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Work Done by Torque Calculator - Savvy Calculator
Work Done by Torque Calculator
About Work Done by Torque Calculator (Formula)
The Work Done by Torque Calculator is an essential tool for engineers and mechanics involved in analyzing rotational motion. Torque is a measure of the force that can cause an object to rotate about
an axis. Understanding how much work is done by this torque is crucial in various applications, from automotive engineering to machinery design. This article will explain the formula used in the Work
Done by Torque Calculator, guide you on how to use it, provide a practical example, and address frequently asked questions.
The formula for calculating the work done by torque is:
Wt = t * d(θ)
• Wt represents the work done by torque.
• t is the torque applied (in Newton-meters).
• d(θ) is the change in angular displacement (in radians).
How to Use
1. Determine the Torque (t): Measure the torque applied in the system, typically in Newton-meters (N·m).
2. Measure Angular Displacement (d(θ)): Determine the change in angular displacement, expressed in radians. You may need to convert degrees to radians if necessary (1 radian = 180/π degrees).
3. Input Values into the Formula: Substitute the values of torque (t) and angular displacement (d(θ)) into the formula: Wt = t * d(θ).
4. Calculate the Work Done: Perform the calculation to find the work done by the torque.
Let’s calculate the work done by torque for a system with the following parameters:
• Torque (t): 15 N·m
• Change in Angular Displacement (d(θ)): 2 radians
Using the formula:
Wt = t * d(θ)
Wt = 15 N·m * 2 radians
Wt = 30 Joules (J)
This result indicates that the work done by the torque in this scenario is 30 Joules.
1. What is torque?
Torque is a measure of the rotational force applied to an object around an axis.
2. Why is work done by torque important?
It helps in analyzing the energy transfer in rotational systems, essential for machinery and engineering applications.
3. What units are used for torque and work?
Torque is measured in Newton-meters (N·m), and work is measured in Joules (J).
4. How do I convert degrees to radians?
To convert degrees to radians, use the formula: radians = degrees × (π/180).
5. Can I use this calculator for non-linear motion?
The calculator is specifically designed for rotational motion and torque.
6. What does d(θ) represent?
d(θ) represents the change in angular displacement, indicating how far an object has rotated.
7. How is torque calculated?
Torque can be calculated using the formula: t = F × r, where F is the force applied, and r is the distance from the pivot point.
8. What happens if the torque is constant?
If the torque is constant, the work done can be calculated using the same formula throughout the displacement.
9. Can I use this for mechanical systems?
Yes, it is widely used in mechanical engineering for analyzing the work done by motors and other rotational devices.
10. What is the significance of work done in physics?
Work done indicates the energy transferred by a force, essential for understanding system efficiency and performance.
11. How often should I calculate work done in a system?
It is advisable to calculate work done whenever there are changes in the system parameters or conditions.
12. What factors can affect torque in a system?
Factors include the magnitude of the force applied, the distance from the pivot point, and the angle of application.
13. Can this calculator be used for complex systems?
Yes, but ensure that all forces and torques are accounted for accurately.
14. What is the difference between work done by torque and linear work?
Work done by torque involves rotational motion, while linear work involves force applied along a straight line.
15. How do I ensure accurate measurements for torque?
Use calibrated tools and measure carefully to ensure precision in your calculations.
16. What tools can help measure torque?
Torque wrenches, dynamometers, and specialized torque measurement devices can provide accurate readings.
17. Can I use this formula for different types of machinery?
Yes, it is applicable to various machinery and applications involving rotational motion.
18. What are some real-world applications of this calculator?
Applications include automotive design, robotics, and any machinery requiring precise torque measurements.
19. Is there a maximum limit for torque calculations?
The limit depends on the material properties and design of the system; exceeding it can lead to failure.
20. What should I do if I receive an unexpected result?
Double-check your input values and calculations for accuracy, and ensure that all units are consistent.
The Work Done by Torque Calculator is a crucial tool for anyone involved in rotational systems, providing insights into energy transfer and system efficiency. By utilizing the formula Wt = t * d(θ),
users can accurately assess the work done by torque in various applications. Regular calculations and adjustments based on system parameters can lead to improved performance and reliability in
engineering practices. Understanding how to use this calculator effectively will enhance your ability to analyze and optimize rotational motion in your projects.
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Engaging Elementary Students in Geometry through Origami
Brief History of Paper
There are a few books out that look at history through objects or food. It is a fascinating way to chart history through time without necessarily charting wars and battles. Mark Kurlansky has
written a few of these. Salt: A World History, Cod: A Biography of the Fish that Changed the World, The Big Oyster: History on the Half Shell, and Milk! and A 10,000-Year Food Fracas are just a few
of his books that look at the history of these items and how the need, distribution, sale, use, and quest for them have changed the world.
Paper: Paging Through History was also written by Kurlansky in 2016. Paper as we know and use it today is a relatively new technology. Paper satisfied a need in society – a need to record
information. As governments grew and changed and as businesses and trade grew and changed, the need to record and keep records of transactions and laws became important.
Kurlansky wrote that the Chinese invented papermaking. That is not to say that a type or form of paper was not being used in other parts of the world around the same time. Evidence shows that
people were writing on materials such as clay, stone, papyrus and parchment.
At one point, the mark of an advanced civilization was one that made paper. As paper became the cheaper option (much cheaper that papyrus, a water plant, which could only be grown along the Nile and
parchment, which is made from the skins of sheep, goats, etc.), its use spread throughout the world. As merchants from the East traveled throughout Asia, paper was traded and eventually paper mills
grew around the world changing the economy of those towns that settled near a good source of running water. Because papyrus was unique to Egypt, it became a valuable commercial product that was
shipped throughout the world.
The papyrus reed was peeled and once the outer layer was removed, there were about twenty inner layers that would be unrolled and laid out flat. The layers were “woven” together, the second set laid
at a 90-degree angle from the first set underneath. Water was used to moisten the sheets and then they were pressed together with weights for a few hours. The reeds, when cut, had a sticky sap that
served as the glue that kept the layers together and if needed, a flour paste was used. The sheets were then rubbed with a stone, piece of ivory, or shell until they were smooth and the layers did
not create grooves that a stylus could move across the sheet.
Parchment was made from the skins of sheep, goats, and cows. Vellum, a finer quality of parchment, was made from the skins of calves. The process was tedious: after being flayed, the skin is soaked
in water for a day. The skin was soaked in a dehairing liquid, which eventually included lime, for eight days. It had to be stirred a couple of times a day and you had to be careful not to soak the
skin too long because it weakened the skin. Next it was stretched out on a stretching frame by wrapping small, smooth rocks in the skins with rope or leather strips. The skin would be scraped to
remove the last of the hair and get the skin the right thickness.
Paper is made by “breaking wood or fabric down into its cellulose fibers, diluting them with water, and passing the resulting liquid over a screen so that it randomly weaves and forms a sheet.”
(Kurlansky xv) Different trees and plants were used to make paper so through the last two to three centuries, paper has evolved from a thicker, coarser paper to the thin, smooth paper that we use
today. Paper was slow to become popular in Europe, it was felt that important and religious books should be written on parchment because they would last longer than books made of paper.
Making paper was not necessarily easier or faster than making papyrus or parchment. In fourteenth century Europe, papermaking was common and wherever there was a river with clean water, a downhill
run or swift moving current, and a town of people who could provide rags, there was a paper mill. As the need for paper grew, workers, who did not have fixed hours, might work all night. Apprentice
paper workers, or children, “who were small enough to crawl into vats, scrubbed the hammers and the equipment clean” during the night when the mill was closed. (Kurlansky 96)
With our new digital technologies, one might think that paper is seeing its last days. Kurlansky would agree that paper might not be here forever. But he does feel it is
more secure than electronic messages. “Electronic messages can be hacked, accessed and reconstructed.” (Kurlansky 334) If origami continues to be popular and used in education, health care and
science, then paper will continue being manufactured in the future.
To give a perspective of how paper has evolved to today, here is a timeline of major events known throughout history taken from Paper: Paging Through History:
3000 BCE Oldest papyrus found – a blank scroll in a tomb at Saqquara, near Cairo
500 BCE Chinese begin writing on silk
252 BCE Dating of the oldest piece of paper ever found in Lu Lan, China
105 BCE Cai Lun of the Chinese Han court is credited with inventing paper
256 CE First known book on paper produced in China
500-600 CE Mayans develop bark paper
610 CE Korean monk takes papermaking to Japan
751 CE Papermaking in Samarkand begins – they are credited for producing
high quality paper exclusively from (linen) rags
1264 CE First record of papermaking in Fabriano, Italy – they are credited
with first using watermarks to identify the papermaker.
1309 CE Paper is first used in England.
1495 CE John Tate establishes the first paper mill in England in Hertfordshire.
1502-20 CE Aztec tribute book lists forty-two papermaking centers. Some villages
produce half a million sheets of paper annually.
1575 CE Spanish build the first paper mill in Mexico.
1729 CE Papermaking in Massachusetts begins.
1833 CE An English patent is granted for making paper from wood.
1863 CE American papermakers start using wood pulp. (Kurlansky 337 – 3446)
History of Origami
The word origami comes from the ori- meaning, folded and –kami meaning, paper. Although the Chinese developed papermaking, the Japanese developed the art of origami. The first Japanese folds date
from the 6^th Century A.D. Since paper was scare and precious at that time, the use of origami was limited to ceremonial occasions. The designs were limited to representations of animals, people,
and ceremonial designs. The designs were passed down from generation to generation, usually from mother to daughter.
Some of the oldest existing directions for paper folding were printed in Japan in 1797, entitled Sembazuru Orikata or Folding of 1000 Cranes. You may be familiar with the story of the young Japanese
girl, who contracted leukemia after World War II from the effects of the Hiroshima atomic blast. The crane is a symbol of good luck in Japan and the tradition was that if you fold 1,000 cranes, you
would be granted one wish. Young Sadako Sasaki decided to fold 1,000 cranes so that her wish to get better would be granted. She died before achieving her goal, 365 short. After she died, her
classmates folded the rest for her and placed them in her coffin.
Akira Yoshizawa is credited with making origami popular again. He was born to dairy farmers on March 14, 1911 in Japan. When he was 13, he had to take a job in a factory in Tokyo. In his early
20s, he was promoted to “technical draftsman,” responsible for teaching new employees basic geometry. He had learned origami as a child so decided to use it as a tool to help these employees
understand geometry.
Yoshizawa quit his job in 1937 to practice origami full time. He lived in poverty for close to twenty years and during World War II, he served in the army medical corps. To cheer up the sick
patients, he made origami models but eventually became sick himself and was sent home. Finally in 1951, a Japanese magazine asked him to fold the twelve signs of the Japanese zodiac. This exposure
basically led to his fame. In 1954, he founded the International Origami Centre in Tokyo and through his travels became a goodwill ambassador for Japan. He died in 2005 at the age of 94.
In the 1960s, two origami societies were established: The Friends of the Origami Center of America and the British Origami Society. With the resurgence of interest in origami, it has evolved into
different forms including modular folds, three-dimensional folds, folds that combine several subjects into a single fold, action figures, and figures that move when tugged. I believe that one reason
that origami has become so popular and so many new ways to fold paper have become popular is because of the variety of paper that is manufactured today. I cannot imagine making action figures or
modular folds with the coarse, thicker paper that was made years ago. It is with new technologies and materials that our paper today can be as think or thick as we desire. Origami paper can be
purchased in square shapes with different colors or patterns on either side to make folding paper much easier and more precise than in times past. Today, the concept of paper folding has also been
used in health care, i.e. cardiac stents, and science, folding lenses to fit into spacecrafts that can be remoting unfolded once in space. Science and technology has taken the concept of paper
folding and used it to fold different materials such as plastic and metals to advance our ability to save a life or see farther into the universe.
Basics of Origami
Origami is the art of folding an uncut sheet of paper into an object and animal. Yoshizawa invented a systematic code of dots, dashes, and arrows that was adopted by western authors Harbin and
Randlett in the early 1960s, which is still used today.
This standardized the techniques and terminology of folds that people around the world use – if you know the system, you can recreate the design even if the book is written in a different language.
The following general rules are given when creating an origami design: students should work on a hard, smooth, and flat surface so that their folds can be accurate, it is important that each fold
and crease be precise and that a pencil or thumbnail is used to move over the fold for exactness, study the diagram/instructions before folding the paper, and if students use colored paper, start
with the colored side facing down at the beginning of their folding.
This system includes instructions using lines, arrows, and terms used to describe these series of lines and arrows. There are five different types of lines: paper edges, either raw or folded, are
drawn with a solid line. Creases are drawn as a thinner line and will often end before the edge of the paper. Valley folds are drawn with a dashed line and mountain folds by a chain of dot-dot-dash
line. The X-ray line or dotted line when shown on a drawing indicates anything hidden behind other layers or represents a hidden edge, fold, or arrow. (Lang 2003, 15).
(Math in Motion, Pearl 41)
Geometry and Origami
Using origami to introduce more complex abstract concepts to elementary students is a way to allow students to comprehend shapes and angles. We may think that all students understand that a square
has four sides of the same length and four right angles, but I have seen the Ah-ha! Moments some fourth graders have when they make one from an 8 ½ by 11” sheet of paper. Using folds or creases;
students can create triangles, such as equilateral triangles. Here again, do students really understand what an equilateral triangle is? They will after completing an activity where they are asked
to create one from a square. Thomas Hull wrote in Project Origami: Activities for Exploring Mathematics that “…when choosing to use origami as a vehicle for more organized mathematics instruction,
an easy choice is to let the students discover things for themselves.” (Hull xi). Origami is a great strategy to use to help students discover properties of two-dimensional shapes. Students can
create hexagons, octagons, and nonagons by using origami.
Students in fourth grade have a difficult time understanding angles. They see two rays coming together at a point. They are shown a protractor and shown how to measure the angle created by the two
rays. Students spend time practicing these measurements with their protractors and terms such as right, acute, obtuse angles. One way to help students to grasp these concepts and discuss where we
use geometry in the world would be use a square sheet of paper is to point that one corner is a right angle and by making creases (folding) the paper, they can create different acute angles.
Students can also fold a square making two creases that intersect at the middle of the square. A circle can be drawn around the intersecting point so that students can see when we divide a square
into four sections, we are creating four right angles each representing 90 degrees or a right angle. If we multiply 90 times 4, we get a product of 360 degrees, which represents the total degrees of
a circle.
Geometry terms such as lines, points, angles, triangles, rectangles, etc. can be modeled and discussed as part of an origami lesson. Asking students to identify these terms and create them will
support students who are visual learners. Another concept that can be modeled is patterning. Students will be able to create patterns through paper folding, especially when creating
three-dimensional shapes. | {"url":"https://theteachersinstitute.org/curriculum_unit/engaging-elementary-students-in-geometry-through-origami/","timestamp":"2024-11-08T22:01:09Z","content_type":"text/html","content_length":"303922","record_id":"<urn:uuid:a4dda39e-023f-4034-a5cc-90617aa553d8>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00160.warc.gz"} |
Resistivity Calculator - Savvy Calculator
Resistivity Calculator
About Resistivity Calculator (Formula)
A Resistivity Calculator is a tool used to calculate the electrical resistivity of a material, which measures its inherent resistance to the flow of electric current. This calculation is crucial in
understanding and predicting the behavior of materials in electrical and electronic applications.
Formula for Resistivity Calculation:
The formula for calculating resistivity (ρ) is:
Resistivity (ρ) = Resistance (R) × (Cross-Sectional Area (A) / Length (L))
• Resistance (R): The resistance of the material, typically measured in ohms (Ω).
• Cross-Sectional Area (A): The area perpendicular to the direction of current flow, typically measured in square meters (m²).
• Length (L): The length of the material along the direction of current flow, typically measured in meters (m).
The unit of resistivity is ohm-meter (Ω·m).
1. Material Selection: Engineers use the Resistivity Calculator to evaluate and select materials for electrical and electronic components based on their resistive properties.
2. Circuit Design: Electronics designers calculate resistivity to ensure proper current flow and voltage distribution in circuits.
3. Wire and Cable Design: Calculating resistivity helps in designing wires and cables for efficient energy transmission.
4. Semiconductor Industry: Resistivity calculations are essential for designing semiconductor materials used in electronic devices.
5. Electrical Installations: Electricians use resistivity calculations to optimize installations and ensure proper conductivity.
In summary, a Resistivity Calculator involves a formula that helps engineers, designers, and professionals calculate the electrical resistivity of materials, which is vital for selecting suitable
materials and designing efficient electrical and electronic systems.
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William Chen
Shapes That Generate Progressively Increasing Number Codes
A polygon is a shape with many sides, it could 3 or 4 or 5 or 6 or more sides. In the same fashion that we can build a sequence of Squared Numbers, like 1-4-9-16-25 etc we can build pentagonal,
hexagonal, heptagonal sequences that carry on forever… Sometimes we need to know a certain number in a certain sequence at a certain position. eg: If I want to know what is the 4th Pentagonal Number,
(the Sequence is 1-5-12-22-35-etc) there exists a special formula that can tell us this. It “t” is the number of sides in a polygon, (in this example t=5), the formula for the nth t-gonal number Poly
(t,n) is [n^2(t-2)-n(t-4)]/2. Now substitute the values of t=5 and n=4 into this formula, giving: [4^2(5-2)-4(5-4)]/2 = [16x3-4x1]/2 = [48-4]/2 = 44/2 = 22 We just established that the 4th Pentagonal
Number is 22 without knowing the full sequence of 1-5-12-22-35… This type of information for Pattern Hunters who need to know such data for their research. Jain 108 Image: (thanks to Tony Foster, who
runs a Facebook Group called Pascals Triangle, that has over 6,000 members, for submitting this artwork on his site) | {"url":"https://www.kaiserxlv.com/post/polygon-sequences","timestamp":"2024-11-14T21:10:40Z","content_type":"text/html","content_length":"1050479","record_id":"<urn:uuid:3e509db9-051c-4c30-8bb3-6ee7458fce92>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00514.warc.gz"} |
Overview of direct torque control of induction machines
Nik Idris, Nik Rumzi (2008) Overview of direct torque control of induction machines. In: Modeling and Control of Power Converters and Drives. Penerbit UTM , Johor, pp. 67-98. ISBN 978-983-52-0648-1
In order to develop the Direct Torque Control (DTC) strategy, it is necessary to construct a dynamic model of the induction machine. The induction machine model used for a high performance control
systems must include all the dynamic effects that occur during the transient states. For simplicity, a few assumptions and conventions are adopted [1][2][3]. Despite the simplification, the model
adopted here is adequate for the design of a control system and is valid for any arbitrary time variation of the machine voltages and currents.
Repository Staff Only: item control page | {"url":"http://eprints.utm.my/24965/","timestamp":"2024-11-02T14:55:54Z","content_type":"application/xhtml+xml","content_length":"17218","record_id":"<urn:uuid:c8307f34-1a45-465d-9342-1ab6b6f73d06>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00308.warc.gz"} |
Mass Measurement - Basculasbalanzas.com
Mass measurement requires a scale of mass and volume. In the United States, a kilogram of water weighs approximately 3 kg. The difference between a kilogram and a gram is about 0.3 grams. The
National Bureau of Standards has done measurements on objects with dissimilar densities, ranging from 2.7 to 16.6 grams per cubic centimeter, in a range of atmospheric conditions from 0.5 to two
atmospheres. This has revealed unsuspected differences of up to 1 milligram per kilogram. Mass measurements require accuracy, and this is especially important when comparing weights of materials that
differ in density.
A balance is one of the most common instruments for mass measurements. A balance is a device used to compare an unknown mass with an object that has a known mass. There are many types of balances,
including beam balances, digital scientific balances, and inertial balances. These devices use springs to measure the mass of an object. The stiffness of the spring and the vibration of the object
help the balance find the unknown mass.
The kilogram was first defined as the mass of one cubic centimeter of water at the melting point. A few centuries later, the kilogram was replaced by the International Prototype Kilogram. The new
metric system took the gram as its unit of mass, but its mass was too small to be commercially useful. To meet this requirement, the kilogram was increased by a factor of 1000. Today, kilogram
weights are used to measure mass in industrial applications.
Gravitational acceleration is another factor that affects mass measurements. Depending on the latitude and altitude, gravitational acceleration varies from place to place. The acceleration at the
Equator is 9.78 m/s2, while at the poles it is 9.832 m/s2 (0.53% difference).
To measure volume in an irregularly shaped object, use the displacement method. To measure a liquid, first fill a measuring vessel with water and then add the object. Next, subtract the first volume
from the second and the difference represents the volume of the object. Using this method, you can calculate the volume of the liquid by adding the volume and finding the density from a density
table. Finally, the mass of a solid can be measured directly in the SI unit of kilogram.
Mass is an important quantity to measure because it represents the force needed to move an object. Because of this, it is important to use standard measurements. For example, a balance must be
calibrated with a standardized mass. Another popular way to measure mass is by using vibrating tube sensors. The mass of an object is equal to its weight multiplied by the acceleration of gravity.
The mass of an object is measured in kilograms, so it is important to remember this when comparing weight and mass.
A kilogram is defined by the Planck constant, which is 6.62607015 x 10-34 J/s. A kilogram mass is equal to the weight of a cubic centimeter. Therefore, the kilogram can be defined by measuring the
mass of a cubic centimeter. The kilogram can also be measured in mm2.
Einstein’s special theory of relativity, introduced in 1905, revolutionized the idea of mass. Einstein’s theory was based on the idea that mass and energy are two of the same. Hence, they are
indistinguishable. However, this does not make them the same thing. This means that the two different units of mass cannot be said to be identical. However, both of them possess mass. This is
important when measuring mass in a system.
A particle is a combination of physical mass and imaginary mass. A particle with an imaginary mass is unstable. It is unstable over time, undergoing a phase transition where it loses its mass. This
process is referred to as tachyon condensation. It is related to Higgs boson and ferromagnetic properties. The exact definition of a particle is still under debate, but these two terms are a good
The inertial mass of an object is the amount of resistance to acceleration when a force is applied to it. This concept was championed by Ernst Mach, and was later developed by Percy W. Bridgman. This
definition of mass differs from the special relativity concept of mass. The inertial mass of an object increases with the amount of acceleration the centre experiences. Therefore, a smaller mass can
accelerate faster than a larger mass.
A kilogram is the standard SI mass unit for measuring an object. However, this unit isn’t always the best choice for measuring the mass of a solid object. It is more accurate than gram and can be
converted into other units that are more appropriate for different applications. For example, a pound of gold weighs more than a kilogram of helium. If you’re looking to buy a gold bar, you should
know the exact weight of a piece of metal before purchasing it. | {"url":"https://www.basculasbalanzas.com/mass-measurement-2/","timestamp":"2024-11-10T16:11:42Z","content_type":"text/html","content_length":"52958","record_id":"<urn:uuid:f6201115-23ca-4bf4-a654-534ec05fa535>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00444.warc.gz"} |
Total variation-based neutron computed tomography
We perform the neutron computed tomography reconstruction problem via an inverse problem formulation with a total variation penalty. In the case of highly under-resolved angular measurements, the
total variation penalty suppresses high-frequency artifacts which appear in filtered back projections. In order to efficiently compute solutions for this problem, we implement a variation of the
split Bregman algorithm; due to the error-forgetting nature of the algorithm, the computational cost of updating can be significantly reduced via very inexact approximate linear solvers. We present
the effectiveness of the algorithm in the significantly low-angular sampling case using synthetic test problems as well as data obtained from a high flux neutron source. The algorithm removes
artifacts and can even roughly capture small features when an extremely low number of angles are used.
The High Flux Isotope Reactor (HFIR) CG-1D cold neutron imaging beamline serves a broad science portfolio such as energy research,^1 nuclear materials research,^2 physics,^3 plant physiology,^4
engineering,^5 and archeology.^6 Neutrons are capable of detecting light elements such as hydrogen and lithium due to their ability to penetrate high-Z materials. Thus, neutron computed tomography
(CT) offers a complement to imaging techniques such as X-ray CT. Multimodal methods (e.g., Refs. 7–9) have been developed for samples with both high- and low-Z materials; however, our current focus
lies in the high-Z case, and so we rely on neutron CT. However, in neutron based tomography, an intrinsic problem is the relatively amounts of neutrons that can be produced as opposed to x-rays and
electrons probes.^10 This results in, for a fixed time, either measurements with relatively few events or measurements requiring longer acquisition times. Our interest lies in the latter scenario.
Parallel to the advancements in the field of neutron computed tomography there have been significant strides in the mathematical reconstruction of tomographic imaging. One of the workhorse algorithms
of (neutron) tomography is the filtered back-projection (FBP) method.^10,11 In recent years, there has been a revolution in the methods used in tomography as a result of the advent of ℓ^
1-regularization based optimization methods.^12–15 It is now possible to recover exactly, using these methods, the benchmark Shepp-Logan phantom problem.^15 The advances in image reconstruction
algorithms for computed tomography^16 have made great strides using iterative reconstruction methods, especially in variations of the Algebraic Reconstruction Technique; see Refs. 17–21 and
references therein. A focus around the development of these techniques has been to reduce the number of projections needed for reconstruction.^22 The current study explores the use of total
variational approaches for neutron tomography, specifically focusing on improving the computational time needed for reconstructions in scenarios where only low sampling measurement rates hold.
The general form of the inverse problem in neutron tomography is given as finding the attenuation coefficient, μ, of an object such that
where δ is the Dirac delta function, the variable σ is the noise in the measurement, I[0] and I are the incoming and final neutron intensity measured as a function of position t, and the angle of
rotation [with respect to the (x, y) Cartesian coordinates] is θ. A schematic of the problem is in shown in Fig. 1. The accuracy in determining μ depends critically upon the number of angles and
positions that can be measured and the signal-to-noise ratio of the measurements. Our goal is to develop an algorithm which can perform the reconstruction—despite under-resolved angular data—by
incorporating recent methods in ℓ^1 image reconstruction.
In practice, given data I/I[0], the inverse problem is written as solving the following minimization problem:
where Φ represents a regularizing term which promotes certain desired structures in the reconstructed attenuation object μ. Here α > 0 is a parameter which balances between the reconstruction’s
fidelity to measured data and its regularity. An equivalent way of writing this problem is as follows:
where the L^2 norm is for functions of the variables θ and t and R[θ] is the Radon transform associated with the angle θ. It has been proven recently that, for piecewise constant images, the best
choice for Φ in Eq. (2) is total variation (TV) based regularization.^15 In a discrete setting (such as practical imaging applications), this means that Φ(μ) takes the form
In contrast to Refs. 23 and 24, we are here using a standard L^2 fidelity penalty term, meaning we are not assuming a Poisson model for the noise in the detector. As noted above, a challenge in
neutron imaging is the trade-off between increased acquisition time for a projection against the relatively low number of counts the detector experiences. Our application is focused on problems where
we allow for increased acquisition times per angular measurement such that a Gaussian model is a reasonable simplification; with a reduction in the number of needed projections for a reconstruction,
this increase does not contribute significantly to the overall time required for measurements. Regarding the second functional Φ in the objective, other penalty terms, such as polynomial
annhilation-based terms, have been developed for imaging with more complicated structures.^25–27 Our present focus is on neutron imaging where a piecewise constant μ is of primary interest. However,
polynomial annihilation (PA)-based methods can be easily included using the methods described in the present work.
In recent years, ℓ^1 regularization has received considerable attention in the design of image reconstruction algorithms from under-sampled and noisy data for images that have some measurable
features with a sparse representation; this is compatible in applications where a relatively small number of measurements are taken.^13,15,28 It is in general still difficult to develop efficient and
robust techniques for solving such image reconstruction problems.
The split Bregman algorithm^29 is a numerically efficient and stable algorithm that has successfully solved the l^1 regularized reconstruction problem for a variety of applications. In this paper, we
extend the split Bregman algorithm as a launching point to develop a new technique for solving the neutron tomography inverse problem using Eq. (4) as a penalty term. We will demonstrate that our
method yields improved accuracy in regions away from discontinuities, especially in the case of under-sampled data. We will adopt the standardizations and terminology from Ref. 29 to describe our
algorithm. Recent work on ℓ^1 regularization for x-ray based computed tomography has relied on a Douglas-Rachford splitting algorithm.^30 While there is an equivalence under suitable reformulations
between exact split Bregman and Douglas-Rachford,^31 the error-forgetting nature of split Bregman^32 gives, in our experience, good convergence behavior despite using very inexact updates which are
relatively inexpensive to compute. Previous work on using sparse-angle neutron-based CT has used split Bregman,^21,22 however, mainly in order to solve exact updates within a hybrid reconstruction
A well-known drawback in using the TV ℓ^1 regularization term is that the reconstructed image defaults to a piecewise constant approximation. While suitable for some applications, in others it is
desirable to see additional details. For example, total generalized variation (TGV), developed in Ref. 33, and multi-wavelets in Ref. 34, allow for finer details in smooth regions of the
reconstruction. The already-mentioned polynomial annihilation (PA) transform was originally designed in Ref. 35 and used in Ref. 36 to demonstrate that encouraging a sparsity of edges in the
underlying image yields improved accuracy for both image reconstruction and edge identification. However, in the present application, we have anecdotally seen that TV provides the best
In order to solve (3) with regularizer ||∇μ||[1] efficiently in multiple dimensions, we will use the split Bregman algorithm, which we present below. The split Bregman algorithm, developed in Ref. 29
, was shown to be equivalent to the Bregman iteration. Its popularity is due to its speed and that its nonlinear steps involve only soft thresholding; meanwhile all other aspects of the algorithm
involve solving invertible linear systems. In recent years, the split Bregman algorithm has been used for solving a broad class of l^1 regularized optimization problems. In particular, the split
Bregman algorithm has been successfully and efficiently used in MRI reconstructions in which context (3) was called the “sparse MRI data reconstruction problem.”^37–39 We now describe how to
efficiently solve (3) with the TV penalty term via an extension of the split Bregman algorithm for TV denoising as described in Ref. 29.
We begin by letting ∇[x] and ∇[y] denote the respective directional differential. Additionally, throughout, for a linear operator A, its adjoint is denoted by A^*. We denote by M the linear operator
which restricts the Radon transform of μ to angles along which measurements have been taken to generate I/I[0]. The split Bregman algorithm becomes a sequence of minimization problems of the form
$μk+1=minμ∑i,j|dx,i,jk|2+|dy,i,jk|2+λ2||dxk−∇xμ−bxk||22+ λ2||dyk−∇yμ−bxk||22+α2||MRθμ−I^k||22,$
$dxk+1,dyk+1=mindx,dy∑i,j|dx,i,j|2+|dy,i,j|2+λ2||dx−∇xμk−bxk||22+ λ2||dy−∇yμk−bxk||22+α2||MRθμk−I^k||22$
Following the structure of Ref. 29, the above can be rewritten as
We note that the update rule for μ^k+1 is simply solving the optimality system for the first L^2 optimization in (5). Additionally, one may modify this as in Ref. 29 so that several updates on
$μk+1,dxk+1,dyk+1,bxk+1,byk+1$ can be performed in an inner loop before updating $I^k+1.$ In practice, the most expensive portion of the algorithm is inverting $(αRθ*M*MRθ−λ(∇x)T∇x−λ(∇y)T∇y)$.
Unlike in Ref. 29, there do not appear any reasonable methods for determining a closed form solution to this linear system. However, the error-forgetting properties of the Bregman iteration discussed
there and in Ref. 32 mean that the inversion in the update rule for μ^k+1 does not need to be computed exactly in order for the iterations to converge. That is, we typically only need an
approximation for the update in order for the split Bregman algorithm to converge to a solution. In light of this property, instead of fully solving
we instead only compute an approximate solution via the conjugate gradient (CG) algorithm. This approximation can be rather inexact without affecting the overall convergence of the algorithm.^32
Therefore, we replace the update rule with
where $CG$ (A, b, x[0], m) denotes the result of the conjugate gradient (CG) algorithm in solving Ax = b with initial guess x[0] and a maximum number of iterations M.
This means that, in total, K split Bregman iterations require only a total of 2K (1 + N) evaluations of the forward and adjoint Radon transform; as these two transforms are the most expensive to
evaluate, we in practice look to reduce N while still maintaining a good convergence. For a square μ, let N[x] be the number of pixels in either direction and N[θ] be the number of angular
measurements taken. For fixed M, then, the complexity of computing rhs^k, μ^k+1, and $I^k+1$ is dominated by the $O(Nx2Nθ)$ Radon transform and its adjoint, as the Laplacian operators are $O(Nx2)$.
The update rules for $sk, dxk, dyk, bxk, byk$ are explicit and $O(Nx2)$. Thus, the overall complexity of an iteration scales with MN[x]N[r]N[θ]. Anecdotally, M can be very small; in the results of
Sec. III, we use M = 5, 10 and still obtain reconstructions which remove the artifacts typically associated with under-resolved angular measurements.
We should briefly remark on the importance of properly constructing R^*. It is common to construct an approximate inverse Radon transform which often includes a filter or mask to improve the
approximate inverse (see, e.g., Ref. 40 for recent work in this direction). In the case where angular measurements fully resolve the data, this approach works reasonably well for our algorithm.
However, we find that it tends to require a rather large number of split Bregman iterations to achieve a good reconstruction. Furthermore, in the case where M corresponds to subselecting angles (that
is, when few angular measurements are taken), this approximation of R^* breaks down. The close connection between R^* and the approximate inverse Radon transform no longer holds, resulting in
Algorithm 1 diverging and leading to arbitrarily large and oscillatory μ. Typically, implementing a discrete Radon transform consists of three separate operations for a series of angles: rotation of
data by the angle, interpolation of this result to an appropriate grid, and summation of the interpolated data over the relevant axis. Construction of R^* requires taking the proper adjoint of each
of these operations. Anecdotally, properly determining the adjoint of the interpolation operation for the particular geometry of the data set is vital to obtaining a convergent and efficient
Algorithm 1.
Algorithm 1
: Initialize k = 0, μ^0 = R^*M^*(I/I[0]), $I^0=(I/I0)$, and$bx0=by0=dx0=dy0=0$
: while ||MR[θ]μ^k − I/I[0]||[2] > σ
: $rhsk=αRθ*MTI^k+λ(∇x)T(dxk−bxk)+λ(∇y)T(dyk−byk)$
: $μk+1=αRθ*M*MRθ−λ(∇x)T∇x−λ(∇y)T∇y−1rhsk$
: $sk=|∇xμk+bxk|2+|∇yμk+byk|2$
: $dxk=max(sk−1/λ,0)∇xμk+bxksk$
: $dyk=max(sk−1/λ,0)∇yμk+byksk$
: $bxk=bxk+1+(∇xfk+1−dxk+1)$
: $byk=byk+1+(∇yfk+1−dyk+1)$
: $I^k+1=I^k+(I/I0)−MRθμk+1$
: k = k + 1
: end
: Initialize k = 0, μ^0 = R^*M^*(I/I[0]), $I^0=(I/I0)$, and$bx0=by0=dx0=dy0=0$
: while ||MR[θ]μ^k − I/I[0]||[2] > σ
: $rhsk=αRθ*MTI^k+λ(∇x)T(dxk−bxk)+λ(∇y)T(dyk−byk)$
: $μk+1=αRθ*M*MRθ−λ(∇x)T∇x−λ(∇y)T∇y−1rhsk$
: $sk=|∇xμk+bxk|2+|∇yμk+byk|2$
: $dxk=max(sk−1/λ,0)∇xμk+bxksk$
: $dyk=max(sk−1/λ,0)∇yμk+byksk$
: $bxk=bxk+1+(∇xfk+1−dxk+1)$
: $byk=byk+1+(∇yfk+1−dyk+1)$
: $I^k+1=I^k+(I/I0)−MRθμk+1$
: k = k + 1
: end
We present here the results of implementing the TV-penalized reconstruction via split Bregman-based methods. After performing some tests on synthetic data—the Shepp-Logan phantom—we present
reconstructions from measurements on two different samples: a steel-aluminum cylinder and a fuel injector. These latter data sets are three-dimensional, as is the quantity of interest: the value of μ
on each voxel. We solve in parallel the reconstruction of 2-dimensional slices where the normal of the slice is the axis of rotation about which angular measurements are taken. This results in
horizontal slices of μ which may then be stacked vertically to generate the volumetric attenuation coefficient. For the data in Secs. III C and III D, neutron computed tomography was performed at the
Oak Ridge National laboratory (ORNL) High Flux Isotope Reactor (HFIR) CG-1D neutron imaging beamline.^41 In order to test the performance of inexact split Bregman with sparse angular measurements, we
compare the reconstructions using a large number of projections against reconstructions using a reduced set of evenly spaced projections. The software implementation of the algorithm was developed to
be deployable on a hybrid high performance computing system. Reconstructions of volumetric data sets of size 2049 × 2049 × 2049 with up to 900 measurements typically take approximately 5 min using
split Bregman on Titan at the Oak Ridge Leadership Computing Facility at ORNL where each slice is processed on a single computational node. Specifically, the software includes a graphical processing
unit (GPU) implementation of the forward and adjoint Radon transforms. On smaller systems with GPU accelerators, similar reconstruction times per slice are observed; the total time then becomes a
function of the number of cores available to run in parallel since each slice requires one core. On a smaller server with 44 cores (and accelerators), reconstructions typically take under half a day.
Thus, even on smaller machines, reconstructions typically take less time to perform than that required for performing a handful of angular measurements.
A. Synthetic data
We first perform the reconstruction algorithm on the 257 × 257 Shepp-Logan phantom using 101 evenly spaced angular measurements; the phantom and the associated sinograms from these measurements are
shown in Fig. 2. We choose this number of angular measurements to firmly be in the undersampled regime, corresponding to a measurement frequency 1/16-th of that obtained from Shannon’s sampling
theorem.^42 First, we focus on establishing the performance of the split Bregman-based reconstruction when we do not fully solve the matrix inversion in (7). To do this, we perform the split Bregman
algorithm with different choices of the maximum number of CG steps allowed in the computing of the μ update. The results for varying the numbers of CG steps used are given in Fig. 3. Regardless of
the number of CG steps used in computing the μ^k+1 update, the algorithm converges. However the quality of the reconstruction after a given number of iterations increases with the number of CG steps.
After 200 iterations, there is a difference of 2 orders of magnitude in the accuracy of the reconstruction when using 10 CG steps and 100 CG steps. However the latter uses roughly 20 times the number
of evaluations of a forward or backward Radon transform. One can see that after approximately 100 iterations using 10 CG steps per iteration one achieves an accuracy of around 10^−4 error; a similar
error can be reached after approximately 50 iterations using 100 steps per iteration. However, the time needed for performing those 50 iterations is nearly 20 times that needed for performing 100
iterations using 10 CG steps. This is due to the cost of repeated evaluations of the Radon transform and its adjoint dominating the costs of the computations in the Bregman algorithm.
Besides choosing the number of CG steps to use per iteration, the other primary parameter to be selected is α. The method of selecting the proper weight between fidelity and regularization terms is
in general often an involved process, beyond the scope of this work; in many problems it remains more of an art relying on the expertise of the user. We show here the results of performing the split
Bregman algorithm with different values of α; the results are shown in Fig. 4. As expected, with too large α, the fidelity term in the objective in (3) dominates and some high frequency artifacts can
begin to appear, as seen in Fig. 4(j). With too small an α, the regularizer dominates the reconstruction, suppressing some features, such as the smaller features in the phantom, as seen in Fig. 4(b).
We also perform the same tests in the case of noised sinograms. The sinograms are additively noised by white noise; the resulting data have a signal-to-noise ratio (SNR) of 40 dB. We see in Fig. 5
that the added noise does not affect the ability of the algorithm to converge regardless of the number of CG steps used in each μ update. In Fig. 6, we see that, again, varying α changes the
reconstruction in the same way as in the noiseless case. However, the additional noise in the given data results in the need to change the scale of α. At lower SNR levels in the sinograms, the range
of α giving reasonable reconstructions becomes narrower and thus harder to find. In such cases, preprocessing of the sinograms becomes necessary, in our experience.
B. Preprocessing
We perform two main preprocessing steps to the non-synthetic data below before solving the tomographic inverse problem (2). In order to reduce streaking in the reconstructed μ, we apply a filter to
ln(I/I[0]) which replaces pixels sufficiently deviating from the mean of their 8 neighbors as in Ref. 43. If a pixel deviates from the mean of its neighbors, we replace its value with that mean. The
second step corrects for possible shifts in the axis of rotation. In the presence of errors in the rotation, reconstructions using filtered back-projection may cause “tuning fork”-type artifacts to
appear.^44 Anecdotally, the algorithm in Sec. II often views such artifacts as features of the data which are to be preserved, instead of as noise to be suppressed. We thus apply to the sinograms the
algorithm for correcting such errors presented in Ref. 44.
C. Cylindrical sample
First, we consider a cylindrical geometry sample made with two metals, aluminum surrounding a stainless steel rod, which have different neutron attenuations. This sample was chosen as its geometry
provides a reasonable benchmark along horizontal slices. A schematic of the sample is shown in Fig. 7. The cylindrical sample was rotated at a step angle of Δθ = 0.5° between 0° and 182.6°; a neutron
radiograph of 90 s was collected. This step angle was chosen in order to obtain a reasonably good reconstruction via filtered back projection using a parallel beam (an approximation that is valid for
a neutron beam with low divergence); despite the undersampling in the angular variable (with respect to the Nyquist rate), the resulting high frequency is relatively minor.
We show in Fig. 8 the attenuation obtained after applying either a filtered back projection, Simultaneous Algebraic Reconstruction Technique (SART) (using the implementation of Ref. 45), or the
algorithm from Sec. II to a horizontal slice. In each case, we expect an outer annulus of low attenuation (corresponding to the aluminum) surrounding a disc of high attenuation (corresponding to the
steel). The reconstructions in the first row are performed using all available portions of ln(I/I[0]). In Fig. 8(a), we show the result of using a filtered back projection, while in Fig. 8(b), we
show the result of using SART. In Fig. 8(c), we show the result of using 200 split Bregman updates and 5 conjugate gradient steps to approximate the update in μ^k+1; we set α = λ = 1 in this case.
The pictures seem to be in agreement, as expected. However, looking at the values of each along the green dashed lines, we see that the filtered back projection gives some high frequency artifacts;
this is shown in Fig. 8(d). Using a spacing of Δθ = 0.5° results in an under-resolved (with respect to the Shannon sampling theorem) regime for the entire image as the sinograms are 2048 pixels per
projection, with the cylindrical sample occupying 900 pixels per projection, thus explaining the presence of small oscillatory artifacts.
The same procedure is performed in the second row of Fig. 8; however, we use an M which selects every 16th angle only, corresponding to uniformly reducing the number of angular measurements taken for
a reconstruction with a spacing of Δθ = 8.8°. This undersampling results in a filtered back projection with significant artifacts, as is seen in Fig. 8(e). The reconstruction from SART, seen in Fig.
8(f), also exhibits significant high frequency artifacts, while also producing blurring at the annular interface; further iterations using this technique continued to produce reconstructions with
such artifacts. However, the total variation penalty nearly completely suppresses such artifacts, resulting in the reconstruction shown in Fig. 8(g). The improvement can be seen even more clearly in
the profiles of each reconstruction in Fig. 8(h). Here we see that the under-resolved angular measurements lead the reconstruction from FBP to suffer from large oscillations in the attenuation
coefficient and the SART reconstruction to blur the interfaces. Meanwhile, the TV reconstruction is again able to suppress these artifacts and recover the expected geometry of the sample.
D. Gasoline direct injector
We also perform reconstruction on a more complex geometry, a gasoline direct injection (GDI) fuel injector, which has seen increased deployment in spark ignition engines.^46,47 The physics of GDI
fluid dynamics is strongly dependent on native injector geometry, manufacturing tolerance, and wear. To improve the design, production methods, and materials of GDI injectors, a strong understanding
of the internal “as-produced” geometries is required. Specifically, internal injector geometry dramatically affects the fuel spray and associated system behavior.^48,49 The unique capability of
neutrons to penetrate the injector and visualize hydrogen-rich fuel flow enables the connection of the internal and external flows to more fully describe the physics of the fuel spray.
In Fig. 9, we show the reconstructed attenuation [via FBP and solving Eq. (4)] for a fuel injector along a horizontal slice close to the base of central rod in the injector. Again, we first use all
angular measurements taken (corresponding to a sampling rate of Δθ = 0.19°) and perform FBP to obtain the attenuation in Fig. 9(a). Again, this step angle is chosen to obtain a reconstruction. We see
in Fig. 9(b) that applying instead TV-based regularization (using the same parameters as above) gives an attenuation which largely agrees with the FBP reconstruction; this is also clear in comparing
the profiles of each in Fig. 9(c). Next, we again consider the reconstructions one gets using subsampled angular data. We now take an angular sampling rate of Δθ = 3.1°. The resulting FBP-derived
attenuation is shown in Fig. 9(d). Here we see again high frequency artifacts, especially near the outer edge of the sample. Although they appear to be of relatively low magnitude, the effect of
these artifacts can be more easily seen in the profile of the attenuation in Fig. 9(f). We also obtain an attenuation via split Bregman; this is shown in Fig. 9(e), which agrees largely with the
attenuations obtained from the full set of angular measurements. We again see the reduction of artifacts, especially the oscillations in the background near the edge of the injector.
We next repeat the reconstruction of the attenuation via FBP and TV across all horizontal slices; this was performed using both the full set of angular measurements and the subsampled (again with Δθ
= 3.1°) data. In Fig. 10, we use all of these slices to generate a volumetric attenuation; a vertical cross section down the center of the resulting reconstruction of μ is shown. We see again
agreement between FBP [in Fig. 10(a)] and TV [in Fig. 10(b)] when all measurements are available. However, in Fig. 10(c), we see that using FBP on the slices in the case of subsampled data results in
the high frequency artifacts noted above throughout the volume. In keeping with the above, the TV regularization suppresses this throughout the volume, as seen in Fig. 10(d).
Finally we demonstrate the ability of TV-based reconstruction to preserve fine detail structures in the case of extremely undersampled data. In Fig. 11, we show the results of reconstructing a
horizontal slice of the fuel injector at its nozzle. We show a detail of this reconstruction of the central 202 × 202 portion of the slice. We are interested in recovering the positions of the
channels in the nozzle of the injector, under extreme subsampling. Our motivation is that in applications where dynamic images are required, acquisition times are increased; it may be possible to
(partially) offset this by reducing the required number of angular measurements.
We show the result of performing FBP using an angular spacing of Δθ = 0.19°, in Fig. 11(a). If we only take 15 angular measurements with spacing of Δθ = 12.8°, the attenuation coefficient from FBP,
shown in Fig. 11(b), suffers from the coarseness of the data. Significantly, the locations and sizes of the channels are relatively obscure due to the large artifacts visible.
We show the result of taking this coarse data and performing TV-based regularization in Fig. 11(c). Letting α = 20, we take a maximum of 10 conjugate gradient steps to calculate each update in μ; 200
Bregman iterations are used. We see that while the TV-based reconstruction exhibits staircasing, seen in the shape of the channels and the outer ring (generated by neutron refraction), the position
of the channels is preserved and given a stronger contrast than using FBP while also removing the significant artifacts that are present in FBP. We note that staircasing is not unexpected due to the
extremely low resolution (a reduction by a factor of 64 from the reference) in the angular measurements resulting in very coarse data for fitting. In general, it is unclear at which point angular
undersampling will result in poor reconstructions. At fine spatial resolutions, the high frequency artifacts–which undersampling introduces–may appear as features near regions where the attenuation
coefficient has a low contrast with the back ground medium. In this case, the low rank of the fidelity term in (3) will lead to poor reconstructions. However, the step angle where this occurs will
depend on the geometry of the sample being measured. Additionally complicating this lack of clarity is that the choice of α can often dramatically alter the quality of the reconstruction. As
mentioned previously, choosing α remains an art; it is possible in some cases that appropriately choosing the parameter will encourage suppressing the high frequency artifacts in without suppressing
low-contrast samples. For more general materials, therefore, the maximum step angle needed for a good reconstruction remains an open question.
Total variation-based reconstruction is suitable in neutron-based computed tomography, reducing and removing artifacts which arise from applying filtered back projection methods to under-resolved
samples. Inexact split Bregman is especially suitable for the efficient solution of the total variation-penalized problem. Only a very small number (less than 10) conjugate gradient steps are needed
to generate an acceptable update direction, reducing the computational overhead needed in iterative procedures by reducing the number of calls to the Radon transform and its adjoint. This
reconstruction technique is especially effective in situations where the number of angular measurements taken is limited due to time constraints or difficulties with the sampling environment;
filtered back projections generate significant high-frequency artifacts in such cases which are not present in the total-variation based reconstructions. This can be seen even in situations where an
extremely limited number of angular measurements are taken. In these cases, staircasing is inevitable, but the ability of TV-based reconstruction to identify the location of features even in the
severely under-resolved case suggests its suitability for future problems in more dynamic neutron-based computed tomography applications.
Part of this research conducted at the High Flux Isotope Reactor was sponsored by the Scientific User Facility Division, Office of Basic Energy Sciences, U.S. Department of Energy. This research used
resources of the Oak Ridge Leadership Computing Facility at the Oak Ridge National Laboratory, which is supported by the Office of Science of the U.S. Department of Energy under Contract No.
This manuscript has been authored by UT-Battelle, LLC under Contract No. DE-AC05-00OR22725 with the U.S. Department of Energy. The United States Government retains and the publisher, by accepting the
article for publication, acknowledges that the United States Government retains a non-exclusive, paid-up, irrevocable, worldwide license to publish or reproduce the published form of this manuscript,
or allow others to do so, for United States Government purposes. The Department of Energy will provide public access to these results of federally sponsored research in accordance with the DOE Public
Access Plan (http://energy.gov/downloads/doe-public-access-plan).
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Final Thoughts on Ratio Analysis
DuPont Equation
The DuPont equationMerges ROA and ROE. Breaks it down into profit on a company’s sales and return on a company’s assets. is a handy way to analyze a company’s financial position by merging the
balance sheet and income statement using measures of profitability. DuPont merges ROA and ROE. ROA is now defined using two other ratios we calculated: net profit margin and total asset turnover. ROA
was calculated as net income divided by total assets.
$Return on Assets= Net IncomeTotal Assets$
ROE was calculated as net income divided by earnings available to shareholders (common equity).
$Return on Equity= Net IncomeCommon Equity$
The DuPont equation defines ROA as follows:
$ROA= Net Profit Margin×Total Asset Turnover$
Remembering that net profit margin is earnings available for shareholders divided by sales and total asset turnover is sales divided by total assets, we can make the following substitutions:
$ROA= earnings available for shareholderssales ×salestotal assets$
From this we see that sales will cancel out and ROA will become earnings available for shareholders divided by total assets.
$ROA= earnings available for shareholderstotal assets$
This will give us the same number for ROA that was calculated using the original formula. However, the DuPont equation breaks it down into two components: profit on a company’s sales and return to
the use of a company’s assets.
Trend Analysis, Comparative Ratios and Benchmarking
Just as important as the actual numbers is the numbers value over time. Trends in ratios tell us a lot about a company and can indicat if a company is trending favorably or unfavorably. Just as time
series data is important—so is cross-sectional. We may have what we consider a fantastic ratio, but it may be low for our industtry. Ratios are particularly useful to compare to other companies and
competitors. Ratios are available for industries and a company can see how it compares to its competitors. | {"url":"https://2012books.lardbucket.org/books/finance-for-managers/s04-06-final-thoughts-on-ratio-analys.html","timestamp":"2024-11-04T21:28:13Z","content_type":"text/html","content_length":"13677","record_id":"<urn:uuid:ad0dd846-5804-4bd2-b9a9-5ba302a65e44>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00438.warc.gz"} |
Does a pure math major need to take ODEs?
• Programs
• Thread starter Konradd
• Start date
In summary, Right now I'm a sophomore at a state uni with hopes of getting into graduate school in pure mathematics. When I was a freshman, I surveyed the three major areas of math - analysis,
algebra, and topology - and I decided that analysis was for me. Although I did very well in Algebra, I found it unbearably boring (is this a problem?) Anyway, as sophomore I've taken measure and
lebesgue theory, complex variables, graduate group theory and rings all of which were graduate courses, and now I'm taking functional analysis, and probability, and algebraic topology. However, I was
told I need to take ODEs - not even a proof heavy one. Just the
Right now I'm a sophomore at a state uni with hopes of getting into graduate school in pure mathematics.
When I was a freshman, I surveyed the three major areas of math - analysis, algebra, and topology - and I decided that analysis was for me. Although I did very well in Algebra, I found it unbearably
boring (is this a problem?) Anyway, as sophomore I've taken measure and lebesgue theory, complex variables, graduate group theory and rings all of which were graduate courses, and now I'm taking
functional analysis, and probability, and algebraic topology.
However, I was told I need to take ODEs - not even a proof heavy one. Just the typical "how to solve ODEs" course. I think this is ridiculous..as I could probably learn the material in a couple of
weeks. Any advice on how to get exempt from this requirement?
Thanks everyone.
I can't comment to specifically on whether or not you should take the class but when I wanted to skip over the first semester of physics I just asked the department chair if I could and he said ok.
Konradd said:
I think this is ridiculous..as I could probably learn the material in a couple of weeks. Any advice on how to get exempt from this requirement?
If it is that easy I don't see why its that big of a problem. If however you are serious about getting rid of it, why don't you see if your school will let you challenge the course/pass out of it,
then you don't have to take it. If not your kind of out of luck, not like its going to be a totally useless course.
Konradd said:
Does a pure math major need to take ODEs?
What do you intend to do with that doctorate in mathematics? The great majority of doctorates become college teachers and, even if you do not intend to do that, you would certainly want to keep that
option open. And if you do teach at a college you will very likely be asked to teach differential equations at some point.
Science Advisor
Homework Helper
calculus is basically about solving differential equations. indeed differential equations are one of the most basic topics in all of mathematics. take a look at the book on ode by arnol'd to find out
how interesting it can be. morse theory in differential topology is basically a consequence of the fundamental theorem of ode. take a look at the little book by wallace.
If you don't know ode it is hard to imagine understanding pde. hodge theory in geometry and topology is basically about solutions of a basic pde, the laplace equation. theta functions in algebraic
geometry are best understood by their fundamental relation to the heat equation another basic pde. deRham cohomology gives the relation between topology and a fundamental differential equation. The
important topic of vector fields on manifolds is the geometric version of differential equations.
The whole subject of linear algebra, in particular the structure of jordan normal forms, is essentially describing the behavior of the simplest differential operator d/dx on a finite dimensional
space of exponential and polynomial functions. the basic existence theorem in ode is usually proved by a beautiful "contraction" mapping technique in a general metric space that is very interesting.
complex analysis is entirely about the solutions of one diff equation ∂/∂zbar = 0.
i suppose someone could teach an ode course that failed to transmit any of these insights, but I think you are advised to learn as much as possible about differential equations in order to understand
pure mathematics, especially analysis. I admit I was like you as a young student and hated ode, viewing it as a plug and chug course with no value. This is the fault of the curriculum which declines
to point out the connections between all these subjects. Linear algebra is one of the worst for refusing to teaching differential operators as indeed the most important linear transformation. Linear
algebra is not about reducing and solving trivial systems of numerical equations.
Last edited:
If it's an easy as you think it will be, just look at it as a chance to reinforce your pure mathematican's prejudices about applied math
Without a doubt you need it. You will learn more than you can imagine.
Teach it to yourself over the summer. Basic no proof odes classes can be really easy. If you teach yourself they will probably let you skip it.
I think you should take the ODE class and feel lucky that their requirements are that loose to begin with! In my university, I'm forced to take classes like Number Theory, Advanced Statistics,
Mathematical Introduction to Fluid Dynamics, etc in my mathematics bachelor degree (for example I asked if I could take Functional Analysis or Measure Theory in place of one of them, but was not
allowed). Not that some of them can't be interesting, but it goes to show how much more demanding a department can be. I think demanding an ODE class is pretty darn basic.
FAQ: Does a pure math major need to take ODEs?
1. What is ODEs and why is it important for a pure math major to take it?
ODEs stands for Ordinary Differential Equations, which are mathematical equations that involve derivatives of a function with respect to one independent variable. It is important for a pure math
major to take ODEs because it is a fundamental course in mathematics that is used in many fields such as physics, engineering, and economics. It also helps develop critical thinking and
problem-solving skills.
2. Can a pure math major succeed without taking ODEs?
While it is possible for a pure math major to succeed without taking ODEs, it is highly recommended to take the course as it provides a strong foundation for higher level math courses and for
real-world applications. Moreover, ODEs is often a prerequisite for many advanced math courses.
3. What topics are typically covered in an ODEs course?
An ODEs course typically covers topics such as first and second-order differential equations, systems of differential equations, existence and uniqueness of solutions, and methods of solving
differential equations (e.g. separation of variables, integrating factors, and series solutions).
4. Are there any math majors that do not require taking ODEs?
It depends on the specific requirements of the math major and the university. Some math majors may have alternative courses that cover similar topics to ODEs. However, ODEs is a commonly required
course for most math majors as it is a fundamental part of the subject.
5. Are there any resources available to help a pure math major with ODEs?
Yes, there are many resources available to help a pure math major with ODEs. These include textbooks, online tutorials, practice problems, and study groups. It is also recommended to seek help from
the professor or teaching assistants if needed. | {"url":"https://www.physicsforums.com/threads/does-a-pure-math-major-need-to-take-odes.585367/","timestamp":"2024-11-05T19:08:38Z","content_type":"text/html","content_length":"117999","record_id":"<urn:uuid:b874a687-f37f-435a-a671-82691f81a796>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00481.warc.gz"} |
Gravitational wave spectroscopy of binary neutron star merger remnants with mode stacking
A binary neutron star coalescence event has recently been observed for the first time in gravitational waves, and many more detections are expected once current ground-based detectors begin operating
at design sensitivity. As in the case of binary black holes, gravitational waves generated by binary neutron stars consist of inspiral, merger, and postmerger components. Detecting the latter is
important because it encodes information about the nuclear equation of state in a regime that cannot be probed prior to merger. The postmerger signal, however, can only be expected to be measurable
by current detectors for events closer than roughly ten megaparsecs, which given merger rate estimates implies a low probability of observation within the expected lifetime of these detectors. We
carry out Monte Carlo simulations showing that the dominant postmerger signal (the =m=2 mode) from individual binary neutron star mergers may not have a good chance of observation even with the most
sensitive future ground-based gravitational wave detectors proposed so far (the Einstein Telescope and Cosmic Explorer, for certain equations of state, assuming a full year of operation, the latest
merger rates, and a detection threshold corresponding to a signal-to-noise ratio of 5). For this reason, we propose two methods that stack the postmerger signal from multiple binary neutron star
observations to boost the postmerger detection probability. The first method follows a commonly used practice of multiplying the Bayes factors of individual events. The second method relies on an
assumption that the mode phase can be determined from the inspiral waveform, so that coherent mode stacking of the data from different events becomes possible. We find that both methods significantly
improve the chances of detecting the dominant postmerger signal, making a detection very likely after a year of observation with Cosmic Explorer for certain equations of state. We also show that in
terms of detection, coherent stacking is more efficient in accumulating confidence for the presence of postmerger oscillations in a signal than the first method. Moreover, assuming the postmerger
signal is detected with Cosmic Explorer via stacking, we estimate through a Fisher analysis that the peak frequency can be measured to a statistical error of ∼4-20 Hz for certain equations of state.
Such an error corresponds to a neutron star radius measurement to within ∼15-56 m, a fractional relative error ∼4%, suggesting that systematic errors from theoretical modeling (â‰100 m) may dominate
the error budget.
All Science Journal Classification (ASJC) codes
• Physics and Astronomy (miscellaneous)
Dive into the research topics of 'Gravitational wave spectroscopy of binary neutron star merger remnants with mode stacking'. Together they form a unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/gravitational-wave-spectroscopy-of-binary-neutron-star-merger-rem","timestamp":"2024-11-09T16:47:33Z","content_type":"text/html","content_length":"58312","record_id":"<urn:uuid:d937b3bf-205e-44f0-bb1a-4b77360b3b13>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00100.warc.gz"} |
Substitution in Cipher
Around 100 BC, Julius Caesar used to send secret messages to his ministers in the war field using a cryptography method that gave him a tremendous strategic advantage.
Even if his messages got intercepted during the event, his enemies could not read them. This oldest cryptography method, named after Julius Caesar, is today known as 'Caesar Cipher.' So are you also
wondering what the code that helped him to hide the messages from enemies was?
Let's now compare this situation to today's era. Nowadays, everyone wants their data to be secured. Cryptography is the process of hiding information. Many cipher algorithms are used for encrypting
and decrypting the data.
Before diving deep into the Substitution in Cipher, let's first quickly understand what a cipher is.
A cipher, also called an encryption algorithm, is a method for encrypting and decrypting data.
A cipher converts the original message into ciphertext using a key. This encrypted, unreadable text is known as ciphertext. The study of cryptographic techniques is known as cryptology or | {"url":"https://www.naukri.com/code360/library/substitution-in-cipher","timestamp":"2024-11-03T19:41:44Z","content_type":"text/html","content_length":"621109","record_id":"<urn:uuid:77881f35-0c75-4411-99b0-85438476aed9>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00387.warc.gz"} |
Contest Results
(Analysis by Dhruv Rohatgi )
Sort the citation counts largest-to-smallest, and consider a bar graph where bar $i$ has width $1$ and height $c_i$. The $h$-index of the papers before the surveys is simply the dimension of the
largest square that fits under this bar graph. We want to determine how much we can increase the $h$-index with $K$ surveys each citing $L$ distinct papers.
Let's binary search on the final $h$-index. Then we just need a way to check whether it's possible to achieve a given $h$-index $h$.
It's clearly optimal to work only with the $h$ papers that start off with the largest citation counts. Note that we have $KL$ total citations to allocate to these papers. If the total citation count
of these papers is less than $h^2 - KL$, then we cannot hope to achieve $h$. This is one failure mode.
Unfortunately, it's not the only failure mode. That is, the converse of the above statement is not true, because we have an added restriction that no survey can cite a paper twice. So for example if
$h=3$, $K = 1$, and $L = 2$, and the top three papers initially have citation counts $(3, 3, 1)$, then we cannot raise the third paper to citation count $3$ since we only have one survey. This
illuminates another possible failure mode: if there is a paper with less than $h-K$ citations initially, then we cannot achieve $h$.
It turns out that these are the only two failure modes. We can prove this by induction on $K$. If $K=0$ then every paper already has at least $h$ citations, so we're certainly happy. Suppose $K>0$.
Every paper needs at most $K$ more citations. There is some set of papers which need exactly $K$ more citations. By the assumption on the sum of citation counts, this set has size at most $L$, so
with one survey we can cite all these papers (plus some others, until we've hit $L$ citations or this survey has cited every paper). Now we're in a situation with $K-1$ remaining surveys, but every
paper needs at most $K-1$ more citations. Moreover, it can be checked that the total number of needed citations is now at most $(K-1)L$. So we can indeed induct, completing the proof.
Below is Danny Mittal's code, implementing the above idea.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Comparator;
import java.util.StringTokenizer;
public class AcowdemiaSilver {
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer tokenizer = new StringTokenizer(in.readLine());
int n = Integer.parseInt(tokenizer.nextToken());
int k = Integer.parseInt(tokenizer.nextToken());
int l = Integer.parseInt(tokenizer.nextToken());
Integer[] publications = new Integer[n];
tokenizer = new StringTokenizer(in.readLine());
for (int j = 0; j < n; j++) {
publications[j] = Integer.parseInt(tokenizer.nextToken());
Arrays.sort(publications, Comparator.reverseOrder());
int upper = n;
int lower = 0;
while (upper > lower) {
int mid = (upper + lower + 1) / 2;
long needed = 0;
for (int j = 0; j < mid; j++) {
needed += (long) Math.max(0, mid - publications[j]);
if (needed <= ((long) k) * ((long) l) && publications[mid - 1] + k >= mid) {
lower = mid;
} else {
upper = mid - 1; | {"url":"https://usaco.org/current/data/sol_prob3_silver_open21.html","timestamp":"2024-11-07T11:06:12Z","content_type":"text/html","content_length":"5945","record_id":"<urn:uuid:e405ceba-5785-491e-9198-bd3ef342520b>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00675.warc.gz"} |
Data Structures and Algorithms
algorithms #
Algorithms are recipes for your computer. They are recipes, usually implemented on a computer following a Turing machine (or a Von Neumann architectured machine), but can also refer to those that are
done on an abstract lambda calculus.
In particular, algorithms are usually clever recipes implemented to solve a particular problem. This can involve things like: how to count the number of ways to make change for a certain number of
dollars; how to find the fastest route from start to finish in a given topology; or how to arrange a set of jumbled numbers into an orderly sequence.
data structures #
Data structures, on the other hand, are abstract structures that allow us to store, retrieve, and manipulate data in convenient ways.
What we mean by “convenient” can vary — maybe it means we can do something quickly. Maybe it means we can yank out a certain element of the dataset quickly. Or maybe it means we can add something to
a collection quickly.
Data structures are closely related to algorithms. For instance, we might need a clever algorithm to set up a data structure in a meaningfully quick way. Or maybe a data structure relies on a clever
algorithm to retrieve or modify an existing data element.
Data structures come in many many flavors:
• maybe we’re dealing with a tree, which is a data structure that has a hierarchical quality to it; this can enable, for instance, fast lookups if we have a balanced binary tree.
• graphs are data structures and can be represented on a computer in many ways; they represent datasets where the individual elements have some sort of connection with other elements.
• We have randomized, or _probabilistic data structures: structures that don’t offer a hard-and-fast guarantee of good performance, but guarantee good performance “almost all the time”, by using
randomness in the underlying methods.
• Another popular structure is the hash table, or just table. These rely on the properties of certain mathematical functions, called hash functions, to provide fast lookups. | {"url":"https://algorithm.land/docs/dsa/","timestamp":"2024-11-03T00:14:48Z","content_type":"text/html","content_length":"10900","record_id":"<urn:uuid:3d437041-5f5d-47c4-9d9b-484fed311478>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00168.warc.gz"} |
A passenger, while boarding the plane, slipped from th... | Filo
Question asked by Filo student
Question A passenger, while boarding the plane, slipped from the stairs and got hurt. The pilot took the passenger in the emergency clinic at the airport for the treatment. Due to this, the plane got
delayed by half an hour. To reach the destination away in time. so that the passengers could catch the connecting flight, the speed of the plane was increased by than the usual speed. Find the usual
speed of the plane What value is depicted in this question?
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Question A passenger, while boarding the plane, slipped from the stairs and got hurt. The pilot took the passenger in the emergency clinic at the airport for the treatment. Due to this,
Question the plane got delayed by half an hour. To reach the destination away in time. so that the passengers could catch the connecting flight, the speed of the plane was increased by than the
Text usual speed. Find the usual speed of the plane What value is depicted in this question?
Updated On Mar 17, 2023
Topic All topics
Subject Mathematics
Class Grade 12
Answer Video solution: 1
Upvotes 113
Avg. Video 18 min | {"url":"https://askfilo.com/user-question-answers-mathematics/question-a-passenger-while-boarding-the-plane-slipped-from-34363333333230","timestamp":"2024-11-07T10:35:39Z","content_type":"text/html","content_length":"162825","record_id":"<urn:uuid:d1599c7f-97f6-4015-844d-997eccd3bd21>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00762.warc.gz"} |
of a complex number
1. n-th root
nth roots of a complex number
This is a complex number nth roots Calculator. Example, to calculate cubic roots of z, enter n = 3.
nth Roots of a complex number
Let z be a complex number which has the following polar form,
`z = r(cos\theta + i * sin\theta)`,
r = |z| is the modulus of z.
`\theta` is the argument of z.
To calculate the polar form, use this calculator:
calculate the polar form of a complex number
z has exactly n nth roots here denoted by `t_k` with `0 <=k<=n-1`,
`t_k = \text{}^nsqrt(r)(cos((\theta+2 \pi k)/n) + i * sin((\theta+2 \pi k)/n))`
To proove that, we use de Moivre's formula which states, for every integer n,
`(cos\alpha+i*sin\alpha)^n = cos(n*\alpha) + i*sin(n*\alpha)`
We apply this formula to `t_k` :
`t_k^n = (\text{}^nsqrt(r))^n(cos(n*(\theta+2 \pi k)/n) + i * sin(n*(\theta+2 \pi k)/n))`
`t_k^n = r(cos(theta) + i * sin(\theta)) = z`
`t_k` is a n-th root of z.
See also
Polar form of a complex number
Modulus of a complex number
Argument of a complex number | {"url":"https://www.123calculus.com/en/complex-number-nth-root-page-1-45-175.html","timestamp":"2024-11-02T04:38:20Z","content_type":"text/html","content_length":"17592","record_id":"<urn:uuid:74631ce2-a314-436b-8ec8-e5b7ce393856>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00312.warc.gz"} |
How to Solve A Matrix Dimension Mismatch In Pytorch?
A matrix dimension mismatch in PyTorch occurs when the dimensions of matrices or tensors do not align properly for a given operation. This can lead to errors such as "RuntimeError: size mismatch".
To solve this issue, you can check the dimensions of your input matrices and make sure they match the expected dimensions for the operation you are trying to perform. You can use the size() or shape
method to inspect the dimensions of your tensors.
If the dimensions do not match, you may need to reshape or transpose one of the matrices so that they align properly. You can use functions such as view(), reshape(), or transpose() to manipulate the
dimensions of your tensors.
It is also important to pay attention to the order of dimensions, especially when dealing with operations such as matrix multiplication or convolution. Make sure that the dimensions are in the
correct order for the operation you are performing.
By properly managing the dimensions of your matrices and tensors, you can avoid matrix dimension mismatch errors in PyTorch and ensure the successful execution of your code.
What are the implications of ignoring matrix dimension mismatches in PyTorch?
Ignoring matrix dimension mismatches in PyTorch can lead to incorrect results or errors in your neural network model. When performing operations such as matrix multiplication or addition, PyTorch
requires that the dimensions of the matrices being operated on are compatible. If these dimensions do not match, PyTorch will throw an error.
