be more efficient. This is particularly used in hybrid algorithms, like Timsort, which use an asymptotically efficient algorithm (here merge sort, with time complexity
{\displaystyle n\log n}
), but switch to an asymptotically inefficient algorithm (here insertion sort, with time complexity
{\displaystyle n^{2}}
) for small data, as the simpler algorithm is faster on small data.
Termination analysis — the subproblem of checking whether a program will terminate at all
Time complexity — includes table of orders of growth for common algorithms
^ "Knuth: Recent News". 28 August 2016. Archived from the original on 28 August 2016.
^ Alfred V. Aho; John E. Hopcroft; Jeffrey D. Ullman (1974). The design and analysis of computer algorithms. Addison-Wesley Pub. Co. ISBN 9780201000290. , section 1.3
^ Juraj Hromkovič (2004). Theoretical computer science: introduction to Automata, computability, complexity, algorithmics, randomization, communication, and cryptography. Springer. pp. 177–178. ISBN 978-3-540-14015-3.
^ Giorgio Ausiello (1999). Complexity and approximation: combinatorial optimization problems and their approximability properties. Springer. pp. 3–8. ISBN 978-3-540-65431-5.
^ Wegener, Ingo (2005), Complexity theory: exploring the limits of efficient algorithms, Berlin, New York: Springer-Verlag, p. 20, ISBN 978-3-540-21045-0
^ Robert Endre Tarjan (1983). Data structures and network algorithms. SIAM. pp. 3–7. ISBN 978-0-89871-187-5.
^ Examples of the price of abstraction?, cstheory.stackexchange.com
^ How To Avoid O-Abuse and Bribes Archived 2017-03-08 at the Wayback Machine, at the blog "Gödel's Lost Letter and P=NP" by R. J. Lipton, professor of Computer Science at Georgia Tech, recounting idea by Robert Sedgewick
^ However, this is not the case with a quantum computer
^ It can be proven by induction that
{\displaystyle 1+2+3+\cdots +(n-1)+n={\frac {n(n+1)}{2}}}
^ This approach, unlike the above approach, neglects the constant time consumed by the loop tests which terminate their respective loops, but it is trivial to prove that such omission does not affect the final result
Sedgewick, Robert; Flajolet, Philippe (2013). An Introduction to the Analysis of Algorithms (2nd ed.). Addison-Wesley. ISBN 978-0-321-90575-8.
Greene, Daniel A.; Knuth, Donald E. (1982). Mathematics for the Analysis of Algorithms (Second ed.). Birkhäuser. ISBN 3-7643-3102-X.
Cormen, Thomas H.; Leiserson, Charles E.; Rivest, Ronald L. & Stein, Clifford (2001). Introduction to Algorithms. Chapter 1: Foundations (Second ed.). Cambridge, MA: MIT Press and McGraw-Hill. pp. 3–122. ISBN 0-262-03293-7.
Sedgewick, Robert (1998). Algorithms in C, Parts 1-4: Fundamentals, Data Structures, Sorting, Searching (3rd ed.). Reading, MA: Addison-Wesley Professional. ISBN 978-0-201-31452-6.
Media related to Analysis of algorithms at Wikimedia Commons
Retrieved from "https://en.wikipedia.org/w/index.php?title=Analysis_of_algorithms&oldid=1081268720"
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Comments on tag 02EJ—Kerodon
Comment #1388 by Paul on May 06, 2022 at 07:44
In Definition 6.2.1.1, there might be a "transformation" missing in the last sentence of the definition, when talking about the counit, right before the
\epsilon
Comment #1412 by Kerodon on May 17, 2022 at 14:19
In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02EJ. The letter 'O' is never used.
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Ozone-oxygen cycle - zxc.wiki
Ozone-oxygen cycle in the ozone layer .
The ozone-oxygen cycle , also known as Chapman's cycle , is the process by which ozone is continually renewed in the ozone layer , with ultraviolet radiation being converted into thermal energy . In 1930 the chemical relationships involved were clarified by Sydney Chapman .
1 Formation of ozone
2 What function ozone fulfills
3 How ozone breaks down
Atomic oxygen is a prerequisite for the formation of ozone. This can occur when ultraviolet sunlight ( wavelength less than 240 nm ) splits an oxygen molecule (O 2 ). These atoms often react with other oxygen molecules to form ozone.
{\ displaystyle \ mathrm {O_ {2} + (radiation <240nm) \ longrightarrow 2O}}
{\ displaystyle \ mathrm {2 \ (O_ {2} + O + M) \ longrightarrow 2 \ (O_ {3} + M)}}
M is a so-called “third collision partner”, a molecule (usually nitrogen or oxygen) that transports away the excess energy of the reaction. Ozone is created slowly because sunlight is not very intense at wavelengths below 240 nm.
What function ozone fulfills
When ozone comes into contact with ultraviolet light in the upper atmosphere , a chemical reaction occurs rapidly . The triatomic ozone molecule becomes diatomic molecular oxygen plus a free oxygen atom:
{\ displaystyle \ mathrm {O_ {3} + radiation \ longrightarrow O_ {2} + O}}
Free atomic oxygen reacts quickly with other oxygen molecules and in turn forms ozone:
{\ displaystyle \ mathrm {O_ {2} + O + M \ longrightarrow O_ {3} + M}}
The chemical energy that is released when O and O 2 combine is converted into kinetic energy of molecular movement. The overall effect is to convert penetrating ultraviolet light into harmless heat. This cycle keeps the ozone layer in a stable equilibrium, while at the same time protecting the lower atmosphere from UV radiation, which is harmful to most living things. It is also one of the two most important sources of heat in the stratosphere (the other is based on the kinetic energy that is released when O 2 is photolyzed to form O atoms ).
How ozone breaks down
When an oxygen atom and an ozone molecule meet, they recombine to form two oxygen molecules:
{\ displaystyle \ mathrm {O_ {3} + O \ longrightarrow 2 \ O_ {2}}}
The total amount of ozone in the stratosphere is determined by a balance between production from solar radiation and decay from recombination.
Free radicals , the most important of which are hydroxyl (OH), nitroxyl (NO) and chlorine (Cl) and bromine (Br) atoms, catalyze the recombination. This reduces the amount of ozone in the stratosphere.
Most of the hydroxyl and nitroxyl radicals are naturally present in the stratosphere, but human influences, especially the decomposition products of chlorofluorocarbons ( CFCs and halons ), have greatly increased the concentrations of chlorine and bromine atoms, which has contributed to the creation of the ozone hole . Each Cl or Br atom can catalyze tens of thousands of decay events before it is removed from the stratosphere.
The ozone-oxygen cycle can be described by the following differential equation:
{\ displaystyle {\ frac {dc_ {1}} {dt}} = {\ frac {c_ {2}} {\ tau _ {2}}} - {\ frac {c_ {1}} {\ tau _ { 1}}}}
Here c 1 is the concentration of ozone and c 2 is the concentration of oxygen. The equation reflects that the production rate of ozone is proportional to the concentration of oxygen, and the decay rate (-) is proportional to the concentration of the ozone itself. The proportionality constants have the dimension of a time. They depend on the known influencing factors of the reactions involved and can therefore change if these boundary conditions vary.
{\ displaystyle \ tau _ {1}, \ tau _ {2}}
The lifespan of ozone depends in particular on the energy flux density of the ozone-destroying UV radiation, on the probability that ozone is destroyed in collision processes (thus on temperature and pressure), and on the concentration of destructive radicals. Under normal conditions, i.e. H. in the absence of competing decay processes caused by light or radicals, the service life is given as about 20 min . With the aforementioned influences, the service life will be correspondingly shorter.
{\ displaystyle \ tau _ {1}}
{\ displaystyle \ tau _ {1} \ approx 10 ^ {3} s}
The time also depends on the energy flow density of the ozone-generating UV radiation.
{\ displaystyle \ tau _ {2}}
The solution of the above differential equation is greatly simplified by the fact that the concentration of diatomic oxygen can be viewed as constant. This means that no corresponding differential equation has to be solved for oxygen. The time constants and the corresponding boundary conditions are also initially viewed as constant. The general solution of the differential equation, as is easily verified by plugging in, is:
{\ displaystyle c_ {1} (t) = {\ frac {\ tau _ {1}} {\ tau _ {2}}} c_ {2} -A \ cdot e ^ {- t / \ tau _ {1 }}.}
The additional constant A only defines the initial concentration of ozone. The following conclusion can be drawn from the result. First of all, it should be noted that the exponential factor tends towards zero for times that are much longer than the decay time of the ozone. The fictitious final state represents the equilibrium concentration of the ozone. The time for the establishment of equilibrium is therefore equal to the chemical life of the ozone, i.e. less than 20 minutes . For the state of equilibrium, a simple form of the law of mass action also applies according to the above equation
{\ displaystyle \ tau _ {1}}
{\ displaystyle {\ frac {c_ {1GG}} {c_ {2}}} = {\ frac {\ tau _ {1}} {\ tau _ {2}}}.}
From this it can be seen in principle that the equilibrium concentration of ozone is lower, the shorter its lifespan. The equilibrium concentration of ozone in the stratosphere is given as around 10 ppm. So it is and with it
{\ displaystyle c_ {1GG} \ approx 10 ^ {- 5} c_ {2}}
{\ displaystyle \ tau _ {2} \ approx 100000 \ tau _ {1}.}
However, this period of time, which is on the order of years, should not be confused with the time until equilibrium is reached. Conversely, it rather represents approximately the time that would be required in the fictitious absence of decay processes to convert all of the oxygen into ozone.
With the same concentrations of ozone and oxygen, the production of ozone would take place much more slowly than the destruction of ozone, which is the direct explanation of the low ozone concentrations in the stratosphere.
According to the standard of the natural decay time of ozone (in the dark), the equilibrium reacts sensitively to additional decay processes caused by free radicals and UV radiation. As a first approximation, the additional decay processes are linearly included in the decay law and thus the inverse lifetime:
{\ displaystyle {\ frac {1} {\ tau _ {1}}} = {\ frac {1} {\ tau _ {push}}} + {\ frac {1} {\ tau _ {radical}}} + {\ frac {1} {\ tau _ {radiation}}}}
This in turn depends on the concentration of free radicals and, accordingly, on the intensity of the destructive radiation. The more free radicals or the more UV radiation there is, the shorter the corresponding service life.
{\ displaystyle \ tau _ {radical}}
{\ displaystyle \ tau _ {radiation}}
Even if z. If, for example, the destruction of ozone by radicals would take 25 times as long as their natural disintegration by impacts and radiation, this would nevertheless result in a 4% shift in the equilibrium, i.e. a 4% reduction in the ozone content. The more free radicals there are in the stratosphere, the lower the ozone content, which is in equilibrium.
Photochemistry of Important Atmospheric Species (PDF file; 866 kB)
This page is based on the copyrighted Wikipedia article "Ozon-Sauerstoff-Zyklus" (Authors); it is used under the Creative Commons Attribution-ShareAlike 3.0 Unported License. You may redistribute it, verbatim or modified, providing that you comply with the terms of the CC-BY-SA.
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Döbereiner's lamp - Simple English Wikipedia, the free encyclopedia
Döbereiner's lamp.
Döbereiner's lamp:
a. glass cylinder
b. open bottle
e. stopcock
f. nozzle
g. platinum sponge
Döbereiner's lamp is one of the first lighters.[1][2][3] It was developed in 1823 by the German chemist Johann Wolfgang Döbereiner.[4][5] The lighter was produced until around 1880.[6][7] The Heidelberg Castle and also the Deutsches Museum (German Museum) still have original Döbereiner lamps to show.
The lighter works based on a chemical reaction between hydrogen and oxygen. In the glass cylinder (a in the picture on the right) is sulfuric acid. In the open bottle (b) is zinc and at the top is a valve. With the lever (f), the bottle will be opened. Then, the sulfuric acid flows into the bottle. When it reaches the Zinc, a reaction starts. It produces hydrogen.
{\displaystyle \mathrm {Zn+H_{2}SO_{4}\longrightarrow Zn^{2+}+SO_{4}^{2-}+H_{2}\uparrow } }
The hydrogen can go out of the bottle at the top. Next to the valve, there is platinum. The hydrogen gas has to bypass the platinum. The hydrogen and the oxygen from the air react because of the catalytic effect of platinum. The gas mixture (called oxyhydrogen) then burns, because the reaction is exothermic, and produces water.
{\displaystyle \mathrm {2\ H_{2}+O_{2}\longrightarrow 2\ H_{2}O} }
To stop this reaction, the bottle only needs to be closed, that means the lever needs to be released. The hydrogen can no longer go out of the bottle and pushes the sulfuric acid back into the glass cylinder.
↑ 1.0 1.1 Hoffman, Roald (2012). "Döbereiner's Lighter » American Scientist". americanscientist.org. Archived from the original on 7 November 2016. Retrieved 26 February 2012.
↑ "BugStores.com - History of Lighters". bugstores.com. 2012. Archived from the original on 22 February 2012. Retrieved 26 February 2012.
↑ "Döbereinersche Feuerzeuge". gnegel.de. 2004. Archived from the original on 4 August 2008. Retrieved 27 February 2012.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Döbereiner%27s_lamp&oldid=8054006"
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Existence of almost-periodic ultra-weak solutions to the equation $u^{\prime }(t) = a(t) Au(t) + f(t)$ in Hilbert spaces
{u}^{\text{'}}\left(t\right)=a\left(t\right)Au\left(t\right)+f\left(t\right)
author = {Zaidman, Samuel},
title = {Existence of almost-periodic ultra-weak solutions to the equation $u^{\prime }(t) = a(t) Au(t) + f(t)$ in {Hilbert} spaces},
TI - Existence of almost-periodic ultra-weak solutions to the equation $u^{\prime }(t) = a(t) Au(t) + f(t)$ in Hilbert spaces
Zaidman, Samuel. Existence of almost-periodic ultra-weak solutions to the equation $u^{\prime }(t) = a(t) Au(t) + f(t)$ in Hilbert spaces. Rendiconti del Seminario Matematico della Università di Padova, Tome 81 (1989), pp. 201-208. http://www.numdam.org/item/RSMUP_1989__81__201_0/
1) A.M. Fink, Almost-Periodic Differential Equations, Lectures Notes in Math., vol. 377, Springer-Verlag, Berlin-Heidelberg-New York (1974). | MR 460799 | Zbl 0325.34039
[2] S. Zaidman, Almost-Periodic Functions in Abstract Spaces, vol.126, Pitman Publishing, R.N.M., Boston-London-Melbourne (1985). | MR 790316 | Zbl 0648.42006
[3] S. Zaidman, Uniqueness of bounded weak solutions to the equation u'(t) = = a(t)Au(t) in Hilbert spaces, Libert. Math. (1988), vol. 8, 65-67. | MR 973126 | Zbl 0669.34039
[4] S. Zaidman, Almost-periodic solutions of differential equations with diagonal operators, Ann. Sc. Math. Québec (1988), vol. 12, no. 2, 287-291. | MR 978460 | Zbl 0671.34054
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Model mixer using intermodulation table (IMT) - Simulink - MathWorks Italia
IMT Mixer
Local oscillator frequency (Hz)
Reference input power (dBm)
Nominal output power (dbm)
Use data file
IMT table
Output signal power (dBm)
Add LO phase noise
Block Parameters When Using Data File
IMT Mixer Icons
IMT Mixer block icon updated
Model mixer using intermodulation table (IMT)
Use the IMT Mixer to perform frequency translation defined in an intermodulation table (see [1], [2], [3], and [4]) for a single tone carrier mixed with a local oscillator (LO) signal. The block includes nonlinear amplification, device and phase noise, and mixer spur visualization. For a single tone carrier Fcar nonlinearly modulated with an LO signal of frequency FLO, the mixer output intermodulation products occur at frequencies:
{F}_{\text{out}}\left(M,N\right)=|M×{F}_{\text{car}}±N×{F}_{\text{LO}}|
Fcar – input RF signal carrier frequency
M and N are nonnegative integers (0,1,…, order of nonlinearity)
For a downconverter, the desired output tone is
|{F}_{\text{car}}-{F}_{\text{LO}}|
, and for an upconverter it is
|{F}_{\text{car}}+{F}_{\text{LO}}|
. All other combinations of M and N represent the spurious intermodulation products.
IMT Mixer block mask icons are dynamic and indicate the current state of the applied noise parameter. For more information, see IMT Mixer Icons.
Carrier frequency (Hz) — Carrier frequency
Carrier frequency, specified as a scalar in hertz. When multiple carriers exist on the input connection, this carrier frequency is selected as the RF signal input. Distance between adjacent carriers must be greater than
\left(1-1{e}^{-8}\right){F}_{\text{c}}
Local oscillator frequency (Hz) — Local oscillator (LO) frequency
Local oscillator (LO) frequency, specified as a scalar in hertz.
Reference input power (dBm) — Reference input power
Reference input power, specified as a scalar in dBm. The expression for the normalized input signal for the specified reference input power is:
{10}^{\left(\frac{\left(-{P}_{\text{rf}}+20\mathrm{log}10\left(\frac{1}{\sqrt{{R}_{\text{in}}}}\right)+30\right)}{20}\right)}
Intermodulation tables assume that the spur levels are measured for a reference input and output power. For example, if you increase the input power by 10 dB, the spurs due to second order non-linearity will increase by 20 dB, the spurs due to third order non-linearity will increase by 30 dB, and so forth.
Nominal output power (dbm) — Nominal LO power
Nominal output power, specified as a scalar in dBm. The expression for the normalized output signal for the specified reference input power is:
2\text{x}{10}^{\left(\frac{\left({P}_{\text{if}}-20\mathrm{log}10\left(\frac{1}{\sqrt{{R}_{\text{out}}}}\right)-30\right)}{20}\right)}
Use data file — Specify data file to use
Select this parameter to specify the data file you want to use to extract the spur table. Clear to specify your own spur values. The data file may contain any combination of IMT table, AC data, and colored spot noise in S2D or P2D format. See [4].
Data file — Data file
samplespur1.s2d (default)
Data file, specified as IMT data, AC data, and colored spot noise. To learn about the combination of inputs a data file can have and how this block extracts and calculates the parameters, see Block Parameters When Using Data File.
To set this parameter, first select Use data file.
Input impedance (Ohm) — Input impedance of mixer
Input impedance of mixer, specified as a real scalar.
Output impedance (Ohm) — Output impedance of mixer
Output impedance of mixer, specified as a real scalar.
Ground and hide negative terminals — Ground RF negative circuit terminals
Select this parameter to internally ground and hide the negative terminals. To expose the negative terminals, clear this parameter. If the terminals are exposed, the input signal is not referenced to the ground.
IMT table — IMT spur visualization
[99 99 99; 99 0 99; 99 99 99] (default) | square matrix
IMT spur visualization, specified as a square matrix.
Output signal power (dBm) — Signal power of desired output tone
Signal power of the desired output tone when plotting intermodulation products, specified as a scalar.
Upconverter (default) | Downconverter
Mixer type, specified as Upconverter or Downconverter.
For a single tone carrier Fcar nonlinearly modulated with an LO signal of frequency FLO, the mixer output intermodulation products occur at frequencies:
{F}_{\text{out}}\left(M,N\right)=|M×{F}_{\text{car}}±N×{F}_{\text{LO}}|
|{F}_{\text{car}}-{F}_{\text{LO}}|
|{F}_{\text{car}}+{F}_{\text{LO}}|
Plot — Visualize IMT table values using specified signal power and mixer type
Visualize IMT table values using specified signal power and mixer type.
Simulate noise — Simulate device or phase noise
Select this parameter to simulate noise as specified in block parameters or in file.
To set this parameter, first select Simulate noise.
White – Spectral density is a single nonnegative value. The power value of the noise depends on the bandwidth of the carrier, and the bandwidth depends on the time step. This is an uncorrelated noise source.
Piece-wise linear – Spectral density is a vector of values [pi]. For each carrier, the noise source behaves like a white uncorrelated noise. The power of the noise source is carrier dependent.
Colored – Depends on both carrier and bandwidth. This is a correlated noise source.
Frequencies (Hz) — Frequency data
Frequency data, specified as a scalar for white noise or vector for piece-wise linear or colored noise in hertz.
To set this parameter, first select Select noise then select Piece-wise linear or Colored in Noise distribution.
To set this parameter, first select Select noise then select Spot noise data in Noise Type.
Add LO phase noise — Add phase noise
Select this parameter to add phase noise to your system with a continuous wave source.
To set this parameter, select Simulate noise.
Phase noise frequency offset with respect to LO signal, specified as a scalar or vector with each element unit in hertz.
The frequency offset values must be bounded by the envelope bandwidth of the simulation. For more information see Configuration.
To enable this parameter, first select Simulate noise then select Add phase noise.
Phase noise level, specified as a scalar or vector or matrix with elements in decibel per hertz.
If you specify a matrix, each column should correspond to a non-DC carrier frequency of the CW source. The frequency offset values must be bounded by the envelope bandwidth of the simulation. For more information see Configuration.
Select this parameter to automatically calculate impulse response for frequency-dependent noises. Clear this parameter to manually specify the impulse response duration using Impulse response duration.
Impulse response duration used to simulate frequency-dependent noise, specified as a scalar in seconds. The time should be an integer multiple of the step size in the configuration block, Tduration = NTstep.
This table shows you how the block extracts the block parameters if you choose to use a data file. The way the block extracts the parameters depends on the contents of the data file.
IMT, AC, and noise data Block extracts value from IMT header of data file Block extracts value from data file Block extracts value from data file Block extracts value from data file
IMT data Block extracts value from IMT header of data file User provides value or block uses default value Block extracts value from data file User provides value or block uses default value
AC and IMT data Block extracts value from IMT header of data file Block extracts value from data file Block extracts value from data file User provides value or block uses default value
Noise and IMT Block extracts value from IMT header of data file User provides value or block uses default value Block extracts value from data file Block extracts value from data file
AC Data User provides value or block uses default value Extracted from the data file User provides value or block uses default value User provides value or block uses default value
AC and Noise User provides value or block uses default value Extracted from the data file User provides value or block uses default value Extracted from the data file
Noise User provides value or block uses default value User provides value or block uses default value User provides value or block uses default value Extracted from the data file
Consider the default data file samplespur1.s2d. This data file contains IMT, AC, and noise data. When you specify the block to use this data file by selecting the Use data file parameter, the block extracts the reference input power, nominal output power, IMT, and noise data from the data file. This image shows the block mask with the extracted Reference input power (dBm) and Nominal output power (dbm) parameters.
To verify the extracted data, inspect the data file using this command
edit(samplespur1.s2d)
This image shows the real value of the mixer gain at the Carrier frequency (Hz) of 1 GHz and the Reference input power (dBm).
The block extracts the Nominal output power (dbm) from the data file using this equation.
This table shows you how the icons on this block will vary based on how you set the Simulate noise and Add LO phase noise parameters on the block.
Not applicable. For more information, see Dependencies.
[1] https://www.mathworks.com/help/rf/examples/visualizing-mixer-spurs.html
[2] https://www.microwavejournal.com/articles/3430-the-use-of-intermodulation-tables-for-mixer-simulations
[3] Faria, Daniel., Lawrence Dunleavy, and Terje Svensen. "The Use of Intermodulation Tables for Mixer Simulations". Microwave Journal, April 2002. https://www.microwavejournal.com/articles/3430-the-use-of-intermodulation-tables-for-mixer-simulations.
[4] "RF Mixing / Multiplication: Frequency Mixers". Electronic Notes. https://www.electronics-notes.com/articles/radio/rf-mixer/rf-mixing-basics.php.
R2021b: IMT Mixer block icon updated
Starting in R2021b, the IMT Mixer block icon has updated. The block icons are now dynamic and show the current state of the noise parameter.
When you open a model created before R2021b containing a IMT Mixer block, the software replaces the block icon with the R2021b version.
Amplifier | S-Parameters
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Convert RF signal to baseband signal - Simulink - MathWorks France
\left(Available\text{ }\text{\hspace{0.17em}}power\text{\hspace{0.17em}}gain+I/Q\text{\hspace{0.17em}}gain\text{\hspace{0.17em}}mismatch\right)
\frac{{R}_{\text{load}}}{{R}_{\text{source}}}>{R}_{\text{ratio}}
\frac{{R}_{\text{load}}}{{R}_{\text{source}}}<\frac{1}{{R}_{\text{ratio}}}
{R}_{\text{ratio}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{\sqrt{1+{\epsilon }^{2}}+\epsilon }{\sqrt{1+{\epsilon }^{2}}-\epsilon }
\epsilon \text{\hspace{0.17em}}=\text{\hspace{0.17em}}\sqrt{{10}^{\left(0.1{R}_{\text{p}}\right)}-1}
\frac{{R}_{\text{load}}}{{R}_{\text{source}}}>{R}_{\text{ratio}}
\frac{{R}_{\text{load}}}{{R}_{\text{source}}}<\frac{1}{{R}_{\text{ratio}}}
{R}_{\text{ratio}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{\sqrt{1+{\epsilon }^{2}}+\epsilon }{\sqrt{1+{\epsilon }^{2}}-\epsilon }
\epsilon \text{\hspace{0.17em}}=\text{\hspace{0.17em}}\sqrt{{10}^{\left(0.1{R}_{\text{p}}\right)}-1}
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Calculate American options prices and sensitivities using Barone-Adesi and Whaley option pricing model - MATLAB optstocksensbybaw - MathWorks Benelux
Compute an American Option Price and Sensitivities Using the Barone-Adesi and Whaley Option Pricing Model
Calculate American options prices and sensitivities using Barone-Adesi and Whaley option pricing model
PriceSens = optstocksensbybaw(RateSpec,StockSpec,Settle,Maturity,OptSpec,Strike)
PriceSens = optstocksensbybaw(___,Name,Value)
PriceSens = optstocksensbybaw(RateSpec,StockSpec,Settle,Maturity,OptSpec,Strike) calculates American options prices using the Barone-Adesi and Whaley option pricing model.
PriceSens = optstocksensbybaw(___,Name,Value) adds optional name-value pair arguments.
Compute the price and sensitivities for the American option.
[Price,Delta,Theta] = optstocksensbybaw(RateSpec,StockSpec,Settle,Maturity,OptSpec,Strike,'OutSpec',OutSpec)
Example: [Price,Delta,Theta] = optstocksensbybaw(RateSpec,StockSpec,Settle,Maturity,OptSpec,Strike,'OutSpec',OutSpec)
PriceSens — Expected prices or sensitivities for American options
Expected prices or sensitivities for American options, returned as a NINST-by-1 matrix.
All sensitivities are evaluated by computing a discrete approximation of the partial derivative. This means that the option is revalued with a fractional change for each relevant parameter. The change in the option value divided by the increment is the approximated sensitivity value.
\mathrm{max}\left(St-K,0\right)
\mathrm{max}\left(K-St,0\right)
optstockbybaw | impvbybaw
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Home : Support : Online Help : Mathematics : Group Theory : AreConjugate
test conjugacy of elements of a permutation group
compute the conjugator of elements of a permutation group
AreConjugate( a, b, G )
Conjugator( a, b, G )
and
b
G
G
if there is an element
g
G
{g}^{-1}·a·g=b
. Any such element
g
is called a conjugator. A conjugator is not generally uniquely determined by
and
b
The AreConjugate( a, b, G ) command returns true if the permutations a and b are conjugate in the permutation group G, and returns false otherwise.
The Conjugator( a, b, G ) command returns an element g in G such that g^(-1) . a . g = b, provided that a and b are conjugate in G. If a and b are not conjugate in G, the value FAIL is returned.
The group G must be an instance of a permutation group, and the permutations a and b must be members of G.
\mathrm{with}\left(\mathrm{GroupTheory}\right):
G≔\mathrm{PermutationGroup}\left(\mathrm{Perm}\left([[2,4,6]]\right),\mathrm{Perm}\left([[1,5],[2,4]]\right),\mathrm{Perm}\left([[1,4],[2,5],[3,6]]\right)\right)
\textcolor[rgb]{0,0,1}{G}\textcolor[rgb]{0,0,1}{≔}〈\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\right)\textcolor[rgb]{0,0,1}{,}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\right)\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)\textcolor[rgb]{0,0,1}{,}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\right)\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\right)〉
a≔\mathrm{Perm}\left([[1,2],[3,4],[5,6]]\right)
\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\right)\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)\left(\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\right)
b≔\mathrm{Perm}\left([[1,2],[3,6],[4,5]]\right)
\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\right)\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\right)\left(\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\right)
\mathrm{AreConjugate}\left(a,b,G\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{AreConjugate}\left(a,b,\mathrm{Symm}\left(6\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
c≔\mathrm{Perm}\left([[1,4],[2,5],[3,6]]\right)
\textcolor[rgb]{0,0,1}{c}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\right)\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\right)
\mathrm{AreConjugate}\left(a,c,G\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
d≔\mathrm{Conjugator}\left(a,c,G\right)
\textcolor[rgb]{0,0,1}{d}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\right)
{d}^{-1}·a·d=c
\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\right)\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\right)\textcolor[rgb]{0,0,1}{=}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\right)\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\right)
The GroupTheory[AreConjugate] and GroupTheory[Conjugator] commands were introduced in Maple 17.
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Honoring Professor Erdogan’s Seminal Contributions to Mixed Boundary-Value Problems of Inhomogeneous and Functionally Graded Materials | J. Appl. Mech. | ASME Digital Collection
Marek-Jerzy Pindera,
Pindera, M., and Paulino, G. H. (July 24, 2008). "Honoring Professor Erdogan’s Seminal Contributions to Mixed Boundary-Value Problems of Inhomogeneous and Functionally Graded Materials." ASME. J. Appl. Mech. September 2008; 75(5): 050301. https://doi.org/10.1115/1.2936920
boundary-value problems, cracks, elastic moduli, elasticity, fracture mechanics, functionally graded materials
Boundary-value problems, Engineering teachers, Fracture mechanics, Functionally graded materials, Fracture (Materials), Elastic moduli, Elasticity
Professor Fazil Erdogan has influenced several generations of applied and solid mechanicians working in the area of mixed boundary-value problems of inhomogeneous media, most notably fracture and contact problems. The analytical approaches that he had developed with his students in the 1960s and 1970s for the formulation and reduction of fracture mechanics problems involving layered media to systems of singular integral equations, and the corresponding solution techniques, have motivated researchers working in this area throughout the entire world. His subsequent work on fracture mechanics problems of inhomogeneous media with smoothly varying elastic moduli has laid the foundation for applying these techniques to functionally graded materials, which played key roles in many technologically important applications (e.g., spatially tailored structures for the new generation of hypersonic aircraft, graded cementitious composites for sustainable infrastructure, high-performance graded components for automobiles, and graded microtools in mechatronics). Professor Erdogan’s continuing leadership role and ceaseless contributions to the fracture and contact mechanics of this new generation of materials provide guidance and motivation for others to follow. This special issue honors Professor Erdogan in recognition of his past and continuing contributions in the area that plays a critical role in the development of engineered material systems for critical technological applications, and builds upon a minisymposium under the above title held at the recent International Conference on Multiscale and Functionally Graded Materials (M&FGM2006) on Oct. 15–18, 2006, Honolulu, HI.
The special issue is comprised of 13 invited papers containing original, previously unpublished contributions in the mechanics of inhomogeneous and functionally graded materials. The invited contributors, including Professor Erdogan who has provided with Dr. Ozturk (coauthor) the lead article summarizing the various types of singularities that may be encountered in contact and fracture mechanics, include selected authors of presentations given at the above conference. Some of these contributors are Professor Erdogan’s ex-students and past or present collaborators, while others are distinguished researchers working in this topical area who did not attend the conference. Analytical, computational, experimental, and theoretical aspects of the mechanics of inhomogeneous media in the broad sense, and functionally graded materials, in particular, are covered by the 13 papers. Topics range from the fundamental aspects of crack propagation in graded materials, construction of elasticity solutions for layered anisotropic media, development of novel computational procedures, and specific problems of technological importance involving graded coatings and cover plates to micromechanics-based calculations involving periodically layered media and functionally graded particulate materials. Examination of the contributed articles reveals the need for a multipronged approach in the modeling and simulation of graded and layered materials, and the important role that locally exact analytical solutions may play in the development of new computational procedures.
Many of us who have been influenced directly or indirectly by Professor Erdogan’s work hope that this will be a lasting issue in an area that continues to grow vigorously. One of the coeditors of this special issue (M.-J. P) recalls his first contact with the work of Professor Erdogan while collaborating some
25years
ago with Dr. Sailon Chatterjee at the Materials Sciences Corporation on fracture mechanics of layered anisotropic materials. “The techniques to which I was being introduced in the course of conducting research on defect criticality of composite laminates for the Naval Air Development Center were based on Professor Erdogan’s now classical papers, and Sailon often telephoned Fazil, who he called his secret weapon, for clarification or guidance to ensure that we were on the right path. It took me a while to realize that this mysterious Fazil was in fact Professor Erdogan whose papers and guidance enabled us to prosper.”
We are indeed grateful that we were given the opportunity to assemble this special issue in order to honor Professor Fazil Erdogan. He continues to be a source of inspiration to the mechanics community in leading the way in the area of mixed boundary-value problems in inhomogeneous and functionally graded media and also in providing selfless guidance to others.
Gradient Elasticity Theory for Mode III Fracture in Functionally Graded Materials—Part II: Crack Parallel to the Material Gradation
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Quantifying Realized Extractable Value | Flashbots
\textrm{MEV}
\textrm{MEV}
Despite much recent discussion about the topic and, in particular, its associated risks for the Ethereum protocol, we still lack a cohesive formal model for quantifying
\textrm{MEV}
extraction. At Flashbots, we have released an
\textrm{MEV}
explorer where we shed light on various aspects of this phenomenon. While we do elaborate on our metrics in the site, they still lack a formal definition. Here, we attempt to provide a unifying operational framework that consolidates
\textrm{MEV}
extraction metrics, focusing on the Ethereum network1.
The first important point to make is that
\textrm{MEV}
is a theoretical quantity that we can only approach asymptotically. Unforseen extraction methods can and will be devised (every new DeFi hack is an
\textrm{MEV}
extraction event). Hence, we will here focus instead on the Realized Extractable Value, notated
\textrm{REV}
\textrm{REV} \leq \textrm{MEV}
\textrm{REV}
is the actual value extracted from the blockchain from
\textrm{MEV}
opportunities2.
We note that there's two classes of actors in the system, searchers and miners. We use the generic label miner for actors that bear the privileged role of transaction inclusion and ordering. Searchers are any non-privileged actors aiming to profit from these opportunities. Crucially, miners can also act as searchers.
Extracting
\textrm{MEV}
incurs externalities, like increasing gas costs for all users of the network, or bloating the chain. We won't dwell into the implications of these externalities here, but we will make them manifest in the model. For more on this and the "
\textrm{MEV}
Crisis", be sure to check our Flashbots introductory post.
We are now ready to begin building our framework, starting with a simple model that ignores direct miner payments and searcher competition and other externalities like opportunity checks. We will revisit the model to account for these at a later stage.
Extraction Model#
We begin with a model where a single searcher performs an
\textrm{MEV}
extraction, sharing proceeds with the block miner via gas fees only. We further assume that miners order transactions by descending gas price, which approximates well their optimal strategy and is the default in Ethereum node software.
We decompose the opportunity's
\textrm{REV}
in two terms:
\textrm{REV} = \textrm{REV}_S + \textrm{REV}_M, \tag{1}
\textrm{REV}_S
is the value that goes to the searcher, and
\textrm{REV}_M
is the value that goes to the miner. Note that, as we expand on below,
\textrm{REV}
already encompasses the opportunity's extraction costs (i.e. the actual
\textrm{REV}
of an opportunity depends on the gas price of the network at extraction time3).
Let's dive into the two different terms. The searcher
\textrm{REV}
is composed by:
\textrm{REV}_S = V_{in} - V_{out} - g_{\textrm{MEV}}.s_{\textrm{MEV}}, \tag{2}
V_{out}
is the value that flows from the searcher to the blockchain in the transactions performing the extraction (excluding gas);
V_{in}
is the value flowing from the blockchain to the searcher;
g_{\textrm{MEV}}
is the gas price of the transactions; and
s_{\textrm{MEV}}
is their size, i.e. the total amount of gas they consumed.
V_{out}
V_{in}
g_{\textrm{MEV}}
are denominated in the base network currency (ETH), while
s_{\textrm{MEV}}
is in units of gas. Separating the gas term from
V_{out}
will be helpful in quantifying the extraction costs, and it is also how we compute
\textrm{REV}_S
in practice.
We use "the blockchain" loosely here to refer to any other addresses (corresponding to smart contracts or EOAs) that are not the EOA of the extracting transactions, or smart contracts controlled by the searcher. Note that identifying these is a heuristic-guided process based on known searcher patterns, and by all means fallible. Also, we stress that any ancillary transactions related to the
\textrm{MEV}
extraction (like the "meat" in a sandwich attack) are not part of the set of transactions that contribute to the above variables.
Turning to the miner side, we have:
\textrm{REV}_M = s_{\textrm{MEV}}.(g_{\textrm{MEV}}-g_{eff}), \tag{3}
g_{eff}
is the effective gas price of the transactions that would have been included in the block had the opportunity not been taken.
\textrm{REV}_M
is thus defined to include the opportunity cost the miner incurs by including the
\textrm{MEV}
-extracting transactions.
As transactions in the mempool are ephemeral, it is impossible to measure
g_{\textrm{eff}}
a posteriori from only blockchain data and logs. We resort to an approximation that also serves as a lower bound for the value realized by the miner:
\textrm{REV}_M \gtrsim s_{\textrm{MEV}}.(g_{\textrm{MEV}}-g_{tail}), \tag{4}
g_{tail}
is the gas price of the last transaction in the block4.
Plugging equations (2) and (4) we get for the total realized value:
\textrm{REV} \gtrsim V_{in} - V_{out} - s_{\textrm{MEV}}.g_{tail}. \tag{5}
In this rendition the miner and searcher roles are blurred, but we can clearly identify the extraction costs of the opportunity, given by the subtrahend
s_{\textrm{MEV}}.g_{tail}
Finally, note that the value split between searcher and miner at this stage is entirely decided by the choice of
g_{\textrm{MEV}}
, which is in turn affected by the nature of the opportunity and the presence of other searchers trying to exploit it.
Including Externalities#
The scenario presented so far ignores the fact that while single-searcher successful value extractions exist, they constitute only a small fraction of all
\textrm{MEV}
activity. In practice, each time an opportunity arises in the network, several searchers will try to take it by engaging in Priority Gas Auctions or submitting multiple competing transactions to backrun a target state change. Moreover, sometimes searchers perform onchain calculations to validate the existence of an opportunity ("preflights"). When these checks fail, there is no value extracted, but miners will still benefit from the collected gas fees.
Note, however, that all of this activity is zero-sum: all costs incurred by searchers by competing with each other, validating state, etc., directly becomes miner profit. From the miner's perspective, the fee paid by a searcher's preflight is no different from, say, the one coming from an unsuspecting token transfer. This goes to say that none of this value amounts to a contribution to the
{\textrm{REV}}
. However, it is of interest of us to quantify it, since it is precisely what generates trouble for regular users of the network not involved in this commerce.
We will then need to take a broader view of
\textrm{MEV}
, and define a new term, the Extractable Value Cost (
{\textrm{EVC}}
), which summarizes all the extra value induced by
\textrm{MEV}
extraction activity, that is not part of the extracted value itself. With this definition, we have:
\textrm{Total MEV Activity} = \textrm{REV}_S + \textrm{REV}_M + {\textrm{EVC}}. \tag{6}
In words, the total
\textrm{MEV}
activity is composed of the extracted
\textrm{MEV}
\textrm{REV}
, divided among searchers and miners), and the global costs of extractable value, going directly to miners and quantified by
{\textrm{EVC}}
{\textrm{EVC}}
only encompasses on-chain activity, ignoring off-chain costs like network bloating due same-nonce transaction replacement.
We can compute the
{\textrm{EVC}}
{\textrm{EVC}} = \sum_{tx\in X} s_{tx}.g_{tx},\quad X=\{\textrm{preflights},\textrm{failures}, \textrm{cancellations} \}.\tag{7}
While altogether eliminating
\textrm{MEV}
is likely impossible5, it is Flashbots' core mission to reduce the negative externalities of
\textrm{MEV}
-related activities by improving the overall efficiency of the extraction process. The present framework allows us to state this mission concisely as that of reducing
{\textrm{EVC}}
Flashbots Bundles#
Flashbots optimizes
\textrm{MEV}
extraction by allowing searchers to submit bundles of ordered transactions, effectively transferring part of the ordering privileges from the miners to the searchers. These bundles are included by the miners provided they get more out of them than by mining regular transactions. While searchers can still set a gas fee, the preferred way to pay the miners for inclusion is via a direct coinbase transfer included in the bundle6.
We thus need to account for this extra term in our miner revenue formula (4):
\textrm{REV}_M^{Flashbots} \gtrsim s_{\textrm{MEV}}.(g_{\textrm{MEV}}-g_{tail}) + V_{direct}, \tag{8}
V_{direct}
is the total value transferred to the coinbase address. Note that this new term doesn't affect the total
\textrm{REV}
as it is accounted for by a corresponding increase of the
V_{out}
Apart from the extra term, we now expand the set of transactions that contribute to the terms in the formula to include all transactions included in the bundle. In particular, if a bundle includes the "meat" in an arbitrage, its size and gas price count towards the extracted value. This ensures we account for the opportunity cost incurred by miners when including bundles instead of regular transactions.
Finally, expression (2) for searcher value stays the same, but also requires expanding the implied transaction set to all transactions in the bundle. The rest of the model remains unchanged by the utilization of Flashbots' bundles. With these changes, the model also encompasses alternative relay services like private mempools.
It is worth noting that Flashbots reducing
{\textrm{EVC}}
does not imply a decrease in miners' revenue, since their share of extracted value might increase as a result of the extraction optimization. Flashbots executions are deterministic and therefore searchers are likely to pay the miners more than if they risked paying for reverts or cancellations. Finding the revenue sharing equilibrium is ultimately a game-theoretic problem which deserves independent treatment.
By unifying various quantities involved in
\textrm{MEV}
extraction, our framework enables us to pose clear-cut questions like "how big are the costs of
\textrm{MEV}
extraction activities for the network?", or "to what extent has Flashbots helped in reducing these costs?". These and other interesting perspectives on
\textrm{MEV}
extraction based on this framework will be included in coming versions of
\textrm{MEV}
-Explore.
Throughout this work, we have swept under the rug the complications involved in actually measuring the "atomic terms" entering the definitions, like
V_{out}
, or the actual set of
\textrm{MEV}
-extracting transactions. This task involves ad hoc heuristics for an ever-growing number of
\textrm{MEV}
extraction strategies, and while we strive for establishing a useful classification, there will always be unforeseen opportunities that will elude detection7.
We pass no judgment as of which types of
\textrm{MEV}
extraction are "good" and which are "bad", although there's certainly ways in which this question could be meaningfully approached ("has an unrelated third party incurred loss due to the extraction?"). We leave this subtle question for future work, but we restate our belief that
{\textrm{EVC}}
is undesirable and we shall strive to minimize it.
Join our Efforts!#
At Flashbots, we perform research and build systems for mitigating the upcoming
\textrm{MEV}
crisis, and we would love your help. We are a distributed organization with the principles of a pirate hacker collective, and have several open positions. We also issue grants to external researchers doing work aligned with ours, please find out more in our Research repository, or join our Discord!
Special thanks to Scott Bigelow, Phil Daian, Stephane Gosselin, Alex Obadia, Tina Zhen, and Jason Paryani for welcoming me onboard and all Flashbots' crew for the work and discussions leading to this post.
Note however that most of what is presented here should apply with minor variations to other blockchains that have actors with transaction-ordering privileges. Work is underway to flesh out these variations.↩
The definitions here are consistent with a theoretical approach to
\textrm{MEV}
to be published in an upcoming paper (sneak peek here).↩
It's up for discussion whether the (non-measurable)
\textrm{MEV}
should also depend on this.↩
One would need to look into adjacent block gas prices if the entire block was taken up by
\textrm{MEV}
-extracting transactions.↩
Note however that it can likely be reduced to some extent with smart protocol design, there's some ongoing efforts to this avail.↩
A direct coinbase transfer could also be present in a standard, non-bundled extraction, but it would make little sense for the searcher to do it since miners typically only look at gas prices for transaction inclusion and ordering.↩
See this paper for a complementary, protocol agnostic approach to
\textrm{MEV}
detection based on flagging certain types of "transaction dynamics" (displacement, insertion, and suppression).↩
« MEV-SGX -- A sealed bid MEV auction design
Flashbots -- Frontrunning the MEV crisis »
Flashbots Bundles
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Scheduling and selection of registers during execution assumes infinite number of registers even though processors only have a limited number of registers, in this article we take a look how variables are mapped onto registers.
Liveness analysis.
Register Allocation by graph coloring.
During machine code generation phase in compiler design we translate variables in the intermediate language one-to-one into registers in the machine language.
A concern is that processors don't have an unlimited number of registers therefore we introduce the concept of register allocation for handing these situations.
More formally, register allocation is the mapping of a large number of variables into a limited number of registers.
This is done by sharing the limited number of registers among several variables and when the variables are too many some variables will be stored in memory temporarily, this is called spilling.
Register allocation can be done in either the intermediate code generation phase or the machine code generation phase.
Register allocation during intermediate code generation has the advantage that the register allocator can be used for several target machines by parameterizing with the available set of registers.
However when intermediate code needs translation, we need extra registers for storage of temporary variables.
Machine code uses symbolic names for registers which are converted to register numbers during register allocation. Register allocation after machine code generation has an advantage which can be seen whereby several instructions are combined into a single instruction hence variables disappear and therefore there is no need for allocation.
Techniques for register allocation into machine code generation or intermediate code generation phase are similar.
A variable is live in the program if the value contained might be potentially used in future computations. Conversely it is dead if there are no ways that it can be used for future computations.
Two variables can only share a register if they are both dead, that is, there is no point in the program where they are both live.
Determining if a variable is live.
It is live if an instruction uses the contents of that variable.(generating liveness)
If a variable is assigned a value in an instruction while it is not used as an operand in that instruction then its is dead at the start of the instruction.(killing liveness)
A variable is live at the end of the instruction if it is live at the start of any of the immediate succeeding instructions.(propagating liveness).
If a variable is live at the end of an instruction and the instruction doesn't assign a value to the variable, then the variable is considered live at the start of the instruction.(propagating liveness).
This is the process of determining if a variable is live and to do this we formalize rules as equations over sets of variables and solve the equations.
Instructions will have successors - instructions that follow the current instruction during execution.
We denote set of successors for i as succ[i].
Rules for finding succ[i]
The instruction j listed after instruction i is in succ[i], unless i is a GOTO of IF-THEN-ELSE instruction.
If instruction i is in the form of GOTO l, the instruction LABEL l is in succ[i]. In a correct program there exists exactly one LABEL instruction with the label used by the GOTO instruction.
If instruction i is IF p THEN
{\mathrm{l}}_{\mathrm{t}}
{\mathrm{l}}_{\mathrm{f}}
are in succ[i].
For correct analysis we assume that the outcomes of an IF-THEN-ELSE instruction are possible.
If the above is not upheld this may result in the analysis claiming a variable is live when in fact it is dead which is worse because we may overwrite a value that could be used later and therefore a program may produce incorrect results.
Correct liveness information is dependent on knowing paths a program takes when executed which is not possible to compute with precision therefore we shall allow imprecise results as long as we err on the side of safety(calling a variable live unless we have proof it is dead).
We have set gen[i] which is a set of variables that an instruction i generates liveness for.
We have kill[i] which is a list of variables that may be assigned values by the instructions.
Examples of gen and kill sets for instructions in intermediate code.
gen[i]
kill[i]
LABEL l ∅ ∅
x := y {y} {x}
x := k ∅ {x}
x := unpop y {y} {x}
x := unop k ∅ {x}
x := y binpop z {y, z} {x}
x := y binpop k {y} {x}
x := M[y] {y} {x}
x := M[k] ∅ {x}
M[x] := y {x, y} ∅
M[k] := y {y} ∅
GOTO l ∅ ∅
IF x relop y THEN
{\mathrm{l}}_{\mathrm{t}}
{\mathrm{l}}_{\mathrm{f}}
{x, y} ∅
x := CALL f(args) args {x}
x, y, and z are possibly identical variables.
k denotes a constant.
For each instruction i, two sets will hold the liveness information.
If in[i] holds variables that are live at the start of i and out[i] holds variables that are live at the end of i, we define them using the following recursive equations.
int[i] = gen[i] ∪ (out[i] \ kill[i])
out[i] =
\stackrel{\cup }{\mathrm{j}\in \mathrm{succ}\left[\mathrm{i}\right]}
in[j].
Since the above equations are recursive we solve them using fixed-point iteration.
First we initialize all in[i] and out[i] to empty sets ∅ and repeatedly calculate new values for these until no further changes occur.
Changes will stop occurring because the sets have a finite number of possible values and adding elements to sets in[i] or out[i] on the right side of the equation cannot reduce elements of the sets in the left side of the equation, therefore after each iteration, some set elements will increase(by a finite number of times) or the set elements will remain unchanged(meaning we are done).
The result set is the solution to the equation.
The equations work under the assumption that all uses of a variable are visible in the analyzed code, that is, if a variable contains e.g the output of a program, the it is used after the termination of the program even if it is not visible in the code of the program itself.
Therefore, we ensure that analysis ends up making this variable live at the end of the program.
An example of Liveness analysis and register allocation for a program that calculates the Nth fibonacci number.
IF n = z THEN end ELSE body
t := a + b
n := n − 1
The succ, gen and kill sets for the above program.
succ[i]
5 6, 13 n, z
7 8 a, b t
9 10 t b
N is given as input by initializing n to N prior execution.
n is live at the start of the program since it is expected to hold the input of the program.
If a variable not expected to hold input at the start of the program is live it may cause problems as it might be used before its initialization.
The program terminates on instruction 13 where a holds the Nth fibnacci number therefore a is live at this point.
We can also note that instruction 13 has non successors succ[13] = ∅ hence set out[13] = {a}.
Other out sets are defined by out[i] =
\stackrel{\cup }{\mathrm{j}\in \mathrm{succ}\left[\mathrm{i}\right]}
In and out sets are initialized as empty sets and iteration happens until a fixed point is reached.
The order in which we treat instructions will determine how quickly we reach the fixed-point but has no effect on the final result of the iteration.
Information in the describe equations flows backwards therefore we calculate out[i] before in[i].
In the above example each iteration calculates sets in the following order.
out[13], in[13], out[12], in[12], ... , out[1], in[1].
The most recent values are used when calculating the right sides of the equations and therefore when a value comes from a higher instruction number, the value from the same column is used.
By the 3rd iteration we have reached a fixed point a the results stop changing after the second iteration.
Below we have a fixed-point iteration for a liveness analysis, we shall use it to explain the concept of interference.
Definition: A variable x interferes with a variable y if x != y and there exists an instruction i such that x ∈ kill[i], y ∈ out[i] and instruction i is not x := y.
Two variables share a register if neither interferes with another.
We can also say that two variables should not be live at the same time with the following differences.
After x := y, x and y may be live simultaneously, but since they contain the same value they share a register. This is an optimization.
Another case is that x is not in out[i] even if it is in kill[i] meaning we have assigned x to a value that won't be read from x. In this case x is not live after instruction i but interferes with y in out[i].
This interference will prevent an assignment to x overwriting y and is important to for preserving correctness.
The image below is an interference graph for the Nth fibonacci program
It is an undirected graph whereby each node is a variable and there is an edge between nodes x and y if x interferes with y or vice versa.
We generate interference for each assignment statement using the definition above
2 b n, a
3 z n, a, b
7 t b, n
8 a t, n
10 n a, b
11 z n, a, b
For register allocation, two variables interfere if they also interfere at any point in the program, interference can also define a symmetric relationship between variables since two involved variables cannot share a register.
A variable cannot interfere with itself and therefore the relation is not reflective.
Two variables share a register if they are not connected by an edge in the interference graph.
For this we assign to each node in the interference graph a register number such that:
Two nodes sharing an edge will have different numbers.
The total number of different registers is less than or equal to the number of available registers.
This is known as graph-coloring(where color = register).
This problem is an NP-complete problem and therefore we need heuristic methods to find solutions where in some cases if a solution doesn't exists the method will give up.
The idea of this heuristic method is as follows; If a node in the graph has strictly fewer N edges, where N is the number of available colors(registers), we set the node aside and color the rest of the graph.
After coloring is done, nodes(n-1) connected by edges to the selected node cannot possibly use all N colors therefore we can pick a color for the selected node from the remaining colors.
Coloring the graph from the previous section is as follows
z has three edges (3 < 4), we remove z from the graph.
a has less than four edges, we remove a from graph.
With the three nodes(b, t, n) left, we give each of them a number e.g 1, 2, 3 respectively.
The three nodes(b, t, n) are connected to a, and they use three colors(1, 2, 3), we select a fourth color, e.g 4.
z is connected to a, b, n therefore we choose a color different from 1, 3, 4 which gives 2, which works.
If there exists a node with less than N edges, place it on the stack along with a list of nodes connected to it and remove it and its edges from the graph.
2.1 If there is a node with less than N edges, select any node and repeat the above step.
2.2 If there are more nodes left in the graph, continue simplifying otherwise start selection.
3. Take a node and its connected nodes from stack an give a color different from other colors of the connected nodes.
3.1. If this is not possible, the coloring fails and we mark it for spilling
3.2. If there are more nodes in the stack process the with selection.
Points unspecified in the algorithm.
From simplification: Node to choose when none of them have less than N edges.
From Selection: Color to choose if there exists several choices.
If both of the above are chosen correctly we have an optimal coloring algorithm, unfortunately computing these choices is costly and therefore we opt to make qualified guesses.
We shall see how this can be done in the heuristics section.
If the selection phase is unable to find a color for a node the the graph-coloring algorithm fails meaning we give up keeping all variables in registers and hence select some variables which will reside in memory. This is spilling.
When we choose variables for spilling, we change the program so that they are kept in memory.
For each spilled variable x:
We choose a memory address.
In every instruction i that reads or assigns x, we rename x to xi.
We insert instruction xi to M[addressx] before the instruction i reads xi.
After the instruction i that assigns xi, we insert instruction to M[addressx] := xi.
If x is live at the start of the program, we add the instruction M[addressx] := x to the start of the program.
If x is live at the end of the program we add instruction x := M[addressx] to the end of the program.
In steps 5 and 6, we use the original name for x.
Following the above, we perform register allocation again.
Additional subsequent register allocations may generate additional spilled variables for the following reasons:
We ignore spilled variables when selecting colors for a node in the selection phase.
We change choices for nodes to remove from the graph in the simplification and selection phase.
Applying the algorithm to the interference graph.
b t, n 3
a b, n, t spill
z a, b, n 2
Removing redundant moves.
An assignment i the form of x := y can be removed from the code if x and y use the same register because the instruction will have no effect.
Most registers will try to allocate x and y in the same register whenever possible so as to remove such redundant move instructions.
Assuming x has been assigned a color at the time we need to select a color for y, we choose the same color for y if it is not used by any variable that y interferes with(including x).
Similarly if x is uncolored, we give it the same color as y if the color is not used for a variable that interferes with x.
This is called biased coloring.
We can also combine x and y into a single node before coloring the graph and only split the combined node if the simplification cannot find a node with less than N edges. This is called coalescing.
We can also perform live-range-splitting a converse of coalescing where by instead of splitting a variable we split its node by giving each occurrence of the variable a different name and inserting assignments between them when necessary.
Use of explicit register numbers.
Some operations require arguments or results stored in specific registers e.g integer multiplication in X86 processors require the first argument to be in the eax register and places the 64-bit result in the eax and edx registers, 32-bits each.
Variables to such operations must be assigned to these registers before register allocation begins. These are pre-colored nodes in the interference graph.
If two pre-colored nodes interfere we cannot make a legal coloring of the graph, a costly solution would be to spill one or both.
An optimal solution is to insert move instructions to move variables to and from the required registers before and after an instruction that requires specific registers.
This specific register must be included as pre-colored nodes in the interference graph but are not removed from it in the simplification phase.
Once pre-colored nodes remain in the graph, the selection phase will begin and when it needs to color a node, it avoids colors used by other neighbors to the node whether pre-colored or colored in the selection phase.
The register allocator removes the inserted moves using the methods described heuristic 1.
Although scheduling and selection of registers assume infinite number of registers, processors only have a limited number of registers
Register allocation is the mapping of a large number of variables into a limited number of registers.
Register allocation aims to minimize space used for spilled values.
Register allocation aims to use fewer registers and minimize loads and stores.
Register Allocation Slides (PDF) by Hugh Leather from University of Edinburgh.
"Basics of Compiler Design" by Torben Ægidius Mogensen Chapter 9.
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(0, 2)
(1, 0)
Plot the points on a graph. Find the slope (rise over run).
m=\frac{\Delta\textit{y}}{\Delta\textit{x}}=\frac{2-0}{0-1}=-2
Find the slope of a line parallel to the given line.
Remember, parallel lines have the same slope.
Find the slope of a line perpendicular to the given line.
Remember, perpendicular lines form a right angle.
The slopes will be the negative reciprocals of each other.
Find the product of the slopes you found in parts (b) and (c).
Use the values from parts (b) and (c).
Make a conjecture about the product of the slopes of any two perpendicular lines. Test your conjecture by creating more examples.
The product of the slopes of two perpendicular lines is always the same number.
Use the eTool below to explore the line.
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Compensate for carrier frequency offset - MATLAB - MathWorks í•œêµ
'OQPSK' or 'QPSK' π/4
'8PSK' π/8
2\mathrm{Ï}
{y}_{\text{n}}={x}_{\text{n}}{e}^{i{\mathrm{λ}}_{\text{n}}}\text{â},
where λn is the output of the direct digital synthesizer (DDS). The DDS is the discrete-time version of a voltage-controlled oscillator and is a core component of discrete-time phase locked loops. In the context of this System object, the DDS works as an integration filter.
{e}_{\text{n}}=\mathrm{sgn}\left(\mathrm{Re}\left\{{x}_{\text{n}}\right\}\right)Ã\mathrm{Im}\left\{{x}_{\text{n}}\right\}â\mathrm{sgn}\left(\mathrm{Im}\left\{{x}_{\text{n}}\right\}\right)Ã\mathrm{Re}\left\{{x}_{\text{n}}\right\}
{e}_{\text{n}}=\mathrm{sgn}\left(\mathrm{Re}\left\{{x}_{\text{n}}\right\}\right)Ã\mathrm{Im}\left\{{x}_{\text{n}}\right\}
{e}_{\text{n}}=\left\{\begin{array}{c}\mathrm{sgn}\left(\mathrm{Re}\left\{{x}_{\text{n}}\right\}\right)Ã\mathrm{Im}\left\{{x}_{\text{n}}\right\}â\left(\sqrt{2}â1\right)\mathrm{sgn}\left(\mathrm{Im}\left\{{x}_{\text{n}}\right\}\right)Ã\mathrm{Re}\left\{{x}_{\text{n}}\right\}\text{,}\text{â}\text{for}\text{â}|\mathrm{Re}\left\{{x}_{\text{n}}\right\}|â¥|\mathrm{Im}\left\{{x}_{\text{n}}\right\}|\\ \left(\sqrt{2}â1\right)\mathrm{sgn}\left(\mathrm{Re}\left\{{x}_{\text{n}}\right\}\right)Ã\mathrm{Im}\left\{{x}_{\text{n}}\right\}â\mathrm{sgn}\left(\mathrm{Im}\left\{{x}_{\text{n}}\right\}\right)Ã\mathrm{Re}\left\{{x}_{\text{n}}\right\}\text{,}\text{â}\text{for}\text{â}|\mathrm{Re}\left\{{x}_{\text{n}}\right\}|<|\mathrm{Im}\left\{{x}_{\text{n}}\right\}|\end{array}
{e}_{\text{n}}=\mathrm{sgn}\left(\mathrm{Re}\left\{{x}_{\text{n-SamplePerSymbol/2}}\right\}\right)Ã\mathrm{Im}\left\{{x}_{\text{n-SamplePerSymbol/2}}\right\}â\mathrm{sgn}\left(\mathrm{Im}\left\{{x}_{\text{n}}\right\}\right)Ã\mathrm{Re}\left\{{x}_{\text{n}}\right\}
{\mathrm{Ï}}_{\text{n}}={g}_{\text{I}}{e}_{\text{n}}+{\mathrm{Ï}}_{\text{n}â1}\text{â},
where ψn is the output of the loop filter at sample n, and gI is the integrator gain. The integrator gain is determined from the equation
{g}_{\text{I}}=\frac{4\left({\mathrm{θ}}^{2}/d\right)}{{K}_{\text{p}}{K}_{\text{0}}}\text{â},
where θ, d, K0, and Kp are determined from the System object properties. Specifically,
\mathrm{θ}\text{â}\text{=}\text{â}\frac{{B}_{\text{n}}T}{\left(\mathrm{ζ}+\frac{1}{4\mathrm{ζ}}\right)}\text{â}\text{and}\text{â}d\text{â}=\text{â}1+2\mathrm{ζ}\mathrm{θ}+{\mathrm{θ}}^{2},
where Bn is the normalized loop bandwidth, and ζ is the damping factor. The phase recovery gain, K0, is equal to the number of samples per symbol. The modulation type determines the phase error detector gain, Kp.
{\mathrm{λ}}_{\text{n}}=\left({g}_{\text{P}}{e}_{\text{n-1}}+{\mathrm{Ï}}_{\text{n-1}}\right)+{\mathrm{λ}}_{\text{n-1}}\text{â},
{g}_{\text{P}}=\frac{4\mathrm{ζ}\left(\mathrm{θ}/d\right)}{{K}_{\text{p}}{K}_{\text{0}}}\text{â}.
The info object function of this System object returns estimates of the normalized pull-in range, the maximum frequency lock delay, and the maximum phase lock delay. The normalized pull-in range, (Δf)pull-in, is expressed in radians and estimated as
{\left(\mathrm{Î}f\right)}_{\text{pull-in}}â\mathrm{min}\left(1,2\mathrm{Ï}\sqrt{2}\mathrm{ζ}{B}_{\text{n}}\right)\text{â}.
The expression for (Δf )pull-in becomes less accurate as
2\mathrm{Ï}\sqrt{2}\mathrm{ζ}{B}_{\text{n}}
{T}_{\text{FL}}â4\frac{{\left(\mathrm{Î}f\right)}_{\text{pull-in}}^{2}}{{B}_{\text{n}}^{3}}\text{â}\text{and}\text{â}{T}_{\text{PL}}â\frac{1.3}{{B}_{\text{n}}}\text{â}.
[2] Zhijie, H., Y. Zhiqiang, Z. Ming, and W. Kuang. “8PSK Demodulation for New Generation DVB-S2.†2004 International Conference on Communications, Circuits and Systems. Vol. 2, 2004, pp. 1447–1450.
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Photosynthesis in Higher Plants - Live Session - NEET & AIIMS 2019
Photosynthesis in Higher Plants - Live Session - NEET & AIIMS 2019Contact Number: 9667591930 / 8527521718
Within the chloroplast, the membranous system consisting
2. Grana lamellae
3. Grana, Stroma lamellae
4. Grana & stroma
Dark reactions are
(1) Directly light-driven & are dependent on NADPH & ATP
(2) Occur in darkness and responsible for the synthesis of NADPH & ATP
(3) Not light-dependent and responsible for the synthesis of NADPH & ATP
(4) Not directly light driven but are dependent on the products of light reactions
T.W. Engelmann used a prism to split light in his experiment on organism named
Which statement is correct regarding the clear division of labor within the chloroplast?
(1) The membranous system is responsible for trapping the light energy & also for the synthesis of ATP and NADPH
(2) The membranous system consisting of grana & stroma lamellae is responsible for enzymatic reactions and synthesis of sugar
(3) Fluid stroma is the site of synthesis of sugar, which in turn forms starch
(4) Both (a) and (c) are correct
The use of [
{}^{14}\mathrm{C}
] radioactive carbon by M. Calvin in algal photosynthesis studies provided
(1) That
{\mathrm{CO}}_{2}
will combine with
{\mathrm{H}}_{2}\mathrm{O}
to produce sugar
(2) That ATP and NADPH were used in the biosynthetic phase
(3) That first
{\mathrm{CO}}_{2}
fixation product was 3carbon organic acid
Choose the correct match of pigments and their colour.
(i) Chlorophyll a Bright green
(ii) Chlorophyll b Blue-green
(iii) Xanthophyll Yellow-green
(iv) Carotenoids Orange green
2. (i) & (ii)
3. (i), (ii) & (iii)
4. (i), (ii), (iii) & (iv)
Hatch and Slack pathway occurs in
1. Maize and tomatoes
2. Sorghum and bell pepper
3. Sugarcanes and tomatoes
4. Sorghum and maize
The photochemical phase of photosynthesis includes
1. Light absorption & formation of high energy chemical intermediates
2. Light absorption, water splitting, and synthesis of carbohydrate
3. Light absorption, oxygen release and formation of high-energy chemical intermediates
4. Light absorption, water splitting, trapping of
{\mathrm{CO}}_{2}
and formation of high energy chemical intermediates
In photosynthesis I [PS-I] and photosynthesis II [PS-II]:
1. Antennae are different
2. Reaction centres are different
3. Light-harvesting complexes are different
In photosystem II, the reaction centre chlorophyll a absorbs
1. 680 nm wavelength of red light
3. 680 nm wavelength of blue light
The stroma lamellae membrane contains
1. PS I & PS II
2. PS I, PS II and NADP reductase enzyme
3. PS I, PS II, NADP reductase & ATP synthase
4. PS I & ATP Synthase
Which is the source of the electron in non-cyclic photophosphorylation?
{\mathrm{CO}}_{2}
{\mathrm{H}}_{2}\mathrm{O}
Oxygen evolving complex is found in
1. PS II
2. PS I
3. Cyt. b
4. ATP synthase
Trough Z-scheme of transfer of electrons
1. Both ATP & NADPH are synthesized
2. Only ATP synthesis occurs
3. Only NADPH synthesis occurs
4. ATP, NADPH and carbohydrate molecule are synthesized
Site of cyclic photophosphorylation is
2. Stroma lamellae
3. Both grana and stroma lamellae
4. Inner membrane of chloroplast
The hypothesis which describes the mechanism of ATP synthesis is
1. Z scheme
2. Non-cyclic photophosphorylation
4. Chemiosmotic
In photosynthesis, the development of proton gradient across a membrane is different from respiration process as
(1) In photosynthesis, proton accumulation is in the inter-membrane space of chloroplast while in respiration, it is inside the membrane of mitochondria
(2) In photosynthesis, protons accumulate inside the thylakoid membrane of chloroplast while in respiration it accumulates in the intermembrane space of mitochondria
(3) In photosynthesis, proton accumulation is in between the stroma and grana lamellae of chloroplast while in respiration it is inside the membrane of mitochondria
(4) All the above said statements have a similar meaning
Since the splitting of water molecule takes place on the inner side of the thylakoid membrane, the protons, accumulate
1. In the stroma
2. In the lumen of the thylakoid
4. In between the outer and inner membrane of chloroplast
NADP reductase enzyme is located on
1. The stroma side of the membrane
2. The lumen side of the thylakoid membrane
3. The inner membrane of chloroplast
4. The intermembranous space of chloroplast
Decreases in pH of the lumen is due to
1. NADP reductase enzyme, present on the membrane
2. Presence of PS I, PSI I and NADP cyt.b on the thylakoid membrane
3. Removal of protons from the lumen to the stroma
4. Accumulation of protons in the lumen
{\mathrm{F}}_{1}
part of ATPase enzyme present on the surface of the membrane on the side that faces the stroma
{\mathrm{F}}_{0}
is embedded in the membrane and forms a transmembrane channel
(3) Conformational change in the
{\mathrm{F}}_{0}
particle makes the enzyme synthesize several molecules of energy-packed ATP
(4) All the above are correct
I. C4 pathway is found in plants adapted to dry tropical regions
II. C3 pathway is the Maine biosynthetic pathway in C4 plants
III. C4 plants have greater productivity of biomass
2. Only II is incorrect
4. Both (II) and (III) are correct
In the given graph what does A, B represent?
(1) A - Absorption spectrum ; B - Action spectrum (Chl. a)
(2) A - Action spectrum ; B - Absorption spectrum (carotenoids)
(3) A - Absorption spectrum ; B - Action spectrum (Ch. b)
(4) A - Action spectrum ; B - Absorption spectrum (Ch.a)
Choose the incorrect option w.r.t. stroma lamellae
1. Presence of PS l
2. Site of cyclic photophosphorylation
3. Perform photosynthesis at wavelength > 680 nm
4. Presence of
{\mathrm{NADP}}^{+}
reductase
Which of the following complex of mitochondrial ETS having two copper centers?
1. Cytochrome b c1 complex
2. NADH dehydrogenase complex
3. Succinate dehydrogenase complex
4. Cytochrome oxidase complex
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Soil (IF5.841), Pub Date : 2022-01-20, DOI: 10.5194/soil-8-49-2022
Rhizosheath size varies significantly with crop genotype, and root exudation is one among its driving factors. Unravelling the relationships between rhizosheath formation, root exudation and soil carbon dynamics may bring interesting perspectives in terms of crop breeding towards sustainable agriculture. Here we grew four pearl millet (C4 plant type: δ13C of −12.8 ‰, F14C = 1.012) inbred lines showing contrasting rhizosheath sizes in a C3 soil type (organic matter with δ13C of −22.3 ‰, F14C = 1.045). We sampled the root-adhering soil (RAS) and bulk soil after 28 d of growth under a semi-controlled condition. The soil organic carbon (SOC) content and δ13C and F14C of soil samples were measured and the plant-derived C amount and Clost
/
Cnew ratio in the RAS were calculated. The results showed a significant increase in δ13C in the RAS of the four pearl millet lines compared to the control soil, suggesting that this approach was able to detect plant C input into the soil at an early stage of pearl millet growth. The concentration of plant-derived C in the RAS did not vary significantly between pearl millet lines, but the absolute amount of plant-derived C varied significantly when we considered the RAS mass of these different lines. Using a conceptual model and data from the two carbon isotopes' measurements, we evidenced a priming effect for all pearl millet lines. Importantly, the priming effect amplitude (Clost
/
Cnew ratio) was higher for the small rhizosheath (low-aggregation) line than for the large rhizosheath (high-aggregation) ones, indicating a better C sequestration potential of the latter.
|
EUDML | Self-adjointness of Schrödinger-type operators with singular potentials on manifolds of bounded geometry. EuDML | Self-adjointness of Schrödinger-type operators with singular potentials on manifolds of bounded geometry.
Self-adjointness of Schrödinger-type operators with singular potentials on manifolds of bounded geometry.
Milatovic, Ognjen. "Self-adjointness of Schrödinger-type operators with singular potentials on manifolds of bounded geometry.." Electronic Journal of Differential Equations (EJDE) [electronic only] 2003 (2003): Paper No. 64, 8 p., electronic only-Paper No. 64, 8 p., electronic only. <http://eudml.org/doc/123539>.
@article{Milatovic2003,
author = {Milatovic, Ognjen},
keywords = {Schrödinger operator; self-adjointness; manifold; bounded geometry; singular potential},
title = {Self-adjointness of Schrödinger-type operators with singular potentials on manifolds of bounded geometry.},
AU - Milatovic, Ognjen
TI - Self-adjointness of Schrödinger-type operators with singular potentials on manifolds of bounded geometry.
KW - Schrödinger operator; self-adjointness; manifold; bounded geometry; singular potential
Ognjen Milatovic, On
m
-sectorial Schrödinger-type operators with singular potentials on manifolds of bounded geometry
Schrödinger operator, self-adjointness, manifold, bounded geometry, singular potential
Articles by Milatovic
|
Ask Answer - Theme 1 From the Beginning of Time, Theme 6 The Three Orders, History, Theme 7 Changing Cultural Traditions, Theme 8 Confrontation of Cultures, Theme 9 The Industrial Revolution, Theme 10 Displacing Indigenous Peoples, Theme 4 The Central Islamic Lands - Popular Questions for School Students
Parth Jakar
What is Puch blade technique? (In 3marks)
compare the conditions of the life of a french serf and roman slave
⇒
what development were made in the field of transport & communication?
what was the position of women in the European society during the 15th and the 16th century...pls provide an answer not the link
are there any similarities or differences in the distribution of settlements shown on maps 1 and 2?
what do you mean by reservation with reference to American natives.
explain the rise of sultanates in arab and iran..
The Harshacharita is a biography of Harshavardhana, the ruler of Kanauj (seep Map 3), composed in Sanskrit by his court poet, Banabhatta (c. seventh century CE). This is an excerpt form the text and extremely rare representation of life in a settlement on the outskirts of a forest in the Vindhyas:
The outskirts being for the most part forest, many parcels of rice-land, threshing ground and arable land were being apportioned by small farmers... it was mainly spade culture... owing to the difficulty of ploughing the sparsely scattered fields covered with grass, with their few clear spaces, their black soil stiff as black iron....
There were people moving along with bundles of bark... countless sacks of plucked flowers,... loads of flax and hemp bundles, quantities of honey, peacocks' tail feathers, wreaths of wax, logs, and grass. Village wives hastened en route for neighbouring villages, all intent on thoughts of sale and bearing on their heads baskets filled with various gathered forest fruits.
⇒
How would you classify the people described in the text in terms of their occupations?
|
EUDML | Twisting gravitational waves and eigenvector fields for on an infinite jet. EuDML | Twisting gravitational waves and eigenvector fields for on an infinite jet.
Twisting gravitational waves and eigenvector fields for
\text{SL}\left(2,ℂ\right)
on an infinite jet.
Finley, J. D. III
Finley, J. D. III. "Twisting gravitational waves and eigenvector fields for on an infinite jet.." Electronic Journal of Differential Equations (EJDE) [electronic only] 2000 (2000): 75-85. <http://eudml.org/doc/226954>.
author = {Finley, J. D. III},
title = {Twisting gravitational waves and eigenvector fields for on an infinite jet.},
AU - Finley, J. D. III
TI - Twisting gravitational waves and eigenvector fields for on an infinite jet.
Classes of solutions; algebraically special solutions, metrics with symmetries
Articles by Finley
|
This problem is a checkpoint for solving linear systems in two variables. It will be referred to as Checkpoint 2B.
Solve the system of linear equations below.
\left. \begin{array} { l } { 5 x - 4 y = 7 } \\ { 2 y + 6 x = 22 } \end{array} \right.
Answers and extra practice for the Checkpoint problems are located in the back of your printed textbook or in the Reference Tab of your eBook. If you have an eBook for CCA2, login and then click the following link: Checkpoint 2B: Solving Linear Systems in Two Variables
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Speed measurement - zxc.wiki
With the speed measurement , technical equipment is used to find out how fast an object is turning. The speed can be measured using the following operating principles:
1 Calculating procedure
1.1 Time measurement of one revolution
1.2 Frequency measurement of vibration or noise
1.3 Angle measurement at fixed time intervals
Time measurement of one revolution
With photocells or other sensors, the switching time is measured which requires the object for one revolution. The speed is calculated with
{\ displaystyle t_ {U}}
{\ displaystyle n}
{\ displaystyle n = {\ frac {1} {t_ {U}}}}
Application e.g. B. in bicycle computers , optical hand-tachometers
Frequency measurement of vibration or noise
The speed can be determined using the sound spectrum of a rotating GyroTwister recorded with a microphone . The humming sound of the device had a frequency of 135 hertz when recording, which corresponds to a speed of 135 revolutions per second or 8100 revolutions per minute.
The speed of rotation of the rotating object is a main component in the frequency spectrum of the mechanical vibrations or acoustic noises generated on the machine. The speed can be determined by measurement (using a strain gauge , microphone , acceleration sensor , etc.).
Angle measurement at fixed time intervals
If the angles φ 1 and φ 2 at two time points , and are well known, the rotational speed calculated by
{\ displaystyle t_ {1}}
{\ displaystyle t_ {2}}
{\ displaystyle n = {\ frac {k \ cdot (\ varphi _ {2} - \ varphi _ {1})} {t_ {2} -t_ {1}}} = {\ frac {k \; \ Delta \ varphi} {\ Delta t}}}
Here k is to be used in such a way that the numerator for a full angle results in the number one for one revolution, i.e. k = 1 / 2π = 1/360 °.
The angle can be measured with an incremental encoder , potentiometer or gyroscope . See also angle measurement .
Application e.g. B. Measuring system for servomotors , yaw rate sensors for navigation
In industry, impulses on gears are often recorded without contact. The field plate sensors used for this supply pulses of up to 20 kHz.
The rotating object is flashed at fixed time intervals. If regular markings on the object rotate exactly by grid widths, the grid seems to stand still. The number of marks and the frequency of the stroboscope can be used to determine the speed.
{\ displaystyle x = 1,2,3 \ dots}
{\ displaystyle N}
{\ displaystyle n = {\ frac {f \ cdot x} {N}}}
Application e.g. B. the turntable .
According to the law of induction , the voltage in a coil is proportional to the speed at which the magnetic flux changes. A rotating permanent magnet generates an alternating voltage in the coil. The voltage level or the frequency can be used to measure the speed.
Application e.g. As the tachometer generator , inductance transmitters , Wiegand sensor
If a permanent magnet rotates in a metal bell, eddy currents occur, which generate a torque that is measured against a spring, see eddy current tachometer .
Rotating weights are pushed outwards by centrifugal force . A spring or the force of gravity create the balance of forces. The resulting movement is mechanically transmitted to a pointer. Today this method is seldom used for measurements, but due to the reliable operation without auxiliary energy, this measuring principle is used for monitoring the maximum speed of elevators , cable cars, etc.
Application e.g. B. with mechanical hand tachometers, centrifugal clutches , historically: gyrometer , centrifugal governor
This page is based on the copyrighted Wikipedia article "Drehzahlmessung" (Authors); it is used under the Creative Commons Attribution-ShareAlike 3.0 Unported License. You may redistribute it, verbatim or modified, providing that you comply with the terms of the CC-BY-SA.
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Nonlinear Curve Fitting with lsqcurvefit - MATLAB & Simulink - MathWorks Benelux
lsqcurvefit enables you to fit parameterized nonlinear functions to data easily. You can also use lsqnonlin; lsqcurvefit is simply a convenient way to call lsqnonlin for curve fitting.
In this example, the vector xdata represents 100 data points, and the vector ydata represents the associated measurements. Generate the data for the problem.
The modeled relationship between xdata and ydata is
{ydata}_{i}={a}_{1}+{a}_{2}\mathrm{exp}\left(-{a}_{3}\phantom{\rule{0.16666666666666666em}{0ex}}{xdata}_{i}\right)+{\epsilon }_{i}.
The goal is to find parameters
{\underset{}{\overset{ˆ}{a}}}_{i}
i
= 1, 2, 3, for the model that best fit the data.
In order to fit the parameters to the data using lsqcurvefit, you need to define a fitting function. Define the fitting function predicted as an anonymous function.
{\underset{}{\overset{ˆ}{a}}}_{i}
Examine the resulting parameters.
The fitted values ahat are within 8% of a = [1;3;2].
If you have the Statistics and Machine Learning Toolbox™ software, use the nlparci function to generate confidence intervals for the ahat estimate.
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Mechanical Properties of Solids - Live Session - NEET & AIIMS 2019
Mechanical Properties of Solids - Live Session - NEET & AIIMS 2019Contact Number: 9667591930 / 8527521718
The deformation of a wire under its own weight compared to the deformation of the same wire subjected a tension equal to the weight of the wire is
The stress-strain graph for two materials is shown in the figure. If the Young's modulus for two materials are
{Y}_{A}
{Y}_{B}
, then-
{Y}_{A}
{Y}_{B}
{Y}_{A}
{Y}_{B}
{Y}_{A}
{Y}_{B}
(4) Can't be predicted from the graph
A wire of length L, cross-sectional area A is made of a material of Young's modulus Y.
The wire is stretched by an amount x, by a constant force which lies well within the
elastic limit. The work done (W) by the force is
W=\frac{Y{x}^{2}}{{L}^{2}}
W=\frac{Y{x}^{2}}{2{L}^{2}}
W=\frac{1}{2} \frac{YA{x}^{2}}{L}
W=\frac{YA{x}^{2}}{L}
\mathrm{\rho }
is the density of the material of a wire and
\sigma
is the breaking stress, the greatest length of the wire that can hang freely without breaking is:
\frac{2}{\mathrm{\rho g}}
\frac{\mathrm{\rho }}{\mathrm{\sigma g}}
\frac{\mathrm{\rho g}}{2\mathrm{\sigma }}
\frac{\mathrm{\sigma }}{\mathrm{\rho g}}
An elastic material of Young's modulus Y is subjected to a stress S. The elastic energy stored per unit volume of the material is:
\frac{SY}{2}\quad \quad
\frac{{S}^{2}}{2Y}
\quad \frac{S}{2Y}\quad
\frac{2S}{Y}
The following four wires are made of the same material. Which of these will have the largest extension, when the same tension is applied?
1. length 50 cm and diameter 0.5 mm
2. length 100 cm and diameter 1 mm
Two wires A and B are of the same material. Their lengths are in the ratio 1 : 2 and the diameters are in the ratio 2 : 1. If they are pulled by the same force, their increase in length will be in the ratio
A material has Poisson's ratio of 0.5. If a uniform rod made of it suffers a longitudinal strain of
2×{10}^{-3}
, what is the percentage increase in volume?
One end of a long metallic rod of length L is tied to the ceilling. The other end is tied to
massless spring of spring constant k. A mass m hangs freely from the free end of the
spring. The area of cross-section and Young's modulus of the wire A and y
respectively. lf the mass is slightly pulled down and released. lf the mass is slightly
pulled down and released . it will oscillate with a time period T equal to
2\mathrm{\pi }\sqrt{\left(\frac{\mathrm{m}}{\mathrm{k}}\right)}
2\mathrm{\pi }{\left[\frac{\left(\mathrm{YA} + \mathrm{kL}\right)\mathrm{m}}{\left(\mathrm{YAk}\right)}\right]}^{1/2}
2\mathrm{\pi }\sqrt{\left(\frac{\mathrm{mYA}}{\mathrm{kL}}\right)}
2\mathrm{\pi }\sqrt{\left(\frac{\mathrm{mL}}{\mathrm{YA}}\right)}
If the ratio of lengths, radii and Young's modulus of steel and brass wires in the figure are a, b and c respectively, then the corresponding ratio of increase in their lengths will be:
\frac{2{a}^{2}c}{b}
\frac{3a}{2{b}^{2}c}
\frac{2ac}{{b}^{2}}
\frac{3c}{2a{b}^{2}}
The bulk modulus of rubber is
9.8 × {10}^{8} N/{m}^{2}
. To what depth a rubber ball be taken in a lake so that its volume is decreased by 0.1%?