Ignoring these dimension mismatches can lead to unexpected behavior in your model. For example, if you attempt to multiply two matrices with incompatible dimensions, you may end up with a matrix that
is filled with NaN values or incorrect results. This can impact the training and accuracy of your model, leading to less reliable predictions.
It is important to always ensure that the dimensions of your matrices match when performing operations in PyTorch to avoid these issues. This can be done by carefully checking and debugging your
code, and making sure that the input data is properly shaped before performing any operations.
How to identify a matrix dimension mismatch in PyTorch?
In PyTorch, a matrix dimension mismatch often occurs when trying to perform operations on tensors that have incompatible shapes. You can identify a matrix dimension mismatch in PyTorch by checking
the shape of the tensors involved in the operation using the .shape attribute.
Here is an example to demonstrate how to identify a matrix dimension mismatch in PyTorch:
1 import torch
3 # create two tensors with incompatible shapes
4 tensor1 = torch.randn(3, 4)
5 tensor2 = torch.randn(2, 3)
7 # try to perform an operation on the tensors
8 result = torch.matmul(tensor1, tensor2)
10 # check the shapes of the tensors involved in the operation
11 print("Tensor 1 shape:", tensor1.shape)
12 print("Tensor 2 shape:", tensor2.shape)
When you run the above code, you will get an error message indicating that there is a matrix dimension mismatch between tensor1 and tensor2. By checking the shapes of the tensors involved in the
operation, you can easily identify the mismatch and adjust the shapes accordingly to fix the issue.
How to utilize PyTorch functions for reshaping tensors to match expected dimensions?
To reshape tensors using PyTorch, you can use the view() function. This function allows you to reshape a tensor to have the specified dimensions. Here is an example of how to use the view() function
to reshape a tensor:
1 import torch
3 # Create a tensor with shape (2, 3, 4)
4 x = torch.randn(2, 3, 4)
6 # Reshape the tensor to have shape (2, 12)
7 reshaped_x = x.view(2, 12)
9 print(reshaped_x.shape)
In this example, we first create a tensor x with shape (2, 3, 4). We then use the view() function to reshape the tensor to have shape (2, 12). Finally, we print the shape of the reshaped tensor to
verify that it has the expected dimensions.
You can also use the reshape() function in PyTorch, which works similarly to the view() function. The main difference between the two functions is that view() returns a view of the original tensor,
whereas reshape() returns a new tensor with the specified dimensions.
1 # Reshape the tensor to have shape (2, 12)
2 reshaped_x = x.reshape(2, 12)
4 print(reshaped_x.shape)
Overall, both the view() and reshape() functions are useful for reshaping tensors in PyTorch to match the expected dimensions.
How to read error messages related to matrix dimension mismatches in PyTorch?
When working with PyTorch and encountering errors related to matrix dimension mismatches, it is important to carefully read and interpret the error message to understand what went wrong. Here are
some tips on how to read these error messages:
1. Look for the specific error message: PyTorch usually provides a clear and specific error message when there is a dimension mismatch. Look for keywords such as "size mismatch" or "inconsistent
size" in the error message to identify the cause of the issue.
2. Check the shapes of the tensors: The error message will often include information about the shapes of the tensors involved in the operation. Check these shapes to ensure that they are compatible
for the operation you are trying to perform.
3. Identify the operation causing the error: The error message will also mention the specific operation that caused the dimension mismatch. This could be a matrix multiplication, addition, or any
other operation. Make sure that the dimensions of the input tensors are compatible for the specified operation.
4. Review your code: Double-check your code to confirm that you are correctly handling the dimensions of the tensors throughout your operations. Avoid hardcoding shapes and use PyTorch's built-in
functions like torch.Size() or tensor.size() to dynamically retrieve tensor dimensions.
5. Use debugging tools: If you are still unable to identify the cause of the dimension mismatch error, you can use debugging tools like pdb or print statements to inspect the shapes of the tensors
at different stages of your code.
By carefully reading and interpreting the error messages related to matrix dimension mismatches in PyTorch, you can quickly identify and resolve the issue in your code. | {"url":"https://stock-market.uk.to/blog/how-to-solve-a-matrix-dimension-mismatch-in-pytorch","timestamp":"2024-11-10T22:07:56Z","content_type":"text/html","content_length":"161667","record_id":"<urn:uuid:b9cdfb90-095c-4b05-b211-4ba82f45099c>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00436.warc.gz"} |
III Theoretical Physics of Soft Condensed Matter - The variational method
4The variational method
III Theoretical Physics of Soft Condensed Matter
4.1 The variational method
The variational method is a method to estimate the partition function
−βF [φ]
is not Gaussian. To simplify notation, we will set
= 1. It is common
to make a notational change, where we replace
] with
We then want to estimate
−F [φ]
We now make a notation change, where we write
, and
] as
instead, called the effective Hamiltonian. In this notation, we write
The idea of the variational method is to find some upper bounds on
terms of path integrals we can do, and then take the best upper bound as our
approximation to F .
Thus, we introduce a trial Hamiltonian H
[φ], and similarly define
We can then write
D[φ] = e
where the subscript 0 denotes the average over the trial distribution. Taking the
logarithm, we end up with
F = F
− loghe
So far, everything is exact. It would be nice if we can move the logarithm inside
the expectation to cancel out the exponential. While the result won’t be exactly
equal, the fact that log is concave, i.e.
log(αA + (1 − α)B) ≥ α log A + (1 − α) log B.
Thus Jensen’s inequality tells us
loghY i
≥ hlog Y i
Applying this to our situation gives us an inequality
F ≤ F
− hH
− Hi
= F
− hH
+ hHi
= S
+ hHi
This is the Feynman–Bogoliubov inequality.
To use this, we have to choose the trial distribution
simple enough to
actually do calculations (i.e. Gaussian), but include variational parameters in
. We then minimize the quantity
− hH
over our variational
parameters, and this gives us an upper bound on
. We then take this to be
our best estimate of
. If we are brave, we can take this minimizing
as an
approximation of H, at least for some purposes. | {"url":"https://dec41.user.srcf.net/h/III_L/theoretical_physics_of_soft_condensed_matter/4_1","timestamp":"2024-11-11T10:52:16Z","content_type":"text/html","content_length":"117472","record_id":"<urn:uuid:9d1bb201-5227-49c2-848d-2df0251d8008>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00086.warc.gz"} |
Tell me - what was your PhD about?
This post is an attempt for me to organise in my own head an answer to a question I’m asked all the time: what was your PhD about?
What did you study?
I studied pure mathematics. Research mathematics is traditionally divided into two areas: applied mathematics, which deals with equations and concepts motivated from other sciences and from modelling
real-world phenomena, and pure mathematics, which deals with studying mathematics for its own sake. Sometimes the two branches cross over, and sometimes mathematics that was once considered pure is
later found to have applications.
My branch of pure mathematics was algebra. Algebra is the study of mathematical structure. When algebraists talk about structure, they mean the kind of structure that numbers and number systems have.
For example, the integers, with their operations of addition and negation, form a type of algebraic structure known as a group. There are many other types of groups, and algebraists (specifically
group theorists) busy themselves with the task of classifying groups: that is, coming up with a list of all possible group structures. Classification is a general mathematical principle that says it
is interesting to completely enumerate or describe all possible structures arising from a certain series of assumptions. The classification of group structures is a monumentally difficult task that
was only (partially) solved in the last decade.
Certain groups are particularly interesting in their own right. One of these groups is called the symmetric group. It has been studied for over a century and can be found in physics, chemistry and
other places in nature. The symmetric group has given rise to its own branch of algebra known as representation theory: this was the sub-branch of algebra that I studied.
Specifically, I was concerned with the classification of representations of a family of algebraic objects related to symmetric groups. Two American professors came up with a groundbreaking result
about six years ago showing that you could understand these objects in a different way, because they were actually the same as another collection of objects, called quantum groups. My thesis extends
their result to a different, related, collection of objects.
If you don’t know anything about matrices this is as far as you can understand I’m afraid. You’ll have to be satisfied with the answer I’ve given above: that I was interested in classifying some
abstract algebraic objects related to centuries-old groups.
If you have studied a first course in linear algebra, we can talk about representations a little bit. Matrices are a way of encoding transformations between linear spaces. Representation theory tries
to make group structures concrete by “representing them” by matrices. The classification of representations then translates into classifying eigenspaces of various matrices. If you are aware of how
ubiquitous linear algebra is in all computational sciences, you should be able to appreciate the usefulness of decomposing eigenspaces into smaller, more tractable, “atomic” pieces: these are known
as “irreducible representations” in representation theory: my goal was to classify irreducible representations of objects I called alternating quiver Hecke algebras.
What is an alternating quiver Hecke algebra?
Let’s break it down into the four words separately:
• alternating: the word alternating comes from the alternating group, which is related to the symmetric group I mentioned earlier. This signifies the objects belong to a family that has been
studied for a very long time.
• quiver: a quiver is a directed graph (a bunch of dots connected with arrows). Quivers are used to encode important information about the objects.
• Hecke: Hecke was a mathematician who first introduced Hecke algebras. Hecke algebras are mathematical objects closely related to symmetric groups.
• algebra: this is the type of mathematical structure I studied.
Why was your thesis result interesting?
The main result in my thesis proved the existence of a new structure in the family of alternating groups. Precisely: I showed that the (modular group algebras of) alternating groups are graded by the
integers. This sheds considerable new light on a family of objects that have been studied for over a hundred years. It also directly connects them with the seemingly disparate world of quantum | {"url":"https://www.clintonboys.com/phd-faq/","timestamp":"2024-11-07T13:43:18Z","content_type":"text/html","content_length":"12669","record_id":"<urn:uuid:f43a688f-3b59-4bdd-8266-cd7234d6203e>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00829.warc.gz"} |
Best orthogonal trigonometric approximations of the Nikol'skii-Besov-type classes of periodic functions of one and several variables
Best orthogonal trigonometric approximations of the Nikol'skii-Besov-type classes of periodic functions of one and several variables
Nikol'skii-Besov-type class, step hyperbolic Fourier sum, best orthogonal trigonometric approximation, orthowidth
Published online: 2022-06-22
We obtained the exact order estimates of the best orthogonal trigonometric approximations of periodic functions of one and several variables from the Nikol'skii-Besov-type classes $B^{\omega}_{1,\
theta}$ ($B^{\Omega}_{1,\theta}$ in the multivariate case $d\geq2$) in the space $B_{\infty,1}$. We observe that in the multivariate case the orders of mentioned approximation characteristics of the
functional classes $B^{\Omega}_{1,\theta}$ are realized by their approximations by step hyperbolic Fourier sums that contain the necessary number of harmonics. In the univariate case, an optimal in
the sense of order estimates for the best orthogonal trigonometric approximations of the corresponding functional classes are the ordinary partial sums of their Fourier series. As a consequence of
the obtained results, the exact order estimates of the orthowidths of the classes $B^{\omega}_{1,\theta}$ ($B^{\Omega}_{1,\theta}$ for $d\geq2$) in the space $B_{\infty,1}$ are also established.
Besides, we note that in the univariate case, in contrast to the multivariate one, the estimates of the considered approximation characteristics do not depend on the parameter $\theta$.
How to Cite
Fedunyk-Yaremchuk, O.; Hembars'ka, S. Best Orthogonal Trigonometric Approximations of the Nikol’skii-Besov-Type Classes of Periodic Functions of One and Several Variables. Carpathian Math. Publ. 2022
, 14, 171-184. | {"url":"https://journals.pnu.edu.ua/index.php/cmp/article/view/5928","timestamp":"2024-11-02T18:04:00Z","content_type":"text/html","content_length":"36620","record_id":"<urn:uuid:f5e1df7a-db7e-4a56-b524-cf0a7fc27f14>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00049.warc.gz"} |
Machine Drawing Symbols
Machine Drawing Symbols - Machine drawings employ standardized symbols and conventions to represent various features, materials, and processes. Here are more commonly used engineering drawing symbols
and design elements as below. Web in learning to read machine drawings, you must first become familiar with the common terms, symbols, and conventions defined and discussed in the following
paragraphs. You can also check out the gd&t symbols and terms on our site. Cut and paste to your drawing. Basic types of symbols used in engineering drawings are countersink, counterbore, spotface,
depth, radius, and diameter. This basic symbol consists of two legs of unequal length. Web this chapter will introduce the five common categories of drawings. Common engineering drawing abbreviations
used in cnc machining. Examples of different types of blueprint symbols used in cnc machining.
Machining Symbols Chart
These symbols include circles, squares, triangles, and other shapes that provide information about the size, shape, and location of different components. Here are more commonly used engineering
drawing symbols and design elements as below. Understanding these symbols is essential for accurate interpretation. Assembling (joining of the pieces) is done by welding, binding with adhesives,
riveting, threaded fasteners, or even yet.
GD&T Symbols Reference Guide from Sigmetrix Mechanical design
By cnccookbook, the leading cnc blog. Gd&t controls variations of size, form, orientation, location and runout individually or in combination. This basic symbol consists of two legs of unequal
length. These symbols and abbreviations are standardized by the american national standards institute (asmi) and the american society of mechanical engineers (asme) in the us. Explanation of how
these symbols help.
List Of Mechanical Drawing Symbols Meaning References Decor
Web may 1, 2022 by brandon fowler. Here are more commonly used engineering drawing symbols and design elements as below. Web basic types of symbols used in engineering drawings are countersink,
counterbore, spotface, depth, radius, and diameter. Calculate the ‘geometric mean of the extreme diameters of each step (d)’ using table 7.8 where d = (product of the diameter steps).
Mechanical Engineering Drawing Symbols Pdf Free Download at
Web you can cut and paste these gd&t symbols into your drawing: Definition of blueprint symbols and their significance in cnc machining. Web the included collection of predesigned mechanical drafting
symbols, machining drawing symbols, and machinist symbols helps in drawing mechanical diagrams and schematics, mechanical drafting symbols chart or mechanical drawing quickly, easily, and
effectively. Work with runsom for your.
Engineering Drawing Symbols And Their Meanings Pdf at PaintingValley
Web you can cut and paste these gd&t symbols into your drawing: Gd&t controls variations of size, form, orientation, location and runout individually or in combination. The given diameter 40 mm lies
in the diameter step 30 mm and 50 mm. This technical drawing shows the machine parts assembly using joining by threaded fasteners. Web the included collection of predesigned.
Mechanical Drawing Symbols Process Flow Diagram Symbols Electrical
Common engineering drawing abbreviations used in cnc machining. Web blueprint symbols, on the other hand, are a type of symbol used in technical drawings that represent various components and
features of a project. Web symbols for indicating surface finish. Web various symbols and abbreviations in engineering drawings give you information about the dimensions, design, and materials used.
Pictures of each.
Mechanical Engineering Drawing Symbols Pdf Free Download at
The following paragraphs cover the common terms most used in all aspects of machine drawings. The symbols also require considerably less space. Pictures of each symbol included. They are 1) piping
and instrument drawings (p&ids), 2) electrical single lines and schematics, 3) electronic diagrams and schematics, 4) logic diagrams and prints, and 5) fabrication, construction, and architectural
drawings. Geometric dimensioning.
M&e Drawing Symbols Back To Basics Komseq
Hence, d = (30x50) = 38.73 mm. These symbols include circles, squares, triangles, and other shapes that provide information about the size, shape, and location of different components. Web various
symbols and abbreviations in engineering drawings give you information about the dimensions, design, and materials used. Common cnc machining blueprint symbols. Web dive into the secrets of surface
finish symbols.
Mechanical Engineering Drawing Symbols Pdf Free Download at
Web dive into the secrets of surface finish symbols and start mastering the pros' techniques today! These symbols include circles, squares, triangles, and other shapes that provide information about
the size, shape, and location of different components. They are 1) piping and instrument drawings (p&ids), 2) electrical single lines and schematics, 3) electronic diagrams and schematics, 4) logic
diagrams and.
Machining Drafting Symbols
Web dive into the secrets of surface finish symbols and start mastering the pros' techniques today! Examples of different types of blueprint symbols used in cnc machining. You can also check out the
gd&t symbols and terms on our site. Cut and paste to your drawing. Assembling (joining of the pieces) is done by welding, binding with adhesives, riveting, threaded.
Hence, D = (30X50) = 38.73 Mm.
These symbols and abbreviations are standardized by the american national standards institute (asmi) and the american society of mechanical engineers (asme) in the us. This list includes
abbreviations common to the vocabulary of people who work with engineering drawings in the manufacture and inspection of parts and assemblies. Symbols take less time to apply on a drawing than would
be required to state the same requirements with words. Download and open the symbol library.
Explanation Of How These Symbols Help In Conveying Precise Instructions And Specifications.
Understanding these symbols is essential for accurate interpretation. How to read an engineering drawing symbol. You can also check out the gd&t symbols and terms on our site. These symbols include
circles, squares, triangles, and other shapes that provide information about the size, shape, and location of different components.
By Cnccookbook, The Leading Cnc Blog.
Web engineering drawing abbreviations and symbols are used to communicate and detail the characteristics of an engineering drawing. Web symbols and conventions: Cut and paste to your drawing. Web
basic types of symbols used in engineering drawings are countersink, counterbore, spotface, depth, radius, and diameter.
Basic Types Of Symbols Used In Engineering Drawings Are Countersink, Counterbore, Spotface, Depth, Radius, And Diameter.
Work with runsom for your cnc programming projects. Common engineering drawing abbreviations used in cnc machining. Web symbols for indicating surface finish. Web may 1, 2022 by brandon fowler.
Related Post: | {"url":"https://participation-en-ligne.namur.be/read/machine-drawing-symbols.html","timestamp":"2024-11-14T11:12:52Z","content_type":"text/html","content_length":"27348","record_id":"<urn:uuid:be314a8f-0c71-4cf8-8860-32e86bff55ed>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00636.warc.gz"} |
Company Time Period Interval Chart Type
Moving Averages
Price Overlays
Bollinger Bands
Fibonacci Retracement
Typical Price
Weighted Close
Technical Indicators
Williams %R
Relative Strength Index
Average True Range
Slow Stochastics
Rate of Change (ROC)
Commodity Channel Index
Moving Average Convergence
Divergence (MACD)
Money Flow Index (MFI)
Moving Averages
Price Overlays
Bollinger Bands
Fibonacci Retracement
Typical Price
Weighted Close
Technical Indicators
Williams %R
Relative Strength Index (RSI)
Average True Range
Slow Stochastics
Rate of Change (ROC)
Commodity Channel Index (CCI)
Moving Average Convergence Divergence (MACD)
Money Flow Index (MFI)
Moving Averages
Price Overlays
Bollinger Bands
Fibonacci Retracement
Typical Price
Weighted Close
Technical Indicators
Williams %R
Relative Strength Index (RSI)
Average True Range
Slow Stochastics
Rate of Change (ROC)
Commodity Channel Index (CCI)
Moving Average Convergence Divergence (MACD)
Money Flow Index (MFI) | {"url":"https://thegroup.com.qa/histchartNew.aspx?lang=en","timestamp":"2024-11-08T04:55:22Z","content_type":"application/xhtml+xml","content_length":"96058","record_id":"<urn:uuid:e3d1df56-f41b-46dd-b217-803898ac367b>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00195.warc.gz"} |
Comparison Operators in Java | 6 Different Comparison Operators in Java
Introduction to Comparison Operators in Java
Operators are considered special characters or symbols used to perform certain operations on variables or values (operands). In Java, there are several operators that are used to manipulate
variables. It includes Arithmetic operators, Bitwise operators, Comparison operators, Logical operators, Misc. operators, Assignment operators, etc. In this article, we will discuss more details on
comparison operators in java.
Comparison Operators in Java
Following are the various comparison operators in Java.
Name of the Operator Operator Example
Equal to = = a= =b
Not equal to != a!=b
Less than < a<b
Greater than > a>b
Less than or equal to <= a<=b
Greater than or equal to >= a>=b
1. Equal to
This operator checks whether the value on the operator’s left side is equal to the value in the right side.
import java.util.Scanner;
public class ComparisonExample {
public static void main(String[] args) {
int x, y;
Scanner sc= new Scanner(System.in);
//take the value of x as input from user and store it in variable x
System.out.print("Enter the value of x : ");
x = sc.nextInt();
//take the value of y as input from user
System.out.print("Enter the value of y : ");
//store the value in variable y
y = sc.nextInt();
//checks whether x and y are equal; Return true if it is same, else returns false
System.out.println(x == y);
Case 1: x = 3; y =5; Returns false as they are not equal
Case 2: x = 4; y =4; Returns true as they are equal
2. Not equal to
This operator checks whether the value on the operator’s left side is not equal to the value on the right side.
import java.util.Scanner;
public class ComparisonExample {
public static void main(String[] args) {
int x, y;
Scanner sc= new Scanner(System.in);
//take the value of x as input from user and store it in variable x
System.out.print("Enter the value of x : ");
x = sc.nextInt();
//take the value of y as input from user
System.out.print("Enter the value of y : ");
//store the value in variable y
y = sc.nextInt();
//checks whether x and y are not equal; Return true if it is not equal, else returns false
System.out.println(x != y);
Case 1: x = 3; y =4; Returns true as they are not equal
Case 2: x = 3; y =3; Returns false as they are equal
3. Less than
This operator checks whether the value on the operator’s left side is less than the value on the right side.
import java.util.Scanner;
public class ComparisonExample {
public static void main(String[] args) {
int x, y;
Scanner sc= new Scanner(System.in);
//take the value of x as input from user
System.out.print("Enter the value of x : ");
//store the value in variable x
x = sc.nextInt();
//take the value of y as input from user
System.out.print("Enter the value of y : ");
//store the value in variable y
y = sc.nextInt();
//Returns true if x is less than y, else false
System.out.println(x < y);
Case 1: x = 4; y =6; Returns true as x is less than y
Case 2: x = 44; y =32; Returns false as x is not less than y
4. Greater than
This operator checks whether the value on the operator’s left side is greater than the value on the right side.
import java.util.Scanner;
public class ComparisonExample {
public static void main(String[] args) {
int x, y;
Scanner sc= new Scanner(System.in);
//take the value of x as input from user
System.out.print("Enter the value of x : ");
//store the value in variable x
x = sc.nextInt();
//take the value of y as input from user
System.out.print("Enter the value of y : ");
//store the value in variable y
y = sc.nextInt();
//Returns true if x is greater than y, else false
System.out.println(x > y);
Case 1: x = 67; y =66; Returns true as x is greater than y
Case 2: x = 43; y =57; Returns false as x is less than y
5. Less than or equal to
This operator checks whether the value on the operator’s left side is less than or equal to the value on the right side.
import java.util.Scanner;
public class ComparisonExample {
public static void main(String[] args) {
int x, y;
Scanner sc= new Scanner(System.in);
//take the value of x as input from user and store it in variable x
System.out.print("Enter the value of x : ");
x = sc.nextInt();
//take the value of y as input from user and store it in variable y
System.out.print("Enter the value of y : ");
y = sc.nextInt();
//Returns true x is less than or equal to y, else false
System.out.println(x <= y);
Case 1: x = 45; y =45; Returns true as x is equal to y
Case 2: x = 45; y =54; Returns true as x is less than y
Case 3: x = 45; y =43; Returns false as x is greater than y
6. Greater than or equal to
This operator checks whether the value on the operator’s left side is greater than or equal to the value on the right side.
import java.util.Scanner;
public class ComparisonExample {
public static void main(String[] args) {
int x, y;
Scanner sc= new Scanner(System.in);
//take the value of x as input from user
System.out.print("Enter the value of x : ");
//store the value in variable x
x = sc.nextInt();
//take the value of y as input from user
System.out.print("Enter the value of y : ");
//store the value in variable y
y = sc.nextInt();
//Returns true x is greater than or equal to y, else false
System.out.println(x >= y);
Case 1: x = 54; y =67; Returns false as x is less than y
Case 2: x = 45; y =36; Returns true as x is greater than y
Case 3: x = 55; y =55; Returns true as x is equal to y
Recommended Articles
This is a guide to Comparison Operators in Java. Here we discuss the introduction and top 6 comparison operators in java with examples and code implementation. You may also look at the following
articles to learn more- | {"url":"https://www.educba.com/comparison-operators-in-java/","timestamp":"2024-11-02T05:32:12Z","content_type":"text/html","content_length":"317351","record_id":"<urn:uuid:fdf349d5-4804-4566-b256-015a3a319385>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00134.warc.gz"} |
Class 10 Maths Sample Paper 2024 with Solution Out, Download PDF
The Central Board of Secondary Education (CBSE) has officially released the Class 10 Maths Sample Paper 2024 with Solution allowing students to better prepare for the half-yearly and final board
exams. Students who are going to appear in the upcoming board exam in 2025, must download and practice the Maths Sample paper Class 10 2024 25 as much as they can. Scroll down this page for the
sample paper class 10 maths with solution 2024 pdf for basis and standard courses.
Class 10 Maths Sample Paper 2024 with Solution
Maths is regarded as the most changeling subject due to its complexity. With proper preparation and strategy, one can successfully achieve the highest score in this paper. The Class 10 Maths Sample
Paper 2024-25 is created in compliance with the revised syllabus and using the most recent updated test pattern. These maths sample papers are essential study materials for students who want to
understand the expected question paper format and typology of questions. These model question papers help students grasp the format of the question paper and the marking scheme.
Pattern of Class 10 Maths Sample Paper 2024-25
Check out the pattern of the CBSE Class 10 Maths Sample Paper 2024-25 discussed below:
1. This question paper contains 38 questions which is divided into 5 Sections A, B, C, D and E.
2. In Section A, Questions no. 1-18 are multiple choice questions (MCQs) and questions no. 19 and 20 are Assertion- Reason based questions of 1 mark each.
3. In Section B, Questions no. 21-25 are very short answer (VSA) type questions, carrying 02 marks
4. In Section C, Questions no. 26-31 are short answer (SA) type questions, carrying 03 marks each.
5. In Section D, Questions no. 32-35 are long answer (LA) type questions, carrying 05 marks each.
6. In Section E, Questions no. 36-38 are case study based questions carrying 4 marks each with
sub parts of the values of 1, 1 and 2 marks each respectively.
7. All Questions are compulsory. However, an internal choice in 2 Question of Section B, 2 Questions of Section C and 2 Questions of Section D have been provided. An internal choice has been
provided in all the 2 marks questions of Section E.
8. Draw neat and clean figures wherever required.
9. Take π =22/7 wherever required if not stated.
Sample Paper Class 10 Maths with Solution 2024 Pdf Download
The sample paper class 10 maths with solution 2024 pdf for basic as well as the standard courses are now available on the official website at cbseacademic.nic.in. To make things simpler here we have
provided the direct link to download the sample paper class 10 maths with solution 2024 pdf for both courses is given below.
Class 10 Maths Sample Paper 2024 with Solutions
Section A
Section A consists of 20 questions of 1 mark each.
1. The graph of a quadratic polynomial p(x) passes through the points (-6,0), (0, -30), (4,-20) and (6,0). The zeroes of the polynomial are
A) – 6,0 B) 4, 6 C) – 30,-20 D) – 6,6
Answer: D) – 6,6
2. The value of k for which the system of equations 3x-ky= 7 and 6x+ 10y =3 is inconsistent, is
A) -10 B) -5 C) 5 D) 7
Answer:B) -5
3. Which of the following statements is not true?
A) A number of secants can be drawn at any point on the circle.
B) Only one tangent can be drawn at any point on a circle.
C) A chord is a line segment joining two points on the circle
D) From a point inside a circle only two tangents can be drawn.
Answer:D) From a point inside a circle only two tangents can be drawn.
4. If nth term of an A.P. is 7n-4 then the common difference of the A.P. is
A) 7 B) 7n C) – 4 D) 4
Answer:A) 7
5. The radius of the base of a right circular cone and the radius of a sphere are each 5 cm in length. If the volume of the cone is equal to the volume of the sphere then the height of the cone is
A) 5 cm B) 20 cm C) 10 cm D) 4 cm
Answer: B) 20 cm
Answer: A) 11/9
7. In the given figure, a tangent has been drawn at a point P on the circle centered at O.
If ∠ TPQ= 110𝑂 then ∠POQ is equal to
A) 110º B) 70º C) 140º D)55º
Answer: C) 140º
8.. A quadratic polynomial having zeroes – √5/2 and √5/2 is
A) 𝑥²− 5√2 x +1 B) 8𝑥²- 20 C) 15𝑥²- 6 D) 𝑥²- 2√5 x -1
Answer: B) 8𝑥²- 20
9. Consider the frequency distribution of 45 observations.
Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 9 15 10 6
The upper limit of median class is
A) 20 B) 10 C) 30 D) 40
Answer: C) 30
10. O is the point of intersection of two chords AB and CD of a circle.
If ∠𝐵𝑂𝐶 = 80𝑂 and OA = OD then 𝛥𝑂𝐷𝐴 𝑎𝑛𝑑 𝛥𝑂𝐵𝐶 are
A) equilateral and similar B) isosceles and similar
C) isosceles but not similar D) not similar
Answer: B) isosceles and similar
11. The roots of the quadratic equation 𝑥²+x-1 = 0 are
A) Irrational and distinct B) not real
C ) rational and distinct D) real and equal
Answer: A) Irrational and distinct
12. If 𝜃 = 30𝑜 then the value of 3tan𝜃 is
A)1 B) 1/√3 C )3/ √3 (D) not defined
Answer: C )3/ √3
13. The volume of a solid hemisphere is 396/7 𝑐𝑚³.The total surface area of the solid hemisphere (in sq. cm) is
B)594/ 7
D) 604/ 7
Answer: B)594/ 7
14. In a bag containing 24 balls, 4 are blue, 11 are green and the rest are white. One ball is drawn at random. The probability that the drawn ball is white in colour is
𝐴) 1/6
B) 3/8
C ) 11/ 24
D) 5/8
Answer: B) 3/8
15. The point on the x- axis nearest to the point (-4,-5) is
A) (0, 0) B) (-4, 0) C ) (-5, 0) D) (√41, 0)
Answer: B) (-4, 0)
16. Which of the following gives the middle most observation of the data?
A) Median B) Mean C) Range D) Mode
Answer: A) Median
17. A point on the x-axis divides the line segment joining the points A(2, -3) and B(5, 6) in the ratio 1:2. The point is
A) (4, 0) B) ( 7/2 ,3/2) C) (3, 0) D) (0,3)
Answer: C) (3, 0)
18. A card is drawn from a well shuffled deck of playing cards. The probability of getting red face card is
𝐴) 3/ 13
B) 1/ 2
C) 3/ 52
D) 3/ 26
Answer: D) 3/ 26
DIRECTION: In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option
A)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
B)Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
C)Assertion (A) is true but reason (R) is false.
D)Assertion (A) is false but reason (R) is true.
19. Assertion (A): HCF of any two consecutive even natural numbers is always 2.
Reason (R): Even natural numbers are divisible by 2.
Answer: (B)
20. Assertion (A): If the radius of the sector of a circle is reduced to half and the angle is doubled then the perimeter of the sector remains the same.
Reason (R): The length of the arc subtending angle θ at the centre of a circle of radius r = 𝛱𝑟𝜃/180.
Answer: (D)
CBSE Class 10 Maths Sample Paper 2024-25: Section B
Section B
Section B consists of 5 questions of 2 marks each.
21. (A)Find the H.C.F and L.C.M of 480 and 720 using the Prime factorisation method.
(A) The H.C.F of 85 and 238 is expressible in the form 85m -238. Find the value of m.
22. (A) Two dice are rolled together bearing numbers 4, 6, 7, 9, 11, 12. Find the probability that the product of numbers obtained is an odd number
(B) How many positive three digit integers have the hundredths digit 8 and unit’s digit 5? Find the probability of selecting one such number out of all three digit numbers.
24. Find the point(s) on the x-axis which is at a distance of √41 units from the point (8, -5).
25. Show that the points A(-5,6), B(3, 0) and C( 9, 8) are the vertices of an isosceles triangle.
Sample Paper Class 10 Maths Standard: Section C
Section C
Section C consists of 6 questions of 3 marks each.
26. (A) In 𝛥ABC, D, E and F are midpoints of BC,CA, and AB respectively. Prove that
△ 𝐹𝐵𝐷 ∼ △ DEF and △ DEF ∼ △ ABC
(B) In 𝛥ABC, P and Q are points on AB and AC respectively such that PQ is parallel to BC. Prove that the median AD drawn from A on BC bisects PQ
27. The sum of two numbers is 18 and the sum of their reciprocals is 9/40. Find the numbers.
28. If 𝛼 and 𝛽 are zeroes of a polynomial 6𝑥²-5x+1 then form a quadratic polynomial whose zeroes are 𝛼² and 𝛽².
29. If cosθ + sinθ = 1 , then prove that cosθ – sinθ = ±1 3
30. (A) The minute hand of a wall clock is 18 cm long. Find the area of the face of the clock described by the minute hand in 35 minutes.
(B) AB is a chord of a circle centered at O such that ∠AOB=60˚. If OA = 14 cm then find the area of the minor segment. (take √3 =1.73)
31. Prove that √3 is an irrational number.
Maths Sample Paper Class 10 2024 25 – Section D
Section D
Section D consists of 4 questions of 5 marks each
32. (A) Solve the following system of linear equations graphically:
x+2y = 3, 2x-3y+8 = 0
(B) Places A and B are 180 km apart on a highway. One car starts from A and another from B at the same time. If the car travels in the same direction at different speeds, they meet in 9 hours. If
they travel towards each other with the same speeds as before, they meet in an hour. What are the speeds of the two cars?
33. Prove that the lengths of tangents drawn from an external point to a circle are equal.
Using the above result, find the length BC of 𝛥ABC. Given that, a circle is inscribed in 𝛥ABC touching the sides AB, BC, and CA at R, P, and Q respectively, and AB= 10 cm, AQ= 7cm, CQ= 5cm.
35. Find the mean and median of the following data:
Class 85-90 90-95 95-100 100-105 105-110 110-115
frequency 15 22 20 18 20 25
The monthly expenditure on milk in 200 families of a Housing Society is given below
Class 10 Maths Sample Paper 2024-25: Section E
Section E
Section E consists of 3 case study-based questions of 4 marks each.
36. Ms. Sheela visited a store near her house and found that the glass jars were arranged one above the other in a specific pattern.
On the top layer there are 3 jars. In the next layer there are 6 jars. In the 3rd layer from the top there are 9 jars and so on till the 8th layer. On the basis of the above situation answer the
following questions.
(i) Write an A.P whose terms represent the number of jars in different layers starting from top . Also, find the common difference.
(ii) Is it possible to arrange 34 jars in a layer if this pattern is continued? Justify your answer.
(iii) (A) If there are ‘n’ number of rows in a layer then find the expression for finding the total number of jars in terms of n. Hence find 𝑆8 .
(iii) (B) The shopkeeper added 3 jars in each layer. How many jars are there in the 5th
layer from the top?
Triangle is a very popular shape used in interior designing. The picture given above shows a cabinet designed by a famous interior designer. Here the largest triangle is represented by △ ABC and
smallest one with shelf is represented by △ DEF. PQ is parallel to EF.
(i) Show that △ DPQ ∼ △ DEF. (1)
(ii) If DP= 50 cm and PE = 70 cm then find 𝑃𝑄/𝐸𝐹.
(iii) (A) If 2AB = 5DE and △ ABC ∼ △ DEF then show that 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 △𝐴𝐵𝐶/𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 △𝐷𝐸𝐹 is constant.
(iii) (B) If AM and DN are medians of triangles ABC and DEF respectively then prove that △ ABM ∼ △ DEN.
38. Metallic silos are used by farmers for storing grains. Farmer Girdhar has decided to build a new metallic silo to store his harvested grains. It is in the shape of a cylinder mounted by a cone.
Dimensions of the conical part of a silo is as follows:
• Radius of base = 1.5 m
• Height = 2 m
• Dimensions of the cylindrical part of a silo is as follows:
• Radius = 1.5 m
• Height = 7 m
On the basis of the above information answer the following questions.
(i) Calculate the slant height of the conical part of one silo.
(ii) Find the curved surface area of the conical part of one silo.
(iii)(A) Find the cost of the metal sheet used to make the curved cylindrical part of 1 silo at the rate of ₹2000 per 𝑚² .
(iii) (B) Find the total capacity of one silo to store grains
How to Use Maths Sample Paper Class 10 2024 25 Answer key
sample paper class 10 maths with solution pdfs are extremely useful resources for preparation for the CBSE Class 10 examination. The accompanying answer keys contain useful information about the
correct responses and can considerably improve your learning. Here’s how to apply them effectively:
• Familiarize Yourself with the Format: Understand the format, question kinds, and marking scheme so that you may prepare accordingly.
• Analyze your performance: Compare your answers to the solutions to identify areas of strength and weakness, and then work on them to improve.
• Identify reoccurring topics: Determine which topics are commonly tested, and concentrate on those.
• Practice different question types: Make sure you’re familiar with MCQs, short answer, and long answer questions.
• Timed Practice: Solve sample papers under exam conditions to mirror the real test experience.
• Practice regularly: Solve sample papers in timed mode to imitate the test setting and improve time management.
• Seek feedback from teachers: Seek expert advice and comments on your performance to identify areas for improvement.
• Consistent practice: Sample papers help to reinforce concepts and enhance problem-solving abilities. Consistent practice, excellent time management, and a positive mindset are crucial for success
in the CBSE Class 10 Mathematics exam. | {"url":"https://www.adda247.com/school/class-10-maths-sample-paper-2024-25/","timestamp":"2024-11-03T02:47:11Z","content_type":"text/html","content_length":"672701","record_id":"<urn:uuid:ed2774c7-75be-4370-a434-b8c6e192a610>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00277.warc.gz"} |
Non-Euclidean Geometries For Our Brain Grid Cells
It took human culture millennia to arrive at a mathematical formulation of non-Euclidean spaces - but that was not because of a limitation of our brains.
Instead, it's likely that even the brains of rodents get there very naturally every day.
Euclidean geometry is the kind of geometry we normally study at school, whereas non-Euclidean geometries are all those that reject one or more of Euclid's five postulates. A geometry that unfolds on
a curved surface is an example. Recent research has investigated how the brain encodes flat spaces and in 2005, Edvard and May-Britt Moser discovered grid cells, neurons of the entorhinal cortex of
rodents that fire in a characteristic way when the animal moves in an arena. The discovery was awarded the Nobel Prize, but all experiments conducted to date have involved flat (Euclidean) surfaces.
So what happens with other types of surface? The starting point is the formation of these brain "maps".
"There are two main classes of theoretical models that attempt to explain it, but both of them assume that our brain contains some kind of "engineer" that has prepared things appropriately," says
SISSA neuroscientist Alessandro Treves. "These models take for granted that the system originates with substantial prior knowledge, and they closely reproduce the behavior of the biological system
under known conditions, since they are constructed precisely on its observation. But what happens in conditions that have yet to be explored experimentally? Are these models able to 'generalize',
that is, to make a genuine prediction to be then confirmed by other experiments? A correct theory should tell us more than what we already know."
Treves and colleagues have been developing a new, radically different model since 2005, and in their recent paper published in the journal Interface have attempted a broad generalization. "Ours is a
self-organizing model, which simulates the behavior of 'artificial' grid cells capable of learning by exploring the environment".
The model is based on mathematical rules and its final characteristics are determined by the environment in which it "learns from experience". In previous studies, the model was tested on flat
surfaces: "in these settings our artificial grid cell shows the same hexagonal symmetrical firing pattern seen in biological cells".
"To apply it to a new situation, we thought of having our model move in a non-Euclidean space, and we chose the simplest setting: a space with a constant curvature, in other words a sphere or
pseudosphere". The recently published study shows the results achieved with the pseudospherical surface, which demonstrate that in this case the firing pattern has a heptagonal, seven-point,
This finding can now easily be compared with the firing of real grid cells, in rodents raised on a pseudospherical surface. "We're waiting for the experimental results of our Nobel Prize-winning
colleagues from Trondheim" explains Treves. "If our results are confirmed, then new theoretical considerations will ensue that will open up new lines of research".
In addition to demonstrating that maps adapt to the environment in which the individual develops (and so are not genetically predetermined), the observation of a heptagonal symmetry in new
experimental conditions - which would show that the brain is able to encode a non-Euclidean space - would also suggest that grid cells might play a role in mapping many other types of space,
"including abstract spaces," adds Treves. "Try to imagine what we might define as the space of movements, or the space of the different expressions of the human face, or shapes of a specific object,
like a car: these are continuous spaces that could be mapped by cells that are not the same but are similar to grid cells, cells that could somehow represent the graph paper on which to measure these | {"url":"https://www.science20.com/news_articles/noneuclidean_geometries_for_our_brain_grid_cells-155374","timestamp":"2024-11-03T10:32:15Z","content_type":"text/html","content_length":"35393","record_id":"<urn:uuid:a2dd1bb9-2694-4423-a177-667a7c628d0c>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00750.warc.gz"} |
Fully-Specified Algorithms for JOSE and COSE
This section describes Two-Layer Encryption in both JOSE and COSE. Each defines multiple ways that a content encryption key can be produced and protected, then later used to decrypt or encrypt
This specification uses the term "Two-Layer Encryption" to refer to what JOSE describes as "Key Encryption" and "Key Agreement with Key Wrapping", and what COSE describes as "Key Transport" and "Key
Agreement with Key Wrap".¶
A distinguishing characteristic of Two-Layer Encryption schemes is that multiple recipients can perform decryptions, using a wide range of algorithms, and that encrypted content encryption keys are
always present.¶
In RSA-OAEP, the content encryption key is encrypted using an asymmetric cryptographic operation. When Key Wrapping without any key establishment is used, the content encryption key is encrypted
using a symmetric cryptographic operation (key wrap). How the content encryption key is generated is out of scope for this discussion.¶
Key wrapping algorithms generally satisfy the following interface:¶
key_encryption_key = \
encrypted_content_encryption_key = \
encrypt(content_encryption_key, key_encryption_key)
content_encryption_key = \
decrypt(encrypted_content_encryption_key, key_encryption_key)
When Key Establishment with Key Wrapping is used, the content encryption key is protected with Key Wrapping, where the Key Encryption Key is derived from an asymmetric cryptographic operation and a
key derivation function.¶
Key Establishment with Key Wrapping algorithms generally satisfy the following interface:¶
private_key, public_key = key_generation(algorithm_identifier)
# ignoring ephemeral/static vs. static/static, etc.
key_encryption_key = \
key_establishment(public_key, private_key)
encrypted_content_encryption_key = \
encrypt(content_encryption_key, key_encryption_key)
content_encryption_key = \
decrypt(encrypted_content_encryption_key, key_encryption_key)
The interface above is consistent with Key Establishment with Direct Encryption. The process of deriving a shared secret and content encryption key is specific to the asymmetric key type used. The
difference is that instead of using the derived content encryption key directly, two-layer encryption always uses a key encryption key, and protects the content encryption key.¶
Regardless of how a Two-Layer Encryption scheme protects the content encryption key, content encryption algorithms generally satisfy the following interface:¶
content_encryption_key = \
unwrap or establish and unwrap or key transport...
ciphertext = encrypt(plaintext, content_encryption_key)
plaintext = decrypt(ciphertext, content_encryption_key)
Depending on the content encryption algorithm, additional parameters such as Additional Authenticated Data (AAD) and/or an Initialization Vector (IV) might be required.¶
Although JOSE and COSE encode Two-Layer Encryptions differently, both rely on a protected content encryption key. The content encryption key is protected using Key Wrapping directly, or through Key
Establishment and then Key Wrapping, or Key Transport, or Key Encryption.¶
When using Two-Layer Encryption, the output of symmetric encryption includes the ciphertext and is accompanied by all parameters necessary for the recipient to decrypt the ciphertext, including
parameters for use with the key establishment algorithm, such as ephemeral or encapsulated keys, any required key derivation functions and their parameters and the key wrapping algorithm. Encrypted
content encryption keys are always present when Two-Layer Encryption is used. Parameters accompanying the ciphertext can include an Initialization Vector (IV), an Authentication Tag, and Additional
Authenticated Data (AAD). Two-Layer Encryption is often used for encrypting the same plaintext to multiple recipients, in contrast with other modes that can only be used to encrypt to a single
Example of a decoded JWE Protected Header for Key Encryption with RSA OAEP and Content Encryption using AES_128_CBC_HMAC_SHA_256:¶
"alg": "RSA-OAEP-256",
"enc": "A128CBC-HS256",
Example of a decoded JWE Protected Header for Key Agreement using ECDH-ES with key wrapping and Content Encryption using AES GCM:¶
"alg": "ECDH-ES+A128KW",
"enc": "A128GCM",
However, despite containing both the key establishment algorithm and a content encryption algorithm, the second example above is not fully specified. In this example, the missing parameter is the
curve name for the ephemeral key used for key agreement.¶
To convey fully-specified Two-Layer Encryption in JOSE, the "alg" value MUST specify all essential parameters for key protection or derive them from the accompanying "enc" value and the "enc" value
MUST be fully specified, specifying all essential parameters for symmetric encryption. For example, ECDH-ES using Concat KDF and P-256 and "A128KW" key wrapping used with AES-GCM.¶
To convey fully-specified Two-Layer Encryption in COSE, the outer "alg" value MUST specify all essential parameters for key protection and the inner "alg" value MUST be fully specified, specifying
all essential parameters for symmetric encryption. For example, ECDH-ES using P-256 w/ HKDF and AES Key Wrap w/ 128-bit key used with AES-GCM.¶
In COSE, preventing cross-mode attacks, such as those described in [RFC9459], can be accomplished in two ways: (1) Allow only authenticated content encryption algorithms. (2) Bind the the potentially
unauthenticated content encryption algorithm to be used into the key protection algorithm so that different content encryption algorithms result in different content encryption keys. Which choice to
use in which circumstances is beyond the scope of this specification.¶
Fully-specified Two-Layer Encryption algorithms enable the sender and receiver to agree on all mandatory security parameters. They also enable a protocol to specify an allow list of algorithm
combinations that does not include polymorphic combinations, such as cross-curve key establishment, cross-mode symmetric encryption, or mismatched KDF size to symmetric key scenarios.¶ | {"url":"https://www.ietf.org/ietf-ftp/internet-drafts/draft-ietf-jose-fully-specified-algorithms-06.html","timestamp":"2024-11-14T18:36:30Z","content_type":"text/html","content_length":"151555","record_id":"<urn:uuid:9c1d6ac2-2b33-4450-80ae-01bf40f4489f>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00346.warc.gz"} |
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Pythagoras was an ancient Greek philosopher. He was born in 570 BC, in Samos (Greece), and died in 495 BC, in Metapontum (Italy). His full name was Pythagoras of Samos.
He was credited with many discoveries in the field of science, mathematics, music, astronomy, and medicine. The discoveries done by him are, Pythagorean or Pythagoras theorem, Pythagorean tuning,
Theory of proportions, Sphericity of the earth, the identity of planet Venus, and five regular solids. He also divided the globe into five climatic zones. He had given the main credit for the
discovery and proof of Pythagoras theorem.
Pythagoras or Pythagorean Theorem
Pythagoras theorem is based on the right-angled triangle or right triangle only. The theorem states that in a right triangle, the sum of the square of base and perpendicular is equal to the square of
the hypotenuse.
In other words ,in a right triangle, the square of the hypotenuse is equal to the sum of the square of the two legs. The legs (base and perpendicular) are the sides of a triangle that forms the
Components of Right Triangle
The following figure represents a right triangle ∆ABC.
• Base: It is a side of the right triangle that is adjacent to the perpendicular and hypotenuse. In ∆ABC, AB is base.
• Perpendicular: It is a side of the right triangle that is adjacent to the base and hypotenuse. It is called the height of the triangle. In ∆ABC, AC is perpendicular.
• Hypotenuse: The side opposite to the right angle is called the hypotenuse. In other words, the longest side of the right triangle is called the hypotenuse. In ∆ABC, BC is the hypotenuse.
• Right-angle: In geometry, the right-angle is an angle that makes an angle of 90°. In ∆ABC, ∠A is a right-angle .
Pythagoras Triples
Pythagoras or Pythagorean triples is a set of three positive integers that satisfies the Pythagoras theorem. The least Pythagorean triple is (3, 4, 5).
In ∆ABC, (a, b, c) is the Pythagoras triples that represent positive integer value and satisfy the theorem.
The following table enlists some Pythagorean triples.
(3, 4, 5) (5, 12, 13) (7, 24, 25) (8, 15, 17)
(9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85)
(16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65)
(36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97)
Facts about Pythagoras Triples
• It always has one even number in a triple.
• The value of c will always be odd.
• It may have two prime numbers.
Pythagoras Theorem Formula
Consider the following figure.
In ∆ABC, AC is perpendicular or height, AB is base, and BC is the hypotenuse. The length of perpendicular, base, and hypotenuse is a, b, and c, respectively. According to the Pythagoras theorem, the
Pythagoras theorem formula can be written as:
Perpendicular^2 + Base^2 = Hypotenuse^2
AC^2+ AB^2 = BC^2
a^2+ b^2 = c^2
Pythagoras Theorem Proof
To Prove: AC^2 = AB^2+ BC^2
Given: A right triangle ∆ABC.
Proof 1:
In the following figure, we have drawn a perpendicular (BD) from point B that meets at point D on the hypotenuse.
The perpendicular divides the triangle into two triangles, i.e., ∆ADB and ∆BDC.
Remember: If we draw a perpendicular from the vertex of the right-angle, the triangles on both sides are equals to each other and also equal to the whole triangle.
According to the above statement, ∆ABC=∆ADB
Adding the equations (i) and (ii), we get:
Hence, the Pythagoras theorem is proved.
Let's see the second way to prove the theorem.
Proof 2:
In the following figure, we have drawn a square ABCD. Inside the square ABCD, we have drawn another square EFGH that forms four triangles ∆AEF, ∆FDG, ∆GCH and ∆HBE.
Now, we will find the area of both squares and triangles, separately.
We know that, the area of square = a^2 (where a is the side of the square)
Area of square ABCD=(a+b)^2
We know that, area of triangle =
Area of a triangle=
There are a total of four triangles, so the area of four triangles will be:
Area of four triangles= 4×2ab
Area of square EFGH=c^2 (where c is the side of the square EFGH)
The total area of the square ABCD will be:
Area of ABCD = Area of Square EFGH + Area of four triangles
Putting the values, we get:
Cancel out the 2ab on both sides, we get:
Hence, the theorem is proved.
Pythagoras Theorem Problems
Example 1: The three sides of a triangle are 5, 12, and 13 cm. Use the Pythagoras theorem and check that the triangle is a right triangle or not.
Given, AB = 12 cm, BC = 5 cm, AC = 13 cm
According to the Pythagoras theorem, AC^2=BC^2+AB^2
Hence, the triangle is a right triangle.
Example 2: Find the value of AC if the length of the base is 3 cm, and the height of the triangle is 4 cm.
In the ∆ABC, given that BC = 3 cm and AB = 4 cm.
According to the Pythagoras theorem, BC^2+AB^2=AC^2
Putting the values of AB and BC in the above formula, we get:
Hence, the length of the hypotenuse is 5 cm.
Example 3: Find the value of the base. If the length of the hypotenuse 10 cm and height of the triangle is 8 cm.
In the ∆ABC, given that AC = 10 cm and BC = 8 cm.
According to the Pythagoras theorem, BC^2+AB^2=AC^2
Putting the values of AC and BC in the above formula, we get:
Hence, the length of the base is 6 cm.
Example 4: The length of the base and hypotenuse is 30 and 50 cm, respectively. Find the height of the triangle.
In the ∆ABC, given that AC = 50 m and AB = 30 m.
According to the Pythagoras theorem, AC^2=BC^2+AB^2
Putting the values of AC and AB in the above formula, we get:
Hence, the height of the triangle is 40 m.
Example 5: If a side of the squared stone is 9 m. Find the length of the diagonal.
In the above figure, we see that there are two triangles ∆ABC and ∆ADC. Let's take the triangle ∆ABC and find the diagonal.
According to the Pythagoras theorem, AC^2=BC^2+AB^2
Hence, the length of the diagonal is 9√2 m.
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Statistics Book By Sher Muhammad Chaudhry Pdf 1653 | Managing analytics
Statistics Book By Sher Muhammad Chaudhry Pdf 1653
Statistics Book By Sher Muhammad Chaudhry Pdf 1653 >>> https://urllio.com/2tzByW
Introduction to Statistical Theory Part-I by Sher Muhammad Chaudhry: A Review
Introduction to Statistical Theory Part-I by Sher Muhammad Chaudhry is a textbook that covers the basic concepts and methods of probability and statistics. The book is intended for undergraduate
students of mathematics, statistics, economics, and other disciplines that require a solid foundation in statistical theory. The book is divided into 14 chapters, each with a set of exercises and
problems for practice and self-assessment. The book also provides solutions to selected questions at the end of each chapter.
The book starts with an introduction to the nature and scope of statistics, followed by a chapter on descriptive statistics that covers measures of central tendency, dispersion, skewness, kurtosis,
and graphical representation of data. The next chapter introduces the concept of probability and its axioms, rules, and theorems. The book then covers various topics in probability theory, such as
random variables, probability distributions, mathematical expectation, moment generating functions, and limit theorems. The book also discusses sampling theory, estimation, hypothesis testing,
analysis of variance, regression, and correlation.
The book is written in a clear and concise style, with examples and illustrations to help the reader understand the concepts and applications of statistics. The book also provides historical notes
and biographical sketches of some prominent statisticians who contributed to the development of the subject. The book is suitable for self-study as well as classroom instruction.
The book can be downloaded as a PDF file from [^1^]. However, the PDF file is not searchable and does not have page numbers. The book can also be purchased from various online platforms.
The book has received mixed reviews from the readers and users. Some have praised the book for its clarity, simplicity, and coverage of the topics. They have found the book helpful for learning and
revising the concepts of statistics. They have also appreciated the historical and biographical notes that enrich the book. Some have also recommended the book for competitive exams and entrance
However, some have criticized the book for its errors, typos, and outdated content. They have complained that the book has many mistakes in formulas, calculations, and solutions. They have also
pointed out that some of the examples and exercises are irrelevant or incorrect. They have suggested that the book needs to be revised and updated to reflect the current trends and developments in
Overall, the book is a useful resource for students who want to learn the basics of statistical theory. However, it is not without its flaws and limitations. The book may not be suitable for advanced
or specialized courses in statistics. The book may also require careful checking and verification of the information and data presented in it. 061ffe29dd | {"url":"https://www.managinganalytics.com/forum/general-discussions/statistics-book-by-sher-muhammad-chaudhry-pdf-1653","timestamp":"2024-11-12T16:35:01Z","content_type":"text/html","content_length":"1050498","record_id":"<urn:uuid:bfe62017-c7bb-4193-8895-311a7f59e5a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00695.warc.gz"} |
Unveiling the Power of Softmax Graph in Machine Learning -
Exploring the significance and applications of the softmax graph in machine learning. Discover how this powerful tool enhances classification accuracy and its role in neural networks.
Softmax graph, a vital component in the field of machine learning, plays a pivotal role in enhancing classification accuracy and improving the performance of neural networks. In this comprehensive
guide, we delve into the depths of the softmax graph, its applications, benefits, and how it contributes to the advancement of artificial intelligence.
Introduction to Softmax Graph
The softmax graph is a mathematical function used primarily in classification problems within the realm of machine learning. It is especially prevalent in tasks where multiple classes need to be
assigned to input data. The primary objective of the softmax function is to convert raw scores, often called logits, into a probability distribution over different classes. This distribution assists
in determining the likelihood of each class being the correct classification for the given input.
The softmax function takes an input vector and transforms it into a vector of probabilities that sum up to 1. This transformation makes it easier to interpret the output and select the class with the
highest probability as the predicted class. The formula for the softmax function is as follows:
Copy code
P(class_i) = e^(logit_i) / sum(e^(logit_j) for all j)
• P(class_i) represents the probability of class i being the correct classification.
• logit_i is the raw score associated with class i.
• sum(e^(logit_j) for all j) is the sum of exponentiated logits over all classes.
Applications of Softmax Graph
The applications of the softmax graph are widespread across the field of machine learning:
Image Classification
In image classification tasks, the softmax graph is a cornerstone. Given an image, a neural network generates logits for various classes. By passing these logits through the softmax function, the
network assigns probabilities to each class, indicating the likelihood of the image belonging to that class. This is instrumental in creating state-of-the-art image recognition systems.
Natural Language Processing
In the realm of natural language processing, the softmax graph is employed for tasks like sentiment analysis, named entity recognition, and language modeling. By applying the softmax function to
output scores generated by a language model, the model can determine the most probable word or phrase, enhancing the accuracy of text generation.
Speech Recognition
Softmax graph finds application in speech recognition systems. When transcribing speech to text, the graph helps identify the most likely words or phrases based on the audio input. This technology
powers voice assistants and transcription services, making them more accurate and reliable.
Reinforcement Learning
Reinforcement learning algorithms also leverage the softmax graph. In scenarios where agents need to make decisions, such as playing games or controlling robotic systems, the softmax function aids in
choosing the most appropriate action from a set of options, optimizing the agent’s performance over time.
Benefits of Softmax Graph
The softmax graph offers several advantages that contribute to its widespread adoption:
• Interpretable Outputs: The transformed probabilities provided by the softmax function are easily interpretable. This allows developers and researchers to understand the model’s decision-making
process and validate its predictions.
• Effective Classification: By assigning probabilities to each class, the softmax graph enables the model to make confident and informed decisions, leading to accurate classifications.
• Regularization: The exponential nature of the softmax function inherently suppresses the impact of outliers, contributing to better generalization and preventing overfitting.
• Gradient Calculation: The softmax function simplifies the calculation of gradients during the training process. This accelerates the convergence of neural network optimization algorithms.
Q: How does the softmax graph handle scenarios where logits are very close to each other?
The softmax function’s exponential nature magnifies small differences between logits. This amplification allows the model to make more distinct predictions, even when the differences between logits
are subtle.
Q: Can the softmax function be applied to binary classification problems?
Yes, the softmax function can be used for binary classification, although it’s less efficient than the sigmoid function in such cases. The softmax function assumes the classes are mutually exclusive,
so it’s best suited for multi-class problems.
Q: Are there alternatives to the softmax graph?
Certainly, alternatives like the sigmoid function are used for binary classification. Additionally, in some cases, models employ hybrid activation functions that combine softmax with other functions
for specialized tasks.
Q: How can I optimize the performance of the softmax graph?
Hyperparameter tuning, increasing the size of your training dataset, and experimenting with different network architectures can enhance the performance of the softmax graph.
Q: Is the softmax graph prone to overfitting?
The softmax graph’s exponential nature helps mitigate overfitting to some extent. However, regular techniques like dropout and early stopping should still be applied to ensure optimal model
Q: Can the softmax function handle a large number of classes?
Yes, the softmax function can handle a large number of classes. However, as the number of classes increases, it becomes essential to ensure a sufficient amount of training data and consider
techniques like hierarchical softmax to manage complexity.
The softmax graph stands as a fundamental tool in the arsenal of machine learning practitioners. Its ability to convert logits into interpretable probabilities fuels accurate classification across
various domains. From image recognition to natural language processing, the softmax graph’s influence is pervasive. As artificial intelligence continues to advance, the softmax graph will remain a
crucial ingredient in building powerful and reliable predictive models. | {"url":"https://winnyoff.com/unveiling-the-power-of-softmax-graph-in-machine-learning/","timestamp":"2024-11-04T05:49:22Z","content_type":"text/html","content_length":"239558","record_id":"<urn:uuid:d4ca267c-7512-4c49-8244-4e5a999fbd6c>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00725.warc.gz"} |
Find all z such that the series (-1)n / (2n+1) • ((1-z) / (1+z))n converges - Stumbling Robot
Find all z such that the series (-1)^n / (2n+1) • ((1-z) / (1+z))^n converges
Find all complex numbers
Then the series converges if
If we write
On the other hand,
and so
hence, the series diverges.
Therefore, the series converges for all
Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): | {"url":"https://www.stumblingrobot.com/2016/03/17/find-z-series-1n-2n1-1-z-1zn/","timestamp":"2024-11-07T09:22:14Z","content_type":"text/html","content_length":"58891","record_id":"<urn:uuid:0c246773-a034-4d74-9e1d-c5adb4f8d520>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00343.warc.gz"} |
Vox in Box
2015 Mock CCC by Alex and Timothy
Problem J4: Vox in Box
Michiru has an interesting Volvox colony which she keeps in a square open-air box (to allow them to photosynthesize). The base of the box is a square with a side length of N units (1 ≤ N ≤ 1000). One
day, she took a picture of it and printed out an ASCII image of the colony. The image is made of the characters ., O, (, ), /, and \. Since Michiru is an inquisitive girl, she would like to know how
many lines of symmetry there are in her Volvox colony.
A line of symmetry can be one of 4 types of lines – either a horizontal line, a vertical line, a diagonal line running top-left to bottom-right, and a diagonal line running top-right to bottom-left.
The horizontal and vertical lines must be parallel to sides of the box, while the diagonal lines must be at 45° angles relative to the sides. Furthermore, a line of symmetry should divide the ASCII
image into two halves of equal area which are symmetrical to each other with respect to that line.
For individual characters on the ASCII image, Michiru has observed that:
• The characters . and O are symmetrical in all 4 directions.
• The characters ( and ) are symmetrical to themselves horizontally, and symmetrical to each other vertically.
• The characters / and \ are symmetrical to themselves diagonally, and symmetrical to each other both vertically and horizontally.
Please help Michiru count the number of lines of symmetry.
Input Format
Line 1 of input will contain a single integer N, the side length of the box.
The next N lines will each have N characters, depicting an ASCII image of the Volvox colony.
Output Format
The output should consist of a single integer, the number of lines of symmetry.
Sample Input 1
Sample Output 1
Explanation of Sample 1
The Volvox is symmetrical horizontally and vertically, but not diagonally.
Sample Input 2
Sample Output 2
Sample Input 3
Sample Output 3
All Submissions
Best Solutions
Point Value: 7
Time Limit: 1.00s
Memory Limit: 64M
Added: Feb 18, 2015
Authors: FatalEagle, Alex
Languages Allowed:
C++03, PAS, C, HASK, ASM, RUBY, PYTH2, JAVA, PHP, SCM, CAML, PERL, C#, C++11, PYTH3 | {"url":"https://wcipeg.com/problem/mockccc15j4","timestamp":"2024-11-13T22:25:59Z","content_type":"text/html","content_length":"11499","record_id":"<urn:uuid:69f62d69-642c-4dc8-a468-5ffe9725cc03>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00352.warc.gz"} |
LIC's Jeevan Saral-Why so much confusion?
LIC’s Jeevan Saral-Why so much confusion?
When I wrote post related to LIC’s recent bonus declaration, I flooded with lot of quarries related to Jeevan Saral plan. Why so much confusion about this plan? Is it due to few new features of this
plan or exaggerated returns shown by agents? I think both. So let us understand what this plan is and how you can calculate your returns.
This plan is different to other plans of LIC in case of accumulation what LIC provides, sum assured calculation and about surrender. Also I think this is highest selling plan in LIC after Jeevan
Anand. I am not going in detail about plan feature but will point out the highlights.
• In this plan you can choose your premium amount which is not possible with other plans. Minimum monthly contribution in this plan is Rs.250.
• Sum Assured will be 250 times of your monthly premium payment. Hence suppose your monthly contribution is Rs.1,000 then sum assured will be Rs.2,50,000. You may call this benefit as “death
benefit sum assured” as this benefit is only meant for deaths. Hence don’t misunderstand that Sum Assured is the amount you get after maturity (which is the case with other plans). Hence I will
call it as Death Benefit SA or DBS. So this option is bit advantages for age old buyers as irrespective of your age this DBS is fixed on your premium payment but not on your age and term you
selected. This is not the case with other plans of LIC. So two persons paying same premium but age different is around 30 yrs then too they will get the same sum assured benefit under this plan.
• In this plan if death occurs during the policy period then your nominee will receive this Death Benefit SA+Return of premiums paid (but excluding 1st year premium)+LA till that period.
• Now the biggest confusion arises is, what you will receive after the maturity. Usually in all LIC policies you will receive Sum Assured+Accrued Bonus+Final Additional Bonus if any. But in this
plan depending on your age and term of the policy your sum assured which is also called as Maturity Sum Assured will change. So during the period of taking this policy you will come to know what
is your Maturity Sum Assured you will receive at the end of policy. This is fixed and will not change during the policy term. Will show you how to calculate it.
In this plan their are two types of premiums. One is called Basic Premium and another is called Net Premium. Basic premium is the base premium without adding any cost, but Net Premium is the premium
which you actually pay to LIC and includes premium mode rebate (rebate of 2% for yearly and 1% for half yearly payment)+charges for providing you the death benefit SA (@1% DBS). Below table will give
you the clear picture about this.
Note-Death Benefit SA charges are arrived as below-
Yearly premium of Rs.3,000 is divided into 12 and then multiplied by 250 times this will be (3,000/12)*250=62,500. This is the DAB and charges for this will be @1 hence Rs.62.
Now notice from above table that basic premium is nothing but base premium but net premium will add rebate and costs of this plan (which includes mode rebate and DAB charges @1%).
If it is so confusing to you then the simple formula to come to net premium will be multiply your basic premium by the factor 1.00083 this will result into net premium which you need to pay to LIC.
Now what you will get after maturity?
In this plan you will receive Maturity Sum Assured which is fixed and known to you in advance during the start of the policy once your age and term you chose. With this Maturity Sum Assured LIC will
also provide you Loyalty Additions. This Loyalty Addition will be declared annually and currently LIC declared Loyalty Addition (LA) for 10 yrs and 11 yrs of policies only and which are Rs.250 and
Rs.300 respectively. Based on this declaration we may presume the returns from this policy which hover around 6-7%. But for 10 yrs policy it is just around 3-4%.
How to calculate Maturity Sum Assured yourself?
LIC provided Basic Maturity Sum Assured list for the premium of Rs.100 per month for all ages. So just you need to select your age and policy term then multiply that to your Basic Premium. I am
working on this chart and soon will upload the whole Basic Maturity Sum Assured Chart.
• Surrender Value-Their are three types of surrender values under this plan. To avail these option policy need to complete atleast 3 years.
a) Guaranteed Surrender Value-In this you will receive 30% of total premiums paid excluding 1st year premium, all extra premiums and accident benefit/term rider premiums.
b) Special Surrender Value-It will be of 1+2 options given below.
1) 80% of MSA will be paid if less than 4 years premium paid, 90% of MSA if between 4 to less than 5 years of premium paid and 100% of MSA if premiums are paid for 5 years or more.
2) Loyalty Addition till that period.
c) Can be made anytime after completion of 3 years or more from the start of the policy provided full premiums are paid.
So surrender will be bit easier once policy completes 5 yrs or more which is not the case with other LIC plans.
Hope this post will make easier in understanding this plan.
1,244 thoughts on “LIC’s Jeevan Saral-Why so much confusion?”
1. s ramana
sir i have jeevan saral polacy 165-15 for sa 62500 term 15 years i have already paid 14 years would it be better to surrender right now or wait till maturity
1. BasuNivesh
Dear Ramana,
As the policy already completed more than 10 years, better to close now itself.
2. Aj U
Hello sir, we had taken a jeevan saral policy for 15 years with an annual premium. In the portal, The death sum assured is showing a number however the maturity sum assured is showing 0.
Please advise if this is any typo or any action that needs to be taken.
1. BasuNivesh
Dear Aj,
Rather than the portal, it is always good to look at the policy bond. Your policy bond is the legal contract.
3. JITENDRA JADHAV
DEAR Basavaraj, SIR
I WAS TAKING LIC JEEVAN SARAL POLICY MARCH 2009 MY DATE OF BIRTH 21.04.1980 HALF YEARLY PREMIUM 4549 TERMS 15 YEARS WHEN I WAS TAKEN POLICY MY AGE IS 29 YEARS SO PLEASE TELL ME HOW MUCH AMOUNT I
WILL GET IN MATURITY PLEASE TELL US
1. BasuNivesh
Dear Jitendra,
Hard for me to tell you the exact amount. Better you visit the nearest branch.
4. Veera
Dear Basavaraj,
This is just a thank you message. I had wanted to surrender my Jeevan saral policy when it was close to 9 years, which would have resulted in a loss. Based on your suggestion on various posts, I
waited for 10 years completion. I surrendered the policy last month and received the amount today. I walk away with 1.6 lacs profit. Though it’s not a handsome return, I’m happy that I’m not at
Your website has been very helpful in making this decision. Thank you wholeheartedly for the good work that you do. God bless!
1. BasuNivesh
Dear Veera,
Great to know and thanks for your kind words 🙂
5. Rohit
Dear Sir,
I took Jeevan Saral in Dec 2011 for Rs 121000 annual premium.
Till Dec 21 I paid 11 installments regularly.
Is it advisable to close this policy now.
What will be the amount, I will be getting now?
Thanks n Regards,
1. BasuNivesh
Dear Rohit,
As the policy completed more than 10 years, it is better to close. Regarding the values, be in touch with LIC branch.
2. Murali
Hi Rohit,
I am a same policy holder as well, just completed 10 years on Dec’21 (48040 annually, Total 4,80,040 for 10 years). when I checked in Feb’22, LIC branch manager told me I would get (4,21,000
instead of total paid 4,80,040). Worried very much with their answer.
Please confirm how much are you getting post checking with LIC as per advice from Mr. Basu sir… Thanks!
6. Bharat Agarwal
LIC Jeevan Saral Plan death sum.assured 150000 premium 72000 yearly started Sept 2013 and insurance paid till March 2021 can you help to find surrender value now and also LIC office showing
surrender value lesser than premium paid I read that after 5 years you get special surrender value how come less can you advise what would be the surrender value
Dear Bharat,
Approach the branch.
7. D.PARTHIBAN
Thank you sir,
I read all your comments,
Myself & my wife have taken this policy for Rs.20lakhs each, for 20yrs, no one gave such clear details of this policy till date properly, but after reading your comments, Now we are clarified.
Dear Pathiban,
My pleasure 🙂
8. K Santhosh Kumar
Hi Team,
I have taken the policy of Jeevan Saral 165, Request to know how much amount I will get the total amount at the time of 25 years (Maturity time). I have no idea of receiving the total amount. Can
you please share the details how much i get the amount?
Installed premium Policy number 1 – 9804 Rupees
Sum assured : 2,00,000
Premium Due From 28/03/2021
Maturity Sum Assured : 2,90,088
Policy Term: 25 Yrs
Installed premium Policy number 2 – 9804 Rupees
Sum assured : 2,00,000
Premium Due From 28/03/2021
Maturity Sum Assured : 2,90,088
Policy Term: 25 Yrs
Questions :
1 How much amount I get if I surrender after 10 years? ( once complete 10 years immediately)
2. Was it interlink with the stock market?
3 After 20 years if we surrender the policies how much I will receive the assuming amount?
4 Most of the suggestions given by you to surrender the policy after 10 years, May I know the reason?
5. Was I get the highlighted amount? Below mentioned
Dear Santhosh,
Approach the branch for amount you get.
9. purnima
Hello sir,
I have invested in jeevan saral policy in the name of my daughter , who was a minor then, with an intention to get it assigned to me later. But after taking the policy i was informed that until
the child attains majority it cannot be assigned. But later i read articles on the policy. Now i am unsure of what to do? Surrender ? Take a loan? Assignment? Or continue till completion of 10
yrs? Can u please guide? the details are as follows:
DOC : sep’2013 (the child was 12 yrs of age)
Premium per annum paid half yearly : Rs.52272
DSA : Rs.11,00,000
MSA : Rs.13,13,1800
PPT : 21 yrs
Premium paid for :14 half years ie upto 2019
Thanks in advance.
Dear Purnima,
Complete 10 years, then surrender it.
10. Bhawna Rana
Hello sir,
I have a lic Jeevan saral policy with following details-
1. Premium – 24020 yrly
2. Sum assured – 500000
3. DOC – 12.12.2011
4. Age at Comm. – 27
Sir, my policy is in force till date. Plz suggest that should I surrender it now or wait till next year?
Thanking u in anticipation.
Dear Bhawna,
Surrender it once it completes 10 years.
11. Rashmi
My jeevan saral.policy was taken in Nov 2011.. quarterly premium is 3062 and Msa is 250000.Can I surrender my policy in Nov 2021 as it will complete my 10 year of premium paying?? Or I have to
wait for one more year. Ie in 2022 to avail loyalty additions?
Dear Rashmi,
Once you pay the 10 year premium, then you can surrender.
12. Rakesh Kumar Rawat
I took LIC jeevan saral plan -165 policy with monthly premium of Rs 2024/- in 14th August 2010 with 21 years period and that time my age was 39 .06 months years. The maturity sum assured is Rs
527860/- and date of maturity is on 14/08/2031. It is learned that the loyalty supplement for this policy is applicable only after the 10 years . When contacted with LIC agent he informed me that
I have to pay regular premium till August 2021 to complete 10 years to enable me to get the loyalty supplement.
As I would like to discontinue the said policy after completion of 10 years , what will be the Full & Final Maturity Value after 10 years.
Your valuable advice will be highly appreciated.
Dear Rakesh,
Yes, LA is applicable only for the policies which complete 10+ years. Hence, you have no option but to continue. Regarding the exact values, better be in touch with the branch (NOT WITH AN
13. G KISHORE KUMAR
Sir My Name is Kishore Kumar
I have 2 Jeevan saral policy
1. policy for 10,00,000 SA taken on 15.12.2011 paying terms 20 years – I am paying 48,000 PA
2. Policy for 12,50,000 SA taken on 26.07.2013 Paying terms 16 years – I am paying 60,000 PA
if I surrender my policy after 10 years of payment what will my returns
Dear Kishore,
Check with branch.
14. Rashmi
Thank you sir . Just need a suggestion shall I continue with my policy term ie. 21 year. Will it
b beneficial?? or shall I surrender once complete 10 year.
Dear Rashmi,
Better to surrender once it completes 10 years.
15. Rashmi
Hi, I have a jeevan saral policy with msa 250000,and quarterly premium of 3062 for 21 year. If I wish to surrender the policy after 10 yrs will I get the loyalty addition?? Also would like to
know till what percentage to total premium paid can be be withdrawn if I want to partially surrender?
Dear Rashmi,
Yes, if you surrender after 10 years, then you get the LA based on 10 years. Better you surrender than partial withdrawal.
16. Sneh Gandhi
Dear Sir, Really appreciate your effort in writing this article and educating people! I have started a home loan at 7% rate and my husband has 4 Jeevan Saral Policies Plan No. 165. We wish to
surrender, to repay our home loan principal amount, but not sure if we should wait or surrender now? Details
of the 4 policies:
Policy 1:
started in yr 2011,
Maturity SA: 2,79,630
Prem paying term and Policy term: 20 yrs
Premium amt: 6065 ; Payment Mode: half yearly
Policy 2 and Policy 3 have exact same features and was started in year 2012.
Policy 4 :
Started in 2013
Maturity SA: 8,43, 856
Prem paying term and Policy term: 35 yrs !!
Premium amount: Rs. 19216 yearly
I know the surrender values against each policy but should I wait for every policy to complete 10 years or give up right away and why- pls share reason. Thank You. God Bless You!
Dear Sneh,
Better to wait for the completion of 10 years, then go ahead for surrender.
1. Sneh
thank you sir. Can you please tell me what is the rate or percentage of Bonus/ loyalty addition in Jeevan Saral Policies..I didn’t find it on LIC website as this policy is discontinued
now. If Bonus is a very small amount, then maybe it is not worth the wait. What is the total Premium I pay in the next 1 or 2 years will probably exceed the Bonus addition that I get
along with surrender value after 10 years.
Dear Sneh,
It is very much available on LIC website under Bonus Rate tab. This policy not offer you bonus but only LA.
17. TANMOY MUKHERJEE
Dear Mr Basavaraj,
I have 2 Jeevan Saral policy. I want to surrender my both policies. What’s your advice…
Details of policy
1) COD – 5/6/2013
Death Sum Assured-625000
Maturity Sum Assured-643675
Term- 20 years(2033)
2) COD – 9/7/2010
Death Sum Assured-62500
Maturity Sum Assured-47325
Term- 15years(2025)
Dear Tanmoy,
Surrender those who completed 10 years.
18. Kavita Kamath
Hello Sir
I am Kavita Kamath
I have a jeevan saral policy which was taken on 24.02.2011 for a total monthly premium of Rs 1531 incl tax (Yearly 18372). The Maturity Sum Assured is Rs 507360. I have been paying it till date.
But now i want to surrender the policy. Can you please let me know what will be my surrender value that i will receive.
Dear Kavita,
It is better you pay it for one more year (completion of 10 years), then opt for surrender.
19. Abhijeet
Hi blogger…thank you for letting us know about Jeevan Saral Plan.
Can you please provide your inputs, as I went to branch for surrendering the policy, but they advised me to pay for 10 yrs and get eligible for additional amount. Should I pay for reaming two
years or not (i.e. 2020 and 2021) ?
Here are the details of policy –
Commencement Date : 2012
SA : 5,00,000
Premium : 24,020
Premium Paying Term (Yrs) :21 i.e. 2033
Maturity Sum Assured: 5,79,500.00
Thanks in advance and appreciate your work !!
Dear Abhijeet,
Better you surrender now.
1. Rahul Garg
Hello Sir,
Thank you for helping so many people from your blog. It is a great help indeed.
I am in an extreme dilemma right now and would appreciate your guidance. Below is my policy details:
I started it in December year 2011 at the age of 18 and the term is 15 years.
Semi-annual Premium: 18,195.
Maturity Sum Assured: 5,88,450.
Till now I made 16 payments of the same which leads to a total payment of 2,91,120 (18195*2 * 16) and when I recently came to know my surrender value which is 2,51,000.
I was planning to surrender the policy but checking this loss stopped me, Initially, I was having high hopes but now my target is to reduce the loss as much as I can. I am also not clear
with the calculation of Loyalty Additions.
Should I continue with this policy for another 6.5 years or another 1.5 years for LA or should I surrender it asap?
Kindly suggest.
Dear Rahul,
The more you try to reduce the loss, the more you will be under the trap.
20. Dinesh
Hi Basuji,
I am confused regarding surrender value clause.
“Special Surrender Value 100% of MSA if premiums are paid for 5 years or more.”
Policy name : Jeevan Saral
Term : 30 years
Premium: 6125 quarterly
Purchased in 2011
Age was 26 at the time of purchase
So, If I calculate the surrender value as per clause as per my understanding it is around Rs. 222590 [=Premium paid till date in years (7.5) x MSA (890360) /policy term in years(30)]. For this
calculation I have taken MSA value which is mentioned on my bond.
But when I have checked with LIC office it is only Rs. 155655 (against Rs.183750 total premiums paid till date).Could you please provide your guidance what is going wrong as I am not getting what
i have paid till date as per clause.
What can be done to get the at least the amount which I have paid till date for premium while surrendering the policy?
Dear Dinesh,
MSA value which you took is for that particular 30 years. You have to reduce the MSA to the 7 years and do the calculation accordingly. This is the catch with this product.
21. vineet wahi
Need advice …!
Policy name : Jeevan Saral
Term : 35 years
Premium: 3062 pm
Purchased in 2011
Age was 29
Should I continue with the same or should I surrender
Dear Vineet,
Continue-If you feel the Life Coverage is sufficient for your family to survive (if death occurs today) and also if you feel 5% to 6% returns are BEST for you.
Discontinue-If you feel both above points are not meeting with this policy.
1. vineet wahi
I do have other lic policy along with 75lac term policy. I want to know if This policy can give me a good return at the time of maturity cuz someone told me to surrender this policy and
switch to a good / high return policy.
What should I do please suggest .
22. KSHIRSAGAR
Monthly premium Rs.1000/- ,Age at the time of policy 34 , term is 10 years Lic calculated maturity sum assured @10889 and loyalty addition @355
pls adv how this figures were calculated .
Why exaggerated returns shown by Lic while introducing policy ?
Dear Kshirsagar,
For your information, such exaggerated figures were not shown by LIC but by few agents.
23. lalit
Hi Tonagatti,
Thanks for your effforts in putting this information in a blog.
I am staying out of india so difficult to visit branch and get things done.
If you dont mind can you please help with the calculation.
I took the policy in dec 2012 at the age of 29 for half yearly premium 9097 (yearly it becomes 18194 Inr) .
I took it as the agent was close relative and now when i review all my financial investment , I find this policy as not a great option and rather i will prefer some MUTUAL FUND + Term policy.
How much roughly I can get now (18194 * 6 years = 109164 INR Premium paid so far) If i surrender the policy. what is the best option to come out or better to stay invested in this ?.if really the
return is 4-5% then definately i am looking to come out of it?
Dear Lalit,
You no need to visit the branch, instead, call them and get the information. However, yes the returns will be around 4% to 5%.
24. Nikhil
Hi Basvaraj,
You said you will are “working on this chart and soon will upload the whole Basic Maturity Sum Assured Chart”.
Can you share the chart or excel sheet.
Dear Nikhil,
Let me share it soon.
25. Sanket Jadhav
Hello Basavaraj,
My plan details are as below.
Plan : Jeevan Saral
DOC : 28/04/2010
Term : 31 Years
Premium : 16844 (Quarterly)
Sum Assured : 13,75,000
1) Can you please let me know if the SA they have mentioned is correct for a policy of 31 yr terms?
2) I wanted it to surrender after 10 year completion, so what will be the amount I’ll get assuming today’s loyalty addition rates.
Dear Sanket,
1) How can I say this without knowing your details (like date of birth and all)? Better to contact LIC branch in this regard.
2) It will be around 4% to 5% returns (It is hard for me to calculate each individual).
1. Sanket+Jadhav
Thank you Basavaraj. I’ll get the details from LIC branch and will share here.
Dear Sanket,
26. Indro Saikia
Dear Basavaraj,
Thanks a lot for the great service to society by educating people with your informative posts.
I have a query w.r.t. my Jeevan Saral policy. Here are the details:
– Monthly premium: 4083/-
– Death benefit: 1000000/-
– Sum assured: 1450000 + Loyalty addition
– Policy term: 25 years
– Policy start: 2009 (9 years)
I have been contemplating since a long time to surrender the policy, but have been advised by agent to continue until completion of 10 years since Loyalty bonuses start getting added after that.
As per details provided in your post, will I be eligible for full MSA (14.5 lacs) since I will have completed 10 years, plus the loyalty addition on the MSA? Based on this, the total value should
be 14.5 lacs + 4.35 lacs(loyalty). Is this calculation correct? Can you please guide?
Dear Indro,
If you go to surrender at 10th year, then your MSA will not be Rs.14.5 lakh. They reduce to the applicable 10 year’s MSA and based on that they will pay the LA. This is where many got cheated
and this is where many got confused.
1. Indro Saikia
Dear Basavaraj,
Thanks for your response. Yes, I called up the LIC customer care number provided in one of the comments below, and got to know that the total amount I will get (surrender value + LA) if I
happen to surrender on 10th year, will be around 6.22 Lacs, which will be a return of 1.32Lacs in 10 years (even less than what an RD at 7% would give). If I surrender now (9th year), my
total surrender value will be around 3.5 lacs, which is less than 1.5 lacs of what I have invested so far. So I believe waiting another year will be more useful at this stage?
However the LIC lady also informed that over the years, this return will increase manyfold, considering many people have already surrendered the policy and ones left will surely get the
benefit. Although I am not sure if that can be believed.
Dear Indro,
It is up to you to wait and see. Regarding LIC lady claim, she is completely misguiding you.
2. Vinod Pandey
Dear Basavaraj, thanks for this blog & info on Jeevan Saral policy. I have done some calculation & tried to compute the returns this policy would give at different time frame & would like
to share that file with you for your inputs. Please guide as to how i can share the file with you.
Dear Vinod,
Send the file to [email protected].
27. Sachin
Dear Sir
I have paid premium for Jeevan Saral for 8 yrs. now.
Wanted to surrender my policy ,however my LIC agent was telling me to wait for it to complete 10 yrs. to avail the loyalty bonus.
Should I wait for Loyalty Additions ,is it a good amount to wait for ?
Dear Sachin, In my view, not a good idea to wait for another 2 years.
1. SRR
In some cases you tell people to wait for 10 years to surrender, and for some you tell them not to pay for 2 more years to complete 10 yrs.
Would be good, if you clear up the matter in just a few lines, about what the matter really is.
I would think along following lines .. when you go buy insurance for your car, you buy it every year. It is only when you meet with an accident, that your losses are covered. You are
paying the insurance company to cover your risk, and not to earn money like in a bank fd. Your car insurance premiums are used by the company for paying for the losses incurred by just
one car involved in an accident out of say a 100 other premium paying cars. The insurance company also recovers it’s own costs and makes profits. Covering risks is an entirely different
matter than investing. This Saral is more of a risk covering thing, as compared to other life products. Higher the age of the insured, higher is the risk covered by LIC. Lower will be
your maturity benefit.
Dear SRR,
If the policy period is about to complete 10 years, then it is better to continue and complete 10 years as LIC will consider 10 years of LA. However, for those who completed more than
10 years, they may discontinue at any point of time.
Now, coming back to your LOGIC of comparing general insurance to life insurance, your comparison itself is illogical. You are comparing Life Insurance with Idemnity insurance. In one
way you preach “Covering risks is an entirely different matter than investing”, by saying so, do you think this policy the risk coverage of this product is far better than even TERM
LIFE INSURANCE? First, decide yourself on what point you wish to defend. Whether you wish to consider this product as risk-mitigating or investment product. If this product was so
fantastic, there may not be cases pending against LIC.
28. PRANOY DEY
sir my agent told me that i will get 45 lakh after 35 years in jeevan saral policy. monthly premium is 2100…. can you tell me is it possible???
Pranoy-Calculate based on above post and arrive at who is right and wrong. Usually, agents exaggerate returns to show some fancy numbers. But the returns from this policy will not cross more
than 5% to 6%.
29. Akash
I have taken policy in year 2007 with annual premium is 24020 for 20 years, Sum assured is 500000. I understand that at maturity time, I will get assured sum + Loyality bonus.
1. I would like to know what will be loyality bonus.
2. I have paid premium till 11 years. IS it ok to complete 20 years or shall i withdraw now. ? if i will withdraw now, what will be surrender value?
Akash-At maturity, you will not receive assured sum but the MATURITY SUM ASSURED+LA. Refer to post and comments for LA assumption. You can continue this policy if you feel 5% to 6% is the
best return for your 20 years of investment.
2. Akash
my DOB is 24-04-1985
Akash-Refer above post.
1. Akash
Thanks for the reply. If I understand correctly, LA will be 500000* 525/1000 after end of 11 year. That will be around 2.6 lakhs. So 2.6 + 5 . Total will be 7.6 after 11 years. Is
calculation correct?
Akash-Here LA will be based on MSA but not on SA.
2. Sagar
Akash please follow below steps and pls pls reply to contribute back to help lacs of other Jeevan Saral buyers as you are one of the unique buyer who has completed 10 years in
Jeevan Saral:
Step 1:- Just call LIC customer care -02227725968..it is quite busy number but try continuously and provide you policy number they will tell you the exact surrender value.
Step2 :-Please share surrender value you get from LIC customer care and also share what surrender value is shown in the chart given to you by LIC agent at time of buying policy.
This info will help lot of people. Thanks in advance Akash
1. Bhuvanesh Rajpurohit
Hi Akash,
Can you share information if you have already done above steps suggested by Sagar?
I have this poilicy which has completed 7 years. I am planning to continue till 10 years thinking about the LA which i will get if i complete 10 years. At-least then I will
have no loss but surely very less returns. I hope my understanding is right 🙁
2. RNagpal
Hi Sagar,
Please advise how much you got after surrendering.
My policy has completed 10 years now and I want to pay the premium and surrender.
Dear Rnagpal,
It depends on many cases. You can’t assume based on what others get.
30. Srinivas R
Hi Basavraj, below are my policy details:
Age: 28 Yrs
Instalment Premium: ? 48,040.00(Yly)
Premium Paying Term: 32 Yrs
Policy Term: 32 Yrs
Sum Assured: ? 10,00,000
Commencement Date: 24/02/2010
Date of Maturity: 24/02/2042
Could you please help me with the Maturity Sum Assured and LA at time of Matiruty?
Srinivas-Please refer above post and also the comments.
31. Sunil Kumar K
Dear Basavaraj,
I have taken the following policy for my life coverage + investment.
Jeevan Saral (Plan-165) , Premium 25,382 p.m. (annual Rs. 3,09,984)
Started when I was 34 years on 04/09/2012. Policy term is 33 years. Maturity is 04/09/2045.
I may not be able to pay this huge monthly premium till its maturity date. If I surrender now how much I will get. OR should I continue this till its 10th year? I dont have a good term policy for
my life cover. What should I do with the policy? Please advice.
Sunil-Better to surrender. Regarding values, contact the LIC Branch.
1. Sunil Kumar K
Thanks Basavaraj for the quick response. Any suggestion on the replacement policy/investments after I stop my Jeevan Saral
Sunil-That depends on many things. Hard to say BLINDLY.
2. Sunil
Basavaraj, can I stop further premiums and wait till it reaches 10 years before surrendering?
Dear Sunil,
Yes you can do so.
32. Devesh
I have taken Jevan Saral Policy (165) with yearly premium – 48040 rs/- on 2011-12 academic year.
Policy name Name : Jevan Saral(165)
premium : 48040 (yearly)
taken at age : 24
Maturity Sum Assured :2109640
Death Benefit Sum Assured : 1000000
Policy Term : 35
When my age will be 59 years old (If I am alive), the policy will matured and I will get 2109640 Rs. Hence, Maturity sum assured is fixed on my policy and it will be approximate 24 % of total
premium I have paid.
48040 x 35 = 1681400 Rs and its 24 % will be my Maturity Sum assured.
You are telling that Maturity sum is not fixed but inmy policy it is fixed and 2109640 r s has been printed in my policy.
Please guide me if my thinking and calculations are right …….
Devesh-MSA not fixed but depends on your age, premium and term.
1. Devesh
Thank you for quick reply. Hence, as per above policy containing my age,premium and term, when my age will be 59 years old (If I am alive), the policy will matured and I will get 2109640
Rs. Am I right?
Devesh-You will receive the applicable MSA+LA.
33. Amit
I have a 5 lac policy which I have taken for 25 years in 2009. I am paying 24000 yearly premium, since last 8.5 years. I am now unable to bear the premiums and want to stop paying it. How much
will I lose if I surrender the policy now, does it make sense to continue paying premiums? At this stage, will I gain more if I were to still surrender it and invest in MF SIP for example.
Amit-If you are unable to bear the premiums then no need continue this policy. Regarding values, visit the concerned branch.
34. Sheldon
So I have a small doubt here.
I have 2 policies of Jeevan Saral.
1) First was taken by my father when I was 12 in the year 2009. The Death Sum Assured is 200000. The Maturity Sum Assured is 224848. There’s a monthly installment of 800 on it. They are not
taking any Accident Benefit Premium from me. So anyway, after calculation, it shows me, I will get approx. 1.2Lakhs, if I mature it after 10 years. So its a profit for me as I am paying 96000 for
10 years [800*12months*10years] and receiving almost 1.2Lakh after 10 years that is in 2019. Am I correct in this calculation?
2) The second and the same policy was taken by my father for himself when he was 55 years old in the same year i.e. 2009. The Death Sum Assured is 200000 for this also. The Maturity Sum Assured
is 69240. There’s a monthly installment of 817 on this one though. So we would pay Rs. 98040 for 10 years [817*12*10] and what we are getting after maturing it after 10 years is Rs. 69505 approx.
So basically this is a pure loss right? We just spoke with our agent regarding surrendering the policy & he said we will get around 76K if we surrender it now which I guess is a lesser loss
comparatively. So is my calculation correct? What do the people here think about it?
I highly appreciate your response. Please guide us with this.
Sheldon-1) Use IRR calculation to arrive at how much PERCENTAGE will be your earning.
2) I am not sure on what basis you are assuming and calculating. Refer above post properly and calculate IRR.
2. Dilip Kulkarni
From my experience with Jeevan Saral policies I can say that if the policy is taken at a young age, preferably below 35 years, then only you can expect to be in the positive. Those who have
taken after age of 35-40 years will be in a loss whether you wait for maturity or surrender it. I suppose you have understood the special surrender and loyalty addition clauses. Main thing in
this policy is understanding the maturity sum assured concept. I am sure that most LIC agents also will not be able to answer any questions on that.
35. Mohit
Sir , still I am v much confused gone through the whole article , which is V nice But need your help.
I have taken this jeevan saral policy for 16 yrs and I am paying 10,208 ra monthly to LIC .
I have started this police since Oct’11 .
Sir how much will i get if i surrender this plan after 10 yrs ?
If after 16 yrs ?
If today ?
The LA which lic declares gets accumulated each year or not .suppose whatever LA lic declared for year 11 -12 and subsequent yrs will be added to my plan or not or will LA will only start after
10 yrs .
Pl help and advise
1. Dilip Kulkarni
Loyalty addition is only one time. At maturity or surrender after 10 years. Do you know what is the maturity sum assured? (not the death sum assured). Moreover the loyalty depends on the
maturity sum assured and the sum changes every progressive year.
2. Dilip Kulkarni
The real problem with this policy is that probably not even one policy holder, including me, knows about the maturity sum assured and neither any agent knows or even if he knows he does not
communicate with the customer. If one knows the maturity sum assured then it is easy to understand the various scenarios about the policy.
1. Sagar
just call LIC customer care -02227725968..it quite busy number but try continuously and provide you policy number they will tell you the exact surrender value.
Basavraj I believe you have to put this number in topic heading to avoid repetitive posting 🙂
Sagar-Ha ha..Sure.
2. Sreenivas
Now you can see the MSA in your policy schedule PDF which is available online on LIC portal. I just downloaded checked today.
1. deepak
Hi Shri..
I dont see that MSA in portal 🙁 ,can you guide me please where to check exactly.
Deepak-NO NEED TO CHECK ANYWHERE. It is written on your LIC Bond itself.
Mohit-LA is one time payment and be considered based on the year in which you will close the plan and depends on the MSA of that particular year. Hence, you have to contact the branch for a
better understanding of exact amount.
36. Ami
Thanks this is the first time I am able to get hold of the terms of the policy , it bad we were trusting and naive at the time of purchase , and the agent happens to be a distant relative.
My take 0-4 years surrender, 9+ years wait for magic 10; thanks and greetings to curator of the blog
Ami-You are now in the right direction.
1. Dilip Kulkarni
Age is an important factor in determining maturity sum surrendered. In my opinion those who have taken the policy below the age of 35 can expect to get some return, though not even
comparable to FD, and those who have taken above age of 35-40 should not expect to get any positive return.
Dilip-It is indeed. But why one has to sacrifice for low negative real return for long term investment?
37. A Mohanty
I’v2 2 Jeevan Saral Policies with combined premium of 10k per month. It’s been 5 years since I’m paying premiums and about to pay the 6th (almost 6 lacs I’ve paid already).
I realise that I’ve committed a mistake going for this. I want to surrender both the policies. But the confusion is whether I should surrender it right now or it’s better to wait for 10 years so
that I can get some LA when I surrender.
Please help me calculate the surrender value now and after 10 years (considering the current loyalty bonus for 10 years). It would be a great help so that I can decide on what to do.
Mohanty-Better to surrender now. Regarding values, better contact the LIC branch.
1. A Mohanty
Thanks a lot. From your reply, one thing I understood that there’s no point holding it for more time.
Thanks again.
38. Vinay Narhari Bhagwat
Sir,I have taken Jeevan Saral Policy and completed 10 years and paid 10 premiums and last 10th Premium paid on July 2017, amount 18.015 INR
Can you please guide me ,should i surrender the policy?
Actually i want to surrender this policy
What will be approximately surrender value?
Vinay-Continue if you feel the returns of around 5% to 6% is BEST. Else better to surrender.
2. sagar
Hello Vinay,
just call LIC customer care -02227725968..it quite busy number but try continuously and provide you policy number they will tell you the exact surrender value.
Please share surrender value you get from LIC customer care and also share what surrender value is shown in the chart given to you by LIC agent at time of buying policy. This info will help
lot of people. Thanks in advance Vinay.
Thanks Basavaraj for this thread
Sagar-Thanks for sharing this 🙂
3. Sagar
Hello Vinay,
Highly Appreciate if you respond for below query, as you are the rare candidate who has completed 10 years with Jevan Saral…I have just completed 7 years and searching for candidate with 10
yrs completion to decide whether it make sense to wait for next 3 yrs.
Please share surrender value you get from LIC customer care and also share what surrender value is shown in the chart given to you by LIC agent at time of buying policy. This info will help
lot of people. Thanks in advance Vinay.
39. Umesh kokitkar
Hi, I have taken Jeevan Saral t.no 165 policy in 2013 and for 35 yrs.
Monthly pay = 3060 (12 policies 255 * 12 = 3060)
Wanted to know maturity amount after end of policy (@35 yrs completion)
Umesh-Refer above post and below comments.
1. Alpesh
Hello Basavaraj ji !!
Part 1
I have taken 3 policies for Jeevan Saral
1 ) Policy 1(self)—335640(Current surrender value)—taken on 09/2009—taken on 12/2013—-Annual premium 48040—paid 7 premiums till date
2 ) Policy 2(sister)—373889(Current surrender value)—taken on 09/2009—-Annual premium 48040—paid 7 premiums till date
3 ) Policy 3(brother)—-333861(Current surrender value)—taken on 09/2009—-Annual premium 48040—paid 7 premiums till date
Policy term is 35 years.
I had asked the advisor to drew for 10 years but he drew for 35 years saying you can stop it whenever you want after 10 years.
I won’t be able to continue this policy for 35 years.
The Insurance advisor who drew this policy said you have to pay premiums till 2018 in order to get more than what you have invested.
Please guide !
Part 2
Jeevan Anand
1 ) Policy 1(self)—57540(Current surrender value)—taken on 12/2013—Annual premium 25191—paid 4 premiums till date
2 ) Policy 2(brother)—56940(Current surrender value)—taken on 12/2013—-Annual premium 24906—paid 4 premiums till date
Policy term is 21 years.
I had asked the advisor to drew for 10 years but he drew for 21 years saying you can stop it whenever you want after 10 years.
I won’t be able to continue this policy for 21 years.
The Insurance advisor who drew this policy said you have to pay premiums till 2022 in order to get more than what you have invested.
Please guide !
When shall I surrender
Part 1— Jeevan Saral policies
Part 2— Jeevan Anand policies
in order to get more than what I have invested ? So that I do not lose on the hard earned money.
Thank you in advance !
Alpesh-You are in trap. Also, in search of getting profit from these, you pay more and more and finally end up with savings account return. You can come out from those policies which
completed 3 years. No point in being ADAMANT that I not BOOK LOSS 🙂 If you close and invest wisely in future, then definitely you can compensate the same.
40. Dilip Kulkarni
Let us all together try to understand some things. Loyalty addition. If someone has purchased a policy of 22 years term having maturity sum assured of Rs. 10 lakhs, then will he get loyalty
addition of (400 X 10,00,000)/1000 i.e. 4 lakhs on surrendering after 10 years? how many think this is correct?
What will be the special surrender value after 10 years? [(10/22) X 10,00,000] i.e. Rs. 4,54,545 plus loyalty addition. Any comments on this?
1. suveen
Dilip- If someone has purchased a policy of 22 years term having maturity sum assured of Rs. 10 lakhs, then will he get loyalty addition of (400 X 10,00,000)/1000 i.e. 4 lakhs on surrendering
after 10 years? how many think this is correct?
In my opinion,its not correct…. Loyalty addition of 4lakh is for complete 22years and for maturity sum assured of 10Lakhs, but for 10years maturity sum is not 10lakhs. It will be somewhere
around 3.2lakhs( for exact MSA u can contact customer care and get it)
Now for (320000*300)/1000 = 96000( loyalty addition for 10years)
Total is LA+MSA=96000+320000 = 416000 ( total surrender value after 10years)
1. Dilip Kulkarni
Can you see the following link about loyalty addition and post your comments on it.
Dilip-What MORE it tells?
1. Dilip Kulkarni
In above example MSA after 10 years is (10/22) X 10lakhs. Is this right? For loyalty addition also the same formula is applied or it is according to the table given in the link.
2. Dilip Kulkarni
Formula to calculate loyalty addition: (Maturity Sum Assured/1000)* Rate of Loyalty Addition as per annual premium band
Note: Maturity Sum Assured on policy bond is written according to the term you have chosen at proposal stage. It will not be used to calculate loyalty addition while surrendering
the policy before term.
Dilip-This is what I too explained. In case of this plan, if policy continued for more than 5 years, then while surrendering they reduce the MSA for how long the policy
2. suveen
Dilip – It is also same as i discussed in above comments…
For the purpose of determining the rate of loyalty addition for exits by death or surrender or maturity the duration in completed years for which premiums have been paid shall be
For example 10years: you have paid so far 30025*10=300250
Now LA calculated as (300250*400)/1000 = 120100
Total = MSA for 10years( 2.8 lakhs appx) + 120100
This is my understanding, please anyone can comment if you guys have surrender policy after 10 years…
1. suveen
Dilip – Exact MSA u can get it from lic customer care service.
LA formula it is according to the table given in the link.
1. Ashish Kumar Roy
Dear Mr.Basuvaraj Tonagatti Sir, I am Ashish Kumar Roy from Nagpur I have a Jeevan Saral Policy (2013 table 165/20)and I am paying 72780/- per (36390+36390) as a premium
which is very difficult for me to deposit. I wanted to know that after 10 yrs. of my policy if I will surrender how much amount I will get as LA as well as paid amount
Kindly calculate and inform me because in that policy its a very worst policy introduce by LIC , so please help me ASAP.
Ashish Kumar Roy
Ashish-Refer my latest LA rates “LIC Bonus Rates for 2017-18 – A complete list“. Using that LA rate, you can calculate on your own.
41. Dilip Kulkarni
I bought a Jeevan Saral policy in 2007 at age of 52. Policy term 18 years. Annual premium Rs. 30025. Maturity sum assured is Rs. 6,25,000. Paid 10 premiums. Want to discontinue. How much special
surrender value (including loyalty additions) should I expect? will it be (10/18)X625000 plus loyalty addition?
Dilip-Contact the branch.
2. suveen
hello Dillip Kulkarni,
can u contact branch and post comments on how much surrender value you will get now (policy inforce for 10years)…
Yours comments will be useful for many of us because even I am waiting my policy to complete 10years. so far i paid 6 premium… when i contacted branch to surrender policy they insisted me to
continue for 10years..
1. Dilip Kulkarni
I have paid the 10th year premium and 10 years will be completed in November. I will contact the LIC office then so that I will get a proper and correct answer. I will post on getting the
information. Meanwhile if anyone else is in a similar position and can comment on the situation, they are welcome.
Dilip-Each individuals values differ and they will come to know when they actually visit the branch. Hence, better you act and share.
1. Sagar
No need to visit branch just call LIC customer care -02227725968..it quite busy number but try continuously and provide you policy number they will tell you the exact surrender
value. Please share surrender value if anyone has completed 10 years…Share committed surrender value time of policy taken and actual you get from talking to above number. I have
completed 6 years and tato wait till completion of 10 years.
Thanks in advance
Suveen-May I know LIC officials login in insisting you continue for 10 years? It is because NO LICian will force you or suggest you to close the policy.
1. suveen
Basavaraj — Lic customer care guy suggested me to continue for 10years so that atleast i will get back 10years premium paid(36030*10=360300) with loyality addition. Altogether with
loyal addition i will get more or less 3.6lakhs… So i am waiting for 10years…
Suveen-Ask the same guy the IRR %. If it is fetching less than the current Bank FD rate, then what is the use of such product?
1. Sagar
Yes but even if it gives less than bank FD rate you will loose some part of your 10-12% premium paid till date (i.e. before completing 10 YRS) so this loss of 10-12% is making
me to wait till completion of 10 years. Your suggestions are welcome on my decision to wait for 10 years(7 years already completed, 3 yrs pending)
1. Anand
I had started the policy in 2007, with annual premium of 24020 in Jeevan saral policy. currently the surrender value revealed is about 357,315/-. Two years back it was
I got info from the customer zone, that new declaration happens in September. So i am thinking right now, what time to exit within this year to get best surrender value.
Any inputs or thoughts?
Anand-The best day is TODAY.
1. Anand
will i get additional LA which may be declared in September? Is there any such thing? will the surrender value vary as on today and one month later?
Since i have completed the “magical 10 year”, then with small wait, i may achieve something better.. unless there is no change at all..
Anand-That magical 10 years or that surprising additional LA will not add any value to you. Rest you are free to decide.
2. Nikunj
I have taken New Jeevan Anand policy in 2015 with sum assured of 2625000 and premium of 122000 annually. Return comes out 6.2% on recurring but problem is how
to calculate Maturity Amount? Even LIC don’t write on website. Agent has written 70L rs after 25 years but based on LIC calculation ( Sum assured + Bonus) it
comes out very less as bonus declared is 49rs. Please help if I should continue it or I will get atleast 70L+.
Please help asap !
Nikung-Your return expectation is correct. Please refer my earlier post where I shared a video of how to calculate returns “Video tutor-How to calculate LIC
policies maturity amount and returns?“.
2. Sagar
Hello Anand,
The surrender value of 357,315/- is matching with the chart given by LIC agent while buying the policy? If not how much is the difference? Please reply, as this will
help me as I am in 8th year of this policy
1. Dilip Kulkarni
Sagar, whatever your decision, please post it on this forum. I think there is no harm in waiting for 10 years to complete. Not 10 premiums but 10 years total. I am
also waiting for my policy to complete 10 years which is October end this year. My annual premium for 18 year policdy is Rs. 30,025. I also was given a quote of
about 1,65.000/ two years ago.
1. Dilip Kulkarni
If you surrender before 10 years you normally get lesser than what you have paid. If you surrender after 10 years at least the loyalty bonus makes the amount
what you get more than what you have paid. But if you calculate rate of return it is very low in both cases.
Dilip-If you surrender it after 12th year, then it will be more than what you surrender in 9th or 10 years.
3. Dilip Kulkarni
I think the question was asked by Anand, not Sagar. So the answer is for Anand. Sagar’s views are also welcome
2. Anand
Hi,Sagar, Chart showed about 349000 for surrendering in year 2017. But after 10 years, returns are just 5-6%. So it’s not worth investment..mixing insurance and
investment is not appropriate..
1. SaralRider
Hi Anand,
The chart showed you 349k but you your value is 357k. I think you did well. Also, to my understanding, mixing insurance and insurance is best option if you
diversify your portfolio properly.
Getting 6% is not bad. Don’t forget that you have 10k deductible from your income which reduces you tax, and you get tax free money when you surrender. So,
when you consider all benefits, your RoR could be 10%.
And don’t forget that it covers the risk 250 times your monthly premium.
I think, this policy is like seed of a fruit tree you plant, you will have to remain patience to bear a fruit.
SaralRider-Getting 6% for long term investment is NOT BAD? God save you and to whom you are suggesting. LET GOD PREVAIL SOME KNOWLEDGE ON YOU.
Sagar-MSA mentioned in BOND holds good if you hold the policy until maturity.
42. Anil Kumar
I have taken Jevan Saral Policy (165) with yearly premium – 48040 rs/- on 2011-12 academic year.
Policy name Name : Jevan Saral(165)
premium : 48040 (yearly)
taken at age : 25
Sum Assured : 1000000
Policy Term : 35
1. As of i have completed 6yrs payment and what is the maximum amount if i continue this policy for 35 years?
2. Is it recommended to continue this policy for that such longer period?
Anil-1) Around 5% to 6%.
2) If you are satisfied with the return I shared in the first answer, then continue. Else, think seriously.
1. Anil Kumar
Thanks quick reply Basavaraj.
When i calculate got the following amount with rate of interest 5 to 6% (4,820,926.32 to 6,043,766.04) I am thinking or calculated right way?.
Do we have any other best policy in market to get more life coverage and returns?
Anil-Whether your concern is buying a right and enough Life Insurance or Investment? Thumb rule is that never combine both.
43. Neeraj Singla
Hello Sir,
One of LIC Agents has given approx 15 policies Jeevan Saral policy to my father. Some of them are monthy,quartely and half yearly plan. Approximately we pay 60000 per year. We have have paid the
premiums for these policies for 5 years. Some of these policies are on my name , my brother,my sister,my mother and fathers name. We want to surrender these policies as its difficult for us to
pay the premiums. So how much surrender value we get? If we continue the policies on my father’s name would it be beneficial and how much sum assured vlaues we get after 10/15 years. Please find
below details:
4 policies half yearly >> Rs1516
5 ploicies quarlety >> Rs766
7 policies yearly >> 2426
Thanks & Regards
Neeraj Singla
Neeraj-After reading below comments, you still asking whether to continue or not? Visit the nearest branch to understand the surrender values.
1. Neeraj Singla
I checked with LIC office and they said we would not get full premiums.We have paid 300000 and they said we would get approx 220000 only. But as per below statement shouldn’t we get Full
premium as we have paid for 5 years? Please clear this doubt.
“b) Special Surrender Value-It will be of 1+2 options given below.
1) 80% of MSA will be paid if less than 4 years premium paid, 90% of MSA if between 4 to less than 5 years of premium paid and 100% of MSA if premiums are paid for 5 years or more.”
Neeraj-I am not said you will get full of what you paid, but I said that MSA will be reduced as per 5 years tenure and can be paid to you. MSA of 5 years will be reduced, which may be
lesser than your premium paid.
1. Neeraj Singla
OK. Thanks alot for your advice.
44. Mark Sabastian
Hi Good Evening Sir,
I have seen quite a few advice from you for others and I do appreciate your time.
Need a help on the policy which I took couple of years back.
Policy name – Jeevan Saral with Profits
Table – 165
Term – 20
Maturity sum assured – 10,00,000
Premium method – Yearly
Premiums paid – 5
First premium – Dec 2012
Premium – Rs. 48,040.00
Next Premium due – Dec 2017
Policy Status – Active
The main reason for me to start this policy was to save money & avoid tax on the other hand so I spoke to an agent who said this policy should suite my needs, I specifically asked him if i need
to close the policy any time will I get my money back? He answered yes however, would need to wait for 3 years.
Post 3 years when I spoke to customer service they said I would loose some money if I cancel, and advised to complete 5 years for some benefits.
Now, I have paid all 5 premiums which is 48,040*5=2,40,200 & when I checked with customer service they advised me that I wouldn’t get the complete amount. I feel like getting fooled as I am
loosing my hard earned money. I think I heard saying I would get only 1,80,000 if i cancel.
Please advise is it best to cancel policy or is there any way for me to get my complete money if i extend for 10 years for loyalty benefits. If I also extend for 10 years I am afraid as i may not
get 48040*10=4,80,400. Kindly advice.
Mark Sabastian
[email protected]
Mark-Better to discontinue. You asked for SAVING and TAX BENEFITS to your agent. So he provided that option, but not an INVESTMENT OPTION. Hence, in my view agent not did anything wrong. But
it is mis-buying than misselling.
1. Mark Sabastian
Thank you so much for your feedback, just one thing I need to ask you sir….,
If I ever continue or wait till I cover my complete tenure, is there any benefits for this policy or will I still lose my money. Please suggest.
Mark-You will not lose money but end up with less returns.
1. Mark Sabastian
Can you tell me how much would I get if I continue for next 5 years, which is ten years in total please.
Sorry to be a pain but your inputs will make me to take decision.
Mark-I can’t tell the exact figures. But returns will be around 5% to 6%.
1. Mark Sabastian
When you say 5 to 6 percent, is it my total premiums paid for ten years with 5 to 6 percent approx additional or am I missing some thing else?
Mark-It is 5% to 6% on your total premium paid over the period of time.
1. Mark Sabastian
Hi Sir,
I have paid premium – 48,040.00 for 6 years which is so far 2,88,240.00
I should be able to manage for another 4 years, which would be 10 years if in total and total would come up to (48040*10) = 4,80,400.00
Hence, if I surrender post 10 years, I should except returns of 4,80,400.00 + 5 or 6 % extra.
Please correct me if I am wrong… Thanks!
Mark-It may be around 3% to 4%.
45. siba dalai
dear sir,
i have bought a jeevan saral policy on 22/2/2011 for a period of 20 years. firstly it was monthly plan of rs. 5104 and thereafter changed to quarterly amount of rs. 15,312. sum assured was
12,5000/- . date of birth is 26/10/1983.
sir, i am really sorry to state that being very poor in calculations, i need your expert advice. my querries are :
1. what amount will i get at the end of 20 years.
2. what amount will i get if i surrender it now ( already 5 years over)
3. should i continue here or invest it in mutual funds.
please advice. no one is helping me out. and you have been so kind in answering everyone’s doubts. the lic agent isnt helping out sincerely. thanks sir in advance.
1. siba dalai
sorry sir, sum assured is 12,50,000/-
Siba-Refer above post and around 1000 comments. You will come to know the answer.
46. M. Saha
my premium is 3062 per quarter ( I think net premium) …. SA is 2.5 lacs ….. term is 20 years… i bought it in Dec 2010 … What would be my return after maturity in year 2030 ? can you elaborate
with a chart or breakup amount.
Saha-Elaborated already in above post and if you go through the comments, then you can understand the return expectation from this plan (around 5% to 6%).
47. Amar
I invested Rs 24020/- for 3 years till Dec 2016.Recently I have taken a term plan also for 50lakhs from LIC. I am investing in PPF also.I am very much confused whether to divert this premium
amount towards ELSS fund or continue with the policy till 11 years or 25 years.My policy term is 25 years.The maturity sum assured is around Rs7.5 lakhs in policy bond.
Amar-What prompted you to buy this product and why you now feeling this to close?
1. Amar
Actually I was not so investment conscious about my investment portfolio at that time as i was not coming in any tax bracket.Now, I am in 20 % tax bracket.At that time I was not aware of
Term plan and also was not comfortable in investing any Private Companies.I used to deposit in only PPF. At that time I was feeling i am short of Insurance and hence bought this policy
without keeping any perspective in mind that I am taking this policy for Investment or Insurance.Now , when I am trying to take advise from others.Some are advising me to surrender and
some say the policy is good you should continue for at least 10 years. I am 34 now and dont want to take any wrong step related to investment and insurance and ruin my later years. I
request you to kindly advise whether to continue investing in this policy or discontinue.I really dont mind investing also if I am supposed to get better returns.As, today I feel I am
adequately insured as far as Insurance is concerned in comparison to my annual package as I am having another Limited endowment policy of 2 lakhs also from LIC and also insured by my
employer for approx 15-18 lakhs.I will be taking your advise as the final call for this policy as I dont wont to linger on with this policy as I have already invested 1.5 lakhs for my
Income tax rebate by investing in PPF, PF , LIC(Limited endowment plan) and LIC e term( Cover of 50 lakhs). I have tried enquiring from lot of people related to its return part also. No
one is able to give a clear picture in this regard also.
Amar-If you have adequate insurance then why you retaining this dummy product? It is your money. Therefore you judge which BEST to you. Hence, never rely on anyone nor listen to those
who suggest you based on their opportunity in your investment (agents). Act fastly and close all such dummy products.
1. Amar
One last question. Should I see this product on investment perspective or Insurance, In case on Investment then what rough percentage can be expected on this product.As this
product is giving me only 5 lakh on insurance front and 10 lakh on accidental.
As I have an option of diverting his fund towards NPS and claim 2 lakh of income tax rebate instead of 1.5 lakh in coming financial year.
Incase I am closing this product this year I will booking loss of around 30000/- which I dont mind if I get returns by investing in long term.
Thanking you in advance.
Amar-For 10 years of investment and return 5% to 6% enough for you? Is so, then continue. If you consider this product as insurance, then think how long your dependents
survive with the death claim amount they receive from this product.
1. Amar
Thanks a ton for your advise.Will be applying for surrendering in this scenario.
1. Wayne
You should seek investments in mutual funds through SIP for wealth generation and buy term insurance for pure insurance need.
48. Kevi
I have a Jeevan Tarang plan of 10 years policy term. I want to surrender this policy on the 9th year after paying 36 premiums (quarterly payment mode). What is the formula to surrender? I want to
know how much I will get.
Kevi-Check with LIC branch.
49. Sachin
Hey mate, you have been doing a fabulous job of explaining the intricacies of the Saral Policy. I have been hurt the same with almost paying through the roof per year premium paying in saral. The
LIC and their agents just care about their bottom line. They gave wrong charts and return values when they were getting people onto buying their policy. Why can’t they be put behind bar for wrong
selling of policies with wrong information.
Sachin-Sadly none come forward and prove that they wrongly sold.
50. shankar baghele
Dear Sir, I was opened policy jeevan saral 165 plan 22 year in sept-2013 , 25473/- Half yearly. at that time misguide & show me higher return by LIC agent & open this policy. Now I want to
surrender this policy after 4 year.
My query is the following question?
1) If continue this policy to full maturity (22year), what is amount I will get?
2) If I surrender this policy after 4 year than how much amount I will get?
3) Can I rest policy after 4 year without pay premium till maturity & after maturity how much I can get.
I want higher return but with this policy I cant get.
please give me proper calculation & suggestion.
Shankar baghele
Shankar-I already replied all such queries in below comments. Please go through them.
51. LICBuyer
Hi Basavaraj,
I have a query regarding my LIC Jeeval Saral Policy Table no 165. Bought this policy on 28/12/2009 (misguided with the chart which agent showed me) at age 26 Policy term -30 years Premium -72780
yearly (approx 6000 per month) Death Benefit Sum Assured 15Lac. I have already paid for 7 years : approx 5 Lacs. I have also taken loan on this policy last year (May 2015) Loan amount 285000. I
have made up my mind to surrender this policy as policy is not good and I am unable to pay premium. Question is should I surrender right now or should i wait for the completion of 10 years for
this policy
1. I want to know if it is good idea to hold this policy for 10 years completion (until Dec 2019) so that I get Loyalty addition.
2. What amount I can expect to get in return from this policy after 10 years completion
3. How loan will be handled in this case. Will LIC deduct Loan and interest at the time of surrender. I found that they have already charged 30K interest for 6 months on loan amount of 2,85,000
which is very disappointing. This interest will be charged every six months. I am confused if it is good idea to hold until Dec2019(as holding also means more interest on associated policy loan).
But even if I surrender now, I have huge loss.
Really need an advice. Please help!
LICBuyer-1) You can surrender now also.
2) Check with LIC.
3) Yes, they deduct all outstanding loan and interest on that before paying you the surrender value.
1. LICBuyer
Thanks for your prompt response. Regarding surrender or holding for 10 year(to get LA) which one will better suit me. What approx LA LIC usually provides?
LICBuyer-I prefer surrender over holding. Check with LIC for LA rates.
1. LICBuyer
Thanks you. you are going a good work in helping people. Genuine advice. Appreciate your efforts!
52. sachin
i have take jeevan saral policy in november 2009, paying premiun of Rs 48040/- yearly.
Clearly understood that the chart agent showed is misguiding.
Currently i have invested 336000 in total whose surrender value is 316000 right now.
What will be the better option—
1) book loss of 20000 and surrender immediately
2) stop paying premiums and keep the policy active for 3 more years and redeem after completing 10 years of policy so that i am eligible for LA/bonus.
3) pay premiums for 3 more years and then surrender the policy after completing 10 years of policy to avail LA/bonus.
In any case if i surrender i am planning to buy jeevan anand policy for approximate same amount for a period of approximate 30 years.
Sachin-First one.
1. sachin
Thank you ver much.
Can u guide me the base for booking loss ?
1. sachin
i mean why did you suggest me to book loss and surrender it..
In case if i reciveLA of rs250 for every 1000, i can get 120000 as LA along with 480000 (total of premium paid)
1. Sachin
Sorry I have paid 386000 and current surrender value is 316000…so lot of loss..
And I have 3 such policies–mine,sisters n cousins..so lot of loss–a total of 2.1lakhs..
Sachin-Your loss will be more when you extend. You can compensate the current loss by wisely investing.
2. Sreenivas
You can cover the loss by investing in mutual funds for long term like 15-20 years.
53. Ranjan
I have taken jeevan saral plan dec 2011. and i wand u surrender now. But LIC Agent is saying that.
25000*10 = 250000
Your balance Amount is 250000-85000 = 165000rs
Thanks & Regs.
Ranjan-Don’t believe agents. Go to LIC branch and enquire.
54. Vaibhav
I have completed 6 years of Jeevan saral 20 yr term plan paying of Rs.6406/- quarterly (i.e. approx Rs. 1,53,744/- till date) as an installment. I want to surrender this policy due to some
financial problem, so I visited to Lic office and I took surrender value statement. Surrender value is of Rs.1,24,000/- ( approx loss of Rs.30,000/- ) I don’t know how they calculated surrender
value but it’s really shocking to loss of Rs 30 k. now what should I do?
Vaibhav-It is the reality. Either surrender or continue.
55. GAURAV
I have taken Jeevan saral policy for 20 years in Feb 2011, yearly premium is 12k, need your help in below queries-
1.)If i want to surrender policy after 10 years i.e in 2021 then will i be eligible for loyalty additions as total tenure is 20 years?
2.)Is LIC declares Loyalty additions every year for Jeevan saral?
3.)I bought this policy in Delhi now i moved to Bangalore can i surrender the policy in Bangalore LIC office, If yes what is the process for the same?
Gaurav-1) Yes LA will be payable but based on 10 years but not on 20 years.
2) As of now YES. Not sure about future.
3) NO.
1. Deepak Shukla
Hello Dear,
Under jeevan saral Im paying 6125 INR as monthly premium. At the age 24 I took this plan. Policy term period is 25 years. Agents show very high figh amount in their chart as maturity. But
as i received my policy it is showing maturity sum assured approx 20-21 lacs which is very low as compared to agent. I took this plan on 15/10/2012. Kindly help me out what i will get
exactly as maturity amount.
Deepak-You will receive Maturity Sum assured and Loyalty Addition. Read the above post fully and below comments. You will get an idea.
2. Rajib Layek
I think the LA will be not given as it is applicable only after 10 years. So if someone continue for 11 years then he will get the LA for last one year not for whole 11 years !!
Rajib-LA is one time payment not for all 10 years or 11 years.
1. Rajib Layek
Yes. I understood. Thanks for confirming !!
56. Ashok Kumar Singh
I purchased Jeeven Saral Policy at the age of 32. I have completed the 6 year of premium. Yearly premium is comes 24500. Agent assured me at the time of purchasing the policy that I get more than
10 Lakhs on the maturity of policy after completion of 16 year. Now it is come to my knowledge through some other agents and friends that the chart shown under this policy is fraud.
Will I continue with this policy or not .What is surrender value of this policy at this time. If I continue, will I get the value actually shown in the chart after 16 years? This question has
become more relevant as the policy purchased for meet the expenditure at the time of my daughter weeding.
If I surrender the policy, Can you suggest me some other policy or investment plant to meet expenditure mention above. Now my age is 39.
Ashok-Better to surrender. For surrender value check with LIC branch. You firmly believe that it is a fraud selling. But asking me whether continue or not. The decision is YOUR’s. Without
knowing much about you, how can I blindly guide you about investment?
1. Ashok Kumar Singh
Thanks for your valuable advise. I am going to surrender this policy. Can you suggest me some other investment option. Am i go with SIP, if yes please suggest some plans.
Thanks for your reply.
Ashok-Without knowing much about you, how can I guide the investments?
57. prakash ghuge
sir i want to surrender my policy
please guide
policy start 28/03/2011
age when started 27 years
monthly preimium 5014
sir pl tell what amout will get me
Prakash-Check with your nearest LIC branch.
1. prakash ghuge
sir what is better option whether i wiil continue the policy or surrender
Prakash-Better to surrender.
1. prakash ghuge
Thank u sir
58. Sheetal Bhat
Hi Basavaraj ji,
I have query regarding my LIC Jeevan Saral policy where i completed 7 years with yearly premium of 36030. When i took this policy he showed a T165 chart and assured that i will take 6.5 lacks
after completing 10 years.
when i enrolled my age was 23 and now i am 30 years of age.
Need your kind and valuable advice what shall i do, because what i understood that i may not get assured amount as per chart. Is this total fraud ? please advice as i am really worried.
Sheetal-He misguided you and it is not assured return. You may expect around 5% to 6% returns. Better to close the policy at the earliest.
59. Joy Mukherjee
Dear Basavaraj,
I have two JEEVAN SARAL policy.
DOB: 30-04-1986
Policy taken – 2011 February,
Yearly premium=Rs.18000/-(for 30 yrs term) and Rs.6000/-(for 25 yrs term)
Please tell me the amount if I surrender the policy after 10 years completion. Thanks in advance.
Joy-Refer below comments. You will find the answer.
60. suveen
Hello Basavaraj Sir,
i planned for Jeevan saral of Rs750000 with annual premium of 36030.
My agent told me i would be getting around 25lakhs after 20 years, i dont know exact MSA or LA.Even though he has given one pamphlet explaining premium & maturity.
Premium – 36030 Rs yearly
term of policy – 20 years
my age 23 years
so pls let me know the actual return after maturity…
2) i have completed 5years, what amount ll i get if i surrender after 5 years..
pls help me out
Suveen-You planned or already bought? If you planned, then you can’t. Because this plan is not available now.
1. suveen
I already bought this policy… I have completed 5 years also…. what amount I ‘ll get if I surrender now?
What ‘ll be my maturity amount if I complete full 20years.
Suveen-Contact your LIC branch or refer the below comments.
2. suveen
Basavaraj sir,
1)I have already bought this policy, completed 5 years also..
Premium – 36030 Rs yearly
term of policy – 20 years
my age at time of enrollment – 23 years
so pls let me know the actual return after maturity…
2) what amount will i get if i surrender now( 5 years completed)?
Pls help me out
Suveen-Regarding exact amount, contact your LIC branch. Returns may be around 4% to 5%.
61. RAMACHANDRAN.A
tax saving instrument was top savings PPF,NSC, FD, ULIPs,ELSS,. The explain with total returs to 3 0r 5 years pls send
Ramachandran-I am unable to understand your requirement.
62. Shakti Prasad
Please help me on my query as it holds important due to my current financial crunch situation.
I took the ‘LIC Jeevan Saral’ plan started in 2010 for 15 yrs with yearly premium as 25000/. I have already paid for 5 premium terms till 2014 for a sum of nearly 1,25,000/-. But due to i lost my
job in 2015 i couldn’t able to pay premium from 2015 & 16. So as i can’t continue this plan due to my bad financial cond. so i want to surrender my policy now after 7 years. So what would be the
payout amount to me if i surrender now. I see the current status showing in LIC portal as: “Reduced pay-out.’ Also let me know if i want to pay premium of last year & this year would i asked to
pay any penalty?
So kindly update me on my above query. Thnx in advance for your support.
Shakti-Please contact your nearest branch for the same. Yes, if want to continue the plan, then you have to pay the penalty.
1. Shakti
Hi Basa,
Thanks for giving quick update. But as i am staying outside of state current for a job search so can’t make to home now.
Its been 6.2 yrs now from the commencement of policy so can you please anticipate what’s the surrender amount i would get where i have paid 1,20000/- for 5 years,i expect to get at par
with my total paid premium or little higher from that but not sure. I believe as a financial analyst you can give me at par update if not concrete. Please update me here.
Shakti-You refer above post and below comments. You will get an idea of how to calculate on your OWN. You no need to be in same state to know the status. Visit nearby LIC branch and
get the status.
2. Angel
Bad blog. You don’t answer straight questions
Angel-Yes, because money is not easy thing.
63. Sandip G
Dear Mr.Basavaraj,
I have taken Jeevan Anand in Feb 2012 with half yearly premium of 52300 approx for 25 yrs with sum assured of 25L at the age of 32yrs.
But after your reviews and lots of search I found that it is better to go for term insurance and balance with other investment as this policy will give max to 6%.
Can you suggest is it better to leave/surrender with Jeevan anand.
Also can you suggest where should I invest the balance.
Sandip-It is better to surrender. But investing the rest of amount depends on your financial goals, which I can’t say plainly.
64. Saurabh Negi
Hellow Basavaraj,
thanks for such an informative post regarding LIC Jeevan Saral Policy. I am myself struggling to get the clearer picture regarding the Maturity Amount. I purchased this policy @ 24yrs , Single
Premium 48040 INR, 5 yrs completed. My policy paper mentions MSA : 1159520 INR (along with Death benifit & Term Rider sumof 10 Lakh each) but in my policy premium receipt its showing MSA = 10Lakh
while that confusing chart showing somewhere around 54 Lakh. I think all of us are having this same confusion. I will b happy if i am getting 5-7% interest overall at the end of policy, if not
please suggest if i should stop paying further and get whatever i can get back.
Saurabh negi
[email protected]
Saurabh-If you have doubt about MSA, then contact the branch.
65. pinkal mahajan
hi my name is pinkal mahajan i am 29 years of age i got this policy at the age of 24. i have been paying 72000rs per year can you please tekk me how much i will get after 35 years. on the chart
(165) it says some 22000000rs. but when i contact branch they say something else. please clarify this.
Pinkar-Refer my above post and below comments. You will not get more than 5% return. Rest is left with you to decide.
66. Chhavi Sinha
Hi Basavaraj,
I had taken Jeevan Saral in June 2011 with a monthly premium of Rs 6123/-, which means i am putting Rs 73476 in a year in my Jeevan Saral account. my policy will be completing 5 years by June
2016 and i was thinking weather to continue with it or not because people were telling me that i am paying a very high premium and the returns are not as expected and it is better off to put the
money into PPF account where i can have better returns.
so there can be few options that i was thinking about and i am confused:
1. to close the policy after 5 years, could you please tell me what is the amount that i should be getting ?
2. secondly , since its a flexible policy i thought of reducing the Premium amount to Rs 1000 or 1500 per month.
3. surrender policy after 10 years and continue to pay what i am paying at the moment.
i am 42 year old and need to have good investment. let me know what you say.
Chhavi-1) Contact LIC branch. 2) What is use? 3) It is of no use.
67. Raja S
My policy was started on March 2013 and have paid till May 2015 without interruption. Its jeevan saral with DAB – Premium monthly 3062.. Almost i had paid the premium for 2 years 3 months.. 9
months i did not pay the premium as per schedule. Almost 3 years completed now.. Now the policy status is lapsed without surrender value. Can i withdraw my amount now?
Raja-No, you can’t withdraw. If you want to withdraw, then you must clear the pending premiums.
68. Ashish Kumar
Hi Mr. Basavaraj,
I have completed 5 years of Jeevan saral paying Rs.5104 monthly (i.e. approx Rs. 3,00,000.00 till date) as an installment. I want to surrender this policy due to some financial problem, how much
amount would I get as the lic website is showing after completion of 5 years if they surrender they will get 100% of the Maturity Sum Assured. So in that case would I get approx 3,00,000.00.
Ashish-Contact the branch for the same. MSA will be reduced to 5 years MSA.
1. Mohsin
Hello Sir…
I have just completed 5 years of jeevan saral policy.
So far I have paid about 1,53,000 RS for 9 jeevan saral policies (Courtesy LIC agent wrong input) which has different maturity dates but started at the same time in May 2011.
My age is 26 years when the investment started.
Now, I want to surrender all the policies. Can I expect to get the money which I have paid ?
I understand that visiting LIC office would be helpful but I am not in a position to visit office….
Mohsin-I think you get.
69. Harsha R
Hi Mr. Basavaraj,
I have completed 5 years of Jeevan saral paying 9187 Rs quarterly. If i surrender now ,will be losing the premiums paid in the 1st year according to LIC people. There is an option called partial
surrender. Can you please throw some light on that.
Harsha-why you want to surrender?
1. Harsha R
Dear Basavraj,
1. Would like to liquidate this amount
2. Will be short of money due to some other financial commitments, so will be difficult to pay premiums of this amount.
Thanks in advance
Thanks in advance
Harsha-In that case better to surrender it.
70. DEBARGHA GANGULY
Hi sir,
I have Jeevan Saral policy policy no #578448787 enrolled in 2011. I want to surrender the plan since I have completed 5 years.The maturity was in 2029 and MSA = 500000/- only.You have stated that
surrendering after 5 years will fetch 100% of MSA value. So am I supposed to get 5lac or (500000/14)*5 = 1.78 lacs? I paid 2042 as monthly premium for 5 years so total premimum would be around
1lac 20k.Can you please tell me teh expected range of the surrender value that I am expected to get.
Debargha-Your MSA will be re-adjusted to 5 years of Jeevan Saral. Based on that they pay. Check with nearest LIC office.
71. k v raghu kumar
sir this is k v Raghu kumar i bought jeevan saral 165 plan in 2009 February,qtly premium 4250. i was paid until 2015 Feb premium. After that i was not paid till now, i have taken loan also amt RS
210000,AND interest RS21000 If i continue this policy i have to pay premiums amount+loan amount+interest.so now i am not able to pay this amount right now , and I asked to LIC officers if I
surround the policy how much amount will i get, they told me approximately 26000 plus or minus, up to Feb 2015 i was paid Rs 306250,if surround this policy i will get loss the amount rs 30000 to
40000, what can i do please give your suggestion, now i am not ready to pay premiums and loan amount and intrests
k v raghu kumar
Raghu-If you are not ready to pay the interest, loan amount and premium, then better to surrender it.
72. prince kumar
Sir I have jeevan anand policy sum assured of 200000. my EMI is 7668 half early for 15 year, how can I get total maturity amount after 15 year. sir plz advice me.
Prince-The returns will be around 5% to 6%.
73. Avinash Kedia
Dear Sir,
I wanted to know one thing that for jeevan saral it is preferred to surrender after 5 yrs if taken.
I pay premium yearly can you please tell 5 yrs will be considered completed after paying 5 premiums or after paying 5 premiums and completing 60 months as well.
Avinash Kedia
Avinash-After completion of 5th year i.e beginning of 6th year.
74. Chellani
Dear Mr Tonagatti,
In one of the post (dated 22 Jan 2016) you have mentioned that LA would be 500 – 600, based on current trend.
Could you please elaborate?
75. Ramakrishna
Hi Basavaraj Tonagatti,
I took Jeevan saral Plan for 25 years with SA: 5,00,000 and Yearly Premium as 24.020 on 17/09/2013.
Till date i have paid three premium’s. After reading this blog, i realised that i had invested in a wrong plan.
So, please suggest me whether i should surrender this plan now or shall i pay two more premiums?? (to complete 5 years)
How much amount will i get if i surrender the plan now?
How much amount will i get if i surrender the plan after completing five years??
Thanks for your help inadvance..
Ramakrishna-Better to surrender after 5 years.
1. Chandru
Please explain.
Why should we surrender for 5 years.
Why not in 3 years.
Chandru-In this policy, if you surrender after 5 years then it is considered as 5 years policy but not the term you opted earlier. This is the uniqueness. Hence, instead of 3 years,
it is best to surrender at 5th year.
76. Sriraj
Hi Basu,
I have purchased Jeevan Saral in 2010. I’m paying half yearly premium of 15466 INR. My MSA is 1336367 INR. Now the policy is in affect for 6 years. I understand that Jeevan Saral has a SSV for >5
years, which is to pay out 100% of MSA . This MSA is calculated for the time frame for which the policy has been in affect i.e 6 years in my case. I also understand that in case of Jeevan Saral,
LA starts from 10th Year onwards. Now I have couple of queries:-
1. If I surrender the Policy on 10th Year completion, would the returns be as calculated below ?
> MSA for 10 year period = 1336367 * 10/35
> LA for 10th year for premium range between 20,000 – 50,000 INR – 375 * MSA for 10 years/1000 = 375 * 1336376 * 10/35 * 1/1000
> Since Policy in affect for > 5 years. As per SSV, MSA Paid out = 100%
> Returns on surrender on 10th Year = MSA + LA
= ( 1336367 * 10/35 ) + ( 375 * 1336367 * 10/35 * 1/1000 )
2. If I opt to not pay any future premiums ( Policy lapsed ), what would be my returns at the end of policy term of 35 years ?
3. What would you recommend, should I hold the policy for its term or should I surrender it right away ?
Sriraj-How you can say that the MSA will be 10/35th of your Rs.13,36,367? You have to check your age while buying policy, premium you paid and arrive at MSA.
1. Sriraj
Hi Basavaraj,
Could you please me understand what would be the surrender value of this Policy on its 10th year completion ?. I need to know the below details actually:
1. MSA on surrender
2. Loyalty Addition on surrender ( I know this is not yet declared. Will be declared on 2019 for 2010 policies. But based on the Loyality declared for the last 3 years, Can I assume it to
be 375 INR per 1000 MSA ? )
3. Total Surrender Value = MSA on surrender ( MSA calculated for 10 years ) + Loyalty Addition
My next premium payment date is on June 2016. If I have to exit from this Bad investment, I have to do it now. If you can guide on the above and answer my above queries related to
calculation, it will be really helpful.
Thanks a million for your help :).
Sriraj-1) MSA on surrender will be known to you once you visit LIC for surrender.
2) You can. But clear picture or shocking picture emerge when you know the actual surrender of this policy.
3) Yes.
First go ahead and surrender it.
1. Rishiraj
Sir I have purchased the jeevan saral policy with half yearly premium INR 3032 in may 2009. I want to surrender this policy due to some reason. Pls tell me about surrender value.
Is it more beneficial to close after 10 yrs?
Rishiraj-Contact LIC branch.
2. Sriraj
Hi Basavaraj,
My Age at purchase is 24.
Premium paid per year = 15466 * 2 = ~31000 INR
Sriraj-Then calculate the MSA based on that.
77. Vimal Gohil
Hi, Basavaraj
I have purchased the LIC Jeevan Saral in May 2010 through LIC agent. Now 5 years are completed for same. Since that agent had made a purchase of policy for 35 years term, can I close or surrender
this policy now ? And, what could be process to do the same.
Let me know so that I can plan for better. Also, I would like to invest for long term in some SIP fashion. What could be the better option ? is it something similar to Tax saving MF.
Vimal-You can close the policy now. For surrender, you have to visit the servicing LIC branch with you original bond and KYC documents. They help you. You refer my earlier post “Top 10 Best
SIP Mutual Funds to invest in India in 2016” and “Top 5 Best ELSS or Tax Saving Mutual Funds to invest in 2016“.
1. Vimal Gohil
KYC is mandatory ? actually, I think I didn’t filled it up. Only bond can work ?
Vimal-Yes, KYC in the sense mainly your bank details to transfer the amount through NEFT.
78. Anil
Hi I have invested 36000 in jeevan saral and 6 years are completed.. Could you advise how much I will get as Lic says I will get 29000rs.and also advise do I need to continue the policy or
surrender it….
Anil-Please read the below around 900 comments patiently. You will get my answer. Because it is repeated many times of what you are asking.
79. krunal patni
Please I want to know I purchase lics jeevan saral policy for 30 years of 1000 monthly premium on 27 age so what will be return i will get at maturity of policy
Krunal-This plan is already closed. You can’t purchase it.
80. Hemanth
Hi Basu, I was calculating MSA, LA and total amount I might receive.
I have jeeval saral policy which I took at the age of 25, half yearly 6065. Sum assured 250000.
Per my calculations.
MSA: 195310.
LA : (195310 * 250*5)/1000 (*5 because I presume from 10th year with 250 there is gradual increase of 50. So, I rounded it off to 250 for all 5 years from 10th to 15th making it 5 times) = 244137
Total amount is 439447.
Am I correct in calculating the LA? And is the total amount correct?
1. Hemanth
Btw, This is for 15 years plan.
Hemanth-No, I don’t think LA will be that high (considering the current trend). I presume around Rs.500 to Rs.600 for 10 years term.
1. Hemanth
Thanks for reply.
If we consider the amount to be 500.
LA calculations for Jeevan Saral would be as below:
LA = (MSA *500)/1000 = (195310*500)/1000 = 97655.
Total Amount would then become = 195310+LA = 292965.
Is this correct or am I still missing anything for LA calculations?
Hemanth-It’s correct. Now calculate the return using my latest post “Video tutor-How to calculate LIC policies maturity amount and returns?“.