The density of metal at normal pressure is
\mathrm{\rho }
. lts density when it is subjected to an excess pressure P is
\mathrm{\rho }
'. lf B is the bulk modulus of the metal, the ratio
\frac{\mathrm{\rho \text{'}}}{\mathrm{\rho }}
1. \(\frac{1}{1-\frac{P}{B}}\)
\quad 1+\frac{B}{P}\quad \quad \quad
\frac{1}{1-{\displaystyle \frac{B}{P}}}\quad
4. \(2+\frac{P}{B}\)
A uniform cylinder rod of length L, cross-sectional area A and Young's modulus Y is acted upon by the forces shown in the figure. The elongation of the rod is:
\frac{3FL}{5AY}
\frac{2FL}{5AY}
\frac{2FL}{8AY}
\frac{8FL}{3AY}
Two wires are made of the same material and have the same volume. However wire 1
has cross sectional area A and Wire 2 has cross-sectional area 3A. lf the length of wire
1 increases by
∆x
on applying force F, how much forces is needed to stretch wire 2 by
A wire of lingth 2 L and radius r is stretched between A and B without the application
of any tension. If Y is the Young's modulus of the wire and it is stretched like ACB,
then the tension in the wire will be (Assume d<<)
\frac{{\mathrm{\pi r}}^{2}{\mathrm{yd}}^{3}}{2{L}^{2}}
\frac{{\mathrm{\pi r}}^{2}{\mathrm{yd}}^{2}}{2{L}^{2}}
\frac{{\mathrm{\pi r}}^{2}\mathrm{Y}.2{\mathrm{L}}^{2}}{{d}^{2}}
\frac{{\mathrm{\pi r}}^{2}Y.2L}{d}
A wire of length L is hanging from a fixed support. The length changes to L1 and L2 when
masses M1 and M2 are suspended respectively from its free end. Then L is equal to-
\frac{{L}_{1}+{L}_{2}}{2}
\sqrt{{L}_{1}{L}_{2}}
\frac{{L}_{1}{M}_{2 }+{L}_{2}{M}_{1}}{{M}_{1}+{M}_{2}}
\frac{{L}_{1}{M}_{2}-{L}_{2}{M}_{2}}{{M}_{2}-{M}_{1}}
Two blocks of masses 1 kg and 2 kg are suspended with the help of two wires having the same
area of cross-section. It the ratio of Young's moduli i.e Y
{}_{1}
: Y
{}_{2}
= 1:2, then the ratio of
extensions produced in wires is:
One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end
of another horizontal thin copper wire of length L and radius R. When the thin copper
wire of length L and radius R. When the arragement is stretched by a applying forces at
two ends, the ratio of the elongation in the thin wire to that in the thick wire is
A wire of length L and cross- sectional area A is made of a material of Young's modulus Y
It is stretched by an amout x, The energy stored is
\frac{YxA}{2L}\quad
\frac{Y{x}^{2}A}{L}\quad \mathbf{\quad }\mathbf{\quad }
\frac{Y{x}^{2}A}{2L}
\quad \frac{2Y{x}^{2}A}{L}
3. Britle
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Topological String Theory Methods of Computer-aided Drug Design/Knots, HOMFLY-PT Polynomial, Chern-Simons Theory and Surgery/Knot theory - Wikibooks, open books for an open world
Topological String Theory Methods of Computer-aided Drug Design/Knots, HOMFLY-PT Polynomial, Chern-Simons Theory and Surgery/Knot theory
< Topological String Theory Methods of Computer-aided Drug Design | Knots, HOMFLY-PT Polynomial, Chern-Simons Theory and Surgery
In this section we are going to review the basics of knot theory. The history of knot theory dates back to ancient times, but the modern mathematical take on the theory started during the early 20th century.
The following video offers a good introduction to knot theory:
Knots[edit | edit source]
In mathematics a knot
{\displaystyle K:S^{1}\hookrightarrow S^{3}}
is a continuous homeomorphism of a circle (
{\displaystyle S^{1}}
) to the
{\displaystyle 3}
-sphere (
{\displaystyle S^{3}=\{(x,y,z,t)\in \mathbb {R} ^{4}|x^{2}+y^{2}+z^{2}+t^{2}=1\}}
{\displaystyle S^{3}}
is also called the one-point compactification of the Euclidean
{\displaystyle 3}
{\displaystyle \mathbb {R} ^{3}=\{(x,y,z)|x,y,z\in \mathbb {R} \}}
), which one can view as the usual Euclidean
{\displaystyle 3}
-dimensional space, but with the points at infinity identified as one, i.e. assuming there are no other barriers, a person inside the manifold can see his back, because the light rays that come from his back goes through the point at infinity and comes back in the opposite direction and reaches his eyes.
Some examples of knots are given below.
Table of knots up to
{\displaystyle 7}
crossings
Unknot (
{\displaystyle 0_{1}}
Trefoil knot (
{\displaystyle 3_{1}}
, also called the
{\displaystyle (2,3)}
-torus knot or
{\displaystyle (3,2)}
-torus knot)
Two knots
{\displaystyle K_{1}}
{\displaystyle K_{2}}
are called the same, denoted
{\displaystyle K_{1}=K_{2}}
, if there is an ambient isotopy, i.e. a continuous mapping
{\displaystyle H:S^{3}\times [0,1]\rightarrow S^{3}}
{\displaystyle t\in [0,1]}
{\displaystyle H(-,t):x\in S^{3}\mapsto H(x,t)\in S^{3}}
is an orientation-preserving homeomorphism.
{\displaystyle H(x,0)=x}
{\displaystyle x\in S^{3}}
{\displaystyle H(K_{1},1)=K_{2}}
In other words, two knots are the same if one can be continuously deformed into the other without crossing each other.
In knot theory, it is natural to generalize the concept of knots to links, where, instead of
{\displaystyle 1}
circle, we consider
{\displaystyle \mu }
copies of the circle in
{\displaystyle S^{3}}
{\displaystyle \mu =0,1,2,...\in \mathbb {Z} _{\geq 0}}
, with no self-intersections in
{\displaystyle S^{3}}
. Each circle in the link
{\displaystyle L}
is called a component, and the number of components is usually denoted by
{\displaystyle \mu (L)}
Examples of links are given below:
{\displaystyle (2,4)}
-torus link
The equivalence of links is similar to that for knots, i.e. by the existence of an ambient isotopy.
Regular diagrams and Reidemeister moves[edit | edit source]
Links are usually visualized using projections onto a plane that have intersection points between only two portions of the curves. The portion at a higher altitude is drawn continuously, and that at a lower altitude is drawn broken by convention. Diagrams obtained this way are called regular diagrams
{\displaystyle D(L)}
of a link
{\displaystyle L}
An important question in knot theory is the so-called recognition problem: given two links
{\displaystyle L_{1}}
{\displaystyle L_{2}}
, prove or disprove that
{\displaystyle L_{1}=L_{2}}
It is therefore natural to ask if one can reduce the problem of showing
{\displaystyle L_{1}=L_{2}}
into that of showing
{\displaystyle D(L_{1})\sim D(L_{2})}
for some equivalence relation
{\displaystyle \sim }
. The answer is yes: we denote
{\displaystyle D(L_{1})\sim D(L_{2})}
{\displaystyle D(L_{1})}
{\displaystyle D(L_{2})}
by a finite sequence of Reidemeister moves, their inverses and planar isotopy.
The Reidemeister moves are given by:
Reidemeister move 1 (
{\displaystyle \Omega _{1}}
{\displaystyle \Omega _{2}}
{\displaystyle \Omega _{3}}
For example, the following shows how to convert a regular diagram into an unknot:
Unknotting a complicated regular diagram
It is Reidemeister's theorem that
{\displaystyle L_{1}=L_{2}\iff D(L_{1})\sim D(L_{2})}
Disproving that two links
{\displaystyle L_{1},L_{2}}
are not the same, i.e. proving
{\displaystyle L_{1}\neq L_{2}}
, is much more difficult, and will be discussed in the next section.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Topological_String_Theory_Methods_of_Computer-aided_Drug_Design/Knots,_HOMFLY-PT_Polynomial,_Chern-Simons_Theory_and_Surgery/Knot_theory&oldid=3815768"
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The Halting Problem!
Need a quick buck? A while ago someone offered $1000 for a developer to “create a debugger program. This program will take as input the source code of another program, and will analyze that other program and determine if it will run to completion, or have an error, or go into an infinite loop.”
What do you think — would you take the contract?
One-time pads 🕵️♀️
Your field operative Alice sends you a top-secret message: hqsoew. Many past spies have been executed after their messages were deciphered. But this time, you and Alice are using a one-time pad — the simplest and most secure cipher that exists!
In this tutorial, you’ll learn how to decipher Alice’s message, and save Alice and your country along the way.
Let’s save the nation! 💪
The quadratic formula: a tutorial
So you need to solve
\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 2x^2 - 3x = 2
. You remember the word ‘quadratic’ and something like
\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \sqrt{4ac - b^2}
… or was it
\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} +b^2
? 😨
Stop trying to memorize the quadratic formula! In this interactive tutorial, you’ll rediscover the quadratic formula with me, and you’ll never have to memorize it again! 💪
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Magnetism and Matter & EMI - Revision Session - NEET & AIIMS 2020
Magnetism and Matter & EMI - Revision Session - NEET & AIIMS 2020Contact Number: 9667591930 / 8527521718
Assertion : Two coaxial conducting rings of different radii are placed in space. The mutual inductance of both the rings maximum if the rings are coplanar.
Reason : For two coaxial conducting rings of different radii, the magnitude of magnetic flux in one ring due to current in other ring is maximum when both rings are coplanar
Assertion: No electric current will be present within a region having a uniform and constant magnetic field.
Reason: Within a region of uniform and constant magnetic field
\overrightarrow{\mathrm{B}}
the path integral of the magnetic field
\mathrm{\varphi }\overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{dl}}
along any closed path is zero. Hence from ampere circuital law
\mathrm{\varphi }\overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{dl}}\quad =\quad {\mathrm{\mu }}_{0}\mathrm{l}
(where the given terms have usual meaning), no current can be present within a region having a uniform and constant magnetic field.
(1) Statement I is True, Statement II is True, Statement II is correct explanation for Statement I.
(2) Statement I is True, Statement II is True, Statement II is NOT a correct explanation for Statement I.
( 3) Statement I is True, Statement II is false.
( 4) Statement I is False, Statement II is true.
Assertion: Two identical circulars closed loops made of copper and aluminium are withdrawn from the magnetic field with equal velocities. The induced emf is same but induced current is different.
Reason : Induced current depends on the resistance of the circuit.
Assertion: A resistance R is connected between the two ends of parallel smooth conducting rails. A conducting rod lies on these fixed horizontal rails and a uniform constant magnetic field B exists perpendicular to the plane of the rails as shown in the figure. If the rod is given a velocity v and released as shown in the figure, it will stop after some time. The total work done by the magnetic field is negative.
Reason: If a force acts opposite to the direction of velocity, its work done is negative.
Assertion : The poles of a magnet cannot be separated by breaking into two pieces.
Reason : The magnetic moment will be reduced to half when magnet is broken into two equal pieces
Assertion: If a compass needle is kept at the magnetic north pole of the earth, the compass needle may stay in any direction.
Reason : Dip needle will stay vertical at the north pole of the earth.
Assertion : The earth's magnetic field is due to iron present in ts core.
Reason : At a high-temperature magnet losses its magnetic property or magnetism.
Assertion : The susceptibility of diamagnetic materials does not depend upon temperature.
Reason : Every atom of a diamagnetic material is not a complete magnet is itself.
A metallic rod of length is tied to a string of length 21 and made to rotate with angular speed won a horizontal table with one end of the string fixed. If there is a vertical magnetic field 'B' in the region, the e.m.f. induced across the ends of the rod is
\frac{2{\mathrm{B\omega l}}^{2}}{2}
\frac{3{\mathrm{B\omega l}}^{2}}{2}
\frac{4{\mathrm{B\omega l}}^{2}}{2}
\frac{5{\mathrm{B\omega l}}^{2}}{2}
A simple pendulum with a bob of mass m and conducting wire of L swings under gravity through an angle 2
\mathrm{\theta }
. The earth's magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is
2\mathrm{BL} \mathrm{sin}\left(\frac{\mathrm{\theta }}{2}\right){\left(\mathrm{gL}\right)}^{1/2}
\mathrm{BL} \mathrm{sin}\left(\frac{\mathrm{\theta }}{2}\right)\left(\mathrm{gL}\right)
\mathrm{BL} \mathrm{sin}\left(\frac{\mathrm{\theta }}{2}\right){\left(\mathrm{gL}\right)}^{3/2}
\mathrm{BL} \mathrm{sin}\left(\frac{\mathrm{\theta }}{2}\right){\left(\mathrm{gL}\right)}^{2}
An inductor (L= 100 mH), a resistor (R = 100
\mathrm{\Omega }
), and a battery (E = 100V) are initially connected in series as shown in the figure. After a long time, the battery is disconnected after short-circuiting the points A and B. The current in circuit 1 ms after the short circuit is
An inductor of inductance L = 400 mH and resistors of resistance of
{\mathrm{R}}_{1} = 2\mathrm{\Omega }
{\mathrm{R}}_{2} = 2\mathrm{\Omega }
are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across Las function of time is
\frac{12}{\mathrm{t}}{\mathrm{e}}^{-3\mathrm{t}}\mathrm{V}
6\left(1 - {\mathrm{e}}^{-\mathrm{t}/0.2}\right)\mathrm{V}
12{\mathrm{e}}^{-5\mathrm{t}}\mathrm{V}
6{\mathrm{e}}^{-5\mathrm{t}} \mathrm{V}
A dip needle lies initially in the magnetic meridian when it shows an angle of dip
\mathrm{\theta }
at a place. The dip circle is rotated through an angle x in the horizontal plane and then it shows an angle of dip
\mathrm{\theta }
'. Then
\frac{\mathrm{tan} \mathrm{\theta }\text{'}}{\mathrm{tan} \mathrm{\theta }}
1. 1/cos x
2. 1/sin x
3. 1/tan x
The figure shows the various positions (labelled by subscripts) of small magnetised needles P and Q. The arrows show the direction of their magnetic moment. Which configuration corresponds to the lowest potential energy among all the configurations shown.
(1) PQ3
Two tangent galvanometers A and B have coils of radii 8 cm and 16 cm respectively and resistance 8
\mathrm{\Omega }
each. They are connected in parallel with a cell of emf 4 V and negligible internal resistance. The deflections produced in the tangent galvanometers A and B are 30° and 60° respectively. If A has 2 turns, then B must have
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Home : Support : Online Help : Mathematics : Factorization and Solving Equations : RegularChains : Rank
rank of a nonconstant polynomial
Rank(p, R)
The function call Rank(p,R) returns the rank of p with respect to the variable ordering of R, that is, the main variable of p raised to its main degree.
This command is part of the RegularChains package, so it can be used in the form Rank(..) only after executing the command with(RegularChains). However, it can always be accessed through the long form of the command by using RegularChains[Rank](..).
\mathrm{with}\left(\mathrm{RegularChains}\right):
R≔\mathrm{PolynomialRing}\left([x,y,z]\right)
\textcolor[rgb]{0,0,1}{R}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{polynomial_ring}}
p≔\left(y+1\right){x}^{3}+\left(z+4\right)x+3
\textcolor[rgb]{0,0,1}{p}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}
\mathrm{MainVariable}\left(p,R\right)
\textcolor[rgb]{0,0,1}{x}
\mathrm{Initial}\left(p,R\right)
\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}
\mathrm{MainDegree}\left(p,R\right)
\textcolor[rgb]{0,0,1}{3}
\mathrm{Rank}\left(p,R\right)
{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{3}}
\mathrm{Tail}\left(p,R\right)
\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}
R≔\mathrm{PolynomialRing}\left([z,y,x]\right)
\textcolor[rgb]{0,0,1}{R}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{polynomial_ring}}
p≔\mathrm{expand}\left(\left(y+1\right){x}^{3}+\left(z+4\right)x+3\right)
\textcolor[rgb]{0,0,1}{p}\textcolor[rgb]{0,0,1}{≔}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}
\mathrm{MainVariable}\left(p,R\right)
\textcolor[rgb]{0,0,1}{z}
\mathrm{Initial}\left(p,R\right)
\textcolor[rgb]{0,0,1}{x}
\mathrm{MainDegree}\left(p,R\right)
\textcolor[rgb]{0,0,1}{1}
\mathrm{Rank}\left(p,R\right)
\textcolor[rgb]{0,0,1}{z}
\mathrm{Tail}\left(p,R\right)
{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}
R≔\mathrm{PolynomialRing}\left([z,y,x],3\right)
\textcolor[rgb]{0,0,1}{R}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{polynomial_ring}}
p≔{\left(x+y\right)}^{3}{z}^{3}+3{z}^{2}+2z+y+4
\textcolor[rgb]{0,0,1}{p}\textcolor[rgb]{0,0,1}{≔}{\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\right)}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}
\mathrm{MainVariable}\left(p,R\right)
\textcolor[rgb]{0,0,1}{z}
\mathrm{Initial}\left(p,R\right)
{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{3}}
\mathrm{MainDegree}\left(p,R\right)
\textcolor[rgb]{0,0,1}{3}
\mathrm{Rank}\left(p,R\right)
{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{3}}
\mathrm{Tail}\left(p,R\right)
\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}
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BidiakisCube - Maple Help
Home : Support : Online Help : Mathematics : Discrete Mathematics : Graph Theory : GraphTheory Package : SpecialGraphs : BidiakisCube
construct Bidiakis cube
BidiakisCube()
The BidiakisCube() command returns the Bidiakis cube, a 3-regular graph with 12 vertices and 18 edges.
The Bidiakis cube is a cubic Hamiltonian graph and its girth is 4.
\mathrm{with}\left(\mathrm{GraphTheory}\right):
\mathrm{with}\left(\mathrm{SpecialGraphs}\right):
B≔\mathrm{BidiakisCube}\left(\right)
\textcolor[rgb]{0,0,1}{B}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Graph 1: an undirected unweighted graph with 12 vertices and 18 edge\left(s\right)}}
\mathrm{IsRegular}\left(B\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{DrawGraph}\left(B\right)
"Bidiakis cube", Wikipedia. http://en.wikipedia.org/wiki/Bidiakis_cube
The GraphTheory[SpecialGraphs][BidiakisCube] command was introduced in Maple 2021.
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Positive Solutions for Discrete Boundary Value Problems to One-Dimensional p-Laplacian with Delay
2013 Positive Solutions for Discrete Boundary Value Problems to One-Dimensional
p
-Laplacian with Delay
Linjun Wang, Xumei Chen
We study the existence of positive solutions for discrete boundary value problems to one-dimensional
p
-Laplacian with delay. The proof is based on the Guo-Krasnoselskii fixed-point theorem in cones. Two numerical examples are also provided to illustrate the theoretical results.
Linjun Wang. Xumei Chen. "Positive Solutions for Discrete Boundary Value Problems to One-Dimensional
p
-Laplacian with Delay." J. Appl. Math. 2013 1 - 8, 2013. https://doi.org/10.1155/2013/157043
Linjun Wang, Xumei Chen "Positive Solutions for Discrete Boundary Value Problems to One-Dimensional
p
-Laplacian with Delay," Journal of Applied Mathematics, J. Appl. Math. 2013(none), 1-8, (2013)
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Class 10 (All in one) - Polynomials
Class 10 (All in one) - PolynomialsContact Number: 9667591930 / 8527521718
For what value of k, 3 is a zero of the polynomial 2x2 + x + k?
If 2 is a zero of polynomial f(x) = ax2 – 3(a – 1)x – 1, then find the value of a.
If 2 and 3 are zeroes of polynomial 3x2 – 2kx + 2m, then find the values of k and m.
What is the geometrical meaning of the zeroes of a polynomial?
The graph of y = p(x) is given, where p(x) is a polynomial. Find the number of zeroes of p(x).
Draw the graph of the linear polynomial x + 5 and also find the zeroes of the polynomial.
Find the zeroes of the quadratic polynomial y2 – 92y + 1920.
If zeroes α and β of a polynomial x2 – 7x + k are such that α – β = 1, then find the value of k.
If α and β are the zeroes of the polynomial 2y2 + 7y + 5, then find the value of α + β + αβ.
If α and β are the zeroes of the quadratic polynomial f(x) = 3x2 – 5x – 2, then evaluate α3 + β3.
If α and β are the zeroes of 4x2 + 3x + 7, then find the value of
\frac{1}{\alpha }+\frac{1}{\beta }.
If the sum and difference of zeroes of quadratic polynomial are –3 and –10, respectively. Then, find the difference of the squares of zeroes.
If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is –1, then find the product of the other two zeroes.
Two zeroes of cubic polynomial ax3 + 3x2 – bx –6 are –1 and –2. Find the third zero and values of a and b.
Find the quadratic polynomial, whose sum of zeroes is 8 and their product is 12. Then, find the zeroes of the polynomial.
Find the quadratic polynomial whose zeroes are
2\sqrt{\text{7}}
-5\sqrt{\text{7}}.
Find the quadratic polynomial whose zeroes are 2 and –6, respectively. Verify the relation between the coefficients and zeroes of the polynomial.
If 1 and –1 are zeroes of polynomial Lx4 + Mx3 + Nx2 + Rx + P, then show that L + N + P = M + R.
How many polynomials will have their zeroes as –2 and 5?
If α and β are zeroes of the quadratic polynomial p(x) = 6x2 + x – 1, then find the value of
\frac{\text{α}}{\text{β}}+\frac{\text{β}}{\text{α}}+2\left(\frac{1}{\text{α}}+\frac{1}{\text{β}}\right)+3\text{αβ}
If α and β are zeroes of the quadratic polynomial f(x) = x2 – 3x – 2, find a polynomial whose zeroes are
\frac{\text{2α}}{\text{β}}\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\frac{\text{2β}}{\text{α}}
\left(\text{2α}+\text{3β}\right)\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\left(\text{3α}+\text{2β}\right)
\frac{{\text{α}}^{\text{2}}}{\text{β}}\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\frac{{\text{β}}^{\text{2}}}{\text{α}}
\frac{\text{1}}{\text{2α}+\text{β}}\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}}\frac{\text{1}}{\text{2β}+\text{α}}
On dividing a polynomial p(x) by 3x + 1, the quotient is 2x – 3 and the remainder is –2. Find p(x).
What will be the quotient and the remainder on division of ax2 + bx + c by px3 + qx2 + rx + 5, p ≠ 0
Divide the polynomial p(x) by the polynomial g(x) and verify the division algorithm in each of the following.
(i) p(x) = 2x4 – 2x3 – 5x2 – x + 8, g(x) = 2x2 + 4x + 3
(ii) p(x) = 10x4 + 17x3 – 62x2 + 30x – 3, g(x) = 2x2 + 7x – 1
Find the value of k, for which polynomial p(x) is exactly divisible by polynomial g(x), in each of the following
(i) p(x) = x3 + 8x2 + kx + 18, g(x) = x2 + 6x + 9
(ii) p(x) = x4 + 10x3 + 25x2 + 15x + k, g(x) = 6x + 7
If the polynomial 6x4 + 8x3 + 17x2 + 21x + 7 is divided by another polynomial 3x2 + 4x + 1, the remainder comes out to be ax + b, then find the values of a and b.
A polynomial g(x) of degree zero is added to the polynomial 2x3 + 5x2 – 14x + 10, so that it becomes exactly divisible by 2x – 3. Find g(x).
If the polynomial f(x) = 3x4 – 9x3 + x2 + 15x + k is completely divisible by 3x2 – 5, then find the value of k and hence the other two zeroes of the polynomial.
The graphs of y = p(x), where p(x) is a polynomial in x are given. Find the number of zeroes of p(x) in each case. For each case, also state whether p(x) is linear or quadratic.
Is the following statement True or False? Justify your answer. ‘If the zeroes of a quadratic polynomial ax2 + bx + c are both negative, then a, b and c all have the same sign.’
If one zero of 2x2 – 3x + k is reciprocal to the other, then find the value of k.
If sum of the squares of zeroes of the quadratic polynomial f(x) = x2 – 4x + k is 20, then find the value of k.
If the zeroes of the quadratic polynomial ax2 + bx + c, where c ≠ 0, are equal, then show that c and a have same sign.
Can (x – l)be the remainder on division of a polynomial, p(x) by (2x + 3)? Justify your answer.
Write whether the following expressions are polynomials or not. Give reasons for your answer.
{x}^{3}+\frac{1}{{x}^{2}}+\frac{1}{x}+1
{x}^{2}+x+3
{y}^{-1/2}-3y+2
\sqrt{2}{y}^{3}+\sqrt{3}y
If one zero of the polynomial (a2 + 9)x2 + 13x + 6a is reciprocal of the other, then find the value of a.
The sum of remainders obtained when x3 + (k + 8) x + k is divided by x – 2 and when it is divided by x + 1, is 0. Find the value of k.
Find the zeroes of the given polynomial by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials
7{y}^{2}-\frac{11}{3}y-\frac{2}{3}.
Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and their coefficients.
q\left(x\right)=\sqrt{3}{x}^{2}+10x+7\sqrt{3}
If the zeroes of the polynomial ax2 + bx + b = 0 are in the ratio m : n, then find the value of
\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}.
If α and β are the zeroes of the quadratic polynomial f(x) = px2 + qx + r, then evaluate
\frac{1}{p\text{α}+q}+\frac{1}{p\text{β}+q}.
If α and β are the zeroes of the quadratic polynomial p(s) = 3s2 – 6s + 4, then find the value of
\frac{\text{α}}{\text{β}}+\frac{\text{β}}{\text{α}}+2\left(\frac{\text{1}}{\text{α}}+\frac{\text{1}}{\text{β}}\right)+3\text{αβ}.
If α and β are the zeroes of the quadratic polynomial f(x) = x2 – px + q, then prove that
\frac{{\text{α}}^{\text{2}}}{{\text{β}}^{\text{2}}}+\frac{{\text{β}}^{\text{2}}}{{\text{α}}^{\text{2}}}=\frac{{p}^{4}}{{q}^{2}}-\frac{4{p}^{2}}{q}+2.
Ajay, Ankit and Vijay respectively calculated the following polynomials with sum of the zeroes as 18 and product of the zeroes as 81.
x2 – 18x + 81, x2 + 18x – 81, 2x2 – 9x – 81
They discussed their solutions among themselves and point out mistakes in the calculations.
(i) Whose calculation is correct?
(ii) What are the values depict here?
Remainder on dividing x3 + 2x3 + kx + 3 by x – 3 is 21. Ahmed was asked to find the quotient. He was little puzzled and was thinking how to proceed. His classmate Vidya helped him by suggesting that he should first find the value of k and then proceed further.
(i) Explain how the question was solved?
(ii) What value is indicated from Vidya's action?
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Bertram_Kostant Knowpia
Bertram Kostant (May 24, 1928 – February 2, 2017)[1] was an American mathematician who worked in representation theory, differential geometry, and mathematical physics.
Bertram Kostant at a workshop on “Enveloping Algebras and Geometric Representation Theory” in Oberwolfach, 2009
Kostant's convexity theorem
Kostant partition function
Kostant polynomial
Kostant–Parthasarathy–Ranga Rao–Varadarajan determinants
Representations of a Lie algebra and its enveloping algebra on a Hilbert space
Moss Sweedler
Kostant grew up in New York City, where he graduated from Stuyvesant High School in 1945.[2] He went on to obtain an undergraduate degree in mathematics from Purdue University in 1950. He earned his Ph.D. from the University of Chicago in 1954, under the direction of Irving Segal, where he wrote a dissertation on representations of Lie groups.
Career in mathematicsEdit
After time at the Institute for Advanced Study, Princeton University, and the University of California, Berkeley, he joined the faculty at the Massachusetts Institute of Technology, where he remained until his retirement in 1993. Kostant's work has involved representation theory, Lie groups, Lie algebras, homogeneous spaces, differential geometry and mathematical physics, particularly symplectic geometry. He has given several lectures on the Lie group E8.[3] He has been one of the principal developers of the theory of geometric quantization. His introduction of the theory of prequantization has led to the theory of quantum Toda lattices. The Kostant partition function is named after him. With Gerhard Hochschild and Alex F. T. W. Rosenberg, he is one of the namesakes of the Hochschild–Kostant–Rosenberg theorem which describes the Hochschild homology of some algebras.[4]
His students include James Harris Simons, James Lepowsky, Moss Sweedler, David Vogan, and Birgit Speh. At present he has more than 100 mathematical descendants.
Kostant's honors include election to the National Academy of Sciences in 1978. In 2012 he became a fellow of the American Mathematical Society.[5] He was an Invited Speaker of the ICM with talk Orbits and quantization theory in 1970 at Nice.[6]
Kostant, Bertram (1955). "Holonomy and the Lie algebra of infinitesimal motions of a Riemannian manifold". Trans. Amer. Math. Soc. 80 (2): 528–542. doi:10.1090/S0002-9947-1955-0084825-8.
Kostant, Bertram (1959). "A formula for the multiplicity of a weight". Trans. Amer. Math. Soc. 93 (6): 53–73. doi:10.1090/S0002-9947-1959-0109192-6. PMC 528626. PMID 16590246.
Kostant, Bertram (1961). "Lie algebra cohomology and the generalized Borel-Weil theorem" (PDF). Annals of Mathematics. 74 (2): 329–387. doi:10.2307/1970237. hdl:2027/mdp.39015095258318. JSTOR 1970237.
Kostant, Bertram (1963). "Lie group representations on polynomial rings". American Journal of Mathematics. 85 (3): 327–404. doi:10.2307/2373130. JSTOR 2373130.
Kostant, Bertram (1969). "On the existence and irreducibility of certain series of representations". Bulletin of the American Mathematical Society. 75 (4): 627–642. doi:10.1090/S0002-9904-1969-12235-4.
Kostant, Bertram (1970). "Quantization and unitary representations". In: Lectures in modern analysis and applications III. Lecture Notes in Mathematics 170. Vol. 170. pp. 87–208. doi:10.1007/BFb0079068. ISBN 978-3-540-05284-5.
with Louis Auslander: Auslander, L.; Kostant, B. (1971). "Polarization and unitary representations of solvable Lie groups". Inventiones Mathematicae. 14 (4): 255–354. Bibcode:1971InMat..14..255A. doi:10.1007/BF01389744. S2CID 122009744.
with Stephen Rallis: Kostant, B.; Rallis, S. (1971). "Orbits and representations associated with symmetric spaces". American Journal of Mathematics. 93 (3): 753–809. doi:10.2307/2373470. JSTOR 2373470.
Kostant, Bertram (1973). "On convexity, the Weyl group and the Iwasawa decomposition". Annales Scientifiques de l'École Normale Supérieure. 6 (4): 413–455. doi:10.24033/asens.1254.
Kostant, Bertram (1977). "Graded manifolds, graded Lie theory, and prequantization". In: Differential Geometrical Methods in Mathematical Physics. Lecture Notes in Math 570. Vol. 570. pp. 177–306. doi:10.1007/Bfb0087788. ISBN 978-3-540-08068-8.
Kostant, Bertram (1978). "On Whittaker vectors and representation theory". Inventiones Mathematicae. 48 (2): 101–184. Bibcode:1978InMat..48..101K. doi:10.1007/BF01390249. S2CID 122765132.
with David Kazhdan and Shlomo Sternberg: Kazhdan, D.; Kostant, B.; Sternberg, S. (1978). "Hamiltonian group actions and dynamical systems of Calogero type". Communications on Pure and Applied Mathematics. 31 (4): 481–507. doi:10.1002/cpa.3160310405.
Kostant, Bertram (1979). "The solution to a generalized Toda lattice and representation theory". Advances in Mathematics. 34 (3): 195–338. doi:10.1016/0001-8708(79)90057-4.
with Shrawan Kumar: Kostant, Bertram; Kumar, Shrawan (1986). "The nil Hecke ring and cohomology of GP for a Kac-Moody group G". Advances in Mathematics. 62 (3): 187–237. doi:10.1016/0001-8708(86)90101-5. PMC 323118. PMID 16593661.
with Shlomo Sternberg: Kostant, Bertram; Sternberg, Shlomo (1987), "Symplectic reduction, BRS cohomology, and infinite-dimensional Clifford algebras", Annals of Physics, 176 (1): 49–113, Bibcode:1987AnPhy.176...49K, doi:10.1016/0003-4916(87)90178-3
"On Laguerre polynomials, Bessel functions, Hankel transform and a series in the unitary dual of the simply-connected covering group of
{\displaystyle {\mbox{SL}}(2,\mathbb {R} )}
". Represent. Theory. 4: 181–224. 2000. doi:10.1090/S1088-4165-00-00096-0.
with Gerhard Hochschild and Alex Rosenberg: Hochschild, G.; Kostant, Bertram; Rosenberg, Alex (2009). "Differential forms on regular affine algebras". In: Collected Papers. New York: Springer. pp. 265–290. doi:10.1007/b94535_14. ISBN 978-0-387-09582-0. (Reprinted from Trans. Amer. Math. Soc., vol. 102, no. 3, March 1962, pp. 383–408)
Kostant, Bertram (2009). "The principal three-dimensional subgroup and the Betti numbers of a complex simple Lie group". In: Collected Papers. pp. 130–189. doi:10.1007/b94535_11. ISBN 978-0-387-09582-0. (Reprinted from the Amer. J. Math., vol. 81, no. 4, Oct. 1959)
Symplectic spinor bundle
^ "Bertram Kostant, professor emeritus of mathematics, dies at 88". MIT News. 16 February 2017.
^ "Professor Kostant's Homepage". MIT Math Department. Retrieved 2007-10-31.
^ Bertram Kostant (2008-02-12). "On Some Mathematics in Garrett Lisi's 'E8 Theory of Everything'". UC Riverside mathematics colloquium. Retrieved 2008-06-15.
^ Porter, Tim (April 8, 2014), "Hochschild-Kostant-Rosenberg theorem", nLab .
^ Kostant, Bertram. "Orbits and quantization theory." Archived 2015-03-26 at the Wayback Machine In Actes, Congrès intern. Math., Nice, vol. 2, pp. 395–400. 1970.
Kostant's home page at MIT
Bertram Kostant at the Mathematics Genealogy Project
Brylinski, Jean-Luc (editor) (1994). Lie Theory and Geometry, In Honor of Bertram Kostant. Boston, Birkhäuser. ISBN 0-8176-3761-3. {{cite book}}: |author= has generic name (help)
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Respiration in Plants - Live Session - NEET & AIIMS 2019Contact Number: 9667591930 / 8527521718
Which of the following cofactors are used during formation of Acetyl-CoA?
(a) CoA, TPP,
{\mathrm{Mg}}^{++}
Mg (b) CoA,
{\mathrm{Mg}}^{++}
{\mathrm{NAD}}^{+}
(c) CoA, TPP, NAD+,
{\mathrm{NAD}}^{+}
(d) CoA, TPP,
{\mathrm{NAD}}^{+}
{\mathrm{Mg}}^{++}
, Lipoic acid
How many Co2 molecules are formed during formation of two molecules of Acetyl CoA in matrix?
Substrate level phosphorylation during TCA cycle occurs:
(a) Once (b) Twice (c) 3 times (d) 6 times
The total number of ATP produced during conversion of succinate to oxaloacetate is
Which is the co-factor of succinic dehydrogenase?
(a) NAD (b) FAD (c) NADP (d) TPP
In Kreb’s cycle FAD participate as electron acceptor during conversion of
(a) Succinyl Co-A to succinic acid (b) α- ketoglutarate to succinyl CoA
(c) Succinic acid to fumaric acid (d) Fumaric acid to malic acid
Mobile carriers of electrons in ETC are
(a) Cytochrome b and Ubiquinone (b) Cytochrome c and Coenzyme Q
(c) Cytochrome a and Cytochrome
{\mathrm{a}}_{3}
(d) FeS complex and Cytochrome
{\mathrm{c}}_{1}
{\mathrm{NADPH}}_{2}
are produced from the oxidation of one glucose molecule during HMP
path way?
Which intermediate of Kreb’s cycle will effect photosynthesis?
(a) Succinyl CoA (b) DHAP (c) PGAL (d) 1, 3 DPGA
All reaction of TCA cycle occur in matrix of mitochondria except
(a) Oxidation of malate (b)Reduction of FAD
(c) Decarboxylation of oxalosuccinic acid (d) Break down of malic acid
The number of dehydrogenation reaction in TCA cycle is
The complete combustion of glucose, which produces i and ii as end products, yields energy most of which is given out as iii .
(a) NADH+
{\mathrm{H}}^{+}
ATP Light
{\mathrm{CO}}_{2}
{\mathrm{H}}_{2}\mathrm{O}
(c) ATP
{\mathrm{CO}}_{2}
(d) NADH+
{\mathrm{H}}^{+}
{\mathrm{CO}}_{2}
Pick out the incorrect option w.r.t glycolysis.
(a) Present in all living organisms.
(b) Glucose undergoes complete oxidation.
(c) Common to both aerobic and anaerobic respiration.
(d) Also called as EMP pathway.
Which statements are true for the fermentation that takes place in yeast?
(i) Pyruvic acid converted into ethanol.
(ii) Takes place in anaerobic conditions.
(iii) Less than 7% of the energy in glucose is released.
(iv) Enzymes involved are pyruvic acid dehydrogenase and alcohol carboxylase.
(v) No CO2 production.
(a) i, ii, iii, v (b) i, ii, iii
(c) ii, iii ,v (d) i, ii, iii, iv
i. Complex V - a. NADH dehydrogenase
ii. Complex I - b. Cytochrome bc1 complex
iii. Complex IV - c. ATP synthase
iv. Complex III - d. Cytochrome c oxidase
(a) i-c, ii-a, iii-d, iv-b (b) i-b, ii-a, iii-d, iv-c
(c) i-c, ii-a, iii-b, iv-d (d) i-b, ii-d, iii-c, iv-a
i. Integral membrane protein complex that forms the channel through which protons cross the inner mitochondrial membrane.
ii. Form of redox-equivalents are removed from PGAL during glycolysis.
iii. Peripheral membrane protein complex that contains the site for ATP synthesis from ADP and inorganic phosphate.
(a) i-
{\mathrm{F}}_{0}
ii- Hydrogen atoms iii-
{\mathrm{F}}_{1}
(b) i-
{\mathrm{F}}_{1}
ii- Oxygen atom iii-
{\mathrm{F}}_{0}
(c) i-
{\mathrm{F}}_{0}
ii- Oxygen molecule iii-
{\mathrm{F}}_{1}
(d) i-
{\mathrm{F}}_{1}
{\mathrm{F}}_{0}
i. In glycolysis, glucose undergoes partial oxidation to form pyruvic acid.
ii. NADH is oxidized to NAD+ rather slowly in fermentation, however the reaction is very vigorous in case of aerobic respiration.
iii. During oxidation within a cell, all the energy contained in respiratory substrates is not released free into the cell, or in a single step.
(a) i, ii, iii are true (b) i is false but ii and iii are true
(c) i and ii are true but iii is false (d) ii is false but i and iii are true
(a) Fermentation takes place in many prokaryotes and unicellular eukaryotes.
(b) Glycerol enters the respiratory pathway after being converted to acetyl CoA.
(c) The complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, leaving molecules of
{\mathrm{CO}}_{2}
(d) Cytochrome c oxidasecomplex contains cytochromes a and
{\mathrm{a}}_{3}
, and two copper centres
The ATP calculations of respiratory pathway can be made only on assumptions like
(a) None of the intermediates in the pathway are utilised to synthesise any other compound.
(b) Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.
(c) There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
Link enzyme in cellular respiration is
(a) citrate synthetase (b) pyruvate dehydrogenase
(c) isocitrate dehydrogenase (d) succinyl thiokinase
Oxidation of one molecule of NADH gives rise to
(a) 3 ATP molecules (b) 12 ATP molecules
(c) 2 ATP molecules (d) 1 ATP molecule
Which one of the following is the terminal electron acceptor?
(a) Molecular
{\mathrm{CO}}_{2}
(b) Molecular
{\mathrm{O}}_{2}
(c) Molecular
{\mathrm{H}}_{2}
{\mathrm{NADPH}}_{2}
The chemiosmotic coupling hypothesis of oxidative phosphorylation proposes that adenosine triphosphate (ATP) is formed because
(a) high energy bonds are formed in mitochondrial proteins
(b) ADP is pumped out of the matrix into the intermembrane space
(c) a proton gradient forms across the membrane
(d) there is a change in the permeability of the inner mitochondrial membrane towards adenosine diphosphate (ADP)
What is the net ATP molecules gain, when 4 molecules of glucose undergo anaerobic respiration in plants?
(a) 8 ATP (b) 20 ATP
(c) 144 ATP (d) 16 ATP
In photosynthesis,
{\mathrm{NADPH}}_{2}
is formed but in respiration it forms during
(a) HMP (b) ETS
(c) Krebs' cycle (d) None of these
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formic acid, sodium salt, sodium hydrocarbon dioxide
387AD98770 Y
For commercial use, sodium formate is produced by absorbing carbon monoxide under pressure in solid sodium hydroxide at 130 °C and 6-8 bar pressure:[1]
Because of the low-cost and large-scale availability of formic acid by carbonylation of methanol and hydrolysis of the resulting methyl formate, sodium formate is usually prepared by neutralizing formic acid with sodium hydroxide. Sodium formate is also unavoidably formed as a by-product in the final step of the pentaerythritol synthesis and in the crossed Cannizzaro reaction of formaldehyde with the aldol reaction product trimethylol acetaldehyde [3-hydroxy-2,2-bis(hydroxymethyl)propanal].[2]
In the laboratory, sodium formate can be prepared by neutralizing formic acid with sodium carbonate. It can also be obtained by reacting chloroform with an alcoholic solution of sodium hydroxide.
Sodium formate may also be created via the haloform reaction between ethanol and sodium hypochlorite in the presence of a base. This procedure is well documented for the preparation of chloroform.
Sodium formate crystallizes in a monoclinic crystal system with the lattice parameters a = 6,19 Å, b = 6,72 Å, c = 6,49 Å and β = 121,7°.[3]
On heating, sodium formate decomposes to form sodium oxalate and hydrogen.[4] The resulting sodium oxalate can be converted by further heating to sodium carbonate upon release of carbon monoxide:[5][4]
{\displaystyle {\ce {2HCOONa->[\Delta ]{(COO)2Na2}+H2\!\uparrow }}}
{\displaystyle {\ce {(COO)2Na2->[{} \atop >\ {\ce {290^{o}C}}]{Na2CO3}+CO\!\uparrow }}}
As a salt of a weak acid (formic acid) and a strong base (sodium hydroxide) sodium formate reacts in aqueous solutions basic:
{\displaystyle {\ce {HCOO^- + H2O <<=> HCOOH + OH^-}}}
A solution of formic acid and sodium formate can thus be used as a buffer solution.