81. Satish
Dear Sir,
Kindly advise, I took Jeevan saral policy plan 165 in 2011 at the age of 28 yrs.
premium of Rs 6125 quarterly.
tenure 15 years
i have paid premium of 4 years till date if i surrender policy now i am getting around Rs 70,000 (with loss of 30000 & 4 years)
what you will suggest should i continue or surrender policy? what would be the surrender value after paying 5 years premium?
Satish-Better to surrender after a year (completion of 5th year). You contact LIC branch regarding the exact values of surrender.
1. Satish
Many thanks Basavaraj.
82. Yogesh Koshti
I have taken Jeevan Saral Policy 4 yaers packa and my monthy sip is Rs 5000 i.e rs 1000 x 5 poiclies. Let me know wheather its is advisabke to continue this policy, are there sufficent returns as
ther sis market rumours that Jeevan saral should not continued as there is case beteween LIC nad IRDA goint on for and less chances of getting assured rtuerns . Please suggest
Yogesh-Jeevan Saral is already a closed plan. Also, in my view, there is no such case between LIC and IRDA. Regarding your investment, better to close once you complete 5 years.
83. PP
I surrendered my Jeevan Saral policy after paying 5 premiums and completing 5 years.
My Sum Assured was 12,62,500. Thus, the monthly premium was 12,62,500 / 250 = 5,050. Yearly premium was 5,050 * 12 = 60,600 minus 2% discount on annual payment = 59,387.50.
According to LIC’s table (I think no. 165), for someone who was 30 years at the time of buying the policy and paid 5 premiums, the “factor” for calculating Maturity Surrender Value (MSA) for the
5th year was 4,726 per Rs 100 of monthly premium.
Thus, the surrender value was calculated as 5,050 * 4,726 / 100 = 238,663. There would have been a reduction of 3-5 thousand rupees from this if the surrender had happened before completing 5
The Rate of Return on the policy was -7.2%. If I had continued with the policy believing that I’ll get BIG loyalty bonuses, the rate of return would not have been more than 5%, typical with any
other LIC policy.
LIC sucks! They have been duping common public for decades, using poor man’s hard earned money for free, making themselves and their agents rich, and returning peanuts to their customers provided
they keep paying premium for ages. It’s a pity that majority in this country do not understand RoR and CAGR, which should be the first thing they should teach in schools.
PP-It is heartening to hear your story. Hope others understand the reality.
84. Sanjay
Hello Sir,
Today I went to LIC office and asked about the policy surrender procedure. My policy start of date is 09/2009 and date of maturity is 09/2029. I am paying Rs.2300 every 3 months as premium. When
I spoke to the LIC officer they asked me why you are surrendering the policy, you have paid premium regularly for 6 years and now why are you surrendering. They asked to to get loan on the
policy. Finally they told till date I have paid premium of Rs.59000 and if I surrender I will be getting only Rs.48000. Then I asked her its mentioned in the policy document like special
surrender value will be 100% of MSA if premiums are paid for 5 years or more. They said there is nothing like that and there will be Rs.11000 loss for you. Hearing this I am very clear that they
are not even promising what is return on the policy document and I am not sure what amount I will be getting in my hand at the end of the policy completion. I am going to fill the form and going
to submit the forms for surrender.
You are really doing a great job here. Thanks for your advice and educating everyone.
85. Kuntal
Hi Basavaraj,
I must say your online platform provides the best information on LIC policies and especially Jivan Saral policy. Thanks for sharing such valuable information on such important investment topics.
I had purchased Jivan Saral policy 8 years ago when I was aged 30 years. My annual premium is Rs. 60k. I have taken this policy for a period of 35 years. After reading your comments here, I am
assuming that this investment will not yield me more than 6-7% returns over a 35 year long period. This is very low compared to the time and the money I am putting in.
I would like to get your advise on whether I should surrender this policy now or wait for the completion of 10 years so that I get some LA.
Your advise will be highly appreciated!
Kuntal-No great benefit in waiting. Hence, better to act and surrender.
86. Anup Banerjee
I have a Jeevan Saral Policy (Table 165) taken in 2012 and will mature in 2037. The sum assured is Rs/-1250000. I pay yearly premium of Rs/-60050. I have paid four premiums so far. My question to
you – If I surrender this policy after 5 years, then as per information shared by you in this blog, I will get 100% of the premium amount paid i.e. 3 lacs. Please correct me if I am wrong.
Also, If I wish to continue with this policy till end than what would be the total amount I get? Any other suggestion regarding this [policy would be really helpful.
Anup-It is slightly positive or negative at 5th year. If you continue this policy till end, then the return will be around 4% to 5%.
87. Tejaswini Kunte
Hello Sir,
I bought LIC Jeevan Saral policy on 7th jan 2010 at the age of 25. I am paying 18015 as yearly premium, policy term is 25 and death sum assured is 375000.
I have paid premium till jan 2014 ie for 5 years .
Please advise, should I go for 25 term or surrender the policy now or pay premium for 10 years and then surrender my policy.
what will be my surrender value after 10 years and now and maturity value after 25 years
Thanks in advance.
Tejaswini-I think we discussed the same in our Facebook chat.
88. jyoti
Dear Sir,
Kindly advise, I took Jeevan saral policy plan 165 in 2011 at the age of 24 yrs.
premium of 29000 half yearly
tenure 15 years
i have paid premium of 4 years till date if i surrender policy now i am getting around 175000 (with loss of 57000 & 4 years)
what you will suggest should i continue or surrender policy? what would be the surrender value after paying 5 years premium?
Jyoti-Better to surrender after 5 years. Regarding values, contact LIC branch.
89. Anuj Agrawal
I started LIC Jeevan Saral Policy 3 years before for a yearly premium of INR 24020.
1. What will be the actual total amount i will receive as per today’s calculation after 35 years? Policy Started at age 26.
2. How can i get out of this Policy immediately or what will be the best time and what will be the charges deducted for same?
Thanks a lot for all your help & valuable information.
Anuj-Your both doubts are already answered in below comments many times. Please go through them.
1. Anuj Agrawal
Can you please let me the formula to calculate it? So that i put the values in and calculate expected outcome from this Policy…
Anuj-You can use the compound interest formula. You search in Google and found many online sites.
1. Anuj Agrawal
Thanks Basavaraj for all your help and replies. i got clarification for my 2nd doubt regarding surrender value. Can you please clarify my first doubt by giving me a rough figure
(What will be the actual total amount after 35 years – Policy Started at age 26 & Annual Premium 24,020.), it will be appreciated. Sorry to bugging you again and again.
Anuj-Considering the current LA trend, the return may be around 5% to 6%. Not more than that. It’s alright and never think by commenting.
1. Anuj Agrawal
Hi Basavaraj,
I am weak in calculating these percentages & opt in for this Policy 🙁 …. that’s why i am asking you that what will the FIGURE (appox.) after calculating? Thanks a ton
Anuj-Basic Maths is enough to do the calculation. If you still have confusion, then please go through the below comments. You notice my calculation and answers.
90. Amit
Sir, what is the income tax implication on surrender value after 5 premium terms?
Amit-Please read my earlier post “Tax Benefits of Life Insurance“.
1. Amit
Thank sir for your kind reply! Is the surrender value received (after paying more than 2 premium terms for traditional policies), required to mention in our ITR of the respective year?
Amit-Read my post about taxation once again.
91. Abhi
Need your expertise here. I’ve paid 5 premiums till date for Jeevan Saral (24020*5) for a measly 5 lacs sum assured (No riders).
The 6th Premium date is approaching in Jan 2016. What would you advise for this plan – Should one:
1. Just stop paying premiums (Make it Paid up) and wait another 10 years for the money
2. Surrender the Policy immediately and collect whatever is left of it
3. Continue with the policy for another 5 premium terms to look at Loyalty Bonus etc
4. Any other advise on this policy?
Do respond! Thx
Abhi-Surrender now itself.
92. vrushali
Dear Basavaraj Tonagatti,
Thanks for your detailed explanation of this policy. I am in dilemma now & appreciate if you help me to solve this.
My DOB is -1989,
JEEVAN SARAL Plan details are as follows:
Policy Date: June-2012,
Half yearly premium=18195, TABLE=165, TERM=35 yrs
As I have completed 3 yrs. of my policy. I went to LICBranch to get surrender amount and Surrender value is Rs 61632/-. But till date I have paid total premium 109170/-. And Now if I surrender
then I will be in Loss of Rs 47000 Approx.
Please suggest me on following option
1) Should I exit at this stage of 3 yrs & take 47K loss. Invest that 61k money in my home loan which will at least reduce my interest component.
2) As in above post you suggested that exit from this plan after 5yr.
If I surrender this policy after 5yr
a) Will I get complete premium back. At least (181950) my 5 yrs paid premium .
b) Or after 5 yr also I have to bear some loss. (May be you have some rough idea on this)
Your suggestion will help me to take further decision.
Vrushali-You may not get the full amount of what you paid in case you surrender after 5 years. But the loss will be minor compare to today’s surrender. Hence, I suggest to surrender after 5
93. Sam
Hi Sir,
I have taken Jeevan Saral policy in 2011 for 16 years with a premium 48040/Year.
My age was 31 years when i enrolled for it, agent has given bond of worth 10 Lac.
It would be great if you explain or guide me where to invest
Please let me know what would be my meturity amount after 16 years?
What is SA for my policy?
How much LA I can get after 10 & during 16th year?
Shall i continue with the policy?
With Regards,
Sam-Your all doubts are explained in above post and below comments. Please go through them.
94. unmesh bagwe
Hi Basu,
I am a regular visitor of your site and find your blog very helpful.
my question is around Jeevan Saral – i have been paying the premium for 7 years and surrendering this policy as it does not suite my needs.
Question : my premium is due next month so do you think i should be paying the pending premium and then surrender to surrender directly ? if you could shed some light in this matter.. a big
thanks in advance for your answer.
Unmesh-Better to surrender before paying premium.
1. unmesh bagwe
thanks basu.
as you suggested i will surrender the policy without paying the pending premium but do you think it will cause any trouble or inconvenience if i surrender the policy when the premium
paying date is passed ?
Unmesh-I don’t think so.
95. Haresh
I have seven years of my LIC Jeevan Saral Policy.
If I will surrender the same now, Shall I get my total premiums paid back or not ?
If yes then shall I get any sort of additional amount ?
Haresh-Please contact LIC branch.
96. Diwakaran
I have taken Jeevan Saral on 2013. Two years paid. But when i look at your blog, i am surprised to see that i was cheated.
My premium details are,
10,000 – per month.
for 35 years.
DOB : 31/07/1984
Please tell me how much i will get it. Will it be good to withdraw after 5 years?
Diwakaran-I will not repeat once again, but you might know my answer 🙂
97. Deepu
Very informative article…
Hi Baswa,
Below are my details
DOB 9 july 1982
premium 9187 quarterly
policy term 20 years
started 24/5/2013
Sum assured 750000
Please help me in finding out total return after 20 years and also what would i get if i surrender in 5 years
Thanks in advance !
Deepu-Go through the above post and older comments. You will get my answer. However, returns will be around 5%.
1. Deepu
Thanks Basu,
I had no interest before reading your blog and not even know the basic of investment but after reading your blog and some interesting question i did some google and found that if i invest
3000 per month in PF I will get near about 1942961 in 20 years (interest rate appx 8.7%), which would be far better then LIC.
Do you agree to this?
The only difference would be that I am not insured with PF but with LIC.
any advise you want to share?
Deepu-I agree. But do you feel the Jeevan Saral insurance coverage is actually fulfills your insurance need? You be underinsured by buying such products.
1. Deepu
No, I dont think that jeevan saral will fulfill the insurance need so I am planning to cancel this and will go for term insurance rather
But again can you give me a rough idea how much will i get if i cancel today (3 years) offcourse i will call LIC as well but thought to check with you as well
hope you wont mid
Deepu-No idea.
98. BHARAT HEMNANI
Have you completed the chart for surrendering.
Bharat-It is not required as the plan itself stopped and you can get the information about surrender in your phone itself.
99. Bikash
Have you completed the chart for surrendering.
100. Wasim
i have a jeevan saral ….premium of 1,40,000 per annum…….i have completed 3 years….when should i terminate to get the whole premium amount back without any interest….or when should i terminate to
suffer least loss in premiums….thanks in advance…
Wasim-Best to complete 5 years.
1. Chandan
For example, we paid Rs. 60,000/- in 5 years under Jeevan Saral Plan,
and we surrender the policy after completion of 5 years, then what amount we get ??
we get paid amount with interest or without interest ??
Also, what will be rate of interest ?
Chandan-Better you contact the LIC Branch.
1. Chandan
I knew
101. C.P.
Dear Basavaraj/ Admin
Today morning I posted my query on your site and it was displayed but later on it deleted by you people but again I am writing to you.
I took Jeevan saral plan (165) on 15-10-2010, that time my age was 19th Years. I pay quarterly payment Rs. 3062/-, and sum assured is 2,50,000/- , Term – 12 years.
Now I am planing to surrender it on 20th Oct-2015 due to some personal reasons.
So please tell me what amount I will get after completion of 5 years, also tell me what rate of interest will be provided to me.
Moreover, if I complete this plan till 12th years what amount I will get ??
Your prompt revert on same will be highly appreciated.
C.P-It was not deleted but need admin approval if you are commenting first time. Once admin approve it then only it is visible. You have to visit LIC office or use the facility they provided
like “Check LIC Policy status without registration by Mobile Phone or SMS” to know the values.
1. C.P.
I am unable to know my query answer
So please give me brief idea what amount I will get after completion of 5 years
C.P-It will be around 4%.
1. C.P.
Hello Sir,
I contacted LIC Branch and came to know that Maturity Value is Rs.50169/- even I paid Rs.64302/- to lic in 21 premium of Rs.3062/- quarterly
so its cleared, we will not get 100% amount even after completion of 5 years
CP-True and that is the reality of traditional plans.
1. C.P.
Hello Sir,
Can you please confirm which form I have to fill for policy surrender
Should I fill form No 5074 ?
Or form no 3510 ?
Or any other form ?
Please confirm asap bcoz I have to send these documents out of station
Note- My policy is not completed
C.P.-Both are one and same (I am not sure why they put two numbers).
2. C.P.
3062 quarterly for 5 years = 61240
I will get more than 61240+ interest = ??
or less amount ??
102. sharad dabir
Dear Mr. Basu,
I have taken jeevan saral policy in 2011 when my age was 60 yr.Annual prem.is Rs.60050.Till today I have paid 4 instalments. Term of the policy is for 10 years. The maturity sum assured is
152950.Tell me how much amount I can get on maturity assuming the current rate of LA.If I decide to surrender the policy , how much will i get now ?
S.T.Dabir, Aurangabad
Sharad-My answer is already available in above post and below comments. Returns will be around 5% to 6%.
1. Sharad Dabir
Dear Basav,
My confusion is regarding the LA payable for a 10 year term policy. It is mentioned in the scheme that LA will be paid only on complition of minimum 10 years. In that sense, the maturity
amount in my case will be – Rs. 152950 (MSA) ) +Rs.45900( LA at the present rate of Rs. 300). = Rs.1198850/-. Is this correct ?
1. Sharad Dabir
Correction in the last line. The amount is.Rs.198850 instead of 1198850
103. satish
Hi Basu,
Some correction is required in your information.
I have surrendered my jeevan saral policy after 5 years and i haven’t received 100% of my premiums. I received Rs 36,000 lower then my premiums paid during 5 years of tenure. But you have
mentioned that if surrendered after 5 years then we will receive 100% premiums.
Please let me know if you need any clarifications.
Though you have made your disclaimer statement clear but as a responsible blogging person we should be very careful in publishing financial related info on web on which people are going to take
Satish-What I said is, LIC consider your policy as 5-year tenure policy. Accordingly they adjust the MSA for 5 years period+LA. If such adjustment is less than of what you paid then it is an
eyewash which LIC created stating that this policy is highly liquid.
104. BIPINKUMAR SINGH
Hi Basavaraj,
I had two policies of jeevan saral for the term of 35 Years as mentioned below:-
Enrollment Date Prm Amt[Monthly] Maturity Amt
28-02-2012 [2,042.00] 5,00,000
14-01-2014 [4,083.00] 10,00,000
Can you tell me the surrender value and paid up value for both policy? AND
what you will suggest to do should I surrender or should i do the paid up for both the policy?
Bipinkumar-Please contact nearest LIC branch for surrender or paid up values. First check the values then decide. You can’t do both the things unless your policy completes 3 years.
1. BIPINKUMAR SINGH
thanks for the reply.:)
105. Kannan
Dear Sir,
Kannan here, I have taken Jeevan saral (165) policy through an agent in April 2009. The Policy details are as follows:-
Jeevan Saral (165) Policy Term – 21 years
Date of commencement: 09/04/2009
My Age at the time of policy: 30 yrs
Premium – Rs. 7656 / quarter
Maturity Sum Assured – Rs 7,24,375
Accident Benefit sum assured- Rs 6,25,000
Term rider Sum Assured – Rs. 6,25,000
Policy Maturity Date: 09/04/2030
My question is
1) The Agent showed me Jeevan Saral ATM Plan chart(Table no 165) that I would get 20 lakhs at the end of 21 yrs after taking a premium of Rs. 7656 / quarter. Will I get 20 lakhs (as per table no
165) at the end of 21yrs?
2) If no, then how much i will get actually at the end of 21 yrs?
K. Kannan
Kannan-Even if you consider Rs.20 lakh as a return then too the return on investment will be around 4.89%. So waiting for such long 30 years and expecting around 5% return is worth?
1. Kannan
Thank you for your quick response…
Please let me know that could I get 20 lacs (as per table 165) at the end of 21yrs?
Kannan-What I am pointing is, Rs.21 lakh you may get it. But this return is suffice and at what cost you are getting? Think seriously.
1. Kannan
As per policy I need to pay premium Rs 7656 per quarter for complete policy term 21 yrs. So as a whole I could have paid around 6.5lacs (Rs 7656 * 4 * 21=Rs 6,43,104 ) at the end
of 21 yrs. As per table 165, LIC will give me lumpsum of 20lacs alongwith loyalty addition after maturity period 21yrs). So I hope it is a good return. Please correct me if I am
1. Kannan
Awaiting for your response
Kannan-I assumed wrongly that the term of policy is 30 years. Now according to your agent, the return on this investment will be 10%. This is totally impossible from this
policy. Considering the expenses involved in traditional plans and the nature of investment LIC do by this amount, it is hard to expect around 5%. Rest is left with you to
1. Kannan
Thank you
106. subhash
Dear Admin and all user,
I want to share my feeling here-
I took Jeevan Saral policy in 2008 for a premium of Rs.20417 annually (my age was 24 years ).During taking the policy my agent show me a detailed chart which show multiple benefit and too much
return. I observed that no agent is showing actual table to customer and formula for calculation of maturity amount. Thanks to Basavraj .
During that time I was only 24 years old and very much eager to take a pension insurance plan for my better future. I am a common man and was not aware about common insurance fraud.
Please note that LIC agent will tell you only high premium plan so that they can get good commission and Jeevan Saral is the best example. We are indian and believe on everybody very easily. We
believe in LIC as our parents teach us that LIC is very good.
After taking this policy I learned many things and invested in other area like Equity, mutual fund, Recurring, FD etc and I want to share my knowledge for you.
1. Insurance- insurance should be used only for covering risk not for investment but most of us try to invest our money in insurance. ULIP is a great example for that. From my past mistake I
learned that everybody should take only pure insurance without any maturity
for eg: Pradhanmantri Suraksha Bima Yozna of Rs.12 premium- it is the best example for pure insurance without maturity but only for accidental insurance
Ask your agent to show life insurance plan without any maturity. Believe me he will try to misguide you because in that case premium amount will be minimum
for eg: if you want a risk cover of Rs.500000 in case of death then you have to pay only 1400 annualy ( if your age is 24 years) so safe your family by taking pure insurance plan
Now come to second option-
2. Investment- you can invest your money in various option as per your wish but i will suggest you to invest in many option rather than in one bucket.
a. Gold
B. Property- Highest Safe Investment and best value
C. Equity Share Market- only for some Tech savy person
D. Mutual Fund- Medium risk and ladder for new investor. Personally I took ELSS mutual fund for 3 years locking period and believe me I got my money double ( may be i choose best fund or luck)
and since then I am investing money every year + tax benefit
E.Kisaan Vikas Patra- we are used to it. we should invest our some % in it.
F. PPF- good for safe side and secure future
G. Local BC or Patti or Kitty- Best for businessman
H. Recurring- Its very easy( if did online) and suggest that every person invest their bank saving in it. You can withdraw anytime.
Question to Admin- Sir, I am feeling that I stuck due to investment in Jeevan Saral plan. I started with Rs.20417 in 2008 for 35 years term plan and now i want to surrender but i want some
return. Through your table i calculate that i will get around 950000 without loyalty which is worst return as per me so please tell me that my calculation is right or wrong and for some return in
which years I should surrender the policy.
Second Question- For securing my future I want to take a pension plan but I don’t have proper knowledge so guide me after surrender this policy which pension plan I should choose.
Third Question- I started my business recently (I worked in various MNC for last 10 years) for which I want to establish a taxation system. I read many books for taxation but got confused and
need your help. For this support I can pay your fee.
My email id [email protected].
Please revert.
Best Regards,
Subhash-I respect your sharing and concerns. However, I have different opinion about investments like Gold, Property, Equity (this is not at all meant for tech savvy, in fact, it is bet for
those who understand the concept of equity) and for local BC. Now regarding your doubts,
1) Your policy already completed 5 years. Hence, visit LIC branch and ask for closure value. Then decide whether it is worth to continue or stop.
2) As of now there is no such pension product, which you may say BEST. You have to accumulate on your own.
3) Sorry, I am not an expert in that matter. I can only handle the personal finance tax but not business or corporate.
107. ANUP PATEL
I have LIC jivan saral policy with premium of 6065 HLY since July 2008.
1. If i surrender this policy than what amount of money i get ?
2. if i surrender after 10 years than what amount of money i get ?
MY MSA is 534990
MY TERM is 35
Anup-Contact LIC branch.
1. Abhishek Saxena
Thanks Basu, I can and will contact LIC nearest Branch.. but what’s the use of this blog then?
Abhishek-I am not here to let all readers to know about the surrender or paid-up values. This is meant to understand you about how it works. Also, if you are so smart to ask me of
WHAT IS THE USE OF THIS BLOG, then I feel it is a pity of you. This blog is meant for making you to understand how these schemes work. You already committed one mistake of investing
without knowing whether it suited to you or not. Nothing comes easily and the same applies to your investment and knowledge.
1. Abhishek Saxena
Thanks Mr Basu.. as far as my knowledge is concerned..it doesn’t need your scale for rating..Instead of reading your blog, people can go to website, article & their policies.. but
if you see most of the comments are asking for your help on MSA.. you should know how to solve it..put a excel which gives indicative , it’s still fine..
I have answer to your offending comments, but I would rather end the discussion with a note that I have got details I was looking for PERIOD
Abhishek-I wrote all information in a simple way. But I can’t spoon feed all. If I do so then what is the difference between me and the agent with whom you bought LIC policy?
108. Abhishek Saxena
I have a Jeevan Saral Policy, premium 9188/- quarterly. Till now I have paid 13 premiums i.e. 3 yrs + 1 quarter.
If I now surrender the policy, what would be the surrender value?
Abhishek-Contact nearest LIC branch.
109. Akarsh Kumar
Hi Sir,
Found your blog very useful. Thanks for the really informative post.
I need a small clarification here. I have a Jeevan Saral policy since 2010. I have been paying Rs.1000/month and now it’s been 5 years and the total premium paid is Rs.60,000. So what is the
amount that I will get back if I surrender now? The special surrender value says, if 5 years’ premium has been paid then 100% of Maturity sum will be assured to us. But guaranteed surrender value
is 30% of the total premium paid excluding 1st year’s premium. Which one of these do I get? And how to avail the Special surrender plan? Please help me out with this.
Akrash-Whichever is higher.
110. prashant
I have started my policy on September 2011 my premium is of 35280 per year I have paid 4 installment. My policy is jeevan saral with profit . how much I will get if I surrender this policy now.
Or if I surrender after 5th installment how much I will get ?
Prashant-Please contact the nearest LIC Branch for surrender value.
111. Rajesh Prabhakar
Sir, Is there any website where we can calculate the maturity amount of different policies?
Rajesh-Sadly no such website available. But by the above method, you can calculate yourself easily.
112. Ambica
Please let me know any plan related to Jeevan Saral benefits
Ambica-As of now no such plans.
113. Praveen Kumar
I purchased a Jeevan Saral Policy in Feb. 2009 and have paid instalments for six years. My quarterly instalment is Rs. 9800/-. Now I want to surrender this policy. How much amount I will get 0ut
of it. I have paid for six years as follows:
9800X24= 235200
Whether I will get some interest on it or not? Kindly help me.
Praveen-Visit nearest LIC branch.
114. sumit saurav
Hello Sir ,
My name is Sumit , I am having a jeevan saral policy taken in 2011 @ the age of 24 . The premium amount is 48000 yearly & policy period is 35 years .I come to know after so many blogs that i did
a mistake of taking this policy . I have paid premium for 4 years & would like to quit after paying the 5th year premium . Can you please let me know the amount i will get aftr completion of 5th
premium under special surrender ?? Please advise . Also please let me now the value after completing ten years of premium . Will get loyalty bonus add on ??
Sumit-Please go through below comments.
115. pradeep singh
i am paying 1500 pm for 15 years and my plan is Jeevan Saral… please calculate the maturity amount and send me the same in my email.
Pradeep-Take the pain and read post and comments 🙂
116. Sameer
hi sir,
please upload the Sum assured chart for LIC jeevan saral policy.
Sameer-It’s already closed plan. Why you need it now?
117. vipul joshi
Hello Sir,
i have bought Jeevan saral in 2011 and paying premium of Rs.30025/- since then. Its for 35 years. I took it when i was 25 years old. I am still confused what i will be getting after 35 years.
Policy date of maturity is in 2046. Sorry for the confusion but i am not good in this type of calculations and lot of confusion about the sum assured and maturity sum etc…
Also let me know how you calculate so I can plan and calculate in future accordingly.
moreover can we take same policy twice or increase premium in between?
Thank you sir
Vipul-Procedure is already explained above. Please go through it.
118. priyanka
thanks for sharing valuable information.
my husband brought a LIC Jeevan Saral Policy (Table 165) for 18 years at the age of 33yr in 2010 , qtly premium 15312. Policy sum assured value 12,50,000/-. Please let me know my total maturity
Priyanka-If you want to surrender within 10 years then it may be less than your savings account interest.
1. Calvin
Why so much partiality. For the guys questions you just told them to read the article and for the girl you answered her question. 🙂
Calvin-Is it? I have not done any such partiality. Initially I tried to reply with their unique quaries. But if the same quaries started to repeat then I requested readers to follow
the earlier comments. For me no differentiation of gender.
119. Ramani
Dear Sir..
Its wonderful the way u explained the calculations. I will be extremely grateful to you if you could tell me the exact surrender value. I am investing Rs 3000/- month in Jeevan saral for 15 years
term starting in March 2011. If I surrender after completing 5 years of premium paid ( i.e in April 2016) what amount I should assume to get ? Pls help. My maturity sum assured is Rs 750000/-
Ramani-Exact surrender value will be available with LIC. So better to contact them.
120. Nitin Masaye
Dear Sir,
I took Jeevan Saral for my Wife on 12-Dec-2011 for 35 yrs term and my annual premium was 12000/.
As one of my friend told that the Maturity Sum assured is not that great in this policy.
So Please guide whether to continue with this policy or surrender it after 5 yrs.
Thanks in advance.
Nitin-Yes your friend is right. If your expectation is more than 6% then better to surrender.
1. sharad
dear sir
i have bought jeevan saral before 5 years back
monthly premium is 1000
how much money i should get after 25 years
121. Pritish Paril
I had taken Jeevan Saral policy in combination with Jeevan Anand in 2013 at the age of 26 for a term of 30 yrs. I pay monthly premium of Rs. 1125 towards Jeevan Anand and Rs. 1021 towards Jeevan
Saral. Please suggest me whether I should surrender Jeevan Saral after lock-in period of 3 yrs ends. My mother also pays Rs. 360000 as annual premium for Jeevan Saral.
Also the premium schedule chart provided by my advisor shows return of 1416000 after 30 yrs from Jeevan Anand policy. Is it true ?
Pritish-Better to surrender after 5th year. What bonus he considered to arrive at that final amount?
122. Chitra
Hello Sir,
My husband has opted for LIC Saral Policy (165) in 2010 at the age of 21. Sum assured-300000, premium-3675 quarterly, Term- 30 years. Location-Pune, Maharashtra.
What would be the amount we will receive after maturity of the policy?
123. Chandra
Dear Sir
Based on your comments, it seems it is advisable to exit this plan after 5 years. I have paid 3 premiums so far. Should I stop paying premiums and surrender after 5 years or pay 2 more premiums
and then exit?
Thank you so much in guiding laymen with this forum.
Chandra-First check with LIC branch how much you get if you stop now or pay for next three years. Then based on that take your decision.
1. kutub
Sir I have a jeevan saral plan
30yrs term
Taken at age 26
Premium paid for 6yrs(rs.1940 monthly)
If I surrender now I will get 100000/-
If surrender after 10yrs will I get rs.400000/-
Is my above calculation true
Sir please help
Kutub-Why can’t confirm the same from LIC branch? Because both calculations are exaggerated.
124. ROHIT HANDA
DEAR SIR,
I HAVE BROUGHT THE PRODUCT OF LIC “JEEVAN SARAL” AT DECEMBER-2013 OF RS. 11, 25,000/- FOR 26 YEARS TERMS & YEARLY PREIMUM PAID OF RS. 54045/- ONLY.
SIR MY QUERY IS THAT IF I HAVE TO COMPLETELY SURRENDER THE POLICY AFTER NEXT FIVE YEARS, SO WHAT I WILL GET ON PRE-MARUITY VALUE,
OR IF I WILL HAVE TO GO AHEAD AFTER NEXT FORTHCOMING ELEVEN YEARS, SO WHAT I WILL GET ON SURRENDER VALUE. IT WILL BE ALSO PAYABLE L.A. OR NOT.
KINDLY HELP ME AT THIS MATTER. I AM VERY GRATEFUL TO YOU.
Rohit-Better you contact LIC branch to know the values of period you mentioned. How can I say those values?
125. raju banerjee
Sir I want to purchase a land worth Rs 12 lacs. as an investment I had given an advance Worth Rs 5 lacs the registration will be on coming March 15. But sir I have a shortage of Rs 3.5 lacs. sir
i am a self employed person running my own computer business and i am an income tax payer. am i eligible for a loan for land purchase. except the advanced deed i Had nothing to give as a security
to bank.
actually i am from west bengal and the property(land) which i want to purchase is in the state of Bihar just 2 hours from my location. if i am eligible, from where do i apply a loan from my
nearest Branch? or in the Branch where I want to purchase the land ?
Raju-Please contact the branches where your land is located.
126. Ramprabhu
I got Jeevan Saral Policy two years before.
My yearly premium is Rs.31,826. ANd the Sum Assured is Rs.6,62,500/-
I just paid two Premium. I am yet to pay premium for this year (Dec 14).
By reading all above comments, I have decided better to quit the policy at this stage and buy some Term Insurance (Jeevan Amulya).
Just wanted your advice. Is it better to quit now or after 5 years?
Please help.
Ramprabhu-It is better to continue if your are satisfied with return I quoted in below comments. Otherwise discontinue immediately after 5 years why to buy offline term insurance of Jeevan
Amulya, when LIC offering cheaper online term plan?
127. Rajinikanth Reddy
Hi Basavaraj Tonagatti,
Excellent comments / reply by you sir,
Keep it,
Rajinikanth Reddy.
128. Deepak
hi Basavaraj,
i have jeeval saral 165 policy my quarterly premium RS 6431
can u explain before 20 year how much my maturity amount .
Deepak-You go through the above review and below comments. You get my answer easily 🙂
Basav for this plan I want more clarity
I have saral plan for 30 Years with 30K yearly prem, 7 year completed, MSA is 162262
LIC office told mi surrender value 159355 for today
Accroding to the product note if I continue till 10th year my surrender value should be MSA i.e. 162262 but my agent told mi that u continue till 10th you will receive the bounus + 100%
of value paid that is 3 lacs + bonus aprox comes to 3.75 lacs
pls suggest what will be surrender value after 10th year,I got for after 7th year
in this situation what should I do
pls note that I don’t want to continue this till 30 years | {"url":"https://www.basunivesh.com/lics-jeevan-saral-why-so-much-confusion/","timestamp":"2024-11-08T07:54:15Z","content_type":"text/html","content_length":"1049885","record_id":"<urn:uuid:c1052496-8461-40ff-93f1-eddfbfedc60a>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00112.warc.gz"} |
Singapore Math: Dimensions Math vs. Primary MathematicsSingapore Math: Dimensions Math vs. Primary Mathematics - StartsAtEight
I’m so excited to share this information with you all! I have been using Singapore Math Primary Mathematics since I started homeschooling in 2006. I used it with all three of my kids, my grandson,
and I recommend it to everyone for elementary math. Now, not only does Singapore Math, Inc. have multiple variations of the Primary Mathematics (including a massive 2022 update), but they have
another great math option for homeschoolers called Dimensions Math.
{I received a free copy of Singapore Math Dimensions Math, and was compensated for my time in writing this review. All views are my own and I was not required to write a positive review. Please see
my full Disclosure Policy for more details.}
About Singapore Math
Singapore Math, Inc. was founded in the late 90’s by Dawn Thomas. She believed there were better opportunities for her daughter in math so she started this company. Dawn’s daughter (Echo Thomas) is
currently the CEO of Singapore Math, Inc.
After years of experience with the Primary Mathematics curriculum the Dimensions Math curriculum was born. Singapore math was striving to create something better and more comprehensive. Something
with more rigor, more extensive educator resources, more content so supplementary material wouldn’t be necessary, and a more engaging student experience.
“Math is just math in Singapore, just like how American football is just football in America. The trademarked term “Singapore math” is coined in the U.S. and is used to differentiate the elementary
math programs that use Singapore’s approach from all the other math curricula in America.” (Wenxi Lee, “The Secrets to Singapore’s World-class Math Curriculum”, p. 13, 2020).
Singapore math curriculum uses a Concrete-Pictorial-Abstract approach. CPA encourages students to understand mathematical concepts by progressing from concrete manipulatives to pictorial
representations and then to abstract symbols. The Singapore math approach also places a strong emphasis on developing number sense and mental math skills. It is mastery based and encourages active
learning and student engagement through hands-on activities and interactive discussions.
About Dimensions Math
I had the pleasure of speaking with the Sales Manager at Singapore Math, Inc. to get the inside scoop on Dimensions Math, as well as the newest Primary Mathematic 2022 edition! I wanted to be sure I
could bring as much light to the amazing Singapore math programs, and help you, my fellow homeschoolers, make an informed choice as to which program would be best for you.
• Dimensions Math is a mastery based program for grades PK-8.
• Includes differentiated learning pathways.
• Has a Home Instructor’s Guide for grades 1-5.
• Is more robust and rigorous program than Primary Mathematics.
• Has more problem and teaching options (various levels and styles of problems to choose from.) Because of this it doesn’t require additional resources.
• Two levels of tests: Test A focuses on key concepts and fundamental problem-solving skills, while Test B focuses on the application of analytical skills and heuristics.
• Dimensions Math contains many practice problems to reduce the need for supplemental material. You shouldn’t expect your kids to do all problems in all lessons. You can pick and choose what’s
• Has an optional add on video subscription which brings a professional Singapore math teacher into your home classroom and provides in-depth instruction covering all the Dimensions Math material
for an entire school year. Available for Grades 1–6.
• There are a ton of Dimensions Math Resources for FREE on the website
About Primary Mathematics
When it comes to Primary Mathematics, I have been using the original U.S. Edition since 2006. A program that is still very popular and one which they are still selling today! BUT, they also have an
updated version, Primary Mathematics 2022 Edition that I got the skinny on so I can share the difference with you!
“Primary Mathematics epitomizes what educators love about the Singapore math approach, including the CPA progression, number bonds, bar modeling, and a strong focus on mental math. It’s a no-fuss,
straightforward program that balances supervised learning and independent practice.” ~Singapore Math, Inc. website
Primary Mathematics U.S. Edition
• The U.S. Edition of Primary Mathematics is the original Singapore math product. It is tried and true and has been used successfully for decades.
• It is not written to any U.S. standards.
• It contains the shortest lessons and workbook exercises. (We always use the Extra Practice Books at each level when we need more problems.)
• The size of the books is smaller than a standard 8.5″ by 11″ piece of paper
Primary Mathematics 2022 Edition
• Primary Mathematics 2022 Edition is the newest Primary Mathematics product.
• What is called the Textbook in the U.S. Edition is called the Student Book.
• What is called the Workbook in the U.S. Edition is called Additional Practice.
• There are Mastery and Beyond Books for each level that offer more distributed practice with more of a spiral approach. These are great for review! (In our house they are used throughout the
summer as a way to help prevent Summer Slide.)
Side Note: There is also a Primary Mathematics Common Core Edition with more closely aligns with Common Core State Standards, and a Standards Edition written to mee the 2006 Mathematics Contents
Standards for California Public Schools which was issued prior to the Common Core State Standards.
Side by Side Comparison of Dimensions Math and Primary Mathematics
Sidenote: Pictured above is a side by side of Dimensions Math and Primary Mathematics U.S. Edition. It demonstrates the distinct size difference between the two. I don’t have a copy of Primary
Mathematics 2022 Edition to hold but wanted you to know it is similar in size and configuration to the Dimensions Math books with colorful images and the larger size book.
All the Singapore Math products provide a solid math foundation for your homeschool. It’s just a matter of which features you feel will best suit your needs. After talking with the Singapore Math,
Inc. Sales Manager, and having had experience with these products I can honestly say you are winning with any of them!
If you are looking for a more simplified, less options to choose from program than I would use Primary Mathematics. If you get easily overwhelmed with having to pick and choose part o a curriculum to
use, then this will be more manageable for you.
If you are looking for an all in one, flexible to adapt to an easy or much more rigorous program than I would used Dimensions Math. This is also a good option if you want the added help of the video
Below I have broken down some of the major similarities and differences.
• Mastery based math
• uses a Concrete-Pictorial-Abstract approach
• Textbooks A and B for each grade correspond to the two halves of the school year.
• Both have Home Instructor’s Guides
• Same larger book format – the Primary Mathematics 2022 Edition and Dimensions Math (8.5″ by 11″) . The original U.S. Edition is smaller (7.5″ by 10″)
• Dimensions Math has two options for tests – Test A focuses on key concepts and fundamental problem-solving skills, while Test B focuses on the application of analytical skills and heuristics.
• Dimensions Math At Home – This video subscription brings a professional Singapore math teacher into your home classroom and provides in-depth instruction covering all the Dimensions Math material
for an entire school year. Available for Grades 1–6.
• Dimensions Math math goes through 8th grade
• Primary Mathematics 2022 Edition has consumable textbooks
• Primary Mathematics 2022 has a Kindergarten option (the U.S. Edition does not)
• Both Dimensions Math and Primary Mathematics 2022 Edition have vibrant illustrations and clean designs (The U.S. Edition is blackline with no real color)
I just want to end by saying, I LOVE Singapore math products of any kind! The Singapore math way is a solid foundation not just in “doing math” but as a way of critical thinking and understanding how
math works. So no matter which one you choose for your family you are sure to give your kids a solid math foundation!
Singapore Math Users Facebook Group
I was so excited to hear about this! It wasn’t a resource I was aware of. This is a great way to connect with other users to share ideas and ask questions.
So to connect with other Singapore math users be sure to check out the Singapore Math Users Facebook Group!
More On Singapore Math
Check out what I have previously written about the Singapore math U.S. Edition here.
See a side by side comparison of the Singapore Math U.S. Edition with CTCMath here. | {"url":"http://www.startsateight.com/singapore-math-dimensions-math/","timestamp":"2024-11-04T13:35:43Z","content_type":"text/html","content_length":"142932","record_id":"<urn:uuid:cc1b258c-073b-480e-8577-5d7a8db59a6b>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00678.warc.gz"} |
About: Script-friendly command-line tools for machine learning and data mining tasks. (The command-line tools wrap functionality from a public domain C++ class library.)
Added support for CUDA GPU-parallelized neural network layers, and several other new features. Full list of changes at http://waffles.sourceforge.net/docs/changelog.html
About: This package is an implementation of a linear RankSVM solver with non-convex regularization.
Initial Announcement on mloss.org.
• Operating System: Agnostic
• Data Formats: Matlab
• Tags: Matlab, Mkl, Classification, Feature Selection, Linear Svm, Convex Optimization, Gradient Based Learning, Ranking, Machine Learning, Optimization, Data Mining, Supervised Learning, Lasso,
Sparsity, Regularization, Algorithm, Discriminant Analysis, Linear Classifier, L1 Minimization, L1 Norm, Gradient Based Optimization, Non Convex
About: PyStruct is a framework for learning structured prediction in Python. It has a modular interface, similar to the well-known SVMstruct. Apart from learning algorithms it also contains model
formulations for popular CRFs and interfaces to many inference algorithm implementation.
Initial Announcement on mloss.org.
About: Efficient implementation of Semi-Stochastic Gradient Descent algorithm (S2GD) for training logistic regression (L2-regularized).
Initial Announcement on mloss.org.
About: This package implements Ideal PCA in MATLAB. Ideal PCA is a (cross-)kernel based feature extraction algorithm which is (a) a faster alternative to kernel PCA and (b) a method to learn data
manifold certifying features.
Initial Announcement on mloss.org.
About: Document/Text preprocessing for topic models: suite of Perl scripts for preprocessing text collections to create dictionaries and bag/list files for use by topic modelling software.
Moved distribution and code across to GitHub. Changed "ldac" format to have 0 offset for word indices. Added "document frequency" (df) filtering on selection of tokens for linkTables. Playing with
linkParse but its still unuseable generally.
About: Big Random Forests
Fetched by r-cran-robot on 2015-11-01 00:00:04.072762
About: SAMOA is a platform for mining big data streams. It is a distributed streaming machine learning (ML) framework that contains a programing abstraction for distributed streaming ML algorithms.
Initial Announcement on mloss.org.
About: deep learning toolkit in R
Fetched by r-cran-robot on 2018-01-01 00:00:07.583485
About: DAL is an efficient and flexibible MATLAB toolbox for sparse/low-rank learning/reconstruction based on the dual augmented Lagrangian method.
• Supports weighted lasso (dalsqal1.m, dallral1.m)
• Supports weighted squared loss (dalwl1.m)
• Bug fixes (group lasso and elastic-net-regularized logistic regression)
About: Estimates statistical significance of association between variables and their principal components (PCs).
Initial Announcement on mloss.org.
• Operating System: Agnostic
• Data Formats: Agnostic, Any Format Supported By R
• Tags: Bioinformatics, Machine Learning, Resampling, Statistics, Principal Component Analysis, Nonparametric Estimation, Unsupervised Learning, Latent Variable Model, Jackstraw, Statistical
About: DRVQ is a C++ library implementation of dimensionality-recursive vector quantization, a fast vector quantization method in high-dimensional Euclidean spaces under arbitrary data distributions.
It is an approximation of k-means that is practically constant in data size and applies to arbitrarily high dimensions but can only scale to a few thousands of centroids. As a by-product of training,
a tree structure performs either exact or approximate quantization on trained centroids, the latter being not very precise but extremely fast.
Initial Announcement on mloss.org.
About: hapFabia is an R package for identification of very short segments of identity by descent (IBD) characterized by rare variants in large sequencing data. It detects 100 times smaller segments
than previous methods.
o citation update
o plot function improved
About: hapFabia is an R package for identification of very short segments of identity by descent (IBD) characterized by rare variants in large sequencing data.
o citation update
o plot function improved
About: A library for calculating and accessing generalized Stirling numbers of the second kind, which are used for inference in Poisson-Dirichlet processes.
Initial Announcement on mloss.org.
About: Evolutionary Learning of Globally Optimal Trees
Fetched by r-cran-robot on 2014-05-01 00:00:05.459097
About: The glm-ie toolbox contains scalable estimation routines for GLMs (generalised linear models) and SLMs (sparse linear models) as well as an implementation of a scalable convex variational
Bayesian inference relaxation. We designed the glm-ie package to be simple, generic and easily expansible. Most of the code is written in Matlab including some MEX files. The code is fully compatible
to both Matlab 7.x and GNU Octave 3.2.x. Probabilistic classification, sparse linear modelling and logistic regression are covered in a common algorithmical framework allowing for both MAP estimation
and approximate Bayesian inference.
added factorial mean field inference as a third algorithm complementing expectation propagation and variational Bayes
generalised non-Gaussian potentials so that affine instead of linear functions of the latent variables can be used
About: ALgebraic COmbinatorial COmpletion of MAtrices. A collection of algorithms to impute or denoise single entries in an incomplete rank one matrix, to determine for which entries this is possible
with any algorithm, and to provide algorithm-independent error estimates. Includes demo scripts.
Initial Announcement on mloss.org.
About: ClowdFlows is a web based platform for service oriented data mining publicly available at http://clowdflows.org . A web based interface allows users to construct data mining workflows that are
hosted on the web and can be (if allowed by the author) accessed by anyone by following a URL of the workflow.
Initial Announcement on mloss.org.
About: [FACTORIE](http://factorie.cs.umass.edu) is a toolkit for deployable probabilistic modeling, implemented as a software library in [Scala](http://scala-lang.org). It provides its users with a
succinct language for creating [factor graphs](http://en.wikipedia.org/wiki/Factor_graph), estimating parameters and performing inference. It also has implementations of many machine learning tools
and a full NLP pipeline.
Initial Announcement on mloss.org.
• Authors: Alexandre Passos, Andrew Mccallum, Luke Vilnis, Sameer Singh, David Belanger, Michael Wick, Brian Martin, Jack Sullivan, Vineet Mundhra, Sam Anzaroot, Ariel Kobren, Karl Schultz, Emma
• License: Apache, Apache 2.0
• Programming Language: Scala | {"url":"https://mloss.org/software/opsys/agnostic/?page=4","timestamp":"2024-11-14T04:42:46Z","content_type":"application/xhtml+xml","content_length":"73635","record_id":"<urn:uuid:ef434909-7c11-4679-95a2-fb559f9c4094>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00334.warc.gz"} |
Simple Interest
Aptitude - Simple Interest
Why should I learn to solve Aptitude questions and answers section on "Simple Interest"?
Learn and practise solving Aptitude questions and answers section on "Simple Interest" to enhance your skills so that you can clear interviews, competitive examinations, and various entrance tests
(CAT, GATE, GRE, MAT, bank exams, railway exams, etc.) with full confidence.
Where can I get the Aptitude questions and answers section on "Simple Interest"?
IndiaBIX provides you with numerous Aptitude questions and answers based on "Simple Interest" along with fully solved examples and detailed explanations that will be easy to understand.
Where can I get the Aptitude section on "Simple Interest" MCQ-type interview questions and answers (objective type, multiple choice)?
Here you can find multiple-choice Aptitude questions and answers based on "Simple Interest" for your placement interviews and competitive exams. Objective-type and true-or-false-type questions are
given too.
How do I download the Aptitude questions and answers section on "Simple Interest" in PDF format?
You can download the Aptitude quiz questions and answers section on "Simple Interest" as PDF files or eBooks.
How do I solve Aptitude quiz problems based on "Simple Interest"?
You can easily solve Aptitude quiz problems based on "Simple Interest" by practising the given exercises, including shortcuts and tricks.
Exercise : Simple Interest - General Questions
• Simple Interest - General Questions
A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned
in 2 years be Rs. 3508, what was the amount invested in Scheme B?
Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).
Then, x x 14 x 2 + (13900 - x) x 11 x 2 = 3508
x - 22x = 350800 - (13900 x 22)
x = 45000
x = 7500.
So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.
Video Explanation: https://youtu.be/Xi4kU9y6ppk
A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?
Principal = Rs. 100 x 4016.25
9 x 5
= Rs. 8925.
How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
Time = 100 x 81 = 4 years.
450 x 4.5
Video Explanation: https://youtu.be/WdBzN0Sj8jc
Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?
Let rate = R% and time = R years.
Then, 1200 x R x R = 432
^2 = 432
^2 = 36
Video Explanation: https://youtu.be/TjjI4iRkzT0
Quick links
Quantitative Aptitude
Verbal (English)
Placement Papers | {"url":"https://www.indiabix.com/aptitude/simple-interest/","timestamp":"2024-11-09T07:41:12Z","content_type":"text/html","content_length":"78482","record_id":"<urn:uuid:66829997-9616-46a0-a36d-a20d9263a119>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00696.warc.gz"} |
Structure of Boundaries of 3-Dimensional Convex Divisible Domains
Geometry Topology Student Seminar
Wednesday, April 3, 2024 - 2:00pm for 1 hour (actually 50 minutes)
Alex Nolte – Georgia Tech
I read Benoist's paper Convexes Divisibles IV (2006, Invent. Math.), and will talk about it. The main result is a striking structural theorem for triangles in the boundaries of 3-dimensional properly
convex divisible domains O that are not strictly convex (which exist). These bound "flats" in O. Benoist shows that every Z^2 subgroup of the group G preserving O preserves a unique such triangle.
Conversely, all such triangles are disjoint and any such triangle descends to either a torus or Klein bottle in the quotient M = O/G (and so must have many symmetries!). Furthermore, this
"geometrizes" the JSJ decomposition of M, in the sense that cutting along these tori and Klein bottles gives an atoroidal decomposition of M. | {"url":"https://math.gatech.edu/seminars-colloquia/series/geometry-topology-student-seminar/alex-nolte-20240403","timestamp":"2024-11-05T04:37:31Z","content_type":"text/html","content_length":"31868","record_id":"<urn:uuid:e309af94-0aff-4fd8-9c7b-66dc4a1839c7>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00830.warc.gz"} |
Fast and Optimal Extraction for Sparse Equality Graphs | TACO Lab
Equality graphs (e-graphs) are used to compactly represent equivalence classes of terms in symbolic reasoning systems. Beyond their original roots in automated theorem proving, e-graphs have been
used in a variety of applications. They have become particularly important as the key ingredient in the popular technique of equality saturation, which has notable applications in compiler
optimization, program synthesis, program verification, and symbolic execution, among others. In a typical equality saturation workflow, an e-graph is used to store a large number of equalities that
are generated by local rewrites during a saturation phase, after which an optimal term is extracted from the e-graph as the output of the technique. However, despite its crucial role in equality
saturation, e-graph extraction has received relatively little attention in the literature, which we seek to start addressing in this paper. Extraction is a challenging problem and is notably known to
be NP-hard in general, so current equality saturation tools rely either on slow optimal extraction algorithms based on integer linear programming (ILP) or on heuristics that may not always produce
the optimal result. In fact, in this paper, we show that e-graph extraction is hard to approximate within any constant ratio. Thus, any such heuristic will produce wildly suboptimal results in the
worst case. Fortunately, we show that the problem becomes tractable when the e-graph is sparse, which is the case in many practical applications. We present a novel parameterized algorithm for
extracting optimal terms from e-graphs with low treewidth, a measure of how “tree-like” a graph is, and prove its correctness. We also present an efficient Rust implementation of our algorithm and
evaluate it against ILP on a number of benchmarks extracted from the Cranelift benchmark suite, a real-world compiler optimization library based on equality saturation. Our algorithm optimally
extracts e-graphs with treewidths of up to 10 in a fraction of the time taken by ILP. These results suggest that our algorithm can be a valuable tool for equality saturation users who need to extract
optimal terms from sparse e-graphs.
In Proc. ACM Program. Lang. (OOPSLA 2024)
🏆 Distinguished Paper | {"url":"https://cse.hkust.edu.hk/~parreaux/publication/oopsla24a/","timestamp":"2024-11-12T04:07:41Z","content_type":"text/html","content_length":"23649","record_id":"<urn:uuid:cc2299dd-8ba8-4888-b133-aee69684f387>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00692.warc.gz"} |
Edit Distance - Edit Distance LeetCode TutorialCup
Edit Distance
Difficulty Level Hard
Frequently asked in Amazon ByteDance Facebook Google Microsoft Palantir Technologies Square
Views 1950
In the edit distance problem we have to find the minimum number of operations required to convert a string X of length n to another string Y of length m.
Operations allowed:
1. Insertion
2. Deletion
3. Substitution
String1 = “abcd”
String2 = “abe”
Minimum operations required is 2 ( one deletion and one substitution )
We can solve this question by solving it’s sub-problem i.e., find the minimum number of operations required to convert a string X[0,…,i] to Y[0,…j].
1. If the last character of X and Y are the same, then find minimum operations required to convert string X[0,…,n-1] to Y[0,…,m-1].
2. If the last character of X and Y are not the same, then we can do one of the three operations:
□ Insertion at last index of X.
□ Deletion of the last index of X.
□ Substitution of the last index of X.
Using the above steps, let’s try to draw the recursion tree for this example:
string X= “ab” and Y = “xy” Therefore n=2 and m=2 ed(i,j) denotes minimum operations required to obtained a string Y[1,…,j] from X[1,…,i]. ( 1-based index).
As none of the characters in X is same as Y, hence at every iteration, we will call recursion three times by applying one of the three allowed operations.
As we can see there are many overlapping sub-problems in the above recursion due to which the worst case complexity of this procedure is O(3^n).
This can be easily reduced using Dynamic Programming which will resolve the issue of re-computation of overlapping sub-problems.
Dynamic Programming Approach
Base conditions
1. To convert a string of length n to the given string of length 0, we require n deletion operations.
2. To convert a string of length 0 to the given string of length n, we require n insertion operations.
At every position we may come across any of these two situations
1. If the last character of both strings is the same. In this case, we don’t need to do any operation, we can simply say that the answer to this problem will be the same as for the sub-problem
editDistance(X[0,…,i-1] , Y[0,….,j-1]).
2. If the last character is not the same, we have to choose one of these operations:
□ Insertion: In this case, we will insert Y[j] character at ith position and add 1 to the answer of editDistance(X[0,….i], Y[0,….,j-1]). By doing this we ensure that now the last character of
both strings is the same.
□ Deletion: In this case, we will delete the ith character of X and add 1 to the answer of editDistance(X[0,….,i-1] , Y[0,…,j]). By doing this we are leaving the last character of X string
i.e., deletion is performed.
□ Substitution: In this case we will substitute X[i] by Y[j] and add 1 to the answer of editDistance(X[0,…,i-1] , Y[0,…,j-1]). This becomes the same case as having equal last character in both
As we require the minimum number of operations hence, we will take a minimum of the three possible operations.
C++ Program
using namespace std;
int min(int a,int b,int c){
return min(a,min(b,c));
int editDistance(string X,string Y){
int n=X.length();
int m=Y.length();
int ed[n+1][m+1];
// to convert a string into null string we need to perform
// deletion operation n number of times where n is length of the string
for(int i=0;i<=n;i++){
// to convert a null string into the given string we need to perform
// insertion operation n number of times where n is length of the string
for(int i=0;i<=m;i++){
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
// no operation required
ed[i][j] = ed[i-1][j-1];
// one of the three operation required
ed[i][j] = 1+ min( ed[i-1][j], //deletion
ed[i][j-1], //insertion
ed[i-1][j-1] //substitution
return ed[n][m];
int main()
string X,Y;
int result = editDistance(X,Y);
cout<<"Minimum operations required to convert string "<<X<<" into string "<<Y<<" is: ";
return 0;
Minimum operations required to convert string programming into string problem is: 7
JAVA Program
import java.util.Scanner;
class Main {
static int min(int x, int y, int z)
return Math.min(x,Math.min(y,z));
static int editDistance(String X, String Y)
int n=X.length();
int m=Y.length();
int ed[][] = new int[n + 1][m + 1];
// to convert a string into null string we need to perform
// deletion operation n number of times where n is length of the string
for(int i=0;i<=n;i++){
// to convert a null string into the given string we need to perform
// insertion operation n number of times where n is length of the string
for(int i=0;i<=m;i++){
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(X.charAt(i - 1)==Y.charAt(j - 1)){
// no operation required
ed[i][j] = ed[i-1][j-1];
// one of the three operation required
ed[i][j] = 1+ min( ed[i-1][j], //deletion
ed[i][j-1], //insertion
ed[i-1][j-1] //substitution
return ed[n][m];
public static void main(String args[])
String X ,Y;
Scanner sc = new Scanner(System.in);
X = sc.nextLine();
Y = sc.nextLine();
int result = editDistance(X,Y);
System.out.println("Minimum operations required to convert string "+X+" into string "+Y+" is: "+result);
Minimum operations required to convert string codingAndChill into string codeAndGrow is: 8
Complexity Analysis
Time complexity: O(n*m)
where n = length of 1st string
m = length of 2nd string
as we are filling a 2D matrix( ed ) using two nested loops. | {"url":"https://tutorialcup.com/interview/string/edit-distance.htm","timestamp":"2024-11-08T05:59:41Z","content_type":"text/html","content_length":"103923","record_id":"<urn:uuid:2797fcb3-f340-4678-9d62-9400533bedac>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00015.warc.gz"} |
Ampere's equation
In physics, more particularly in electrodynamics, Ampère's equation describes the force between two infinitesimal elements of electric-current-carrying wires. The equation is named for the early
nineteenth century French physicist and mathematician André-Marie Ampère.
Rather than giving Ampère's original infinitesimal equation, which is not without problems,^[1]^[2] we will describe two common cases obtained by integration: a system consisting of two straight
wires and a system of two closed loops. Since the integrals over disputed terms in Ampère's infinitesimal equation vanish, the equations for these integrated systems are generally accepted and,
moreover, are in full agreement with experiment.
Electromagnetic units
Equations will be given in two common systems of electromagnetic units (SI and Gaussian units) and to that end we define the constant k as follows,^[3]
${\displaystyle k={\begin{cases}{\displaystyle {\frac {\mu _{0}}{4\pi }}}&{\hbox{for SI units}}\\\\{\displaystyle {\frac {1}{c^{2}}}}&{\hbox{for Gaussian units}}.\end{cases}}}$
Here μ[0] is the magnetic constant (also known as vacuum permeability). The quantity c is the speed of light in vacuum, in SI units a defined value denoted by c[0] = 299792458ms^−1 (exactly).
Two straight, infinite, and parallel wires
Consider two wires, one carrying an electric current i[1], the other i[2]. Both currents are constant in time; the wires are infinite, straight, and parallel, and a distance r apart.
The magnetic flux density B[2] due to the wire with current i[2] is directed in circles about wire 2, and at the distance r has magnitude (in SI units):^[4]
${\displaystyle B_{2}={\frac {\mu _{0}i_{2}}{2\pi r}}\ .}$
The force F[1] on wire 1 due to this magnetic flux density is found regarding the current as a movement of charge. A wire carrying a current i[2] in time dt moves a charge q = i[2] dt. In this time
the charge moves a distance dℓ = v dt (with v the speed of the electrons traveling down the wire), suggesting i[2] dℓ = q v. The Lorentz force upon the charge subject to magnetic flux density B
perpendicular to the flow of charge is then radially directed with magnitude F[] = qvB = i[2]B dℓ, which is the force on each element of length dℓ of the wire. Thus, the force per unit length upon
wire 1 is:
${\displaystyle F_{1}={\frac {\mu _{0}i_{1}i_{2}}{2\pi r}}\,}$
which is Ampère's law for the force per unit length l of wire. The force exerted by wire 1 on wire 2 has the same magnitude, so the subscript on the force is unnecessary. In general units, the force
per unit length is:
${\displaystyle F=2k{\frac {i_{1}i_{2}}{r}}.\,}$
The force F is attractive if the currents run in the same direction and repulsive if they flow in opposite direction.
Definition of the ampere
This force between straight, parallel wires is used to define the SI unit of current, the ampere, symbol A.^[5] Take two infinitely long wires in vacuum at a distance r = 1 m, consider the force that
one meter of these wires exert on one another (l = 1 m) and let this force be F = 2⋅10^−7 N (newton). Then for i[1] = i[2] the current strengths are by definition both equal to one ampere (1 A). In
SI units this implies that k = 10^−7 N/A^2 and hence that the magnetic constant is:
μ[0] = 4π k = 4π⋅10^−7 N/A^2.
Two loops
Let ${\displaystyle \scriptstyle i_{1}}$ and ${\displaystyle \scriptstyle i_{2}}$ be electric currents constant in time. They run in separate loops (closed curves) C[1] and C[2], see figure on the
right, where all quantities are defined. The total force between two loops is given by the double path integral over the loops
${\displaystyle \mathbf {F} _{12}=-ki_{1}i_{2}\oint _{C_{1}}\oint _{C_{2}}{\frac {(d{\boldsymbol {l}}_{1}\cdot d{\boldsymbol {l}}_{2})\,\mathbf {r} _{12}}{|\mathbf {r} _{12}|^{3}}}.\qquad \qquad
\qquad (1)}$
Here r[12] = r[1]−r[2] is the vector locating element 1 from the location of element 2, pointing from 2 to 1.
Alternative expression
One often finds the following expression for the force between two electric-current-carrying loops:^[6]
${\displaystyle \mathbf {F} _{12}=ki_{1}i_{2}\oint _{C_{1}}\oint _{C_{2}}{\frac {d{\boldsymbol {l}}_{1}\times (d{\boldsymbol {l}}_{2}\,\times \mathbf {r} _{12})}{|\mathbf {r} _{12}|^{3}}},\qquad
\qquad \ \ (2)}$
instead of the simpler expression in Eq. (1). Here the multiplication signs (×) indicate vector products. The integrand [expression under the integral of Eq. (2)] follows from the Biot-Savart-Laplace
expression for the magnetic induction B(r[1]) due to a segment of the second loop. Insertion of B(r[1]) into the Lorentz force that acts on the current in segment dl[1] gives the integrand of Eq.
The labeling of the segments being arbitrary, one would expect the same force (in absolute value) when the labels 1 and 2 are interchanged, or in other words, one would expect Newton's third law ${\
displaystyle \scriptstyle \mathbf {F} _{12}=-\mathbf {F} _{21}}$ (action is minus reaction) to hold. This is not the case, the integrand in equation (2) is non-symmetric under interchange of labels 1
and 2 and hence the integral also appears to be non-symmetric. However, after integration the expression becomes antisymmetric (changes sign under interchange of 1 and 2) and hence satisfies Newton's
third law.
To see this we note that the force in Eq. (2) has in fact two contributions, as follows from a result well-known in vector analysis,
${\displaystyle {\frac {d{\boldsymbol {l}}_{1}\times (d{\boldsymbol {l}}_{2}\,\times \mathbf {r} _{12})}{|\mathbf {r} _{12}|^{3}}}={\frac {(d{\boldsymbol {l}}_{1}\cdot \mathbf {r} _{12})\,d{\
boldsymbol {l}}_{2}}{|\mathbf {r} _{12}|^{3}}}-{\frac {(d{\boldsymbol {l}}_{1}\cdot d{\boldsymbol {l}}_{2})\,\mathbf {r} _{12}}{|\mathbf {r} _{12}|^{3}}}.}$
The second contribution gives an integral that is obviously equal to Eq. (1), which is manifestly antisymmetric under interchange of labels (the dot product does not change, the vector r[12] changes
sign). We will show that the first contribution vanishes after integration over a closed curve.
${\displaystyle {\frac {\mathbf {r} _{12}}{|\mathbf {r} _{12}|^{3}}}=-{\boldsymbol {abla }}_{1}\left({\frac {1}{|\mathbf {r} _{12}|}}\right)\equiv -\mathbf {A} ,}$
where A is a short hand notation. Applying Stokes' theorem, we find that the first contribution becomes
${\displaystyle -ki_{1}i_{2}\oint _{C_{2}}\left[\oint _{C_{1}}(d{\boldsymbol {l}}_{1}\cdot \mathbf {A} )\,\right]d{\boldsymbol {l}}_{2}=-ki_{1}i_{2}\oint _{C_{2}}\left[\iint _{S}({\boldsymbol
{abla }}_{1}\times \mathbf {A} )\cdot d\mathbf {S} _{1}\,\right]d{\boldsymbol {l}}_{2}.}$
It is well-known in vector analysis that
${\displaystyle {\boldsymbol {abla }}_{1}\times {\boldsymbol {abla }}_{1}\Phi =0}$
for any scalar function Φ and in particular for Φ ≡ 1/|r[12]|. Hence the first contribution to the force of Eq. (2) vanishes.
Ampère's original law for the force exerted upon element 1 by element 2 is (û[12] = −û[21]):^[7]
${\displaystyle d^{2}\mathbf {F_{21}} =-d^{2}\mathbf {F_{12}} =-i_{1}i_{2}{\frac {\mu _{0}}{4\pi }}\ {\frac {\mathbf {{\hat {u}}_{12}} }{r_{12}^{2}}}\left(2\ (d{\boldsymbol {l_{1}\cdot }}d{\
boldsymbol {l_{2}}})-3(\mathbf {{\hat {u}}_{12}\cdot } d{\boldsymbol {l_{1}}})(\mathbf {{\hat {u}}_{12}\cdot } d{\boldsymbol {l_{2}}})\right)\ ,}$
with û[12] a unit vector pointing along the line joining element 2 to element 1 and r[12] the length of this line. The force element is second order because it is a product of two infinitesimals.
This force law leads to the same force between closed current loops as the more commonly used Grassmann's law presented above:
${\displaystyle d^{2}\mathbf {F_{21}} =-i_{1}i_{2}{\frac {\mu _{0}}{4\pi }}\ {\frac {1}{r_{12}^{2}}}\left(\ (d{\boldsymbol {l_{1}\cdot }}d{\boldsymbol {l_{2}}})\mathbf {{\hat {u}}_{12}} -(\mathbf
{{\hat {u}}_{12}\cdot } d{\boldsymbol {l_{1}}})d{\boldsymbol {l_{2}}}\right)\ ,}$
or, interchanging 1 and 2, the force on element 2 by element 1 is:
${\displaystyle d^{2}\mathbf {F_{12}} =i_{1}i_{2}{\frac {\mu _{0}}{4\pi }}\ {\frac {1}{r_{12}^{2}}}\left(\ (d{\boldsymbol {l_{1}\cdot }}d{\boldsymbol {l_{2}}})\mathbf {{\hat {u}}_{12}} -(\mathbf
{{\hat {u}}_{12}\cdot } d{\boldsymbol {l_{2}}})d{\boldsymbol {l_{1}}}\right)\ ,}$
which is not simply opposite in sign to the previous expression. Unlike Ampère's law, Grassmann's law is not antisymmetric under exchange of the indices 1 and 2, and violates Newton's law of action
opposite to reaction. Grassmann's law is readily derived from the Biot-Savart law, and is consistent with space-time symmetry. Its violation of Newton's law of action and reaction stems from its
basis relying not upon action-at-a-distance, but being based instead upon a force mediated by a field, which itself has physical properties to take into account.^[8] There exists some debate over the
ultimate form of the force law between current elements.^[9] | {"url":"https://en.citizendium.org/wiki/Ampere%27s_equation","timestamp":"2024-11-08T18:44:57Z","content_type":"text/html","content_length":"111545","record_id":"<urn:uuid:5e235bc9-0fb0-4d36-8e93-5f82daacd5ea>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00008.warc.gz"} |
Mathematical Biology Seminar
Event Type
Department of Mathematics
Nov 6, 2024 1:00 pm
Gökçe Dayanıklı (University of Illinois Urbana-Champaign)
Daniel Cooney
Originating Calendar
Speaker: Gökçe Dayanıklı (University of Illinois Urbana-Champaign)
Title: Finding Optimal Policies for Large Populations: An Application to Epidemic Control
Abstract: The COVID-19 pandemic showed us that regulators need to find optimal mitigating policies for a large population of interacting agents who optimize their own objectives in a game
theoretical framework instead of following these policies perfectly. However, it is well known that finding an equilibrium in a game with a large number of agents is a challenging problem because
of the increasing number of interactions among the agents, and adding a principal to the game escalates the challenges further. In this talk, in order to approximate the game between the
principal and the large number of agents, we consider a Stackelberg mean field game model, motivated by the modeling of the epidemic control in large populations. The agents play a
non-cooperative game in which they can control their transition rates between states to minimize an individual cost. The principal can influence the resulting Nash equilibrium through incentives
to optimize her own objective. Later, we propose an application to an epidemic model of SIR type in which the agents control their interaction rate, and the principal is a regulator acting with
non-pharmaceutical interventions. To compute the solutions, we propose an innovative numerical approach based on Monte Carlo simulations and machine learning tools for stochastic optimization.
Finally, we briefly discuss another game formulation for a continuum of non-identical players evolving on a finite state space where their interactions are represented by a limit of graph. | {"url":"https://calendars.illinois.edu/detail/7421?eventId=33501988","timestamp":"2024-11-08T23:19:22Z","content_type":"text/html","content_length":"46389","record_id":"<urn:uuid:6c4ab8be-9ace-4643-a6b5-7968c7ba201e>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00259.warc.gz"} |
Relations between Combinatorics and Other Parts of Mathematicssearch
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Softcover ISBN: 978-0-8218-1434-5
Product Code: PSPUM/34
List Price: $139.00
MAA Member Price: $125.10
AMS Member Price: $111.20
eBook ISBN: 978-0-8218-9322-7
Product Code: PSPUM/34.E
List Price: $135.00
MAA Member Price: $121.50
AMS Member Price: $108.00
Softcover ISBN: 978-0-8218-1434-5
eBook ISBN: 978-0-8218-9322-7
Product Code: PSPUM/34.B
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• Proceedings of Symposia in Pure Mathematics
Volume: 34; 1979; 378 pp
MSC: Primary 05
Brings into focus interconnections between combinatorics on the one hand and geometry, group theory, number theory, special functions, lattice packings, logic, topological embeddings, games,
experimental dsigns, and sociological and biological applications on the other hand.
• Permission – for use of book, eBook, or Journal content
• Book Details
• Table of Contents
• Requests
Volume: 34; 1979; 378 pp
MSC: Primary 05
Brings into focus interconnections between combinatorics on the one hand and geometry, group theory, number theory, special functions, lattice packings, logic, topological embeddings, games,
experimental dsigns, and sociological and biological applications on the other hand.