Sodium formate is slightly water-hazardous and inhibits some species of bacteria but is degraded by others.
In structural biology, sodium formate can be used as a cryoprotectant for X-ray diffraction experiments on protein crystals,[6] which are typically conducted at a temperature of 100 K to reduce the effects of radiation damage.
Sodium formate plays a role in the synthesis of formic acid, it is converted by sulfuric acid via the following reaction equation:
{\displaystyle \mathrm {2\ HCOONa+H_{2}SO_{4}\longrightarrow 2\ HCOOH+Na_{2}SO_{4}} }
Sodium formate is converted with sulfuric acid to formic acid and sodium sulfate.
Solid sodium formate is used as a non-corrosive agent at airports for de-icing of runways in mix with corrosion inhibitors and other additives, which rapidly penetrate solid snow and ice layers, detach them from the asphalt or concrete and melt the ice rapidly. Sodium formate was also used as a road deicer in the city of Ottawa from 1987 to 1988.[7]
The high freezing point depression e.g. in comparison to the still frequently used urea (which is effective but problematic due to eutrophication) effectively prevents the re-icing, even at temperatures below −15 °C. The thawing effect of the solid sodium formate can even be increased by moistening with aqueous potassium formate or potassium acetate solutions. The degradability of sodium formate is particularly advantageous with a chemical oxygen demand (COD) of 211 mg O2/g compared with the de-icing agents sodium acetate (740 mg O2/g) and urea with (> 2,000 mg O2/g).[8]
Saturated sodium formate solutions (as well as mixtures of other alkali metal formates such as potassium and cesium formate) are used as important drilling and stabilizing aids in gas and oil exploration because of their relatively high density. By mixing the corresponding saturated alkali metal formate solutions any densities between 1,0 and 2,3 g/cm3 can be set. The saturated solutions are biocidal and long-term stable against microbial degradation. Diluted, on the other hand, they are fast and completely biodegradable. As alkali metal formates as drilling aids make it unnecessary to add solid fillers to increase the density (such as barytes) and the formate solutions can be recovered and recycled at the drilling site, formates represent an important advance in exploration technology.[9]
^ W. H. Zachariasen: "The Crystal Structure of Sodium Formate, NaHCO2" in J. Am. Chem. Soc., 1940, 62(5), S. 1011–1013. doi:10.1021/ja01862a007
^ a b T. Meisel, Z. Halmos, K. Seybold, E. Pungor: "The thermal decomposition of alkali metal formates" in Journal of Thermal Analysis and Calorimetry 1975, 7(1). S. 73-80. doi:10.1007/BF01911627
^ T. Yoshimori, Y. Asano, Y. Toriumi, T. Shiota: "Investigation on the drying and decomposition of sodium oxalate" in Talanta 1978, 25(10) S. 603-605. doi:10.1016/0039-9140(78)80158-1
^ Bujacz, G.; Wrzesniewska, B.; Bujacz, A. (2010), "Cryoprotection properties of salts of organic acids: a case study for a tetragonal crystal of HEW lysozyme", Acta Crystallographica Section D: Biological Crystallography, vol. 66, no. 7, pp. 789–796, doi:10.1107/S0907444910015416, PMID 20606259
^ Frank M. D'Itri (1992). Chemical Deicers and the Environment. Google Books. p. 167. ISBN 9780873717052.
^ "Deicer Anti-icing Snow melting Thawing Chemicals Manufacturers". Archived from the original on 2018-08-05. Retrieved 2022-03-02.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Sodium_formate&oldid=1074826901"
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Determine fixed-point types for matrix solution of complex-valued A'AX=B using QR decomposition - MATLAB fixed.complexQlessQRMatrixSolveFixedpointTypes - MathWorks Italia
Determine fixed-point types for matrix solution of complex-valued A'AX=B using QR decomposition
T = fixed.complexQlessQRMatrixSolveFixedpointTypes(m,n,max_abs_A,max_abs_B,precisionBits)
T = fixed.complexQlessQRMatrixSolveFixedpointTypes(m,n,max_abs_A,max_abs_B,precisionBits,noiseStandardDeviation,p_s,regularizationParameter)
T = fixed.complexQlessQRMatrixSolveFixedpointTypes(m,n,max_abs_A,max_abs_B,precisionBits) computes fixed-point types for the matrix solution of complex-valued A'AX=B using QR decomposition. T is returned as a struct with fields that specify fixed-point types for A and B that guarantee no overflow will occur in the QR algorithm transforming A in-place into upper-triangular R, where QR=A is the QR decomposition of X, and X such that there is a low probability of overflow.
T = fixed.complexQlessQRMatrixSolveFixedpointTypes(m,n,max_abs_A,max_abs_B,precisionBits,noiseStandardDeviation,p_s,regularizationParameter) computes fixed-point types for the matrix solution of complex-valued
{\left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]}^{\text{'}}\cdot \left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]X=\left({\lambda }^{2}{I}_{n}+A\text{'}A\right)X=B
{A}^{\prime }AX=B
A
m
n
m\ge n
B
n
p
X
n
p
{A}^{\prime }AX=B
m
A
n
m\gg n
A
A
m=n
x=a/b
a,b\in \left[-1,1\right]
x
b
0
\text{rank}\left(A\right)
|1+1i|=\sqrt{2}
-50
A
A
A
A
{\sigma }_{\text{noise}}={\sigma }_{\text{thermal}\text{ }\text{noise}}
\mathit{A}
\mathit{R}={\mathit{Q}}^{\prime }\mathit{A}
X=\left({A}^{\prime }A\right)\B
X=\left({A}^{\prime }A\right)\B
{\left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]}^{H}\left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]X=\left({\lambda }^{2}{I}_{n}+{A}^{H}A\right)X=B
A
m
n
m\ge n
B
n
p
X
n
p
{I}_{n}=\text{eye}\left(n\right)
\lambda
\text{rank}\left(A\right)
|1+1i|=\sqrt{2}
-50
A
A
A
A
{\sigma }_{\text{noise}}={\sigma }_{\text{thermal}\text{ }\text{noise}}
\left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]
R={Q}^{H}\left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]
X=\left({\left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]}^{H}\left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]\right)\B
X=\left({\left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]}^{H}\left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]\right)\B
{X}_{\text{double}}=\left({\left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]}^{H}\left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]\right)\B
{\sigma }_{q}=\left({2}^{-\mathrm{precisionBits}}\right)/\left(\sqrt{6}\right)
Probability that estimate of lower bound s is larger than actual smallest singular value of matrix, specified as a scalar. Use fixed.complexSingularValueLowerBound to estimate the smallest singular value, s, of A. If p_s is not specified, the default value is
{p}_{s}=\left(1/2\right)\cdot \left(1+\text{erf}\left(-5/\sqrt{2}\right)\right)\approx 3\cdot {10}^{-7}
{\left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]}^{\text{'}}\cdot \left[\begin{array}{c}\lambda {I}_{n}\\ A\end{array}\right]X=\left({\lambda }^{2}{I}_{n}+A\text{'}A\right)X=B
The QR algorithm transforms A in-place into upper-triangular R where QR=A is the QR decomposition of A.
Use fixed.complexQlessQRMatrixSolveFixedpointTypes to compute fixed-point types for the inputs of these functions and blocks.
Complex Partial-Systolic Matrix Solve Using Q-less QR Decomposition
\mathrm{max}\left(|R\left(:\right)|\right)\le \sqrt{m}\mathrm{max}\left(|A\left(:\right)|\right).
\mathrm{max}\left(|X\left(:\right)|\right)\le \frac{n\cdot \mathrm{max}\left(|B\left(:\right)|\right)}{\mathrm{min}{\left(\text{svd}\left(A\right)\right)}^{2}}.
Computing the singular value decomposition to derive the above bound on X is more computationally intensive than the entire matrix solve, so the fixed.complexSingularValueLowerBound function is used to estimate a bound on min(svd(A)).
fixed.complexQuantizationNoiseStandardDeviation | fixed.complexSingularValueLowerBound | fixed.qlessqrFixedpointTypes | fixed.qlessQRMatrixSolve
Complex Burst Matrix Solve Using Q-less QR Decomposition | Complex Partial-Systolic Matrix Solve Using Q-less QR Decomposition | Complex Partial-Systolic Matrix Solve Using Q-less QR Decomposition with Forgetting Factor
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EUDML | Removable singularities for spaces of analytic functions, . EuDML | Removable singularities for spaces of analytic functions, .
Removable singularities for
{H}^{p}
spaces of analytic functions,
0<p<1
Björn, Anders. "Removable singularities for spaces of analytic functions, .." Annales Academiae Scientiarum Fennicae. Mathematica 26.1 (2001): 155-174. <http://eudml.org/doc/122670>.
@article{Björn2001,
author = {Björn, Anders},
title = {Removable singularities for spaces of analytic functions, .},
AU - Björn, Anders
TI - Removable singularities for spaces of analytic functions, .
Capacity and harmonic measure in the complex plane
Articles by Björn
|
EUDML | The absolute anabelian geometry of canonical curves. EuDML | The absolute anabelian geometry of canonical curves.
The absolute anabelian geometry of canonical curves.
Mochizuki, Shinichi. "The absolute anabelian geometry of canonical curves.." Documenta Mathematica Extra Vol. (2003): 609-640. <http://eudml.org/doc/126234>.
keywords = {correspondences; Grothendieck conjecture, canonical lifting, -adic Teichmüller theory; Grothendieck conjecture, canonical lifting, -adic Teichmüller theory},
title = {The absolute anabelian geometry of canonical curves.},
TI - The absolute anabelian geometry of canonical curves.
KW - correspondences; Grothendieck conjecture, canonical lifting, -adic Teichmüller theory; Grothendieck conjecture, canonical lifting, -adic Teichmüller theory
correspondences, Grothendieck conjecture, canonical lifting,
p
-adic Teichmüller theory, Grothendieck conjecture, canonical lifting,
p
-adic Teichmüller theory
Articles by Mochizuki
|
EUDML | Strongly nonlinear elliptic problem without growth condition. EuDML | Strongly nonlinear elliptic problem without growth condition.
Strongly nonlinear elliptic problem without growth condition.
Anane, Aomar; Chakrone, Omar
Anane, Aomar, and Chakrone, Omar. "Strongly nonlinear elliptic problem without growth condition.." Electronic Journal of Differential Equations (EJDE) [electronic only] 2002 (2002): 41-47. <http://eudml.org/doc/127125>.
author = {Anane, Aomar, Chakrone, Omar},
keywords = {-Laplacian; growth condition; -Laplacian},
title = {Strongly nonlinear elliptic problem without growth condition.},
AU - Anane, Aomar
AU - Chakrone, Omar
TI - Strongly nonlinear elliptic problem without growth condition.
KW - -Laplacian; growth condition; -Laplacian
p
-Laplacian, growth condition,
p
|
EUDML | Sheaves and -bicompactifications of mappings. EuDML | Sheaves and -bicompactifications of mappings.
Sheaves and
𝔗a
-bicompactifications of mappings.
Ulyanov, V.M.
Ulyanov, V.M.. "Sheaves and -bicompactifications of mappings.." Sibirskie Ehlektronnye Matematicheskie Izvestiya [electronic only] 4 (2007): 504-546. <http://eudml.org/doc/53842>.
@article{Ulyanov2007,
author = {Ulyanov, V.M.},
keywords = {fan product; inverse limit; algebra of functions; completely regular mapping},
title = {Sheaves and -bicompactifications of mappings.},
AU - Ulyanov, V.M.
TI - Sheaves and -bicompactifications of mappings.
KW - fan product; inverse limit; algebra of functions; completely regular mapping
fan product, inverse limit, algebra of functions, completely regular mapping
Articles by Ulyanov
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Wallace–Bolyai–Gerwien theorem - Wikipedia
Wallace–Bolyai–Gerwien theorem
When can a polygon can be formed from another by cutting it into a finite number of pieces
By the Wallace–Bolyai–Gerwien theorem, a square can be cut into parts and rearranged into a triangle of equal area.
In geometry, the Wallace–Bolyai–Gerwien theorem,[1] named after William Wallace, Farkas Bolyai and Paul Gerwien, is a theorem related to dissections of polygons. It answers the question when one polygon can be formed from another by cutting it into a finite number of pieces and recomposing these by translations and rotations. The Wallace–Bolyai–Gerwien theorem states that this can be done if and only if two polygons have the same area.
Wallace had proven the same result already in 1807.
According to other sources, Bolyai and Gerwien had independently proved the theorem in 1833 and 1835, respectively.
3 Notes about the proof
4 Degree of decomposability
There are several ways in which this theorem may be formulated. The most common version uses the concept of "equidecomposability" of polygons: two polygons are equidecomposable if they can be split into finitely many triangles that only differ by some isometry (in fact only by a combination of a translation and a rotation). In this case the Wallace–Bolyai–Gerwien theorem states that two polygons are equidecomposable if and only if they have the same area.
Another formulation is in terms of scissors congruence: two polygons are scissors-congruent if they can be decomposed into finitely many polygons that are pairwise congruent. Scissors-congruence is an equivalence relation. In this case the Wallace–Bolyai–Gerwien theorem states that the equivalence classes of this relation contain precisely those polygons that have the same area.
The theorem can be understood in a few steps. Firstly, every polygon can be cut into triangles. There are a few methods for this. For convex polygons one can cut off each vertex in turn, while for concave polygons this requires more care. A general approach that works for non-simple polygons as well would be to choose a line not parallel to any of the sides of the polygon and draw a line parallel to this one through each of the vertices of the polygon. This will divide the polygon into triangles and trapezoids, which in turn can be converted into triangles.
Secondly, each of these triangles can be transformed into a right triangle and subsequently into a rectangle with one side of length 1. Alternatively, a triangle can be transformed into one such rectangle by first turning it into a parallelogram and then turning this into such a rectangle. By doing this for each triangle, the polygon can be decomposed into a rectangle with unit width and height equal to its area.
Since this can be done for any two polygons, a "common subdivision" of the rectangle in between proves the theorem. That is, cutting the common rectangle (of size 1 by its area) according to both polygons will be an intermediate between both polygons.
Notes about the proof[edit]
First of all, this proof requires an intermediate polygon. In the formulation of the theorem using scissors-congruence, the use of this intermediate can be reformulated by using the fact that scissor-congruences are transitive. Since both the first polygon and the second polygon are scissors-congruent to the intermediate, they are scissors-congruent to one another.
The proof of this theorem is constructive and doesn't require the axiom of choice, even though some other dissection problems (e.g. Tarski's circle-squaring problem) do need it. In this case, the decomposition and reassembly can actually be carried out "physically": the pieces can, in theory, be cut with scissors from paper and reassembled by hand.
Nonetheless, the number of pieces required to compose one polygon from another using this procedure generally far exceeds the minimum number of polygons needed.[2]
Degree of decomposability[edit]
Consider two equidecomposable polygons P and Q. The minimum number n of pieces required to compose one polygon Q from another polygon P is denoted by σ(P,Q).
Depending on the polygons, it is possible to estimate upper and lower bounds for σ(P,Q). For instance, Alfred Tarski proved that if P is convex and the diameters of P and Q are respectively given by d(P) and d(Q), then[3]
{\displaystyle \sigma (P,Q)\geq {\frac {d(P)}{d(Q)}}.}
If Px is a rectangle of sides a · x and a · (1/x) and Q is a rectangle of size a, then Px and Q are equidecomposable for every x > 0. An upper bound for σ(Px,Q) is given by[3]
{\displaystyle \sigma (P_{x},Q)\leq 2+\left\lceil {\sqrt {x^{2}-1}}\right\rceil ,\quad {\text{for }}x\geq 1.}
Since σ(Px,Q) = σ(P(1/x),Q), we also have that
{\displaystyle \sigma \left(P_{\frac {1}{x}},Q\right)\leq 2+\left\lceil {\frac {\sqrt {1-x^{2}}}{x}}\right\rceil ,\quad {\text{for }}x\leq 1.}
The analogous statement about polyhedra in three dimensions, known as Hilbert's third problem, is false, as proven by Max Dehn in 1900. The problem has also been considered in some non-Euclidean geometries. In two-dimensional hyperbolic and spherical geometry, the theorem holds. However, the problem is still open for these geometries in three dimensions.
^ Gardner, R. J. (1985-02-01). "A problem of Sallee on equidecomposable convex bodies". Proceedings of the American Mathematical Society. 94 (2): 329–332. doi:10.1090/S0002-9939-1985-0784187-9. ISSN 0002-9939. JSTOR 2045399.
^ "Dissection".
^ a b McFarland, Andrew; McFarland, Joanna; Smith, James T. (2014). Alfred Tarski. Birkhäuser, New York, NY. pp. 77–91. doi:10.1007/978-1-4939-1474-6_5. ISBN 9781493914739.
Scissors Congruence - An interactive demonstration of the Wallace–Bolyai–Gerwien theorem.
Video showing a sketch of the proof
An Example of the Bolyai–Gerwien Theorem by Sándor Kabai, Ferenc Holló Szabó, and Lajos Szilassi, the Wolfram Demonstrations Project.
A presentation about Hilbert's third problem at College of Staten Island CUNY - Abhijit Champanerkar.
Optimal dissection of a unit square in a rectangle
Retrieved from "https://en.wikipedia.org/w/index.php?title=Wallace–Bolyai–Gerwien_theorem&oldid=1058646205"
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Understanding Euler's Number
Euler's Number in Finance: Compound Interest
Why Is Euler's Number Important?
How Is Euler's Number Used in Finance?
What Is Euler's Number Exactly?
Euler's number is a mathematical expression for the base of the natural logarithm. It is usually represented by the letter e and is commonly used in problems relating to exponential growth or decay.
Another way to interpret Euler's number is as the base for an exponential function whose value is always equal to its derivative. In other words, e is the only possible number such that ex increases at a rate of ex for every possible x.
Euler's number is an important constant that is found in many contexts and is the base for natural logarithms.
An irrational number denoted by e, Euler's number is 2.71828..., where the digits go on forever in a series that never ends or repeats (similar to pi).
Euler's number is used in everything from explaining exponential growth to radioactive decay.
In finance, Euler's number is used to calculate how wealth can grow due to compound interest.
Although commonly associated with Leonhard Euler, the constant was first discovered in 1683 by the mathematician Jacob Bernoulli. Bernoulli was trying to determine how wealth would grow if interest were compounded more often, instead of on an annual basis.
Imagine lending money at a 100% interest rate, compounded every year. After one year, your money would double. But what if the interest rate were cut in half, and compounded twice as often? At 50% every six months, your money would grow by 225% in one year. As the interval gets smaller, the total returns get slightly higher. Bernoulli found that if interest is calculated n times per year, at a rate of 100%/n, the total accreted wealth at the end of the first year would be slightly greater than 2.7 times the initial investment if n is sufficiently large.
However, the key work surrounding the constant was not performed until several decades later, by Leonhard Euler. In his Introductio in Analysin Infinitorum (1748), Euler proved that the constant was an irrational number, whose digits would never repeat. He also proved that the constant can be represented as an infinite sum of inverse factorials:
e = 1 + \frac{ 1 }{ 1 } + \frac { 1 }{ 2 } + \frac { 1 }{ 1 \times 2 \times 3 } + \frac {1 }{ 1 \times 2 \times 3 \times 4 } + ... + \frac { 1 }{ n! }
e=1+11+21+1×2×31+1×2×3×41+...+n!1
Euler used the letter e for exponents, but the letter is now widely associated with his name. It is commonly used in a wide range of applications from population growth of living organisms to radioactive decay of heavy elements like uranium by nuclear scientists. It also has applications in trigonometry, probability, and other areas of applied mathematics.
The first digits of Euler's number are 2.71828..., although the number itself is a non-terminating series that goes on forever, like pi (3.1415...).
Compound interest has been hailed as a "miracle" of finance, whereby interest is credited not only initial amounts invested or deposited, but also on previous interest received. Continuously compounding interest is achieved when interest is reinvested over an infinitely small unit of time—and while this is practically impossible in the real world, this concept is crucial for understanding the behavior of many different types of financial instruments from bonds to derivatives contracts.
Compound interest in this way is akin to exponential growth, and is expressed by the following formula:
\begin{aligned}&\text{FV} = \text{PV} e ^ {rt} \\&\textbf{where:} \\&\text{FV} = \text{Future value} \\&\text{PV} = \text{Present value of balance or sum} \\&e = \text{Euler's constant} \\&r = \text{Interest rate being compounded} \\&t = \text{Time in years} \\\end{aligned}
FV=PVertwhere:FV=Future valuePV=Present value of balance or sume=Euler’s constantr=Interest rate being compoundedt=Time in years
Therefore, if you had $1,000 paying 2% interest with continuous compounding, after 3 years you would have:
\$1,000 \times 2.71828 ^ { ( .02 \times 3 ) } = \$1,061.84
$1,000×2.71828(.02×3)=$1,061.84
Note that this amount is greater than if the compounding period were a discrete period, say on a monthly basis. In this case, the amount of interest would be computed differently: FV = PV(1+r/n)nt, where n is the number of compounding periods in a year (in this case 12):
\$1,000 \Big ( 1 + \frac { .02 }{ 12 } \Big ) ^ { 12 \times 3 } = \$1,061.78
$1,000(1+12.02)12×3=$1,061.78
Here, the difference is only a matter of a few cents, but as our sums get larger, interest rates get higher, and the amount of time gets longer, continuous compounding using Euler's constant becomes more and more valuable relative to discrete compounding.
Euler's number (e) should not be confused with Euler's constant, denoted by the lower case gamma (γ). Also known as the Euler-Mascheroni constant, the latter is related to harmonic series and has a value of approximately 0.5772....
Euler's number is one of the most important constants in mathematics. It frequently appears in problems dealing with exponential growth or decay, where the rate of growth is proportionate to the existing population. In finance, e is also used in calculations of compound interest, where wealth grows at a set rate over time.
Euler's number frequently appears in problems related to growth or decay, where the rate of change is determined by the present value of the number being measured. One example is in biology, where bacterial populations are expected to double at reliable intervals. Another case is radiometric dating, where the number of radioactive atoms is expected to decline over the fixed half-life of the element being measured.
Euler's number appears in problems related to compound interest. Whenever an investment offers a fixed interest rate over a period of time, the future value of that investment can easily be calculated in terms of e.
To put it simply, Euler's number is the base of an exponential function whose rate of growth is always proportionate to its present value. The exponential function ex always grows at a rate of ex, a feature that is not true of other bases and one that vastly simplifies the algebra surrounding exponents and logarithms. This number is irrational, with a value of approximately 2.71828....
Correction–December 5, 2021: An earlier version of this article incorrectly conflated Euler's number with Euler's constant.
Popular Mechanics. "What's the Big Deal with Euler's Number?" Accessed Dec. 5, 2021.
School of Mathematics and Statistics, University of St. Andrews. "The Number e." Accessed Dec. 5, 2021.
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Classification of links up to self pass-move
October, 2003 Classification of links up to self pass-move
Tetsuo SHIBUYA, Akira YASUHARA
A pass-move and a
♯
-move are local moves on oriented links defined by L. H. Kauffman and H. Murakami respectively. Two links are self pass-equivalent (resp. self
♯
-equivalent) if one can be deformed into the other by pass-moves (resp. #-moves), where none of them can occur between distinct components of the link. These relations are equivalence relations on ordered oriented links and stronger than link-homotopy defined by J. Milnor. We give two complete classifications of links with arbitrarily many components up to self pass-equivalence and up to self
♯
-equivalence respectively. So our classifications give subdivisions of link-homotopy classes.
Tetsuo SHIBUYA. Akira YASUHARA. "Classification of links up to self pass-move." J. Math. Soc. Japan 55 (4) 939 - 946, October, 2003. https://doi.org/10.2969/jmsj/1191418757
Keywords: #-move , Arf invariant , link-homotopy , pass-move
Tetsuo SHIBUYA, Akira YASUHARA "Classification of links up to self pass-move," Journal of the Mathematical Society of Japan, J. Math. Soc. Japan 55(4), 939-946, (October, 2003)
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ChiSquareRandomVariable - Maple Help
Home : Support : Online Help : Education : Student Packages : Statistics : Random Variable Distributions : ChiSquareRandomVariable
ChiSquareRandomVariable(nu)
The chi-square random variable is a continuous probability random variable with probability density function given by:
f\left(t\right)={\begin{array}{cc}0& t<0\\ \frac{{t}^{\frac{\mathrm{\nu }}{2}-1}{ⅇ}^{-\frac{t}{2}}}{{2}^{\frac{\mathrm{\nu }}{2}}\mathrm{\Gamma }\left(\frac{\mathrm{\nu }}{2}\right)}& \mathrm{otherwise}\end{array}
0<\mathrm{\nu }
2
\mathrm{with}\left(\mathrm{Student}[\mathrm{Statistics}]\right):
X≔\mathrm{ChiSquareRandomVariable}\left(\mathrm{\nu }\right):
\mathrm{PDF}\left(X,u\right)
{\begin{array}{cc}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{<}\textcolor[rgb]{0,0,1}{0}\\ \frac{{\textcolor[rgb]{0,0,1}{u}}^{\frac{\textcolor[rgb]{0,0,1}{\mathrm{\nu }}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{u}}{\textcolor[rgb]{0,0,1}{2}}}}{{\textcolor[rgb]{0,0,1}{2}}^{\frac{\textcolor[rgb]{0,0,1}{\mathrm{\nu }}}{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{\Gamma }}\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{\mathrm{\nu }}}{\textcolor[rgb]{0,0,1}{2}}\right)}& \textcolor[rgb]{0,0,1}{\mathrm{otherwise}}\end{array}
\mathrm{PDF}\left(X,0.5\right)
\frac{\textcolor[rgb]{0,0,1}{0.7788007831}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{0.5}}^{\textcolor[rgb]{0,0,1}{0.5000000000}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{\nu }}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1.}}}{{\textcolor[rgb]{0,0,1}{2.}}^{\textcolor[rgb]{0,0,1}{0.5000000000}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{\nu }}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{\Gamma }}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{0.5000000000}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{\nu }}\right)}
\mathrm{Mean}\left(X\right)
\textcolor[rgb]{0,0,1}{\mathrm{\nu }}
\mathrm{Variance}\left(X\right)
\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{\nu }}
Y≔\mathrm{ChiSquareRandomVariable}\left(3\right):
\mathrm{PDF}\left(Y,x,\mathrm{output}=\mathrm{plot}\right)
\mathrm{CDF}\left(Y,x\right)
{\begin{array}{cc}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{<}\textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\left(\frac{\sqrt{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{x}}{\textcolor[rgb]{0,0,1}{2}}}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\frac{\sqrt{\textcolor[rgb]{0,0,1}{\mathrm{\pi }}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{erfc}}\textcolor[rgb]{0,0,1}{}\left(\frac{\sqrt{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{x}}}{\textcolor[rgb]{0,0,1}{2}}\right)}{\textcolor[rgb]{0,0,1}{2}}\right)}{\sqrt{\textcolor[rgb]{0,0,1}{\mathrm{\pi }}}}& \textcolor[rgb]{0,0,1}{\mathrm{otherwise}}\end{array}
\mathrm{CDF}\left(Y,6,\mathrm{output}=\mathrm{plot}\right)
The Student[Statistics][ChiSquareRandomVariable] command was introduced in Maple 18.
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EUDML | Torsion, curvature and deflection -tensors on . EuDML | Torsion, curvature and deflection -tensors on .
Torsion, curvature and deflection
d
-tensors on
{J}^{1}\left(T,M\right)
Neagu, Mircea; Udrişte, Constantin
Neagu, Mircea, and Udrişte, Constantin. "Torsion, curvature and deflection -tensors on .." Balkan Journal of Geometry and its Applications (BJGA) 6.1 (2001): 29-44. <http://eudml.org/doc/122208>.
author = {Neagu, Mircea, Udrişte, Constantin},
keywords = {-jet fibre bundle; -linear connection; nonlinear connection; torsion and curvature; deflection -tensors; -jet fibre bundle; -linear connection; deflection -tensors},
title = {Torsion, curvature and deflection -tensors on .},
TI - Torsion, curvature and deflection -tensors on .
KW - -jet fibre bundle; -linear connection; nonlinear connection; torsion and curvature; deflection -tensors; -jet fibre bundle; -linear connection; deflection -tensors
l
-jet fibre bundle,
\text{Γ}
-linear connection, nonlinear connection, torsion and curvature, deflection
d
-tensors,
l
{\Gamma }
-linear connection, deflection
d
|
EUDML | Decomposing four-manifolds up to homotopy type. EuDML | Decomposing four-manifolds up to homotopy type.
Decomposing four-manifolds up to homotopy type.
Cavicchioli, Alberto, Ruini, Beatrice, and Spaggiari, Fulvia. "Decomposing four-manifolds up to homotopy type.." Beiträge zur Algebra und Geometrie 44.1 (2003): 189-201. <http://eudml.org/doc/233429>.
@article{Cavicchioli2003,
author = {Cavicchioli, Alberto, Ruini, Beatrice, Spaggiari, Fulvia},
keywords = {four-manifolds; homotopy type; decomposition; intersection form},
title = {Decomposing four-manifolds up to homotopy type.},
AU - Cavicchioli, Alberto
AU - Ruini, Beatrice
AU - Spaggiari, Fulvia
TI - Decomposing four-manifolds up to homotopy type.
KW - four-manifolds; homotopy type; decomposition; intersection form
four-manifolds, homotopy type, decomposition, intersection form
{E}^{4}
4
Articles by Cavicchioli
Articles by Ruini
Articles by Spaggiari
|
Probability integral transform - Wikipedia
In probability theory, the probability integral transform (also known as universality of the uniform) relates to the result that data values that are modeled as being random variables from any given continuous distribution can be converted to random variables having a standard uniform distribution.[1] This holds exactly provided that the distribution being used is the true distribution of the random variables; if the distribution is one fitted to the data, the result will hold approximately in large samples.
The result is sometimes modified or extended so that the result of the transformation is a standard distribution other than the uniform distribution, such as the exponential distribution.
One use for the probability integral transform in statistical data analysis is to provide the basis for testing whether a set of observations can reasonably be modelled as arising from a specified distribution. Specifically, the probability integral transform is applied to construct an equivalent set of values, and a test is then made of whether a uniform distribution is appropriate for the constructed dataset. Examples of this are P–P plots and Kolmogorov–Smirnov tests.
A second use for the transformation is in the theory related to copulas which are a means of both defining and working with distributions for statistically dependent multivariate data. Here the problem of defining or manipulating a joint probability distribution for a set of random variables is simplified or reduced in apparent complexity by applying the probability integral transform to each of the components and then working with a joint distribution for which the marginal variables have uniform distributions.
A third use is based on applying the inverse of the probability integral transform to convert random variables from a uniform distribution to have a selected distribution: this is known as inverse transform sampling.
Suppose that a random variable X has a continuous distribution for which the cumulative distribution function (CDF) is FX. Then the random variable Y defined as
{\displaystyle Y=F_{X}(X)\,,}
has a standard uniform distribution.[1][2]
Equivalently, the distribution of X on
{\displaystyle \mathbb {R} }
is the pushforward measure of the uniform measure on
{\displaystyle [0,1]}
, pushforwarded by
{\displaystyle F_{X}^{-1}}
Given any random continuous variable
{\displaystyle X}
{\displaystyle Y=F_{X}(X)}
{\displaystyle y\in [0,1]}
{\displaystyle F_{X}^{-1}(y)}
exists (i.e., if there exists a unique
{\displaystyle x}
{\displaystyle F_{X}(x)=y}
{\displaystyle {\begin{aligned}F_{Y}(y)&=\operatorname {P} (Y\leq y)\\&=\operatorname {P} (F_{X}(X)\leq y)\\&=\operatorname {P} (X\leq F_{X}^{-1}(y))\\&=F_{X}(F_{X}^{-1}(y))\\&=y\end{aligned}}}
{\displaystyle F_{X}^{-1}(y)}
does not exist, then it can be replaced in this proof by the function
{\displaystyle \chi }
{\displaystyle \chi (0)=-\infty }
{\displaystyle \chi (1)=\infty }
{\displaystyle \chi (y)\equiv \inf\{x:F_{X}(x)\geq y\}}
{\displaystyle y\in (0,1)}
, with the same result that
{\displaystyle F_{Y}(y)=y}
{\displaystyle F_{Y}}
is just the CDF of a
{\displaystyle \mathrm {Uniform} (0,1)}
random variable, so that
{\displaystyle Y}
has a uniform distribution on the interval
{\displaystyle [0,1]}
For an illustrative example, let X be a random variable with a standard normal distribution
{\displaystyle {\mathcal {N}}(0,1)}
. Then its CDF is
{\displaystyle \Phi (x)={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{x}{\rm {e}}^{-t^{2}/2}\,{\rm {d}}t={\frac {1}{2}}{\Big [}\,1+\operatorname {erf} {\Big (}{\frac {x}{\sqrt {2}}}{\Big )}\,{\Big ]},\quad x\in \mathbb {R} ,\,}
{\displaystyle \operatorname {erf} (),}
is the error function. Then the new random variable Y, defined by Y=Φ(X), is uniformly distributed.
If X has an exponential distribution with unit mean, then its CDF is
{\displaystyle F(x)=1-\exp(-x),}
and the immediate result of the probability integral transform is that
{\displaystyle Y=1-\exp(-X)}
has a uniform distribution. The symmetry of the uniform distribution can then be used to show that
{\displaystyle Y'=\exp(-X)}
also has a uniform distribution.
^ a b Dodge, Y. (2006) The Oxford Dictionary of Statistical Terms, Oxford University Press
^ Casella, George; Berger, Roger L. (2002). Statistical Inference (2nd ed.). Theorem 2.1.10, p.54.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Probability_integral_transform&oldid=1087652419"
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Detect Vehicles in Lidar Using Image Labels - MATLAB & Simulink - MathWorks France
3-D Region Proposal
Estimate the Distance of Vehicles from the Ego Vehicle
This example shows you how to detect vehicles in lidar using label data from a co-located camera with known lidar-to-camera calibration parameters. Use this workflow in MATLAB® to estimate 3-D oriented bounding boxes in lidar based on 2-D bounding boxes in the corresponding image. You will also see how to automatically generate ground truth as a distance for 2-D bounding boxes in a camera image using lidar data. This figure provides an overview of the process.
This example uses lidar data collected on a highway from an Ouster OS1 lidar sensor and image data from a front-facing camera mounted on the ego vehicle. The lidar and camera data are approximately time-synced and calibrated to estimate their intrinsic and extrinsic parameters. For more information on lidar camera calibration, see Lidar and Camera Calibration.
Note: The download time for the data depends on the speed of your internet connection. During the execution of this code block, MATLAB is temporarily unresponsive.
lidarTarFileUrl = 'https://www.mathworks.com/supportfiles/lidar/data/WPI_LidarData.tar.gz';
imageTarFileUrl = 'https://www.mathworks.com/supportfiles/lidar/data/WPI_ImageData.tar.gz';
imageDataTarFile = fullfile(outputFolder,'WPI_ImageData.tar.gz');
if ~exist(outputFolder,'dir')
disp('Downloading WPI Lidar driving data (760 MB)...')
websave(lidarDataTarFile,lidarTarFileUrl)
untar(lidarDataTarFile,outputFolder)
if ~exist(imageDataTarFile,'file')
disp('Downloading WPI Image driving data (225 MB)...')
websave(imageDataTarFile,imageTarFileUrl)
untar(imageDataTarFile,outputFolder)
if ~exist(fullfile(outputFolder,'imageData'),'dir')
imageDataLocation = fullfile(outputFolder,'imageData');
images = imageSet(imageDataLocation);
% Load downloaded lidar data into the workspace
lidarData = fullfile(outputFolder,'WPI_LidarData.mat');
load(lidarData);
% Load calibration data
if ~exist('calib','var')
load('calib.mat')
% Define camera to lidar transformation matrix
camToLidar = calib.extrinsics;
intrinsics = calib.intrinsics;
Alternatively, you can use your web browser to first download the datasets to your local disk, and then uncompress the files.
This example uses prelabeled data to serve as ground truth for the 2-D detections from the camera images. These 2-D detections can be generated using deep learning-based object detectors like vehicleDetectorYOLOv2 (Automated Driving Toolbox), vehicleDetectorFasterRCNN (Automated Driving Toolbox), and vehicleDetectorACF (Automated Driving Toolbox). For this example, the 2-D detections have been generated using the Training Image Labeler app. These 2-D bounding boxes are vectors of the form:
\left[\mathit{x}\text{\hspace{0.17em}}\mathit{y}\text{\hspace{0.17em}}\mathit{w}\text{\hspace{0.17em}}\mathit{h}\right]
\mathit{x}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\mathit{y}
represent the xy-coordinates of the top-left corner, and
\mathit{w}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\mathit{h}
represent the width and height of the bounding box respectively.
Read a image frame into the workspace, and display it with the bounding boxes overlaid.
load imageGTruth.mat
im = imread(imageFileNames{50});
imBbox = imageGTruth{50};
showShape('rectangle',imBbox)
To generate cuboid bounding boxes in lidar from the 2-D rectangular bounding boxes in the image data, a 3-D region is proposed to reduce the search space for bounding box estimation. The corners of each 2-D rectangular bounding box in the image are transformed into 3-D lines using camera intrinsic parameters and camera-to-lidar extrinsic parameters. These 3-D lines form frustum flaring out from the associated 2-D bounding box in the opposite direction of the ego vehicle. The lidar points that fall inside this region are segmented into various clusters based on Euclidean distance. The clusters are fitted with 3-D oriented bounding boxes, and the best cluster is estimated based on the size of these clusters. Estimate the 3-D oriented bounding boxes in a lidar point cloud, based on the 2-D bounding boxes in a camera image, by using the bboxCameraToLidar function. This figure shows how 2-D and 3-D bounding boxes relate to each other.
The 3-D cuboids are represented as vectors of the form:
\left[\mathit{xcen}\text{\hspace{0.17em}}\mathit{ycen}\text{\hspace{0.17em}}\mathit{zcen}\text{\hspace{0.17em}}\mathit{dimx}\text{\hspace{0.17em}}\mathit{dimy}\text{\hspace{0.17em}}\mathit{dimz}\text{\hspace{0.17em}}\mathit{rotx}\text{\hspace{0.17em}}\mathit{roty}\text{\hspace{0.17em}}\mathit{rotz}\right]
\mathit{xcen},\text{\hspace{0.17em}}\mathit{ycen},\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\mathit{zcen}
represent the centroid coordinates of the cuboid.
\mathit{dimx},\mathit{dimy},\mathrm{and}\text{\hspace{0.17em}}\mathit{dimz}
represent the length of the cuboid along the x-, y-, and z-axes, and
\mathit{rotx},\mathit{roty},\mathrm{and}\text{\hspace{0.17em}}\mathit{rotz}
represent the rotation ,in degrees, of the cuboid along the x-, y-, and z-axes.
Use ground truth of the image to estimate a 3-D bounding box in the lidar point cloud.
pc = lidarData{50};
% Crop point cloud to process only front region
roi = [0 70 -15 15 -3 8];
ind = findPointsInROI(pc,roi);
pc = select(pc,ind);
lidarBbox = bboxCameraToLidar(imBbox,pc,intrinsics, ...
camToLidar,'ClusterThreshold',2,'MaxDetectionRange',[1,70]);
showShape('Cuboid',lidarBbox)
view([-2.90 71.59])
To improve the detected bounding boxes, preprocess the point cloud by removing the ground plane.
Use the helperLidarCameraObjectsDisplay class to visualize the lidar and image data. This visualization provides the capability to view the point cloud, image, 3-D bounding boxes on the point cloud, and 2-D bounding boxes on the image simultaneously. The visualization layout is consists of these windows:
Image — Visualize an image and associated 2-D bounding boxes
Perspective View — Visualize the point cloud and associated 3-D bounding boxes in a perspective view
Top View — Visualize the point cloud and associated 3-D bounding boxes from the top view
% Initialize display
display = helperLidarCameraObjectsDisplay;
initializeDisplay(display)
% Update display with point cloud and image
updateDisplay(display, im, pc)
Run bboxCameraToLidar on 2-D labels over first 200 frames to generate 3-D cuboids
% Load point cloud and image
im = imread(imageFileNames{i});
pc = lidarData{i};
% Load image ground truth
imBbox = imageGTruth{i};
groundPtsIndex = segmentGroundFromLidarData(pc,'ElevationAngleDelta',15, ...
'InitialElevationAngle',10);
nonGroundPts = select(pc,~groundPtsIndex);
if imBbox
[lidarBbox,~,boxUsed] = bboxCameraToLidar(imBbox,nonGroundPts,intrinsics, ...
camToLidar,'ClusterThreshold',2,'MaxDetectionRange',[1, 70]);
% Display image with bounding boxes
im = updateImage(display,im,imBbox);
% Display point cloud with bounding box
updateDisplay(display,im,pc);
updateLidarBbox(display,lidarBbox,boxUsed)
Detected bounding boxes by using bounding box tracking, such as joint probabilistic data association (JPDA). For more information, see Track Vehicles Using Lidar: From Point Cloud to Track List.
For vehicle safety features such as forward collision warning, accurate measurement of the distance between the ego vehicle and other objects is crucial. A lidar sensor provides the accurate distance of objects from the ego vehicle in 3-D, and it can also be used to create ground truth automatically from 2-D image bounding boxes. To generate ground truth for 2-D bounding boxes, use the projectLidarPointsOnImage function to project the points inside the 3-D bounding boxes onto the image. The projected points are associated with 2-D bounding boxes by finding the bounding box with the minimum Euclidean distance from the projected 3-D points. Since the projected points are from lidar to camera, use the inverse of camera-to-lidar extrinsic parameters. This figure illustrates the transformation from lidar to camera.