Permission – for use of book, eBook, or Journal content
Please select which format for which you are requesting permissions. | {"url":"https://bookstore.ams.org/PSPUM/34","timestamp":"2024-11-05T04:10:42Z","content_type":"text/html","content_length":"99536","record_id":"<urn:uuid:93637214-5ea0-443b-a5f7-289a04123a18>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00651.warc.gz"} |
The 24^th European Conference on Iteration Theory (ECIT 2024) dedicated to the memory of Oleksandr Sharkovsky will take place in Vimeiro, Portugal, a small village in front of the sea, on the west
coast of Portugal, 50 km from Lisbon, between Monday, 27^th May 2024 and Friday, 31^st May 2024. The conference will be held in association with the International Society of Difference Equations
In Memoriam: Oleksandr Mykolayovych Sharkovsky – A Mathematical Genius
(1936 – 2022)
In this conference, we gather to pay tribute to the memory of Олекса́ндр
Миколайович Шарко́вський, Oleksandr Mykolayovych Sharkovsky, a
distinguished figure in the world of mathematics. Born on December 7,
1936, in Kyiv, Ukraine, Sharkovsky’s journey through the realm of
mathematics commenced in his youth, marked by an unwavering curiosity
for numbers and patterns.
Alexander Sharkovsky’s technical acumen was truly remarkable. His
contributions to the field of dynamical systems and chaos theory stand
as enduring milestones in the annals of mathematical thought. Navigating
the intricacies of mathematical structures, he unveiled the profound
insights encapsulated in the Sharkovsky theorem, an epochal contribution
to the study of periodic points in continuous dynamical systems.
With remarkable lucidity, Sharkovsky’s bridged the chasm between
theoretical constructs and real-world applications in the case of his
studies in turbulence. His work bestowed upon mathematicians and
scientists indispensable tools for comprehending the intricate tapestry
of interwoven systems that govern our universe. He shed light on the
concealed order lurking within ostensibly random phenomena in dynamic
Through his mentorship, he nurtured generations of mathematicians…
Amidst his mathematical endeavors, Sharkovsky championed the imperative
of fostering international collaboration and camaraderie among
mathematicians. He believed fervently in the universality of the
language of mathematics, transcending geographical boundaries to unite
minds across the globe.
As we reflect upon the legacy of this mathematical luminary, let us not
only mourn his departure but also celebrate the profound heritage he
bequeaths to us. Oleksandr Mykolayovych Sharkovsky’s mathematical
brilliance will continue to inspire and guide generations of
mathematicians yet unborn. His work has flung open the doors to
uncharted territories of comprehension, igniting creativity and
curiosity in countless hearts and minds.
In this poignant moment, let us remember not just the equations and
theorems but the man who embodied the spirit of mathematical
exploration, who uncovered the radiance concealed within the deepest
recesses of mathematical enigma. Oleksandr Sharkovsky’s legacy shall
endure; his contributions to the realm of mathematics shall resound
through the corridors of academic history.
As we bid farewell in this conference to this towering mathematical
figure and amiable man, let us honor his memory by perpetuating our
quest to explore the boundless wonders of mathematics, just as he did
throughout his illustrious life. To quote the great Carl Sagan,
“Somewhere, something incredible is waiting to be known.” Sharkovsky has
shown us that this pursuit of discovery is noble and everlasting. Rest
in peace, esteemed mathematician, colleague, and friend, for your legacy
shall forever shine as a guiding star in the mathematical firmament.
And in his honor, we shall not forget the remarkable order of natural
numbers encapsulated in the Sharkovsky ordering, a testament to his
enduring impact on the world of mathematics.
As its name states the main topic of this conference is iteration theory. In particular among the main topics are discrete dynamical systems, functional equations and difference equations. Questions
of this type arise for example in biology, physics, economics and engineering.
History of ECIT
The first European Conference on Iteration Theory took place 1973 in Toulouse (France). It was continued 1977 in Graz (Austria), 1980 in Marburg (Germany), 1982 in Toulouse (France), 1984 in Lochau
(Austria), 1987 in Caldes de Malavella (Spain), 1989 in Batschuns (Austria), 1991 in Lisboa (Portugal), 1992 in Batschuns (Austria), 1994 in Opava (Czechia), 1996 in Urbino (Italy), 1998 in Muszyna
(Poland), 2000 in La Manga del Mar Menor (Spain), 2002 in Évora (Portugal), 2004 in Batschuns (Austria), 2006 in Gargnano (Italy), 2008 in Yalta (Ukraine), 2010 in Nant (France), 2012 in Ponta
Delgada (Portugal), 2014 in Łagów (Poland), 2016 in Innsbruck (Austria), 2018 in Murcia (Spain), 2022 in Reichenau an der Rax (Austria).
The conference has the support of the national funds through the FCT – Fundação para a Ciência e a Tecnologia, under the scope of the projects UIDB/04459/2020 and UIDP/04459/2020 (CAMGSD-IST) and
UIDB/04674/2020 (CIMA) | {"url":"https://ecit2024.isel.pt/","timestamp":"2024-11-08T08:12:32Z","content_type":"text/html","content_length":"43844","record_id":"<urn:uuid:3f734bf8-66d9-4e14-837b-f7600e5cc4a8>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00671.warc.gz"} |
[QSMS Seminar 14,16 Dec] A brief introduction to differential graded Lie algebras I, II
• Date : 12월 14일(화), 16일(목) 16:00-17:30
• Place : Zoom (ID: 642 675 5874)
• Speaker : 조창연 (QSMS, SNU)
• Title : A brief introduction to differential graded Lie algebras I, II
• Abstract : The importance of differential graded Lie algebras goes back at least to Quillen’s rational homotopy theory, which also motivated their applications to deformation theory. Later, such
an idea was developed further by Deligne, Drinfeld, and Feigin, and influenced many including Kontsevich and Soibelman. The purpose of these talks is to give a short introduction to the notion of
differential graded Lie algebras and its relationship to deformation theory. These talks are intended to be an elementary introduction to the subject, but due to the current nature of it, I’ll
say something about the theory of infinity-categories. The first talk will be devoted to exploring some of the fundamentals of differential graded Lie algebras and infinity-categories, and the
application to deformation theory will be covered in the later half of the second talk. | {"url":"https://qsms.math.snu.ac.kr/index.php?mid=board_sjXR83&listStyle=viewer&order_type=desc&l=en&sort_index=title&page=5&document_srl=2055","timestamp":"2024-11-09T04:18:24Z","content_type":"text/html","content_length":"21872","record_id":"<urn:uuid:b387ad67-38a0-475b-8cfe-b079c0746724>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00261.warc.gz"} |
Your Name Will Be In This Puzzle – Can You Find It?
Home Life Your Name Will Be In This Puzzle – Can You Find It?
Your Name Will Be In This Puzzle – Can You Find It?
If you want to solve this puzzle, just look at this block of text we have provided you with seemingly random letters and try and find your name. This puzzle is 100% a lot harder than most people will
expect it to be so don’t try to solve it quickly.
Can you find your name in this Puzzle Picture?
Here is the answer:
You see what we did there?
Most people are tricked with the very title of the article. Even though you’re looking for your name, you should be looking for “your name”.
Can you solve this Math Puzzle?
Can you solve this Math Puzzle? You should share this puzzle with your friends if you enjoyed it.
If your sock drawer has 6 black socks,
4 brown socks, 8 white socks, and 2 tan socks, how many socks would you have to pull out in the dark to be sure you had a matching color pair?
Here is the answer!
The Answer is 5, there are only four colors, so five socks guarantee that two will be the same color.
Facebook Comments | {"url":"https://us.wakeupyourmind.net/name-will-puzzle-can-find-2.html","timestamp":"2024-11-14T05:03:28Z","content_type":"text/html","content_length":"107305","record_id":"<urn:uuid:e2f98642-d844-4a11-8d15-f3f0f833f34d>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00492.warc.gz"} |
mp_arc 04-276
04-276 Artur Avila, David Damanik
Generic Singular Spectrum For Ergodic Schr\"odinger Operators (20K, LaTeX) Sep 7, 04
Abstract , Paper (src), View paper (auto. generated ps), Index of related papers
Abstract. We consider Schr\"odinger operators with ergodic potential $V_\omega(n)=f(T^n(\omega))$, $n \in \Z$, $\omega \in \Omega$, where $T:\Omega \to \Omega$ is a non-periodic homeomorphism. We
show that for generic $f \in C(\Omega)$, the spectrum has no absolutely continuous component. The proof is based on approximation by discontinuous potentials which can be treated via Kotani
Files: 04-276.src( 04-276.keywords , Avila_Damanik.TEX ) | {"url":"http://kleine.mat.uniroma3.it/mp_arc-bin/mpa?yn=04-276","timestamp":"2024-11-05T03:53:21Z","content_type":"text/html","content_length":"1662","record_id":"<urn:uuid:977e015b-3558-4f88-b926-3a316f747da1>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00211.warc.gz"} |
Alibaba SQL Interview Question | DataLemur
Compressed Mean Alibaba SQL Interview Question
Compressed Mean
Alibaba SQL Interview Question
You're trying to find the mean number of items per order on Alibaba, rounded to 1 decimal place using tables which includes information on the count of items in each order ( table) and the
corresponding number of orders for each item count ( table).
Column Name Type
item_count integer
order_occurrences integer
Example Input:
item_count order_occurrences
There are a total of 500 orders with one item per order, 1000 orders with two items per order, and 800 orders with three items per order."
Example Output:
Let's calculate the arithmetic average:
Total items =
Total orders =
Mean =
The dataset you are querying against may have different input & output - this is just an example!
Step 1: Calculate the weighted average of items per order
To calculate the weighted average of items per order, we multiply each with the corresponding number of occurrences , calculate the sum using , and finally divide it by the total number of orders
using .
However, it's important to note that both and are of integer type by default, which means that division will return an integer result. To ensure that the output is rounded to 1 decimal place, we can
cast either column to a decimal type using or .
Step 2: Round results to 1 decimal place
To round the result to 1 decimal place, we can use the function.
Sourced from | {"url":"https://datalemur.com/questions/alibaba-compressed-mean","timestamp":"2024-11-01T23:42:41Z","content_type":"text/html","content_length":"81768","record_id":"<urn:uuid:e7288370-91e5-40a8-b375-3562a0283c57>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00541.warc.gz"} |
Data Analytics, Statistics, Chemometrics, and Artificial Intelligence | Page 4
Data Analytics, Statistics, Chemometrics, and Artificial Intelligence
Latest News
Revolutionizing Tissue Analysis: Femtosecond Double-Pulse LIBS with Machine Learning Breaks New Ground
November 12th 2024
A recent study presents a new technique that combines femtosecond double-pulse laser-induced breakdown spectroscopy (fs-DP-LIBS) with machine learning (ML) algorithms to significantly enhance tissue
discrimination and signal quality, paving the way for more precise biomedical diagnostics.
More News | {"url":"https://www.spectroscopyonline.com/topic/data-analysis-statistics-chemometrics-artificial-intelligence?page=4","timestamp":"2024-11-13T10:50:31Z","content_type":"text/html","content_length":"306055","record_id":"<urn:uuid:22602fa6-9a1f-41a5-8bff-d19b1b1f42ca>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00115.warc.gz"} |
30++ Vector Addition Worksheet
2 min read
30++ Vector Addition Worksheet. Vector components and vector addition worksheet answers: Note that p is the angle with the horizontal axis.
Vector Addition Worksheet Answers Math Worksheets Printable from mathworksheetprintable.com
This account has been suspended. Graphically add each pair of vectors shown below in its box, making sure to show the vector addition as well as the resultant with a dotted line and. Its velocity is
represented by the vector v = 500, 500.
4,000+ Vectors, Stock Photos & Psd Files.
Multiply a vector by a scalar. Vector components and vector addition worksheet answers: Find & download free graphic resources for addition worksheet.
Vector Addition Worksheet With Answers Download Print.
On a separate piece of paper, use the following individual vectors to graphically find the resultant vector in the first three problems. This worksheet with vector addition of each of the sum is the.
The sum or difference of two vectors.
Determine The Magnitude (In Centimeters) And Direction (In Standard Form) Of The Resultant Vector B + A For Each Of The Combinations Below.
Free for commercial use high quality images 4,000+ vectors, stock photos & psd files. Note that p is the angle with the horizontal axis.
Remember, The Resultant Vector Must Have Both.
Once you have done this for all vectors to be added, draw the final vector with its tail at the tail of the first vector and head at the head of the last vector. Let’s look at some examples of
vectors, vector. Its velocity is represented by the vector v = 500, 500.
Free For Commercial Use High Quality Images.
Download all files as a compressed.zip. This is the sum of all the smaller. Worksheets are scalars and vectors, addition and subtraction of geometric vectors, assignment date period, vectors. | {"url":"https://worksheets.decoomo.com/vector-addition-worksheet/","timestamp":"2024-11-02T13:59:31Z","content_type":"text/html","content_length":"198133","record_id":"<urn:uuid:b4413d47-e7c5-40d3-9ec6-eccd40d1096a>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00861.warc.gz"} |
Find the correct sharpening angle
Are you just getting started with sharpening with a whetstone, and do you need help finding the correct angle? We can help. First of all, consider which sharpening stone to start with. The right
angle determines the sharpness of the knife.
Determine the sharpening angle for your knife
The cutting edge of a knife can have different angles. The smaller the angle, the sharper the knife. Therefore, the cutting edge of a very sharp knife, is very narrow. Japanese knives are made of
very hard steel (starting from 60 HRC) and can be made extremely sharp, when sharpened under a small angle. Other, for example German knives (55-58 HRC), often are made of slightly softer types of
steel and are sharpened under a wider angle.
The angle of the cutting edge of Japanese knives is usually 30°. For normal knives you can use an angle between 36° and 40°. The sharpening angle for your knife is half the angle of the cutting edge
(after all, you will sharpen on 2 sides). The sharpening angle of a Japanese knife (please find a few examples below) is therefore 15° (30°/2) and of other knives 18° to 20°.
Does your knife become dull or does it damage quickly? Then maybe you are sharpening the knife too thin. In this case, we recommend using a larger sharpening angle.
Examples of brands that carry knives made of hard steel (sharpening at an angle of 15°) Böker, Eden, Kai, Miyabi by Zwilling, Robert Herder and Sakai Takayuki.
If it's still unclear, there's always the 'pen trick'.
How high do you keep the back of your blade when sharpening
The grinding angle which you will achieve, is determined by the height of the back of your knife against the stone. You can determine the correct height with a simple formula. If you have determined
the required grinding angle, measure the width of the knife. Then look in the table below how high you must keep your knife against the stone.
Width of the blade Sharpening angle 12° Sharpening angle 15° Sharpening angle 18° Sharpening angle 20° Sharpening angle 23°
10 mm 2.08 mm 2.59 mm 3.09 mm 3.42 mm 3.91 mm
15 mm 3.12 mm 3.88 mm 4.64 mm 5.13 mm 5.86 mm
20 mm 416 mm 5.18 mm 6.18 mm 6.84 mm 7.82 mm
25 mm 5.20 mm 6.47 mm 7.73 mm 8.55 mm 9.77 mm
30 mm 6.24 mm 7.77 mm 9.27 mm 10.26 mm 11.72 mm
35 mm 7.28 mm 9.06 mm 10.82 mm 11.97 mm 13.68 mm
40 mm 8.32 mm 10.35 mm 12.36 mm 13.68 mm 15.63 mm
45 mm 9.36 mm 11.65 mm 13.91 mm 15.39 mm 17.58 mm
50 mm 10.40 mm 12.94 mm 15.45 mm 17.10 mm 19.54 mm
55 mm 11.44 mm 14.24 mm 17 mm 18.81 mm 21.49 mm
60 mm 12.47 mm 15.53 mm 18.54 mm 20.52 mm 23.44 mm
Helpful tools to sharpen with the correct height
If you don't want to work with a ruler, there are two sharpening rail set helpers that we can recommend. The sharpening rail is clamped on the back of the knife. The small sharpening rail creates a
fixed height of 4.5mm. The bigger sharpening rail creates a fixed height of 6.5mm. This means, that when using a sharpening rail your are less flexible with your angle than if you choose the angle
manually. In this table below we show the possible heights:
Height of the blade Sharpening angle 12° Sharpening angle 15° Sharpening angle 18° Sharpening angle 20° Sharpening angle 23°
10 mm 2.08 mm 2.59 mm 3.09 mm 3.42 mm 3.91 mm
15 mm 3.12 mm 3.88 mm 4.64 mm 5.13 mm 5.86 mm
20 mm 4.16 mm 5.18 mm 6.18 mm 6.84 mm 7.82 mm
25 mm 5.20 mm 6.47 mm 7.73 mm 8.55 mm 9.77 mm
30 mm 6.24 mm 7.77 mm 9.27 mm 10.26 mm 11.72 mm
If you use a sharpening rail, we recommend to apply a strip of adhesive tape on the back of the knife before clamping on the sharpening rail, to prevent damage of the blade. | {"url":"https://www.knivesandtools.co.uk/en/ct/find-the-correct-sharpening-angle-in-three-steps.htm","timestamp":"2024-11-08T10:54:58Z","content_type":"text/html","content_length":"105048","record_id":"<urn:uuid:f981f837-6b5b-4e2f-9ab9-a738c69a1e8d>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00200.warc.gz"} |
How Much Money Is 10 Dimes Worth in Dollars? Answer + Conversion Calcu
How Much Money Is 10 Dimes Worth in Dollars? (Answer, Dimes To Dollars Calculator, and Coin Conversion Formula)
Do you want to know the answer to ‘How much is 10 dimes worth in dollars?’ 10 dimes are worth 1 dollar, which is equivalent to 100 cents.
What if you don’t have precisely 10 dimes? How do you calculate how many dollars you have in dimes? That’s simple! Use our dimes to dollars converter to turn your dimes into dollars.
Please continue reading to use our converter, find out how to convert dimes to dollars with a conversion formula, and learn more about dimes.
10 Dimes to dollars converter (Conversion calculator)
Use our free 10 dimes to dollars converter to quickly calculate how much your dimes are worth in dollars. Just type in how many dimes you have, and our converter does the rest for you!
Looking at the converter, you will see that we already entered 10 dimes, giving us an answer of $1.00. That answers our question about ‘how much is 10 dimes worth in dollars?’ 10 dimes equals 1
Now it’s your turn! Type in how many dimes you have, and our dimes to dollars calculator will tell you how much that is in dollars. We make converting from dimes to dollars easy, no matter how many
dimes you have. Whether you have 10 dimes or 100 dimes, we can solve it all.
10 Dimes to dollars conversion table (Fast unit conversion method)
One popular method to convert dimes into dollars is to use our conversion table. Conversion tables, also called conversion charts, list the number of dimes in one column with the corresponding number
of dollars in the second column.
To use our conversion table to find the number of dollars in 10 dimes, find 10 dimes in the first column and note that the matching number of dollars in the second column is $1. There is 1 dollar in
10 dimes.
Number of Dimes Dollars $
10 $1
20 $2
30 $3
40 $4
50 $5
100 $10
500 $50
1000 $100
How to convert dimes to dollars (Calculate the answer using a conversion formula)
Use our converter or the following conversion formula to convert dimes to dollars.
Dimes to dollars conversion formula:
Dollars = Dimes x 0.10 dollars per dime
The formula says that we can determine the number of dollars we have by multiplying the number of dimes by 0.10, which is the number of dollars in one dime.
For example, to determine how many dollars are in 10 dimes, we multiply 10 dimes by 0.10, as shown below.
Dollars = 10 dimes x 0.10 dollars per dime = 1 dollar
How much is 20 dimes?
20 dimes is equal to $2.00.
How much is 102 dimes?
102 dimes is equal to $10.20.
How much money is 10,000 dimes?
10,000 dimes is equal to $1,000. This is because there are 10 dimes in $1. So, if you have 10,000 dimes, you have $1,000.
How much is 10 million dimes?
10 million dimes are worth one million dollars.
How much is 10 dimes and 25 quarters worth?
10 dimes and 25 quarters are worth $7.25.
To answer this question yourself, follow these steps:
Step 1: Convert the 10 dimes into dollars.
10 dimes x 0.10 dollars/dime = $1
Step 2: Convert the 25 quarters into dollars
25 quarters x 0.25 dollars/quarters = $6.25
Step 3: Add the value of the 10 dimes and the 25 quarters together to find the answer.
$1 + $6.25 = $7.25
How much is 10 dimes in cents?
10 dimes is equal to 100 cents.
How much is 10 silver dimes worth?
When the silver price is $19 per troy ounce, the melt value of 10 silver dimes is $13.75.
How many dimes is $5?
There are 50 dimes in $5.
What is 2 dollars in dimes?
There are 20 dimes in 2 dollars.
How many dimes are in a roll?
A roll of dimes contains 50 dimes worth five dollars.
How much is 10 coins on TikTok?
Ten coins on TikTok are worth 0.11 dollars. TikTok coins are a virtual currency that you can use to purchase virtual gifts or send to other users within the TikTok app. Users can buy coins with real
money through the app or on TikTok’s website.
Frequently asked questions: Converting dimes to dollars
People often have specific questions about converting dimes to dollars. Here are the answers to some of the most common questions people ask about dimes to dollars conversions.
Is a dime less than a cent?
No, a dime is worth more than a cent. A dime is worth 10 cents, while a cent is worth 1 cent.
How many cents are in a dime?
There are ten cents in a dime.
Is a dime a tenth of a dollar?
Yes, a dime is worth one-tenth of a dollar.
How many pennies are in a dime?
There are 10 pennies in a dime.
How many nickels is 2 dollars?
There are 40 nickels in 2 dollars.
What are the dimensions of a dime?
A dime is 0.705 inches (1.791 centimeters) in diameter and has a thickness of 0.053 inches (0.135 centimeters).
How thick is a dime?
Modern-day dimes are 1.35 mm (0.135 cm) thick, equal to 0.061 inches.
What is the volume of a dime?
The volume of a single dime is 0.020755 cubic inches, equivalent to 0.34011 cubic centimeters.
How much does a dime weigh? (Mass)
One dime weighs 2.268 grams, which is equal to 0.08 ounces.
How many ridges on a dime?
According to the U.S. Mint, there are 118 ridges on a dime. Ridges, also called grooves, are a physical security feature that makes dimes challenging to counterfeit.
When dimes were made from silver, the reeded ridges also helped prevent coin clipping, which is a form of fraud. What is coin clipping, and how do ridges deter this fraudulent practice? Imagine that
you have a bucket full of silver dimes. While each dime is worth 10 cents, the metal is also valuable.
By shaving off a small amount of the silver from each coin in the bucket, a scammer could amass a sizeable amount of valuable silver to sell for scrap. The preferred target of silver scrapers has
always been the edges of coins. It is much easier, in comparison, to notice if the face or back of a coin has been scraped. We would be much more likely to see if someone had defaced Franklin D.
Roosevelt’s face than the edge of the dime!
What are dimes made of?
Dimes are primarily copper but also are made of nickel. To be precise, our modern-day dimes are 91.67% copper and 8.33% nickel. It wasn’t always this way, though! Before 1965, dimes were 90% silver
and 10% copper. If you have 200 dimes ($20), you are holding 415.82 grams of copper and 37.78 grams of nickel in your hand. If you prefer to work in pounds, that’s 0.917 pounds of copper and 0.083
pounds of nickel.
What hasn’t changed is that dimes are still worth 10 cents, even though they are no longer made from silver.
Are dimes magnetic?
No, dimes are not magnetic, even though they are made from a nickel-copper alloy.
Nickel-copper alloys are only magnetic when the nickel content exceeds 56%. Since dimes are only 8.33% nickel, they are not magnetic.
How many different coins are used in the United States?
There are currently six different coins in circulation in the United States: the penny, nickel, dime, quarter, half-dollar, and dollar coin.
How many different dollar coins are there?
There have been thirteen different types of dollar coins minted in U.S. history, including early dollar coins, the 1804 dollar coin, Seated Dollar Liberty coins, Gold dollar coins, Trade dollar
coins, Morgan dollar coins, Peace dollar coins, Eisenhower dollar coins, Susan B. Anthony dollar coins, American Silver Eagle, Sacagawea dollar coins, Presidential Dollar coins, and American
Innovation Dollar Coins.
How many different denominations of U.S. paper currency are there?
There are currently six denominations of U.S. paper currency in circulation: $1, $2, $5, $10, $20, and $100 bills. The first five denominations are frequently called “federal reserve notes”, while
the $100 bill is known as a “century note”. All six denominations are produced by the Bureau of Engraving and Printing (BEP).
The $1 and $2 bills are the oldest designs still in circulation, last updated in 1963. The $5 bill was last redesigned in 2008, while the $10 and $20 bills were last updated in 2006. The $100 bill is
the most recent design, having been released in 2013. All U.S. currency is printed on special paper that contains a unique blend of cotton and linen fibers.
In conclusion, 10 dimes are worth 1 dollar, which is equivalent to 100 cents.
To convert between 10 dimes and dollars yourself, use our converter or read the steps in our ‘How to convert dimes to dollars’ section. | {"url":"https://www.tipwho.com/article/how-much-is-10-dimes-in-dollars-answer-conversion-calculator/","timestamp":"2024-11-14T07:51:09Z","content_type":"text/html","content_length":"151046","record_id":"<urn:uuid:7321b6a5-d7bd-4eaa-b633-d6bc33ab7f6c>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00301.warc.gz"} |
Venn Diagrams Part 1
Venn diagrams are a means to display categories of data graphically. They are more flexible than contingency tables, allowing complex reasoning, and have many applications in set theory.
Typically each set is illustrated by a bubble, allowing intersection, with one or more of the other sets, and an intersection simultaneously of all the sets. All the regions are a Venn diagram are
mutually exclusive and exhaustive, since each element may only appear once on a Venn diagram. This does not mean that the sets are mutually exclusive since, for example, on a Venn diagram
illustrating which newspapers people read, with each set representing a different newspaper, a person may read more than one newspaper, hence be in more than one set.
Example: 100 people were asked which newspapers they read. The results showed that 30 read Daily Trash, 26 read The Honest Untruth, 21 read The Dirty Digger, 5 read both Daily Trash and The Honest
Untruth, 7 read both The Honest Untruth and The Dirty Digger, 6 read both The Dirty Digger and Daily Trash and 2 read all three.
We have to fill out the diagram above. It is best to work out from the centre. 2 people read all three newspapers, so the entry in the central region is 2. Complete the regions in the order shown:1)
2) 3) 4) 5) 6). At the end, of the 100 people asked, 61 read a newspaper an 39 don't. This number, 39, goes out side any set as shown. | {"url":"https://astarmathsandphysics.com/a-level-maths-notes/s1/3746-venn-diagrams-1.html","timestamp":"2024-11-09T20:15:55Z","content_type":"text/html","content_length":"31744","record_id":"<urn:uuid:4e0bb46d-157a-4e91-8161-f3b69988984e>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00174.warc.gz"} |
Indefinite quadratic polynomials of small signature
Cook, R. J ; Raghavan, S (1984) Indefinite quadratic polynomials of small signature Monatshefte fur Mathematik, 97 (3). pp. 169-176. ISSN 0026-9255
Full text not available from this repository.
Official URL: http://www.springerlink.com/content/l08427hk280567...
Related URL: http://dx.doi.org/10.1007/BF01299144
Let F(X)=Q(X)+L(X) be a real quadratic polynomial with no constant term. Suppose that the quadratic part Q(X) is indefinite of type (r, n-r). For an integer k≥4 we show that if min (r, n-r) ≥k there
exists a function f(n, k)=–1/2+3/(4k+2)+O[k] (1/n) with the following property. For any η>0 and all large enough X there is an integer vector χ≠0 such that |χ| ≤X and |F(X)|«X^ƒ(n,k)+n.
Item Type: Article
Source: Copyright of this article belongs to Springer.
ID Code: 37828
Deposited On: 25 Apr 2011 10:41
Last Modified: 25 Jun 2012 15:07
Repository Staff Only: item control page | {"url":"https://repository.ias.ac.in/37828/","timestamp":"2024-11-04T01:40:13Z","content_type":"application/xhtml+xml","content_length":"15890","record_id":"<urn:uuid:b0f5b9c9-d253-4744-b91f-7d0786398767>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00454.warc.gz"} |
Important Formulas of Cube Calculators | List of Important Formulas of Cube Calculators
List of Important Formulas of Cube Calculators
Important Formulas of Cube calculators give you a list of online Important Formulas of Cube calculators. A tool perform calculations on the concepts and applications for Important Formulas of Cube
These calculators will be useful for everyone and save time with the complex procedure involved to obtain the calculation results. You can also download, share as well as print the list of Important
Formulas of Cube calculators with all the formulas. | {"url":"https://www.calculatoratoz.com/en/-important-formulas-of-cube-Calculators/CalcList-10319","timestamp":"2024-11-07T07:43:56Z","content_type":"application/xhtml+xml","content_length":"123236","record_id":"<urn:uuid:2923bb84-6508-4906-9a5e-5c9de1268deb>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00195.warc.gz"} |
Thorpe Park Discussion Thread
Wow! Those are very realistic animatronics.
Top Posters In This Topic
This ride looks huge from certain angles!
Looks good. Those soldiers look amazing had to look twice to see they weren't real, because first thing I thought they where actors. It looks better en better every time I watch.
^I think they are actors. Unless I am very much mistaken...
^I think they are actors. Unless I am very much mistaken...
I don't think he was being serious........
99.99% of it looks fantastic, the wheels on the operator's booth look ridiculous. It's a little thing but for something most riders are going to look at, what with it being right infront of the lift
hill that draws your attention when you've boarded, it seems strange that it's one of the few things that hasn't been designed/themed realistically.
the wheels on the operator's booth look ridiculous.
But it's themed to an upside down vehicle that's smashed through the roof....
Overall, the ride seems a bit short for me. Missing that last "oomph" to get it past "it's already over?" standards.
But it's themed to an upside down vehicle that's smashed through the roof....
They are way too small and too close together.
LOVE the swarm "scream"
also! Loving this
I spy a water effect in the 6th shot. This is going to be so awesome... Although, I think it's funny that the two shortest rides in the park are right next to each other...
Just got back from Thorpe Park. I managed to get four rides in, two front row, one back and one middle. Got to say that the front row is totally worth it. Also its awesome when you sit on the outside
seat. You can't actually touch the water which is a bit of a shame.
The ride operations were good. The front row queue needs work as the ops kept getting confused about who was front row and who wasn't. Also toward the end of the day, it kept shutting because of
technical difficulties. Engineers seemed to head to the brakes each time.
Oh and the actors were really really good. I don't know if they are, but it would be really cool to keep them in all the time.
Edit: Photos!
Soldiers letting you into the Park
I agree the wheels look a bit daft
Stuck on the Lift Hill
This girl was sobbing
This girl was screaming and shouting at people
This guy was really good. He made the day!
My friend getting interviewed by the "news" channel
My friend getting properly creeped out by the Priest
Edited by PoisonedPirate
99.99% of it looks fantastic, the wheels on the operator's booth look ridiculous. It's a little thing but for something most riders are going to look at, what with it being right infront of the
lift hill that draws your attention when you've boarded, it seems strange that it's one of the few things that hasn't been designed/themed realistically.
But it's themed to an upside down vehicle that's smashed through the roof....
They are way too small and too close together.
I think it's supposed to be a Police "Trailer", similar to the one pictured below. Either way, I think the theming is awesome on this thing!
This thing looks fantastic! I especially love how at certain points during the ride, the Swarm 'scream' is played as the train passes. I really cannot wait to try this out for myself in the next
couple of weeks.
Is that scream supposed to be the plane crashing maybe? It kind of sounds like it to me - but then it also just sounds like a really awesome scream/noise that I could sit around and listen and watch
all day.
Also they did a *darn* good job of theming this thing - the pictures are convincing enough that it's the end of the world! The theme music and the actors really add to it, too.
^The scream is the sound of The Swarm attacking. There's an ad that uses the same sound a couple pages back. The Swarm attacks a kid on a bike and as it flies over it makes that noise. It's quite the
cool sound if you ask me.
The ride's theming is great. And what's nice is that letting the theming go over time will just make it look more and more post-apocalyptic. Seems like a win win if you ask me
A few observations from my visit today:
The fire effect on the fire engine wasn't working yet, and I didn't notice the Manta like water spray working yet either. The water effect you see in one of the pictures is just a spray coming from
the fire engine.
The TV's in the queue line showed news reports which were quite authentic, thought it was a little controversial when they mentioned the area being near heathrow (regarding the plane crash theming)
but cool.
Actors were great and hope they stay, at least for the opening weeks.
The ride is a bit slow. I don't know if it's standard with this design, but I did notice the ride dragging a little. For example the zero-g roll has very little zero-g. Break-in period or
intentional? The drop is great fun, and would agree the front (right) is quite something, very intimidating being slowly turned upside down at 127ft with nothing below or in front of you. Don't get
me wrong, I really loved the ride, but it's graceful nature caught me out a little as well as it feeling quite low to the ground for most of the ride.
I feel a little jealous of Raptor in regards to near miss elements. The aircraft wing is great but it's passed under so fast you barely notice it, and I wasn't too thrilled by the church near miss. I
would have liked something along the lines of Raptor's 3 (or 4?) consecutive misses towards the end.
A fully enclosed (church) station would have been nice to increase the tension but it does look great. A seperate front row queue in the station doesn't exist so as previously mentioned you get a
ticket for the front and have to effectively push through the regular line and wait to the side at the gate load position.
The coaster is near silent but the area was audibly rich, from the news reports to sounds effects to that great Swarm sound triggered at the bottom of the first drop, the area was buzzing!
All I can say is that if you're off to Dollywood for Wild Eagle, you're in for a very graceful, and very comfortable ride. And I'm a little jealous as I think the height and location will make it a
great experience.
Where do you get the front row tickets from?
I think the problem with the Silly Little Tires is that they weren't scaled up along with the overturned truck. If they got some slightly larger tires (can't they just sort of root around in a scrap
yard or something? They don't have to be immaculate) and recessed them into the 'truck' like all tires are everywhere, it would look sufficiently convincing.
That being said, HOLY HELL THIS LOOKS AWESOME. I also really really hope they keep the actors. Or at least some of them.
Where do you get the front row tickets from?
The queue splits toward the end and you pick whether or not you want to ride at the front. Kind of like Stealth but the queues then rejoin, hence the need for tickets.
That being said, HOLY HELL THIS LOOKS AWESOME. I also really really hope they keep the actors. Or at least some of them.
I really want them to keep the actors. The Priest and the General guy were pretty terrifying.
I didn't notice the Manta like water spray working yet either. The water effect you see in one of the pictures is just a spray coming from the fire engine.
It was definately working, but just a bit hard to spot unless you are looking for it!
Waiting around outside the park for my friends with the tickets to arrive;
Looks good as you approach it then;
I liked the station fly-over bit (both on ride and off ride);
And the themeing on the control box was just great!
Its a very quiet ride (themed sound effects apart)
And I didn't realise there was a little manta-like splash thingy... it was a bit hard to see to be honest but it was working.
And the near-misses are good;
Instead of closing the park at 4pm, they did a evactuation drill at 4pm - fair enough really.
Now everybody wave at the nice random girl waving to you in the picture.
Edited by davidmorton
I didn't notice the Manta like water spray working yet either. The water effect you see in one of the pictures is just a spray coming from the fire engine.
It was definately working, but just a bit hard to spot unless you are looking for it!
What about this one?
No one seems to mention it, did they not use it all day?
^ I didn't see it. I didn't even know about til just now!
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Linear Inequalities Worksheet
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Frequency Converter
Last updated:
Frequency Converter
Our frequency converter will instantly convert one frequency value into another unit of frequency.
Frequency informs us how often something is repeated and is a parameter commonly used in electricity, waves, or rotational motion.
Keep reading, and we'll tell you more about what frequency is, what is the unit of frequency, and the frequency symbol.
What is frequency?
At its most basic, frequency describes physical actions that cyclically repeat many times, such as oscillations and rotational motion. Frequency determines the number of cycles per unit time.
Frequency is measured in many applications, which include:
• Electrical circuits — Frequency is the number of repetitions of the sine wave during a positive-to-negative cycle.
• Waves — The frequency gives us the number of waves that pass through a point at a selected time. Check here how to find sound frequency.
• Light waves — The visible spectrum color depends on the frequency of the wave, e.g., 400 THz (4 × 10^14 Hz) is red light, and 800 THz (8 × 10^14 Hz) is violet light.
• A healthy person's heart beats at a frequency of 45 to 220 beats per minute.
What is the unit of frequency?
The symbol for frequency is the letter f, and the SI unit of frequency is hertz (Hz), where 1 hertz equals one cycle per second. Hence, it follows that the more cycles per second, the higher the
In rotational motion, we use traditional units referred to as rotational frequency, such as revolutions per minute (rpm) and degrees per second (deg/s). One hertz is equal to 60 rpm: 1 Hz = 60 rpm.
Angular frequency (denoted as the Greek letter omega, ω) describes the angular displacement of a body per unit of time. Since a body moves along a circular path and its displacement involves an
angle, the unit of angular frequency is usually radians per second (rad/s). Learn more from this angular frequency calculator.
Other SI units of frequency
Omni's frequency converter uses the following units of frequency converted into hertz, including the angular frequency units:
Frequency unit Value in hertz [Hz]
picohertz (pHz) 1 x 10⁻¹²
nanohertz (nHz) 1 x 10⁻⁹
microhertz (μHz) 0.000001
millihertz (mHz) 0.001
centihertz (cHz) 0.01
decihertz (dHz) 0.1
decahertz (daHz) 10
hectohertz (hHz) 100
kilohertz (KHz) 1000
megahertz (MHz) 1,000,000
gigahertz (GHz) 1 x 10⁹
terahertz (THz) 1 x 10¹²
revolutions per minute (RPM) 0.0167
revolutions per hour (RPH) 2.7 x 10⁻⁴
radians per second (rad/s) 0.159155
degrees per second (deg/s) 2.7 x 10⁻³
Examples of how to calculate frequency
There are three basic formulas for how to calculate the frequency, and they are discussed in detail in our frequency calculator. The frequency is related to the period T, i.e., the time required for
one cycle of oscillation or rotation, by the following relation:
$\small{f = \frac{1}{T}}$
The frequency formula in terms of wavelength λ and wave speed ν is as follows (read more about this relationship in our frequency to wavelength calculator):
$\small{f = \frac{u}{\lambda}}$
Finally, frequency and angular frequency ω are related in this way:
$\small{f = \frac{\omega}{2 \pi}}$
How to use this frequency converter
Using our frequency converter is pretty simple and intuitive. Just follow these steps:
1. Type frequency value which you want to convert. The default unit is hertz, but you can freely modify the frequency units. For example, convert MHz to Hz by entering 5 MHz in the field. The result
is 5 x 10⁶ Hz.
2. Now let us convert from Hz to rad/s. Type 60 Hz (utility frequency of AC in your home) and convert from Hz to rad/s. You will get 376.99 radians per second. 🙌
What does Hz mean?
The Hz (hertz) is the SI unit of frequency. It can be expressed as s⁻¹, meaning something happens once in a second. The unit is named after Heinrich Rudolf Hertz, the first person to provide
conclusive proof of the existence of electromagnetic waves.
How can I find the frequency of a wave?
Use frequency formulas to calculate the wave frequency:
1. Divide wave speed by the wavelength. For instance, if a wave is traveling 5 m/s and it has a 4-meter wavelength, the frequency will be 5 m/s / 4 m = 1.25 Hz.
2. Divide 1 by the time it takes for the wave to complete a full cycle (period). If the wave needs 0.2 seconds to complete one full cycle, then the wave frequency is 5 Hz.
What is the frequency of light?
The visible light frequency is approximately 400 THz (red light) to 700 THz (violet light). A THz (terahertz) is a unit of frequency equal to 1 x 10¹² Hz. The frequency of a light wave determines the
color we see. Lower frequencies (longer wavelengths) correspond to "warmer" colors, while higher frequencies (and shorter wavelengths) correspond to "cold" colors.
How many Hz is 100 MHz?
100,000,000 Hz or 1 x 10⁸ Hz. To convert MHz to Hz:
1. Find the conversion factor for MHz: 1,000,000.
2. Multiply 100 MHz by 1,000,000.
3. Get the result in Hz: 100,000,000. | {"url":"https://www.omnicalculator.com/physics/frequency-converter","timestamp":"2024-11-06T15:51:36Z","content_type":"text/html","content_length":"383256","record_id":"<urn:uuid:9bfbcd98-f1ed-4840-af71-9630d77493fc>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00221.warc.gz"} |
Proposed projects for the 2002 DIMACS and DIMACS/DIMATIA REU Programs
Project #: DIMACS2002-01
Parallel Planarity Algorithms
Mentor: Eric Allender, Computer Science Department
There are many clever algorithms to check if a graph is planar. Hopcroft and Tarjan [1] gave a linear-time algorithm almost 30 years ago, but it was much harder to find efficient parallel algorithms.
Ja'Ja' and Simon [2] gave NC algorithms, and a log-time PRAM algorithm with essentially optimal processor and time bounds was given by Ramachandran and Reif [3]. Still, it remained unknown which
complexity class planarity is in, until Allender and Mahajan [4] showed that the Ramachandran and Reif algorithm can be reduced to the reachability problem in undirected graphs, showing that
planarity is in the class SL. The Ramachandran and Reif algorithm is complicated, because much effort is spent minimizing processors. In this project we will study their algorithm and see if a
simpler algorithm can be found that can show that planarity is in SL. Prerequisites: A course in graph theory.
[1] Hopcroft, J., and R. Tarjan, R., ``Efficient Planarity Testing,'' Journal of the ACM, 21, (1974), 549--568.
[2] Ja'Ja', J., and Simon, J., ``Parallel Algorithms in Graph Theory: Planarity Testing,'' SIAM Journal on Computing, 11, (1982), 314--328.
[3] Ramachandran, V., and Reif, J., ``Planarity Testing in Parallel, Journal of Computer and System Sciences, 49, (1994), 517--561.
[4] Allender, E. and Mahajan, M., ``The Complexity of Planarity Testing,'' Proc. 17th Symposium on Theoretical Aspects of Computer Science (STACS) 2000, Lecture Notes in Computer Science 1770, 2000,
Project #: DIMACS2002-02
Tightest Convex Majorants of Multilinear Polynomials in a Small Number of Binary Variables.
Mentor: Endre Boros, RUTCOR
Nonlinear binary optimization is one of the most general forms of many combinatorial optimization problems (see e.g., [1]). Applications are widespread and include VLSI design, statistical mechanics,
portfolio selection, various production and scheduling problems, game theoretical problems, reliability problems, etc. It is a very hard problem to solve, in general. Finding efficiently an
approximate solution is very important, and this research direction has become very active in the past 10 years. The best methods, however, are based on formulations in much higher dimensions than
the original problem, posing great computational difficulties when these methods are applied. A promising approach to find approximation without increase in the dimension of the problem, introduced
in [2], is based on finding tight convex majorants of the objective function. This leads to the question proposed here: given a multilinear polynomial f(x), find a convex function g(x) such that f(x)
less than or equal to g(x) for all x in the unit cube [0,1]^n for which the maximum difference (g(x)-f(x)) over the binary vectors x in {0,1}^n is as small as possible. For instance, if f(u,v)=uv in
n=2 dimensions, then we know that a best convex majorant of f is g(u,v)=((u+v)/2)^2. It is, however, not known what are the tightest convex majorants, even for 3-variable multilinear polynomials,
e.g., f(u,v,w)=2uv-5vw. Best linear majorants were considered in the literature, leading to good bounds [3,4] and to several interesting open questions about polyhedral combinatorics. Possible
formulations for finding a tightest convex majorant include linear programming based on row generation, or semidefinite programming, both of which are interesting and somewhat challenging already for
small (n=3,4) dimensions. Prerequisite: An introductory course in operations research or linear programming would help, but an entire course is not needed, since the necessary background concerning
LP, semidefinite models, etc. can be learned to the extent needed. We will probably only consider problems involving 3-4 variables.
[1] Padberg, M., ``The Boolean Quadric Polytope: Some Characteristics, Facets and Relatives,'' Mathematical Programming, 45, (1989), 139-172.
[2] Badics, T., ``Approximation of Some Nonlinear Binary Optimization Problems,'' Ph.D. Thesis, RUTCOR, Rutgers University, 1996.
[3] Boros, E., Lari, I., and Simeone, B., ``Block Linear Majorants in Quadratic 0-1 Optimization,'' RUTCOR Research Report, RRR 18-2000, Rutgers University, March, 2000.
[4] Hammer, P.L., Hansen, P., and Simeone, B., ``Roof Duality, Complementation and Persistency in Quadratic 0-1 Optimization,'' Mathematical Programming, 28, (1984), 121-155.
Project #: DIMACS2002-03
Patterns in the Logical Analysis of Data
Mentor: Peter Hammer, RUTCOR
This project is devoted to a frequently encountered problem of data analysis, in which a set of ``observations'' is given, with each of the observations being represented as a vector of binary
attribute values. The observations in the data set are of two types, and the type of each observation (e.g., positive or negative) is known. Typical data analysis problems related to such data sets
include classification (i.e., identification of the type of a new observation not included in the data set), determination of characteristic properties of observations of the same type, analysis of
the role of various attributes, etc. The logical analysis of data (LAD) ([1, 2, 3, 4, 5]) is a methodology addressing the above kinds of problems. The mathematical foundation of LAD is in discrete
mathematics, with a special emphasis on the theory of Boolean functions. Patterns are the key building blocks in LAD [4], as well as in many other rule induction algorithms (such as C4.5 rules CN2,
AQ17-HCI, RISE, RIPPER and SLIPPER -- see for example [6, 7]). Since a typical data set has an exceedingly large number of patterns, all these algorithms are limited to the consideration of small
subsets of patterns. In most algorithms, the choice of such a subset of patterns is not explicitly analyzed, in spite of the fact that it has been observed in empirical studies and practical
applications that some patterns are more "suitable" than others for use in data analysis. The goal of this project is to model various such suitability criteria and to analyze the relevance of
notions of suitability for algorithms for LAD. Prerequisites: A little knowledge of graph theory, linear programming, and any elements of discrete mathematics are all useful, but none is absolutely
[1] Boros, E., Hammer, P.L., Ibaraki, T., and Kogan, A., ``Logical Analysis of Numerical Data,'' Math. Programming, 79, (1997), 163-190.
[2] Boros, E., Hammer, P.L., Ibaraki, T., Kogan, A., Mayoraz, E., and Muchnik, I., ``An Implementation of Logical Analysis of Data,'' IEEE Trans. on Knowledge and Data Engineering, 12, (2000),
[3] Boros, E., Ibarakai, T., and Makino, K., ``Logical Analysis of Binary Data with Missing Bits,'' Artificial Intelligence, 107, (1999), 219-263.
[4] Crama, Y., Hammer, P.L., and Ibaraki, T., ``Cause-effect Relationships and Partially Defined Boolean Functions,'' Annals of Oper. Res., 16, (1988), 299-326.
[5] Ekin, O. Hammer, P.L., and Kogan, A., ``Convexity and Logical Analysis of Data,'' Theoretical Computer Science, 244, (2000), 95-116.
[6] Cohen, W.W., and Singer, Y., ``A Simple, Fast, and Effective Rule Learner,'' in Proc. of the Sixteenth National Conference on Artificicial Intelligence, AAAI Press, Menlo Park, CA, 1999, 335-342.
[7] Domingos, P., ``Unifying Instance-based and Rule-based Induction,'' Machine Learning, 24, (1996), 141-168.
Project #: DIMACS2002-04
Text Categorization
Mentor: David Madigan, Department of Statistics
Text categorization is the automated assigning of documents to predefined content-based categories. Applications of text categorization include controlled vocabulary indexing (in both traditional
bibliographic databases and newer Web-based directories), content filtering of email and Web access, subsetting information feeds, and data mining on records with both textual and nontextual fields.
This project will involve developing and evaluating novel Bayesian statistical methods for text categorization and applying them to very large collections of documents. Prerequisites for this work
are a basic proficiency in computer programming as well as an undergraduate course in probabiliity and statistics.
Key References include:
Gelman, A., Carlin, J.B., Stern, H.S., and Rubin, D.B., Bayesian Data Analysis, Chapman and Hall, 1995.
Keim, M., Madigan, D., and Lewis, D., ``Bayesian Information Retrieval,'' Proceedings of the Sixth International Workshop on Artificial Intelligence and Statistics, 1997, 303--310.
Lewis, D.D., Schapire, R.E., Callan, J.P., and Papka, R. (1996). ``Training Algorithms for Linear Text Classifiers,'' SIGIR '96: Proceedings of the 19th Annual International ACM SIGIR Conference
on Research and Development in Information Retrieval, Konstanz. Hartung-Gorre Verlag, 1996, 298--306.
Project #: DIMACS2002-05
Protein-induced DNA Looping
Mentor: Wilma Olson, Department of Chemistry
Many genetic processes are mediated by proteins that bind at separate, often widely spaced, sites on the double helix, tethering the intervening DNA into a loop [1, 2]. Examples of these
processes include gene expression and its control, DNA replication, genetic rearrangements, multi-site cutting by restriction enzymes, and DNA packaging. A DNA loop stabilized by such long-range
protein-protein contacts constitutes a discrete topological unit. As long as the ends of the DNA stay in place and the duplex remains unbroken, the linking number, Lk, or number of times the two
strands of the double helix wrap around one another, is conserved. This constraint in Lk underlies the well known supercoiling of DNA: the stress induced by positioning the ends of the polymer in
locations other than the natural (relaxed) state perturbs the overall coiling of the chain axis and/or the twisting of successive base-pairs in the intervening parts of the chain [3]. As a first
step in understanding the effects of specific proteins and drugs on DNA looping, we propose to study the imposed bending and twisting of neighboring base pairs [4] in known complexes of proteins
and drugs with double helical DNA stored in the Nucleic Acid Database [5]. By subjecting a DNA segment of the same chain length as that found in a given complex to the observed orientation,
displacement, and superhelical stress and setting the elastic moduli to sufficiently high values, we can use existing programs, e.g., [6], to simulate the presence of a rigidly bound molecule at
arbitrary positions on circular DNA molecules or to model specific systems in which DNA looping plays an important role, e.g., the lac repressor-operator assembly in EscherichiaA0coli [7]. One
student could devote a summer project to the collection of geometric information. Other students could apply this information to simulations of DNA loops and folds in subsequent years.
Prerequisites: Students should have an interest (but not necessarily formal training) in molecular biology, familiarity with linear algebra in order to understand the parameters used to describe
DNA structure, and knowledge of basic chemistry and physics to understand the nature of the DNA molecules and the elastic treatment of the double helix.
[1] Halford, S. E., Gowers, D. M., and Sessions, R. B., ``Two are Better than One,'' Nature Struct. Biol., 7, (2000), 705-707.
[2] Schleif, R., ``DNA Looping,'' Annu. Rev. Biochem., 61, (1992), 199-223.
[3] Bauer, W. R. ``Structure and Reactions of Closed Duplex DNA,'' Annu. Rev. Biophys. Bioeng., 7, (1978), 287-313.
[4] Olson, W. K., Gorin, A. A., Lu, X.-J., Hock, L. M., and Zhurkin, V. B., ``DNA Sequence-dependent Deformability Deduced from Protein-DNA Crystal Complexes,'' Proc. Natl. Acad. Sci. USA, 95,
(1998), 11163-11168.
[5] Berman, H. M., Olson, W. K., Beveridge, D. L., Westbrook, J., Gelbin, A., Demeny, T., Hsieh, S.-H., Srinivasan, A. R., and Schneider, B., ``The Nucleic Acid Database: A Comprehensive
Relational Database of Three-dimensional Structures of Nucleic Acids,'' Biophys. J., 63, (1992), 751-759.
[6] Westcott, T. P., Tobias, I., and Olson, W. K., ``Modeling Self-contact Forces in the Elastic Theory of DNA Supercoiling,'' J. Chem. Phys., 107, (1997), 3967-3980.
[7] Muller-Hill, B., The Lac Operon, Walter de Gruyter, Berlin, 1996.
Project #: DIMACS2002-06
Algorithms for machine learning
Mentors: Kobbi Nissim and Ofer Melnik, DIMACS
There are many problems that lie on the frontier between machine learning and algorithmics. As an example one may look at issues of dimensionality- as the number of input variables or model
parameters increase, problems become computationally harder and more opaque. Thus the intrinsic challenges revolve around finding ways to address issues of high-dimensionality. Approaches to date
have involved approximation, dimensionality reduction, re-representation, complexity analysis, generalization/model assumptions or exploiting regularity.
This research project will involve some programming, design and analysis. A background in computer science, programming, math/ stats with an interest in machine learning is desired.
Project #: DIMACS2002-07
Mentor: Leonid Khachiyan, Department of Computer Science
Project #: DIMACS2002-08
Mentor: Vladimir Gurvich, RUTCOR
Dr. Gurvich has three problems to work on, students selected to work with Dr. Gurvich may work on one or more of these problems.
I. It is known that, given two m x n matrices A and B, there exist m + n potentials p[1],...,p[m]; q[1],...,q[n]. Such that
max a[ij] + p[i] - q[j] = min b[ij] - p[i] + q[j] = v for all i = 1,...,m; j = 1,...,n.
This theorem has important applications in game theory. But how to find such potentials? No polynomial algorithm is known yet, though the problem is in the intersection of NP and co-NP.
II. Difference graphs. Given n finite sets S[1],...,S[n] , let us define a graph G with n vertices v[1],...,v[n] in which 2 vertices v[i] and v[j] are connected by an edge IFF both
differences S[i] \ S[j] and S[j] \ S[i] have at least k elements each. EVERY graph can be realized in such a way if k is arbitrarily large. But is it still true if k is bounded by a constant?
Even for k=2 no counter-example is known. If k=1 then only co-occurrence graphs can be realized.
III. In 1912 Zermelo proved that games with perfect information (e.g. Chess) can be solved in pure stratagies, that is without randomizing. A strategy is called STATIONARY if in every
position the move may only depend on this position but not on the preceding moves. Does Zermelo's Theorem generalizes this case? Is it true that games with perfect information can be solved
in PURE STATIONARY strategies? For acyclic games the answer is "YES". But for games with cycles the problem is open
Project #: DIMACS2002-09
Modeling Bid Preparation Costs in Auctions
Mentor: Michael Rothkopf, RUTCOR
Most models of auctions ignore bid preparation costs. Most if not all of those that don't ignore them treat them as a fixed amount. I am interested in a more nuanced model of bid preparation
costs in which bidders may invest more or less effort in bid preparation and in which high effort implies better value estimation. While many approaches are possible, I envision the following
class of models: Before an auction of a given type (standard sealed bid, sealed second-price, progressive), each of n a priori symmetric bidders has to decide how much to invest in gathering
information about her value for what is to be sold. The value is, in the general case, composed of a private value component and a value component common to all bidders, and the information
gathering investment can be allocated between these components.
After each bidder makes her investment and gets the information, the bidders then participate in the auction. Several issues are of interest. How much is spent on evaluation? If there is both
private and common information, how is the effort allocated between seeking information on the common value and how much on the private value? How inefficient is the allocation of the item
compared to the case when information is free? How do these answers depend upon the number of bidders, the cost of information, and the type of auction?
I envision this analysis being done, at least initially, on a simply, stylized model in which random variables have independent uniform distributions. The student working on this should have had
a course in mathematical statistics or probability theory. Some background in microeconomics or game theory would be a plus.
Project #: DIMACS2002-10
Distributed cryptography and computer security
Mentor: Rebecca Wright, DIMACS
The exact problem will be chosen from general topic areas that include design and analysis of cryptographic protocols, distributed cryptography, privacy, secure multiparty computation, and
fault-tolerant distributed computing. Possible goals include designing protocols, systems, and services that perform their specified computational or communication functions even if some of the
participants or underlying components behave maliciously, and that balance individual needs such as privacy with collective needs such as network survivability and public safety.
Depending on the interests and expertise of the student, this project can be more focused on foundational mathematical results, on implementation and experimentation, or on written analysis of
software usage and its security implications.
NOTE: Dr. Wright is currently as AT&T Labs. She will be a Visiting Research Associate at DIMACS during the term of the REU program. Background information on her interests can be found at her
website: www.kiskeya.net/rwright
Project #: DIMACS2002-11
Computational experimentation with augmented Lagrangian methods for nonlinear optimization
Mentor: Jonathan Eckstein, RUTCOR
Contacting the Center
Document last modified on February 21, 2002. | {"url":"http://dimacs.rutgers.edu/archive/REU/2002/proposed.html","timestamp":"2024-11-12T05:46:46Z","content_type":"text/html","content_length":"21252","record_id":"<urn:uuid:8d8e458a-c4ea-4b80-9594-d9a76ccf68a1>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00085.warc.gz"} |
Angle of Elevation Word Problems Worksheet
(1) Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10√3 m. Solution
(2) A road is flanked on either side by continuous rows of houses of height 4 √3 m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of
elevation from the pedestrian to the top of the house is 30° . Find the width of the road. Solution
(3) To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the
wall, what is the height of the window? (√3 = 1.732) Solution
(4) A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top
of the pedestal is 40° . Find the height of the pedestal. (tan 40° = 0.8391, √3 = 1.732)
(5) A flag pole ‘h’ meters is on the top of the hemispherical dome of radius ‘r’ meters. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from
the dome and seeing the bottom of the pole at an angle 30° . Find (i) the height of the pole (ii) radius of the dome. (√3 = 1.732)
(6) A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m
away from the base of the pole, what is the height of the pole? Solution
(7) The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric
pole? Solution
(8) A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are 4° and 8° . What is the
height of the peak if the distance between consecutive milestones is 1 mile. (tan 4° = 0.0699, tan 8° = 0.1405) Solution
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
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CLASS 2 - Learn CBSE
NCERT Class 2 Joyful-Mathematics Book Solutions Lesson 3 Fun with Numbers Questions and Answers, We have compiled the NCERT Class 2 Joyful-Mathematics Book Solutions for all topics in a comprehensive
way to support students who are preparing effectively for the exam. You will discover both numerical and descriptive answers for all Chapter 3 Joyful-Mathematics, concepts … Read more
NCERT Class 2 Joyful-Mathematics Chapter 2 Shapes Around us
NCERT Class 2 Joyful-Mathematics Book Solutions Lesson 2 Shapes Around us Questions and Answers, We have compiled the NCERT Class 2 Joyful-Mathematics Book Solutions for all topics in a comprehensive
way to support students who are preparing effectively for the exam. You will discover both numerical and descriptive answers for all Chapter 2 Joyful-Mathematics, concepts … Read more
NCERT Class 2 Joyful-Mathematics Chapter 1 A Day at the Beach
NCERT Class 2 Joyful-Mathematics Book Solutions Lesson 1 A Day at the Beach Questions and Answers, We have compiled the NCERT Class 2 Joyful-Mathematics Book Solutions for all topics in a
comprehensive way to support students who are preparing effectively for the exam. You will discover both numerical and descriptive answers for all Chapter 1 … Read more
NCERT Class 2 English Mridang Chapter 13 We are all Indians
Learn NCERT Class 2 English Book Solutions Lesson 13 We are all Indians Questions and Answers, from this page for free of cost. We have compiled the NCERT Class 2 English Book Solutions for all
topics in a comprehensive way to support students who are preparing effectively for the exam. You will discover both numerical … Read more
NCERT Class 2 English Mridang Chapter 12 Little Drops of Water
NCERT Class 2 English Book Solutions Lesson 12 Little Drops of Water Questions and Answers, from this page for free of cost. We have compiled the NCERT Class 2 English Book Solutions for all topics
in a comprehensive way to support students who are preparing effectively for the exam.
NCERT Class 2 English Mridang Chapter 11 The Smart Monkey
Learn NCERT Class 2 English Book Solutions Lesson 11 The Smart Monkey Questions and Answers, from this page for free of cost. We have compiled the NCERT Class 2 English Book Solutions for all topics
in a comprehensive way to support students who are preparing effectively for the exam. You will discover both numerical and … Read more
NCERT Class 2 English Mridang Chapter 10 The Crow
Learn NCERT Class 2 English Book Solutions Lesson 10 The Crow Questions and Answers, from this page for free of cost. We have compiled the NCERT Class 2 English Book Solutions for all topics in a
comprehensive way to support students who are preparing effectively for the exam.
NCERT Class 2 English Mridang Chapter 9 My Name
Learn NCERT Class 2 English Book Solutions Lesson 9 My name Questions and Answers, from this page for free of cost. We have compiled the NCERT Class 2 English Book Solutions for all topics in a
comprehensive way to support students who are preparing effectively for the exam.
NCERT Class 2 English Mridang Chapter 8 A Show of Clouds
NCERT Class 2 English Book Solutions Lesson 8 A Show of Clouds Questions and Answers, from this page for free of cost. We have compiled the NCERT Class 2 English Book Solutions for all topics in a
comprehensive way to support students who are preparing effectively for the exam.
NCERT Class 2 English Mridang Chapter 7 This is My Town
NCERT Class 2 English Book Solutions Lesson 7 This is My Town Questions and Answers from this page for free of cost. We have compiled the NCERT Class 2 English Book Solutions for all topics in a
comprehensive way to support students who are preparing effectively for the exam. | {"url":"https://learncbse.org.in/category/class-2/","timestamp":"2024-11-13T08:43:25Z","content_type":"text/html","content_length":"169456","record_id":"<urn:uuid:2e2a3066-b14c-4cad-8f76-72d0ae8e8f67>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00240.warc.gz"} |
Determining the Axis That Is Parallel to the Straight Line with the Given Equation
Question Video: Determining the Axis That Is Parallel to the Straight Line with the Given Equation Mathematics • Third Year of Preparatory School
Which axis is the straight line π ₯ = 5 parallel to?
Video Transcript
Which axis is the straight line π ₯ equals five parallel to?
Letβ s think about the line with the equation π ₯ equals five. And now the way of expressing this is the π ₯-coordinate is five, which means that every point that lies on this line has an π
₯-coordinate which is equal to five. Letβ s think about what that would look like. Here we have a set of axes. The point with coordinates five, zero would be on this line. It has an π ₯-coordinate
of five and a π ¦-coordinate of zero. The point with coordinates five, one would also be on this line. This point has an π ₯-coordinate of five, and it has a π ¦-coordinate of one. In the same
way, the point with coordinates five, two also lies on the line.
Below the π ₯-axis, the point with coordinates five, negative one will also lie on the line with equation π ₯ equals five. And we can continue in this way in both directions. Connecting these
points together, we see that the line with equation π ₯ equals five is a vertical line passing through every point in the coordinate plane that has an π ₯-coordinate of five. As this line is
vertical, it is parallel to the π ¦-axis.
In general, we can recall that lines with equations of the form π ₯ equals π for some constant value of π are vertical lines which are parallel to the π ¦-axis. On the other hand, lines with
equations of the form π ¦ equals π for some constant π are horizontal lines which are parallel to the π ₯-axis. Our answer to the question β Which axis is the straight line π ₯ equals five
parallel to?β is the π ¦-axis. | {"url":"https://www.nagwa.com/en/videos/763152457909/","timestamp":"2024-11-04T09:22:33Z","content_type":"text/html","content_length":"248936","record_id":"<urn:uuid:a7b79380-d07e-4135-be28-7507b57d058c>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00415.warc.gz"} |
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Exercise 15.4
Ex 15.4 Class 7 Maths Question 1.
A bulb is kept burning just right above the following solids. Name the shape of the shadows obtained in each case. Attempt to give a rough sketch of the shadow. (You may try to experiment first and
then answer these questions).
When the light falls just above the solids.
(i) A ball; the shadow looks like a circle.
(ii) A cylindrical pipe; the shadow looks like nearly rectangular.
(iii) A book; the shadow looks like nearly rectangular.
Ex 15.4 Class 7 Maths Question 2.
Here are the shadows of some 3-D objects, when seen under the lamp of an overhead projector. Identify the solid(s) that match each shadow. (There may be multiple answers for these!)
(i) The given shadow corresponds to a sphere.
(ii) The given shadow corresponds to a cube.
(iii) The given shadow corresponds to a pyramid.
(iv) The given shadow corresponds to a cuboid or a cylinder.
Ex 15.4 Class 7 Maths Question 3.
Examine if the following are true statements:
(i) The cube can cast a shadow in the shape of a rectangle.
(ii) The cube can cast a shadow in the shape of a hexagon.
(i) The given statement is true.
(ii) The given statement is false.
NCERT SolutionsMathsScienceSocialEnglishSanskritHindiRD Sharma | {"url":"https://www.cbselabs.com/ncert-solutions-for-class-7th-maths-chapter-15-visualising-solid-shapes-exercise-15-4/","timestamp":"2024-11-10T14:17:54Z","content_type":"text/html","content_length":"192258","record_id":"<urn:uuid:e6ca5c9a-80c3-424b-90e1-a488f47635ab>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00607.warc.gz"} |
3-D Plotting
Tim R. Colvin
How many times have you seen beautiful three-dimensional graphics in the ads for video monitors and printers? Now, with these easy-to-use programs, you can create three-dimensional images of your
own. Versions are included for the Commodore 64, Atari, Apple, IBM PC and PCjr.
These two programs, "Rectan" and "Spheri," will plot three-dimensional figures using information which you provide.
You don't really need to delve into the mathematics which produce the images. You can just fiddle with the examples given to produce many effective displays. Let's look at some graphic examples.
First type in each program and SAVE it to tape or disk.
Then LOAD Rectan. To have Rectan draw a hyperbolic paraboloid, or "saddle function" (it resembles a riding saddle), replace line 790 with:
790 Z-X*X/4-Y*Y/9
and give the following inputs:
For another interesting design, use:
790 Z=-l/(X*X+Y*Y+.5)
and give the following inputs:
The program will print SCREEN SCALING IN PROGRESS. The program is scaling the image to fit on the screen, which can require a lot of time. The rule is: The more complicated the description of the
surface, the longer this step takes.
The Plotting Begins
When the previous step is completed, the screen will clear and turn cyan. The high-resolution plotting now begins. When the plot is finished, the color of the top left corner of the screen will
change color. The program is locked in a loop so you can look at your creation. When you have finished looking at the display hold down RUN/ STOP and hit RESTORE.
A Spheri Demonstration
To see a torus (doughnut shape), type NEW to clear memory. Then LOAD Spheri, replace lines 820-840 with:
820 XT=*(4+C1)*C2
830 YT=(4+C1)*S2
840 ZT=S1
and give the following inputs:
For a sphere, use:
820 XT=C1*C2
830 YT=C1*S2
840 ZT=S1
and give the following inputs:
An Illusion Of Depth
These programs use rectangular and spherical coordinate systems to create an illusion of depth in the screen image. You're probably familiar with the X-Y coordinate system used to specify the
location of a point on a flat surface. For example, in Figure 1 the point is located five units over on the X axis and six units up on the Y axis. The point is said to be at location 5,6.
This simple system works well for specifying the location of a point in a two-dimensional design on a flat surface, but for 3-D plotting you need a third coordinate.
Several coordinate systems are commonly used to plot three-dimensional surfaces. The particular coordinate system you should use depends on the shape you want to draw. Any system can be used, but if
you choose the right system, you can simplify your calculations considerably.
A Simple Solution
The easiest system to understand is just an extension of the rectangular (X-Y) coordinates you are already familiar with. All you need to add is a third coordinate (Z) for the third dimension. For
example, the point in Figure 2, below, is located five units out on the X axis, six units over on the Y axis, and four units up on the Z axis. The point is said to be at location 5,6,4.
A System For The Stars
On the other hand, if the design you wish to draw is roughly the shape of a sphere, you should use spherical coordinates. In that system, a point is described by two angles and a distance from the
origin. For example, astronomers use spherical coordinates to describe the position of a star relative to the earth. The azimuthal angle of the star, designated by the Greek letter "theta", is the
direction you must face to view the star. If north is taken to be zero degrees, then a star that lies due east has an azimuthal angle of 90 degrees. The elevation angle, designated by the Greek lette
"phi", specifies how much you must tilt your head back to look directly at the star. If the horizon is taken to be zero degrees, a star that is directly overhead has an elevation angle of 90 degrees.
Finally, the radial distance, designated by the letter r, is the distance between the earth and the star.
Using spherical coordinates, the point shown in Figure 2 has an azimuthal angle of 50.2 degrees, an elevation angle of 33.7 degrees, and a radial distance of 8.77 units, as shown in Figure 3.
Despite the fine graphics they produce, these programs have a couple of limitations. Screen pixels are taller than they are wide, which makes spheres look slightly less round than they should Also,
we see the surface as if it were transparent and contour lines were drawn on it. A more advanced program (such as those available commercially) would remove lines that we couldn't see if the surface
were not transparent.
The Mathematics Of 3-D Plotting
"Rectan" plots surfaces using rectangular coordinates (x,y,z). The values for x and y are specified; the value of z is then given by z = f(x,y) for some function f.
To use Rectan, specify the function f(x,y) in line 790. For example, z = x*x/4-y*y/9 defines a hyperbolic paraboloid.
"Spheri" plots surfaces using spherical coordinates. This method describes a point on the surface using three parameters: radial distance from the origin, r; azimuthal angle, theta; and elevation
angle, phi.
To use Spheri, specify x,y, and z (called XT,YT, and ZT in lines 820-840) as functions of r, theta, and phi in lines 820-840.
Parameters And Slices
Both programs are structured the same. You specify parameter ranges. In Rectan these are for x and y; in Spheri, for [theta] and [phi].
Next enter the number of slices for the parameters. Each slice corresponds to a contour line on the surface. A contour line is where one of the parameters is held constant.
Finally, you specify an observation angle. This is the angle which allows you to see a three-dimensional surface on a two-dimensional video screen. The most commonly used angle is 45 degrees.
If you'd like any technical information, or if you have a particular surface in mind but don't know how to write an equation for it, please write to:
[S:Tim R. Colvin 1414 San Remo Dr. Pacific Palisades, CA 90272:S] | {"url":"https://www.atarimagazines.com/compute/issue48/3d_plotting.php","timestamp":"2024-11-14T07:12:46Z","content_type":"text/html","content_length":"23415","record_id":"<urn:uuid:8c9448cc-0059-423e-a5d8-8dbce1123bcd>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00874.warc.gz"} |
-Laplace equations through
geometric Sobolev type inequalities
Published Paper
Inserted: 22 apr 2016
Last Updated: 31 oct 2018
Journal: Journal of the European Mathematical Society
Year: 2015
Links: JEMS
In this paper we prove a Sobolev and a Morrey type inequality involving the mean curvature and the tangential gradient with respect to the level sets of the function that appears in the inequalities.
Then, as an application, we establish \textit{a priori} estimates for semi-stable solutions of $-\Delta_p u= g(u)$ in a smooth bounded domain $\Omega\subset \mathbb{R}^n$. In particular, we obtain
new $L^r$ and $W^{1,r}$ bounds for the extremal solution $u^\star$ when the domain is strictly convex. More precisely, we prove that $u^\star\in L^\infty(\Omega)$ if $n\leq p+2$ and $u^\star\in L^{\
frac{np}{n-p-2}}(\Omega)\cap W^{1,p}_0(\Omega)$ if $n>p+2$. | {"url":"https://cvgmt.sns.it/paper/3008/","timestamp":"2024-11-13T05:27:16Z","content_type":"text/html","content_length":"8794","record_id":"<urn:uuid:f6a9b4b9-f2fb-4308-8083-8cfff986a8bd>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00876.warc.gz"} |
Desmos: A Definitive Guide on Graphing and Computing | Math Vault (2024)
Think you’re fond of of graphing and computing stuffs? Great! Because you might remember this thing called theTexas Instrument TI-83from the old days. Sure, while programmable calculators in general
are still pretty muchpopular these days, the graphing calculators from the21^st-century are also coming in waves as we speak—potentiallydisrupting the market of scientific computing and educational
In particular, there’s a certain education startup out there, relentlessly seeking to hijack our Internet browsers and mobile devicesinto a — should we say —graphing extravaganza. And it comes with a
funky name as well:Desmos.
Yep. You’ve heard it right. As Greek-mythology-inspired as it sounds, Desmos actually has nothingto do with thegiant monster responsible for turningMount Olympus into rubble through the wrath of
infernal flames. Instead, it is probably better known as an innocent-looking mathtool for the scientifically-minded, making applied math ever more palatableand entertaining!
Oh. Don’t believe in the mighty power of Desmos? Good job! Because it’ll then beour duty tobeg to differ, and attempt to convince you otherwise. Indeed, if at the end of the module you still find the
scope of Desmos’ functionalitiesunappealing, then — and only then — shall we concede defeat and return to our ivory tower for more advanced Buddhist meditation training! On the other hand, if you’re
just way toolazy to read the 12-page Desmos user manual, and are looking for more concrete examples to kick-start the creative process, then this oneis for you too!
Graphing with Desmos
What isthe single word peoplethink of when theyhear online graphing calculator? Graphing of course! Indeed, this is something that Desmos does incredibly well — despite having a user interface that
appears to bedeceptively simple. In what follows, we will seehow to wecan useDesmos to graph equations, functions and inequalities of different forms, beforeintroducing some bonus features such as
graph segmentation,simultaneous graphing and animations!
When it comes to graphing, equation is one of those first words that comes to mind. On the other hand, it can also be construed as an umbrella term encompassinga plethora of mathematical objects such
as explicit functions, implicit functions, parametric equations and polar curves. In what follows, we will illustrate how each of them can be graphed in Desmos — one stepat a time.
Before doing any graphing though, we need to first learn how totype out a few mathsymbols that are frequently sought for. To that end, we have provideda partial list of common symbolssupportedin
Desmos — along with their associated commands:
• The Multiplication symbolcan be obtained by typing * (Shift+8on US keyboards).
• The Division line can be obtained by typing / (i.e., backslash).
• $\pi$ can be obtained by typing pi, and $\theta$ by theta. The same goes other Greek Letters such as $\alpha$,$\beta$, $\tau$ and $\phi$.
• $\sqrt{}$ by sqrt, and $\sqrt[n]{}$ by nthroot (after which the value of $n$ can be adjusted manually).
• Use the key^(Shift+6 on US keyboards) to createa superscript (e.g., $2$ in $x^2$), and the key _(Shift+Hyphen on US keyboards)fora subscript(e.g., $1$ in $a_1$).
• The ceiling function by ceil(), the floor function by floor(), and the sign function by sign().
• The absolute value symbol by | (i.e., Shift+\ on US keyboards).
For some commands (e.g., division, superscript, subscript), typing them out can move the cursor to the upper or lower area. In which case, youcan use the arrow keysto help younavigate around
theexpression and type out what you want. Other times, you might need to use parentheses to group some of yourexpressions together — before they can be properly interpreted by Desmos.
Defining Functions
Traditionally, the most popular functions are the ones expressed in terms of $x$, which can be typed into a command line as follows:
\begin{align*} y = \text{some algebraic expressions in terms of }x \end{align*}
By reversing the roles of $x$ and $y$, functions in terms of $y$ can also be typed out into a command line as follows:
\begin{align*} x = \text{some algebraic expressions in terms of }y \end{align*}
So, what kinds of functions are supported in Desmos? Basically, all of the elementary functions you can think of! These include the 6 basic trigonometricfunctions, exponential/logarithmic functions
(through the commandsexp, lnandlog), polynomials, and rational functions. In addition, the 6 basicinverse trigonometric functions— along with the 6 basic hyperbolic functions—are readilysupported by
simply typing out the function’s name as well.
Now, if a function in terms of $x$ is meant to be referred on a repeated basis, then a name can be assignedto it by replacing the $y$ on the left-hand side with, say, $f(x)$. In a similar manner, a
function in terms of $y$ can also be assigned a name as well, from whichwe can define and name acompoundfunctionin a command line — without having to resort to rewriting the expressions again and
• $y = 2f(3x-3)+e$
• $g(y) = \dfrac{\pi}{f(y)}$
• $f_1(x)= 55f(x)+43g(x) – 100 h(x)$
• $y = g(x)^2 \cdot \left( g(x)^{f(x)} \right)$
• $h(x) = 2 g(f(x)) + 3 g(f(x))$
And in case you’re wondering, the polar functions are equally supported in Desmos as well. However, do note that in this case, the functions will have to be typed out using$r$ and $\theta$ as the
designated variablesinstead (e,g., $r=\frac{1}{2} \theta^2$).
Calculus-Related Functions
As it turns out, Desmos is remarkably receptive to calculus-based expressions as well. For example, once a function in terms of $x$ is given, by attaching $\dfrac{d}{dx}$ in front of it, the graph of
its derivative function can be readily produced — without having to resort to the explicit formula of the derivative. Similarly, the derivative of a function in terms of $y$ can also be graphed by
appending $\dfrac{d}{dy}$ in front of it. Prettyneat, right?
Alternatively, if a function has already been assigned a name, we can alsouse the prime notation to denote itsderivative function (as in $f'(x)$). Apart from being easy to implement, the prime
notationalso has the advantage ofbeing able to refer tohigher derivativesby simply adding a few extra $’$ (e.g., $g”$), which —comparing tothe repeated use of $\dfrac{d}{dx}$ — is definitely a plus
to have.
Of course, if Desmos is good with the derivative functions, thenit shouldn’t a surprise that italso supports theintegral functionsas well. For the record, theintegral operator symbol $\int$ can be
obtained by typing int into the command line, afterwhich you will have use the arrow keys to navigate around thelower/upper limitsand the integrand.
A note of caution though: when setting up an integral function —especially the ones involving multiple integral operators—you want totake that extra care to make sure that the variable of integration
is different fromthe variable(s) in the limits. In doing so, you are in effect removing yourself from the need ofpunching Desmos for real, in casethe expression doesn’tturn out to be exactly
Desmos-interpretable. 🙂
Piecewise Functions
To define a piecewise function in Desmos, wecanuse the following syntax on a command line:
\begin{align*} y \ (\text{or }x) = \{ \text{condition 1}: \text{definition 1}, \text{condition 2}: \text{definition 2}, \ldots \} \end{align*}
As with other functions, a piecewise function can also be given a name by simply replacing the leftmost variable with the name of the function. This way,we can recycle the function repeatedlywithout
having to redefine it ever again.
For example, let’s say that we are interested in defining the function $f$ as the sign function. That is:
\begin{align*} f(x) = \begin{cases} -1 &x<0 \\ 0 & x=0 \\ 1 & x>0 \end{cases} \end{align*}
then this would be the expression we want to type into thecommand line:
\begin{align*} f(x)=\left\{x<0:-1, x=0:0, x>0:1\right\} \end{align*}
which should produce the following figure:
In some occasions, it might be necessary to use the $\le$ and$\ge$ symbols to define a function accurately. In which case, just know thatthe $\le$ symbol can be obtained by typing out < and =in that
order, and the$\ge$ symbol by typing out > and =, again in that order.
Other Equations
Since Desmos has its interface in Cartesian coordinates by default, it’s only natural that one woulduse it to plotequations expressed in terms of $x$ and $y$. In fact,implicit functions such asthat
of a circle, anellipseor a hyperbolaare all very good candidates for this.
However, what is less clear is that Desmos can also interpret parametric equationsas well, provided that we type inthe equations for $x$ and $y$as if they werethe coordinates of thepoints instead (as
in$(2\cos t, 3 \sin t)$ ), and that the variable $t$ — the designated variable for parametric equations in Desmos — is used throughout the expression. For some reason, Desmos simply won’ttake
anyothervariable such as $x$ or $\theta$ for thispurpose.
On a more optimistic side, if you manage to type in aparametric equation the rightway, then you should be able see an inequality about the domain popping up right underneath the command line. Don’t
take this for granted by the way, because this is whereyou get to configure the range of the variable $t$, which in essencedeterminesthe portion of the parametric equations that is actually displayed
on the graph.
All right. So far so good?Here’s a figure illustratinghow implicit functions and parametric equations work out in Desmos:
Think that wasquite a list of equations already? Know this: almost all equations can be turned into inequalities by replacing the $=$ sign with $<$, $>$, $\le$ or $\ge$, thereby easilydoubling the
amount of graphs onecan plot in Desmos. Just a caveat though: don’tdoitin a command line wherenaming the function is precisely the primary goal.
To get it started, here’s a list of inequalities you can tryout andget the juice flowing:
• Plane: $y<2x$
• Circle: $x^2+y^2 \le 3^2$
• Inside of a Polar Curve: $r<3\cos ^2\theta$
• Area Between Two Functions(e.g., $f(x) \le y \le g(x)$)
• Spacein Between Implicit Functions: $2xy+y^2<4$
And here’s what happened when we throw inthese inequalities into Desmos.
which is pretty cool. And that’s just astart!
Graph Segmentation
When a valid equation/inequality is entered into a command line, Desmos will — by default — plot its graph by assuming the full domain under which the equation/inequality is satisfied. However,
thisdoesn’t always yield the desired effects, and there are occasions where it’s preferable notto do so. In which case, one can always choose tosegment the graphby imposing restrictions on
theequation/inequality in question — as long as the following syntax rule is being adhered to:
\begin{align*} \text{equation/inequality} \, \{ \text{condition 1}\}\{ \text{condition 2} \} \ldots \end{align*}
For example, to graph the function $x^2$ withthe domain restricted to only the positive numbers, the following line would do:
\begin{align*} y=x^2 \, \{ x>0 \} \end{align*}
Of course, the restriction could have been an inequality on$y$ as well, as in the graph of the function $x^2$ where $y$ is restricted to the interval $[5,15]$:
\begin{align*} y=x^2 \, \{ 5 \le y \le 15 \} \end{align*}
or perhaps an inequality concerning both $x$ and $y$:
\begin{align*} y=x^2 \, \{ x+y<5 \}\{x>0\}\end{align*}
In fact, the restriction(s) could have been any number of equations/inequalities, involving anycombination of thedesignated variables(e.g., $x$, $y$, $r$, $\theta$), with thecaveatbeing that the
variables in the restrictions need to becompatible with the equation/inequality they are restricting in the first place (e.g., a polar equationcan only be restricted usingthepolar variables $r$ and $
Sounds a bit obscure? Here’s a picture for more info:
Simultaneous Graphing
Tired of plotting very similar graphs one by one? Well, despairnot, for there is a way out when you’re with Desmos. Indeed, if an equation/inequality is expressed in terms of some parameter(s), which
is itself definedasa list of multiple numbers, thenmultiple graphs can be created simultaneously for each of these numbers once and for all.
Hmm, what are we talking about here? Here’s an accompanying picture to help us out:
As you can see here, by defining a function with the dummy parameter $a$, which is itself defined as a list with the numbers $1$, $2$, $3$ and $4$, we were ableto graph a total of four functions
simultaneously with a mere two lines. Talk about laziness and efficiency!
Andif you’re feeling a bit adventurous, you can certainly try entering an equation/inequality with multiple parameters, each of which isdefined as a list on its own with equal sizes. In fact, this
“graphing-in-bulk” method has proved to be a formidable techniqueindrastically slowing down / paralyzing ourcomputers andmobile devices!
On a more serious note though, if you’re looking to exploit the functionalities of Desmos toits fullest, this one is a must. Not only willit automate a graphing process that would otherwise be
annoyingly tedious tocarry out, but it also prevent us fromreinventing the wheel when a template for the graphs isalready readily available. Oh, here comes another one!
Animation with UndefinedParameters
When you enter a line of mathematical expression containing anundefined parameter, Desmos will give you to option to include a slider for that parameter, which has the capability of allowingyou
toadjust ofits value manually — or even better yet — create an animationout of it byallowing the value of the parameter to increase/decrease automatically.
In particular, theslider will allow you to control the rangeand the increment size of the parameter in question, along withthe speedand thedirection with which the parameter changes. And the reason
why it matters, is becauseby having sliders controlling the behaviors of our parameters of interest, it becomes possible for us to create various animated objects such asmovable points,
rotatingequationsandanimated regions/boundaries.
For example, to modela particle moving along a sine-wavetrajectory, all we need to do is toembed an undefined parameterintothe coordinates of the sine wave(as in$(a, \sin 2\pi a)+3$ ), before
activatingthe slider for the parameter usingthe “play” button. And if the point is meant to be moving along a circular path instead, putting an expression such as $(3\cos a – 5, 3\sin a + 6)$ into
the command line would do.Furthermore,ifthe trajectory of a movable point actually defines a function, then Desmos will automatically make the pointdraggable across the $x$-axis, making it easier to
manipulate than ever!
But then of course, movable points are just one of the many features animation has to offer.For example,if wewish toinvestigatethe contribution of the different parameters of a depressed cubic
equation, then wecan do so by entering $y=x^3+ax^2+bx+c$ into the command line, beforeactivating the sliders for $a$, $b$ and $c$ in any combination we please.Actually, why nottoy around withthe
sliders a bit first and observe how the shape of the graph changes during this process!
Andif instead of having one function change its shape, you are interested in seeing several functions under a common parameter moving all at once, then you can — for example— try typing out$y=ax$, $y
=ax^2$, $y=ax^3$ into three command lines, activate the slider, and watch as the polynomials move in tandem with pride!
Computing with Desmos
While the interface of Desmos is primarily composed of agraphing grid than anything else, the fact still remains that itwas built fundamentally for computing purpose — and will probably always be.In
fact, wewill soon see that Desmos — while obviously well-equipped toperform basic computations — can be hijacked into doinga whole bunch of non-graph-related stuffs such as calculating apartial sum,
estimating therootsof a function, determining the value of adefinite integral, or even finding thegreatest common factors froma list of integers!
Elementary Calculations
In Desmos, any mathematical expression involving addition, subtraction, multiplication (*), division (/) and exponentiation (^) can be put into a command line,so that ifthe expression entered
contains no variable whatsoever, then the output can becalculated and returned right back to you — usually in a blink of aneye.
In addition, if some functions such as $f$ and $g$ were already being defined in the command lines and we wish to evaluate an expression involvingtheirfunction values (e.g., $4f(15)+2g(0)+5$), then
we are warranted to type in that same expression into a command line as well, after which Desmos would be more than happy to comply with our request. What’s more, by using the prime notation,we can
even get Desmosto evaluate the derivative of a function at a specific point(e.g., $f'(3)$).
By default, when an equation is entered into a command line, Desmos will give usthe option to display the key points(e.g., intercepts, local maxima, local minima) on the graph — whenever applicable.
Furthermore, if wechoose to display these key points, then Desmos will give awaythe coordinates of these points for free as well — usually up to the firstfourdecimal digits if wejust zoom in the
Alternatively, if we are givena function $f$ and weare interestedin making a more accurate estimationfor the coordinates of akey points (e.g., root finding), we can always try typing in $f(a)$ in a
new line and activate the slider for $a$, which in turn allows usto control the value of $a$ manually, and see its effect on $f$ in real-time. In fact, if we just configure the step of the slider to
$10^{-6}$ (i.e., the smallest valid increment), then weshould be able score a few extradecimals this way.
Furthermore, by naming the mathematical expression we want to compute, we can pass down the output of the computation into a new variable, subsequently using it for other fancier purposes such as
building elaborate computations — or graphicalizing the outputof the computations intofiguresand animations.
When it comes to large-scale computations, it is sometimes more cost-efficient to take the time to constructalist of numbers first, than to manually write down the mathematical expressions one after
another. In which case, just know that in Desmos, we can use thesquare-bracket keys to createa list, whose members could eitherenumerated explicitly, or implicitly by typing three dots (as in $[7,8,
\ldots ,22]$).
In fact, one can also create a list where the members are listed implicitly but with a non-trivial increment (as in$[2,7, \ldots, 42]$), wherethe increment can be made to be as big as $5635$, or as
small as $0.01$ — a flexibility which makesDesmos a powerful tool for defining alargeset of numbers.
In Desmos, the square-bracket keys are specifically conceived for the purpose ofenumerating the members of a list.For delimiters whichserve to group an algebraic expression together, use parentheses
So why exactly do we even bother with creating a list?Because we can use it to run “bulk computations” foreach of the members in the list! And if we intend to use a list on a repeatedbasis,thenwe can
choose to assign a name to the list,and pass the name down into thecommand lines for even fancier purposes which include, among other:
• Performing atransformationon the list.
• Simultaneously graphing several functions of the same family.
• Plotting a seriesof points/line/mathematical objectson the graph.
• Calculating thesum of a list — or any other computation which depends on numerically manipulating the members in thelist.
All of which are exemplified in the figure below:
In some occasions, you might it easier to embed the listing and the computations into onesingle command line, but as the task complexitygrows, you might want to consider writing them down in several
lines instead to improve legibility and facilitate future references.
Computational Commands
In addition to the standard features offered by a scientificcalculator, Desmos has a bit of extra commands available to a typical programmable calculator as well. For example, we can use Desmos to
compute the greatest common factorusing the gcd command — as long as we adhere to the following syntax:
\begin{align*} \gcd (\text{number 1}, \ldots, \text{number n}) \end{align*}
In a similar manner, we can also compute:
• The least common multiple(through the lcm command)
• The minimum of multiple numbers (through the min command)
• The maximum(through the max command)
• The least positive residue of $a$ under the modulus $b$ (by entering mod(a,b))
• The number of combinations available whenchoosing $r$ objects from $n$ objects (by entering nCr(n,r))
• The number of permutationswhen choosing $r$ objects from $n$ (by entering nPr(n,r))
• The factorial of $n$ (by entering n!)
For example, if your grandpa offers you the chance of choosing 5 giant gummy bears from the 15 that he has to offer, thenyou can use Desmos to figure out how many ways you can choose the gummies by
typingnCr(15,5) into a command line, and be delighted to find out that you actuallyhave $3003$ choices that youcan exercise freely. That is, $20+$ times more choices than gummies!
As for thefactorial, we’ve got an interesting fact for you:Desmos can calculatethe factorial of a number up to $170!$, which iseasily more than double of what can be achieved usinga TI-83 — or any of
itsvariants for that matter!
Actually, we are not quite done with the computational commands yet. Here’s five more interesting unary operations— for your pleasure:
• ceil(): The ceiling function. It allows you to return the ceiling of a number.
• floor(): The floor function. It returns the floor of a number. Both ceil and floor are relevant in building thestaircase functions.
• round(): The rounding function which rounds a number to an integer.
• abs(): The absolute value function. Using the vertical bar key| is preferred though.
• sign(): The sign function discussed a bit earlier, which comes in handy in simplifying some obscure mathematical expressions.
If you look closely at the above figure with the Desmos keypad, you mightnotice something that looks like a summation sign, along with the Pi product symbol right next to it. Yep. That’s definitely
something Desmos can handle for sure!
For example, let’s say that youwant to estimate the value of the infinite series $\displaystyle \frac{1}{1^2}+\frac{1}{2^2}+\cdots$, then one of the things you can do is to type something like this
into the command line:
\begin{align*} \sum_{i=1}^{1000} \frac{1}{i^2} \end{align*}
For the record, the $\sum$ symbol can betyped out using the sum command, after which you will have to use the arrow keys to navigate around and specify the lower/upperlimits. By default, Desmos uses
$n$ (instead of $i$, for example) as the main index variable, but you can always change it to another variable at your own discretion — as long as you choose a variable that is not pre-defined for
other purposes.
Here, out of curiosity, we decided totoyaround with the upper limit a bit, and found that Desmos seems to top out at $9,999,999$ — just one unit short $10$ millions. That’san astoundingimprovement
from the age of programmable calculators!
And it gets even better: by using an undefined parameter as the upper limit and configuring the slider accordingly. we get toturn the output of the computations into an animation — asyou can see in
the figure on the left-hand side.Watch asDesmos showcases itsimpressivenumber-crunching powerin real-time!
Alternatively, wecan alsoevaluate the product of a sequence using the prod command, and — if needed — stack up as many summations and products as we like. By nesting one operator inside another, we
can also evaluate complexmathematicalexpressions such as those involving double summation or triple product — for instance.
In fact, with just a bit of imagination, we can eventurn the result of computations into ananimated graph, by convertingthe numbers into an animated line segment, or — if we so prefer — convert the
numbers into a comparative diagram of our liking!
What`s more, the sum and prod commands are not restricted to the computations of numbers either. For instance, we could try plotting out the $n$^th-degree Taylor polynomials of $\cos x$ by leaving$n$
asanundefined parameter, so that once we turn on slider with $1$ as the step-size, we should start to see a bunch of functions flashinglike crazy!
Definite Integrals
Looking to calculate the area below a strange-looking curve? Strugglingto model the total distance traveled by an annoying fly? Or maybe you just want to project your annual revenue for the next
year? Either way, the definite integral should have you covered!
In Desmos, the integral symbol $\int$ can be typed out using the int command, after which you can use the arrow keys to navigate around and enter the upper/lower limits. Of course, when defining a
definite integral, it’s necessary to include thedifferentialin such a way that the variable of the differential corresponds to the variablein the function you wish tointegrate on.
Similar to the cases with thesummation and product operators, you are also free to stack up as many integral symbols as you like, and use Desmos to evaluate, say, a double integral (e.g., the volume
of a geometrical figure) or even a triple integral (e.g., the total mass of a metal rod). But whichever the case, it’s always a good idea to get into the habitofusing parentheses judiciously, so that
Desmos understands your expression correctly at each stage of the construction. Otherwise, it’s not our fault if Desmos refuses to parse your expressions like it’s crazy!
So what else can you do with the integral operator? For one,you can trymixing itwith summation and product operators, since they are after all the same kind of operator anyway. Alternatively, you can
also use integrals to define an equation or an inequality (e.g., integral function), and do all kinds of stuffs with it such as verifying the FirstFundamental Theorem of Calculus!
Statistics with Desmos
In addition to the standard graphing/computing features available to most graphing calculators, Desmos also has some neat, nativestatistical featuresfor the analytically-driven, data-minded
individuals. While it’s certainly no replacement for specialized statistical softwares such as R or SPSS, its capabilities in creating/manipulating tables, computing statistical measuresand providing
regression modelsshould be more than enough tocover the basics for the non-statisticians among us!
While a list of data forms the basis of univariate statistics, having a table of data with multiple columns opens us to the whole new world of multivariate statistics,a branch of science which is
becoming increasingly difficultto ignore inthis day and age — wherebig data and information mining proliferates.
And while you’ll most likely not use Desmos for processing big data involving giant tables and fancy metrics, the fact remains you can still createa table in Desmosthroughthe “add item” menu($+$ sign
in the upper-left corner), and enter the data manually as you see fit. Being primarily a graphing calculator in nature, Desmos seems to prefer presentinga new table withtwo humble columns by default:
the first column $x_1$ presumably for the $x$-values, and the second column $x_2$ for the $y$-values.
Naturally, this setup wouldlead to the use of table as a way of plotting multiplepointsof a function, by first filling out a list ofinput values in the $x_1$ column, followed by redefining the name
of the second column as a function of $x_1$ — so thatDesmos can learn to automatically fill in the second column all on its own.
If you intend to havethevalues under the first column to be equally spaced, but are too lazy to enter the values yourself manually, then you can auto-fill the first column by pressingEnterrepeatedly
— after entering the first two values into thecolumn.
In fact, by creating as many columns as we want, we can extend this point-plotting feature a bit tothree or four functions simultaneously, All we have to do is to make sure that “function columns”
are labeledas a function of $x_1$, and watch as Desmos auto-fills the rest of the table and graph for us. No need to ever redo the $x$-values over and over again!
Now, does that mean that you always have to produce a table in Desmos from scratch? Not really. In fact, if you’re familiar any spreadsheet software, you couldimport the table to Desmos via a simple
keyboard copy-and-paste shortcut (Ctrl+C to copy, Ctrl+V to paste). While for a small two-column table, importingdata can sometimes be an overkill.For a giant, multi-dimensional table with a dozen of
variables, the copy-pasting shortcut can be a true life-saver.
By the way, did you notice the circle icons on the top of mostcolumns? Not only are these icons created every time wecreate a new column, but they are also our only gateway towardscustomizing the
appearance of the points on the graph as well. By clicking on the gear iconfollowed by the relevantcircle icon, Desmos will give us the option tochange thecolor of the points under the entire column.
Alternatively, we can alsoadjustthegraphing stylebetween the points here, by choosing to have eitherline segments orcurvespassing through them— a feature which comes in handyfordrawingfigures or
makingpolygon plots.
To be sure, if you name a column as a function of another pre-existingcolumn, then the values underthe new column will be determined by those under the old column. However, if you choose to labelthe
newcolumn as a new variable, then you are in effect making it possible for the entriesunder the column to assume any numerical value.
(for the record, a new variable doesn’t includeany of the pre-defined variablessuch as $x$ and $y$.However, when you attach a new subscript to a pre-defined variable, it does become a new variable as
a result.)
Moreover, for each of the columns that are labeled as a new variable, you can makethe points underneath itdraggable throughthe drag setting (accessible again via the gear and circleicons). By
configuring the points so that they are either draggable in the horizontal directions, vertical directions, or in every directions, youare essentially giving yourself the choice of manipulating the
data visually — which in manycases is more effective thanmanipulating the data numerically.
Descriptive Statistics
Every timewe are givena collection ofnumbers — either in the form ofa list or a columnfrom atable — wecan computesome statistical measuresbased on them. Here’s a list of native, unary statistical
commands supported in Desmos:
• Mean: Use the mean() commandto obtain the arithmetic mean of a list of numbers, which can be defined explicitly (e.g., $\mathrm{mean}([2.5,5.7,15,3]) $), or referred to via a variable name (e.g.,
$\mathrm{mean}(a_1) $).
• Median: Use the median() command to return the middle number (for a set with an odd number of data) or the mean of the two middle numbers (for a set with an even number of data).
• Standard Deviation: use the stdev() command forsample standard deviation, and stdevp()for population standard deviation.
• Mean Absolute Deviation: use the mad() command to find the mean of all absolute deviations of a list.
• Variance: use the var() command to obtain the sample variance.
• Total: use the total() command to find the sum of all numbers in the list.
• Length: use the length() command to find out the number of data in the list.
• Extrema: use the min() command for the minimum, and max() for the maximum.
Additionally, here’s a list ofbinary statistical commands supported in Desmos, with slightly more delicate syntax:
• Quantile: To obtain the $p$-quantile of a list $a$, type in quantile(a,p) into a command line. Note that $p$ must be presented as a decimal between $0$ and $1$.
• Covariance: To obtain the sample covariance between the lists $a$and$b$, type in cov(a,b) or cov(b.a) into a command line.
• Correlation: To obtain the Pearson correlation coefficient between the lists $a$ and $b$, type in corr(a,b) or corr(b,a) into a command line.
To be sure, the fact that these are about the only native commands supported in Desmos doesn’t mean that these are the only statistical measures we are restricted to. In fact, we could , for example,
evaluate the range of a list $a$ by using the command max(a) - min(a),and calculate the Inter-Quartile Range by using quantile(a,0.75) - quantile(a,0.25).
In fact, with just a bit of imagination and ingenuity, it is possible to make out some line charts and bar graphsin Desmos as well. All that is required is some ability in drawing multiplevertical
bars and line segments— and perhaps the ability in doing so at the right spots as well. Here’s afigure to keepthe idea of data visualization in the loop!
All in all, producing charts and diagram in Desmos is more of an art than an actual science!Click to Tweet
When lookingata seriesof points and itsassociatedscatter plot, it’s only natural that we seek to further understandthe nature of relationship — if any — between the two variables in question. In
which case, it makes sense that we useregression models to explain and predict the behaviors of one variable from another. In fact, if weapply the “right” regression model to the data, wecan — more
often than not —uncoversurprising insights that would have been hardto obtainotherwise.
In Desmos, if you havea table with even just two columns — say $x_1$ and $y_1$ — thenyoucan createa regression model for the associated points to your heart’s content. For a linear regression model,
just type in $y_1 \sim mx_1 + b$ into a new line, and Desmos will be more than happy to provide a best-fitting line for you.For a quadratic model, you can type insomething along the line of $y_1 \sim
a{x_1}^2+b x_1 + c$, and a parabolashould be ready for you in a blink of aneye. Additionally,if you prefer fittinga higher-degreepolynomial instead, or maybe you just find anexponentialor a logistic
modelto be more appropriate,then you arecertainly welcome try out thoseas well — as long as you followthe “regression syntax” outlined a bit above.
What’s more,while the graphing interface of Desmos is built primarily for 2D graphing, that actually doesn’t deter us from doingmultivariate regressioneither. Indeed, if we are given agiant table
with4 columns — says $x_1$, $x_2$, $x_3$ and $y_1$ — then wecanfita linear model for these variablesby simplytyping something along the line of $y_1 \simax_1+bx_2+cx_3 +d$ into the command line, so
thateven ifDesmos is not properly equipped to plot graphs involving multiple independent variables, we can continue to run multivariate regression as if nothing happened in the first place!
While avariable name usually takes the form ofa single letter in Desmos, we are still free to use as much subscripts as we want to. For example, instead of using P as a variable for population size,
we might just aswelluse P[opulation] instead, thereby eliminating the need of guessing what the variable stands for in the near future.
At the end of the day,whichever regression model wechoose, Desmos will always be glad to present us withthe least-square line for thedata, along its parameters (e.g., slope,intercept) and the
correlation/$r^2$of the model. Additionally, we can also use the grey “plot” button(under “residuals” tag) to include the residuals into both the graph and the table — and if we are adventurous
enough —pass on the residual variable into the command line again for more detailed analysis!
All right. With most of the main features now settled, let’s move on to some other miscellaneous functionalities whichyou might find useful from time to time. In particular, we’ll see how to use
Desmos to tweak around thegraph setting, create notes, organizing lines into folders, and make use ofthe saving and sharing functionalities!
Graph Setting
If you look at the upper-right corner of Desmos’ user interface, you should be able to see an icon with a wrenchon it. Why bothers? Because this is the place where you can have access to the Graph
Setting menu, which contains a plethora of global settingthat onecan tweak around for practical and not-so-practical purposes.These include, among others:
• Toggling the grid from rectangular form to circular form— or disabling it entirely.
• Labeling the $x$ and $y$-axes.
• Changing the step size of each axis (e.g., using $\dfrac{\pi}{2}$ as step-size when graphing trigonometricfunctions).
• Interpreting the angles in either degree or radian.
In addition, underneath the wrench icon, you should be able to see a $+$ and a $-$ icon. As one would expect, theseicons are used forzooming inandzooming outthe graph, respectively, and when either
of the features is exercised, youwill see beneath the $-$ iconan additional home icon, which allows you to return to the default zoom level. Think of it as some sort ofsafety net in case you blow up
the graph out of proportion (literally)!
A note is exactly that: a note for yourself and others looking atyourgraph and command lines. To create a new note, click on thecommandline where you want the note to appear, thentype a" (i.e., the
double-quote symbol) to turn the command line into a line for note. While admittedly a non-computationalfeature, a notecan still be used to include anycomment, instruction orexplanations deemed
relevant to the graph and thesurrounding computations thereof. In fact, wecan even add a few hyperlinks here and there to spice up the discussion a bit. Just remember tostart the link with ahttpor
https prefix though.
Collapsible/Expandable Folders
While primarilya cosmetic feature, a folder is integral tool for organizing your command lines into a coherent set of groups — the latter of which can collapsed or expanded upon demand. Similar to
the case with a note, a folder can be created by first clicking on the command line where the folder should appear, and then by accessing the Add Item menuviathe $+$ icon nearthe upper-left
corner.Once afolder is created, itcan be given a label, after which a command line can be dragged in or out of the folder with ease, and the triangular arrow iconnext to it can be used to expand/
collapse thefolder as one wishes.
Whilegraphical calculators are excellent tools for creating geometrical figures, there are certain times where animage goes beyond simple geometry and needs to be imported from somewhere else
instead. Fortunately, this is a feature wellsupported in Desmos, where animage can be added to the graph the sameway a folder is addedtoa command line.
By default, Desmos likes to place an image so that its centeris located at the origin, althoughwe can still move/resize the image by dragging along itscenter or the 8 boundary points surrounding it.
In addition, an image’s opacity can be adjusted through the Gear Icon, and itsdimensional values /coordinates of the centercan also be configurednumerically. Heck, we can evensneak insome undefined
parametersin them, therebyturning a static image into a flying picture— where the dimension changes from one place to another!
Saving / Sharing Graphs
Findthe mighty power of Desmos appealing and intend to use itextensively in the near future? Then it makes sense that you create an account, and work on your graphswhile logged into the account. Why?
Because then you can actually savethese graphs for real, andshare them with other like-minded individuals!
In fact, working on a graph while logged in allows you to give a title to the graph, so that if you decide to save it for later, a simple Ctrl + S will do. On the other hand, any saved graph can also
be deleted — if needed to — by hitting the $\displaystyle \times$ button next tothe name of the graph to be deleted.
And if you’re feeling generous enough, you can always share your workwith others by generating a link for the graph — through the greenShare Graph icon near the upper right corner. Remember, Desmos
is a cloud-based application after all, which means that every time you save a graph and publish it somewhere online, you are in effect contributing to an ever-growing database ofDesmos modules— all
in the name of science and technology!
Closing Words
Ouf! That’s quite a bit on an innocent-lookingonline graphing calculator isn’t it? By the way, if you’ve manage to make it this far, you are already a true hero on your way tobecoming a power user of
Desmos! At the end of the day, whether you decide to use Desmos forgraphing, computing, statistics or other purposes, thehope is that you would find a way toleverage thesefunctionalities and adapt
them to your own needs. Who knows, maybe you can even turn some of yourinspiration into afruitful, creative process — with perhaps a bit of technical twist along the way!
All right.Ready for the recap? Here’s ajam-packedinteractive table on what we’ve covered thus far:
• Graphing
• Computing
• Statistics
• Miscellaneous
Greek Letters ($\pi, \theta, \alpha, \beta$)
Roots ($\sqrt{}, \sqrt[n]{x}$)
Ceiling Function
Floor Function
Sign Function
Absolute Value
Arrow Keys
Defining Functions
Functions (in terms of $x$)
Functions (in terms of $y$)
Trig Functions
Exponential Functions
Logarithmic Functions
Polynomial Functions
Inverse Trig Functions
Hyperbolic Functions
Function Naming
Compounding Functions (via $+$, $-$,$\times$, $\div$, $\hat{}$ and $\circ $)
Calculus-Related Functions
Derivative Operator
Prime Notation
Integral Operator
Lower/Upper Limits (Integral)
Piecewise Functions
Syntax (Piecewise Functions)
Naming a Piecewise Functions
Sign Function
$\le, \ge$ Symbols
Other Equations
Cartesian Coordinates
Implicit Equations (e.g., circle, ellipse, hyperbola)
Parametric Equations
Designated Variable
Adjusting the Domain (Parametric Equations)
Circles (Inside or Outside)
Polar Curves (Inside or Outside)
Area (between two functions)
Space(in between the implicit functions)
Restriction (Syntax)
Restriction on $x$
Restriction on $y$
RestrictionInvolving Both $x$ and $y$
Restriction Involving Other Designated Variables
Using Multiple Parameters
Undefined Parameter
Adjusting Range (Parameter)
Increment Size (Parameter)
Speed (Parameter)
Direction (Parameter)
“Play” Button (Slider)
Movable Point
Multiple Parameters (On a Command Line)
Common Parameter (On Multiple Command Lines)
Function Values (Evaluation)
Derivatives (Evaluation)
Key Points (Intercepts, Local Maxima, Local Minima)
Displaying Coordinates (Key Points)
Root Finding (Using Slider)
Naming Expressions
Creating a List
Explicit Enumeration
Implicit Enumeration
List Naming
List Transformation
Simultaneous Graphing (Using List)
Plotting Multiple Mathematical Objects (Using List)
List-Related Calculations
Greatest Common Factor
Least Common Multiple
Least Positive Residue
Combination (Counting)
Permutation (Counting)
Ceiling Function
Floor Function
Rounding Function
Absolute Value Function
Sign Function
$\sum$ (Operator)
$\prod$ (Operator)
Arrow Keys (Navigation)
Dummy Variable
Upper/Lower Limits (Summation)
Animating Series Calculation
Nesting Operators
Animated Graph (Series)
Taylor Polynomial
Integral Operator
Upper/Lower Limits
Double/Triple Integrals
Integral Function
Table vs. List
Creating a Table
Plotting Multiple Points on aFunction (Table)
Creating Multiple Columns (Table)
Importing Spreadsheet Data
Adjusting Color (for Points in the Table)
Graphing Style (Points in the Table)
New Variable (as Column Name)
Draggable Points
Statistical Measures
Standard Deviations
Mean Absolute Deviation
(Sample) Covariance
Inter-Quartile Range
Line Charts /Bar Graphs
Scatter Plot
Regression Model
Linear Regression
Quadratic Regression
Higher-Polynomial Model
Logistic Model
Multivariate Regression
Use of Subscripts(Variable Naming)
Parameter (Least-Square Line)
Correlation / $r^2$ (Least-Square Line)
Residual (Least-Square Line)
Grid Setting (Rectangular vs. Circular)
Labeling Axes
Configuring Step Size
Angle Setting (Degree vs. Radian)
Zoom In / Zoom Out
Default Zoom Level
Creating a Note (Keyboard Shortcut)
Creating a Folder
Naming a Folder
Adding / Removing Lines from aFolder
Expanding / Collapsing a Folder
Importing an Image
Adjusting Opacity
Coordinates of the Center
Dimensional Values
Moving / Resizing Images (Visually vs. Numerically)
Embedding Undefined Parameters
Creating an Account
Title (Graph)
Saving a Graph
Deleting a Graph
Generate a Link for the Graph
And with that, this definitive guide on Desmos has now come to an end. Of course, that doesn’t mean that it’s over yet — for there is yet another way to use Desmos which has great aesthetic and
pedagogical value, and that is in the creation of computational drawings!
How's your higher math going?
Shallow learning and mechanical practices rarely work in higher mathematics. Instead, use these 10 principles to optimize your learning and prevent years of wasted effort.
About the author
Math Vault and its Redditbots enjoy advocating for mathematical experience through digital publishing and the uncanny use of technologies. Check out their 10-principle learning manifesto so that you
can be transformed into a fuller mathematical being too.
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1. Yeah. It’s pretty cool. Figured out the one with the p-series. 🙂
1. Yep. You’ve got it. 🙂
2. As a middle-grade student, my knowledge is basic.
How can I add an x-intercept in Desmos? I need an x-intercept, but I only know how to do a y-intercept and Google is failing me. XD
1. Hi Rayray. There is no innate function for x-intercept for good reason — because a function or an equation can have 0 or multiple copies of them.
However, all is not lost though, as Desmos will display the coordinates of the x-intercepts if you click on them on the graph. The coordinates are up to a few decimal points of accuracy, and
you can type them into the field if you want to.
3. Impressive way presenting Mathematical Ideas so that more & more of target group can grasp them
1. Thank you! Graphing and computing using Desmos is quite a topic all on its own, so we thought that keeping everything under one hood would make people life’s easier!
4. How can I get more than one decimal place for a slider? I tried to set the step to 0.01, but it does not work.
1. Hi Yif. It probably depends on how you’re using the parameter. You can adjust the size of step up to 0.000001 by clicking on the cog icon, and you might have to fill in the upper and lower
bounds for that to happen.
1. You can do this by making a slider on the graph.
Define the variable you want to use for the slider, eg "a"
Set lower, upper limit and stepsize eg 0, 10, 0.001
Now define a slideable point: (a,1) and check the label box
The point should now appear in your graph. It might not show all the decimals but if you zoom in on the point to have more accuracy in positioning it, additional decimals will appear. I
use this for sliders and switches (0-1). That way I can close the sidepanel and still cotnrol the graph.
2. Neat idea Frederik. Thanks for sharing!
5. Great Job!! It's my first time to use it but extremely satisfied with such explanation. This wonderful to us the mathematics teachers/students.
However, what isn't clear enough is how to copy your graph to other areas of work like Microsoft word.
1. Hi William. Glad that you find the tool useful. If you’re merely looking to copy an unanimated graph to a word processor, you can either use the share icon on the upper right of the Desmos
interface to export the graph as an image, or to simply take a screenshot (which can be done natively in both Mac and Windows). | {"url":"https://picketfencesrealtyllc.net/article/desmos-a-definitive-guide-on-graphing-and-computing-math-vault","timestamp":"2024-11-02T05:01:16Z","content_type":"text/html","content_length":"151112","record_id":"<urn:uuid:a306f108-21dd-43e6-9c93-b21573ce6208>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00839.warc.gz"} |
Positioning performance of the NTCM model driven by GPS Klobuchar model parameters
Issue J. Space Weather Space Clim.
Volume 8, 2018
Space weather effects on GNSS and their mitigation
Article Number A20
Number of page(s) 10
DOI https://doi.org/10.1051/swsc/2018009
Published online 27 March 2018
J. Space Weather Space Clim. 2018, 8, A20
Research Article
Positioning performance of the NTCM model driven by GPS Klobuchar model parameters
German Aerospace Center (DLR), Institute of Communications and Navigation, Kalkhorstweg 53, 17235 Neustrelitz, Germany
^* Corresponding author: mainul.hoque@dlr.de
Received: 29 June 2017
Accepted: 31 January 2018
Users of the Global Positioning System (GPS) utilize the Ionospheric Correction Algorithm (ICA) also known as Klobuchar model for correcting ionospheric signal delay or range error. Recently, we
developed an ionosphere correction algorithm called NTCM-Klobpar model for single frequency GNSS applications. The model is driven by a parameter computed from GPS Klobuchar model and consecutively
can be used instead of the GPS Klobuchar model for ionospheric corrections. In the presented work we compare the positioning solutions obtained using NTCM-Klobpar with those using the Klobuchar
model. Our investigation using worldwide ground GPS data from a quiet and a perturbed ionospheric and geomagnetic activity period of 17 days each shows that the 24-hour prediction performance of the
NTCM-Klobpar is better than the GPS Klobuchar model in global average. The root mean squared deviation of the 3D position errors are found to be about 0.24 and 0.45m less for the NTCM-Klobpar
compared to the GPS Klobuchar model during quiet and perturbed condition, respectively. The presented algorithm has the potential to continuously improve the accuracy of GPS single frequency mass
market devices with only little software modification.
Key words: GNSS / positioning / range error / ionospheric correction / modelling
© M.M. Hoque et al., Published by EDP Sciences 2018
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1 Introduction
The ionospheric delay is considered as one of the biggest errors for single frequency use of space based Global Navigation and Satellite System (GNSS). At GNSS operating frequencies in the order of
1–2GHz the ionospheric delay may cause link related range errors of up to 100m. Thus, GNSS single frequency operations need ionospheric delay information for mitigating ionospheric propagation
errors. The ionospheric propagation delay is inversely proportional to the square of the signal frequency and directly proportional to the integral of the electron density along the ray path called
total electron content (TEC) which is commonly expressed in units of TEC termed TECU (1TECU=10^16 electrons/m^2). Therefore, single frequency GNSS positioning needs either TEC or equivalent
ionospheric delay information for mitigating ionospheric propagation errors.
Global Positioning System (GPS) utilizes the Ionospheric Correction Algorithm (ICA), also known as Klobuchar model, for correcting ionospheric signal delay or range error (Klobuchar, 1987;
IS-GPS-200G, 2012). In order to do this, GPS transmits 8 ionospheric correction coefficients in the navigation message on a daily basis as driving parameter set for the Klobuchar model. The Klobuchar
model gives a representation of the mean vertical delay at the GPS L1 frequency as a half-cosine function with varying amplitude and period. The peak of the cosine is fixed at 14 hour local time (LT)
and during night-time hours the vertical ionospheric delay is fixed at a constant value of 5ns or 9.24TECU. The amplitude and period of the cosine function are modelled by 3rd order polynomials
whose coefficients are broadcasted in the GPS navigation message.
The European satellite navigation system Galileo uses a three dimensional time dependent ionospheric electron density model called NeQuick (Nava et al., 2008) for single frequency ionospheric
correction. The original NeQuick model uses a monthly averaged solar radio flux index F10.7 as a proxy measure of the solar activity level. However, to achieve higher accuracy, the Galileo
ionospheric correction model uses an effective ionization level called Az as a primary input parameter for the NeQuick model. The Az approach whose polynomial coefficients are derived from dual
frequency measurements at selected ground stations takes implicitly into account the daily variation of the solar activity and the user's local geomagnetic conditions. The Galileo satellites
broadcast Az coefficients via the navigation message for computing Az at user level all over the globe.
In our former study (Jakowski et al., 2011a) we developed an empirical global TEC model called Neustrelitz TEC Model (NTCM) for estimating trans-ionospheric radio wave propagation errors. The NTCM
approach explicitly describes the TEC dependencies on local time, geographic/geomagnetic location and solar irradiance and activity using only 12 model coefficients. The model coefficients were
computed by nonlinear fitting of global TEC data covering one decade, i.e. about one solar cycle to the model approach in least squares sense. So we used TEC data obtained from Center for Orbit
Determination System (CODE) at the University of Bern (http://aiuws.unibe.ch/ionosphere/) for the years 1998–2007. The only driving parameter of the model is the daily solar radio flux proxy F10.7.
Our subsequent study (Hoque et al., 2015) shows that NTCM model can successfully be used in GNSS applications for ionospheric correction estimation.
In a recent study (Hoque & Jakowski, 2015) we simplified the original NTCM model in order to use it as a broadcast ionosphere model for future satellite navigation systems and called it NTCM-BC. In
contrast to the original NTCM, the NTCM-BC coefficients are not constant because they are adapted to the current ionospheric conditions on a daily base. Being valid for a period of typically 24h
worldwide, NTCM-BC reacts to global ionospheric dynamics and therefore can achieve a higher accuracy than NTCM which uses fixed coefficients.
In a more recent study (Hoque et al., 2017) we recomputed the NTCM model coefficients using more recent (2002–2015) TEC data from CODE. Additionally, we substituted the daily solar radio flux index
F10.7 that drives NTCM by providing a proxy of the solar activity level, by a new quantity called Klobpar. The Klobpar is computed from the GPS Klobuchar model driven by the broadcasted coefficients
as pointed out in section 2 in detail. Since F10.7 is not available to the GPS receiver without additional data link, the user cannot use F10.7-driven-NTCM model in operational purposes. Keeping this
in mind we proposed a new NTCM model driven by a parameter computed from the GPS Klobuchar model. Therefore, GPS users can easily compute the value of Klobpar using broadcasted coefficients and then
use it as a primary input parameter for the NTCM-Klobpar model. Our research, using post processed reference TEC data from more than one solar cycle, showed that on average the RMS modelled TEC
errors are up to 40% less for the proposed NTCM model compared to the Klobuchar model during high solar activity period, and about 10% less during low solar activity period (Hoque et al., 2017).
In the work presented here, we evaluate for the first time the performance of the NTCM-Klobpar model in the position domain by comparing position estimates for selected quiet and perturbed
ionospheric conditions with those obtained by using the mother Klobuchar model. Additionally, we compare position solution estimates for original F10.7 driven NTCM and NTCM-BC models. We calculated
numerous test user positions using a standard Single Point Positioning (SPP) approach (IS-GPS-200F, 2012) in which the ionospheric correction is provided either by the GPS Klobuchar or NTCM-Klobpar
or NTCM or NTCM-BC. The actual 3D user positions are known and subtracted from each SPP solution to obtain associated position errors. Then, the models are compared in terms of estimated position
errors. In addition, the actual 3D positions are compared with ionosphere uncorrected SPP solutions.
2 Data sources and data processing
To compare the different models we selected a period of 17 days from 15–31 January, 2011 (day of year or DOY 15–31) during which the solar activity level proxy F10.7 lies between 77–84 solar flux
units (sfu, 1sfu=10^−22Wm^−2Hz^−1) and the geomagnetic activity proxy Kp does not exceed the 3.0 level (see Fig. 1a) indicating quiet ionospheric and geomagnetic conditions. Again, we selected a
period of 17 days from 23rd October–8th November, 2011 (DOY 296–312) during which F10.7 lies between 121 and 178sfu and Kp exceeds the 4.0 threshold for two times (see Fig. 1a) indicating
ionospheric perturbed conditions.
The daily GPS Klobuchar coefficients which are driving parameters for the Klobuchar model were downloaded from the NOAA's (National Oceanic and Atmospheric Administration) National Geodetic Survey
(NGS) archive ftp://www.ngs.noaa.gov/cors/rinex/ The only driving parameter of NTCM which is the daily F10.7 is obtained from the Space Physics Interactive Data Resource (SPIDR). The NTCM-BC
coefficients are computed each day using dual-frequency GPS data from globally distributed International GNSS Service (IGS, Dow et al., 2009) stations downloaded from the NASA's Crustal Dynamics Data
Information System (CDDIS) archive (Noll, 2010; ftp://cddis.gsfc.nasa.gov/gnss/data/). The procedures used for inter-frequency satellite and receiver bias estimation and TEC calibration, NTCM-BC
optimization are discussed by Jakowski et al. (2011a), Hoque & Jakowski (2015) and Hoque et al. (2015).
The only driving parameter of NTCM-Klobpar is the Klobpar which is defined as the sum of TEC obtained by the GPS Klobuchar model at two specific geodetic locations A(latitudeϕ=10°N, longitudeλ=
90°W) and B(ϕ=10°S, λ=90°W) at 14 Universal Time (UT). The GPS Klobuchar model is used to compute the GPS L1 ionospheric delay T[iono] at points A and B in seconds. The vertical delays
are then converted to TEC to compute the daily Klobpar parameter in TECU as $Klobpar=TECA+TECB=(TionoA+TionoB)⋅c⋅F,$(1) where speed of light c=299792458m/s and factor $F=(1575.42 × 106)240.3 × 1016
=6.1587$ (for details we refer to Hoque et al., 2017).
As discussed in Hoque & Jakowski (2015), the NTCM-BC can be optimized on a daily basis in the GNSS control segment using TEC data of the previous day from monitor stations like the GPS Klobuchar or
Galileo NeQuick. To accomplish the model comparisons as close as possible to conditions applicable to an operational GNSS, a set of globally distributed IGS stations is selected as sensor/monitor
stations for NTCM-BC optimization and a set of test stations is chosen from remaining IGS stations for the analysis of modelling errors. In the present study, we followed the same procedure. The
number of monitor and test stations selected each day during quiet and perturbed period is plotted in Figure 1b. The selected test stations are used to perform a global analysis of model performance
in position solution estimates. The considered models are Klobuchar, NTCM, NTCM-BC and NTCM-Klobpar.
As an example, on 16th January (DOY 16) and 24th October (DOY 297) 2011, we selected 35 and 31 test stations worldwide, respectively. The location of test stations (marked with red crosses) as well
as monitor stations (marked with green dots) on global map is shown in Figure 2a, b for DOY 16 and DOY 297, respectively.
We used the GPS P1 pseudorange measurements from RINEX (Receiver Independent Exchange) observation files for SPP solution computation. The RINEX observation files from worldwide IGS stations are
downloaded from the archive ftp://cddis.gsfc.nasa.gov/gnss/data/ for the selected quiet and perturbed period. As reference user position we used IGS weekly solutions of station coordinate given in
the SINEX (Solution Independent Exchange) file. The SINEX data are downloaded from the same archive mentioned above.
Fig. 1
(a) F10.7 and Kp variation during selected quiet and perturbed days and (b) Number of monitor and test stations used in NTCM-BC optimization and all models validation, respectively.
Fig. 2
Location of monitor stations marked as green dots and test stations marked as red crosses. The red line indicates the geomagnetic equator whereas magenta lines at±20 of the red line bound
equatorial anomaly regions.
3 Single-point positioning approach
We implemented a single-point positioning SPP algorithm based on DOD SPS (2008) and IS-GPS-200F (2012) to validate the performance of different ionospheric corrections. We used the GPS P1 pseudorange
measurements recorded at 30-second interval as input to the SPP module. The pseudorange is a measure of the distance between the satellite and the receiver which is different from the geometric
distance between them due to the errors of the both clocks and the influences of the signal propagation mediums. Taking the satellite and receiver clock errors δt[s] and δt[r], the ionospheric
effects δ[ion], tropospheric effects δ[tro] into account, the pseudorange $Rrs$ can be written as $Rrs(tr,te)=ρrs(tr,te)−(δtr−δts)c+δion+δtro+ϵ,$(2) where c is the speed of light, $ρrs$ is the
geometric distance between the satellite and receiver, t[e] and t[r] are the GPS signal emission and reception time of the satellite and receiver, respectively. The term ε represents errors due to
effects that are not modelled such as multipath, receiver noise, etc.
During the signal transmission, the receiver rotates with the Earth, therefore the so-called Sagnac corrections need to be considered. At the equator, the Earth rotation effect is equivalent to about
31 meters position displacement (Xu, 2007). To mitigate the Sagnac effect we corrected the original satellite coordinates using the method described by Seeber (2003). The GPS satellite broadcasts
navigation messages for satellite clock correction coefficients and orbit parameters. Using these parameters we corrected the satellite clock error as described in IS-GPS-200F (2012). The correction
accounts for clock error characteristics of bias, drift and aging, as well as for the satellite implementation characteristics of group delay bias and mean differential group delay.
3.1 Tropospheric correction
The troposphere is a non-ionized medium extending from the Earth's surface up to about 10–13km of altitude. The tropospheric delay on GNSS signal can be separated into a dry or hydrostatic and a wet
component. The hydrostatic component in zenith direction (ZHD) amounts to about 2.3m whereas zenith wet delay (ZWD) is only in the range of about 0.15m in global average. The dry (hydrostatic)
component contributes about 90% of the total tropospheric error while the wet component contributes about 10% of the error. Typical tropospheric correction models can calculate the zenith total delay
(ZTD=ZHD+ZWD) from measured, predicted or long-time mean values of pressure, temperature and humidity on the earth surface and derive the delay in other elevation angles by multiplying with so
called mapping functions.
For tropospheric correction we used the ESA blind model (Martellucci & Blarzino, 2003) which is an extension to the approach suggested by the RTCA-MOPS (2001) model by using input parameters derived
from Numerical Weather Predictions (NWP) spatial fields. The use of spatial fields produced by the European Centre for Meteorological Weather Forecast (ECMWF) permits to increase the spatial
resolution and the temporal resolution of the meteorological parameters to be used as input to the propagation model of tropospheric delay (Galileo, 2004). The ZTD is calculated by summing up the ZWD
and ZHD, and the Niell mapping function (Neill, 1996) is used for zenith to slant delay conversion.
3.2 Ionospheric correction
The variability of the Earth's ionosphere is much larger than that of the troposphere, and it is more difficult to model. The ionospheric range error can vary from a few meters to a few tens of
meters at the zenith, whereas the tropospheric range error at the zenith is generally between two to three meters. The range error of the ionosphere frequently changes by at least one order of
magnitude during the course of each day.
As already mentioned, the ionospheric correction is provided either by the GPS Klobuchar, NTCM, NTCM-BC or NTCM-Klobpar. All these models are 2 dimensional vertical delay models indicating that the
ionospheric delay at real slant ray paths has to be derived by means of an appropriate mapping function.
The GPS Klobuchar model uses an elevation dependent mapping function to convert vertical to slant delay at user level. The used vertical delay corresponds at the ionospheric pierce point (IPP)
location of the satellite-receiver ray path at the height of 350km. In case of NTCM models, a thin-shell ionosphere mapping function is used (e.g., Jakowski et al., 2011b). As already mentioned,
NTCM and NTCM-Klobpar model coefficients are computed based on CODE TEC maps which represent vertical delays at an IPP height of 450km. Therefore, the vertical delay corresponds at the IPP height of
450km is used for both versions of NTCM driven by F10.7 and Klobpar. However, NTCM-BC model coefficients are computed based on GPS data from worldwide IGS stations and in the ionosphere estimation
procedure an IPP height of 400km is used. So for the NTCM-BC we used the IPP height of 400km for mapping function.
We computed receiver position by first linearizing the range observation equations and then using ordinary least-squares principle. We estimated residuals, i.e., the difference between the actual
observations and the new estimated model for the observations. Several iterations are done before obtaining the final solution. The basic approach is given in associated matlab routines published
along with the paper by Borre (2003).
4 Evaluation of models and discussion
In order to analyze and discuss the ionospheric delay correction effectiveness of the models described in the previous section, the following analysis method has been applied to compare the SPP
results obtained by each of the models.
The approximate 3D user positions (X, Y, Z coordinates in Earth Centered Earth Fixed system) are given in RINEX observation files. However, their accuracy is not known. Therefore, we looked for
station coordinate provided by International GNSS Service (IGS) community. We found that the IGS routinely generates a number of weekly station coordinates among other products by combining
independent estimates from at least seven IGS Analysis Centers (ACs) and distributes in SINEX file format. The combined solutions are aimed to align to an IGS realization (IGS05) of the International
Terrestrial Reference Frame (ITRF, 2005). A measure of the internal coordinate consistency is given by Ferland & Piraszewski (2009) by analyzing the residual standard deviations between the ACs and
the IGS weekly combination solution. They found the station coordinate consistency during the GPS weeks 1400–1480 as about 2–3mm for the horizontal components and about 7mm for the vertical
component which are representative of the coordinate's accuracy. The SINEX provided station coordinates are assumed as reference values and subtracted from each SPP solution to calculate positioning
errors. For simplicity the weekly station coordinates are assumed to be constant during the given GPS week.
The hourly mean 3D position errors over the selected quiet and perturbed period at different test stations are computed and plotted in Figures 3–5. The figures show the variation of the hourly mean
3D position error as a function of local time (LT) at the corresponding station location. Whereas the Figure 3 shows model performance at high latitude stations kir0 (67.9°N, 21.1°E), mar6
(60.6°N, 17.3°E) and mdvj (56°N, 37.2°E), Figure 4 shows model performance at middle and low latitude stations wtzz (49.1°N, 12.9°E), rabt (34°N, −6.9° E), kokb (22.1°N, −159.7°E), and
Figure 5 shows model performance at the southern hemisphere stations mtwa (−10.3°N, 40.2°E), tah2 (−17.6°N, −149.6°E), suth (−32.4°N, 20.8°E) and chat (−44.0°N, ‑176.6°E). Each sub plot
contains positioning results for Klobuchar model (blue marked with square), NTCM (magenta marked with circle), NTCM-BC (black dotted curve) and NTCM-Klobpar models (green marked with circle).
Additionally, the position solution without any ionosphere correction is plotted (No Corr − red dotted curve).
As already mentioned the reference station coordinates are taken from SINEX files. However, for few stations the SINEX station coordinates were not available. In such cases, we used approximate
station coordinates from RINEX files. As for example in Figure 3 for “kir0”, SINEX station coordinates were not available for the quiet period. However, during the perturbed period coordinates were
available and we compared the results obtained using both SINEX and RINEX station coordinates and found very similar results. So for “kir0” station, we presented the results obtained using RINEX
station coordinates during the quiet period. Similarly, for “mtwa” station, SINEX station coordinates were not available for both periods. However, we found similar performance results for stations
“tah2”, “suth” and “chat” when using SINEX and RINEX station coordinates and therefore, for “mtwa” station we presented the results obtained using RINEX station coordinates. In the title of each sub
plot (see Figures 3–5) the used reference station coordinate source is mentioned as either “sinex” or “rinex”.
Figure 3 shows significant improvement in the position solution when using NTCM-Klobpar instead of the mother Klobuchar model. During the ionospheric perturbed period the improvement is more evident
(see right panel plots). We found that all three NTCM models show similar performances during both periods. We found that at kir0 station (see Fig. 3a) the hourly mean 3D position estimates obtained
by the Klobuchar model are worse than those without using ionosphere correction model especially at early and evening hours during quiet period. At mdvj station (see Fig. 3e) we found similar trend
during early hours. Left panel plots (b), (d) and (f) show that at high latitude stations the peak of diurnal variation is shifted to about 18 LT for Klobuchar model whereas the peak for No Corr case
is at about 14 LT. The exact reason for this deviation is not known. However, the maximum of mismodelling at 18 LT may be related to the perturbations. During the perturbation period we see this
behavior also at other latitudes and also in our NTCM models. It disappears in the analysis for the quiet period.
Figure 4 shows positioning results obtained at middle and low latitude stations during the selected quiet and perturbed periods. It is evident that the NTCM-Klobpar model performs better than the
Klobuchar model at middle latitudes. As before, NTCM and NTCM-BC models perform very similar to the NTCM-Klobpar model. Like high latitudes, at middle and low latitudes the Klobuchar model performs
worse than No Corr solution during night time hours especially during quiet period (see Figure 4a, c and e). This is may be due to the reason that during night-time hours the vertical ionospheric
delay of Klobuchar model is fixed at a constant value of 5ns or 9.24TECU. However, during high solar activity condition, it seems that the period of the cosine function of the Klobuchar model is
better modelled and we found better results compared to the No Corr solution.
Figure 5 compares positioning results at several southern hemisphere stations. We found that all models perform very similar; that means we don't have significant improvement using NTCM models over
the Klobuchar model. One reason may be that the lack of sufficient number of IGS stations in southern hemisphere which are used in CODE TEC map generation as well as NTCM-BC coefficients updates.
Since the CODE TEC maps from previous and current solar cycles are used in NTCM and NTCM-Klobpar coefficients generation (Jakowski et al., 2011a; Hoque et al., 2017), they inherit inaccuracy due to
disperse data distribution in the southern hemisphere from CODE TEC maps.
For global analysis of model comparisons, we computed the mean, standard deviation (STD), Root Mean Squared (RMS) of 3D position errors as well as 65 and 95 percentiles of position errors. A
percentile is a measure of the position error below which a given percentage of observations in the data set fall. The statistical estimates are computed over the data from test stations during quiet
and perturbed periods separately. However, to obtain statistically representative results, the data are arranged into the following groups:
• i)
geographic latitude range 0≥ϕ≥ |90°| and local time LT range 0–24;
• ii)
0≥ϕ≥|90°| and LT 6–18h;
• iii)
0≥ϕ≥|90°| and LT 0–6 and 18–24h;
• iv)
0≥ϕ≥|30°| and LT 0–24h;
• v)
30>ϕ≥|60°| and LT 0–24h;
• vi)
60>ϕ≥|90°| and LT 0–24h.
We performed statistical analysis for each case and the results are given in Table 1 and 2. The reference station coordinate values are taken from SINEX files. So we only used stations for which
SINEX data were available. The number of test stations exceeds 30 for each day during quiet and perturbed periods. It should be noted the statistical estimates may change if another set of test
stations is considered.
Comparing RMS, mean, STD, 65 and 95 percentile errors at the same row in Table 1 and 2 we found that for case:
• i)
all samples: NTCM-Klobpar gives smaller values showing better performance than the Klobuchar model for both quiet and perturbed ionospheric conditions. As for example, the corresponding RMS
errors are 2.6, 5.8m for NTCM-Klobpar whereas they are 2.8, 6.3m for Klobuchar model. Comparing RMS values we found that NTCM-BC (2.5, 5.6m) performs better than NTCM (2.6, 5.7m) and
NTCM-Klobpar (2.6, 5.8m) during both periods. Corresponding bar plots are given in Figure 6.
• ii)
daytime condition: we see similar performances as mentioned in case i;
• iii)
nighttime condition: we see that for all models the values are less compared to daytime condition (i.e., case for both quiet and perturbed periods;
• iv)
low latitude region: The RMS error is the same about 3.1m during quiet period whereas about 0.4m less for the NTCM-Klobpar compared to the Klobuchar model during perturbed period. We noted that
the 65 percentile error is slightly higher for NTCM-Klobpar during perturbed period than those for Klobuchar model. However, the RMS, mean as well as STD and 95 percentile errors are less for the
• v)
mid latitude region: the RMS error is about 0.3 and 0.5m less for the NTCM-Klobpar compared to the Klobuchar model during quiet and perturbed condition, respectively. The mean, STD, 65 and 95
percentile values are less for the NTCM-Klobpar model. We found that NTCM and NTCM-BC perform better than the Klobuchar model.
• vi)
high latitude region: the RMS error is about 0.6 and 0.8m less for the NTCM-Klobpar compared to the Klobuchar model during quiet and perturbed condition. The mean, STD, 65 and 95 percentile
values are significantly less for the NTCM-Klobpar model. NTCM and NTCM-BC perform better than the Klobuchar model during both periods. Like middle latitudes, NTCM and NTCM-BC perform better than
Klobuchar model at high latitudes.
Bar plots in Figure 6a–d show the mean, STD, RMS, 65 and 95 percentile of 3D position errors considering all samples (case i) during quiet and perturbed period separately. We found that NTCM-Klobpar
error estimates are less compared to those of the Klobuchar model. It is also evident that the F10.7-driven-NTCM and NTCM-BC also perform better than the Klobuchar model. Among all three NTCM
versions, the NTCM-BC performs the best.
Fig. 3
Hourly mean 3D position error at high latitude stations kir0, mar6 and mdvj during quiet (left panel- plots a, c, e) and perturbed (right panel- plots b, d, f) period.
Fig. 4
Hourly mean 3D position error at middle and low latitude stations wtzz, rabt and kokb during quiet (left panels- plots a, c, e) and perturbed (right panels- plots b, d, f) period.
Fig. 5
Hourly mean 3D position error at southern hemisphere stations mtwa, tah2, suth and chat during quiet (left panels- plots a, c, e, g) and perturbed (right panels- plots b, d, f, h) period.
Table 1
Statistical estimates of 3D position errors for the Klobuchar model and NTCM driven by Klobpar during quiet and perturbed ionospheric conditions.
Table 2
Statistical estimates of 3D position errors for NTCM and NTCM-BC during quiet and perturbed ionospheric conditions.
Fig. 6
Model comparison bar plots of mean, STD, RMS and 95 percentile of 3D position error during quiet and perturbed period.
5 Conclusion
Summarizing the evaluation results, it can be stated that all three different NTCM model approaches NTCM, NTCM-BC and NTCM-Klobpar achieve a better performance in the position domain than the
Klobuchar model regularly provided by GPS. Focusing on estimating the accuracy of the NTCM-Klobpar approach we found an improvement in the order of 0.24m and 0.45m in global average for unperturbed
low solar activity and perturbed medium solar activity conditions, respectively. It can be expected that the improvement is even more pronounced at high solar conditions and during ionospheric
perturbations. This will be shown in further studies that include severe storm and high solar activity conditions. Since the NTCM-Klobpar approach uses the Klobuchar coefficients, regularly provided
in the GPS navigation message, for estimating the current solar activity level, this model could further improve single frequency positioning performed by mass market devices. To reach this goal
NTCM-Klobpar must be implemented in mass market GPS receivers. Since the model approach is very compact, the required technology modification is rather easy to handle.
We would like to give thanks to sponsors and operators of NASA's Earth Science Data Systems and the CDDIS for archiving and distributing the IGS data. We would like to acknowledge the support of the
organizations contributing to the IGS by providing GNSS data to the CDDIS for the international science community. We would like to give thanks NOAA's NGDC for disseminating historical solar and
magnetic data via SPIDR. Also thanks to SOPAC service for making available daily GNSS satellites ephemeris data. The editor thanks two anonymous referees for their assistance in evaluating this
Cite this article as: Hoque MM, Jakowski N, Berdermann J. 2018. Positioning performance of the NTCM model driven by GPS Klobuchar model parameters. J. Space Weather Space Clim. 8: A20
All Tables
Table 1
Statistical estimates of 3D position errors for the Klobuchar model and NTCM driven by Klobpar during quiet and perturbed ionospheric conditions.
Table 2
Statistical estimates of 3D position errors for NTCM and NTCM-BC during quiet and perturbed ionospheric conditions.
All Figures
Fig. 1
(a) F10.7 and Kp variation during selected quiet and perturbed days and (b) Number of monitor and test stations used in NTCM-BC optimization and all models validation, respectively.
In the text
Fig. 2
Location of monitor stations marked as green dots and test stations marked as red crosses. The red line indicates the geomagnetic equator whereas magenta lines at±20 of the red line bound
equatorial anomaly regions.
In the text
Fig. 3
Hourly mean 3D position error at high latitude stations kir0, mar6 and mdvj during quiet (left panel- plots a, c, e) and perturbed (right panel- plots b, d, f) period.
In the text
Fig. 4
Hourly mean 3D position error at middle and low latitude stations wtzz, rabt and kokb during quiet (left panels- plots a, c, e) and perturbed (right panels- plots b, d, f) period.