% Get lidar to camera matrix
lidarToCam = invert(camToLidar);
% Loop first 200 frames. To loop all frames, replace 200 with numel(imageGTruth)
[distance,nearestRect,idx] = helperComputeDistance(imBbox,nonGroundPts,lidarBbox, ...
intrinsics,lidarToCam);
% Update image with bounding boxes
im = updateImage(display,im,nearestRect,distance);
updateLidarBbox(display,lidarBbox)
updateDisplay(display,im,pc)
helperComputeDistance
function [distance, nearestRect, index] = helperComputeDistance(imBbox, pc, lidarBbox, intrinsic, lidarToCam)
% helperComputeDistance estimates the distance of 2-D bounding box in a given
% image using 3-D bounding boxes from lidar. It also calculates
% association between 2-D and 3-D bounding boxes
numLidarDetections = size(lidarBbox,1);
nearestRect = zeros(0,4);
distance = zeros(1,numLidarDetections);
index = zeros(0,1);
for i = 1:numLidarDetections
bboxCuboid = lidarBbox(i,:);
% Create cuboidModel
model = cuboidModel(bboxCuboid);
% Find points inside cuboid
ind = findPointsInsideCuboid(model,pc);
pts = select(pc,ind);
% Project cuboid points to image
imPts = projectLidarPointsOnImage(pts,intrinsic,lidarToCam);
% Find 2-D rectangle corresponding to 3-D bounding box
[nearestRect(i,:),idx] = findNearestRectangle(imPts,imBbox);
index(end+1) = idx;
% Find the distance of the 2-D rectangle
distance(i) = min(pts.Location(:,1));
function [nearestRect,idx] = findNearestRectangle(imPts,imBbox)
numBbox = size(imBbox,1);
ratio = zeros(numBbox,1);
% Iterate over all the rectangles
for i = 1:numBbox
bbox = imBbox(i,:);
corners = getCornersFromBbox(bbox);
% Find overlapping ratio of the projected points and the rectangle
idx = (imPts(:,1) > corners(1,1)) & (imPts(:,1) < corners(2,1)) & ...
(imPts(:,2) > corners(1,2)) & (imPts(:,2) < corners(3,1));
ratio(i) = sum(idx);
% Get nearest rectangle
[~,idx] = max(ratio);
nearestRect = imBbox(idx,:);
function cornersCamera = getCornersFromBbox(bbox)
cornersCamera = zeros(4,2);
cornersCamera(1,1:2) = bbox(1:2);
cornersCamera(2,1:2) = bbox(1:2) + [bbox(3),0];
cornersCamera(3,1:2) = bbox(1:2) + bbox(3:4);
cornersCamera(4,1:2) = bbox(1:2) + [0,bbox(4)];
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Characterization for Rectifiable and Nonrectifiable Attractivity of Nonautonomous Systems of Linear Differential Equations
2013 Characterization for Rectifiable and Nonrectifiable Attractivity of Nonautonomous Systems of Linear Differential Equations
Yūki Naito, Mervan Pašić
We study a new kind of asymptotic behaviour near
t=\mathrm{0}
for the nonautonomous system of two linear differential equations:
\mathbf{x}\mathrm{\text{'}}\left(t\right)=A\left(t\right)\mathbf{x}\left(t\right)
t\in \left(\mathrm{0},{t}_{\mathrm{0}}\right]
, where the matrix-valued function
A=A\left(t\right)
has a kind of singularity at
t=\mathrm{0}
. It is called rectifiable (resp., nonrectifiable) attractivity of the zero solution, which means that
\parallel \mathbf{x}\left(t\right){\parallel }_{\mathrm{2}}\to \mathrm{0}
t\to \mathrm{0}
and the length of the solution curve of
\mathbf{x}
is finite (resp., infinite) for every
\mathbf{x}\ne \mathbf{0}
. It is characterized in terms of certain asymptotic behaviour of the eigenvalues of
A\left(t\right)
t=\mathrm{0}
. Consequently, the main results are applied to a system of two linear differential equations with polynomial coefficients which are singular at
t=\mathrm{0}
Yūki Naito. Mervan Pašić. "Characterization for Rectifiable and Nonrectifiable Attractivity of Nonautonomous Systems of Linear Differential Equations." Int. J. Differ. Equ. 2013 (SI2) 1 - 11, 2013. https://doi.org/10.1155/2013/740980
Received: 2 April 2013; Accepted: 21 May 2013; Published: 2013
Yūki Naito, Mervan Pašić "Characterization for Rectifiable and Nonrectifiable Attractivity of Nonautonomous Systems of Linear Differential Equations," International Journal of Differential Equations, Int. J. Differ. Equ. 2013(SI2), 1-11, (2013)
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While watering her outdoor plants, Maura noticed that the water coming out of her garden hose followed a parabolic path. Thinking that she might be able to model the path of the water with an equation, she quickly took some measurements. The highest point the water reached was
8
feet, and it landed on the plants
10
feet from where she was standing. Both the nozzle of the hose and the top of the flowers were
4
feet above the ground. Help Maura write an equation that describes the path of the water from the hose to the top of her plants. What domain and range make sense for the model?
Refer to problem 2-69 in Lesson 2.1.5.
The domain is all the possible
x
-values that make sense in this problem. The range is all the possible
y
-values that make sense in the problem. For example, there won’t be any negative
y
-values here.
Use the eTool below to check your work.
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Biotechnology : Principles and Processes - Live Session - NEET 2020
Biotechnology : Principles and Processes - Live Session - NEET 2020Contact Number: 9667591930 / 8527521718
An important limitation to the use of Agrobacterium tumefaciens is that it can not:
1. infect dicots
2. be genetically modified
3. be cultured on a nutrient medium
4. infect crop plants such as wheat and corn
A cloning vector has two antibiotic resistance genes- for tetracycline and ampicillin. A foreign DNA
was inserted into the tetracycline gene. Non-recombinants would survive on the medium containing :
The technique not used for transformation of plant cells in recombinant procedures is:
1. Biolistics
2. Agrobacterium mediation
3. Use of viruses
4. Micro-injection
In the screening process during rDNA experiments, clones that metabolize -gal turn:
The plasmid pBR322 does not contain:
2. A gene that encodes for restrictor of plasmid copy number
3. Gene for Ampicillin and Streptomycin resistance
4. Sites for many restriction enzymes
The process of separation and purification of expressed protein before marketing is called
Credit for construction of first recombinant DNA may be given to
(1) Charles Darwin and Alfred Wallace
(2) Stanley Cohen and Herbert Boyer
(3) Meselson and Stahi
(4) Esther and Joshua Lederberg
Which of the following represents a correct palindromic sequence recognised by EcoRl?
↓\phantom{\rule{0ex}{0ex}}5\text{'}-GAATTC-3\text{'}\phantom{\rule{0ex}{0ex}}5\text{'}-CTTAAG-3\text{'}\phantom{\rule{0ex}{0ex}} ↑
↓\phantom{\rule{0ex}{0ex}}5\text{'}-CCCGGG-3\text{'}\phantom{\rule{0ex}{0ex}}3\text{'}-GGGCCC-5\text{'}\phantom{\rule{0ex}{0ex}} ↑
↓\phantom{\rule{0ex}{0ex}}5\text{'}-GAATTC-3\text{'}\phantom{\rule{0ex}{0ex}}3\text{'}-CTTAAG-5\text{'}\phantom{\rule{0ex}{0ex}} ↑
↓\phantom{\rule{0ex}{0ex}}5\text{'}-ATGCCG-3\text{'}\phantom{\rule{0ex}{0ex}}3\text{'}-TACGGC-5\text{'}\phantom{\rule{0ex}{0ex}} ↑
Restriction in Restriction endonuclease enzyme refers to
(1) cleaving of phosphodiester bond in DNA by the enzyme
(2) Cutting of DNA at specific position only
(3) Prevention of bacteriophage multiplication in bacteria
(4) Cutting each of the two strands of DNA at specific points in sugar phosphate backbone
Which of the following steps is/are catalysed by Taq polymerase in a PCR?
(1) Denaturation of template DNA
(2) Annealing of primers to template DNA
(3) Extension of primer end on template DNA
In case of Bam HI, H represents
(3) Name of scientist
Normal E. coil cells carry resistance against which of the following antibiotics?
Which of the following statements is correct in the context of observing DNA separated by agarose gel electrophoresis?
(1) DNA can be seen in visible light
(2) DNA can be seen without staining in visible light
(3) Ethidium bromide stained DNA can be seen in visible light
(4) Ethidium bromide stained DNA can be seen under exposure to UV light
In agarose gel electrophoresis, DNA molecules of different lengths are separated on the basis of their
(1) Charge only
(2) Size only
(3) Charge to size ratio
Significance of treating bacterial cells with calcium chloride before transformation is to facilitate
(1) Binding of DNA to the cell surface
(2) Uptake of DNA through membrane transport proteins
(3) Uptake of DNA by creating transient pores in the bacterial cell wall
(4) Expression of antibiotic resistance gene
Pure DNA precipitated by addition of chilled ethanol can be removed from solution by
(1) Elution
(2) Gel electrophoresis
(3) Spooling
The optimum temperature for polymerisation in PCR is _____________ while the enzyme responsible for the mentioned step can tolerate temperatures upto __________. Select the correct option according to the blanks.
°
°
°
°
°
°
°
°
Stirrer in stirred tank type bioreactor facilitates
(1) Oxygen delively from outside to inside
(2) Mixing and aeration
(4) Foam control
Separation and purification by filtration, centrifugation of desired compound produced in bioreactor is a part of
(1) Downstream processing only
(2) Scaling up and downstream processing
(4) Screening for recombinants and downstream processing
Each restriction endonuclease functions by inspecting the length of DNA sequence. It cleaves _______________.
(1) Only the master strand to produce sticky end
(2) Sense strand of DNA to produce sticky ends
(3) Each of the two strands of the double helix at specific points in their sugar phosphate backbones
(4) Messenger RNA to remove exons
A set of bacterial clones, each containing a plasmid or phage, is called
(1) Gene library
(2) Gene pool
(3) Genophore
(4) Genome
Select the option that excludes characteristics applicable to plasmids
(a) Circular DNA
(b) Linear DNA
(c) Present in all bacteria
(d) Contain essential genes
(e) Extra chromosomal self-replicating
(1) b & d only
(2) b, c & d only
(3) d b, e & c only
(1) Reversing transcribing RNA into DNA
(2) Digesting DNA
(3) Amplifying DNA
(4) Amplifying proteins and separating DNA
The correct order of steps in Polymerase Chain Reation (PCR) is
(1) Extension Denaturation, Annealing
A gene whose expression helps to identify transformed cell is known as
What is the criterion for DNA fragments movement on agarose gel during gel electrophoresis?
(2) Taq polymerase
(3) Polymerase III
(1) sal I
(2) Eco RV
(3) Xho I
(4) Hind III
(1) Bacterial artificial chromosome
(2) Yeast artificial chromosome
(4) Cosmid
Which of the following is not correctly matched for the oranism and its cell wall degrading enzyme?
(1) Plant cells - Cellulase
(2) Algae - Methylase
(3) Fungi - Chitinase
(4) Bacteria - Lysozyme
The colonies of recombinant bacteria appear white in contrast to colonies of non-recombinant bacteria because of
(1) Insertional inactivation of alpha-galactosidase in non-recombinant bacteria
(2) Insertional inactivation of beta-galactosidase in recombinant bacteria
(3) Inactivation of glycosidase enzyme in recombinant bacteria
(4) Non-recombinant bacteria containing beta-galactosidase
The figure below is the diagrammatic representation of the E.coli vector pBR 322. Which one of the given options correctly identifies its certain component(s)?
am{p}^{R}, te{t}^{R}
- antibiotic resistance genes
(2) ori-original restriction enzyme
(3) rop-reduced osmotic pressure
(4) Hind III, EcoRI - selectable markers
The figure below shows three steps (A, B, C) of Polymerase Chain Reaction (PCR). Select the option giving correct identification together with what it represents?
(1) A - Annealing with two sets of primers
(2) B - Denaturation at a temperature of about
98°C
separating the two DNA strands
(3) A - Denaturation at a temperature of about
50°
(4) C - Extension in the presence of heat stable DNA polymerase
Continuous addition of sugars in 'fed batch' fermentation is done to
(1) Degrade sewage
(2) Produce methane
(3) Obtain antibiotics
(4) Purify enzymes
(1) Bacterial cell can carry out the RNA splicing reactions
(2) The human chromosome can replicate in bacterial cell
(3) The mechanism of gene regulation is identical in humans and bacteria
(4) The genetic code is universal
Which one of the following hydrolyses internal phosphodiester bonds in a polynucleotide chain?
(1) Lipase
All the following statements about stanley Cohen and Herbert Boyer are correct but one is wrong. Which one is wrong?
(1) They discovered recombinant DNA (r-DNA) technology which marked the birth of modern biotechnology.
(2) They first produced, healthy sheep clone, a Finn Dorset lamb, Dolly, form the differentiated adult mammary cells.
(3) They invented genetic engineering by combining a piece of foreign DNA containing a gene from a bacterium with a bacterial (E.coli) plasmid using the enzyme restriction endonuclease.
(4) They isolated the antibiotic resistance gene by cutting out a piece of DNA from a plasmid which was responsible for conferring antibiotic resistance.
Extraction of bands of DNA from the agarose gel is temed as
(1) Down stream processing
(4) Insertional inactivation
In plant biotechnology, PEG is used in
(1) Protoplast isolation
(2) Cell culture preparation
Restriction enzymes cut the strand of DNA a little away from the centre of the palindromic site, but between the same two bases on the opposite single stranded strands, these overhanging stretches formed on each strand are called __________.
(1) Blunt ends
(2) Sticky ends
(3) Staggered end
Which of the following is considered as molecular glue?
What would happen if a recombinant DNA is inserted within the coding sequence of an enzyme,
\beta -
galactosidase?
(1) This will result in the inactivation of the enzyme
(2) The presence of chromogenic substrate will give blue coloured colonies
(3) The recombinant colonies do not produce any colour
(4) Both (1) & (3) are correct
You suspect your patient to be suffering from a bacterial disease, however the number of bacteria in the patient's body is very less. which method can help you detect these pathogens in the laboratory?
(1) Hybridoma technology
(3) Somatic hybridization
cDNA is
(1) Formed by reverse transcriptase
(2) Cloned DNA
(3) Circular DNA
(4) Recombinant DNA
A: Restriction enzymes belong to a larger class of enzymes called nucleases.
R: Each restriction enzyme recognises a specific palindromic nucleotide sequence in the DNA.
A: Taq Polymerase is involved in PCR technique.
R: This enzyme remain active during the high temperature including denaturation of double stranded DNA.
A: Pst I generates single stranded over hanging stretches in DNA after digestion that facilitate ligation.
R: These short extensions can form phosphodiester bonds with their complementary counterparts.
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Bifurcation Analysis in a Kind of Fourth-Order Delay Differential Equation
Xiaoqian Cui, Junjie Wei, "Bifurcation Analysis in a Kind of Fourth-Order Delay Differential Equation", Discrete Dynamics in Nature and Society, vol. 2009, Article ID 235038, 20 pages, 2009. https://doi.org/10.1155/2009/235038
Xiaoqian Cui1 and Junjie Wei1,2
2Department of Mathematics, Harbin Institute of Technology (Weihai), Weihai, Shandong 264209, China
A kind of fourth-order delay differential equation is considered. Firstly, the linear stability is investigated by analyzing the associated characteristic equation. It is found that there are stability switches for time delay and Hopf bifurcations when time delay cross through some critical values. Then the direction and stability of the Hopf bifurcation are determined, using the normal form method and the center manifold theorem. Finally, some numerical simulations are carried out to illustrate the analytic results.
Sadek [1] has considered the following fourth-order delay differential equation:By constructing Lyapunov functionals, it was given a group of conditions to ensure that the zero solution of (1.1) is globally asymptotically stable when the delay is suitable small, but if the sufficient conditions are not satisfied, what are the behaviors of the solutions? This is a interesting question in mathematics. The purpose of the present paper is to study the dynamics of (1.1) from bifurcation. We will give a detailed analysis on the above mentioned question. By regarding the delay as a bifurcation parameter, we analyze the distribution of the roots of the characteristic equation of (1.1) and obtain the existence of stability switches and Hopf bifurcation when the delay varies. Then by using the center manifold theory and normal form method, we derive an explicit algorithm for determining the direction of the Hopf bifurcation and the stability of the bifurcating periodic solutions.
We would like to mention that there are several articles on the stability of fourth-order delay differential equations, we refer the readers to [1–8] and the references cited therein.
The rest of this paper is organized as follows. In Section 2, we firstly focus mainly on the local stability of the zero solution. This analysis is performed through the study of a characteristic equation, which takes the form of a fourth-degree exponential polynomial. Using the approach of Ruan and Wei [9], we show that the stability of the zero solution can be destroyed through a Hopf bifurcation. In Section 3, we investigate the stability and direction of bifurcating periodic solutions by using the normal form theory and center manifold theorem presented in Hassard et al. [10]. In Section 4, we illustrate our results by numerical simulations. Section 5 with conclusion completes the paper.
2. Stability and Hopf Bifurcation
In this section, we will study the stability of the zero solution and the existence of Hopf bifurcation by analyzing the distribution of the eigenvalues. For convenience, we give the following assumptions:with and are both continuous functions and those three-order differential quotients at origin are existent. We rewrite (1.1) as the following form:It is easy to see that is the only trivial solution to the system (2.1) and the linearization around is given byIts characteristic equation is
Lemma 2.1. Suppose () andare satisfied. Then the trivial solution is asymptotically stable when .
Proof. When , (2.3) becomesBy Routh-Hurwitz criterion, all roots of (2.4) have negative real parts if and only ifThe conclusion follows from () and ().
Let be a root of (2.3), then we haveSeparating the real and imaginary parts givesAdding up the squares of both equations yieldsLet , and denoteThen (2.8) becomesSetThen we haveConsiderLet . Then (2.13) becomeswhereDefineThen by Lemma 2.2 in Li and Wei [11], we have the following results on the distribution of the roots of (2.10).
Lemma 2.2. (i) If , then (2.10) has positive roots if and only if and .
(ii) If , then (2.10) has positive roots if and only if there exists at least one , such that and .
Without loss of generality, we assume that equation has four positive roots denoted by , and , respectively. Then (2.8) also has four positive roots, say .
From (2.7), and conditions () and (), we have thatHence, we definewhenwhenLetbe the root of (2.3) satisfying .
Lemma 2.3. Suppose . If , then is a pair of simple purely imaginary roots of (2.3); and when ; and when .
Proof. Substituting into (2.3) and differentiating with respet to givesThenwhereand for , and , we can know that when ; and when . This completes the proof.
From and , it is easy to know that: if satisfies , if the equation has positive roots, then the number of the roots must be even; and from Lemma 2.3, we have that the sign of changes as varies, and then the stability switches may happen.
From Lemmas 2.1–2.3 and the theory in [9], we have the following.
Lemma 2.4. Suppose that (), () and are satisfied.
(i)If conditions (i) and (ii) in Lemma 2.2 are not satisfied, then all the roots of (2.3) have negative real parts for all .(ii)If one of conditions (i) and (ii) in Lemma 2.2 is satisfied, let then all roots of (2.3) have negative real parts when ; and there may exist an integer such that , and all the roots of (2.3) have negative real parts when , and (2.3) has at least a pair of roots with positive real parts when , where .
From Lemma 2.4 and applying the Hopf bifurcation theorem for functional differential equations [12, Chapter 11, Theorem 1.1], we have the following results.
Theorem 2.5. Suppose (), (), and are satisfied.
(i)If conditions (i) and (ii) in Lemma 2.2 are not satisfied, then the trivial solution of system (2.1) is asymptotically stable when .(ii)If one of conditions (i) and (ii) in Lemma 2.2 is satisfied, let , then the trivial solution of system (2.1) is asymptotically stable when ; and there may exist an integer such that , and the trivial solution of system (2.1) is asymptotically stable when , and is unstable when , where .(iii)The system (2.1) undergoes a Hopf bifurcation at the origin when , with
In this section, we will study the direction, stability, and the period of the bifurcating periodic solution. The method we used is based on the normal form method and the center manifold theory presented by Hassard et al. [10].
We first rescale the time by to normalize the delay so that system (2.1) can be written as the formThe linearization around is given byand the nonlinear term isThe characteristic equation associated with (3.2) isComparing (3.4) with (2.3), one can find out that , and hence, (3.4) has a pair of imaginary roots , when for , , and the transversal condition holds.
Let , where , , , Then is the Hopf bifurcation value for (3.1). Let be the root of (3.4).
For , letBy the Riesz representation theorem, there exists a matrix whose components are bounded variation functions in such thatIn fact, we choosewhere
For , defineHence, we can rewrite (3.1) in the following form:where , for .
For , defineFor and , define the bilinear formwhere . Then and are adjoint operators, and are eigenvalues of . Thus, they are also eigenvalues of .
By direct computation, we obtain thatis the eigenvector of corresponding to , andis the eigenvector of corresponding to . Moreover,where
Using the same notation as in Hassard et al. [10], we first compute the coordinates to describe the center manifold at . Let be the solution of (3.1) when .
DefineOn the center manifold , we havewhere and are local coordinates for center manifold in the direction of and . Note that is real if is real. We consider only real solutions.
For solution in of (3.1), since ,We rewrite this aswhere Compare the coefficients of (3.20) and (3.21), noticing (3.23), we haveBy (3.10) and (3.21), it follows thatwhereExpanding the above series and comparing the coefficients, we obtainNotice thatthat is,Thusand we haveThen we haveSo we only need to find out , , , and to obtain .
When , we haveComparing the coefficients with (3.26), we getFrom (3.27), (3.32), (3.33), and (3.34), we deriveThen we can getNotice thatWe obtainwhere HenceConsequently, from (3.32),Substituting , , , and intowe can obtain . Then we obtain the sign of
By the general theory due to Hassard et al. [10], we know that the quantity of determines the stability of the bifurcating periodic solutions on the center manifold, and determines the direction of the bifurcation; and we have the following.
Theorem 3.1. (i) If , then the Hopf bifurcation at the origin of system (1.1) is supercritical (subcritical).
(ii) If , then the bifurcating periodic solutions of system (1.1) are asymptotically stable (unstable).
4. An Example and Numerical Simulations
In this section, we give an example and present some numerical simulations to illustrate the analytic results.
Example 4.1. Consider the following equation:Clearly,
By direct computation, we know () and () are satisfied. That is, the data satisfy the conditions of Lemma 2.1. The characteristic equation isand we can obtain As shown in Figure 1, the equation has four roots asand HenceFor , we obtain that the zero solution of system (4.1) is asymptotically stable when .
The curve of function
h\left(V\right)={V}^{4}-27{V}^{3}+232{V}^{2}-648V+324
According to the formula given in Section 3, we can obtain thatThen we haveHence, when , we have
Conclusion of (4.1)
The zero solution of system (4.1) is asymptotically stable when . The Hopf bifurcation at the origin when is supercritical, and the bifurcating periodic solutions are asymptotically stable.
The following is the results of numerical simulations to system (4.1).
(i)We choose , then the zero solution of system (4.1) is asymptotically stable, as shown in Figure 2.(ii) We choose being near to , a periodic solution bifurcates from the origin and is asymptotically stable, as shown in Figure 3.
(a) y-x phase plots
(b) y waveform
(c) x waveform
(d) v-u phase plots
(e) v waveform
(f) u waveform
The zero solution of system (4.1) is asymptotically stable when
\tau =0.4
For (4.1) with
\tau =0.64>{\tau }_{2}^{0}
and sufficiently near
{\tau }_{2}^{0}=0.596
, the bifurcating periodic solution from zero solution occurs and is asymptotically stable.
In this paper, we consider a certain fourth-order delay differential equation. The linear stability is investigated by analyzing the associated characteristic equation. It is found that there may exist the stability switches when delay varies, and the Hopf bifurcation occurs when the delay passes through a sequence of critical values. Then the direction and the stability of the Hopf bifurcation are determined using the normal form method and the center manifold theorem. Finally, an example is given and numerical simulations are carried out to illustrate the results. By using Lyapunov's second method, Sadek [1] investigated the stability of system (1.1). The main result is as the following.
Theorem 5.1. Suppose that the following hold.
(i)There are constants and such thatfor all .(ii) as , andfor all , where is a positive constant such thatwith .(iii) and for all , and for all . Then the zero solution of (1.1) is asymptotically stable, provided thatwith .
Comparing Theorem 5.1 with Theorem 2.5 obtained in Section 2, one can find out that if the sufficient conditions to ensure the globally asymptotical stability of system (1.1) given in [10] are not satisfied, we can also get the stability of system (1.1), but here the stability means local stability, and the system undergoes a Hopf bifurcation at the origin. Otherwise, here we just need to give the condition on the origin of and , the condition is relatively weak.
This work was supported by the National Natural Science Foundation of China (No10771045), and Program of Excellent Team in Harbin Institute of Technology.
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Copyright © 2009 Xiaoqian Cui and Junjie Wei. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Geodetic reference system 1980 - zxc.wiki
The Geodetic Reference System 1980 ( GRS 80 , English Geodetic Reference System 1980 ) is an earth model that includes the most important parameters of the earth's shape , the earth's rotation and the gravitational field .
The reference ellipsoid of the GRS 80 serves the European spatial reference system ETRS89 as a geometrical calculation and mapping surface and with it forms the geodetic datum for uniform national surveying . The compliant UTM mapping is used as the mapping system. In Germany, the Gauss-Krüger coordinates related to the Bessel ellipsoid ( Potsdam date ) and the Krassowski ellipsoid ( Pulkovo date ) are converted to GRS80 and UTM.
GRS 80 also defines the normal gravity and thus replaces GRS 67 . WGS 84 practically uses the ellipsoid of the GRS 80 but contains more data about the gravitational field .
GRS 80 was adopted in 1979 at the general assembly of the IUGG , the international umbrella organization for geodesy and geophysics , and published in detail in 1980 by the head of the study group, Helmut Moritz , after consultation with the International Astronomical Union (IAU) .
The four defining geometric and physical parameters are:
Major semi-axis / equator radius
{\ displaystyle a = 6 {,} 378 \, 137 \ cdot 10 ^ {6} \, {\ textrm {m}}}
Gravitational constant times earth mass
(more precisely measurable than the individual factors)
{\ displaystyle GM _ {\ oplus} = 3 {,} 986 \, 005 \ cdot 10 ^ {14} \, {\ textrm {m}} ^ {3} / {\ textrm {s}} ^ {2} \ ,}
Difference in the main moments of inertia related to M ⊕ a 2
(without the influence of tidal forces )
{\ displaystyle J_ {2} = 1 {,} 08263 \ cdot 10 ^ {- 3}}
(including information on the orientation of the axis of rotation and the prime meridian )
{\ displaystyle \ omega = 7 {,} 292 \, 115 \ cdot 10 ^ {- 5} \, {\ textrm {rad / s}} \,}
The derived parameters for describing the shape of the Earth's ellipsoid are:
Small semi-axis / pole half-axis
{\ displaystyle b \ approx 6 {,} 356 \, 752 \, 314 \ cdot 10 ^ {6} \, {\ textrm {m}}}
Parameters of Earth flattening
{\ displaystyle f ^ {- 1} \ approx 2 {,} 98257222 \ cdot 10 ^ {2}}
List of geodetic reference points in Germany
Bernhard Heckmann: Introduction of the ETRS89 / UTM position reference system when switching to ALKIS. In: Communications of the DVW Hessen-Thüringen, 1/2005, p. 17 ff.
Ralf Strehmel: Official reference system for the situation - ETRS89. Surveying Brandenburg, 1/1996 ( PDF ).
Bernhard Heck: Calculation methods and evaluation models for national surveying. Karlsruhe 1987.
Hans-Gert Kahle : Introduction to Higher Geodesy . 2nd, revised and expanded edition. Verlag der Fachvereine, Zurich 1988.
International Association of Geodesy (IAG): Geodesist's Handbook 1980.
Helmut Moritz: Geodetic Reference System 1980 (MS Word document, if the formulas are not displayed: Ctrl-A Alt-F9; DOC file; 63 kB)
This page is based on the copyrighted Wikipedia article "Geod%C3%A4tisches_Referenzsystem_1980" (Authors); it is used under the Creative Commons Attribution-ShareAlike 3.0 Unported License. You may redistribute it, verbatim or modified, providing that you comply with the terms of the CC-BY-SA.
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EUDML | On complete convergence for -mixingales. EuDML | On complete convergence for -mixingales.
On complete convergence for
{L}^{p}
-mixingales.
Hu, Yijun. "On complete convergence for -mixingales.." International Journal of Mathematics and Mathematical Sciences 24.11 (2000): 737-747. <http://eudml.org/doc/48953>.
author = {Hu, Yijun},
keywords = {complete convergence; -mixingale; Banach space-valued random variables; strong law of large numbers; random sums; -mixingale},
title = {On complete convergence for -mixingales.},
AU - Hu, Yijun
TI - On complete convergence for -mixingales.
KW - complete convergence; -mixingale; Banach space-valued random variables; strong law of large numbers; random sums; -mixingale
complete convergence,
{L}^{p}
-mixingale, Banach space-valued random variables, strong law of large numbers, random sums,
{L}^{p}
-mixingale
|
Optimization Theory Overview - MATLAB & Simulink - MathWorks India
Optimization techniques are used to find a set of design parameters, x = {x1,x2,...,xn}, that can in some way be defined as optimal. In a simple case, this process might be the minimization or maximization of some system characteristic that is dependent on x. In a more advanced formulation, the objective function f(x), to be minimized or maximized, might be subject to constraints in one or more of these forms:
Equality constraints, Gi(x) = 0 ( i = 1,...,me)
Inequality constraints, Gi( x) ≤ 0 (i = me + 1,...,m)
Parameter bounds, xl, xu, where xl ≤ x ≤ xu, some xl can be –∞, and some xu can be ∞
A General Problem (GP) description is stated as
\underset{x}{\mathrm{min}}f\left(x\right),
\begin{array}{l}\begin{array}{cc}{G}_{i}\left(x\right)=0& i=1,...,{m}_{e}\\ {G}_{i}\left(x\right)\le 0& i={m}_{e}+1,...,m\end{array}\\ {x}_{l}\le x\le {x}_{u},\end{array}
where x is the vector of length n design parameters, f(x) is the objective function (which returns a scalar value), and the vector function G(x) returns a vector of length m containing the values of the equality and inequality constraints evaluated at x.
An efficient and accurate solution to this problem depends not only on the size of the problem in terms of the number of constraints and design variables, but also on characteristics of the objective function and constraints. When both the objective function and the constraints are linear functions of the design variable, the problem is known as a Linear Programming (LP) problem. Quadratic Programming (QP) concerns the minimization or maximization of a quadratic objective function that is linearly constrained. For both the LP and QP problems, reliable solution procedures are readily available. More difficult to solve is the Nonlinear Programming (NP) problem in which the objective function and constraints can be nonlinear functions of the design variables. A solution of the NP problem generally requires an iterative procedure to establish a direction of search at each major iteration. This solution is usually achieved by the solution of an LP, QP, or unconstrained subproblem.
All optimization takes place in real numbers. However, unconstrained least-squares problems and equation solving can be formulated and solved using complex analytic functions. See Complex Numbers in Optimization Toolbox Solvers.
|
(Redirected from Character string)
This article is about the data type. For other uses, see String.
Find sources: "String" computer science – news · newspapers · books · scholar · JSTOR (March 2015) (Learn how and when to remove this template message)
1 String datatypes
1.4.1 Null-terminated
1.4.2 Byte- and bit-terminated
1.4.3 Length-prefixed
1.4.4 Strings as records
2 Literal strings
3 Non-text strings
4 String processing algorithms
5 Character string-oriented languages and utilities
6 Character string functions
7.1 Concatenation and substrings
7.5 Lexicographical ordering
See also: Comparison of programming languages (string functions)
Main article: Null-terminated string
Main article: String literal
"Stringology" redirects here. For the physical theory, see String theory.
See also: Tuple
{\displaystyle \Sigma ^{*}=\bigcup _{n\in \mathbb {N} \cup \{0\}}\Sigma ^{n}}
{\displaystyle L:\Sigma ^{*}\mapsto \mathbb {N} \cup \{0\}}
{\displaystyle L(st)=L(s)+L(t)\quad \forall s,t\in \Sigma ^{*}}
Retrieved from "https://en.wikipedia.org/w/index.php?title=String_(computer_science)&oldid=1087579196"
|
Evaluate standard errors for multivariate normal regression model - MATLAB ecmmvnrstd - MathWorks Australia
ecmmvnrstd
Compute Standard Errors for Multivariate Normal Regression
StdParameters
Evaluate standard errors for multivariate normal regression model
[StdParameters,StdCovariance] = ecmmvnrstd(Data,Design,Covariance)
[StdParameters,StdCovariance] = ecmmvnrstd(___,Method,CovarFormat)
[StdParameters,StdCovariance] = ecmmvnrstd(Data,Design,Covariance) evaluates standard errors for a multivariate normal regression model with missing data. The model has the form
Dat{a}_{k}\sim N\left(Desig{n}_{k}×Parameters,\text{\hspace{0.17em}}Covariance\right)
[StdParameters,StdCovariance] = ecmmvnrstd(___,Method,CovarFormat) adds an optional arguments for Method and CovarFormat.
This example shows how to compute standard errors for a multivariate normal regression model.
[Mean,Covariance] = ecmnmle(Data);
% Estimate the sample standard errors for model parameters for each asset.
StdParam = ecmmvnrstd(TestData, TestDesign, Covar,'hessian')
StdParam = 2×1
Covariance — Estimates for covariance of regression residuals
Estimates for the covariance of the regression residuals, specified as an NUMSERIES-by-NUMSERIES matrix.
Method — Method of calculation for the information matrix
'hessian' (default) | character vector
(Optional) Method of calculation for the information matrix, specified as a character vector defined as:
'hessian' — The expected Hessian matrix of the observed log-likelihood function. This method is recommended since the resultant standard errors incorporate the increased uncertainties due to missing data.
'fisher' — The Fisher information matrix.
If Method = 'fisher', to obtain more quickly just the standard errors of variance estimates without the standard errors of the covariance estimates, set CovarFormat = 'diagonal' regardless of the form of the covariance matrix.
StdParameters — Standard errors for each element of Parameters
Standard errors for each element of Parameters, returned as an NUMPARAMS-by-1 column vector.
StdCovariance — Standard errors for each element of Covariance
Standard errors for each element of Covariance, returned as an NUMSERIES-by-NUMSERIES matrix.
|
EUDML | Subspaces of for that are admissible as kernels. EuDML | Subspaces of for that are admissible as kernels.
{L}_{p}
0\le p<1
that are admissible as kernels.
Allis, James T.
Allis, James T.. "Subspaces of for that are admissible as kernels.." The New York Journal of Mathematics [electronic only] 9 (2003): 137-140. <http://eudml.org/doc/123571>.
@article{Allis2003,
author = {Allis, James T.},
keywords = {-spaces; Banach space; rigid subspace; kernel; admissible kernel; -spaces},
title = {Subspaces of for that are admissible as kernels.},
AU - Allis, James T.
TI - Subspaces of for that are admissible as kernels.
KW - -spaces; Banach space; rigid subspace; kernel; admissible kernel; -spaces
{L}_{p}
-spaces, Banach space, rigid subspace, kernel, admissible kernel,
{L}_{p}
{L}^{p}
Articles by Allis
|
New Representations of the Group Inverse of 2×2 Block Matrices
2013 New Representations of the Group Inverse of
2×2
Block Matrices
Xiaoji Liu, Qi Yang, Hongwei Jin
This paper presents a full rank factorization of a
2×2
block matrix without any restriction concerning the group inverse. Applying this factorization, we obtain an explicit representation of the group inverse in terms of four individual blocks of the partitioned matrix without certain restriction. We also derive some important coincidence theorems, including the expressions of the group inverse with Banachiewicz-Schur forms.
Xiaoji Liu. Qi Yang. Hongwei Jin. "New Representations of the Group Inverse of
2×2
Block Matrices." J. Appl. Math. 2013 1 - 10, 2013. https://doi.org/10.1155/2013/247028
Xiaoji Liu, Qi Yang, Hongwei Jin "New Representations of the Group Inverse of
2×2
Block Matrices," Journal of Applied Mathematics, J. Appl. Math. 2013(none), 1-10, (2013)
|
WordCount - Maple Help
Home : Support : Online Help : Programming : Names and Strings : StringTools Package : English Text : WordCount
Words( s )
WordCount( s )
The Words(s) command splits a string into its constituent words. A list of the words in the input string s, in order, is returned. The non-word characters in the string are elided from the output.
The WordCount(s) command counts the number of words in a given string s.
A word is defined to be a maximal contiguous sequence of alphanumeric characters or the apostrophe character ('). Words are delimited by non-word characters.
8
\mathrm{with}\left(\mathrm{StringTools}\right):
\mathrm{Words}\left("This is a\nsentence."\right)
[\textcolor[rgb]{0,0,1}{"This"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"is"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"a"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"sentence"}]
\mathrm{WordCount}\left("This is a\nsentence."\right)
\textcolor[rgb]{0,0,1}{4}
\mathrm{Words}\left("The Word command correctly parses a word with an apostrophe, like isn\text{'}t, as a single word."\right)
[\textcolor[rgb]{0,0,1}{"The"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"Word"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"command"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"correctly"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"parses"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"a"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"word"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"with"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"an"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"apostrophe"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"like"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"isn\text{'}t"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"as"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"a"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"single"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"word"}]
\mathrm{WordCount}\left("The WordCount command correctly counts a word with an apostrophe, like isn\text{'}t, as a single word."\right)
\textcolor[rgb]{0,0,1}{16}
\mathrm{Words}\left("The Words command ignores punctuation, like colons \left(:\right); this is very useful!"\right)
[\textcolor[rgb]{0,0,1}{"The"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"Words"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"command"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"ignores"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"punctuation"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"like"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"colons"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"this"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"is"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"very"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"useful"}]
\mathrm{WordCount}\left("The WordCount command also ignores punctuation, like colons \left(:\right); this is very useful!"\right)
\textcolor[rgb]{0,0,1}{12}
|
Post Video Test: Periodic Classification of elements
Post Video Test: Periodic Classification of elementsContact Number: 9667591930 / 8527521718
For Avogadro's no. of 'X' atoms, if half of the atoms of 'X' transfer one electron to the other half of 'X' atoms, 409 KJ must be added. If these
{\mathrm{X}}^{-}
ions are subsequently converted to
{\mathrm{X}}^{+}
, an additional 733 KJ must be added. The IP and EA of X (in eV) are :
1. IP = 11.83 eV, EA = 3.35 eV
2. IP = 3.35 eV, IP = 11.83 eV
3. IP = 11.83 eV, EA = 11.83 eV
4. IP = 3.35 eV, IP = 3.35 eV
The least stable cation is :
{\mathrm{Li}}^{+}
{\mathrm{K}}^{+}
{\mathrm{Al}}^{2+}
{\mathrm{Si}}^{2+}
Which of the following pairs show reverse trend on moving along a period from left to right and from up to down in a group?
1. Nuclear charge and electron affinity
3. Atomic radius and electron affinity
Which will show maximum non-metallic character?
A mixture contains F and Cl atoms. The removal of an electron from each atom of the sample requires 284 KJ while addition of an electron of each atom of mixture releases 68.8 KJ energy.
If IE per atom for F and Cl are
27.91 \mathrm{x} {10}^{-22}
KJ and
20.77 \mathrm{x} {10}^{-22}
KJ.
{\mathrm{EA}}_{1}
for F and Cl are
-5.53 \mathrm{x} {10}^{-22}
-5.78 \mathrm{x} {10}^{-22}
KJ repectively. The % composition of mixture is :
1. 38% Cl, 62% F
2. 38% F and 62% Cl
4. can't be calculated
In which the following arrangements, the sequence is not strictly according to the property written against it?
{\mathrm{Co}}_{2} < {\mathrm{SiO}}_{2} < {\mathrm{SnO}}_{2} < {\mathrm{PbO}}_{2}
: increasing oxidising power
2. HF < HCl <HBr < HI : increasing acid strength
{\mathrm{NH}}_{3} < {\mathrm{PH}}_{3} < {\mathrm{AsH}}_{3} < {\mathrm{SbH}}_{3}
: increasing basic strength
4. B < C < O < N : increasing first ionisation enthaply
Which is not correct order for the stated property?
1. Ba > Sr > Mg : Atomic radius
2. F > O > N : First ionisation energy
3. Cl > F > I : Electron affinity
4. O > Se > Te : Electronegativity
The electronic confuguration of the element with maximum electron affinity is :
1{\mathrm{s}}^{2}, 2{\mathrm{s}}^{2}2{\mathrm{p}}^{3}
1{\mathrm{s}}^{2}, 2{\mathrm{s}}^{2}2{\mathrm{p}}^{5}
1{\mathrm{s}}^{2}, 2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}, 3{\mathrm{s}}^{2}3{\mathrm{p}}^{5}
1{\mathrm{s}}^{2}, 2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}, 3{\mathrm{s}}^{2}3{\mathrm{p}}^{3}
In which element shielding effect is not possible?
Which transition involves maximum amount of energy :
{\mathrm{M}}_{\left(\mathrm{g}\right)}^{-} \to {\mathrm{M}}_{\left(\mathrm{g}\right)} + \mathrm{e}
{\mathrm{M}}_{\left(\mathrm{g}\right)}^{-} \to {\mathrm{M}}_{\left(\mathrm{g}\right)}^{+} + 2\mathrm{e}
{\mathrm{M}}_{\left(\mathrm{g}\right)}^{+} \to {\mathrm{M}}_{\left(\mathrm{g}\right)}^{2+} + \mathrm{e}
{\mathrm{M}}_{\left(\mathrm{g}\right)}^{2+} \to {\mathrm{M}}_{\left(\mathrm{g}\right)}^{3+} + \mathrm{e}
The ionisation energy of nitrogen is more than oxygen because of :
1. more attraction of electron by the nucleus
2. the extra stability of half filled p-orbitals
3. the size of nitrogen atom is smaller
4. more penetrating effect
The first ionisation potential of aluminium is smaller than that of magnesium because :
1. atomic size of Al > Mg
2. atomic size of Al < Mg
3. Al has one electron in p-orbital
A sudden jump between the values of second and third ionization energies of an element would be associated with the electronic configuration :
1{\mathrm{s}}^{2}, 2{\mathrm{s}}^{2} 2{\mathrm{p}}^{6}, 3{\mathrm{s}}^{1}
1{\mathrm{s}}^{2}, 2{\mathrm{s}}^{2} 2{\mathrm{p}}^{6}, 3{\mathrm{s}}^{2} 3{\mathrm{p}}^{1}
1{\mathrm{s}}^{2}, 2{\mathrm{s}}^{2} 2{\mathrm{p}}^{6}, 3{\mathrm{s}}^{2} 3{\mathrm{p}}^{2}
1{\mathrm{s}}^{2}, 2{\mathrm{s}}^{2} 2{\mathrm{p}}^{6}, 3{\mathrm{s}}^{2}
The IP
{}_{1}
, IP
{}_{2}
{}_{3}
{}_{4}
and IP
{}_{5}
of an element are 7.1, 14.3, 34.5, 46.8, 162.2 eV repectively. The element is likely to be :
When a
{\mathrm{F}}_{\left(\mathrm{g}\right)}
atom is changed to
{\mathrm{F}}_{\left(\mathrm{g}\right)}^{-}
, its size increases from 72 pm to 136 pm. The % increase in terms of volume during the formation of F to
{\mathrm{F}}^{-}
5.75×{10}^{2}
5.75×{10}^{6}
6.75×{10}^{2}
1.96 ×{10}^{4}
kJ energy is supplied to 1 mol
{\mathrm{Li}}_{\left(\mathrm{g}\right)}
, it ionised to
{\mathrm{Li}}^{3+}
\mathrm{I}.\mathrm{E}{.}_{1}
for Li is 520 kJ
{\mathrm{mol}}^{-1}
\mathrm{I}.\mathrm{E}{.}_{1}
for H is
2.18×{10}^{-18}
{\mathrm{atom}}^{-1}
\mathrm{I}.\mathrm{E}{.}_{2}
for Li will be :
{\mathrm{mol}}^{-1}
11.81×{10}^{3}
{\mathrm{mol}}^{-1}
6.54×{10}^{-3} \mathrm{kJ} {\mathrm{mol}}^{-1}
{\mathrm{Cs}}^{+}
ions will be obtained if 1 joule energy is supplied to Cs-atoms, if
{\mathrm{IE}}_{1}
for Cs is 376
\mathrm{kJ} {\mathrm{mol}}^{-1}
1.60×{10}^{18}
6.023×{10}^{23}
3.76×{10}^{23}
∆{\mathrm{H}}_{{\mathrm{EG}}_{1}}
value for chlorine is 3.7 eV. The amount of energy released when 2g chlorine is completely converted to
{\mathrm{Cl}}^{-}
1. 2.4 kcal
For chlorine atom,
{\mathrm{IE}}_{1}
∆{\mathrm{H}}_{{\mathrm{EG}}_{1}}
values are 13.0 eV and 3.60 eV respectively. If Avogadro number of atoms are converted into
{\mathrm{Cl}}^{-}
by the reaction
\mathrm{Cl}+\mathrm{e}\to {\mathrm{Cl}}^{-}
. If the energy released in this reaction is used to ionise Cl-atoms as
\mathrm{Cl}\to {\mathrm{Cl}}^{+}+{\mathrm{e}}^{-}
. Then the number of Cl-atoms ionised is :
1.667×{10}^{23}
6.023×{10}^{23}
3.2×{10}^{23}
In the long form of periodic table, the elements having lowest ionisation potential are placed in :
1. I group
2. IV group
3. VII group
4. zero group
|
Electric Charges And Fields, Popular Questions: CBSE Class 12-science SCIENCE, Science - Meritnation
\mu C
\stackrel{\to }{{r}_{0}}=\left(\stackrel{^}{i}+\stackrel{^}{j}\right)m
\stackrel{\to }{{r}_{1}}=\left(4\stackrel{^}{i}+5\stackrel{^}{j}\right)m
900\frac{V}{m}
9\frac{kV}{m}
90\frac{V}{m}
\stackrel{⇀}{E}=a\sqrt{x}\stackrel{^}{i}
\stackrel{\to }{E}
\mathcal{l}
\mathcal{l}
\mathcal{l}
\mathcal{l}
Charges of + 15pC, - 15pC,+ 20pC and – 50pC are placed in order at each of the corners of a square of a side 20cm. Calculate the intensity of electric field at the point of intersection of the diagonals.
Deepthi S asked a question
Using Gauss's theorem,obtain an expression for the force between 2 point charges?
|
EUDML | Heegaard splittings of (surface) x I are standars. EuDML | Heegaard splittings of (surface) x I are standars.
Heegaard splittings of (surface) x I are standars.
Martin Scharlemann; Abigail Thompson
Scharlemann, Martin, and Thompson, Abigail. "Heegaard splittings of (surface) x I are standars.." Mathematische Annalen 295.3 (1993): 549-564. <http://eudml.org/doc/165057>.
@article{Scharlemann1993,
author = {Scharlemann, Martin, Thompson, Abigail},
keywords = {Heegaard splittings of a closed orientable genus surface crossed with an interval; compression bodies; spines},
title = {Heegaard splittings of (surface) x I are standars.},
AU - Scharlemann, Martin
TI - Heegaard splittings of (surface) x I are standars.
KW - Heegaard splittings of a closed orientable genus surface crossed with an interval; compression bodies; spines
Claire Renard, Gradients de Heegaard sous-logarithmiques d’une variété hyperbolique de dimension trois et fibres virtuelles
Heegaard splittings of a closed orientable genus
g
surface crossed with an interval, compression bodies, spines
3
Articles by Martin Scharlemann
Articles by Abigail Thompson
|
Quantization API Reference — PyTorch 1.11.0 documentation
Quantization API Reference
Quantization API Reference¶
This module contains Eager mode quantization APIs.
Top level APIs¶
Converts a float model to dynamic (i.e.
prepare_qat
Converts submodules in input module to a different module according to mapping by calling from_float method on the target module class.
fuse_modules
QuantWrapper
add_quant_dequant
add_observer_
swap_module
propagate_qconfig_
default_eval_fn
get_observer_dict
Traverse the modules and save all observers into dict.
torch.quantization.quantize_fx¶
This module contains FX graph mode quantization APIs (prototype).
prepare_fx
Prepare a model for post training static quantization
prepare_qat_fx
Prepare a model for quantization aware training
convert_fx
Convert a calibrated or trained model to a quantized model
fuse_fx
Fuse modules like conv+bn, conv+bn+relu etc, model must be in eval mode.
torch (quantization related functions)¶
This describes the quantization related functions of the torch namespace.
torch.Tensor (quantization related methods)¶
Quantized Tensors support a limited subset of data manipulation methods of the regular full-precision tensor.
q_zero_point
q_per_channel_scales
q_per_channel_zero_points
q_per_channel_axis
resize_
torch.quantization.observer¶
This module contains observers which are used to collect statistics about the values observed during calibration (PTQ) or training (QAT).
Base observer Module.
MovingAverageMinMaxObserver
PerChannelMinMaxObserver
MovingAveragePerChannelMinMaxObserver
The module records the running histogram of tensor values along with min/max values.
PlaceholderObserver
RecordingObserver
NoopObserver
get_observer_state_dict
Returns the state dict corresponding to the observer stats.
load_observer_state_dict
Given input model and a state_dict containing model observer stats, load the stats back into the model.
default_observer
default_placeholder_observer
Default placeholder observer, usually used for quantization to torch.float16.
default_debug_observer
Default debug-only observer.
default_weight_observer
default_histogram_observer
default_per_channel_weight_observer
default_dynamic_quant_observer
default_float_qparams_observer
torch.quantization.fake_quantize¶
This module implements modules which are used to perform fake quantization during QAT.
FakeQuantizeBase
Base fake quantize module Any fake quantize implementation should derive from this class.
Simulate the quantize and dequantize operations in training time. The output of this module is given by::.
FixedQParamsFakeQuantize
Simulate quantize and dequantize with fixed quantization parameters in training time.
FusedMovingAvgObsFakeQuantize
Fused module that is used to observe the input tensor (compute min/max), compute scale/zero_point and fake_quantize the tensor.
default_fake_quant
default_weight_fake_quant
default_per_channel_weight_fake_quant
default_histogram_fake_quant
default_fused_act_fake_quant
default_fused_wt_fake_quant
default_fused_per_channel_wt_fake_quant
Disable fake quantization for this module, if applicable. Example usage::.
Enable fake quantization for this module, if applicable. Example usage::.
Disable observation for this module, if applicable. Example usage::.
Enable observation for this module, if applicable. Example usage::.
torch.quantization.qconfig¶
This module defines QConfig objects which are used to configure quantization settings for individual ops.
default_qconfig
default_debug_qconfig
default_per_channel_qconfig
default_dynamic_qconfig
float16_dynamic_qconfig
float16_static_qconfig
per_channel_dynamic_qconfig
float_qparams_weight_only_qconfig
default_qat_qconfig
default_weight_only_qconfig
default_activation_only_qconfig
default_qat_qconfig_v2
This module implements the combined (fused) modules conv + relu which can then be quantized.
This is a sequential container which calls the Conv1d and ReLU modules.
This is a sequential container which calls the Linear and ReLU modules.
This is a sequential container which calls the Conv 1d and Batch Norm 1d modules.
This is a sequential container which calls the Conv 1d, Batch Norm 1d, and ReLU modules.
BNReLU2d
This is a sequential container which calls the BatchNorm 2d and ReLU modules.
update_bn_stats
freeze_bn_stats
This module implements the quantized implementations of fused operations like conv + relu. No BatchNorm variants as it’s usually folded into convolution for inference.
A BNReLU2d module is a fused module of BatchNorm2d and ReLU
torch.nn.intrinsic.quantized.dynamic¶
This module implements the quantized dynamic implementations of fused operations like linear + relu.
A LinearReLU module fused from Linear and ReLU modules that can be used for dynamic quantization.
torch.nn.qat.dynamic¶
This module implements versions of the key nn modules such as Linear() which run in FP32 but with rounding applied to simulate the effect of INT8 quantization and will be dynamically quantized during inference.
A linear module attached with FakeQuantize modules for weight, used for dynamic quantization aware training.
This module implements the quantized versions of the nn layers such as ~`torch.nn.Conv2d` and torch.nn.ReLU .
This is the quantized equivalent of LeakyReLU.
This is the quantized equivalent of Sigmoid.
FXFloatFunctional
module to replace FloatFunctional module before FX graph mode quantization, since activation_post_process will be inserted in top level module directly
torch.nn.quantized.functional¶
This module implements the quantized versions of the functional layers such as ~`torch.nn.functional.conv2d` and torch.nn.functional.relu . Note: relu() supports quantized inputs.
kH \times kW
sH \times sW
kD \ times kH \times kW
sD \times sH \times sW
y = xA^T + b
\text{LeakyReLU}(x) = \max(0, x) + \text{negative\_slope} * \min(0, x)
This is the quantized version of hardtanh().
Applies the quantized version of the threshold function element-wise:
This is the quantized version of elu().
This is the quantized version of hardsigmoid().
float(input, min_, max_) -> Tensor
Dynamically quantized Linear, LSTM, LSTMCell, GRUCell, and RNNCell.
A dynamic quantized linear module with floating point tensor as inputs and outputs.
A dynamic quantized LSTM module with floating point tensor as inputs and outputs.
Note that operator implementations currently only support per channel quantization for weights of the conv and linear operators. Furthermore, the input data is mapped linearly to the the quantized data and vice versa as follows:
\begin{aligned} \text{Quantization:}&\\ &Q_\text{out} = \text{clamp}(x_\text{input}/s+z, Q_\text{min}, Q_\text{max})\\ \text{Dequantization:}&\\ &x_\text{out} = (Q_\text{input}-z)*s \end{aligned}
\text{clamp}(.)
is the same as clamp() while the scale
s
z
are then computed as decribed in MinMaxObserver, specifically:
\begin{aligned} \text{if Symmetric:}&\\ &s = 2 \max(|x_\text{min}|, x_\text{max}) / \left( Q_\text{max} - Q_\text{min} \right) \\ &z = \begin{cases} 0 & \text{if dtype is qint8} \\ 128 & \text{otherwise} \end{cases}\\ \text{Otherwise:}&\\ &s = \left( x_\text{max} - x_\text{min} \right ) / \left( Q_\text{max} - Q_\text{min} \right ) \\ &z = Q_\text{min} - \text{round}(x_\text{min} / s) \end{aligned}
[x_\text{min}, x_\text{max}]
denotes the range of the input data while
Q_\text{min}
Q_\text{max}
are respectively the minimum and maximum values of the quantized dtype.
s
z
implies that zero is represented with no quantization error whenever zero is within the range of the input data or symmetric quantization is being used.
torch.quantization.quantize_fx
torch (quantization related functions)
torch.Tensor (quantization related methods)
torch.quantization.observer
torch.quantization.fake_quantize
torch.quantization.qconfig
torch.nn.intrinsic.quantized.dynamic
torch.nn.qat.dynamic
|
EUDML | Hyperbolic structure on a complement of tori in the 4-sphere. EuDML | Hyperbolic structure on a complement of tori in the 4-sphere.
Hyperbolic structure on a complement of tori in the 4-sphere.
Ivanšić, Dubravko
Ivanšić, Dubravko. "Hyperbolic structure on a complement of tori in the 4-sphere.." Advances in Geometry 4.1 (2004): 119-139. <http://eudml.org/doc/123622>.
@article{Ivanšić2004,
author = {Ivanšić, Dubravko},
keywords = {hyperbolic structure; 4-manifold; discrete subgroup},
title = {Hyperbolic structure on a complement of tori in the 4-sphere.},
AU - Ivanšić, Dubravko
TI - Hyperbolic structure on a complement of tori in the 4-sphere.
KW - hyperbolic structure; 4-manifold; discrete subgroup
hyperbolic structure, 4-manifold, discrete subgroup
{E}^{4}
4
Articles by Ivanšić
|
EUDML | Proximinal subspaces of of finite codimension. EuDML | Proximinal subspaces of of finite codimension.
A\left(K\right)
Rao, T. S. S. R. K.. "Proximinal subspaces of of finite codimension.." International Journal of Mathematics and Mathematical Sciences 2003.39 (2003): 2501-2505. <http://eudml.org/doc/50735>.
author = {Rao, T. S. S. R. K.},
title = {Proximinal subspaces of of finite codimension.},
AU - Rao, T. S. S. R. K.
TI - Proximinal subspaces of of finite codimension.
Best approximation, Chebyshev systems
Articles by Rao
|
Molecular basis of inheritance (6 Nov)- live session - NEET 2020 Contact Number: 9667591930 / 8527521718
Which of the following process is related to reverse transcription?
1. DNA dependent DNA synthesis
2. RNA dependent DNA synthesis
3. DNA dependent RNA synthesis
4. RNA dependentv polypeptide synthesis
Which of the following structires are present in core particle of nucleosome?
1. Octamer of histone proteins
2. 200 bp of DNA
3. Non-histone proteins
4. Linker DNA
Which of the following radioactive isotopes were utilised for labelling protein and DNA in transduction experiment respectively?
{}_{32}\mathrm{P}, {}_{35}\mathrm{P}
{}_{35}\mathrm{S}, {}_{35}\mathrm{P}
{}_{35}\mathrm{S}, {}_{32}\mathrm{P}
{}_{32}\mathrm{S}, {}_{35}\mathrm{P}
Unwinding of DNA creates tension which is related by enzyme
1. Helicase
2. Topoisomerase
Out of two strands of DNA one is carrying genetic information for transcription and it is called
1. Coding strand
2. Non template strand
3. Sense starnd
4. Template strand
Poly A tail is present in
1. mRNA of bacteria
2. tRNA of eukaryotes
3. Promotor of bacteria
4. mRNA of eukaryotes
In eukaryotes, RNA polymerase III catalyses the synthesis of
1. 5 S rRNA, tRNA & SnRNA
2. mRNA, HnRNA & SnRNA
3. 28 S rRNA, 18 S rRNA & 5 S rRNA
4. All types of rRNA & tRNA
In lac operon, the lac mRNA
1. Has several initiation and termination codons
2. Forms four different enzymes
3. Is not transcribed in the presence of lactose
4. Is involved in an anabolic reaction
In the technique of DNA fingerprinting digestion of DNA is followed by
If the sequence of one strand of DNA is 5' A T G C A T C G 3', find the sequence of the complementary strand in 5'
\to
3' direction
1. T A C G T A G C
2. C G A T G C A T
3. A T G C A T C G
4. A T C G T A C G
NHC structure proteins are
1. Basic proteins rich in lysine, arginine
2. Regulatory proteins
3. Catalytic proteins rich in tryptophan and arginine
4. Required for packaging of chromatin at higher levels.
Which of the following bond is not present in DNA?
\mathrm{\beta }-1\text{'}-9-\mathrm{N}-
glycosidic bond
3\text{'}-5\text{'}
phosphodiester bond
\mathrm{\beta }-1\text{'}-1-\mathrm{N}-
\mathrm{\beta }-1\text{'}-2-\mathrm{N}-
How many trypes of DNA polymerases are present in bacteria?
For the strand seperation and stabilisation during DNA replication which of the following set of enzymes and proteins are required?
1. SSBP, gyrase, primase
2. Topoisomerase, helicase, ligase
3. Gyrase, ligase, primase
4. Topoisomerase, helicase, SSBP
What is correct for bacterial transcription?
1. mRNA requires processing to become active
2. Translation can begin when mRNA is fully transcribed
3. Transcription and translation takes place in the same compartment
4. Rho factor initiates the process
In-vitro template independent RNA synthesis is a feature of
3. Ochoa enzyme
In protein synthesis, which of the following are required for the synthesis of charged tRNA?
1. Amino acid, GTP, initiation codon, ribosome
2. Amino acid, ATP,
{\mathrm{MG}}^{++}
, enzyme, tRNA
{\mathrm{K}}^{+}
, enzyme, mRNA
4. Aminoacyl tRNA, ribosome, initiation codon, release factor
The accessibility of promoter regions of bacterial DNA in many cases regulated by the interaction of proteins with sequences termed
2. structural genes
3. Inhibitor genes
1. Third base of a codon lacks vibrating capacity
2. Third base can establish H-bonds even with the non-complementary anticodon
3. Specificity of a anticodon is particularly determined by first two codon
4. Major cause of degeneracy is the first two N-bases of codon
If there are 81 million bases in RNA of human cell, then calculate the total number of introns present in cDNA
3. Equal to ribonucleotides
4. Half the number of ribonucleotides
Majority of unusual bases are found in tRNA,
\mathrm{T\psi C}
loop is
1. First loop from 5' - end of tRNA
2. AA - tRNA synthetase binding loop
3. Ribosomal binding loop
4. Nodoc site
How many amino acids will be coded by the mRNA sequence - 5'CCCUCAUAGUCAUAC3' if a adenosine residue is inserted after 12th nucleotide?
1. Five amino acids
2. Six amino acids
3. Two amino acids
4. Three amino acids
Identification and binding of RNA polymerase to the promoter sequence is a function of
1. Rho factor
2. Sigma factor
Pribnow box is a consensus of ____________ bases, forming a binding site for E. coli RNA polymerase at promotor.
2. AGGAGG
When the genomes of two people are cut using the same restriction enzyme, the length and number of fragments obtained are different, this is called
Which of the following does not code for any proteins?
1. Micro-satellites
2. Exons
3. Mini-satellites
In which step of DNA profiling, nitrocellulose membrane is used?
4. an enhancer
1. AGGUAUCGCAU
2. UGGTUTCGCAT
3. UCCAUAGCGUA
4. ACCUAUGCGAU
1. Alec Jeffreys - Streptococcus pneumoniae
2. Alfred Hershey and Martha Chase - TMV
3. Francois Jacob and Jacques Monod - Lac operon
4. Matthew Meselson and F. Stahl - Pisum sativum
1. Groffith
3. Avery, Mcleod and McCarty
DNA replication in bacteria occurs
1. During S-phase
2. Within nucleolus
3. Prior to fission
4. Just before transcription
1. r-RNA
2. t-RNA
3. m-RNA
4. mi-RNA
During DNA replication, Okazaki fragments are used to elongate
1. The leading strand towards replication fork
2. The lagging strand towards replication fork
3. The leading strand away from replication fork
4. The lagging strand away from the replication fork
1. Vince rosea
2. Visia faba
2. Criston
Which of the following rRNAs acts as structural RNA sa well as ribozyme in bacterial?
1. 5 S rRNA
2. 18 s rRNA
4. 5.8 S rRNA
1. Lactose and Galactose
Which of the following is not required for any of the techniques of DNA fingerprinting available at present?
1. DNA - DNA hybridization
3. Zinc finger analysis
Satellite DNA is important because it
1.code for enzymes needed for DNA replication
2. Codes for proteins needed in cell cycle
3. Shows high degree of polymorphism in population and also the same degree of polymorphism in an individual, which is heritable form parents to children
4. Does not code for proteins and is same in all members of the population.
Gene regulation governing lactose operon of E. coli that involves the lac 1 gene product is
1. Feedback inhibition because excess of
\mathrm{\beta }-
galactosidase can switch off transcription
2. Positive an inducible because it can be induced by lactose
3. Negative and inducible because repressor protein prevents transcription
4. Negative and repressible because repressor protein prevents transcription
Which enzyme/s will be produced in a cell in which there is a nonsence mutation in the lac Y gene?
1. Lactose permease
2. Transacetylase
3. Lactose permease and transacetylase
\mathrm{\beta }-
galactosidase
Which one of the following is not a part of a transcription ubit in DNA?
2. The structural gene
3. The inducer
Read the following four statement (A-D)
A. In transcription, adenosine pair with uracil
B. Regulation of lac operon by repressor is reffered to as positive regulation
C. The human genome has approximately 50,000 genes
D. Haemophilia is a sex-linked recessive disease
PCR and Restriction Fragment Length Polymorphism are the methods for
3. Study of enzymes
Which one of the following does not follow the central dogma of molecular biology?
4. Chalmydomonas
One gene-one enzyme relationship was established for the first time in
1. Diploccus pneumoniae
Telomere repetitive DNA sequence control the function of eukaryote chromosomes because they:
1. Prevent chromosome loss
2. Act as replicons
3. Are RNA transcription initiator
4. Help chromosome pairing
During transcription, RNA polymerase holoenzyme binds to a gene promoter and assumes a saddle - like structure. What id it's DNA - binding sequence?
2. TTAA
3. AATT
The length of DNA molecule greatly exceeds the dimensions of the nucleus in eukaryotic cells. How is this DNA accommodated?
1. Through elimination of repetitive DNA
2. Deletion of non-essential genes
3. Super-coiling in nucleosomes
4. DNA se digestion
In an inducible operon, the genes are
1. Always expressed
2. Usually not expressed unless a signal turns them 'on'
3. Usually expresses unless a signal turns them 'off'
4. Never expressed
A single strand of nucleic acid tagged with a radioactive molecule is called
The reaction, Amino acid + ATP
\to
Aminoacyl - AMP + P-P depicts
1. Amino acid assimilation
2. Amino acid transformation
3. Amino acis activation
4. Amino acid translocation
There are special proteins that help to open up DNA double helix in front of the replication fork. These proteins are
2. DNA topoisomerase I
3. DNA gyrase
In split genes, the coding sequences are called
2. Cristons
3. Introns
4. Operons
DNA elements, which can switch their position, are called
Irregularity is found in Drosophila during the organ differentiation. For example, in place of wing long legs are formed. Which gene is responsible?
1. Double dominant gene
2. Homeotic gene
3. Complimentary gene
4. Plastid gene
Which of the following reunites the exon segments after RNA splicing?
2. RNA primase
3. RNA ligase
4. RNA proteoses
During transcription, the DNA site at which RNA polymerase binds is called
Which antibiotics inhibits interaction between tRNA and mRNA during bacterial protein synthesis?
E. coli about to replicate was placed in a medium containing radioactive thymidine for five minutes. Then it was made to replicate in a normal medium. Which of the following observation shall be correct?
1. Both the strands of DNA will be radioactive
2. One strand radioactive
3. Each strand hald radioactive
4. None is radioactive
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Post Video Test - Hydrocarbons
Post Video Test - HydrocarbonsContact Number: 9667591930 / 8527521718
\left({\mathrm{CH}}_{3}{\right)}_{3}\mathrm{COH}\to
The reaction of toluene with Cl2 in the presence of FeCl3 gives 'X' and reaction in presence of presence of light gives 'Y'. Thus 'X' and 'Y' are
1. X= benzal chloride, Y=o-chlorotoluene
2. X=m-chlorotoluene, Y=p-chlorotoluene
3. X=o and p-chlorotoluene, Y= trichloromethyl benzene
4. X= benzyl hloride, Y=m-chlorotoluene
Benzene reacts with n-propyl chloride in the presence of anhydrous AlCl3 to give
1. 3-propyl-1-chlorobenzene
2. n-propyl benzene
4. iso-propyl benzene
The reaction of toluene with
{\mathrm{Cl}}_{2}
{\mathrm{FeCl}}_{3}
gives 'X' and reaction in presence of light gives 'Y'. Thus, 'X' and 'Y' are
1. X = benzal chloride, Y= o-chlorotoluene
2. X = m-chlorotoluene, Y= p-chlorotoluene
3. X = o and p-chlorotoluene, Y = trichloromethyl benzene
4. X = benzal chloride, Y = m-chlorotoluene
Ethylene forms ethylene chlorohydrin by the action of:
1. dry chlorine gas
2. dry hydrogen chlorine gas
3. solution of chlorine gas in water
A mixture of ethane, ethene and ethyne is passed through an ammoniacal
{\mathrm{AgNO}}_{3}
solution. The gases that remain unreacted are-
1. Ethane and Ethene
2. Ethane and Ethyne
3. Ethene and Ethyne
4. Ethane only
Ozonolysis of will give:
OC-C{H}_{2}-C{H}_{3}-C{H}_{3}-CO\phantom{\rule{0ex}{0ex}} | |\phantom{\rule{0ex}{0ex}}C{H}_{2}— C{H}_{2} — C{H}_{2}
An octane number 100 is given to:
2. Iso-octane
3. neo-pentane
4. neo-octane
Which of the following molecules/species are aromatic in character?
Xylene on oxidation with acidic
{\mathrm{KMnO}}_{4}
2. Isophthalic acid
1-butyne on reaction with hot alkaline
{\mathrm{KMnO}}_{4}
{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}{\mathrm{CH}}_{2}\mathrm{COOH}
{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}\mathrm{COOH}
{\mathrm{CO}}_{2}
{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}\mathrm{COOH}
{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}\mathrm{COOH}
+ HCOOH
What is the end product of the following sequences of operations?
{\mathrm{CaC}}_{2}\stackrel{{\mathrm{H}}_{2}\mathrm{O}}{\to }
\underset{{\mathrm{Hg}}^{2+}}{\overset{\mathrm{Dil}.{\mathrm{H}}_{2}{\mathrm{SO}}_{4}}{\to }}
\underset{{\mathrm{H}}_{2}}{\overset{\mathrm{Ni}}{\to }}
{\mathrm{C}}_{2}{\mathrm{H}}_{5}\mathrm{OH}
{\mathrm{C}}_{2}{\mathrm{H}}_{4}
\mathrm{R}-\mathrm{CH}={\mathrm{CH}}_{2}\underset{{\mathrm{C}}_{2}{\mathrm{H}}_{5}\mathrm{OH}}{\overset{\mathrm{Na}/{\mathrm{NH}}_{3}\left(\mathrm{l}\right)}{\to }}
{\mathrm{RCH}}_{2}{\mathrm{CH}}_{3}
2. Fisher-Spier reduction
4. Arndt-Eistert reduction
The cylindrical shape of an alkyne is due to
1. three sigma
\mathrm{C}-\mathrm{C}
\mathrm{\pi }
\mathrm{C}-\mathrm{C}
3. two sigma
\mathrm{C}-\mathrm{C}
\mathrm{\pi }
\mathrm{C}-\mathrm{C}
4. one sigma
\mathrm{C}-\mathrm{C}
\mathrm{\pi }
\mathrm{C}-\mathrm{C}
In Friedel-Craft's alkylation, besides
{\mathrm{AlCl}}_{3}
the other reactants are
{\mathrm{C}}_{6}{\mathrm{H}}_{6} + {\mathrm{NH}}_{2}
{\mathrm{C}}_{6}{\mathrm{H}}_{6} + {\mathrm{CH}}_{4}
{\mathrm{C}}_{6}{\mathrm{H}}_{6} + {\mathrm{CH}}_{3}\mathrm{Cl}
{\mathrm{C}}_{6}{\mathrm{H}}_{6} + {\mathrm{CH}}_{3}\mathrm{COCl}
{\mathrm{CH}}_{3}\mathrm{C}\equiv \mathrm{C}.{\mathrm{CH}}_{3}
\underset{\left(\mathrm{ii}\right) {\mathrm{H}}_{2}\mathrm{O}/\mathrm{Zn}}{\overset{\left(\mathrm{i}\right) \mathrm{X}}{\to }}
C{H}_{3}-C-C-C{H}_{3}\phantom{\rule{0ex}{0ex}} || ||\phantom{\rule{0ex}{0ex}} O O
In the above reaction X is:
{\mathrm{HNO}}_{3}
{\mathrm{O}}_{2}
{\mathrm{O}}_{3}
{\mathrm{KMnO}}_{4}
+ Alc.KOH
\to
2. But-2-ene
4. But-1-yne
Alkene-1 on hydroboration followed with action of
{\mathrm{H}}_{2}{\mathrm{O}}_{2}
1. alkanol-2
3. alkanal
4. alkanone
{\mathrm{CH}}_{3}-\mathrm{CHCOOK}\phantom{\rule{0ex}{0ex}} |\phantom{\rule{0ex}{0ex}}{\mathrm{CH}}_{3}-\mathrm{CHCOOK}
\stackrel{\mathrm{Electrolysis}}{\to }
A (Major)
The product "A" is -
The most stable alkene is
\mathrm{CH}\equiv \mathrm{CH}\stackrel{{\mathrm{O}}_{3}/\mathrm{NaOH}}{\to }
\stackrel{\mathrm{Zn}/{\mathrm{CH}}_{3}\mathrm{COOH}}{\to }
{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}\mathrm{OH}
{\mathrm{CH}}_{3}\mathrm{COOH}
{\mathrm{CH}}_{3}\mathrm{OH}
\stackrel{\mathrm{NBS}}{\to }
Products formed are
1. (A) and (B) only 2. (A) and (C) only
3. (B) and (C) only 4. (A), (B), (C) and (D)
Observe the following sequence of reactions.
The product R is :
'X' is :
\stackrel{{\mathrm{H}}_{2}\left(1\mathrm{mole}\right)/\mathrm{Pt}}{\to }
Double bond equivalent (degree of unsaturation) of (A) is:
The correct order of reactivity of I,II and III towards addition reactions is:
(1) I>III>II
(2) I>II>III
(3) III>II>I
(4) III>I>II
On catalytic reduction by one mole H2/Pt ,how many alkenes will give n-butane?
Taking into account the stability of various carbocations and, as well as the rules governing mechanisms of carbocation rearrangements, which reaction is most likely to occur during the given reaction ?
\stackrel{1\mathrm{eq}.\mathrm{NBS}}{\to }
\stackrel{{\mathrm{PhO}}^{-}}{\to }
Y (major product)
The product Y is -
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Microbes in Human Welfare - Live Session - NEET & AIIMS 2019
Microbes in Human Welfare - Live Session - NEET & AIIMS 2019Contact Number: 9667591930 / 8527521718
Correctly identify the enzymes (A), (B), and (C) of yeast during the leavening process of bread making:
\mathrm{Wheat} \mathrm{flour} \mathrm{starch} \stackrel{\mathrm{A}}{\to } \mathrm{Maltose} \stackrel{\mathrm{B}}{\to } \mathrm{Glucose} \stackrel{\mathrm{C}}{\to } \mathrm{Ethyl} \mathrm{alcohol} + {\mathrm{CO}}_{2}
(1) Zymase Amylase Maltase
(2) Zymase Maltase Amylase
(3) Amylase Maltase Zymase
(4) Maltase Amylase Zymase
A. Curd (i) LAB
B. Yoghurt (ii) Penicillium
C. Buttermilk (iii) Lactobacillus bulgaricus and
D. Cheese (iv) LAB and Streptococcus cremoris
(1) A = (i), B = (iii), C = (iv), D = (ii)
(2) A = (i), B = (iv), C = (i), D = (iii)
(3) A = (iv), B = (ii), C = (i), D = (iii)
(4) A = (ii), B = (iv), C = (iii), D = (i)
(i) Statins are produced by some yeasts
(ii) Statins are blood cholesterol-lowering agents
(iii) Statins resemble mevalonate
(iv) Statins are competitive inhibitor of \beta-hydroxy- -methyl glutaryl CoA reductase (HMG )
(v) Statins are immunosuppressive agents.
(a) ii) , iii and (iv)
© and
Which of the following are end products during fermentation caused by yeasts?
(1) Ethanol
+{\mathrm{CO}}_{2}+
+{\mathrm{H}}_{2}\mathrm{O}+
+{\mathrm{H}}_{2}\mathrm{S}+
+{\mathrm{H}}_{2}\mathrm{S} +{\mathrm{H}}_{2}\mathrm{O}+
Enzyme having thrombolytic effect is:
In a sewage treatment plant (STP), the primary treatment is a (A) which involves removal of small and large particles from sewage through _(B) and (C)
1. Physical process Filtration Sedimentation
2. Chemical process Filtration Sedimentation
3. Biological Filtration process Sedimentat sedimentation
4. Biological process Microbial digestion Sedimentation
In sewage treatment plant (STP), the flocs (aerobic bacteria+fungal filament) settle in settling tank. This settled sediment is called:
1. Activated sludge/ secondary sludge
2. Activated sludge/ primary sludge
3. Primary effluent
4. Secondary effluent
In a sewage treatment plant (STP), the primary sludge is obtained by a:
1. Physico-chemical process
4. Physico-biological process
Select the incorrect choice w.r.t STP:
1. Biological treatment requires aeration
2. Primary treatment requires aeration.
3. Tertiary treatment requires aeration.
Biogas production is a type of:
1. Anaerobic action
2. Aerobic action
In a STP, the sewage water is treated till:
1. BOD of sewage water rises significantly at the end of secondary treatment.
2. BOD of sewage water is reduced significantly at the end of secondary treatment.
3. Anaerobic oxidation becomes very fast.
4. Reduction of organic wastes becomes complete.
Biogas production occurs with the help of:
2. Anaerobic breakdown of organic matter
3. Aerobic breakdown of organic matter
4. More than one option is correct
The technology of biogas production was developed in India mainly by the contribution of:
1. KVIC and IARI
2. NEERI and RRI
3. KVIC and NEERI
4. IARI and NEERI
Gobar gas is produces by:
Choose the odd one out w.r.t biopesticide
2. Nucleopolyhedovirus (NPV)
Butterfly caterpillars can be controlled by the biocontrol agent like:
Organic farming involves:
1. Integrated pest management,i.e., IPM
2. use of biopesticides and biofertilizers
3. Use of pest resistant varieties
4. More than one option is involved
A non-symbiotic biofertilizer is:
1. Rhizobium and Frankia
2. Azospirillum and Azobacter
3. Rhizobium and Azospirillum
4. Frankia and Azobacter
Which of the following is an important biofertilizer in rice fields?
4. neurospora
The symbiotic association of plant roots and fungi:
1. Is called mycorrhizal involving fungi Glomus
2. Is called lichen
3. Is called mycorrhizal involving fungi Neurospora
4. Is called mycoherbicide
1. Aulosira is the most active, non-symbiotic nitrogen fixer in the three rice fields
2. Trichoderma is a free-living fungi common in root ecosystems and effective against several plant pathogens to prevent fields
4. 'College' is obtained from blue-green algae.
|
Write a possible equation for each of these graphs. Assume that one mark on each axis is one unit. When you are in class, check your equations on a graphing calculator and compare your results with your teammates.
Notice that this is a hyperbola
(y=\frac{1}{x})
Notice that the vertex is
(0,−5)
Notice that this is a cubic graph.
Notice that this is an exponential graph.
y
(0,−6)
Hint (f)-(i):
See the hints from parts (a)-(e) to help you solve parts (f)-(i).
|
Copy the Venn diagram below on your paper. Then show where each person described should be represented in the diagram. If a portion of the Venn diagram remains empty, describe the qualities a person would need to belong there.
Carol: “I rarely study and enjoy braiding my long hair.”
Bob: “I never do homework and have a crew cut.”
Pedro: “I love joining after school study teams to prepare for tests and I like being bald!”
Write items that satisfy #1 in that circle. Write items that satisfy #2 in that circle. Any items that satisfy both #1 and #2 go where the circles intersect. Any items satisfy neither conditions go outside the circles.
For each part (a) through (c) think about whether the person described studies a lot for class and has long hair.
If they have long hair, they belong in circle #
2
If they study a lot, they should be in circle #
1
If both, they should be in the overlap of the two circles.
If neither, they should be outside both circles.
Carol is in circle #
2
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1. Vessels are unicellular and with narrow lumen
2. Vessels are multicellular and with wide lumen
3. Tracheids are unicellular and with wide lumen
4. Tracheids are multicellular and with narrow lumen
The endodermis is absent or indistinct in the:-
1. Roots of all plants
2. Stems of pteridophytes
3. Leaves of gymnosperms
4. Stems of woody plants
The term 'bark' includes :
IV. Secondary phloem
4. I, II, III and IV only
In the dicot root, the vascular cambium:-
1. is absent
2. is completely secondary in origin
3. does not form a continuous ring
4. originates from the tissue just above the phloem bundles
Summer wood is:
1. the same as heartwood.
2. found to the outside of each annular ring of xylem.
3. found to the inside of each annular ring of xylem.
4. formed throughout the growing season.
The annular and spirally thickened conducting elements generally develop in the protoxylem when the root or stem is
1. maturing 2. elongating
3. widening 4. differentiating
Palisade parenchyma is absent in leaves of :-
Reduction in vascular tissue, mechanical tissue and cuticle is characteristic of
4. hydrophytes
Vascular tissues in flowering plants develop from
2. plerome
3. periblem
4. dermatoge
For a critical study of secondary growth in plants, whic one of the following pairs is suitable:
1. Sugarcane and sunflower
2. Teak and pine
3. Deodar and fern
4. Wheat and maiden hair fern
A common structural feature of vessel elements and sieve tube elements is :-
1. thick secondary walls
2. pores on lateral walls
3. presence of P-protein
4. enucleate condition
Epidermis is derived from:-
3. procambium
Passage cells are found in
1. Dicot stem
Bark refers to
1. Phellem + Phellogen + Phelloderm
2. Periderm + Cortex
3. Phellem + Phelloderm + Secondary phloem
4. Periderm + Cortex + Pericycle + Secondary phloem
The branched sclereids present in hydrophytes are:-
1. osteosclereids
2. trichosclereids
3. macrosclereids
4. astrosclereids
As secondary growth proceeds, in a dicot stem, the thickness of.
1. sapwood increases
2. heartwood increases
3. both sapwood and heartwood increases
4. both sapwood and heartwood remains the same
Glycolate induces opening of stomata in
{\mathrm{CO}}_{2}
{\mathrm{CO}}_{2}
(d) Absence of
{{\mathrm{CO}}_{2}}_{}
Two adjacent vessel and tracheid exchange sap through
1. Perforated end walls
3. Intercellular spaces
4. Intracellular spaces and parenchyma
1. fusiform initial cells
3. protoderm
A stem without vessel possessing prominent sieve tubes would belong to
4. Trochodendron
While removing the skin of a potato tuber, we remove
In monocot stem, Trichomes are ___________ and help in reduction of transpiration.
2.absent
3. sometimespresent
Pith is ___________ in monocot stem.
1.differentiated
2.undifferentiated
3. not formed
Dicotyledonous leaf is also known as
1. Dorsiventral leaf
2. Bi-facial leaf
4. Isobilateral leaf
The tissues present between the adaxial and abaxial epidermis are
1.mesophyll
3.vascular bundle
Type of vascular bundle present in dicot leaf is
1. conjoint, collateral and closed
2. conjoint, collateral and open
3. conjoint, bicollateral and closed
4. radial, collateral and open
In grasses, bulliforms are formed from
1. epidermis only
2. epidermis and veins
3. epidermis and hypodermis
Dicot leaves are ____________and monocot leaves are _______________.
1.hypostomatic and amphistomatic
2.amphistomatic and hypostomatic
3. Astomatic and amphistomatic
4. epistomatic and hypostomatic
The cells which are formed from vascular cambium towards the centre are____________and towards the periphery are _________________
1. secondary xylem, secondary phloem
2. bast, wood
3. wood, bast
Primary xylem remains less or more intact towards the __________
1. around the centre
2. around the periphery
3. outer to the pith
Annual rings are bands
1.secondary xylem and xylem rays
2.secondary phloem and phloem rays
3. secondary phloem and xylem rays
Least distinct annual rings are present in regions like
1. tropical areas
How many of the followings constitute the bark?