In the text
Fig. 5
Hourly mean 3D position error at southern hemisphere stations mtwa, tah2, suth and chat during quiet (left panels- plots a, c, e, g) and perturbed (right panels- plots b, d, f, h) period.
In the text
Fig. 6
Model comparison bar plots of mean, STD, RMS and 95 percentile of 3D position error during quiet and perturbed period.
In the text
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Spectral Parameters for Scattering Amplitudes in N=4 Super Yang-Mills Theory
Planar N=4 Super Yang-Mills theory appears to be a quantum integrable four-dimensional conformal theory. This has been used to find equations believed to describe its exact spectrum of anomalous
dimensions. Integrability seemingly also extends to the planar space-time scattering amplitudes of the N=4 model, which show strong signs of Yangian invariance. However, in contradistinction to the
spectral problem, this has not yet led to equations determining the exact amplitudes. We propose that the missing element is the spectral parameter, ubiquitous in integrable models. We show that it
may indeed be included into recent on-shell approaches to scattering amplitude integrands, providing a natural deformation of the latter. Under some constraints, Yangian symmetry is preserved.
Finally we speculate that the spectral parameter might also be the regulator of choice for controlling the infrared divergences appearing when integrating the integrands in exactly four dimensions.
Dive into the research topics of 'Spectral Parameters for Scattering Amplitudes in N=4 Super Yang-Mills Theory'. Together they form a unique fingerprint. | {"url":"https://researchprofiles.herts.ac.uk/en/publications/spectral-parameters-for-scattering-amplitudes-in-n4-super-yang-mi","timestamp":"2024-11-05T17:01:29Z","content_type":"text/html","content_length":"50496","record_id":"<urn:uuid:fd85c6cc-c180-431b-b714-b5e3902e8b48>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00571.warc.gz"} |
What is the partial pressure of water at 30 degrees Celsius?
Vapor Pressure of Water from 0 °C to 100 °C
T °C P (torr)
27 26.7
28 28.4
29 30.0
30 31.8
What is the partial pressure of water at 25 C?
0.0313 atm
This equilibrium partial pressure of vapor above a liquid is known as the equilibrium vapor pressure of the substance. The vapor pressure of water at room temperature (25° C) is 0.0313 atm, or 23.8
mm of mercury (760 mm Hg = 1 atm).
How do you find the partial pressure of water?
When you collect a gas by bubbling it thru water to a graduated cylinder, this gas is saturated with water vapour. Thus Plaboratory=Pgas+PSVP . At 25 ∘C,SVP=23.8 mm Hg. So you have to subtract this
SVP from the laboratory pressure in order to find Pgas , the pressure exerted by whatever gas you are collecting.
What is the value of partial pressure of water?
The partial pressure that water molecules exert to escape through the surface is called thevapor pressure of the water. At normal body temperature, 37°C (98.6°F), this vapor pressure is 47 mm Hg.
How do you find the vapor pressure of water at 25 C?
Simple formula simple_pressure = e^(20.386 – (5132 / (temperature + 273)) , where vapor pressure is expressed in mmHg and temperature in kelvins.
What is the partial pressure of water at 26 degrees Celsius?
Water vapor pressure in mm Hg between 15.0°C and 30.8°C
Temperature (°C) 0.0 0.2
24 22.377 22.648
25 23.756 24.039
26 25.209 25.509
What is partial pressure formula?
As has been mentioned in the lesson, partial pressure can be calculated as follows: P(gas 1) = x(gas 1) * P(Total); where x(gas 1) = no of moles(gas 1)/ no of moles(total). As you can see the above
formulae does not require the individual volumes of the gases or the total volume.
How do I calculate partial pressure?
There are two ways to calculate partial pressures: 1)Use PV = nRT to calculate the individual pressure of each gas in a mixture. 2)Use the mole fraction of each gas to calculate the percentage of
pressure from the total pressure assignable to each individual gas.
What is the vapor pressure of water at 20 C?
17.535 17.753
Water vapor pressure in mm Hg between 15.0°C and 30.8°C
Temperature (°C) 0.0 0.2
19 16.477 16.685
20 17.535 17.753
21 18.650 18.880
What is the vapor pressure of water at 68 F?
0.33889 psia
For the first version of 68 F water, the specific gravity is 1.0 and the vapor pressure is 0.33889 psia.
What is the vapor pressure of A at 35 C?
Vapor pressure of water
T [°C] T [°F] P [atm]
20 68 0.0231
25 77 0.0313
30 86 0.0419
35 95 0.0555 | {"url":"https://www.blfilm.com/2021/01/02/what-is-the-partial-pressure-of-water-at-30-degrees-celsius/","timestamp":"2024-11-07T11:03:26Z","content_type":"text/html","content_length":"65872","record_id":"<urn:uuid:77b24358-9b99-4b6a-91e5-2ce305dfe578>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00449.warc.gz"} |
Surprisingly Rational Circle Segments
(DRAFT: Liable to change)
26 Jun 2018
Last updated:
13 Jul 2018
This paper is
http://www.cs.bham.ac.uk/research/projects/cogaff/misc/multicirc.html http://www.cs.bham.ac.uk/research/projects/cogaff/misc/multicirc.pdf
A partial index of discussion notes in this directory is in
I'll state a problem, and present a solution, below. Some readers may prefer to confront the problem without any hints regarding a solution. There is a separate web page that states this problem
without presenting a solution here:
multicirc-problem.html http://www.cs.bham.ac.uk/research/projects/cogaff/misc/ multicirc-problem.html
It makes some comments about the nature of the problem and its relationship to Immanuel Kant's ideas about mathematics, and mathematical limitations of AI systems based on statistical/probabilistic
reasoning, e.g. Deep Learning mechanisms.
------------------------------------ + ------------------------------------
Touching Circles
Fig 1 Two circles with centres at A and B of radius R touching as shown.
If two circles touch at a point (i.e. each is tangent to the other at that point)
then their centres and that point are co-linear. How do you know that's true?
I'll use two touching circles, and a third circle passing through the point of contact, to formulate a problem, then show how to solve the problem, in a surprising way, by embedding the three circles
in a larger, more complex four-circle structure, in which some relationships become "obvious", thereby revealing the solution to the problem.[Thanks]
Initial problem statement
Fig 2 Add a third circle with centre C, also of radius R, above the line through A and B,
with C placed symmetrically in relation to circles A and B, and passing through the point
of contact of circles A and B, as shown. The line AB must then be a tangent to C. Why?
Note: The symmetry specified in
Fig 2
implies that the centre of circle C is perpendicularly above the line joining the centres of the circles A and B. The centre of each circle must be a distance
from the intersection point, since they all have the same radius:
What is the area of the portion of circle C in Fig 2 that is outside the circles A and B, i.e. the area of the darker region?
That looks like a difficult question to answer because of the peculiar shape of the darker region. It is bounded by a convex curved portion at the top of circle C, and two concave portions below,
meeting at a pointed cusp, where circles A, B and C intersect.
We'll adopt a round-about strategy for working out the area, using a new drawing, Fig 3, below, which has a new circle D, also of radius R, embedded in it, but without the shading in Fig 2.
Circle D is placed below Circle C, symmetrically in relation to circles A and B, and passing through the point of contact of circles A, B, and C as shown in Fig 3.
Fig 3 There are several ways of thinking about this figure, derived from Fig 2.
The original two circles A and B and the two new circles C and D interact to produce new intersection points, in addition to the intersection point at the center where they all meet. In Fig 3 blue
lines have been added joining intersection points common to pairs of circles.
Because the construction has vertical, horizontal and diagonal symmetries, the four short blue lines form two co-linear pairs, forming two longer blue lines intersecting at the point where the
circles intersect.
By reasoning about features of the new figure we'll find a simple way to calculate the shaded area in Fig 2. (You may already find it obvious.)
Note on creation of Fig 3 from Fig 2
One way to create Fig 3 from Fig 2 is to produce Circle D and the lower two blue lines by reflecting region C through the horizontal line joining A and B in Fig 2.
Another way is to construct Circle D in Fig 3 in the same way as Circle C was constructed in Fig 2: i.e. draw a new circle of radius R, with center at distance R perpendicularly below, not above,
the point of contact of circles A and B, then add the lower two blue lines joining points of intersection of new and old circles. Here the process of construction is the "mirror image" of the
process of construction of Fig 2.
Whichever way Fig 3 is created, it now has four symmetrically located regions similar to the shaded region in Fig 2, each with a sharp cusp pointing at the centre of the whole figure, where all four
circles meet.
If L is the length of each blue line in Fig 3, then either of the above constructions produces four blue lines of the same length L, each joining a pair of intersection points of two overlapping
The blue line segment of length L passing between A and D, and the blue line segment of length L between C and B must be collinear and of equal length. (Why?)
Likewise the two blue line segments passing between A and C, and between D and B must be collinear and of equal length L.
Consequently, there is a pair of blue lines, each of length 2xL, intersecting at right angles, and at their midpoints, in the center of Fig 3, where all the circles intersect.
Fig 2 was constructed to be symmetric about a vertical line. So reflecting it to make it symmetric about a horizontal line, produces a new figure with both vertical and horizontal symmetry. We'll use
this as the basis for constructing the next figure, with horizontal and vertical lines through the centre of the figure, shown as dashed black lines in Fig 4, below.
We then add red lines joining pairs of end points of the dashed lines. Because of the symmetries in the construction process, the red lines form a square as shown in Fig 4.
There are now eight circle sectors outside the red square, but inside the circles, in addition to the eight sectors inside the red square, previously visible as areas of overlap in Fig 3, above,
where they were separated by the blue lines, also shown in Fig 4, below.
Fig 4 Extra dashed lines, and a red square joining their endpoints, added to the previous figure.
The new red square must also pass through the ends of the blue lines, where the circles intersect.
Each pair of blue half-lines of length L meeting at right angles can be taken as two sides of a square. Completing each square with two red lines of length L is another way to produce Fig 4 above,
with four axes of symmetry, two diagonal, shown in blue, as previously, and one horizontal and one vertical axis of symmetry shown as dashed black lines in Fig 4.
Each dashed line has two halves, where each half is a diameter of length 2xR of one of the four circles. In Fig 4 the symmetries (reflections) imply that the four blue lines meeting at the center,
meet at right angles and that the four quadrilaterals formed by adding two red lines to each pair of blue lines are all squares, as shown in the figure.
Answering the question about areas
We can now answer our original question about the shaded area in Fig 2 (replicated below on the left of Fig 5).
Figures 3 and 4 show that the shaded area in Fig 2 (=Fig 5) is formed by removing four regions from circle C, where each of the four regions is bounded by a straight line and a circular arc in the
new figure on the right of Fig 5.
But the symmetries in Fig 4 show that the remaining area after removing the four regions from a circle is the same as the area of the square inscribed in the circle, that's because four such segments
surround the square in each circle.
Figures 3 and 4 show that the shaded area in Fig 2 (or Fig 5 below) in the circle C is produced by removing from the circle four portions each equivalent to one of the four sectors surrounding the
square inscribed in the circle.
So the "strangely shaped" shaded area must be the same as the area of the square that would be obtained by moving two of the shaded circular sectors at the circumference of the circle, into the
central region of Circle C, as shown in Fig 5:
Fig 5 Transforming the shaded region of Fig 2 (on the left) to a square by
moving two sectors of the shaded region at a1 and b1 into the unshaded
regions at a2 and b2. In this process regions a2 and b2 become shaded,
and regions a1 and b1 become unshaded, leaving a square region shaded.
So the shaded region on the left has the same area as the square on the right.
Merely inspecting the original figure (on the left of Fig 5 does not reveal the equality between regions that is achieved by embedding the figure in the larger construction in Fig 4, whose symmetries
allow equalities of regions to be guaranteed without using any explicit calculation of areas. Neither does inspection of the left side of Fig 5 reveal that the blue chord common to circles A and C,
and blue chord common to circles B and C, shown on the right side of Fig 5 meet at a right angle.
The derivation of Fig 3 and Fig 4 from Fig 2 can be seen as using a combination of Euclidean Geometry (e.g. drawing circles) and Origami Geometry, using paper folds, for which axioms are given in
Is there is a more direct, purely Euclidean, way of showing that the two blue lines in Fig 5 must meet at right angles, without drawing the extra circles and lines?
Now we know that the area requested in the initial problem statement, above is the same as the area of the inscribed square, on the right of Fig 5. We can use Pythagoras' theorem to derive the area
of the square, using Fig 6, below.
Fig 6 A circle of radius R with an inscribed square of side L
A square bounded by a circle of radius
through the corners of the square is made of four right-angled triangles each of which has two sides of length
meeting at right angles, and a third side, the hypotenuse, of length
. Given
we can compute
using Pythagoras' theorem.
L² = R² + R² = 2xR²
So the area of the square, and therefore the area of the shaded portion of Fig 2 is simply
L² = 2xR².
This result is so simple that the derivation given above is probably unnecessarily complex. Please let me know if you find a simpler derivation.
I showed Steve Vickers the problem statement at the top of this file
using this figure, copied from above:
Fig 2a
After which he thought about it then a short time later wrote to me saying (with my inserts in red):
"Here's my solution to your problem of the three circles.
I was trying to think of a slick geometric argument, and failed, so in the train I just went for the calculation.
The area required is the circle C less two lenses.
Half a lens is a quarter circle less half a square, so has area
πR^2/4 - R^2/2
where R is the common radius of the circles."
At first I failed to understand because I thought Steve was referring to half the blue square of side L in
Figure 6, copied here as Figure 6a.
Fig 6a A circle of radius R with an inscribed square of side L
Then I realised he was talking about the four smaller squares of side R. However, only half of each smaller square is drawn in the figure.
The blue square of side L is made up of four right-angled triangles each of which is half of a smaller square of side R, not drawn.
As Figure 6a shows, the big square of side L has area composed of 4 half squares of side R. So the total area of the big square is twice the area of one small square, i.e. the area of the big
blue square is:
Continuing Steve's reasoning:
"Half a lens is a quarter circle less half a square (of side R), i.e. πR^2/4 - R^2/2
where R is the common radius of the circles.
(It is not hard to see that circle C, in the original figure, is the same size as the other two, and that the intersection lenses subtend angles of 90 degrees at the centre of C.)
Hence the required area is: πR^2 - 4(πR^2/4 - R^2/2) = 2R^2
A problem:
What exactly justifies the claim that the two chords produced by the intersections between Circle C and the two other circles A and B are of the right size to form the side of an inscribed
square, rather than an inscribed quadrilateral of some other shape, e.g. with two short and two long sides meeting?
It is obvious from the symmetry between circles A and B that the inscribed quadrilateral must have a vertical axis of symmetry joining the top and bottom corners. But the assumpton that the
quadrilateral is a square requires a proof that the horizontal line joining the intersection between A and C and the intersection between B and C is also an axis of symmetry.
In the proof based on Fig 4 the required symmetry came from the construction of circle D adding a new axis of symmetry. Is something equivalent implicit in Fig 5a ?
Steve continued with his second solution:
Having calculated that, I immediately saw a geometric argument. If you slice each lens into two, you can rearrange the four halves around the circumference of circle C and they leave a square
hole in the middle. Its area (circle less four half lenses) is the same as the answer we are looking for. The side of the square is √2*R by Pythagoras, so the area is 2*R^2.
This is essentially the the same as the reasoning given just below Figure 6a, above.
.... It would be interesting to know how other people do it. For myself, I couldn't see the geometric argument until the algebra had given me clues about what to go for. I think I realized
straight away that since the πs had gone I could look for a rearrangement with no curved sides, and then the actual answer was easy.
Thanks to Manfred Kerber Manfred Kerber for drawing my attention to this problem, after learning of it from Colin Rowat. So thanks also to Colin Rowat.
What sort of brain is capable of discovering and solving problems like this, or understanding the reasoning used above?
What can we learn from this about mathematical consciousness and the mechanisms it uses in humans?
In particular, since I am sure I am not the only person able to work out the proof by thinking about the diagrams presented above, what does that imply about mechanisms of human visual/spatial
What kind of robot brain would enable a robot to discover and solve a problem like this one?
Can any existing neural net model of brain function accommodate such mathematical discovery processes?
This is part of the Meta-Morphogenesis project:
Mathematical phenomena, their evolution and development
(Examples and discussions on this web site.)
(or ...../.pdf)
Some (Possibly) New Considerations Regarding Impossible Objects
Meta-Morphogenesis and Toddler Theorems: Case Studies
(or ...../.pdf)
The Triangle Sum Theorem
Old and new proofs concerning the sum of interior angles of a triangle.
(More on the hidden depths of triangle qualia.)
Using Apollonius' construction to find the solution to a "maximum angle" problem.
(or ...../.pdf)
Reasoning About Continuous Deformation of Curves on a torus and other things.
A 17 month toddler discovers and solves a 3D topology problem
(or ...../.pdf)
A Super-Turing (Multi) Membrane Machine for Geometers
(Also for toddlers, and other intelligent animals)
Maintained by Aaron Sloman
School of Computer Science
The University of Birmingham | {"url":"https://www.cs.bham.ac.uk/research/projects/cogaff/misc/multicirc.html","timestamp":"2024-11-13T02:48:50Z","content_type":"text/html","content_length":"24382","record_id":"<urn:uuid:e5b22ded-db81-459e-8e45-75bde6783b56>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00650.warc.gz"} |
Contracting gas
Some time ago, professor Oak bought an old flat that had no gas contract. What follows is a simpliflied model of the nightmare that he had to suffer.
To acquire gas, you need two papers: one from the gas distributor, and another from the gas marketer. Initially, you have none of them. When you try to get a paper from the distributor, you will get
it with probability p[d]. However, if you already have a paper from the marketer, you will lose it with probability q[m] (the distributor will decide that it is not good enough). Simetrically, when
you try to get a paper from the marketer, you will get it with probabilityp[m]. However, if you already have a paper from the distributor, you will lose it with probability q[d]. You spend a whole
day every time that you try to get a paper. You win this stupid game when you first manage to have a valid paper from both the distributor and the marketer.
Given all this information, and assuming an optimal strategy, what is the expected number of days to get both papers and therefore gas?
Input consists of several cases, each with p[d], q[m], p[m] and q[d] in this order. All the probabilities are real numbers with at most two digits after the decimal point. Additionally, p[d] and p[m]
are at least 0.1, and q[m] and q[d] are at most 0.9.
For every case, print with four digits after the decimal point the optimal expected number of days to get gas. The input cases have no precision issues. | {"url":"https://jutge.org/problems/P33955_en","timestamp":"2024-11-13T01:25:02Z","content_type":"text/html","content_length":"24414","record_id":"<urn:uuid:f8d205ac-8033-417e-852a-6a62a906d834>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00182.warc.gz"} |
Time history analysis of seismic response of through CFST non isolated and isolated arch bridges
To explore the difference in the impact of transverse bracing on the seismic effect of through concrete-filled steel tube arch bridges with non-isolated and earthquake-isolated, nine non-isolated and
earthquake-isolated structural models under different cross-bracing arrangements were established, and Elcentro seismic waves were selected. The internal force, displacement, velocity, absolute
acceleration, relative acceleration, and separation of arch ribs of each model were compared and analyzed under uniform excitation along the bridge, transverse and vertical directions,
multi-dimensional combined excitation, and multipoint excitation considering the traveling wave effect. Based on the shear force and displacement of the earthquake support, it is concluded that the
internal force response of different excitations of various models is more complicated. The installation of transverse bracing on the upper part of the arch rib can reduce the vertical displacement
of the arch rib of the nonseismic structure. The “X”-shaped cross brace at the top of the arch rib and the “K”-shaped cross brace at the lower part help to reduce the transverse acceleration of the
arch rib. The absolute acceleration and relative acceleration of the seismic structure arch ribs are significantly reduced.
• Time-history response of arch foot axial force
• Time-history response of vault displacement under various conditions of non-seismic isolation
• Time-history response of vault lateral velocity in each case of seismic isolation
• Time-history response of lateral acceleration of nine vaults under working conditions
• Hysteresis curve
1. Introduction
In recent years, there have been frequent earthquakes around the world. As a lifeline project for postdisaster reconstruction and disaster relief, bridges have always received extensive attention. A
large number of concrete-filled steel tube arch bridges have been built, and research on related cross-bracing arrangements is also ongoing.
Dong Rui et al. [1] studied the effectiveness of new L-shaped cross-braces in the stability of long-span concrete-filled steel tube truss arch bridges. Hejiang Third Bridge was taken as the
engineering background, using a combination of numerical calculation and theoretical analysis to compare and analyze its mechanical performance and stability, and use orthogonal experiment and
variance analysis methods to evaluate the significance of L-shaped cross braces in the stability of long-span CFST truss arch bridges Zhang Sumei and Yundi [2] analyzed and compared the possible
layout schemes of cross braces and X braces for a 360-meter-span half-through concrete-filled steel tube arch bridge, and proposed the rationality of X braces and cross braces accordingly. According
to the principle of equal bracing area and similar material consumption of transverse bracing system, four bracing schemes were proposed and analyzed for ultimate bearing capacity respectively; Wan
Peng et al. [3] designed the Guangzhou Xinguang Bridge with a main span of 428 meters in plan, the large-scale finite element software ANSYS was used to establish a three-dimensional finite element
model of the full bridge, and the influence of the number and position of the transverse braces on the elastic stability and the ultimate bearing capacity of the plane was analyzed. Jin Bo et al. [4]
used the finite element method to analyze the influence of transverse bracing on the overall stability of a cable-stayed concrete-filled steel tube arch bridge; Chen Baochun et al. [5] found arch and
arch-girder composite bridges are the main ones; Liu Zhao et al. [6] derived the analytical calculation formula for the lateral elastic stability bearing capacity of arch bridges with transverse
braces based on the energy principle, and verified the proposed finite element numerical solution through a numerical example. The correctness of the analytical formula and finally discussed the
influence of structural parameters on the stability of bearing capacity; Wu Meirong et al. [7] stepped into the non-thrust half-through concrete-filled steel tube arch bridge in terms of rise-span
ratio, width-span ratio, main arch rib stiffness, transverse bracing Changes in the dynamic characteristics of the bridge structure when the layout mode, suspender failure, and support layout are
changed; Kong Dandan et al.[8] took a steel truss arch bridge in a certain city as the research object and showed that increasing the number of wind bracing structures can significantly improve the
structure’s performance stability; but when the number of wind bracing is sufficient, the continue to increase the number of wind bracing structures, the stability of the structure cannot be greatly
improved, and the setting of diagonal braces has a great influence on the overall stability, especially “K” and “X” diagonal braces have a significant impact on the structural stability; Li Xiayuan
et al. [9] relying on a certain through-type steel tube concrete arch bridge, based on the original bridge wind bracing form, using the MIDAS Civil finite element analysis software to establish the
“-” The calculation model for the through-type steel tube concrete arch bridge with “X”-shaped wind bracing, “K”-shaped wind bracing, “m”-shaped wind bracing, and “X”-shaped wind bracing, extracts
the first 20-order natural frequency and The vibration mode types of the first 6 steps were compared and analyzed with the original bridge; Zheng Xiaoyan et al. [10] studied the stability of the tied
arch bridge during the construction phase and the influence of temporary transverse bracing on the structural stability.
In this paper, nine non-isolated and earthquake-isolated structural models under different cross-bracing arrangements were established, and Elcentro seismic waves were selected. The internal force,
displacement, velocity, absolute acceleration, relative acceleration, and separation of arch ribs of each model were compared and analyzed under uniform excitation along the bridge, transverse and
vertical directions, multi-dimensional combined excitation, and multipoint excitation considering the traveling wave effect.
The layout position and layout of the transverse bracing have different effects on the through-type concrete-filled steel tube seismic arch bridge and the seismic isolation arch bridge. The article
will conduct comparative analysis and research to provide the necessary references for the design and construction of similar arch bridges.
2. Principles of time history analysis
The vibration equation for dynamic time history analysis is:
where ${M}_{S}$, ${C}_{S}$, ${K}_{S}$ denote the mass matrix, damping matrix and stiffness matrix of the corresponding structural non-supporting position, respectively, use ${M}_{b}$, ${C}_{b}$, ${K}
_{b}$ to denote the mass matrix, damping matrix and stiffness matrix of the corresponding structural support position, respectively, and use ${\stackrel{¨}{y}}_{s}$, ${\stackrel{˙}{y}}_{s}$, ${y}_{s}
$ to denote the structural non-supporting position under earthquake action, the acceleration, velocity and absolute displacement of the support, with ${\stackrel{¨}{y}}_{b}$, ${\stackrel{˙}{y}}_{b}$,
${y}_{b}$, respectively represent the acceleration, velocity and absolute displacement vector of the structural support position under the action of an earthquake. ${F}_{b}$ is the reaction force of
the support under the action of an earthquake. Then the vibration equation can be expressed in the following form:
$\left[\begin{array}{c}{M}_{\begin{array}{l}ss\\ \end{array}}\\ 0\end{array}\right\begin{array}{c}{0}_{\begin{array}{l}\\ \end{array}}\\ {M}_{bb}\end{array}]\left\{\begin{array}{l}\stackrel{..}{{y}_
{s}}\\ \stackrel{..}{{y}_{b}}\end{array}\right\}+\left[\begin{array}{l}{C}_{\begin{array}{l}ss\\ \end{array}}\\ {C}_{bs}\end{array}\right\begin{array}{l}{C}_{\begin{array}{l}sb\\ \end{array}}\\ {C}_
{bb}\end{array}]\left\{\begin{array}{l}\stackrel{.}{{y}_{s}}\\ \stackrel{.}{{y}_{b}}\end{array}\right\}+\left[\begin{array}{l}{K}_{\begin{array}{l}ss\\ \end{array}}\\ {K}_{bs}\end{array}\right\begin
{array}{l}{K}_{\begin{array}{l}sb\\ \end{array}}\\ {K}_{bb}\end{array}]\left\{\begin{array}{l}{y}_{\begin{array}{l}s\\ \end{array}}\\ {y}_{b}\end{array}\right\}=\left[\begin{array}{c}0\\ {F}_{b}\end
3. Finite element model
Taking an actual through arch bridge as the background, nine non-seismic and seismic finite element models of different transverse bracing arrangements are established. The transverse bracing
arrangement and finite element model are shown in Table 1 and Fig. 1. The seismic isolation model is equipped with lead-core rubber seismic isolation bearings, and the bearing parameters are shown in
Table 2. The bridge has a main span of 127 m and a bridge deck width of 31 m. The arch rib cross-section is dumbbell-shaped. The diameter of the upper and lower arch ribs is 1.2 m, and the diameter
of the cross brace is 1.3 m.
Table 1The layout of transverse bracing in various working conditions
Working Working Working
condition 1 Working condition 2 Working condition 3 condition Working condition 5 Working condition 6 Working condition 7 Working condition 8 condition
The vault has one Five
A cross brace in Three “-”-shaped cross Three “-”- shaped cross Five-way One “-”- shaped cross brace on One “-”- shaped cross brace on One “-”- shaped cross
the shape of “-” braces on the vault and braces on the vault and "-" cross the vault, and two “K” cross the vault, two “K” cross braces “-”- shaped cross brace on the vault and “X”-
on the vault the middle and upper the middle and lower parts brace braces in the middle and upper in the middle and lower part brace and four four “X” cross braces shaped
parts part “K”-shaped cross cross
braces braces
Table 2Parameter table of lead rubber bearing
Support plane size (mm×mm) Lead core yield (kN) Rigidity before yielding (kN/mm) Rigidity after yielding (kN/mm) Horizontal equivalent stiffness (kN/mm)
1320×1320 964 25.6 3.9 6.4
Fig. 1Finite element model diagram
4. Analysis of dynamic characteristics
Through the finite element software analysis of the dynamic characteristics, the frequency and mode shape of the non-isolated and isolated models under nine working conditions are obtained. The first
three orders are shown in Table 3, and the frequency comparison is shown in Fig. 2. It can be seen that the first-order modes of the two models under nine working conditions are all arch rib lateral
inclination, and the first-order frequencies of working conditions 1, 2, 3, and 4 have little difference, while the first-order frequencies of working conditions 8 and 9 are relatively different.
Large “K”-shaped cross braces and “X” cross braces can increase the fundamental frequency, and the effect of being close to the lower part of the arch rib is obvious. The “X” cross brace on the dome
actually reduces the fundamental frequency. The second and third order frequencies and modes of the two models are quite different, and the influence of the cross bracing of the non-isolated model is
more obvious than that of the isolated model.
Table 3The first third-order frequency and mode shape of each working condition
Working condition Types Frequency and mode shape First order Second order Third order
Mode shape
Frequency 0.167 0.520 0.521
Working condition 1
Mode shape
Frequency 0.160 0.253 0.283
Mode shape
Frequency 0.167 0.517 0.647
Working condition 2
Mode shape
Frequency 0.160 0.252 0.284
Mode shape
Frequency 0.168 0.518 0.647
Working condition 3
Mode shape
Frequency 0.161 0.252 0.284
Mode shape
0.168 0.515 0.645
Working condition 4
Mode shape
Frequency 0.161 0.252 0.284
Mode shape
Frequency 0.188 0.522 0.645
Working condition 5
Mode shape
Frequency 0.177 0.252 0.290
Mode shape
Frequency 0.213 0.546 0.645
Working condition 6
Mode shape
Frequency 0.195 0.252 0.297
Mode shape
Frequency 0.231 0.548 0.642
Working condition 7
Mode shape
Frequency 0.205 0.252 0.306
Mode shape
Frequency 0.332 0.619 0.640
Working condition 8
Mode shape
Frequency 0.233 0.251 0.363
Mode shape
Frequency 0.330 0.640 0.691
Working condition 9
Mode shape
Frequency 0.232 0.251 0.365
5. Selection of seismic wave and apparent wave speed
The seismic fortification intensity of the area where the bridge is located is 8 degrees (0.2 g), and the site category is Type II. The El Centro seismic wave is selected, and the peak acceleration
value of the seismic wave is multiplied by a coefficient of 0.339 for adjustment. The adjusted seismic wave is shown in Fig. 3, and the action time is taken as 20 s, the excitation direction is
uniform excitation along the bridge direction, uniform excitation across the bridge direction, uniform excitation vertical direction, multi-dimensional combination one (long bridge direction + 0.3
horizontal bridge direction + 0.3 vertical) excitation, multi-dimensional combination two (0.3 forward bridge direction + Transverse bridge direction + 0.3 vertical direction) excitation,
multi-dimensional combination three (0.3 along bridge direction + 0.3 transverse bridge direction + vertical direction) excitation and the apparent wave speed is 100 m/s, 200 m/s, 300 m/s, 400 m/s,
Multi-point excitation of 500 m/s, 1000 m/s, 1500 m/s, 2000 m/s.
Fig. 2Frequency comparison
Fig. 3El-centro seismic wave adjusted
6. Earthquake response analysis
6.1. Internal force of arch rib
See Table A1 for the maximum internal force and damping rate of arch ribs in different models under uniform excitation. See Table A2 for the maximum internal force and damping rate of arch ribs in
different models under multi-dimensional combined excitation. Under multi-point excitation considering traveling wave effect, the maximum internal force and shock absorption rate of arch ribs in
different models under various working conditions are shown in Table A3. The time-history response of partial arch foot axial force is shown in Fig. 4.
Through the comparison of Table A1 to Table A3 and Fig. 4, we can get:
(1) Under the action of seismic waves with different wave speeds in the bridge direction, transverse bridge direction, combination 1 and bridge direction, the main internal force of the seismic
isolation structure arch rib in each working condition is significantly reduced;
(2) Under the action of vertical earthquake, the main internal forces of the seismic isolation structure arch ribs in various working conditions increased, the shear force ${F}_{Z}$ increased by more
than twice, and the bending moment ${M}_{y}$ increased by more than three times;
(3) Under the action of the second combination earthquake, the arch rib axial force of each working condition of the seismic isolation structure decreases, the shear force ${F}_{z}$ increases, the
bending moment ${M}_{z}$ in working condition 8 and 9 increase, and the rest decrease. Under the action of the combination three earthquakes. The main internal force of the arch rib of the seismic
isolation structure in the working condition increased, the shear force ${F}_{Z}$ increased more than doubled, and the bending moment ${M}_{y}$ increased more than doubled;
(4) Under the effects of lateral earthquake and combination, the main internal force of the seismic isolation structure arch ribs in working conditions 8 and 9 increase significantly.
Fig. 4Time history response of arch foot axial force
a) Working condition1 along the bridge
b) Condition 7 along the bridge
c) Condition 1 Cross-bridge direction
d) Working condition 5 Combination one
6.2. Arch rib displacement
The maximum displacement of the arch rib under transverse excitation is shown in Table 4, and the time-history response of the DY time history of the vault displacement under non-seismic conditions
is shown in Fig. 5.
Fig. 5DY time-history response of vault displacement under various conditions of non-seismic isolation
Through the comparative analysis of Table 4 and Fig. 5, we can get:
(1) Under the action of transverse bridge seismic wave, the arch ribs of non-seismic and isolation models mainly undergo lateral displacement. The lateral displacements of working conditions 1, 2, 3,
and 4 are not much different. The lateral displacements of working conditions 5, 6, and 7 are more than other, the working condition is small, and it is concluded that the “K”-shaped cross brace is
better than the “-” cross brace and the “meter” cross brace in reducing the lateral displacement of the arch rib;
(2) Comparing various working conditions, it can be concluded that setting up transverse bracing on the upper part can reduce the vertical displacement of the arch rib of the non-seismic model.
Table 4Arch rib displacement (unit: cm)
Working condition
Incentive direction Displacement direction Model
Non-isolated 0.220588 0.217772 0.218886 0.216088 0.19172 0.215155 0.190403 0.234662 0.238253
Along the bridge Horizontal
0.095618 0.095142 0.095793 0.095229 0.08672 0.090369 0.097304 0.14055 0.138715
Non-isolated 12.178108 12.190638 12.160166 12.177714 10.963965 9.490335 8.852437 11.574302 11.437827
Cross bridge
Cross bridge Horizontal
13.456979 13.433771 13.436001 13.414679 12.387384 11.74146 10.855834 17.104378 16.846004
Non-isolated 0.575183 0.569438 0.572487 0.566709 0.505177 0.567788 0.505335 0.55977 0.564412
Isolated 0.225666 0.225264 0.225162 0.224671 0.19714 0.210035 0.216076 0.3417 0.344369
6.3. Arch rib speed
The maximum speed of arch ribs under transverse excitation is shown in Table 5. The time-history response of the transverse velocity of the vault under each condition of seismic isolation is shown in
Fig. 6. Through the comparative analysis of Table 5 and Fig. 6, we can get:
(1) Under the action of transverse bridge seismic waves, the lateral velocity of arch ribs in non-seismic and seismic isolation models basically increases in working conditions 1 to 8, while working
condition 9 decreases slightly;
(2) Under the action of transverse bridge seismic waves, the longitudinal and vertical speeds of arch ribs in non-seismic and seismic models are relatively small in condition five;
(3) The speed of the arch ribs of the seismic isolation structure in each working condition is reduced.
Table 5Arch rib speed (unit: cm/s)
Working condition
Incentive direction Speed direction Model
Non-isolated 1.572553 1.573233 1.55233 1.555467 1.481314 1.592136 1.495981 1.442885 1.443201
Along the bridge
Isolated 0.8401 0.842009 0.831214 0.836434 0.801512 0.875925 0.845275 0.88414 0.885579
Non-isolated 25.172737 25.105125 25.468387 25.328813 27.179484 29.70583 31.043626 39.780206 38.68521
Cross bridge Cross bridge
Isolated 19.793487 19.782502 19.837443 19.837609 18.940581 23.446982 26.028057 39.182534 38.69969
Non-isolated 3.808268 3.734763 3.786084 3.709999 3.400459 3.806481 3.455232 3.459467 3.521407
Isolated 1.711642 1.729262 1.698642 1.715102 1.625263 1.767564 1.702519 1.688422 1.690779
Fig. 6Time-history response of vault lateral velocity in each case of seismic isolation
6.4. Absolute acceleration of arch rib
The maximum absolute acceleration of the arch rib under transverse excitation is shown in Table 6, and the time-history response of the lateral acceleration of the nine vaults under working
conditions is shown in Fig. 7. Through the comparative analysis of Table 6 and Fig. 7, it can be obtained:
(1) Under the action of the transverse bridge seismic wave, the non-isolated and isolated model arch rib lateral acceleration, the non-seismic structure working condition 5 and working condition 7
are smaller, the seismic isolation structure working condition 7 is relatively small, and the working condition 9 is relatively small. Working condition 8 is reduced, it can be inferred that the “米
”-shaped cross brace at the top of the arch rib, and the “K”-shaped cross brace at the lower part will help reduce the absolute acceleration of the arch rib.
(2) The absolute acceleration of the arch rib of the seismic isolation structure in each working condition is significantly reduced.
Table 6Absolute acceleration of arch ribs (unit: cm/s2)
Working condition
Incentive direction Absolute acceleration direction Model
Non-isolated 302.069666 298.588074 314.635446 306.619751 291.123396 315.517481 291.100165 343.150259 338.930472
Cross bridge Cross bridge
Isolated 83.273905 82.935068 82.92821 83.290524 82.154905 81.068172 79.804811 91.825414 88.816718
Fig. 7Time-history response of lateral acceleration of nine vaults under working conditions
6.5. Relative acceleration of arch rib
The maximum relative acceleration of arch ribs under transverse excitation is shown in Table 7.
Table 7Relative acceleration of the arch ribs (unit: cm/s2)
Working condition
Incentive direction Relative acceleration direction Model
Non-isolated 350.799422 353.664181 349.011316 349.716454 351.965781 330.543943 325.231323 348.104201 346.062742
Cross bridge Cross bridge
Isolated 157.707118 157.621842 157.721081 157.612437 156.244506 156.104435 154.734779 150.697906 151.129868
Through the comparative analysis of Table 7, we can get:
(1) Under the action of the transverse bridge seismic wave, the relative acceleration of the arch ribs of the non-seismic and isolation models is relatively small for the non-seismic structure
working conditions 6 and 7, and the seismic isolation structure working conditions 1 to 7 basically show a decreasing trend;
(2) The relative acceleration of the arch rib of the seismic isolation structure in each working condition is significantly reduced.
6.6. Shear force and displacement of seismic isolation support
See Table 8 for the maximum shear force and displacement of the seismic isolation support. See Fig. 8 for the shear force comparison of some supports. See Fig. 9 for the displacement comparison of
some supports.
Table 8Maximum shear force and displacement of seismic isolation support
Incentive direction Working condition 1 2 3 4 5 6 7 8 9
Shear force / kN 940.21 939.78 939.62 940.56 940.73 940.49 939.69 938.43 938.22
Along the bridge
Displacement / cm 4.87 4.87 4.86 4.90 4.90 4.90 4.89 4.84 4.84
Shear force / kN 873.87 873.54 873.76 873.44 861.16 844.86 843.41 762.70 753.41
Cross bridge
Displacement / cm 4.34 4.33 4.34 4.33 4.23 4.09 3.95 3.40 3.35
Shear force / kN 190.50 190.67 190.48 190.66 190.81 190.56 190.93 191.25 191.63
Displacement / cm 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.76
Shear force / kN 934.84 937.98 937.77 936.80 934.97 936.03 937.64 938.89 938.63
Combination one
Displacement/cm 4.97 4.99 5.00 5.01 4.98 4.98 4.96 4.94 4.91
Shear force / kN 852.10 851.64 855.63 852.19 837.90 822.17 815.99 745.03 741.08
Combination two
Displacement / cm 4.37 4.38 4.41 4.37 4.22 4.18 3.91 3.45 3.34
Shear force / kN 448.11 450.55 448.51 452.70 440.66 442.84 446.90 442.42 442.31
Combination three
Displacement / cm 1.88 1.84 1.84 1.81 1.88 1.85 1.85 1.86 1.86
Fig. 8Comparison of bearing shear force
a) Along the bridge and combination one
b) Cross-bridge direction and combination two
Fig. 9Comparison of bearing displacement
a) Along the bridge and combination one
b) Cross-bridge direction and combination two
See Figure 10 for the shear response time history of some supports. Hysteresis curve for some supports. See Fig. 11. From the analysis of Table 8 and Fig. 8 to Fig. 11, we can get:
(1) The maximum shear force of the seismic isolation support under the excitation of working conditions 1 to 7 is greater than that of combination 1, and the maximum shear force of the seismic
isolation support under the excitation of working conditions 8 and 9 is greater than the excitation of the forward bridge;
(2) Under the action of the transverse bridge direction and combination two, the maximum shear force of the seismic isolation support of working conditions 1 to 9 shows a decreasing trend, and the
linear direction is basically similar, and the transverse bridge excitation of each working condition is greater than the combination two excitation;
(3) The maximum displacement of each working condition is that the excitation of combination one is greater than the excitation along the bridge direction, and the cross-bridge direction and
combination two are basically the same, and there is a decreasing trend from working condition 1 to working condition 9;
Under the vertical excitation, the shear force and displacement of all working conditions are basically the same.
Fig. 10Time history response of bearing shear force
a) Working condition 5, along the bridge direction excitation
b) Working condition 1 cross-bridge excitation
Fig. 11Hysteresis curve
a) Condition 1 along the bridge direction excitation
b) Working condition 9 combination one incentive
7. Conclusions
Nine non-isolated and earthquake-isolated structural models under different cross-bracing arrangements were established, and Elcentro seismic waves were selected. The internal force, displacement,
velocity, absolute acceleration, relative acceleration, and separation of arch ribs of each model were compared and analyzed under uniform excitation along the bridge, transverse and vertical
directions, multi-dimensional combined excitation, and multipoint excitation considering the traveling wave effect.
Through the above comparative analysis, we can get:
1) The main internal force of the arch ribs of the seismic isolation structure in each working condition decreases significantly under the action of the bridge direction, the horizontal bridge
direction, the combination one, and the seismic waves with different wave speeds. Under the vertical earthquake action, the arch of the seismic isolation structure, the main internal force of the rib
increases. Under the action of the second combination earthquake, the axial force of the arch rib in each working condition of the seismic isolation structure decreases, the shear force ${F}_{z}$
increases, the bending moment ${M}_{z}$ working conditions eight and nine increase, and the rest decrease, and the combination three under the action of an earthquake, the main internal forces of the
seismic isolation structure arch ribs in various working conditions have increased;
2) Under the action of transverse bridge seismic waves, the arch ribs of non-seismic and isolation models mainly undergo lateral displacement. The “K”-shaped cross brace is better than the “-” cross
brace and the “meter” shape in reducing the lateral displacement of the arch rib. Transverse bracing, setting transverse bracing on the upper part of the arch rib can reduce the vertical displacement
of the arch rib of the non-seismic model;
3) Under the action of transverse seismic waves, the lateral velocity of the arch ribs of the non-seismic and isolation models basically increased, and the velocity of the arch ribs of the seismic
isolation structure under various working conditions decreased;
4) The “meter”-shaped cross brace at the top of the arch rib and the “K”-shaped cross brace at the lower part help reduce the lateral acceleration of the arch rib. The absolute acceleration and
relative acceleration of the arch rib of the seismic isolation structure under various working conditions are significantly reduced;
5) Under the action of the maximum shear force of the seismic isolation support in the transverse direction and the combination two, working conditions 1 to 9 show a decreasing trend, and the linear
directions are basically similar. In all conditions, the excitation of combination one is greater than the excitation along the bridge direction, and the cross-bridge direction and combination two
are basically the same, and there is a decreasing trend from working condition one to working condition nine.
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Edition), Vol. 39, No. 1, pp. 108–112, 2009.
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About this article
Seismic engineering and applications
transverse bracing
seismic isolation
through concrete-filled steel tube arch bridge
time history analysis
This research was funded by the Fundamental Research Funds for the Central Universities (31920210078), the National Natural Science Foundation of China (51868067).
Data Availability
The datasets generated during and/or analyzed during the current study are available from the corresponding author on reasonable request.
Conflict of interest
The authors declare that they have no conflict of interest.
Copyright © 2022 Zhonghu Gao, et al.
This is an open access article distributed under the
Creative Commons Attribution License
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That the worm falls off the end of the rope depends on the fact that the incredible
harmonic series
1 + 1/2 + 1/3 + 1/4 + . . .
diverges to infinity, growing as large as you please!
Standard Podcast [ 5:28 ] Play Now | Play in Popup | Download
Dana Richards, editor of Martin Gardner’s Colossal Book of Short Puzzles and Problems explains why the worm makes it, in only about 15,092,688,622,113,788,323,693,563,264,538,101,449,859,497 steps!
(Give or take a few.) This incredible fact depends on the mysterious Harmonic Series, discussed a little more in our next post.
Standard Podcast [ 6:10 ] Play Now | Play in Popup | Download
Dana Richards, editor of The Colossal Book of Short Puzzles and Problems discusses the amazing Martin Gardner and his legacy!
What follows after 0, 1, 2, … , once you’ve managed to list every counting number?
Around 1875, Georg Cantor created — or discovered if you like — the transfinite ordinals : the list continues 0, 1, 2, …, then ω , ω + 1, ω + 2, etc, for quite a long long way. John H. Conway tells
us about his Surreal Numbers , which add in such gems as
1 / √ ω
Check out Knuth’s Surreal Numbers, Conway & Guy’s Book of Numbers , or for more advanced users, Conway’s On Numbers and Games.
Standard Podcast [ 10:04 ] Play Now | Play in Popup | Download
Niclas Hedell, a listener, poses a problem from his days in the Swedish military: given two trees in the forest, and a rope twice as long as the distance between the trees, how do you find a third
tree so that all three make a right triangle.
And we explain how the Stork can catch the Frog.
Standard Podcast [ 8:36 ] Play Now | Play in Popup | Download
Amusingly, this problem has exactly the same solution as the proof that there are as many rational numbers as there are counting numbers. And the proof generalizes: one stork can catch three frogs,
or ten or fifty.
Here are some bonus problems:
1. The stork can catch the frog even if it can start at any rational number and hop any fixed rational distance each step.
2. However, if the frog can start at any real number or hop any real distance, the stork has no strategy that guarantees a catch. This is, in effect, the same as proving that the real numbers are
not countable.
A contestant for our Million-Dollar-Give-Away sent in Rayo’s Number, hitherto the largest number ever used for any real purpose: to wit, winning the
Check out the article by Scot Aaronson that inspired them to duke it out! And this thread on the math forum is quite interesting as well.
Standard Podcast [ 15:21 ] Play Now | Play in Popup | Download
With enough time and patience and bananas, can we go as far as we please?
[ 6:50 ] Play Now | Play in Popup | Download
EO. Spaghetti Loops
Follow Up: The Harmonic Series
EH. The Worm Makes It!
Permalink Comments
EG. The Colossal Book of Short Puzzles and Problems
Permalink Comments
CW. The Surreal Numbers
Permalink Comments
CK. The Third Tree
Follow-up: The Stork and The Frog
CH. Rayo’s Number!
BL. Eternally diminishing returns
Permalink Comments | {"url":"http://mathfactor.uark.edu/category/the-mathfactor-podcast/infinity/","timestamp":"2024-11-11T23:50:24Z","content_type":"application/xhtml+xml","content_length":"118055","record_id":"<urn:uuid:bb230022-ca68-4361-9170-4e059b3e53a8>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00731.warc.gz"} |
A Clique-Based Approach to the Identification of Common Gene Association Sub-Networks
A Clique-Based Approach to the Identification of Common Gene Association Sub-Networks ()
1. Introduction
Gene and gene products interact with each other due to biochemical interactions and regulatory activities [1]. Many methods have been developed to analyze these networks. Popular methods include
weighted correlation network analysis (WGCNA) [2], Bayesian networks [3], autoregressive models [4], state-space models [5] and graphical Gaussian models [6]. Few studies however, have been devoted
to analyze networks from multiple species simultaneously. In this study, we focus on the identification of common gene association sub-networks from multiple species. First we need to derive gene
network from individual species. We chose Graphical Lasso model [6] for this task because it can handle large covariance/correlation matrices of mathematically deficient rank which is often the case
for genomic data.
Identification of common gene association sub-networks is related to the subgraph isomorphism problem. The subgraph isomorphism problem can be reduced to finding maximal clique from merged graph [7]
which can be constructed following graph product rules. There are a number of heuristics for finding maximal cliques. Local search may be the simplest greedy strategy that starts with some initial
solution and moves from neighbor to neighbor as long as possible while increasing the clique number. The main problem with this strategy is its inability to escape local maxima where the search
cannot find any further neighborhood solution. Battiti and Protasi [8] proposed reactive local search that allows the search to explore solutions that do not decrease the clique number by dynamically
changing some of the parameters [8]. Another widely used heuristic is replicator equation [7]. This method is based on a continuous formulation of the maximal clique problem as quadratic programming
The paper is organized as follows. In Section 2, we will describe graphical Lasso method to construct gene association networks. We will also describe graph merging and how to find maximal cliques
using replicator equations. The experiments and results will be discussed in Sections 3 & 4. We offer a conclusion in Section 5.
2. Methods
To find common gene association sub-networks from two species, we need to perform ortholog mapping from two species. Orthologs are genes in different species that originated by vertical descent from
a single gene of the last common ancestor. We will then construct gene association network of the orthologous genes for the two species respectively. This is followed by graph merging and maximal
clique searching of the merged graph. Finally the common sub-networks are recovered for each species. Figure 1 shows the overview of the approach.
2.1. Construction of Gene Association Network
The inverse covariance matrix
Figure 1. Clique-based approach to identification of common gene association sub-network from two species.
if ^−^1 is the solution to the following optimization problem:
where tr denotes trace and [1] norm, and
This is a box-constrained quadratic program and can be solved using an interior-point procedure [12]. By permuting the rows and columns so the target column is always the last, they solve a similar
problem like (3) for each column, updating their estimate of
The solution for problem (3) is
2.2. Identification of Conserved Gene Association Sub-Networks
Detecting common sub-network is a challenging task. However, we are approaching this problem from a graph product point of view. We will merge the two graphs by mapping corresponding orthologous
genes and create the edges for merged graph
In other words, an edge in
We can avoid spurious solutions by substituting
The optimization problem
The simplex
To find all large local maximizers
For a local maximizer
3. Experimental Data
3.1. Graph Benchmark Data
We used the DIMACS benchmark data set [20] to validate the effectiveness of the replicator equations to find maximal cliques.
3.2. Gene Expression Data
We applied the method to find common stress responsive gene association networks for two related bacteria Deinococcus radiodurans and Thermus thermophilus. We downloaded the gene expression data sets
GSE 29516 for D. radiodurans from gene expression omnibus [21]. GSE29516 consists of microarray data from transcription profiling of D. radiodurans treated with 0.3 M NaCl or 2 M salt. We downloaded
the gene expression series GSE21289 for T. thermophilus. GSE21289 contains the gene expression data of T. thermophilus HB8 wild-type strain in response to high salt stress.
3.3. Ortholog Mapping
Ortholog mapping is done via multi-genome homology comparison tools available from the Comprehensive Microbial Resource web site [22]. In the case of multiple genes in a cluster, we used the one with
the highest score, resulting in 744 one-to-one ortholog pairs. Gene expression data and ortholog mapping table is stored in an inhouse relational database for easy retrieval and crossreferencing
across the two species Gene expression data for orthologous genes are retrieved through database queries.
4. Results and Discussions
4.1. Benchmark Results on DIMACS Challenge Sets
We recorded running time and clique found using our replicator equations. Table 1 shows the results on some DIMACS challenge instances.
In general, our implementation is able to find maximal cliques in reasonable time. And we were able to find cliques that are close to their corresponding maximum clique numbers for each benchmark
data set.
4.2. Identification Common Gene Association Sub-Networks
Two common gene association sub-networked were identified using the described procedure (Figure 2).
Annotations of the genes in Figure 2 are given in Table 2.
Table 1. The performance of clique finding algorithm on some DIMACS challenge instances.
Figure 2. Common gene association sub-networks identified for D. radiodurans and T. thermophilus.
Among these genes, ABC transporter ATP-binding protein has been previously reported to be osmo-regulated [23]. Two of the genes are not fully annotated (hypothetical protein in Table 2), we think
they are worthy
Table 2. Annotation of genes in the identified common subnet-networks.
of further investigation because they seem to be related to osmosis stress response based on our study on two different species. A ribosomal protein has been found by Schmalisch et al. to be general
stress protein in Bacillus subtilis [24]. We found three ribosomal proteins that are related to the stress response in both D. radiodurans and T. thermophilus. This is consistent with the finding
from Schmalisch et al. [24].
5. Conclusions
In this study, we developed an efficient computational framework that combines graphical lasso model, graph product and replicator equation based clique solver to identify common gene association
sub-network from multiple species. Our method provides an approach to identifying conserved pathway components.
We applied our method and identified common gene association sub-networks for two related bacterial species D. radiodurans and T. thermophilus subjected to similar environmental stress. We confirmed
some stress responsive genes with previous studies. Our method also demonstrated how these genes interact with other genes and these interactions potentially are conserved because they are discovered
via simultaneous study of two related species.
Our method is not limited to finding common gene association sub-network across multiple species. It can also be adapted to identify core interaction network for the same species subjected to
different environmental stresses. It can also be employed to identify common gene/protein sub-networks for related diseases such as diabetes and hypertension.
6. Acknowledgements
This work was supported by a grant from North Carolina Space Consortium (2010-1435-CE-02). Financial support from the National Science Foundation (NSF) through a grant to the Quality Education for
Minorities Networks (QEM/HBCU-UP Faculty Professional Development and Mentoring Program) are highly acknowledged. The work was also supported in part by NSF CREST HRD- 0833184 and by the US Army
Research Office (ARO) W911NF-0810510. | {"url":"https://www.scirp.org/journal/paperinformation?paperid=32501","timestamp":"2024-11-04T20:59:52Z","content_type":"application/xhtml+xml","content_length":"120941","record_id":"<urn:uuid:86ce8998-5efc-46a2-8c3c-4c04c1fada36>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00371.warc.gz"} |
PCA(Principal Component Analysis) on MNIST Dataset
This article was published as a part of the Data Science Blogathon.
Introduction to PCA
Hello Learners, Welcome!
In this article, we are going to learn about PCA and its implementation on the MNIST dataset. In this article, we are going to implement the Principal Component Analysis(PCA) technic on the MNIST
dataset from scratch. but before we apply PCA technic to the MNIST dataset, we will first learn what is PCA, the geometric interpretation of PCA, the mathematical formulation of PCA, and the
implementation of PCA on the MNIST dataset.
So the PCA is the technic of dimensionality reduction. dimensionality reduction is nothing but the reduction of n dimension data to n’ dimension data, where n > n’. there are several types of
datasets that have lots of features and this feature is nothing but the extent of data points or datasets. Also, in the dataset, several features have less impact on the final result and increase the
processing time of the machine learning models. and we humans can visualize the data only in 2D and 3D but we can’t imagine the higher dimensions. Solving this visualization and optimization problem,
there are lots of technics available in machine learning such as PCA, t-SNE, Random Forests, kernel PCA, Truncated SVD, etc.
So the dataset we are going to use in this article is called the MNIST dataset, which contains the information of handwritten digits 0 to 9. in this dataset the information of single-digit is stored
in the form of 784*1 array, where the single element of 784*1 array represents a single pixel of 28*28 image. here the value of single-pixel varies from 0 to 1, where the black colour is represented
by 1 and white by 0 and middle values represent the shades of grey.
Geometric Interpretation of PCA:
So basically the work of PCA is to reduce the dimensions of a given dataset. which means if we were given the dataset which has d-dimensional data then our task is to convert the data into
d’-dimensional data where d > d’. so for understanding the geometric interpretation of PCA we will take an example of a 2d dataset and convert it into 1d data set because we can’t imagine the data
more than 3d. but anything we learn from 2d interpretation, we can also do it to higher dimensions.
Now let’s take an example, Suppose we have a DxN dimensional dataset called X, where the d = 2 and n = 20. and the two features of the dataset is f1 and f1,
Now let’s see that we make the scatter plot with this data and its data distribution is look like the figure shown below,
After seeing the scatter plot, you can easily say that the variance of feature f1 is much more than the variance of feature f2. The variability of f2 is unimportant compared to the variability of f1.
if we have to choose one feature between f1and f1, we can easily select the feature f1. now let’s suppose that you cannot visualize 2d data and for visualizing the data you have to convert your 2d
data into 1d data then what do you do? so the simple answer is you directly keep those features that have the highest variance. and remove those features which have less impact on the overall result.
and that’s what PCA internally does.
So first of all we ensure that our data is standardized because performing the PCA on standardized data becomes much easier than original data.
So now again let’s see that we have a d*n dimensional dataset called X, where the d = 2 and n = 20. and the two features of the dataset are f1 and f2. and remember we standardized the data. but in
this case, the scatter plot looks like this.
In this case, if we have to decrease dimensions from 2d to 1d then we can’t clearly select feature f1 or f2 because this time the variance of both features is almost the same both the features seem
important. so how does PCA do it?
In this situation, PCA tries to draw the vector of line in the direction where the variance of data is very high. which means instead of projecting the data or measuring the variance in the f1 or f1
axis what if we quantify the variance in the f1′ or f2′ direction because measuring the variance in the f1′ or f2′ direction makes much more sense.
So PCA tries to find the direction of vector or line where the variance of data is very high. the direction of vector where the variance of data is highest is called PC1 ( Principal Component 1 ) and
second-highest is called PC2 and third is PC3 and so on.
Mathematical Formulation of PCA:
So we show the geometric intuition of PCA, we show that how does PCA reduces the dimensions of data. so PCA simply finds the direction and draws the vector where the variance of data is very high,
but you might wonder how the PCA does it and how it finds the right direction of vector where the variance of data is very high. how the PCA calculates the angle and gives us the accurate slope. so
PCA uses two techniques to find the angle of a vector. the two methods are Variance maximization and Distance Minimization. so let’s learn about them in brief
1. Variance Maximization: In this method, we simply project all the data points on the unit vector u1 and find the variance of all projected data points. We select that direction where the variance
of projected points is maximum.
So let’s assume that we have two-dimensional datasets and the features of the dataset are f1 and f2, and xi is datapoint and u1 is our unit vector. and if we project the data point xi on u1 the
projected point is xi’,
u1 = unit vector
|| u1 || = 1 (length of unit vector)
f1 and f2 = features of dataset
xi = data point
xi’ = projection of xi on u1
now assume that D = { xi } (1 to n) is our dataset
and D’ = { xi’ } (1 to n) is our dataset of projected point of xi on u1.
xi’ = (u1 * xi)/||u1||
assumption that u1 is unit vector so, length of unit vector ||u1|| = 1
=> xi’ = u1 * xi
=> xi’ = u1T * xi …….(1)
now x^’ = u1T * x^ ……..(2) [ x^ = mean of x ]
so find u1 such that the variance{ projection of xi on u1 } is maximum
var {u1T * xi} (i is 1 to n)
if data is columns standardized then mean = 0 and variance = 1
so x^ = [0, 0, 0… .. . . . .0]
=> u1T * x^ = 0
we want to maximize the variance.
2. Distance Minimization: So in this technique of PCA we are trying to minimize the distance of data point from u1 ( unit vector of length 1)
|| xi ||2 = di2 + (u1T * xi)2 [ Pythagoras theorem ]
di2 = || xi ||2 – (u1T * xi)2
=> di2 = xiT * xi – (u1T * xi )2
we want to minimize the sum of all distance squared.
Implementing PCA on MNIST dataset:
So as we talked about the MNIST dataset earlier and we just complete our understanding of PCA so it is the best time to perform the dimensionality reduction technique PCA on the MNIST dataset and the
implementation will be from scratch so without wasting any more time lets start it,
So first of all we import our mandatory python libraries which are required for the implementation of PCA.
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
Now let’s load our MNIST dataset from our computer which is stored in .csv format. we only imported 20000k rows for simplicity you can download the MNIST dataset from this link: https://
df = pd.read_csv('mnist_train.csv', nrows = 20000)
print("the shape of data is :", df.shape)
Hit Run to see the output
import numpy as np
import pandas as pd
# import matplotlib.pyplot as plt
# import seaborn as sns
df = pd.read_csv('train.csv', nrows = 20000)
print("the shape of data is :", df.shape)
Extracting label column from the dataset
label = df['label']
df.drop('label', axis = 1, inplace = True)
ind = np.random.randint(0, 20000)
plt.figure(figsize = (20, 5))
grid_data = np.array(df.iloc[ind]).reshape(28,28)
plt.imshow(grid_data, interpolation = None, cmap = 'gray')
Plotting a random sample data point from The dataset using matplotlib imshow() method
Column standardization of our dataset using StandardScalar class of sklearn.preprocessing module. because after column standardization of our data the mean of every feature becomes 0 (zero) and
variance 1. so we perform PCA from the origin point.
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
std_df = scaler.fit_transform(df)
Now Find the Co-Variance matrix which is AT * A using NumPy matmul method. after multiplication, the dimensions of our Co-Variance matrix is 784 * 784 because AT(784 * 20000) * A(20000 * 784).
covar_mat = np.matmul(std_df.T, std_df)
Finding the top two Eigen-values and corresponding eigenvectors for projecting onto a 2D surface. The parameter ‘eigvals’ is defined (low value to high value), the eigh function will return the
eigenvalues in ascending order and this code generates only the top 2 (782 and 783) eigenvalues.
converting the eigenvectors into (2,d) form for easiness of further computations
from scipy.linalg import eigh
values, vectors = eigh(covar_mat, eigvals = (782, 783))
print("Dimensions of Eigen vector:", vectors.shape)
vectors = vectors.T
print("Dimensions of Eigen vector:", vectors.shape)
here the vectors[1] represent the eigenvector corresponding 1st principal eigenvector
here the vectors[0] represent the eigenvector corresponding 2nd principal eigenvector
If we multiply the two top vectors to the Co-Variance matrix, we found our two principal components PC1 and PC2.
final_df = np.matmul(vectors, std_df.T)
print("vectros:", vectors.shape, "n", "std_df:", std_df.T.shape, "n", "final_df:", final_df.shape)
Now we vertically stack our final_df and label and then Transpose them, then we found the NumPy data table so with the help of pd.DataFrame we create the data frame of our two components with class
final_dfT = np.vstack((final_df, label)).T
dataFrame = pd.DataFrame(final_dfT, columns = ['pca_1', 'pca_2', 'label'])
Now let’s visualize the final data with help of the seaborn FacetGrid method.
sns.FacetGrid(dataFrame, hue = 'label', size = 8)
.map(sns.scatterplot, 'pca_1', 'pca_2')
So you can see that we are successfully converted our 20000*785 data to 20000*3 using PCA. So this is how PCA is used to convert big extent to smaller ones.
What do we learn in this article? We took a brief intro about the PCA and mathematical intuition of PCA. This was all from me thank you for reading this article. I am currently pursuing a b.tech in
CSE I loved to write articles in data Science. Hope you like this article.
Thank you.
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Repeating/Terminating Decimals Calculator - Online Fraction Converter
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Repeating Decimals
Tool to find the period of a fraction or a decimal number with repeating decimals. The period is a set of digits that is repeated at infinity in the decimals of the number (usually a rational number
or a periodic fraction).
Repeating Decimals - dCode
Tag(s) : Arithmetics
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Repeating Decimals
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Fraction Finder
Answers to Questions (FAQ)
What are repeating decimals in a fraction? (Definition)
The periodic decimal expansion/development of a rational number or a fraction (numerator over denominator) is the sequence of numbers that are repeated at infinity in the decimal writing of the
Example: 1/3 = 0.3333333333… The digit 3 is repeated to infinity
Example: 1/27 = 0.037037037037037… The digits 037 are repeated to infinity
All fractions do not have a repeating decimal form, some have a terminating decimal form.
What are terminating decimals in a fraction? (Definition)
A terminating decimal indicates that no sequence of numbers repeats infinitely in the decimal writing of the number.
Example: 4/25 = 0.16 the development is finished and does not continue
Any number that is written in decimal form with a finite number of digits (after the decimal dot) has is a terminating decimal number.
How to write repeating decimals?
Multiple notations are possible.
The first uses … points of suspension, but does not define the part that repeats. It is practical but not rigorous and therefore not recommended.
Example: $ 37/300 = 0.12333333333\dots $
Notation with a bar above the repeated part.
Example: $ 37/300 = 0.12\overline{3} $
Notation with a bar below the repeated part.
Example: $ 37/300 = 0.12\underline{3} $
Notation between brackets
Example: $ 37/300 = 0.12[3] $
NB: For clarity, it is better to write the fraction in an irreducible form.
How to find decimals from a fraction?
Divide the numerator by the denominator. Lay the Euclidean division by hand or use a calculator.
How to find the fraction from decimals?
Take $ x $ a number, and $ n $ the size (the number of digits) of the periodic part of the decimal expansion. To get a fractional writing, solve $ x \times 10^n - x $.
Example: $ x = 0.1\overline{6} = 0.1666666\dots $, the repeating portion is $ 6 $, a single digit so $ n=1 $. Calculate $ 10^1 \times x = 1.\overline{6} = 1.6666666\dots $ and solve $ 10x−x = 9x = 1.
\overline{6}−0.1\overline{6}=1.5 \iff 9x=1.5 \iff x=1.5/9 = 15/90 = 1/6 $
What are the most known decimal developments?
The inverses of prime numbers provide long and interesting periodic decimal developments.
Example: $ 1/3 = 0.333333\dots $
Example: $ 1/7 = 0.142857142857\dots $
Is there an infinite decimal expansion with a series of digits that never repeats?
Any rational number (any fraction) has a finite developpement or a periodic decimal expansion with a finite number of digits that repeat themselves ad infinitum.
But there are real numbers that are not rational numbers (which are not fractions) which have decimals without repetition
Example: $ \pi = 3.14159265\dots $ has no known repetition to date.
Example: Champernowne's constant will never have any repetition, it is a universe number.
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Quantum Technologies for Fundamental Physics
Via Guarnotta, 26 - 91016 ERICE (Sicily) - Italy The conference will take place at the San Domenico Institute in the Dirac Lecture Hall: Piazzetta S. Domenico, 1, 91016 Erice TP: ttps://goo.gl/maps/
The event will take place at EMFCSC 's premises in Ericehttp://www.ccsem.infn.it
The workshop will start each morning at 9:00am and will take place in the San Domenico Institute in the Dirac Lecture Hall: Piazzetta S. Domenico, 1, 91016 Erice TP: https://goo.gl/maps/
The Poster session will take place on Saturday starting from 4:30pm at San Rocco. | {"url":"https://agenda.infn.it/event/34296/timetable/?view=standard_numbered","timestamp":"2024-11-04T01:43:02Z","content_type":"text/html","content_length":"321507","record_id":"<urn:uuid:8f726be0-55f6-42f5-9441-500f4d228d3c>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00280.warc.gz"} |
union t1 t2 makes the class of t1 and the class of t2 be the same (if they are already equal, then nothing changes). The value of the combined class is the value of t1 or t2; it is unspecified which.
After union t1 t2, it will always be the case that same_class t1 t2. | {"url":"https://ocaml.janestreet.com/ocaml-core/113.33/doc/core/Core/Std.mod/Union_find.mod/","timestamp":"2024-11-11T03:17:16Z","content_type":"text/html","content_length":"4532","record_id":"<urn:uuid:e6035fb6-ddc8-4a92-8d5f-9d4c2761d954>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00637.warc.gz"} |
Maple, Geometer's Sketchpad and MatLab Related Sources
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Maple Related Sources (from Stat/Math Center, The Indiana University)
Maple Application Center (from Waterloo Maple Software Inc)
Maple Worksheets (Saint Louis University)
Cryptography and coding theory with MAPLE page
Maple bilingual Web Sites (University of Texas, San Antonio)
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Maple Advisor Database (by Prof. Robert B. Israel at University of British Columbia)
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Educational Material: Computer Algebra Software-CAIN Europe
Maple Worksheets on Differential Geometry and Multivariable Calculus
Maple Web Sites
Computer - Algebra in Baden-Wurttemberg Projekt MathCom
Mathematik Mit Maple im World Wide Web
ACE-An Algebraic Combinatorics Environment for the computer system MAPLE
AIM(Alice Interactive Mathematics)- a web-based system designed to administer graded tests with mathematical content.
Geometer's Sketchpad Related Sources
Geometer's Sketchpad Related Resources (from Swarthmore)
Geometer's Sketchpad Resource Center (Technical Support, On Line Activity Guide, JavaSketchpad DR3, Bibliography)
Cabri ¿Í GSPÀÇ ºñ±³ (¼öÇлç¶û¿¡¼ Á¦°øÇÑ ÀÚ·á)
MatLab Related Sources
MatLab Related Sources (from Stat/Math Center, The Indiana University)
Last updated : 5/13/99 | {"url":"http://mathedu.or.kr/source/Library/MSM.htm","timestamp":"2024-11-11T18:18:08Z","content_type":"text/html","content_length":"8907","record_id":"<urn:uuid:c04ce060-0ca1-4f49-b3d8-406ef32e669f>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00100.warc.gz"} |
How do you simplify and write (6.5times10^3)(5.2times10^-8) in standard form? | HIX Tutor
How do you simplify and write #(6.5times10^3)(5.2times10^-8)# in standard form?
Answer 1
To simplify and write ((6.5 \times 10^3)(5.2 \times 10^{-8})) in standard form, you multiply the coefficients and add the exponents of the powers of 10.
(6.5 \times 10^3) is (6,500) and (5.2 \times 10^{-8}) is (0.000000052).
Multiplying (6,500) by (0.000000052) results in (0.000338).
Therefore, ((6.5 \times 10^3)(5.2 \times 10^{-8})) simplifies to (0.000338).
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Answer 2
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Answer 3
To simplify and write (6.5 × 10^3)(5.2 × 10^-8) in standard form, you first multiply the numerical parts and then add the exponents of 10.
(6.5 × 10^3)(5.2 × 10^-8) = (6.5 × 5.2) × 10^(3 + (-8))
= 33.8 × 10^(-5)
Therefore, in standard form, (6.5 × 10^3)(5.2 × 10^-8) = 3.38 × 10^(-4).