( secondary phloem, secondary cortex, cork cambium, complimentary cells, epidermis, periderm )
Complimentary cells are formed by the activity of
3. phelloderm
If bark is removed then plants becomes_________ due to excess of water________
1. thick, gain
2. thin, loss
3. dead, loss
4. strong, intake
In dicot root the complete ring of vascular bundle is formed by
1. conjunctive tissue
Lenticels are present in
1. stem only
2. root only
4. root, stem and leaf
The main mechanical tissue of the plant is
Screlenchymatous fibres are
1. thick walled, elongated and pointed cells
2. contains simple and bordered pits
3. longest cells in plant body
1. cork cambium-periderm
2. ground meristem- cortex and pith of a dicot stem
3. procambium-primary vascular tissue
4. protoderm-vascular tissue of a monocot stem
I. Gymnosperms lack albimunous cells and sieve celles
II. The companion cells are specialized parenchymatous cells
III. Phloem parenchyma is absent in most of the monocotyledons
IV. Phloem fibers are generally absent in the primary phloem
Read the different components from I to IV in the list given below and tell the correct order of the components with reference to their arrangement from outer side to inner side in a woody dicot stem
I. Secondary cortex
IV. Phellem
1. III,IV,II,I
3. IV,I,III,II
4. IV,III,I,II
Which of the following facilitates opening of stomatal aperture:-
1. Contraction of outer wall of guard cells
2. Decrease in turgidity of guard cells
3. Radial orientation of cellulose microfibrils in the cell wall of guard cells
4. Longitudinal orientation of cellulose microfibrils in the cell wall of guard cells
A monocot characterstic of the monocot root is the presence of
2. scattered vascular bundles
3. vasculature without cambium
4. cambium sandwiched between phloem and xylem along the radius
In a ring girdled plant?
1. the shoot dies first
2. the root dies first
3. the shoot and root die together
4. neither root nor shoot will die
If the stoma remains surrounded by a limited number of cells which cannot be distinguished from other epidermal cells, then this type of stomata is known as:
1. perigynous type
2. paracytic type
3. anisocytic type
4. anomocytic type
1. cork lacks stomata, but lenticels carry out transpiration
2. Passage cells help in transfer of food from cortex to phloem
3. Sieve tube elements possess cytoplasm but no nuclei.
4. The shoot apical meristem has a quiescent centre.
In the sieve elements, which one of the following is the most likely function of P-proteins?
1. Deposition of callose on sieve plates
2. Providing energy for active translocation
3. Autolytic enzymes
4. Sealing mechanism on wounding
In succulent plants the stomata open in night and close in a day. Which among the following would be the best hypothesis to explain the mechanism of stomatal action in night only?
1. CO2 accumulates, reduces pH, stimulate enzymes resulting in accumulation of sugars.
2. Increase in CO2 concentration, conversion of organic acids into starch resulting in the increased conversion into sugars resulting in K+ transport.
3. Low CO2 concentration accumulates organic acids resulting in the increased concentration of cell sap.
4. CO2 used up, increase pH results in accumulation of sugars
Tyloses are balloon-like outgrowths in the lumen of secondary xylem tracheids and vessel of duramen region. These are actually
1. outgrowth of vessels of xylem
2. ingrowth of vessels
3. outgrowth of parenchymatous cell
4. swelling of xylem fibres for no function
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findsym - Maple Help
Home : Support : Online Help : Connectivity : MTM Package : findsym
findsym(M)
findsym(M,n)
The findsym(M) function finds symbolic variables in M. The sequence of symbols are returned in alphabetical order.
The findsym(M,n) function finds the n symbolic variables in an expression nearest in alphabetical order to the letter x.
Symbolic constants like the imaginary unit and Pi are not considered variables, so they will not be returned from a call to findsym even when present in the given expression.
\mathrm{with}\left(\mathrm{MTM}\right):
M≔\mathrm{Matrix}\left(2,3,'\mathrm{fill}'=a\left(x+y\right)\right):
\mathrm{findsym}\left(M\right)
[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{a}& \textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{y}\end{array}]
\mathrm{findsym}\left(\mathrm{\alpha }+a+b\right)
[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{a}& \textcolor[rgb]{0,0,1}{\mathrm{\alpha }}& \textcolor[rgb]{0,0,1}{b}\end{array}]
\mathrm{findsym}\left(\mathrm{cos}\left(\mathrm{\alpha }\right)b\mathrm{x1}+14y,2\right)
[\begin{array}{cc}\textcolor[rgb]{0,0,1}{\mathrm{x1}}& \textcolor[rgb]{0,0,1}{y}\end{array}]
\mathrm{findsym}\left(y\left(4+3I\right)+6\right)
\textcolor[rgb]{0,0,1}{y}
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Niobium–tin - Wikipedia
Niobium–tin is an intermetallic compound of niobium (Nb) and tin (Sn), used industrially as a type II superconductor. This intermetallic compound has a simple structure: A3B. It is more expensive than niobium–titanium (NbTi), but remains superconducting up to a magnetic flux density of 30 teslas [T] (300,000 G),[1] compared to a limit of roughly 15 T for NbTi.
Nb-Sn phase diagram
Nb3Sn was discovered to be a superconductor in 1954. The material's ability to support high currents and magnetic fields was discovered in 1961 and started the era of large-scale applications of superconductivity.
The critical temperature is 18.3 kelvins (−254.8 °C; −426.7 °F). Application temperatures are commonly around 4.2 K (−268.95 °C; −452.11 °F), the boiling point of liquid helium at atmospheric pressure.
In April 2008 a record non-copper current density was claimed of 2,643 A mm−2 at 12 T and 4.2 K.[2]
1 Niobium tin composite wire
1.1 Strain effects
4 Developments and future uses
Niobium tin composite wireEdit
Unit cell of the A3B phases of Nb3Sn
Mechanically, Nb3Sn is extremely brittle and thus cannot be easily drawn into a wire, which is necessary for winding superconducting magnets. To overcome this, wire manufacturers typically draw down composite wires containing ductile precursors. The "internal tin" process includes separate alloys of Nb, Cu and Sn. The "bronze" process contains Nb in a copper–tin bronze matrix. With both processes the strand is typically drawn to final size and coiled into a solenoid or cable before heat treatment. It is only during heat treatment that the Sn reacts with the Nb to form the brittle, superconducting niobium–tin compound.[3] The powder-in-tube process is also used.[2][4]
The high field section of modern NMR magnets are composed of niobium–tin wire.
Strain effectsEdit
Inside a magnet the wires are subjected to high Lorentz forces as well as thermal stresses during cooling. Any strain in the niobium tin causes a decrease in the superconducting performance of the material, and can cause the brittle material to fracture. Because of this, the wires need to be as stiff as possible. The Young’s modulus of niobium tin is around 140 GPa at room temperature. However, the stiffness drops down to as low as 50 GPa when the material is cooled below 50 K (−223.2 °C; −369.7 °F).[5] Engineers must therefore find ways of improving the strength of the material. Strengthening fibers are often incorporated in the composite niobium tin wires to increase their stiffness. Common strengthening materials include Inconel, stainless steel, molybdenum, and tantalum because of their high stiffness at cryogenic temperatures.[6] Since the thermal expansion coefficients of the matrix, fiber, and niobium tin are all different, significant amounts of strain can be generated after the wire is annealed and cooled all the way down to operating temperatures. This strain is referred to as the pre-strain in the wire. Since any strain in the niobium tin generally decreases the superconducting performance of the material, a proper combination of materials must be used to minimize this value. The pre-strain in a composite wire can be calculated by the formula
{\displaystyle \varepsilon _{m}={\frac {V_{c}E_{c}\{{\frac {\Delta L}{L_{c}}}-{\frac {\Delta L}{L_{f}}}\}-\sigma _{cu,y}V_{cu}-\sigma _{bz,y}V_{bz}}{V_{f}E_{f}+V_{c}E_{c}}}.}
where εm is the pre-strain, ΔL/Lc and ΔL/Lf are changes in length due to thermal expansion of the niobium tin conduit and strengthening fiber respectively; Vc, Vf, Vcu, and Vbz are the volume fractions of conduit, fiber, copper, and bronze; σcu,y, and σbz,y are the yield stresses of copper and bronze; and Ec, and Ef are the Young’s modulus of the conduit and the fiber.[7] Since the copper and bronze matrix deforms plastically during cooldown, they apply a constant stress equal to their yield stress. The conduit and fiber, however, deform elastically by design. Commercial superconductors manufactured by the bronze process generally have a pre-strain value around 0.2% to 0.4%. The so-called strain effect causes a reduction in the superconducting properties of many materials including niobium tin. The critical strain, the maximum allowable strain over which superconductivity is lost, is given by the formula
{\displaystyle \varepsilon _{c}=\varepsilon _{co}\{1-{\frac {B}{B_{c2m}}}\}.}
where εc is the critical strain, εco is a material dependent parameter equal to 1.5% in tension (−1.8% in compression) for niobium tin, B is the applied magnetic field, and Bc2m is the maximum [Type-II superconductor|upper critical field] of the material.[8] Strain in the niobium tin causes tetragonal distortions in the crystal lattice, which changes the electron-phonon interaction spectrum. This is equivalent to an increase in disorder in the A15 crystal structure.[9] At high enough strain, around 1%, the niobium tin conduit will develop fractures and the current carrying capability of the wire will be irreversibly damaged. In most circumstances, except for high field conditions, the niobium tin conduit will fracture before the critical strain is reached.
Nb3Sn was discovered to be a superconductor in 1954, one year after the discovery of V3Si, the first example of an A3B superconductor.[10] In 1961 it was discovered that niobium–tin still exhibits superconductivity at large currents and strong magnetic fields, thus becoming the first known material to support the high currents and fields necessary for making useful high-power magnets and electric power machinery.[11][12]
Nb3Sn wire from the ITER fusion reactor, which is currently under construction.
The central solenoid and toroidal field superconducting magnets for the planned experimental ITER fusion reactor use niobium–tin as a superconductor.[13] The central solenoid coil will produce a field of 13.5 teslas (135,000 G). The toroidal field coils will operate at a maximum field of 11.8 T. Estimated use is 600 metric tons (590 long tons) of Nb3Sn strands and 250 metric tonnes of NbTi strands.[14][15]
At the Large Hadron Collider at CERN, extra-strong quadrupole magnets (for focussing beams) made with niobium–tin are being installed in key points of the accelerator between late 2018 and early 2020.[16] Niobium tin had been proposed in 1986 as an alternative to niobium–titanium, since it allowed coolants less complex than superfluid helium,[clarification needed][citation needed] but this was not pursued in order to avoid delays while competing with the then-planned US-led Superconducting Super Collider.
Developments and future usesEdit
Hafnium or zirconium added to niobium–tin increases the maximum current density in a magnetic field. This may allow it to be used at 16 tesla for CERN’s planned Future Circular Collider.[17]
Niobium–titanium, more ductile than Nb-Sn
^ Godeke, A.; Cheng, D.; Dietderich, D. R.; Ferracin, P.; Prestemon, S. O.; Sa Bbi, G.; Scanlan, R. M. (September 1, 2006). "Limits of NbTi and Nb3Sn, and Development of W&R Bi–2212 High Field Accelerator Magnets". Office of Science, High Energy Physics, U.S. Department of Energy. Retrieved December 26, 2015. {{cite journal}}: Cite journal requires |journal= (help)
^ a b "Record current with powder-in-tube superconductor". laboratorytalk.com. Archived from the original on October 8, 2008. Retrieved September 6, 2008.
^ Scanlan, R.; Greene, A. F.; Suenaga, M. (May 1986). Survey Of High Field Superconducting Material For Accelerator Magnets. 1986 ICFA Workshop on Superconducting Magnets and Cryogenics. Upton, NY. Report LBL-21549.
^ Lindenhovius, J. L. H.; Hornsveld, E. M.; den Ouden, A.; Wessel, W. A. J.; ten Kate, H. H. J. (2000). "Powder-in-tube (PIT) Nb3Sn conductors for high-field magnets". IEEE Transactions on Applied Superconductivity. 10 (1): 975–978. Bibcode:2000ITAS...10..975L. doi:10.1109/77.828394. S2CID 26260700.
^ Bussiére, J. F.; LeHuy, H.; Faucher, B. (1984). Clark, A. F; Reed, R. P (eds.). "ELASTIC BEHAVIOR OF POLYCRYSTALLINE Nb3Sn, V3Ga AND Nb3Ge". Advances in Cryogenic Engineering Materials. Springer, Boston, MA. 30: 859–866. doi:10.1007/978-1-4613-9868-4. ISBN 978-1-4613-9870-7. Retrieved May 20, 2020.
^ Flükiger, R.; Drost, E.; Specking, W. (1984). Clark, A. F; Reed, R. P (eds.). "EFFECT OF INTERNAL REINFORCEMENT ON THE CRITICAL CURRENT DENSITY OF Nb 3 Sn WIRES". Advances in Cryogenic Engineering Materials. Springer, Boston, MA. 30: 875–882. doi:10.1007/978-1-4613-9868-4. ISBN 978-1-4613-9870-7. Retrieved May 20, 2020.
^ Steeves, M. M.; Hoenig, M. O.; Cyders, C.J. (1984). Clark, A. F; Reed, R. P (eds.). "EFFECTS OF INCOLOY 903 AND TANTALUM CONDUITS ON CRITICAL CURRENT IN Nb3Sn CABLE-IN-CONDUIT CONDUCTORS". Advances in Cryogenic Engineering Materials. Springer, Boston, MA. 30: 883–890. doi:10.1007/978-1-4613-9868-4. ISBN 978-1-4613-9870-7. Retrieved May 20, 2020.
^ Ekin, J. W. (1984). Clark, A. F; Reed, R. P (eds.). "STRAIN EFFECTS IN SUPERCONDUCTING COMPOUNDS". Advances in Cryogenic Engineering Materials. Springer, Boston, MA. 30: 823–836. doi:10.1007/978-1-4613-9868-4. ISBN 978-1-4613-9870-7. Retrieved May 20, 2020.
^ Godeke, A. (2008). "A Review of the Properties of Nb 3 Sn and Their Variation with A15 Composition, Morphology and Strain State". Superconductor Science and Technology. IOP Publishing Ltd. 19 (8): 68–80.
^ Matthias, B. T.; Geballe, T. H.; Geller, S.; Corenzwit, E. (1954). "Superconductivity of Nb3Sn". Physical Review. 95 (6): 1435. Bibcode:1954PhRv...95.1435M. doi:10.1103/PhysRev.95.1435.
^ Geballe, Theodore H. (1993). "Superconductivity: From Physics to Technology". Physics Today. 46 (10): 52–56. Bibcode:1993PhT....46j..52G. doi:10.1063/1.881384.
^ Godeke, A. (2006). "A review of the properties of Nb3Sn and their variation with A15 composition, morphology and strain state". Supercond. Sci. Technol. 19 (8): R68–R80. arXiv:cond-mat/0606303. Bibcode:2006SuScT..19R..68G. doi:10.1088/0953-2048/19/8/R02. S2CID 73655040.
^ "Results of the first tests on the ITER toroidal magnet conductor". Commissariat à l'Énergie Atomique. September 10, 2001. Retrieved September 6, 2008.
^ Grunblatt, G.; Mocaer, P.; Verwaerde, Ch.; Kohler, C. (2005). "A success story: LHC cable production at ALSTOM-MSA". Fusion Engineering and Design (Proceedings of the 23rd Symposium of Fusion Technology). 75–79: 1–5. doi:10.1016/j.fusengdes.2005.06.216. S2CID 41810761.
^ "Alstom and Oxford Instruments Team Up to Offer Niobium–tin Superconducting Strand". Alstrom. June 27, 2007. Retrieved September 6, 2008.
^ Rossi, Lucio (October 25, 2011). "Superconductivity and the LHC: the early days". CERN Courier. CERN. Retrieved December 10, 2013.
^ MagLab awarded $1.5M by U.S. Department of Energy to develop better superconductors July 2020
Wikimedia Commons has media related to Niobium-tin.
European Advanced Superconductors
Retrieved from "https://en.wikipedia.org/w/index.php?title=Niobium–tin&oldid=1086702726"
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Moving Charges and Magnetism (17 Nov) - Live Session - NEET 2020
Moving Charges and Magnetism (17 Nov) - Live Session - NEET 2020 Contact Number: 9667591930 / 8527521718
A circular coil of radius R carries an electric current. The magnetic field due to the coil at a point on the axis of the coil located at a distance r from the centre of the coil, such that r>>R varies as
1. 1/r
2. 1/r3/2
3. 1/r2
The magnetic field due to a straight conductor of uniform cross-section of radius a and carrying a steady current is represented by :
Two parallel beams of positrons moving in the same direction will :
1. repel each other
2. will not interact with each other
3. attract each other
4. be deflected normal to the plane containing the two beams
A proton and an
\mathrm{\alpha }
-particle, moving with the same velocity, enter a uniform magnetic field, acting normal to the plane of their motion. The ratio of the radii of the circular paths described by the proton and
\mathrm{\alpha }
-particle is :
Circular loop of a wire and a long straight wire carry currents Ic and Ie, respectively as shown in figure. Assuming that these are placed in the same plane, the magnetic fields will be zero at the centre of the loop when the separation H is :
\frac{{\mathrm{I}}_{\mathrm{e}}\mathrm{R}}{{\mathrm{I}}_{\mathrm{c}}\mathrm{\pi }}
\frac{{\mathrm{I}}_{c}\mathrm{R}}{{\mathrm{I}}_{e}\mathrm{\pi }}
\frac{{\mathrm{\pi I}}_{\mathrm{c}}}{{\mathrm{I}}_{\mathrm{e}}\mathrm{R}}
\frac{{\mathrm{I}}_{\mathrm{e}}\mathrm{\pi }}{{\mathrm{I}}_{\mathrm{c}}\mathrm{R}}
What si the magnetic field at a distance R from a coil radius r carrying current I?
\frac{{\mathrm{\mu }}_{0}{\mathrm{IR}}^{2}}{2{\left(\mathrm{R}}^{}}
2 + r2)32
\frac{{\mathrm{\mu }}_{0}{\mathrm{Ir}}^{2}}{2{\left(\mathrm{R}}^{}}
\frac{{\mathrm{\mu }}_{0}\mathrm{I}}{2\mathrm{r}}
\frac{{\mathrm{\mu }}_{0}\mathrm{l}}{2\mathrm{R}}
A long straight wire of radius
\mathrm{\alpha }
carries a steady current i. The current is uniformaly distributed across its cross section. The ratio of the magnetic field at a/2 and 2a is
In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential V and then made to decribe semicircular paths of radius R using a magnetic field B. If V and B are kept constant, the ratio
\left(\frac{\mathrm{charge} \mathrm{on} \mathrm{the} \mathrm{ion}}{\mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{ion}}\right)
will be proportional to
\frac{1}{\mathrm{R}}
\frac{1}{{\mathrm{R}}^{2}}
{\mathrm{R}}^{2}
Two concentric coils each of radius equal to 2
\mathrm{\pi }
cm are placed at right angles to each other. 3 ampere and 4 ampere are the currents flowing in each coil respectively. The magnetic induction in Weber/
{\mathrm{m}}^{2}
at the centre of the coils will be
\left({\mathrm{\mu }}_{0} = 4\mathrm{\pi } \mathrm{x} {10}^{-7} \mathrm{Wb}/\mathrm{A}.\mathrm{m}\right)
{10}^{-5}
12 \mathrm{x} {10}^{-5}
7 \mathrm{x} {10}^{-5}
5 \mathrm{x} {10}^{-5}
The magnetic field due to a square loop of side a carrying a current I at its centre is
\frac{{\mathrm{\mu }}_{0}\mathrm{i}}{2\mathrm{a}}
\frac{{\mathrm{\mu }}_{0}\mathrm{i}}{\sqrt{2}\mathrm{\pi a}}
\frac{{\mathrm{\mu }}_{0}\mathrm{i}}{2\mathrm{\pi a}}
\sqrt{2}\frac{{\mathrm{\mu }}_{0}\mathrm{i}}{\mathrm{\pi a}}
Electron of mass m and charge q is travelling with a speed along a circular path of radius r at right angles to a uniform magnetic field of intensity B. If the speed of the electron is doubled anf the magnetic field is halved the resulting path would have a radius
\frac{\mathrm{r}}{4}
\frac{\mathrm{r}}{2}
Electron moves at right angles to a magnetic field of 1.5 x 10
{}^{-2}
tesla with speed of 6 x 10
{}^{7}
m/s. If the specific charge of the electron is 1.7 x 10
{}^{11}
C/kg. The radius of circular path will be
An electron beam passes through a magnetic field of 2 x 10
{}^{-3}
Wb/m
{}^{2}
and an electric field of 1.0 x 10
{}^{4}
V/m both acting simultaneously. The path of electron remains undeviated. The speed of electron if the electric field is removed, and the radius of electron path will be respectively
1. 10 x 10
{}^{6}
m/s, 2.43 cm
2. 2.5 x 10
{}^{6}
3. 5 x 10
{}^{6}
A charged particle is released from rest in a region of uniform and magnetic fields which are parallel to each other. The particle will move on a:
Four wires, each of length 2.0 m, are bent into four loops P, Q, R and S and then suspended in a uniform magnetic field. if the same current is passed each other, then the torque will be maximum on the loop
A square coil of side a carries a current I. The magnetic field at the centre of the coil is
\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{I}}{\mathrm{a\pi }}
\frac{\sqrt{2}{\mathrm{\mu }}_{0}\mathrm{I}}{\mathrm{a\pi }}
\frac{{\mathrm{\mu }}_{0}\mathrm{I}}{\sqrt{2}\mathrm{a\pi }}
\frac{2\sqrt{2}{\mathrm{\mu }}_{0}\mathrm{I}}{\mathrm{a\pi }}
A charged particle moves through a magnetic field in a direction perpendicular to it. Then the
1. velocity remains unchanged
2. speed of the particle remains unchanged
3. directoin of the particle remains unchanged
4. acceleration remains unchanged
Wires 1 and 2 carrying currents
{\mathrm{i}}_{1}
{\mathrm{i}}_{2}
respectively are inclined at an angle
\mathrm{\theta }
to each other. What is the force on a small element d/ of wire 2 at a distance of r wire 1 (as shown in figure) due to the magnetic field of wire 1?
\frac{{\mathrm{\mu }}_{0}}{2\mathrm{\pi r}}{i}_{1}{i}_{2} dl \mathrm{tan} \theta
\frac{{\mathrm{\mu }}_{0}}{2\mathrm{\pi r}}{i}_{1}{i}_{2} dl \mathrm{sin}\theta
\frac{{\mathrm{\mu }}_{0}}{2\mathrm{\pi r}}{i}_{1}{i}_{2} dl \mathrm{cos} \theta
\frac{{\mathrm{\mu }}_{0}}{4\mathrm{\pi r}}{i}_{1}{i}_{2} dl \mathrm{sin} \theta
If we double the radius of a coil keeping the current through it unchanged, then the magnetic field at any point at a large distance from the centre becomes approximately
A portion of a conductive wire is bent in the form of a semicircle of radius r as shown nelow in fig. At the centre of semicircle, the magnetic induction will be
\frac{{\mathrm{\mu }}_{0}}{4 \mathrm{\pi }}.\frac{\mathrm{\pi i}}{r}gauss
\frac{{\mathrm{\mu }}_{0}}{4 \mathrm{\pi }}.\frac{\mathrm{\pi i}}{r}\mathrm{tesla}
A coil of circular cross-section having 1000 turns and 4 cm2 face area is placed with its axis parallel to a magnetic field which decreases by 10-2 Wb m-2 in 0.01 s. The emf induced in the coil is :
3. 4 mV
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EUDML | Heegner Points and Derivatives of L-Series. II. EuDML | Heegner Points and Derivatives of L-Series. II.
Heegner Points and Derivatives of L-Series. II.
D. Zagier; W. Kohnen; B. Gross
Zagier, D., Kohnen, W., and Gross, B.. "Heegner Points and Derivatives of L-Series. II.." Mathematische Annalen 278 (1987): 497-562. <http://eudml.org/doc/164302>.
@article{Zagier1987,
author = {Zagier, D., Kohnen, W., Gross, B.},
keywords = {derivatives of -series; modular curve; Jacobian; Heegner divisor; conjectures of Birch and Swinnerton-Dyer},
title = {Heegner Points and Derivatives of L-Series. II.},
AU - Zagier, D.
AU - Kohnen, W.
TI - Heegner Points and Derivatives of L-Series. II.
KW - derivatives of -series; modular curve; Jacobian; Heegner divisor; conjectures of Birch and Swinnerton-Dyer
E. Kowalski, P. Michel, A lower bound for the rank of
{J}_{0}\left(q\right)
Winfried Kohnen, Estimates for Fourier coefficients of Siegel cusp forms of degree two
Nils-Peter Skoruppa, Heegner cycles, modular forms and jacobi forms
derivatives of
L
-series, modular curve, Jacobian, Heegner divisor, conjectures of Birch and Swinnerton-Dyer
L
Articles by D. Zagier
Articles by W. Kohnen
Articles by B. Gross
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Augmented Dickey–Fuller test - Wikipedia
Time series statistical test
In statistics and econometrics, an augmented Dickey–Fuller test (ADF) tests the null hypothesis that a unit root is present in a time series sample. The alternative hypothesis is different depending on which version of the test is used, but is usually stationarity or trend-stationarity. It is an augmented version of the Dickey–Fuller test for a larger and more complicated set of time series models.
The augmented Dickey–Fuller (ADF) statistic, used in the test, is a negative number. The more negative it is, the stronger the rejection of the hypothesis that there is a unit root at some level of confidence.[1]
The testing procedure for the ADF test is the same as for the Dickey–Fuller test but it is applied to the model
{\displaystyle \Delta y_{t}=\alpha +\beta t+\gamma y_{t-1}+\delta _{1}\Delta y_{t-1}+\cdots +\delta _{p-1}\Delta y_{t-p+1}+\varepsilon _{t},}
{\displaystyle \alpha }
{\displaystyle \beta }
the coefficient on a time trend and
{\displaystyle p}
the lag order of the autoregressive process. Imposing the constraints
{\displaystyle \alpha =0}
{\displaystyle \beta =0}
corresponds to modelling a random walk and using the constraint
{\displaystyle \beta =0}
corresponds to modeling a random walk with a drift. Consequently, there are three main versions of the test, analogous to the ones discussed on Dickey–Fuller test (see that page for a discussion on dealing with uncertainty about including the intercept and deterministic time trend terms in the test equation.)
By including lags of the order p the ADF formulation allows for higher-order autoregressive processes. This means that the lag length p has to be determined when applying the test. One possible approach is to test down from high orders and examine the t-values on coefficients. An alternative approach is to examine information criteria such as the Akaike information criterion, Bayesian information criterion or the Hannan–Quinn information criterion.
The unit root test is then carried out under the null hypothesis
{\displaystyle \gamma =0}
against the alternative hypothesis of
{\displaystyle \gamma <0.}
Once a value for the test statistic
{\displaystyle \mathrm {DF} _{\tau }={\frac {\hat {\gamma }}{\operatorname {SE} ({\hat {\gamma }})}}}
is computed it can be compared to the relevant critical value for the Dickey–Fuller test. As this test is asymmetrical, we are only concerned with negative values of our test statistic
{\displaystyle \mathrm {DF} _{\tau }}
. If the calculated test statistic is less (more negative) than the critical value, then the null hypothesis of
{\displaystyle \gamma =0}
is rejected and no unit root is present.
The intuition behind the test is that if the series is characterised by a unit root process then the lagged level of the series (
{\displaystyle y_{t-1}}
) will provide no relevant information in predicting the change in
{\displaystyle y_{t}}
besides the one obtained in the lagged changes (
{\displaystyle \Delta y_{t-k}}
). In this case the
{\displaystyle \gamma =0}
and null hypothesis is not rejected. In contrast, when the process has no unit root, it is stationary and hence exhibits reversion to the mean - so the lagged level will provide relevant information in predicting the change of the series and the null of a unit root will be rejected.
A model that includes a constant and a time trend is estimated using sample of 50 observations and yields the
{\displaystyle \mathrm {DF} _{\tau }}
statistic of −4.57. This is more negative than the tabulated critical value of −3.50, so at the 95 percent level the null hypothesis of a unit root will be rejected.
Critical values for Dickey–Fuller t-distribution.
Without trend With trend
Sample size 1% 5% 1% 5%
T = 25 −3.75 −3.00 −4.38 −3.60
T = 100 −3.51 −2.89 −4.04 −3.45
T = ∞ −3.43 −2.86 −3.96 −3.41
Source[2]: 373
There are alternative unit root tests such as the Phillips–Perron test (PP) or the ADF-GLS test procedure (ERS) developed by Elliott, Rothenberg and Stock (1996).[3]
In R, there are various packages supplying implementations of the test. The forecast package includes a ndiffs function (which handles multiple popular unit root tests),[4] the tseries package includes an adf.test function[5] and the fUnitRoots package includes an adfTest function.[6] A further implementation is supplied by the "urca" package.[7]
Gretl includes the Augmented Dickey–Fuller test.[8]
In Matlab, the adfTest function [9] is part of the Econometrics Toolbox,[10] and a free version is available as part of the 'Spatial Econometrics' toolbox[11]
In SAS, PROC ARIMA can perform ADF tests.[12]
In Stata, the dfuller command is used for ADF tests.[13]
In EViews, the Augmented Dickey-Fuller is available under "Unit Root Test."[14][15][16][17]
In Python, the adfuller function is available in the Statsmodels package[18] and the ARCH package[19] also provides an Augmented Dickey–Fuller test.
In Java, the AugmentedDickeyFuller class is included in SuanShu[20] available under the com.numericalmethod.suanshu.stats.test.timeseries.adf package.
In Julia, the ADFTest function is available in the HypothesisTests package.[21]
Kwiatkowski–Phillips–Schmidt–Shin (KPSS) test
^ Fuller, W. A. (1976). Introduction to Statistical Time Series. New York: John Wiley and Sons. ISBN 0-471-28715-6.
^ Elliott, G.; Rothenberg, T. J.; Stock, J. H. (1996). "Efficient Tests for an Autoregressive Unit Root". Econometrica. 64 (4): 813–836. doi:10.2307/2171846. JSTOR 2171846. S2CID 122699512.
^ "ndiffs {forecast} | inside-R | A Community Site for R". Inside-r.org. Archived from the original on 2016-07-17. Retrieved 2020-02-23.
^ "R: Augmented Dickey-Fuller Test". Finzi.psych.upenn.edu. Retrieved 2016-06-26.
^ "Comparing ADF Test Functions in R · Fabian Kostadinov". fabian-kostadinov.github.io. Retrieved 2016-06-05.
^ https://cran.r-project.org/web/packages/urca/urca.pdf[bare URL PDF]
^ "Introduction to gretl and the gretl instructional lab" (PDF). Spot.colorado.edu. Retrieved 2016-06-26.
^ "Augmented Dickey-Fuller test - MATLAB adftest". Mathworks.com. Retrieved 2016-06-26.
^ "Econometrics Toolbox - MATLAB". Mathworks.com. Retrieved 2016-06-26.
^ "Econometrics Toolbox for MATLAB". Spatial-econometrics.com. Retrieved 2016-06-26.
^ David A. Dickey. "Stationarity Issues in Time Series Models" (PDF). 2.sas.com. Retrieved 2016-06-26.
^ "Augmented Dickey–Fuller unit-root test" (PDF). Stata.com. Retrieved 2016-06-26.
^ "Memento on EViews Output" (PDF). Retrieved 17 June 2019.
^ "EViews.com • View topic - Dickey Fuller for Multiple Regression Models". Forums.eviews.com. Retrieved 2016-06-26.
^ "Augmented Dickey-Fuller Unit Root Tests" (PDF). Faculty.smu.edu. Retrieved 2016-06-26.
^ "DickeyFuller Unit Root Test". Hkbu.edu.hk. Retrieved 2016-06-26.
^ "statsmodels.tsa.stattools.adfuller — statsmodels 0.7.0 documentation". Statsmodels.sourceforge.net. Retrieved 2016-06-26.
^ "Unit Root Testing — arch 4.19+14.g318309ac documentation". arch.readthedocs.io. Retrieved 2021-10-18.
^ "SuanShu | Numerical Method Inc". Numericalmethod.com. Archived from the original on 2015-08-15. Retrieved 2016-06-26.
^ "Time series tests". juliastats.org. Retrieved 2020-02-04.
Greene, W. H. (2002). Econometric Analysis (Fifth ed.). New Jersey: Prentice Hall. ISBN 0-13-066189-9. [page needed]
Said, S. E.; Dickey, D. A. (1984). "Testing for Unit Roots in Autoregressive-Moving Average Models of Unknown Order". Biometrika. 71 (3): 599–607. doi:10.1093/biomet/71.3.599.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Augmented_Dickey–Fuller_test&oldid=1080037958"
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Class 10 (All in one) – Quadratic Equations
Class 10 (All in one) – Quadratic EquationsContact Number: 9667591930 / 8527521718
Solve the equation by factorization.
21x2 – 2x +
\frac{1}{21}
3{x}^{2}+2\sqrt{5}x-5=0
{x}^{2}-\left(\sqrt{2}+1\right)x+\sqrt{2}=0
\sqrt{3}{x}^{2}+11x+6\sqrt{3}=0
Solve the quadratic equation by method of factorization.
5x+\frac{1}{x}=6;\text{\hspace{0.17em}\hspace{0.17em}}x\ne 0
\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6};\text{\hspace{0.17em}\hspace{0.17em}}x\ne 3,-5
\frac{21}{{x}^{2}}-\frac{29}{x}-10=0;\text{\hspace{0.17em}\hspace{0.17em}}x\ne 0
Using factorization method, solve the quadratic equation.
\frac{x+3}{x+2}=\frac{3x-7}{2x-3}
2\left(\frac{x+2}{2x-3}\right)-9\left(\frac{2x-3}{x+2}\right)=3
Write the discriminant of the following
\sqrt{3}{x}^{2}-2\sqrt{2}x-2\sqrt{3}=0
(iii) 2x2 + 3x + m = 0
(iv) x2 + x + 7 = 0
(v) (4x – 3)2 + 20x = 11
(vi) a2x2 + 8abx + 4b2 = 0
Determine whether the roots of the quadratic equation are real or not.
3\sqrt{3}{x}^{2}+10x+\sqrt{3}=0
Find the value of K for which following equation has real and equal roots
K{x}^{2}-2\sqrt{5}x+4=0
{x}^{2}-2x\left(1+3K\right)+7\left(3+2K\right)=0
Determine the value(s) of m for which the equation x2 + m(4x + m – 1) + 2 = 0 has real roots.
For what value(s) of n, the equation (n + 3)x2 – (5 – n)x + 1 = 0 has coincident roots?
Find the roots of the following quadratic equation by applying the quadratic formula.
abx2 + (b2 – ac)x – bc = 0, a, b ≠ 0
If two numbers differ by 2 and their product is 360, then find the numbers.
If the product of two consecutive natural numbers is 210, then determine the numbers.
The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is
\frac{29}{10}
. Find the fraction.
The product of Ram’s age 5 year back and 9 year hence (in year) is 15. Find the present age of Ram.
Divide 29 into two parts, so that the sum of the squares of the parts is 425.
If the product of two consecutive odd numbers is 143, then find the numbers.
Some students planned a picnic. The total budget for food was ₹2000. But 5 students failed to attend the picnic and thus the cost of food for each member increased by ₹20. How many students attended the picnic and how much did each student pay for the food?
A person on tour has ₹360 for his expenses. If he extends his tour for 4 days, then he has to cut down his daily expenses by ₹3. Find the original duration of the tour.
A rectangular field is 20 m long and 14 m wide. There is a path of equal width all around it, having an area of 111 sq. m. Find the width of the path.
In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/h and the time of flight increased by 30 min. Find the duration of the flight.
The area of a right angled triangle is 480 cm2. If the base of the triangle is 8 cm more than twice the height (altitude) of the triangle, then find the sides of the triangle.
A two-digit number is such that the product of its digit is 35. When 18 is added to the number, the digits interchange their places. Find the number.
Twenty seven years hence Sanjay’s age will be square of what it was 29 years ago. Find his present age.
If the discriminant of the equation 6x2 – bx + 2 = 0 is 1, then find the value of b.
Check whether the following statement is true or false. Justify your answer.
“Every quadratic equation has atleast one real root.”
If the product of two consecutive integers is 306, then write the quadratic representation of this situation.
If a number is added to twice its square, then the resultant is 21. Write the quadratic representation of this situation.
Which constant must be added and subtracted to solve the quadratic equation
9{x}^{2}+\frac{3}{4}x-\sqrt{2}=0
by the method of completing the square?
A quadratic equation with integral coefficients has integral roots. Justify your answer.
x:\sqrt{6x+7}-\left(2x-7\right)=0
Find the roots of the equation ax2 + a = a2x + x
Show that (x2 + l)2 – x2 = 0 has no real roots.
Find the roots of the quadratic equation a2b2x2 + b2x – a2x – 1 = 0
\sqrt{2x+9}+x=13
Find the numerical difference of the roots of equation x2 – 7x – 18 = 0
\frac{1}{2}
{x}^{2}+kx-\frac{5}{4}=0,
In a cricket match. Harbhajan took three wickets less than twice the number of wickets taken by Zaheer. The product of the numbers of wickets taken by these two is 20. Represent the above situation in the form of a quadratic equation.
Find the quadratic equation, if
x=\sqrt{5+\sqrt{5+\sqrt{5+...\infty }}}
and x is a natural number.
x: \frac{16}{x}-1=\frac{15}{x+1}; x\ne 0,-1
Find the value of p, when
p{x}^{2}+\left(\sqrt{3}-\sqrt{2}\right)x-1=0\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}x=\frac{1}{\sqrt{3}}
is one root of this equation.
The difference of two numbers is 4. If the difference of their reciprocals is
\frac{4}{21},
then find the two numbers.
The sum of two number is 11 and the sum of their reciprocals is
\frac{11}{28}.
Find the numbers.
At t min past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 min less than
\frac{{t}^{2}}{4}
min. Find t.
Seven years ago, Varun's age was five times the square of Swati's age. Three years hence, Swati's age will be two-fifth of Varun's age. Find their present ages.
Find the roots of the equation a2x2 – 3abx + 2b2 = 0 by the method of completing the square.
The sum of a number and its positive square root is
\frac{6}{25}.
Find the number.
The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by
\frac{1}{15}.
Find the fraction.
Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of other two by 60. Find the numbers.
x: \frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x},
where a, b, x ≠ 0 and a + b + x ≠ 0
If the roots of the equation x2 + 2cx + ab = 0 are real and unequal, then prove that the equation x2 – 2(a + b)x + a2 + b2 + 2c2 = 0 has no real roots.
\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}=\frac{2}{3};\text{\hspace{0.17em}\hspace{0.17em}}x\ne 1,\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}3
Find the roots of 5(x + 1) + 5(2 – x) = 53 + 1 by factorisation method.
If –5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, then find the value of k.
A piece of cloth costs ₹ 200. If the piece was 5 m longer and each metre of cloth costs ₹ 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per metre?
₹ 6500 were divided equally among a certain number of persons. If there had been 15 more persons, each would have got ₹ 30 less. Find the original number of persons.
A shopkeeper buys a number of books for ₹ 1200. If he had bought 10 more books for the same amount, each book would have cost him ₹ 20 less. How many books did he buy?
If the equation (1 + m2)x2 + (2mc)x + (c2 – a2) = 0 has equal roots, then prove that c2 = a2 (1 + m2).
A factory kept increasing its output by the same percentage every year. Find the percentage, if it is known that the output doubles in the last two years.
x=\frac{1}{2-\frac{1}{2-\frac{1}{2-x}}};x\ne 2.
If the roots of the equation (a – b)x2 + (b – c)x + (c – a) = 0 are equal, then prove that 2a = b + c.
There is a square field whose side is 44 m. A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. The total cost of laying the flower bed and gravelling the path at ₹ 2.75 and ₹ 1.50 per m2 respectively, is ₹ 4904. Find the width of gravel path.
Due to some technical problems, an aeroplane started late by one hour from its starting point. The pilot decided to increase the speed of the aeroplane by 100 km/h from its usual speed to cover a journey of 1200 km in same time.
(i) Find the usual speed of the aeroplane.
(ii) What values (qualities) of the pilot are represented in the question?
|
Conditionally defined expression or function - MATLAB piecewise - MathWorks Switzerland
\mathit{y}=\left\{\begin{array}{ll}-1& \mathit{x}<0\\ 1& \mathit{x}>0\end{array}
\left\{\begin{array}{cl}-1& \text{ if }x<0\\ 1& \text{ if }0<x\end{array}
\left(\begin{array}{ccc}-1& \mathrm{NaN}& 1\end{array}\right)
\mathit{y}\left(\mathit{x}\right)=\left\{\begin{array}{ll}-1& \mathit{x}<0\\ 1& \mathit{x}>0\end{array}
\left\{\begin{array}{cl}-1& \text{ if }x<0\\ 1& \text{ if }0<x\end{array}
\left(\begin{array}{ccc}-1& \mathrm{NaN}& 1\end{array}\right)
y=\left\{\begin{array}{cc}-2& x<-2\\ 0& -2<x<0\\ 1& otherwise.\end{array}
\left\{\begin{array}{cl}-2& \text{ if }x<-2\\ 0& \text{ if }x\in \left(-2,0\right)\\ 1& \mathrm{ otherwise}\end{array}
\left(\begin{array}{ccccc}-2& 1& 0& 1& 1\end{array}\right)
y=\left\{\begin{array}{cc}-2& x<-2\\ x& -2<x<2\\ 2& x>2\end{array}.
1
1
\mathit{y}=\left\{\begin{array}{ll}1/\mathit{x}& \mathit{x}<-1\\ \mathrm{sin}\left(\mathit{x}\right)/\mathit{x}& \mathit{x}\ge -1\end{array}
\left\{\begin{array}{cl}-\frac{1}{{x}^{2}}& \text{ if }x<-1\\ \frac{\mathrm{cos}\left(x\right)}{x}-\frac{\mathrm{sin}\left(x\right)}{{x}^{2}}& \text{ if }-1<x\end{array}
\left\{\begin{array}{cl}\mathrm{log}\left(x\right)& \text{ if }x<-1\\ \mathrm{sinint}\left(x\right)& \text{ if }-1\le x\end{array}
1
\mathrm{sin}\left(1\right)
-1
\left\{\begin{array}{cl}-3& \text{ if }x<-1\\ -1& \text{ if }x\in \left[-1,0\right)\\ 3& \text{ if }0\le x\end{array}
\left\{\begin{array}{cl}1& \text{ if }x<-1\\ 3& \text{ if }x\in \left[-1,0\right)\\ -1& \text{ if }0\le x\end{array}
\left\{\begin{array}{cl}2& \text{ if }x<-1\\ -2& \text{ if }x\in \left[-1,0\right)\\ 2& \text{ if }0\le x\end{array}
\left\{\begin{array}{cl}\frac{1}{2}& \text{ if }x<-1\\ -\frac{1}{2}& \text{ if }x\in \left[-1,0\right)\\ \frac{1}{2}& \text{ if }0\le x\end{array}
\left\{\begin{array}{cl}-1& \text{ if }x<0\\ 1& \text{ if }0<x\end{array}
\left\{\begin{array}{cl}\frac{1}{x}& \text{ if }5<x\\ -1& \text{ if }x<0\\ 1& \text{ if }0<x\end{array}
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he works on information processing in neural systems, information theoretic optimization in collective behavior, urban organization and dynamics, and the development of science and technology. Luis obtained his PhD from Imperial College, University of London for work on critical phenomena in the early Universe, and associated mathematical techniques of Statistical Physics, Field Theory and Non-linear Dynamics. He held postdoctoral positions at the University of Heidelberg, Germany, as a Director’s Fellow in the Theoretical Division at LANL, and at the Center for Theoretical Physics at MIT. In 2000 he was awarded the distinguished Slansky Fellowship at Los Alamos National Laboratory for excellence in interdisciplinary research. He has been a scientist at LANL since the spring of 2003, first at the Computer and Computational Sciences Division (CCS), and since September 2005 in the Theoretical Division (T-5: Mathematical Modeling and Analysis). He is also External Professor at the Santa Fe Institute.
[[Image:Ellison.jpg|200px|left]]
'''Christopher Ellison''', Graduate Student, Complexity Sciences Center, Physics Department, University of California at Davis
[[Image:dpf.jpg|200px|left]]
"Dominos, Ergodic Flows": We present a model, developed with Norman Packard, of
a simple discrete open flow system. Dimers are created at one edge
of a two-dimensional lattice, diffuse across, and are removed at the opposite side.
A steady-state flow is established, under various kinetic rules.
In the equilibrium case, the system reduces to the classical monomer-dimer tiling problem,
whose entropy as a function of density is known. This entropy density is reproduced
locally in the flow system, as shown by statistics over local templates. The goal is to
clarify informational aspects of a flowing pattern.
'''Susanne Still''', Assistant Professor of Computer Science, Department of Information and Computer Sciences, University of Hawaii at Manoa.
[[Image:Tononi.jpg|200px|left]]
'''Giulio Tononi''', Professor, School of Medicine, Department of Psychiatry, University of Wisconsin - Madison
[[Image:Trabesinger.jpg|200px|left]]
'''Andreas Trabesinger''', Senior Editor, Nature Physics
{\displaystyle Omega}
|
\mathrm{store}[=\mathrm{name}]
\mathrm{storeall}[=\mathrm{name}]
\mathrm{ezcriteria}=[\mathrm{quantity},\mathrm{size ratio}]
\mathrm{tosolve}=n
\mathrm{stl}='\mathrm{time}'
\mathrm{itl}='\mathrm{time}'
\mathrm{ranking}=[[\mathrm{ind weights}1,\mathrm{dep weights}1],[\mathrm{ind weights}2,\mathrm{dep weights}2],\mathrm{...}]
\mathrm{casesplit}=[\mathrm{<>},\mathrm{`=`},\mathrm{<>}]
\mathrm{mindim}=...
\mathrm{mindim}='n'
\mathrm{mindim}
\mathrm{pivselect}=\mathrm{choice}
\mathrm{nopiv}=[\mathrm{var1},...]
\mathrm{faclimit}=n
Typically the pivot chosen is the coefficient of the leading indeterminate in the equation. In the event that the leading indeterminate is itself a factor of the equation, and this same leading indeterminate factor occurs in
\mathrm{factoring}=\mathrm{desc}
\mathrm{desc}=\mathrm{none}
\mathrm{desc}=\mathrm{nolead}
\mathrm{desc}=\mathrm{all}
\mathrm{checkempty}
\mathrm{grobonly}
\mathrm{grob_rank}=[[1,\mathrm{deg},\mathrm{none}],[1,\mathrm{ilex}]]
|
Circular motion - Physics
Circular motion (12576 views - Physics)
In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with constant angular rate of rotation and constant speed, or non-uniform with a changing rate of rotation. The rotation around a fixed axis of a three-dimensional body involves circular motion of its parts. The equations of motion describe the movement of the center of mass of a body. Examples of circular motion include: an artificial satellite orbiting the Earth at a constant height, a ceiling fan's blades rotating around a hub, a stone which is tied to a rope and is being swung in circles, a car turning through a curve in a race track, an electron moving perpendicular to a uniform magnetic field, and a gear turning inside a mechanism. Since the object's velocity vector is constantly changing direction, the moving object is undergoing acceleration by a centripetal force in the direction of the center of rotation. Without this acceleration, the object would move in a straight line, according to Newton's laws of motion.
{\displaystyle {\vec {F}}=m{\vec {a}}}
Examples of circular motion include: an artificial satellite orbiting the Earth at a constant height, a ceiling fan's blades rotating around a hub, a stone which is tied to a rope and is being swung in circles, a car turning through a curve in a race track, an electron moving perpendicular to a uniform magnetic field, and a gear turning inside a mechanism.
1.1.1 In polar coordinates
1.1.2 Using complex numbers
1.1.4 Relativistic circular motion
2 Non-uniform
In the case of rotation around a fixed axis of a rigid body that is not negligibly small compared to the radius of the path, each particle of the body describes a uniform circular motion with the same angular velocity, but with velocity and acceleration varying with the position with respect to the axis.
{\displaystyle \omega ={\frac {2\pi }{T}}=2\pi f={\frac {d\theta }{dt}}\ }
and the units are radians/second
{\displaystyle v={\frac {2\pi r}{T}}=\omega r}
{\displaystyle \theta =2\pi {\frac {t}{T}}=\omega t\,}
{\displaystyle \alpha ={\frac {d\omega }{dt}}}
{\displaystyle a={\frac {v^{2}}{r}}=\omega ^{2}r}
{\displaystyle F_{c}={\dot {p}}\ {\overset {{\dot {m}}=0}{=}}\ ma={\frac {mv^{2}}{r}}}
{\displaystyle \mathbf {v} ={\boldsymbol {\omega }}\times \mathbf {r} \ ,}
{\displaystyle \mathbf {a} ={\boldsymbol {\omega }}\times \mathbf {v} ={\boldsymbol {\omega }}\times \left({\boldsymbol {\omega }}\times \mathbf {r} \right)\ ,}
During circular motion the body moves on a curve that can be described in polar coordinate system as a fixed distance R from the center of the orbit taken as origin, oriented at an angle θ(t) from some reference direction. See Figure 4. The displacement vector
{\displaystyle {\vec {r}}}
is the radial vector from the origin to the particle location:
{\displaystyle {\vec {r}}(t)=R{\hat {u}}_{R}(t)\ ,}
{\displaystyle {\hat {u}}_{R}(t)}
is the unit vector parallel to the radius vector at time t and pointing away from the origin. It is convenient to introduce the unit vector orthogonal to
{\displaystyle {\hat {u}}_{R}(t)}
as well, namely
{\displaystyle {\hat {u}}_{\theta }(t)}
. It is customary to orient
{\displaystyle {\hat {u}}_{\theta }(t)}
to point in the direction of travel along the orbit.
{\displaystyle {\vec {v}}(t)={\frac {d}{dt}}{\vec {r}}(t)={\frac {dR}{dt}}{\hat {u}}_{R}(t)+R{\frac {d{\hat {u}}_{R}}{dt}}\ .}
Because the radius of the circle is constant, the radial component of the velocity is zero. The unit vector
{\displaystyle {\hat {u}}_{R}(t)}
has a time-invariant magnitude of unity, so as time varies its tip always lies on a circle of unit radius, with an angle θ the same as the angle of
{\displaystyle {\vec {r}}(t)}
. If the particle displacement rotates through an angle dθ in time dt, so does
{\displaystyle {\hat {u}}_{R}(t)}
, describing an arc on the unit circle of magnitude dθ. See the unit circle at the left of Figure 4. Hence:
{\displaystyle {\frac {d{\hat {u}}_{R}}{dt}}={\frac {d\theta }{dt}}{\hat {u}}_{\theta }(t)\ ,}
where the direction of the change must be perpendicular to
{\displaystyle {\hat {u}}_{R}(t)}
(or, in other words, along
{\displaystyle {\hat {u}}_{\theta }(t)}
) because any change
{\displaystyle d{\hat {u}}_{R}(t)}
{\displaystyle {\hat {u}}_{R}(t)}
would change the size of
{\displaystyle {\hat {u}}_{R}(t)}
. The sign is positive, because an increase in dθ implies the object and
{\displaystyle {\hat {u}}_{R}(t)}
have moved in the direction of
{\displaystyle {\hat {u}}_{\theta }(t)}
. Hence the velocity becomes:
{\displaystyle {\vec {v}}(t)={\frac {d}{dt}}{\vec {r}}(t)=R{\frac {d{\hat {u}}_{R}}{dt}}=R{\frac {d\theta }{dt}}{\hat {u}}_{\theta }(t)=R\omega {\hat {u}}_{\theta }(t)\ .}
{\displaystyle {\begin{aligned}{\vec {a}}(t)&={\frac {d}{dt}}{\vec {v}}(t)={\frac {d}{dt}}\left(R\omega {\hat {u}}_{\theta }(t)\right)\\&=R\left({\frac {d\omega }{dt}}{\hat {u}}_{\theta }(t)+\omega {\frac {d{\hat {u}}_{\theta }}{dt}}\right)\ .\end{aligned}}}
{\displaystyle {\hat {u}}_{\theta }(t)}
is found the same way as for
{\displaystyle {\hat {u}}_{R}(t)}
{\displaystyle {\hat {u}}_{\theta }(t)}
is a unit vector and its tip traces a unit circle with an angle that is π/2 + θ. Hence, an increase in angle dθ by
{\displaystyle {\vec {r}}(t)}
{\displaystyle {\hat {u}}_{\theta }(t)}
traces an arc of magnitude dθ, and as
{\displaystyle {\hat {u}}_{\theta }(t)}
{\displaystyle {\hat {u}}_{R}(t)}
{\displaystyle {\frac {d{\hat {u}}_{\theta }}{dt}}=-{\frac {d\theta }{dt}}{\hat {u}}_{R}(t)=-\omega {\hat {u}}_{R}(t)\ ,}
where a negative sign is necessary to keep
{\displaystyle {\hat {u}}_{\theta }(t)}
{\displaystyle {\hat {u}}_{R}(t)}
. (Otherwise, the angle between
{\displaystyle {\hat {u}}_{\theta }(t)}
{\displaystyle {\hat {u}}_{R}(t)}
would decrease with increase in dθ.) See the unit circle at the left of Figure 4. Consequently, the acceleration is:
{\displaystyle {\begin{aligned}{\vec {a}}(t)&=R\left({\frac {d\omega }{dt}}{\hat {u}}_{\theta }(t)+\omega {\frac {d{\hat {u}}_{\theta }}{dt}}\right)\\&=R{\frac {d\omega }{dt}}{\hat {u}}_{\theta }(t)-\omega ^{2}R{\hat {u}}_{R}(t)\ .\end{aligned}}}
{\displaystyle {\vec {a}}_{R}(t)=-\omega ^{2}R{\hat {u}}_{R}(t)\ ,}
{\displaystyle {\vec {a}}_{\theta }(t)=R{\frac {d\omega }{dt}}{\hat {u}}_{\theta }(t)={\frac {dR\omega }{dt}}{\hat {u}}_{\theta }(t)={\frac {d\left|{\vec {v}}(t)\right|}{dt}}{\hat {u}}_{\theta }(t)\ .}
Circular motion can be described using complex numbers. Let the x axis be the real axis and the
{\displaystyle y}
axis be the imaginary axis. The position of the body can then be given as
{\displaystyle z}
, a complex "vector":
{\displaystyle z=x+iy=R(\cos[\theta (t)]+i\sin[\theta (t)])=Re^{i\theta (t)}\ ,}
where i is the imaginary unit, and
{\displaystyle \theta (t)}
is the argument of the complex number as a function of time, t.
{\displaystyle {\dot {R}}={\ddot {R}}=0\ ,}
{\displaystyle v={\dot {z}}={\frac {d\left(Re^{i\theta [t]}\right)}{dt}}=R{\frac {d\left(e^{i\theta [t]}\right)}{dt}}=Re^{i\theta (t)}{\frac {d(i\theta [t])}{dt}}=iR{\dot {\theta }}(t)e^{i\theta (t)}=i\omega Re^{i\theta (t)}=i\omega z}
{\displaystyle {\begin{aligned}a&={\dot {v}}=i{\dot {\omega }}z+i\omega {\dot {z}}=\left(i{\dot {\omega }}-\omega ^{2}\right)z\\&=\left(i{\dot {\omega }}-\omega ^{2}\right)Re^{i\theta (t)}\\&=-\omega ^{2}Re^{i\theta (t)}+{\dot {\omega }}e^{i{\frac {\pi }{2}}}Re^{i\theta (t)}\ .\end{aligned}}}
{\displaystyle v=r{\frac {d\theta }{dt}}=r\omega }
{\displaystyle {\vec {u}}\cdot {\vec {a}}=0.}
{\displaystyle \alpha ^{2}=\gamma ^{4}a^{2}+\gamma ^{6}({\vec {u}}\cdot {\vec {a}})^{2},}
{\displaystyle \alpha ^{2}=\gamma ^{4}a^{2}.}
{\displaystyle \alpha =\gamma ^{2}{\frac {v^{2}}{r}}.}
{\displaystyle a=v{\frac {d\theta }{dt}}=v\omega ={\frac {v^{2}}{r}}}
In non-uniform circular motion, normal force and weight may point in the same direction. Both forces can point down, yet the object will remain in a circular path without falling straight down. First let's see why normal force can point down in the first place. In the first diagram, let's say the object is a person sitting inside a plane, the two forces point down only when it reaches the top of the circle. The reason for this is that the normal force is the sum of the tangential force and centripetal force. The tangential force is zero at the top (as no work is performed when the motion is perpendicular to the direction of force applied. Here weight force is perpendicular to the direction of motion of the object at the top of the circle) and centripetal force points down, thus normal force will point down as well. From a logical standpoint, a person who is travelling in the plane will be upside down at the top of the circle. At that moment, the person's seat is actually pushing down on the person, which is the normal force.
The reason why the object does not fall down when subjected to only downward forces is a simple one. Think about what keeps an object up after it is thrown. Once an object is thrown into the air, there is only the downward force of earth's gravity that acts on the object. That does not mean that once an object is thrown in the air, it will fall instantly. What keeps that object up in the air is its velocity. The first of Newton's laws of motion states that an object's inertia keeps it in motion, and since the object in the air has a velocity, it will tend to keep moving in that direction.
{\displaystyle {\begin{aligned}F_{net}&=ma\,\\F_{net}&=ma_{r}\,\\F_{net}&={\frac {mv^{2}}{r}}\,\\F_{net}&=F_{c}\,\end{aligned}}}
{\displaystyle F_{net}=F_{c}\,}
, we can draw free body diagrams to list all the forces acting on an object then set it equal to
{\displaystyle F_{c}\,}
. Afterwards, we can solve for what ever is unknown (this can be mass, velocity, radius of curvature, coefficient of friction, normal force, etc.). For example, the visual above showing an object at the top of a semicircle would be expressed as
{\displaystyle F_{c}=n+mg\,}
{\displaystyle {\sqrt {a_{r}^{2}+a_{t}^{2}}}=a}
Radial acceleration is still equal to
{\displaystyle v^{2}/r}
. Tangential acceleration is simply the derivative of the velocity at any given point:
{\displaystyle a_{t}=dv/dt\,}
. This root sum of squares of separate radial and tangential accelerations is only correct for circular motion; for general motion within a plane with polar coordinates
{\displaystyle (r,\theta )}
, the Coriolis term
{\displaystyle a_{c}=2(dr/dt)(d\theta /dt)}
should be added to
{\displaystyle a_{t}}
, whereas radial acceleration then becomes
{\displaystyle a_{r}=-v^{2}/r+d^{2}r/dt^{2}}
V8 engineDiesel engineDifference engineElectrical engineeringEngineEngine configurationEngine tuningEngineeringFour-stroke engineHeat engineHistory of the steam engineInternal combustion engineMechanical engineeringPetrol enginePhysicsRadial engineReciprocating engineShaft (mechanical engineering)Steam engineStirling engineTwo-stroke engineV12 engineAmmeter
This article uses material from the Wikipedia article "Circular motion", which is released under the Creative Commons Attribution-Share-Alike License 3.0. There is a list of all authors in Wikipedia
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Magnetism And Matter, Popular Questions: CBSE Class 12-science SCIENCE, Science - Meritnation
\left(1\right) 1 : \sqrt{2} : 2\sqrt{2} \left(2\right) 1 : \sqrt{2} : \sqrt{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(3\right) 1 : 2 : 2\sqrt{2} \left(4\right) \sqrt{2} : 2 : 1
{10}^{4}
{10}^{-4}
{s}^{2}
{s}^{2}
{s}^{2}
{s}^{2}
\left|{T}_{1}-{T}_{2}\right|
\left(1\right) 2\mathrm{\pi }\sqrt{\frac{61}{5g}} \left(2\right) 2\mathrm{\pi }\sqrt{\frac{5l}{4\mathrm{g}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(3\right) 2\mathrm{\pi }\sqrt{\frac{4 l}{5\mathrm{g}}} \left(4\right) 2\mathrm{\pi }\sqrt{\frac{5 l}{6\mathrm{g}}}
c{m}^{2}
°
\sqrt{13}
\sqrt{13}
\sqrt{3}
Which substance obeys Curie's Law ; feromagnetic or paramagnetic or both of them? give reason ,why they obey this law?
Sk The Beatstar asked a question
here, its written that "The North magnetic pole is located at latitude of 79.74° N and a longitude of 71.8° W, a place somewhere in North Canada. The magnetic South Pole is at 79.74° S and 108.22° E in Antarctica." But in the video its showing south pole and mentioning the location as 79.74° N and 71.8° W...
What is faradays right hand rule and how does it help to assign a direction ie whether the current is flowing from north or south?
Kiran Rana & 1 other asked a question
which topic is important in magnetism and matter?
how does coercivity and retentivity depend on the width of hysteresis curve?
what is 1/4IIEo
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EUDML | General Selmer groups and critical values of Hecke L-functions. EuDML | General Selmer groups and critical values of Hecke L-functions.
General Selmer groups and critical values of Hecke L-functions.
Guo, Li. "General Selmer groups and critical values of Hecke L-functions.." Mathematische Annalen 297.2 (1993): 220-234. <http://eudml.org/doc/165128>.
keywords = {Selmer groups; Hecke -functions; elliptic curve; complex multiplication; Grössencharacter; Birch and Swinnerton-Dyer conjecture; Tate-Shafarevich group},
title = {General Selmer groups and critical values of Hecke L-functions.},
TI - General Selmer groups and critical values of Hecke L-functions.
KW - Selmer groups; Hecke -functions; elliptic curve; complex multiplication; Grössencharacter; Birch and Swinnerton-Dyer conjecture; Tate-Shafarevich group
𝐩
Selmer groups, Hecke
L
-functions, elliptic curve, complex multiplication, Grössencharacter, Birch and Swinnerton-Dyer conjecture, Tate-Shafarevich group
Articles by Li Guo
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Legacy Congestion Management - ccie.nyquist.eu
Input and Output Queues
In the incoming direction,there is only one software queue that is always serviced using a FIFO algorithm. To only configuration available for the input queue is the size. You can specify it with:
R(config-if)# hold-queue IN-SIZE in
!Default:75
To see information about the hold-queue, use:
R# show interface INTERFACE | i Input queue
You can enable Selective Packet Discard to have more control over the Ingress Queue.
When a device needs to send packets out an interface, it sends them to the Hardware Queue, also known as the TX Ring. The Hardware Queue is always serviced using FIFO. The size of the outgoing Hardware Queue can be configured using:
R(config-if)# tx-ring-limit HQ-SIZE
To see the size of the hardware queue, use:
R# show controllers INTERFACE | i tx_limited
! In this example, 128 is the HQ-SIZE
When the Hardware Queue is full, the packets are queued in the Outgoing Software Queue. The size of this queue can also be configured using:
R(config-if)# hold-queue OUT-SIZE out
This queue is serviced by default using FIFO (high speed interfaces) or WFQ (Serial interfaces with bandwidth lower than E1 – 2048kbps). However, the outgoing software Queue can be configured to use one of the available queueing algorithims: FIFO, Fair Queuing (WFQ or CBWFQ), Priority Queueing (PQ) or Custom Queuing (CQ). To see information about the hold-queue, use:
R# show interface INTERFACE | i Output queue
! The display differs, according to the algorithm used
Legacy Queuing algorithms
By default, all interfaces, except serial interfaces at E1(2048 kbps) or lower speeds, use the FIFO queuing. This means that packets are only served in the order of their arrival. A single queue is used for FIFO.
On serial interfaces with lower speeds (<2048 kbps), the default queuing mechanism used is Fair Queuing, but if you want, you can disable it and use FIFO instead:
R(config-if)# no fair-queue
To verify that FIFO is the queueing algorithm of an interface, use one of the following commands:
R# show interface INTERFACE | i Queue
! For FIFO it will say:
Interface FastEthernet0/0 queueing strategy: none
'Show queue' not supported with FIFO queueing.
The main disadvantage of FIFO is that there is no difference between important traffic and less important traffic, so applications that need a certain level of QoS will not get it. Also, smaller flows, may end up being delayed or dropped altogether because the high bandwidth consumers are using all the available bandwidth.
Fair Queuing is a technique that offers a chance for the smallest flows to send traffic. At each time interval, the available bandwidth is split among all flows, based on the minimum request. That is the smallest flow will get the required bandwidth to send all traffic, and the other flows will get the same amount. The remaining bandwidth will be split again, also based on the minimum request, but this time there will be less flows. Cisco implements Fair Queuing with the possibility to prioritize traffic according to their importance. You can use fair queueing on an interface (called Flow-Based WFQ) or in a class-map inside a policy (called Class-Based WFQ).
Flow-Based Weighted Fair Queuing is the default queuing mechanism used on slow speed (<2048 kbps) serial interfaces. A flow is identified as all the packets that share the same source and destination IP, source and destination TCP port and the same IP ToS Byte. The weight for a flow is calculated using the formula:
W_{flow}= \frac{32384}{IPPrecedence + 1}
In each time unit, a number of bytes inversely proportional to the weight or directly proportional to IP_Precedence+1 is sent for each flow. Let’s define the following variables:
Bi = the bandwidth assigned to a flow with IP Precedence i.
Ni = the number of flows with IP Precedence i
B = available bandwidth
The next two equations are true:
BW = \sum_{i=0..7}N_iB_i
BW = k\sum_{i=0..7}(i+1)N_i = k(1N_0 +2N_1+3N_2+4N_3+5N_4+6N_5+7N_6+8N_7)
If you know the values for each Ni, then you know how much of the bandwidth will be assigned to a flow with a certain priority:
B_i = \frac{B*(i+1)}{\sum_{i=0..7}(i+1)N_i}
To enable flow-based WFQ if not already on, use:
R(config-if)# fair-queue [CDT [CONVERSATIONS [RSVP-QUEUES]
! CDT = Congestive Discard Threshold. Default:64
! CONVERSATIONS = number of Queues used - default: 256
! RSVP-QUEUES = number of queues available for RSVP Reservation - default:0
WFQ assigns a hash value to each flow. Based on this hash, a flow is assigned to a CONVERSATION. Different flows can be assigned to the same Conversation.
WFQ also uses a flavor of RED to drop packets when the Congestive Discard Threshold (CDT) is reached. It will drop packets from the flow with the highest scheduled time calculated as
Time_{scheduled} = Size * Weight + Time_{previous}
For a new flow, the previous Time is considered as the scheduled time of the last packet sent, therefor new flows are penalised just as old flows.
In addition to the queues assigned for each conversation (Queue ID: 1->CONVERSATIONS), there are 8 Link Queues which are reserved for network control and routing packets (Queue ID:CONVERSATIONS+1->CONVERSATIONS+8). The weight assigned for these queues is 1024 – which is lower than the lowest weight for normal traffic (32384/8 = 4048).
Then, if RTP Priority is used, it will use a Queue ID of CONVERSATIONS+9. This queue will have a weight of 0 and will act like a Priority Queue.
In the end, Queues for the RSVP flows are assigned and they will have a weight of 128.
To verify WFQ configuration, use:
R#show interface INTERFACE | b Input queue
! The result is the same as:
R# show queueing interface INTEFACE
The fallowing command displays information about all interfaces configured for fair-queueing:
R# show queueuing fair
CQ allows the use of up to 16 queues, with an assigned byte count. CQ cycles through each queue in a round-robin manner and allows each queue to send packets until the assigned byte count is exceeded. There is a default system queue (Queue 0) that cannot be configured, which will contain packets used for L2 keepalives. All other traffic, including routing updates are assigned to Queue 1 by default.
CQ uses a Deficit Round Robin algorithm, meaning that each time interval, a queue is allowed to send traffic until it exceeds the assigned Byte Count. If the Byte Count is exceeded, the packets are sent anyway, but the execess will be deducted from the Byte Count in the next round. The Byte Count also accounts for the L2 overhead. Let’s define for each queue:
BCq = Byte Count for Queue q
PSq = Average Packet Size for Queue q
TSq = Traffic Share for Queue q
BC_q = k_q * PS_q
\frac{BC_{q1}}{BC_{q2}} = \frac{TS_{q1}}{TS_{q2}} = \frac{k_{q1}PS_{q1}}{k_{q2}PS_{q2}}
We can then chose a convenient value for one of the k factors and find the Byte Count for each Queue so that our traffic is shared accordingly.
We can define different configurations for CQ on each interface. All configurations applied to an interface are grouped under a Queue-list. To configure CQ, first set up the queue length and byte count for each queue, using:
R(config)# queue-list CQ-LIST queue Q-ID limit LENGTH
R(config)# queue-list CQ-LIST queue Q-ID byte-count BYTES
We will then classify packets and assign them to each queue. The packets can be assigned by interface or by protocol:
R(config)# queue-list CQ-LIST protocol PROTOCOL Q-ID [PROTOCOL-OPTIONS]
R(config)# queue-list CQ-LIST interface INTERFACE Q-ID
A default Queue can be assigned for unmatched traffic:
R(config)# queue-list CQ-LIST default Q-ID
! Default: Q-ID=1
By default, Queue 0 is used as a Priority Queue. You can set more queues to be served as priority queues, using:
R(config)# queue-list CQ-LIST lowest-custom Q-ID
!Default: Q-ID=1
This command actually defines what is the lowest Queue that is served in a round-robin fashion. Queues from 0 to the LowestCustomQueue-1 are considered Priority Queues, and the others are considered Custom Queues. As you can see, by default only Queue 0 is a Priority Queue.
In the end, apply CQ-LIST configuration on the interface:
R(config-if)# custom-queue-list CQ-LIST
You can verify configuration with: To verify the configuration, use:
R1#show interface INTERFACE | b Queueing
Queueing strategy: custom-list 1
0: 0/20/0 1: 0/30/0 2: 0/20/0 3: 0/20/0 4: 0/20/0
10: 0/20/0 11: 0/20/0 12: 0/20/0 13: 0/20/0 14: 0/20/0
15: 0/20/0 16: 0/20/0
To see the configured CQ-LISTS, use:
R#show queueing custom
List Queue Args
1 2 interface Serial1/0
1 2 protocol ip
1 1 limit 30
1 2 byte-count 128
Priorty Queueing (PQ)
Priorty Queuing is a Queuing mechanism wich uses 4 queues knwon as HIGH, MEDIUM, NORMAL and LOW. They are serviced in the order of their priority, with a higher priority queue being always served before a lower priority queue. As in CQ, the configuration that is applied on an interface is grouped in a LIST.
Each queue has a defined length and when it is full, packets are dropped. The length of each queue can be set using:
R(config)# priority-list PQ-LIST queue-limit HIGH-LEN [MEDIUM-LEN [NORMAL-LEN [LOW-LEN]]]
!defaults: HIGH = 20, MEDIUM = 40, NORMAL: 60, LOW: 80
As with CQ, packets are assigned to a queue based on a classification process:
R(config)# priority-list PQ-LIST protocol PROTOCOL {high|medium|normal|low} [PROTOCOL-OPTIONS]
R(config)# priority-list PQ-LIST interface INTERFACE {high|medium|normal|low}
By default, all unclassified packets are assigned to the NORMAL queue, but this can be changed with:
R(config)# priority-list PQ-LIST default {high|medium|normal|low}
In the end, apply the PQ-LIST configuration on the interface:
R(config-if)# priority-group PQ-LIST
R# show interface INTERFACE | b Queue
To see the existing CQ-LISTS:
R# show queueing priority
2 low default
2 high limit 10
2 medium limit 20
2 normal limit 30
2 low limit 40
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Quadratic equations - August Crash Course Live Session
Quadratic equations - August Crash Course Live Session Contact Number: 9667591930 / 8527521718
The position of a particle moving along the x-axis is given by
\mathrm{x}=\frac{{\mathrm{t}}^{3}}{3}-2{\mathrm{t}}^{2}+3\mathrm{t}+25.
The time (s) when particle comes to an instantaneous stop is/are
1. t = 1 sec
A gun is mounted on a plateau 960 m away from its edge as shown. Height of plateau is 960 m. The gun can fire shells with a velocity of 100 m/s at any angle. Of the following choices, what is the minimum distance (OP)x from the edge of plateau where the shell of gun can reach?
Two particles, one with constant velocity 50 m/s and the other with a uniform acceleration 10 m/s2, start moving simultaneously from the same place in the same direction. They will be at a distance of 125 m from each other after:
5\sqrt{2} \mathrm{sec}
10\left(\sqrt{2}+1\right) \mathrm{sec}
The sum of the magnitude of two force acting at a point is 16 N. The resultant of these forces is perpendicular to the smaller force and has a magnitude of 8 N. If the smaller force is of magnitude x, then the value of x is:
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Locomotion and Movement - Live Session - NEET 2020Contact Number: 9667591930 / 8527521718
During muscle contraction, the length of all the following gets reduced except :
The human ribs are termed as bicephalic because :
Joint between the carpals
Joint between metacarpal of thumb and phalange
The combination of a neuron and the muscle fiber it associates with is called a/an ____
2. motor end plate
4. myoneural junction
The upper ends of the forearm bones articulate with each other in a :
2. ellipsoid joint
Sliding filament theory can be best explained as
1. when myofilaments slide pass each other actin filaments shorten while myosin filament do not shorten
2. actin and myosin filaments shorten and slide pass each other
3. actin and myosin filaments do not shorten but raher slide pass each other
4. when myofilament slide pass each other myosin filament shorten while actin filaments do not shorten
The process by which most of the cartilaginous tissue is transformed into bones is
2. intermembranous ossification
1. contract with a larger force
2. contract with a smaller force
3. contract with a same force
4. undergo tetany
1. tendon reflex
2. ipsilateral reflex
3. flexor reflex
4. patellar reflex
1. ligament of neck
2. ligament of ankles
3. ligament of palm
4. ligament of face
Oxygen content reduction makes the glycolyse (glycogenesis) intensity increased due to
1. increase of ADP concentration in cell
2. increase of NAD+ concentration in cell
3. increase in ATP concentration in cell
4. increase of concentration of peroxides and free radicals
The action potential that triggers a muscle contraction travels deep within the muscle cell by means of :
2. transverse tubules
4. motor end plates
A motor unit is best described as :
1. All the nerve fibres and muscle fibres in a single muscle bundle
2. One muscle fibre and its single nerve fibre
3. A single motor neuron and all the muscle fibres that it innervates
4. It is the neuron which carries the message from muscle ot CNS
Put the following phrases in proper order to describe what occurs at the neuromuscular junction to trigger muscle contraction :
I. Receptor sites on sarcolemma
II. Nerve impulse
III. Release of Ca+2 from sarcoplasmic reticulum
IV. The neurotransmitter acetylcholine is released
V. Sarcomere shortem
VI. Synaptic cleft
VII. Spread of impulses over sarcolemma on T-tubules
1. II,IV,I,VI,VII,III,V
2. II,IV,VI,I,VII,III,V
3. I,II,III,IV,V,VI,VII
4. VII,VI,V,IV,III,II,I
Pelvicc girdle (hip girlde ) is composed _____coxal (hip) bones :
Assertion : Muscle twitch is related with tetanus
Reason : Tetanus is caused by many nerve impulses at a time
Assertion : Ball and socket joints are the most mobile joints
Reason : Synovial fluid is present here
Assertion : Triceps is said to be an extensor muscle for elbow joint
Reason : Triceps relaxes durings extension of forearm at the elbow joint
1. Movement is one of the significant features of living beings
2. Streaming of protoplasm in the unicellular organisms like Ameoba is a simple form of movement
3. Animals and plants exhibit a wide range of movements
4. Movement of cilia, flagella and tantacles is limited to lower organisms
Which of the following holds all the muscle bundle together in a muscle ?
2. Collegenous connective tissue layer
Actin and Myosin in muscle fibril are arranged
1. Parallel to each other and perpendicular to the longitudinal axis of the myofibrils
2. Perpendicular to each other and parallel to the longitudinal axis of the myofibrils
3. Parallel to each and parallel to the longitudinal axis of the myofibrils
4. Perpendicular to each other perpendicular to the longitudinal axis of the myofibrils
How many filament are there in thin filament ?
1. Two actin filaments and two tropomyosin filaments
2. Two actin filaments and one tropomyosin filaments
3. One actin filament and one tropomyosin
4. One actin and two tropomyosin filament
Subunit of troponin masks the
1. Active binding sites for myosin on actin
2. Active binding sites for actin on Myosin
3. in resting state, active binding sites for Myosin on actin
4. In resting state, active binding sites for actin on Myosin
The HMM of muscle fibre has not got
The binding of calcium to troponin actually indicates
1. Unmasking of Myosin binding site over meromyosin
2. Unmasking of Myosin binding site over actin
3. Unmasking of actin binding site over myosin
4. Unmasking of actin binding site over tropolosin
____ projects outwards at regular ____ from each other from the surface of polymerised Myosin filaments and is also known as ___
1. Head and short arm, angle, cross arm
2. Head and short arm, distance, cross arm
4. Head and short arm, distance, LMM
Which of the following step during sliding Filament theory is the last step before relaxing state ?
1. ATP utilization
2. Release of ADP and inorganic phosphate
3. Cross bridge formation
4. Binding of calcium with troponin
1. A new ATP binds and cross bridge is broken with next cycle of contraction
2. The process of contraction continues till the calcium ions are pumped back to the sarcoplasmic cisternae and resulting in unmasking of actin filaments
3. The reaction time of the fibres can vary in different muscles
Which of the following is a cranial bone and is not single in number ?
Which of the following can not be true ?
1. All the mammals have seven cervical vertebrae
2. Vertebral column protects the spinal cord, supports the head
3. Vertebral coulmn serves as the point of attachment for the ribs and musculature
4. Sternum is a flat bone on the ventral midline of thorax
1. Force generated by muscle is used to carry out movement through joints
2. The movability of joints vary depending on different factor
3. Joints have been classified into three major functional forms
Which of the following seen in Gout ?
1. Inflammation of joints due to accumulatiton of uric acid crystals
2. Inflammation of bones due to accumulatiton of uric acid crystals
3. Inflammation of joints due to accumulatiton of calcium
4.Inflammation of joints due to accumulatiton of estrogens
Contractile unit of muscle is part of myofibril between
1. Z-line and I-band
2. Z-line and Z-line
3. Z-line and A-band
4. A-band and I-band
A sarcomere consists of
1. One A-band and one I-band
2. Half A-band and two half I-band
3. Half A-band and one I-band
4. One A-band and two half I-band
The monomeric protein which polymerises to form myosin is
2. Meromyosin
Type of vertebrae in case of human is
1. Amphiplatyan
2. Procoelus
3. Amphicoelus
4. Heterocoelus
Cori cycle involves
3. Liver and muscles both
Phalangeal formula for the hand is
What would happen if ATP suddenly were not available after the sarcomere had started to shorten ?
1. Cross bridges would not be able to detach from actin
2. Muscle would remain in a state if rigidity
3. Muscle would relax immediately
Which of the following is not applicable to red muscle fibres when compared to white muscle fibres ?
1. Sustained contraction for long periods
2. Rich in myoglobin
3. Faster in contraction rate
4. Rich in mitochondria
Stimulation of muscle fiber by a motor neuron occurs at
2. The transverse tubules
3. The myofibril
4. The sarcoplasmic reticulum
Number of anterior curves present with human vertebral column is
One of the following ions is essential for muscular contraction
{\mathrm{Na}}^{+}, {\mathrm{Ca}}^{++}
{\mathrm{Mg}}^{++}, {\mathrm{Ca}}^{++}
{\mathrm{Mg}}^{++}, {\mathrm{K}}^{+}
{\mathrm{K}}^{+},{\mathrm{Na}}^{+}
80 % of the lactic acis is converted to ____ in the liver
The sheath covering the bundle of muscle fibres is
Which bone is keystone of the cranial floor ?
Anaerobic work becomes painful due to accumulation of
1. Ca2+ ions
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One-time pads 🕵️♀️ | TigYog
Your first spy, Bob, used a different cipher. It worked like this:
Could not load <object> SVG
Every letter in his secret message was replaced by the letter two places before it in the alphabet. For example, the message dead becomes the ciphertext bcyb.
The last ciphertext you received from Bob was mpylecq. What was the original message Bob sent?
No, but that sounds like something a spy would send, doesn’t it? Bob sent the word oranges, which was encoded as follows. First, he encodes the letter o. Write out the letters of the alphabet: you will find that o is the
15^{th}
letter. He subtracts
2
15-2=13
. Then he looks up the
13^{th}
letter of the alphabet, which is m. So he sends the letter m.
He does this for each letter in succession. The main tricky letter is a. For this, we ‘wrap around’ back to the end of the alphabet. See the picture above, where a goes to y.
Yes, Bob loved his oranges! Sadly, it was his downfall. Counter-intelligence deciphered Bob’s message, and he was arrested at the local orange grove.
You probably worked out the method to decode Bob’s secret message. Beforehand, you had agreed with Bob that he would shift each letter backwards by
2
places. We say you agreed on the shift
\red{S_{bob}} = -2
. This is the secret amount that Bob adds to each letter. To decode the message, you subtract
\red{S_{bob}}
from each letter. With
\red{S_{bob}} = -2
, you shift each letter forward by
2
But how did the enemy decode Bob’s message? The number
\red{S_{bob}} = -2
was secret between you and Bob, so the enemy had to figure out the value of
\red{S_{bob}}
in some other way.
To see the weakness, consider a ciphertext you intercepted from an enemy spy, Jon. The ciphertext is cbobobt. But what does it say? You suspect the word is another fruit 🍎🍇🍌🍍🍓, but you don’t know Jon’s secret shift value,
\pink{S_{jon}}
. Can you figure it out?
\pink{S_{jon}} = -3
\pink{S_{jon}} = -1
\pink{S_{jon}} = 1
\pink{S_{jon}} = 4
No, Jon’s secret shift is
\pink{S_{jon}} = 1
. The secret word was bananas, which was shifted forward by
1
to produce cbobobt.
Not quite! It’s
1
-1
\pink{S_{jon}}
is the number of places that Jon shifts forward, not backward. The secret word was bananas, which was shifted forward by
1
Yes, the word was bananas. You arrest Jon at the banana plantation, and have him executed in the main square.
Repeated letters in the message become repeated letters in the ciphertext. Look at those three as in bananas, all shifted forwards to b in the ciphertext cbobobt. Patterns in the message become patterns in the ciphertext. Cryptanalysts can use these patterns to figure out what the original message was!
Let’s make a more secure cipher 🔐
The weakness in this cipher is that every letter is shifted by the same secret value,
\red{S_{bob}}=-2
. That is, the secret
\red{S_{bob}}
is used repeatedly. To keep Alice safe, you have agreed to shift every letter by a different amount. Surely this will keep enemy prying eyes away!
You and Alice agree on the following sequence of shifts:
\green{P_{alice}} = 7,1,3,3,0,4,8,5, \ldots
To encode the first letter, Alice will shift it by
7
; to encode the second letter, Alice will shift it by
1
; and so on. You receive a ciphertext from Alice: hqsoew. What does it say?
No, that’s not it. Alice sent the word apples, encoded as follows. For the first letter, a, she took the first number from the list
\green{P_{alice}}
. That number is
7
. She then shifted a right
7
places, to get h. For the second letter, p, she took the second number from the list above, which is
1
. She shifted p right
1
place, to get q. And so on, using the shift sequence
7,1,3,3,0,4
Yes, she needs to meet in the apple orchard! 🍏
You and Alice are now using a one-time pad. This cipher, used correctly, is unbreakable! 🔐
In the next lesson, you’ll learn why this cipher is unbreakable, some ways the one-time pad can go wrong, and how to avoid those. Stay tuned!
\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 2x^2 - 3x = 2
\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \sqrt{4ac - b^2}
\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} +b^2
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