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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Practical Calculation (3h and 30 min Running Time)
Practical Calculation
We are excited to introduce the latest course by Grandmaster Renato Quintiliano, titled "Practical Calculation." Unlike traditional courses that focus on opening repertoires, this unique offering
dives into the art and science of calculation in chess, a crucial skill for tournament players and enthusiasts alike.
Course Structure: The course is meticulously divided into five comprehensive sections, each designed to enhance your calculation skills through structured lessons and practical exercises:
1. Candidate Moves:
□ Introduction: Understand the importance of identifying potential moves to consider in various positions.
□ Test Positions & Solutions: 7 challenging positions to test your skills, followed by detailed annotated solutions.
2. Improving Variations:
□ Introduction: Learn how to refine and optimize your lines of thought during calculation.
□ Test Positions & Solutions: 7 positions to practice improving variations, complete with in-depth explanations.
3. Calculating Long Lines:
□ Introduction: Master the technique of visualizing and calculating extended sequences of moves.
□ Test Positions & Solutions: 7 long-line scenarios to solve, with thorough analysis provided.
4. Calculating in the Endgame:
□ Introduction: Focus on the critical endgame phase, where precise calculation can make or break the game.
□ Test Positions & Solutions: 7 endgame positions to hone your skills, accompanied by comprehensive solutions.
5. Studies:
□ Introduction: Explore composed positions that highlight specific themes and challenge your calculation abilities.
□ Test Positions & Solutions: 7 studies to test and improve your calculation, with detailed annotations.
In addition to the theoretical and practical content, the course also features a Video Version with a running time of 3 hours and 30 minutes. This video complements the written material by providing
visual explanations and insights directly from GM Renato Quintiliano.
Why This Course?
Calculation is often the deciding factor in chess, separating good players from great ones. This course aims to provide practical insights and examples that demonstrate how calculation works in real
games. GM Renato Quintiliano shares valuable experiences from his own tournament practice, emphasizing the non-linear and complex nature of chess calculation. By engaging with this course, you will
not only improve your calculation techniques but also gain a deeper understanding of the thought processes behind them.
The course is designed to be interactive and engaging, with critical moments highlighted in red as additional exercises. These marked positions serve as focal points for practice, allowing you to
test and measure your progress.
Introduction by GM Renato Quintiliano
For the first time, I've decided to make a database that is not focused on an opening repertoire. Instead, I wanted to talk about something that could be more useful and effective for everyone,
especially tournament players. So, calculation technique seemed a logical subject to exploit. This database is divided into five sections: Candidate moves, Improving variations, Calculating long
lines, Calculating in the endgame, and Studies. Although all these topics have been well examined by many books and better players/trainers than me, I still wanted to contribute by sharing my view
and some interesting examples from my tournament practice.
Therefore, instead of selecting my best tactical games, my idea is to show how calculation works in practice, in a real game. It's never so clear-cut as textbook examples; our opponents often have a
good continuation, both players miss resources along the lines, etc... but so is chess: a difficult game, in which you almost always only win with a little help from the opponent. Nevertheless, the
examples selected still look good enough - at least I hope - and I tried to explain the thinking process in the best way possible, as I think this can be the most useful thing for improving players.
For practical purposes, I marked some positions in the examples as red. These are critical moments that can be used as additional exercises if you want.
I also offer this first diagram as a warm-up for the upcoming positions.
White to move
If you are not able to find the best sequence here, I hope that after studying the database you can return to this position and make progress. Then we will know if the database is really helpful! | {"url":"https://www.modern-chess.com/practical-calculation-3h-and-30-min-running-time-483","timestamp":"2024-11-07T17:03:12Z","content_type":"text/html","content_length":"43209","record_id":"<urn:uuid:e3eb982f-7639-44f8-83cf-d9c57d711ab1>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00237.warc.gz"} |
Multiplication Chart 1 72 2024 - Multiplication Chart Printable
Multiplication Chart 1 72
Multiplication Chart 1 72 – If you are looking for a fun way to teach your child the multiplication facts, you can get a blank Multiplication Chart. This will likely enable your little one to fill
out the information by themselves. You will discover empty multiplication graphs for many different product can vary, such as 1-9, 10-12, and 15 merchandise. You can add a Game to it if you want to
make your chart more exciting. Here are several tips to obtain your kid started off: Multiplication Chart 1 72.
Multiplication Graphs
You should use multiplication maps as part of your child’s student binder to enable them to memorize math specifics. While many children can memorize their mathematics specifics normally, it will
require many others time to accomplish this. Multiplication graphs are an effective way to strengthen their boost and learning their self confidence. As well as being instructional, these maps can be
laminated for additional toughness. Allow me to share some beneficial methods to use multiplication charts. You may also have a look at these web sites for valuable multiplication fact resources.
This training includes the basic principles from the multiplication dinner table. In addition to discovering the principles for multiplying, individuals will fully grasp the very idea of elements and
patterning. By understanding how the factors work, students will be able to recall basic facts like five times four. They is likewise able to use the house of one and zero to eliminate more advanced
products. Students should be able to recognize patterns in multiplication chart 1, by the end of the lesson.
In addition to the normal multiplication chart, students might need to build a chart with increased variables or less variables. To create a multiplication graph or chart with a lot more elements,
pupils must make 12 furniture, every with twelve rows and 3 posts. All 12 desks have to match using one sheet of document. Lines should be attracted with a ruler. Graph document is right for this
project. Students can use spreadsheet programs to make their own tables if graph paper is not an option.
Online game suggestions
If you are instructing a newbie multiplication lesson or concentrating on the mastery of the multiplication dinner table, you are able to think of fun and interesting game ideas for Multiplication
Chart 1. Several enjoyable suggestions are listed below. This video game requires the individuals to be work and pairs on the same issue. Then, they are going to all hold up their charge cards and
explore the perfect solution for a minute. If they get it right, they win!
When you’re training children about multiplication, among the best instruments you can allow them to have is a computer multiplication graph or chart. These printable bedding arrive in a range of
styles and might be published using one web page or several. Little ones can understand their multiplication facts by copying them from your memorizing and chart them. A multiplication chart may help
for a lot of good reasons, from helping them find out their math information to instructing them utilizing a calculator.
Gallery of Multiplication Chart 1 72
Multiplication Chart 0 12 Pdf PrintableMultiplication
Multiplication Tables Free Printable Multiplication Multiplication
72 Times Tables Chart Times Tables Worksheets
Leave a Comment | {"url":"https://www.multiplicationchartprintable.com/multiplication-chart-1-72/","timestamp":"2024-11-13T22:18:12Z","content_type":"text/html","content_length":"52185","record_id":"<urn:uuid:bd4547b0-fece-4a1e-811f-63e799180070>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00372.warc.gz"} |
NEC Learning Management Systems
National Engineering College
(An Autonomous Institution Affiliated to Anna University, Chennai), Kovilpatti
M.E. – Computer Science and Engineering CURRICULUM & SYLLABUS Regulations 2019
19CT17E DEEP LEARNING L T P C QP 3 0 0 3 A
COURSE OUTCOMES Upon completion of this course, the student will be able to
CO1: Understand the basis of Machine Learning (K2)
CO2: Explore various Deep Learning Networks (K2)
CO3: Implement Convolutional and Recurrent Neural Algorithms (K3)
CO4: Analyze optimization and generalization in deep learning (K4)
CO5: Explore the deep learning applications (K3)
Introduction to machine learning - Linear models (SVMs and Perceptrons, logistic regression). Learning Algorithms – Capacity, Overfitting and underfitting – Hyperparameters and Validation Sets –
Estimators, Bias and Variance – Maximum Likelihood Estimation – Bayesian Statistics – Supervised Learning Algorithms – Unsupervised Learning Algorithms – Stochastic Gradient Descent – Building a
Machine Learning Algorithm – Challenges Motivating deep learning.
UNIT II DEEP NETWORKS 9
History of Deep Learning- A Probabilistic Theory of Deep Learning- Backpropagation and other Differentiation Algorithms – Regularization: Dataset Augumentation – Noise Robustness -Early Stopping,
Bagging and Dropout - batch normalization- VC Dimension and Neural Nets-Deep Vs Shallow Networks- Convolutional Networks- Generative Adversarial Networks (GAN), Semisupervised Learning
Convolutional Neural Networks: The Convolution Operation – Motivation – Pooling – Variants of the basic Convolution Function – Structured Outputs – Data Types – Efficient Convolution Algorithms.
Recurrent Neural Networks: Bidirectional RNNs – Deep Recurrent Networks – Recursive Neural Networks.
Optimization in deep learning– Non-convex optimization for deep networks- Stochastic Optimization- Generalization in neural networks- Spatial Transformer Networks- Recurrent networks, LSTM -
Recurrent Neural Network Language Models- Word-Level RNNs & Deep Reinforcement Learning - Computational & Artificial Neuroscience
Imagenet- Object Detection – Object Tracking - Audio WaveNet - Natural Language Processing Word2Vec - Joint Detection - Face Recognition - Scene Understanding - Gathering Image Captions. L:45; TOTAL:
45 PERIODS REFERENCES 1. Bengio, Yoshua, Ian J. Goodfellow, and Aaron Courville. "Deep learning." An MIT Press book in preparation,2016. 2. Dr.Adrian Rosebrock, ―Deep Learning for Computer Vision
with Python: Starter Bundle‖, PyImage Search, 1st edition, 2017. 3. Deng & Yu, Deep Learning: Methods and Applications, Now Publishers, 2013. 4. Michael Nielsen, Neural Networks and Deep Learning,
Determination Press, 2015. | {"url":"https://moodle.nec.edu.in/","timestamp":"2024-11-12T23:39:36Z","content_type":"text/html","content_length":"1049811","record_id":"<urn:uuid:46dc9a1c-6875-4b06-a337-81fe2afff59e>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00146.warc.gz"} |
PROG-A-THON March 2021 solution sketches - via
PROG-A-THON March 2021 solution sketches
Competition Committee - March 8, 2021
We hope you had fun during the contest! On this page, you can find the solution sketches to the problems in the PROG-A-THON. You can also download the testdata that we used to judge your solutions
here. So if you want to find out which testcase your solution failed on, you can test that out yourself with the data. If you still have any questions about the solutions to these problems or you
want to know for example why your solution got a Wrong Answer or Time Limit Exceeded verdict, you can of course always send us an email at competitie@svia.nl.
1: BBQ playlist
Testdata: download
You were given $n$ song lengths and a crossfade of $c$ and you should determine how long it takes to listen to the complete playlist.
The first step is to convert the m:ss input format into just seconds. Then we subtract (n-1)*c seconds for the crossfades. Finally, we convert the seonds back to the hh:mm:ss format.
You should only do the conversion back to minutes at the latest step. Otherwise, 1:00 minus 10 seconds might become 1:-10 instead of 0:50!
2: Fun with fractions
Testdata: download
To solve this problem, suppose you are given a fraction in the form x/y. Then you can simply convert it to a mixed fraction by calculating z = (x // y) and w = (x % y). The answer is then equal to z
3: Archery
Testdata: download
Per testcase, for every (x,y) one should find the smallest p such that x*x + y*y <= 20*20*p*p holds. This can be done naively in a for/while loop, as well as with integer division (both are fast
enough). Add 11 - p to the score counter and print it at the end of each test case, but only if p < 11.
Pseudocode to calculate points for a shot at (x,y) python for p in 1..11: if x*x + y*y <= 20 * 20 * p * p: points += 11 - p break
4: Hospital routing software
Testdata: download
The problem here was to find the shortest path from A to B in a 'half-circular grid'. One way you could do this (that would be within the time limit) was to simply run Dijkstra.
However, if you think a little bit more about the problem, there is a shorter, faster solution possible. The first observation is that every shortesdt route consists of at most three parts: first,
walk towards the center, then walk along a circular arc and finally walk away from the center.
So you could also loop over all the relevant circle arcs and find the shortest path among them. In fact, because you are optimizing something linear, the shortest path is always either the outermost
or the innermost route.
Example pseudocode:
def route(x, y, r, x1, y1, x2, y2): center_route_length = (y1 + y2) * (r / y) inner_route_length = abs(y1 - y2) * (r / y) + \ abs(x1 - x2) * (pi * min(y1, y2) * (r / y) / x) return min
(center_route_length, inner_route_length)
Complexity: O(n), O(1)
5: Cheating the system
Testdata: download
For each Program Committee simulate how many votes are needed to achieve a majority. This can be done by taking one vote at a time from the currently highest voted party and adding it to the party of
via. This can (but doesn't nessecaryily have to) be speed up by taking enough votes at a time from the currently highest voted parties until they are equal to the next highest voted party.
Then, (greedily) take the districts that need the fewest votes until party via has won the majority of Program Committees. Print the sum of the votes of the districts you have picked.
6: Trampoline
Testdata: download
The general idea to solving this problem is to convert the problem into a graph. You create a node for every trampoline and the start and end point. You can easily calculate the shortest path between
any two of these nodes: either walk directly from start to end, or (if you start at a trampoline) use the trampoline and walk the remaining distance (always launch in the same direction as where you
are headed).
On the resulting graph you can run a shortest path algorithm such as Dijkstra to determine the shortest path.
Complexity: O(n^2 log(n))
7: Movie maker
Testdata: download
Suppose that you want to have item $i$ lexicographically at the $p_i$th place by value. Compose every name from three parts:
'1' + i + '0' * p_i
Where the number i is padded to a fixed width. '0' * p_i means that we pad the number with p_i zeroes.
8: Perseverance's traveled path
Testdata: download
Problem: given an path of n vertices v_i and arrival times a_i, and a sampling time interval t, find the path recorded by the navigation system and compute the relative loss of distance.
The first position of the estimated path is always v_1 at time 0. Then for each time interval, find the position on the actual path for the next vertex. Compute the distance for both paths and
compute the difference.
Example pseudocode:
``` s_real = 0 s_record = 0 p = v_1
for g in 0,t,2t,...,a_n Find the next vertex v_i having g < a_i. Compute p1, the position at time g, by interpolating between v_{i-1} and v_i. s_record = s_record + ||p-p1|| p = p1
Print percentage (s_real - s_record) / s_real ```
9: Meshed security
Testdata: download
Implement your favorite path finding algorithm from top to bottom, implementing a system to prevent passing sensors (or rather; circles). Then, one by one add the sensors, until you can no longer
reach the top.
10: A Farewell to the Table Football Table
Testdata: download
The problem is to find a subset of a given sequence of positive integers a_1 ... a_n such that a_i >= 2a_{i-1} that sums to t. Thus we know that a_i > sum(a_1 ... a_{i-1}). Therefore the
solution must use the largest a_i such that a_i <= t. To solve this problem we must substract greedily the largest number that is smaller or equal to t, until t = 0. If this turns out to be
impossible we print 0 and exit, otherwhise we print the names that correspond with the chosen numbers.
It is important to note that a_i can be a very large number. Therefore a language must be used that has big integers or you should implement big integers yourself. Surprisingly, for this reason it is
the easiest to solve the problem in python, because python supports big integers out-of-the-box.
11: Fairy Park
Testdata: download
The problem is the following: given a graph G = (V, E), per node a time tv and a cost cv , and a time t∗for all edges. You are traversing the graph starting at 1 and at every node you pass you have
stay kt_v time and spend kc_v money with k ≥ 1. What is the least amount of money you need to spend such that you arrive at 1 after X time?
The problem is a variant of Knapsack. The time is the “weight”, X is the size of the bucket, and cv is the “gain” of an edge. You should extend the standard Knapsack-DP to handle the graph structure.
Use DP over the current node v and the time t you can move freely from e. dp(1, X - t_1) + c_v.
12: Completing the puzzle
Testdata: download
The problem here is to assemble a puzzle consisting of pieces that all have unique edges. A possible approach you could take is the following:
• Choose an arbitrary puzzle piece and place it at the origin (0,0)
• Perform a breadth-first search from this piece and fix the coordinates and orientation of the other puzzle pieces as is needed to fit these pieces.
• Check that this solution is indeed valid (i.e. there are no gaps, there is no overlap, the puzzle is rectangular)
Note that each step can be done in at most linear time.
Complexity: O(n)
13: Memory management
Testdata: download
The general idea of this problem is to find, given a sequence of numbers $a_1, \dots, a_n$, a contiguous subsequence that maximizes the greatest common divisor of that subsequence multiplied by the
length of the subsequence.
Because of the bounds, a quadratic solution that simply calculates this number for every contiguous subsequence is too slow.
Enumerate the right boundary j from left to right. Store the smallest left boundaries i where i <= j that give a distinct GCD and calculate the maximum.
14: Climbing wall
Testdata: download
Problem: scale a climbing wall without running out of stamina
We can observe that the required stamina is at most h * w * 9 = 5625. Solution outline: * Create a set of graph nodes from all reachable holds on the wall. * Copy every node s times to count the
remaining stamina at that hold. * For an edge from hold u to v taking t amount of stamina, insert an edge from every copy of node u to every node v with exactly t less stamina remaining. * Use
Dijkstra's algorithm to find the shortest path from the bottommost hold to the topmost hold, and compute the total distance of the path.
15: Ordering student IDs
Testdata: download
The essence of this problem comes down to finding indices $i < j$ that describe a minimal-length substring of the IDs such that the order with respect to the substring is identical to the original
Note first of all that if any two subsequent strings $s_k$ and $s_{k+1}$ are ordered correctly w.r.t. substring $(i,j)$, then all strings are ordered correctly.
For each index i, we compute $\alpha_{ki}$, the smallest index such that $s_{k}$ and $s_{k+1}$ are sorted correctly w.r.t the interval $(i, \alpha_{ki})$. We can determine all $\alpha_{ki}$ for two
subsequent strings in O(l) if we use dynamic programming.
We then determine for each index $i$ the maximum $\alpha_{ki}$ over all $k$ and we finally output the sortest interval among all $(i, max_k {\alpha_{ki}})$
Complexity: O(n*l)
16: Mirrors
Testdata: download
The problem here is to find out how many orientations there are in which we can send a light particle so that it reaches the end point (possibly making use of the one-time mirrors).
Note that because a mirror disappears after reflecting a particle, there can be only one solution for every subset permutation. Because N is smaller or equal than 8, we can iterate over all subsets
and permutations.
For a subset permutation, we can mirror the plane and its elements for each mirror to model the reflected world. Then finally we can simply find the angle of the line between start and end.
Furthermore, you need to ensure that you handle corners correctly: if the particle would travel too close to a corner, you should ignore it.
Complexity: O(N^2 * N!) | {"url":"https://svia.nl/progathon/2021/solutions","timestamp":"2024-11-07T00:17:57Z","content_type":"text/html","content_length":"51949","record_id":"<urn:uuid:ee109d56-610d-4b86-b07f-1af99ad5d4e1>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00107.warc.gz"} |
The total value of your portfolio is $60,000 and the portfolio
has three stocks, A, B...
The total value of your portfolio is $60,000 and the portfolio has three stocks, A, B...
The total value of your portfolio is $60,000 and the portfolio has three stocks, A, B and C, with betas of 1, 1.3 and .8, respectively. And, you have invested in $16,000, $24,000 on stocks A and B,
and the rest on C. What would be the beta of your portfolio?
Multiple Choice
pORTFOLIO bETA = Weighted Avg Beta of securities in that portfolio.
Security Beta Investment Weight Wtd Beta
A 1.00 $ 16,000.00 0.2667 0.267
B 1.30 $ 24,000.00 0.4000 0.520
C 0.80 $ 20,000.00 0.3333 0.267
Portfolio Beta 1.053
OPtion B is correct | {"url":"https://justaaa.com/finance/1307157-total-value-of-your-portfolio-is-60000-and","timestamp":"2024-11-04T11:13:40Z","content_type":"text/html","content_length":"41694","record_id":"<urn:uuid:0b540955-a386-43b0-9ee4-7f56eb85c6d8>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00418.warc.gz"} |
How to Calculate Applied Force: A Comprehensive Guide - The Tech EdvocateHow to Calculate Applied Force: A Comprehensive Guide
How to Calculate Applied Force: A Comprehensive Guide
If you’ve ever wondered how to calculate the applied force on an object, you’re in luck. In this article, we will provide a step-by-step guide on how to determine the applied force, making use of
fundamental physics concepts like Newton’s Laws of Motion.
Requirements to Calculate Applied Force
Before diving into the calculations, it’s essential to first understand the basic principles of applied force. Applied force can be defined as the external force exerted on an object that causes it
to move or change its state of motion. To calculate applied force, you’ll need to be familiar with several concepts:
1. Newton’s Second Law of Motion: This law states that the net force acting on an object is equal to its mass multiplied by its acceleration (F = ma).
2. Mass: The mass (m) of an object is a measurement of how much matter it contains.
3. Acceleration: Acceleration (a) is a measure of how quickly an object’s velocity changes over time.
4. Gravity: The force of gravity acting on an object can also affect the applied force calculations.
Step-By-Step Guide to Calculate Applied Force
Now that you have a basic understanding of the concepts involved, follow these steps to calculate the applied force:
1. Identify the mass and acceleration: First, determine the mass and acceleration values for your problem. Mass is typically given in kilograms (kg), while acceleration should be in meters per second
squared (m/s²).
2. Determine any additional forces: Sometimes, other forces such as gravity, friction, or tension come into play when calculating applied force. If there are any other forces involved in your
problem, identify them and note their respective values.
3. Apply Newton’s Second Law: Use Newton’s Second Law of Motion (F = ma) as a foundation for calculating the total net force acting on an object.
4. Add or subtract additional forces: If your problem includes any other forces, like gravity or friction, calculate their effects on the net force and add or subtract them as necessary.
5. Calculate applied force: After accounting for all the forces acting on an object, you can now calculate the applied force itself. To do this, simply rearrange Newton’s Second Law equation (F = ma)
to isolate applied force (F).
Example Calculation
To help illustrate these steps, let’s consider an example: a 10 kg object is being pushed across a horizontal surface with an acceleration of 2 m/s². How much force is being applied?
1. Mass and acceleration are given as m = 10 kg and a = 2 m/s².
2. Since there are no other forces mentioned in the problem, we can move forward to step three.
3. Apply Newton’s Second Law: F = ma = (10 kg)(2 m/s²).
4. Calculate the total net force: F = 20 N (Newton).
5. In this case, since there were no additional forces involved, the applied force is equal to the total net force: F_applied = F_net = 20 N.
Calculating applied force is essential for understanding various physics principles and real-world applications. By following these steps and using Newton’s Second Law of Motion as a foundation, you
can accurately determine the applied force in any given scenario. Remember to consider all relevant forces and carefully account for their effects when making your calculations. | {"url":"https://www.thetechedvocate.org/how-to-calculate-applied-force-a-comprehensive-guide/","timestamp":"2024-11-11T00:53:58Z","content_type":"text/html","content_length":"130294","record_id":"<urn:uuid:e81c80a0-6689-4cae-bbac-ab91cfeb4610>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00112.warc.gz"} |
Inverse square law
In Coulomb’s Law, the distance between charges appears in the equation as $1/r^2$. That makes Coulomb’s Law an example of an inverse square law. Another well-known inverse square law is Newton’s Law
of Gravitation.
Written by Willy McAllister.
It makes intuitive sense that the electric force between two objects goes down if the distance between them gets bigger. But why is the drop off precisely related to the square of the distance? Is
the exponent $2$ a coincidence?
No, it is not.
An inverse square law is characteristic of anything that spreads out from a point in straight lines. Both gravity and electric force act like this. We illustrate the idea with a fable.
The fable of the butter gun
Suppose a restaurant has the problem of buttering toast. They want to be very modern and do toast buttering with a machine. The restaurant owner invents a Butter Gun, with melted butter in the
handle. Butter can squirt out in straight lines from a point at the business end of the butter gun.
Here is a piece of toast, and the lines of butter go out and hit it all over.
Now instead of one toast, the butter lines might continue on. You can put the toast farther back, at twice the distance. Two pieces of toast wide, and two toasts high.
Altogether, four pieces of toast to intercept the butter. The butter will be a quarter as thick when the toast is twice as far away. This is the inverse square law (of buttering).
Extending the idea: At triple the distance, you can arrange $3$ toasts by $3$ toasts to fit within the spray lines, for $9$ total toasts. You get $1/9$th the thickness of butter, for “economy”
The lines of force coming out from a point charge spread out exactly the same as the lines of butter from the tip of the butter gun. At twice the distance, the electric force is a quarter as strong.
This is the inverse square law (of electric force).
The Fable of the Butter Gun was inspired by this marvelous 3-minute video clip by Professor Eric Rogers, Princeton University Department of Physics, in 1959.
The Fable of the Butter Gun
(In addition to this YouTube version, this video can also be found here.)
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To share something privately: Contact me. | {"url":"https://spinningnumbers.org/a/inverse-square-law.html","timestamp":"2024-11-05T05:30:10Z","content_type":"text/html","content_length":"24085","record_id":"<urn:uuid:5cc32a27-3382-4c91-9f09-6571b6c14bfe>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00859.warc.gz"} |
Who Invented Math? The History of Mathematics
Who invented math? It’s a deceptively complex question—a lot harder than 2+2. Math has been around forever, but we are always learning more about it.
Short answer: Many different people invented math, including ancient societies and many famous mathematicians who came along later.
The long answer: It depends on what kind of math you’re asking about. Below is a look at the history of mathematics and the people who contributed to developing math as we know it today.
Jump to:
What is math?
According to Britannica Kids, math is the study of numbers. It’s a kind of language that we use every day to calculate distances, tell time, build things, and so on.
Mathematicians think about math in two areas: pure and applied. Pure math is studying math for its own sake. Figuring out how to solve a particular algorithm or tackling a theory, for example.
Applied math is using math to solve real-life problems, like building a house or predicting an earthquake.
There are lots of different types of math: arithmetic, algebra, geometry, trigonometry, statistics, and more.
So, since math is already a part of the world, the first question is, can math be invented at all?
Was math discovered or invented?
Some mathematicians think that math is invented, as people name aspects of math or create new ways of solving problems. Other people think that math is always there—the concepts and ideas exist in
nature, just waiting for us to discover them.
So, who invented math?
Here’s a look at the history of math and many of the societies and people who contributed to its development.
Early Societies
Jeff Dahl, public domain, via Wikimedia Commons
Math has evolved over thousands of years, with input from thousands of mathematicians. We don’t know exactly how prehistoric humans dealt with math problems (like counting how many berries they
picked, or figuring out the distance between two places), but researchers believe that people were using addition, multiplication, and other math concepts in early China, India, and Mesopotamia.
In fact, the oldest clay tablets we have with math inscribed on them are more than 4,000 years old. They’re from Mesopotamia. We also have Egyptian papyrus sheets with math written on them. So,
there’s evidence of math from the two oldest societies in the world.
Around 1800 B.C.E., the ancient Babylonians developed a number system based on the number 60 (it’s still used today to think about angle measurement). They were the first people we know of to use
actual numbers to represent amounts.
It’s clear that, considering the pyramids and their society, the Egyptians used math. They definitely understood geometry and even had a formula for calculating the volume of a truncated pyramid.
The Ancient Greeks
Anderson, CC0, via Wikimedia Commons
There’s more information about who invented (or discovered) math concepts as human society evolved. The Greeks, more than 2,500 years ago, started doing more advanced math. Plato, Euclid, and
Archimedes are still remembered for their mathematical achievements. For example, Pythagoras studied triangles and he invented what we learn about triangles, called the Pythagorean theorem.
We also know that in ancient Greece, math became something to study, and mathematicians started thinking about specific theories and building on one another’s work.
After Ancient Greece
Godfrey Kneller, public domain, via Wikimedia Commons
After ancient Greece, mathematicians continued making new discoveries and new theories and solving new problems. In 17th-century England, Sir Isaac Newton developed the field of calculus on his own.
At the same time, in Germany, Gottfried Leibniz was also involved in developing calculus. Some mathematicians have created problems and hypotheses that have never been solved, like Bernhard Riemann,
who created the Riemann hypothesis, which has been attempted but never proven.
And throughout history, women have also studied math and invented math concepts. For example, Emmy Noether gained recognition for her innovations in advanced algebra, and Katherine Johnson calculated
and analyzed flight paths for spacecraft that sent astronauts to the moon. Mathematicians of color who have made significant contributions to mathematics include Fern Hunt, who created math models to
describe different kinds of movement, and Mark Dean, a mathematician and computer scientist who holds patents on the computer that all PCs are based upon.
As math has evolved, people are building on what we know to create new types of math and new ways to use math, like applying math to build computers and create game theory, a branch of applied
mathematics. So, maybe the question isn’t who invented math, but what will math invent next?
Videos About the Invention of Math
Use these videos to explore how different math concepts came about.
The Origin of Numbers
How Old Is Zero?
Where Do Math Symbols Come From?
Who Invented Algebra?
Who Invented Geometry?
Who Invented Trigonometry?
More Teaching Resources
Plus, get all the latest teaching tips and tricks when you sign up for our newsletters! | {"url":"http://pbcclothing.com/index-271.html","timestamp":"2024-11-08T13:35:27Z","content_type":"text/html","content_length":"101041","record_id":"<urn:uuid:a7b9f88a-9f13-4bee-8f9b-56d673be1649>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00808.warc.gz"} |
This project studies the problem of visualizing large-scale and high-dimensional data in a low-dimensional (typically 2D or 3D) space. Much success has been reported recently by techniques that first
compute a similarity structure of the data points and then project them into a low-dimensional space with the structure preserved. These two steps suffer from considerable computational costs,
preventing the state-of-the-art methods such as the t-SNE from scaling to large-scale and high-dimensional data (e.g., millions of data points and hundreds of dimensions). We propose the LargeVis, a
technique that first constructs an accurately approximated K-nearest neighbor graph from the data and then layouts the graph in the low-dimensional space. Comparing to t-SNE, LargeVis significantly
reduces the computational cost of the graph construction step and employs a principled probabilistic model for the visualization step, the objective of which can be effectively optimized through
asynchronous stochastic gradient descent with a linear time complexity. The whole procedure thus easily scales to millions of high-dimensional data points. Experimental results on real-world data
sets demonstrate that the LargeVis outperforms the state-of-the-art methods in both efficiency and effectiveness. The hyper-parameters of LargeVis are also much more stable over different data sets.
The LargeVis can be very useful in understanding the distributed representations of data learned by neural networks, which are usually hundreds and thousands of dimensions and hard to interpret by
Jian Tang, Microsoft Research, jiatang@microsoft.com, tangjianpku@gmail.com | {"url":"https://jian-tang.com/selected_projects/visualization/","timestamp":"2024-11-07T13:29:02Z","content_type":"text/html","content_length":"11507","record_id":"<urn:uuid:c3504f72-599e-4910-b1e1-4035c0cd6b6d>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00149.warc.gz"} |
How to make an empty tensor in pytorch?
You can create an empty tensor in PyTorch using the torch.empty() function.
Here is an example of how to create an empty tensor with a shape of (3, 4):
1 import torch
3 empty_tensor = torch.empty(3, 4)
4 print(empty_tensor)
This will create a tensor with the specified shape but with uninitialized values. If you want to initialize the tensor with zeros, you can use the torch.zeros() function instead:
1 zeros_tensor = torch.zeros(3, 4)
2 print(zeros_tensor)
These are two ways to create an empty tensor in PyTorch. | {"url":"https://devhubby.com/thread/how-to-make-an-empty-tensor-in-pytorch","timestamp":"2024-11-03T16:00:58Z","content_type":"text/html","content_length":"113339","record_id":"<urn:uuid:a8f6d0d0-7c85-48a7-8631-6fdab5c82a63>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00585.warc.gz"} |
Line Bundles on Complex Tori (Part 1)
Posted by John Baez
A complex abelian variety is a group in the category of smooth complex projective varieties. They’re called that because — wonderfully — they turn out to all be abelian! I’ve been studying
holomorphic line bundles on complex abelian varieties, which is a really nice topic with fascinating connections to quantum physics, Jordan algebras and number theory. This is the book that’s helped
me the most so far:
• Christina Birkenhake and Herbert Lange, Complex Abelian Varieties, Springer, Berlin, 2004.
But the subject is so rich that it can be hard to see the forest for the trees! So for my own benefit I’d like to describe the classification of holomorphic line bundles on an abelian variety — or
more generally, any ‘complex torus’.
A complex torus is the same as the quotient of a finite-dimensional complex vector space by a lattice. Every abelian variety is a complex torus, but not every complex torus is an abelian variety: you
can’t make them all into projective varieties.
I will avoid saying a lot of things people usually say about this subject, in order to keep things short.
Complex line bundles on a topological space $X$ are classified up to isomorphism by elements of $H^2(X,\mathbb{Z})$.
This becomes particularly nice when $X$ is a real torus. Let’s equip $X$ with the structure of an abelian Lie group. Then the universal cover $\tilde{X}$ is a finite-dimensional real vector space,
and the kernel of the projection
$p \colon \tilde{X} \to X$
is a lattice
$L = \ker p$
in the vector space $\tilde{X}$ — that is, a free abelian group of rank equal to the dimension of that vector space. This lets us write
$X = \tilde{X}/L$
In this situation $H^2(X,\mathbb{Z})$ is isomorphic to the group of alternating bilinear maps
$A \colon L \times L \to \mathbb{Z}$
There’s nothing magic about the number $2$ and the word ‘bilinear’ here — the same kind of description works for $H^i(X, \mathbb{Z})$ for any $i$, using alternating multilinear maps.
So, complex line bundles on a real torus $X$ are classified up to isomorphism by alternating bilinear maps $A \colon L \times L \to \mathbb{Z}$. And I should be clear: we could use isomorphism of
topological line bundles, or of smooth line bundles: it makes no difference here.
But now suppose $\tilde{X}$ is equipped with the structure of a complex vector space. This makes $X = \tilde{X}/L$ into a complex manifold, and indeed a compact abelian complex Lie group. We call $X$
a complex torus. In fact, every compact connected complex Lie group is a complex torus.
Complex tori are a lot subtler than real tori. All real tori of a given dimension are isomorphic. This is not true for complex tori! For example, complex tori of complex dimension 1 — hence real
dimension 2, so they look like doughnuts — are called elliptic curves. There’s an interesting space of different isomorphism classes of elliptic curves, called the moduli space of elliptic curves.
The story gets even more complicated in higher dimensions.
Now for the real point of this article. If $X$ is a complex torus, it’s an interesting question to classify holomorphic complex line bundles on $X$. How does it work?
This question breaks up nicely into two parts:
Question 1. Which alternating bilinear $A \colon L \times L \to \mathbb{Z}$ come from holomorphic complex line bundles?
Question 2. If $A \colon L \times L \to \mathbb{Z}$ comes from a holomorphic complex line bundle, how many different such bundles give this particular $A$?
A nice thing is that the answer to Question 2 turns out not to depend on the choice of $A$, as long as there’s any holomorphic line bundle at all giving that $A$.
Answer to Question 1. The answer to this question is simple, though it takes work to show it’s correct. Using the fact that every vector in $\tilde{X}$ is a real linear combination of vectors in the
lattice $L \subseteq \tilde{X}$, you can show any alternating bilinear map
$A \colon L \times L \to \mathbb{Z}$
extends uniquely to an alternating real-bilinear map
$\tilde{A} \colon \tilde{X} \times \tilde{X} \to \mathbb{R}$
It then turns out that $A$ comes from a holomorphic complex line bundle if and only if
$\tilde{A}(i v, i w) = \tilde{A}(v,w)$
for all $v,w \in \tilde{X}$.
Answer to Question 2. This answer is also simple. Suppose $A$ comes from some holomorphic complex line bundle. Then the set of isomorphism classes of holomorphic complex line bundles giving this $A$
is $X^*$, the ‘dual’ of torus $X$.
What’s dual of a torus? In this game it’s another torus — not the usual Pontryagin dual of the torus. Remember, for any real torus $X$ we can write
$X = \tilde{X}/L$
The real vector space $\tilde{X}$ has a dual $\tilde{X}^*$, defined in the usual way. Sitting inside this dual vector space we have the so-called dual lattice of the lattice $L$, defined like this:
$L^\ast = \{ f \in \tilde{X}^\ast : \; f(v) \in \mathbb{Z} \; for \; all \; v \in L \}$
So, the quotient ${\tilde{X}}^\ast / L^\ast$ is a torus, called the dual torus of the torus $X$ and again denoted with a star:
$X^\ast = {\tilde{X}}^\ast / L^\ast$
Now if $X$ is a complex torus, the vector space $\tilde{X}$ is a complex vector space, and thus so is its dual ${\tilde{X}}^\ast$. Beware: ${\tilde{X}}^\ast$ is still defined in the usual way: it’s
the real dual of the underlying real vector space of $\tilde{X}$. But it gets the structure of a complex vector space from that of $\tilde{X}$ — this is a bit tricky, and I’ll talk about it next
time. So the dual torus $X^\ast$ is a complex torus if $X$ is.
People like to package up the answers to Question 1 and Question 2 into a single theorem, the Appell–Humbert theorem. And the usual statement of this theorem involves lots of extra jargon — which you
have to learn if you want to study this subject:
• The set of isomorphism classes of holomorphic complex line bundles on a complex torus $X$ is called its Picard group, $\mathrm{Pic}(X)$. This is actually an abelian group, because you can tensor
line bundles. In fact, it’s a complex Lie group.
• The identity component $\mathrm{Jac}(X)$ of the Picard group is called the Jacobian of $X$, or, as if there weren’t enough jargon already, the Picard variety of $X$. This is a complex torus, and
an abelian variety when $X$ is.
• The quotient $\mathrm{Pic}(X)/\mathrm{Jac}(X)$ is called the Néron–Severi group of $X$ and denoted $\mathrm{NS}(X)$. This is a finitely generated free abelian group.
So we have a short exact sequence
$0 \to \mathrm{Jac}(X) \to \mathrm{Pic}(X) \to \mathrm{NS}(X) \to 0$
You should think of it this way:
• The Néron–Severi group $\mathrm{NS}(X)$ describes the ‘discrete degrees of freedom’ required to choose a holomorphic complex line bundle over $X$. That’s because it’s the group of connected
components of $\mathrm{Pic}(X)$. It’s a finitely generated free abelian group.
• The Jacobian $\mathrm{Jac}(X)$ describes the ‘continuous degrees of freedom’ required to choose a holomorphic complex line bundle over $X$. That’s because it’s the identity component of $\mathrm
{Pic}(X)$. It’s a complex torus.
But even better:
• $\mathrm{NS}(X)$ is naturally isomorphic to the subgroup of $H^2(X,\mathbb{Z})$ coming from holomorphic complex line bundles — which we described in our answer to Question 1.
• $\mathrm{Jac}(X)$ is naturally isomorphic to the dual torus $X^*$, which we described in our answer to Question 2.
Putting all these facts together, we get this:
Appell–Humbert Theorem. If $X$ is a complex torus, we have a short exact sequence
$0 \to \mathrm{Jac}(X) \to \mathrm{Pic}(X) \to \mathrm{NS}(X) \to 0$
• $\mathrm{NS}(X)$ is isomorphic to the abelian group whose elements are alternating bilinear forms $A \colon L \times L \to \mathbb{Z}$ whose real-linear extension $\tilde{A}$ obeys $\tilde{A}(i
v, i w) = \tilde{A}(v,w)$. (The set of these becomes an abelian group under addition.)
• $\mathrm{Jac}(X)$ is isomorphic to the dual torus $X^\ast$.
The dimension of the complex torus $\mathrm{Jac}(X)$ depends only on the dimension of $X$, since it’s just the dual torus. But the rank of the free abelian group $\mathrm{NS}(X)$ depends on more than
just the dimension of $X$. Indeed, a generic complex torus has no holomorphic complex line bundles on them except those that are topologically trival. For such complex, $\mathrm{NS}(X)$ has just one
element! But in general $\mathrm{NS}(X)$ is a subgroup of $H^2(X,\mathbb{Z})$ whose rank depends on the complex structure of $X$, not just its dimension. It’s bigger for complex tori where the
lattice $L \subseteq \tilde{X}$ gets along better with the complex structure on $\tilde{X}$, and these are the ones worth focusing on.
There’s a lot more to say, and I’m proud of myself for having not said it. Maybe I’ll say more later. I’m having tons of fun looking at examples of the theorems I just stated.
Posted at March 13, 2022 10:59 PM UTC
There’s a lot more to say, but I’m proud for having not said it.
I saw a paper of Witten’s recently, String Theory & Noncommutative Geometry on the Arxiv. And in the comments under the abstract, he said:
100 pages, sorry!
I think long papers ought not to be called papers, like a long short story is not called a short story but a novella …
Posted by: Mozibur Ullah on March 14, 2022 2:31 PM | Permalink | Reply to this
Re: naming of long papers
We do have words like “monograph”.
And “book”.
Electronic distribution has somewhat eroded these distinctions, since a 500-page book can go on the arXiv just as easily as a 30-page paper (or a 5-page “note”), but maybe we should make more of an
effort to use them.
Posted by: Mike Shulman on March 21, 2022 7:05 AM | Permalink | Reply to this
Re: naming of long papers
Mathematicians (I wont dare to discuss physicists) have to learn a discipline all there own in writing mathematics.
• When you have written a piece, try and write again in half the length/number of pages – it will be better.
• You do not have to tell all that you know on a topic, just enough to make for people in your area to know what you are doing, let people work some of it out.
• Your readers will not want to read 100+ page articles, they have not got time – fight your vanity. Seeing an introduction alone which 30 pages alone will make many readers give up.
• Do you believe your rersults are so much better than those of M. Atiyah, JP Serre, R. Thom, D. Quillen, that you need to write a 100 page article when they transformative mathematics in 30 pages
?(due to obvious technological reasons.)
• Write fewer papers. Think harder and longer and deeper.
Posted by: Simon Scott KCL Math on April 13, 2024 11:21 PM | Permalink | Reply to this
Re: Holomorphic Line Bundles on Complex Tori
Posted by: Allen Knutson on March 15, 2022 3:34 PM | Permalink | Reply to this
Re: Holomorphic Line Bundles on Complex Tori
A complex abelian variety is a group in the category of smooth complex projective varieties. They’re called that because — wonderfully — they turn out to all be abelian!
Sorry to comment on something so trivial, but is that really why they’re called abelian?
I ask because once upon a time, a long time ago, I went to the first couple of lectures of a course on abelian varieties, and I thought the lecturer began by saying something like “They’re not called
abelian varieties because the group structure is abelian. But fortunately, it is abelian, otherwise the terminology would be very confusing”.
This is the only thing I retain from those lectures. If someone stopped me in the street and made me tell them everything I know about abelian varieties, I’d say “they turn out to be abelian groups,
but that’s not why they’re called abelian! (Now please let me go.)” If it turns out that I misremembered even that, then what I retain falls to zero. So, I cling on to the hope that I’m right.
Posted by: Tom Leinster on March 15, 2022 9:41 PM | Permalink | Reply to this
Re: Holomorphic Line Bundles on Complex Tori
Hmm, I think you’re right: maybe the fact that they’re abelian groups in the category of varieties is not why they were originally called abelian varieties!
Like many other great mathematicians of the 1800s, Abel did a lot of work on elliptic integrals and their generalizations, which are now called abelian integrals. This was later systematized using
abelian varieties.
Here’s what Wikipedia says:
The theory of abelian integrals originated with a paper by Abel published in 1841. This paper was written during his stay in Paris in 1826 and presented to Augustin-Louis Cauchy in October of the
same year. This theory, later fully developed by others, was one of the crowning achievements of nineteenth century mathematics and has had a major impact on the development of modern
mathematics. In more abstract and geometric language, it is contained in the concept of abelian variety, or more precisely in the way an algebraic curve can be mapped into an abelian varieties.
Abelian integrals were later connected to the prominent mathematician David Hilbert’s 16th Problem, and they continue to be considered one of the foremost challenges to contemporary mathematical
A slick modern way of summarizing some of this work is described here:
• John Baez, Two miracles of algebraic geometry, August 10, 2016.
Namely, at least if we work over $\mathbb{C}$, the category of abelian varieties is monadic over the category of pointed projective varieties.
This may at first seem similar to how the category of abelian groups is monadic over the category of pointed sets. But there’s a big difference: for pointed varieties, we get an idempotent monad.
That is, being an abelian group is really just a property of a pointed projective variety, not an extra structure!!!
Posted by: John Baez on March 15, 2022 10:42 PM | Permalink | Reply to this
Re: Holomorphic Line Bundles on Complex Tori
I’m not a historian, but I think there might be a connection between abelian varieties being abelian, and the naming of abelian groups, which is better than coincidence.
Abelian functions (in the modern terminology) are meromorphic functions $\mathbb{C}^g \to \mathbb{C}$, periodic with respect to a lattice $L \subset \mathbb{C}^g$ of rank $2g$. From a modern
perspective, these are meromorphic functions on the quotient $A:=\mathbb{C}^g/L$. The quotient $A$ is an abelian variety.
For any positive integer $n$, multiplication by $n$ is a degree $n^{2g}$ map $\mu: A \to A$. This map can be written as some very complicated algebraic transformation from abelian functions on $A$ to
other abelian functions on $A$. If we want to invert this transformation, then we need to solve a polynomial of degree $n^{2g}$. However, this polynomial is unusually nice: It’s abelian, of the form
$(\mathbb{Z}/n \mathbb{Z})^{2g}$. (I’m glossing over some details here; we need to assume that the kernel $\mu^{-1}(0)$ is defined over our ground field.)
Now, what Cox says is that Abel was studying polynomials whose Galois group was what we now call abelian. Why was he studying such equations? Here I don’t know, but it sure seems reasonable to guess
that he was trying to invert the transformations of abelian functions. (And Cox’s quote at the bottom of page 144 suggests I am right.) Why does inverting transformations of abelian functions involve
solving equations with abelian Galois group? From a modern perspective, because these transformations encode multiplication by $n$ in the abelian group $A$.
So I think that abelian groups were named because they are the sort of Galois groups that coming up when inverting transformations of abelian functions, and the reason those Galois groups are abelian
is because abelian varieties are abelian.
Posted by: David Speyer on March 17, 2022 7:41 PM | Permalink | Reply to this
Re: Holomorphic Line Bundles on Complex Tori (Part 1)
Here is a link to an article by Steve Kleiman that reviews some of the history of Abelian integrals and Abelian varieties.
Posted by: Jason Starr on March 22, 2022 11:56 AM | Permalink | Reply to this
Re: Holomorphic Line Bundles on Complex Tori (Part 1)
That looks fun, especially the first part!
We develop in detail most of the theory of the Picard scheme that Grothendieck sketched in two Bourbaki talks and in commentaries on them. Also, we review in brief much of the rest of the theory
developed by Grothendieck and by others. But we begin with a twelve-page historical introduction, which traces the development of the ideas from Bernoulli to Grothendieck, and which may appeal to
a wider audience.
Posted by: John Baez on March 22, 2022 11:28 PM | Permalink | Reply to this
Re: Holomorphic Line Bundles on Complex Tori (Part 1)
If you want to waste a little time, try to fill in the missing details in the proof of the Appell-Humbert Theorem presented in Griffiths-Harris. It is possible to do with some group cohomology, but I
have no idea how a student with no such background is expected to finish that proof. I sometimes run a seminar on mathematical writing, and I present that proof next to the proof from Mumford’s
“Abelian varieties” as one of the lectures.
Posted by: Jason Starr on March 22, 2022 11:25 AM | Permalink | Reply to this
Re: Holomorphic Line Bundles on Complex Tori (Part 1)
I may give it a try! Personally I’m having a lot of fun learning this material from Birkenhake and Lange’s Complex Abelian Varieties, and they give a fairly low-tech proof of the Appell–Humbert
theorem early on. Since I’m taking a lowbrow computational approach to this material, what matters most to me is the explicit construction of a holomorphic line bundle from a pair $(H, \chi)$ where
$H \in NS(X)$ and $\chi$ is a ‘semicharacter’ for $H$. This gives a map from $NS(X) \times X^\ast$ to $Pic(X)$, and so far I’m less concerned with the proof that this is one-to-one and onto.
I like ‘abstract nonsense’, including group cohomology, but since I’ll never be a full-fledged algebraic geometer, I’m focusing on playing around with some specific highly symmetrical examples of
abelian varieties, and line bundles on those.
Posted by: John Baez on March 22, 2022 11:24 PM | Permalink | Reply to this | {"url":"https://classes.golem.ph.utexas.edu/category/2022/03/holomorphic_line_bundles_on_co.html","timestamp":"2024-11-04T04:24:46Z","content_type":"application/xhtml+xml","content_length":"71557","record_id":"<urn:uuid:970e083e-1a97-46d0-a942-70b79fced745>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00774.warc.gz"} |
1858: [09NOIP improvement group] target Sudoku
[Title Description]
Xiaocheng and Xiaohua are both good students who love mathematics. Recently, they have become addicted to Sudoku games. They want to compete with Sudoku. But ordinary Sudoku was too simple for them,
so they asked Dr. Z for advice. Dr. Z took out his recently invented "target Sudoku" as the topic of the competition between the two children.
The square of target Sudoku is the same as that of ordinary Sudoku, which is 9 squares wide × Nine of the nine high squares are three wide × Three grid high small Jiugong grid (separated by thick
black lines). In this big Jiugong grid, some numbers are known. According to these numbers, use logical reasoning to fill in the numbers from 1 to 9 in other spaces. Each number cannot be repeated in
each small nine palace, and each number cannot be repeated in each row and column. However, target Sudoku is different from ordinary Sudoku in that each square has a score, and like a target, the
closer to the center, the higher the score. (as shown in the figure)
The specific score distribution in the above figure is: the innermost grid (yellow area) is 10 points, the circle outside the yellow area (red area) is 9 points, the outer circle (blue area) is 8
points, the circle outside the blue area (brown area) is 7 points, and the outermost circle (white area) is 6 points, as shown in the above figure. The requirement of the competition is that everyone
must complete a given Sudoku (each given Sudoku may have different filling methods), and strive for a higher total score. The total score is the sum of the product of the score on each grid and the
number filled in the corresponding grid when completing the Sudoku. As shown in the figure, in the following target Sudoku game, which has filled in numbers, the total score is 2829. The game
stipulates that the winner will be determined by the total score.
Eager to win, the town found you who are good at programming and asked you to help him find the highest score for a given target Sudoku.
Nine lines altogether. 9 integers per line (each number is in the range of 0-9), indicating an unfilled Sudoku square, and the unfilled space is represented by "0". Every two numbers are separated by
a space.
1 line in total. Output the highest score of target Sudoku. If the Sudoku has no solution, the integer - 1 is output.
[input example]
[output example]
[input / output example 2]
[data range]
40% of the data, and the number of non-zero numbers in Sudoku is not less than 30.
80% of the data, and the number of non-zero numbers in Sudoku is not less than 26.
100% of the data, and the number of non-zero numbers in Sudoku shall not be less than 24.
for the first time:
Search down from the first one to find the one with the highest score.
Result: 40 points # TLE
The second time:
Search from the most in each row and column to find the one with the highest score.
Results: 75 points # there are 5 test points TLE
third time:
Search from the most in each row and column to find the one with the highest score.
In the above search, try the number a[x][y] from 1 to 9 all the time, and judge whether it can be filled every time from 1 to 9. Therefore, the judgment check in this code is actually very
time-consuming. You can create an array cando [] [] (a value of 0 means that it can be filled, and a value of 1 means that it cannot be filled). Then check the number of this row and column and the
number of this Jiugong lattice (assuming that it is row N and column m). If the number is not 0, the number cannot be filled in and c[a[n][m]]=1. After checking, carry out the cyclic search of 1 ~ 9:
if the number c[i]=1, then continue to find the number i+1; Otherwise, search for the next number.
The AC code is as follows:
#include<bits/stdc++. h> / / Sudoku
using namespace std;
int a[11][11];
int b[11];//Maximum number of known per line
int bid[11];//id of b
int bbid;//bid number
int c[11];//Maximum known number per column
int cid[11];//id of c
int ccid;//cid number
int maxx;//Maximum score
int cando[1001][11];//Determine what number can be filled in the search
int candoid=1;//Number
bool fff=false;//Is there a solution
void checkx(int x) {//Judge which line
for(int i=1; i<=9; i++) {
void checky(int y) { //Determine which column
for(int i=1; i<=9; i++) {
void jggfz(int djgg) {//Change the ninth grid into the starting point (x,y) and judge the ninth grid
int x,y;
else if(djgg==2)
else if(djgg==3)
else if(djgg==4)
else if(djgg==5)
else if(djgg==6)
else if(djgg==7)
else if(djgg==8)
for(int k=x; k<x+3; k++) {
for(int i=y; i<y+3; i++) {
int fzfz(int x) { //Auxiliary score calculation
return 10*a[5][5];
int sum=0;
for(int i=x; i<=10-x; i++)
for(int i=x+1; i<=10-(x+1); i++)
for(int i=x+1; i<=10-(x+1); i++)
for(int i=x; i<=10-x; i++)
return sum;
int fz() { //Calculation of score
int sum=0;
for(int i=1; i<=5; i++)
return sum;
void dfs(int na,int nb) { //Line number
if(bbid==10) {
int sum=fz();
if(a[na][nb]!=0) {
int bbidd=bbid,ccidd=ccid;
//Just judge once and tell what you can fill in
for(int i=1;i<=9;i++)
int djgg=(na-1)/3*3+(nb-1)/3+1;//Use the formula to calculate the ninth house of X and y
for(int i=1; i<=9; i++) {
int bbidd=bbid,ccidd=ccid;
int main() {
for(int i=1; i<=9; i++) {
for(int j=1; j<=9; j++) {
//Which row has the most known numbers, which row to fill in, and which column has the most known numbers, which column to fill in
for(int i=1; i<=9; i++)
for(int i=1; i<=9; i++) {
for(int j=1; j<=9-i; j++) {
return 0;
The code is written for a long time and runs slowly, but it is easy to understand. | {"url":"https://programming.vip/docs/1858-09noip-improvement-group-target-sudoku.html","timestamp":"2024-11-07T16:55:56Z","content_type":"text/html","content_length":"13971","record_id":"<urn:uuid:3b110bb9-2828-490e-9421-bdfce351cae2>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00713.warc.gz"} |
Why Cosmic Inflation’s Last Great Prediction May Fail
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Travel the universe with Dr. Ethan Siegel as he answers the biggest questions of all
And what it means if we don’t see gravitational waves from inflation in the next 5–10 years.
“The paradigm of physics—with its interplay of data, theory and prediction—is the most powerful in science.” –Geoffrey West
One of the greatest scientific achievements of the early 20th century was the discovery of the expanding Universe: that as time goes on, distant galaxies are receding from us, as the space between us
expands according to Einstein’s General Relativity. In the mid-20th century, a great idea was put forth, that if the Universe is getting bigger and cooler today, then it was smaller, hotter and
denser in the past: the Big Bang. The Big Bang made a few extra predictions:
• there would be a great cosmic web of structure, with small, medium and large-scale structures clumped together in certain patterns,
• there would be a leftover glow of radiation from the early Universe, that’s cooled to just a few degrees above absolute zero,
• and there would be a specific set of ratios for the lightest elements in the Universe, for the different isotopes of hydrogen, helium and lithium.
Image credit: NASA / WMAP science team, of the discovery of the CMB in 1965 by Arno Penzias and Bob Wilson.
In the 1960s and 1970s, these predictions were all confirmed to varying degrees of accuracy, and the Big Bang became overwhelmingly accepted as the leading theory of where everything we can perceive
and detect in the Universe originated. But there were a few questions that were unanswered when it came to the Big Bang, a few phenomena that were completely unexplained within this framework.
1. Why was the Universe the exact same temperature everywhere?
2. Why was the Universe so spatially flat; why did the expansion rate and the matter/energy density balance each other so perfectly well?
3. If the Universe achieved such high energies early on, why haven’t we seen the stable relics that should be spread throughout the Universe from it?
Image credit: E. Siegel, from his book Beyond The Galaxy. If these three different regions of space never had time to thermalize, share information or transmit signals to one another, then why are
they all the same temperature?
If the Universe were expanding according to the rules of General Relativity, there’s no reason to expect that regions of space separated by distances greater than the speed of light were connected,
much less the same exact temperature. If you take the Big Bang all the way back to its logical conclusion—to an infinitely hot, dense state—there’s no way to come up with answers to these
questions. You just have to say, “it was born this way,” and from a scientific point of view, that’s wholly dissatisfying.
But there’s another option. Perhaps, instead of the Universe just being born at the moment of the Big Bang with these conditions, there existed an early stage that set up these conditions and the
hot, dense, expanding and cooling Universe that gave rise to us. This would be a job for theorists: to figure out what possible dynamics could set the stage for the Big Bang with these conditions to
occur. In 1979/1980, Alan Guth put forth the revolutionary idea that would change the way we thought about our Universe’s origins: cosmic inflation.
Image credit: Alan Guth’s 1979 notebook, tweeted via @SLAClab, from https://twitter.com/SLAClab/status/445589255792766976.
By postulating that the Big Bang was preceded by a state where the Universe wasn’t filled with matter-and-radiation, but rather by a huge amount of energy inherent to the fabric of space itself, Guth
was able to solve all of these problems. In addition, as the 1980s progressed, further developments occurred that made it clear that, in order for inflationary models to reproduce the Universe we
• to fill it with matter-and-radiation,
• to make the Universe isotropic (the same in all directions),
• to make the Universe homogeneous (the same in all locations),
• and to give it a hot, dense, expanding state,
there were quite a few classes of models that could do it, as developed by Andrei Linde , Paul Steinhardt, Andy Albrecht, with additional details worked out by people like Henry Tye, Bruce Allen,
Alexei Starobinskii, Michael Turner, David Schramm, Rocky Kolb and others. But the simplest ones—the ones that solved the problem and had the fewest free parameters—fell into just two categories.
Images credit: Ethan Siegel, with google’s graph tool. The two simplest classes of inflationary potentials, with chaotic inflation (L) and new inflation (R) shown.
There was new inflation, where you had a potential that was very flat at the top and that the inflaton field could “roll down, slowly” to reach the bottom, and there was chaotic inflation, where you
had a U-shaped potential that, again, you’d roll down slowly.
In both these cases, your space would expand exponentially, be stretched flat, have the same properties everywhere, and when inflation came to an end, you’d get back a Universe that very much
resembled our own. In addition, you’d also get six extra, new predictions out, all of which had not yet been observed at the time.
1. A Perfectly Flat Universe. Because inflation causes this rapid, exponential expansion, it takes whatever shape the Universe happened to be and stretches it to tremendous scales: to scales much,
much larger than what we can observe. As a result, the part that we see looks indistinguishable from flat, the same way that the ground outside your window may look flat, but it’s actually part
of the entire, curved Earth. We just can’t see enough to know what the true curvature actually is.
2. A Universe with fluctuations on scales larger than light could’ve traveled across. Inflation—by causing the space of the Universe to expand exponentially—causes what happens on very small
scales to get blown up to much larger ones. This includes quantum fluctuations, which normally fluctuate in-place in empty space. But during inflation, thanks to the rapid, exponential expansion,
these small-scale energy fluctuations get stretched across the Universe onto gigantic, macroscopic scales that should wind up spanning the entire visible Universe!
3. A Universe with a maximum temperature that’s not arbitrarily high. If we could take the Big Bang all the way back to arbitrarily high temperatures and densities, we’d find evidence that the
Universe once reached at least the temperature scale at which the laws of physics break down: the Planck scale, or around energies of 10^19 GeV. But if inflation occurred, it must have occurred
at energy scales lower than that, with the result that the maximum temperature of the Universe post-inflation must be some energy scale lower than 10^19 GeV.
4. A Universe whose fluctuations were adiabatic, or of equal entropy everywhere. Fluctuations could have come in different types: adiabatic, isocurvature, or a mixture of the two. Inflation
predicted that these fluctuations should have been 100% adiabatic, which means that detailed measurements of the types of quantum fluctuations the Universe started off with should reveal
signatures in the microwave background and in large-scale cosmic structure.
5. A Universe where the spectrum of fluctuations was just slightly less than having a scale invariant (n_s < 1) nature. This is a big one! Sure, inflation generically predicts that these
fluctuations should be scale-invariant. But there’s a slight caveat, or a correction to that: the shape of the inflationary potentials that work—their slopes and concavities—affect how the
spectrum of fluctuations departs from perfect scale invariance. The two simplest classes of inflationary models, new inflation and chaotic inflation, give predictions for n_s that typically cover
the range between 0.92 and 0.98.
6. And finally, a Universe with a particular spectrum of gravitational wave fluctuations. This is the last one, and the only major one that hasn’t yet been confirmed. Some models—like the simple
chaotic inflation model—give large-magnitude gravitational waves (the kind that could’ve been seen by BICEP2), while others, like the simple new inflation model, can give very small-magnitude
gravitational waves.
Image credit: ESA and the Planck Collaboration.
Over the past 35 years, we’ve made incredible, all-sky measurements of the fluctuations in the cosmic microwave background, from scales as large as the entire visible Universe down to angular
resolutions of a mere 0.07°. As space-based satellites became more and more capable over time—COBE in the 1990s, WMAP in the 2000s, and now Planck in the 2010s—we’ve gained incredible insight
into the Universe when it was less than 0.003% its current age.
Image credit: Sloan Digital Sky Survey (SDSS), including the current depth of the survey.
Similarly, large-scale structure surveys have become incredibly ubiquitous, with some covering the entire sky and others covering huge patches at even greater depths. With the Sloan Digital Sky
Survey providing the best modern data sets, we’ve been able to confirm the first five of these six predictions, placing inflation on a very firm footing.
1. The Universe is observed to be exactly spatially flat—with a curvature of 1, exactly—to a precision of 1.0007 ± 0.0025, as best shown by the large-scale structure of the Universe.
2. The fluctuations in the cosmic microwave background show a Universe with scales that extend up to and beyond the horizon of the observable Universe.
3. The maximum temperature that our Universe ever could have achieved, as shown by the fluctuations in the cosmic microwave background, is only ~10^16 GeV, or a factor of 1,000 smaller than a
non-inflationary Universe.
4. The types of fluctuations the Universe was born with, to the best of our measurements, are 100% adiabatic, and 0% isocurvature. The correlations between the cosmic microwave background and the
large-scale structure of the Universe show this, although this wasn’t confirmed until the early 2000s.
5. And from the latest data from the most advanced cosmic microwave background satellite, Planck, gives us a scalar spectral index (which comes from the densityfluctuations) that’s not only less
than 1, it’s precisely measured to be n_s = 0.968 ± 0.006.
That last number, n_s, is really, really important if we want to look for the sixth and final prediction of inflation: gravitational wave fluctuations.
Image credit: NASA / WMAP science team.
The spectrum of fluctuations in the microwave background looks like the squiggled line, above, today, but it grew out of the interplay of all the different forms of energy over time, from the end of
inflation until the Universe was 380,000 years old. It grew from the density fluctuations at the end of inflation: the horizontal line. Only, that line isn’t quite horizontal; there’s a slight tilt
to the line, and the slope represents the departure of spectral index, n_s, from 1.
The reason this is important is that inflation makes a specific prediction for a special ratio (r), where r is the ratio of the gravitational wave fluctuations to the scalar spectral index, n_s. For
the two main classes of inflationary models—as well as in other models—there is a huge disparity in what r is predicted to be.
Image credit: Kamionkowski and Kovetz, to appear in ARAA, 2016, from http://lanl.arxiv.org/abs/1510.06042. Results presented at AAS227.
For chaotic models, r is typically very large: no smaller than about 0.01, where 1 is the maximum conceivable value. But for the new inflation models,r can vary from as large as about 0.05 down to
tiny, minuscule numbers like 10^–60! But these various r values are often correlated with specific values for ns, as you can see above. If n_s turns out to actually be the value that we’ve best
measured it to be right now—0.968—then the simplest models you can write down for both chaotic inflation and new inflation only give values of r that are bigger than about 10^–3.
As reported by Mark Kamionkowski in his talk at AAS (and based on his paper here), all the simple models one can write down, for the measured value of n_s, means that r can’t range from 10^–60 to 1;
it can only range from 10^–3 to 1. And this could be very, very problematic in short order, because there are a whole host of ground-based surveys that are measuring the type of signal that can
measure r, already constrained to be less than 0.09, if it’s greater than or equal to ~10^–3.
Image credit: Kamionkowski and Kovetz, to appear in ARAA, 2016, from http://lanl.arxiv.org/abs/1510.06042. Results presented at AAS227.
The gravitational wave fluctuations produced by inflation cause both E-mode and B-mode polarizations, but the density fluctuations (and ns) show up in only the E-modes. So if you measure the B-mode
polarizations, you can learn about the gravitational wave fluctuations and determine r!
This is what experiments such as BICEP2, POLARBEAR, SPTPOL and SPIDER, among others, are working to measure right now. There are B-mode polarization signals caused by lensing effects, but if the
inflationary fluctuations are larger than r ~ 0.001, they’ll be able to be seen in 5–10 years by the experiments running and planned to run over that time.
Image credit: Planck science team.
If we find a positive signal for r, either a chaotic inflation (typically if r > 0.02) or a new inflation (typically for r < 0.04, and yes, there’s overlap) model could be strongly, strongly favored.
But if the measured value for n_s stays what it’s thought to be right now, and after a decade we’ve constrained r < 10^–3, then the simplest models for inflation are all wrong. It doesn’t mean
inflation is wrong, but it means inflation is something more complicated than we first thought, and perhaps not even a scalar field at all.
If nature is unkind to us, the last great prediction of cosmic inflation—the existence of primordial gravitational waves—will be elusive to us for many decades to come, and will continue to go
This article was partially based on information obtained during the 227th American Astronomical Society meeting, some of which may be unpublished.
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How Much Does 125 Gallons of Water Weigh? Unveiling the Weight of 125 Gallons of Water - The Techy Life
How Much Does 125 Gallons of Water Weigh? Unveiling the Weight of 125 Gallons of Water
Water is a vital resource that sustains life on Earth. Whether it is for drinking, cooking, or other everyday activities, the weight of water can often be underestimated. Curiosity about the weight
of water becomes even more intriguing when larger quantities are involved. It is a common misconception that liquid substances do not possess significant weight, leading one to wonder: how much does
125 gallons of water actually weigh? In this article, we will unveil the surprising weight of 125 gallons of water, shedding light on the magnitude of this essential substance and furthering our
understanding of its impact on various aspects of our lives.
Definition and Measurement of Gallons
2.1 What is a gallon?
In this section, we will provide an explanation of what a gallon is and its common usage. A gallon is a unit of measurement used for liquid capacity. It is primarily used in the United States and
some other countries. One gallon is equivalent to 128 fluid ounces or 3.785 liters. It is widely used for measuring liquids such as milk, gasoline, and water.
2.2 Conversion of gallons to other units of measurement
In this subsection, we will discuss the conversion of gallons to other units of measurement for better comprehension. Understanding the conversion between gallons and other units can help individuals
grasp the weight of water more easily. For instance, one gallon is approximately equal to 8.34 pounds or 3.785 kilograms. By knowing these conversions, one can easily estimate the weight of a given
volume of water.
Understanding Water Density
3.1 Definition of water density
In this section, we will define water density and explain its relevance to determining water weight. Water density refers to the amount of mass contained in a given volume of water. It is commonly
expressed in kilograms per cubic meter (kg/m³) or grams per milliliter (g/mL). Understanding water density is crucial for accurately calculating the weight of a specific volume of water.
3.2 Effects of temperature on water density
In this subsection, we will elaborate on how temperature can affect water density. Water is unique in that its density changes with temperature. As water gets colder, it becomes denser, and as it
gets warmer, it becomes less dense. This temperature-dependent variation in density can have significant implications when calculating the weight of water, as it is essential to account for the water
temperature to obtain accurate results.
IWater Weight Calculation
4.1 Formula for calculating water weight
In this section, we will provide the formula for calculating water weight. The formula is simple: water weight equals water volume multiplied by water density. By using the appropriate units and
considering the water temperature, individuals can easily calculate the weight of a given volume of water.
4.2 Considerations for accurate weight calculation
In this subsection, we will discuss the considerations for accurately calculating the weight of water. Factors such as water temperature, pressure, and impurities can affect water density and,
consequently, the calculated weight. It is crucial to take these considerations into account to ensure accurate weight calculations for any given volume of water.
Understanding Water Density
Water density plays a crucial role in determining the weight of water, including 125 gallons of it. Density is defined as the mass per unit volume of a substance. In the case of water, its density is
approximately 1 gram per cubic centimeter (g/cm³) at standard atmospheric pressure and temperature.
It is important to understand that water density can vary with changes in temperature. As the temperature increases or decreases, the molecules in water expand or contract, respectively, causing a
change in density. This means that the weight of water can fluctuate depending on its temperature.
To calculate the weight of a given volume of water, including 125 gallons, the formula is simple: volume multiplied by density. Therefore, to determine the weight of 125 gallons of water, we multiply
the volume, measured in gallons, by the density, measured in grams per cubic centimeter.
However, when performing this calculation, it is essential to account for the temperature of the water. As mentioned earlier, water density changes with temperature, so it is crucial to measure the
water’s temperature accurately. The density of water at different temperatures is well-documented, and tables or charts can be consulted to obtain the correct density value for the given temperature.
By using the accurate density value, which corresponds to the temperature of the water, the weight of 125 gallons of water can be calculated with precision. This calculation provides important
information, especially in situations where an accurate understanding of the weight is necessary, such as transportation or construction projects.
Understanding the weight of water has practical applications in various scenarios. For instance, when transporting large quantities of water, such as in tankers or water trucks, knowing the weight
helps ensure that vehicles are not overloaded and that weight limits are not exceeded. Similarly, in construction, water weight can impact structural integrity and determine the required support or
In everyday life, knowledge of water weight can influence decisions related to water usage. For example, understanding that water is relatively heavy can help individuals manage their water
consumption and reduce unnecessary waste. Additionally, it can provide insight into the weight-bearing capacity of plumbing systems, preventing potential damage from excessive water pressure.
In conclusion, understanding the weight of 125 gallons of water is essential due to its practical implications and relevance in various contexts. Water density, influenced by temperature, plays a
crucial role in calculating the weight accurately. Knowing the weight of water enables individuals to make informed decisions about water usage, as well as ensuring safe transportation and
construction practices.
IWater Weight Calculation
Formula for Calculating Water Weight
When it comes to determining the weight of water, a specific formula is used. The formula for calculating water weight is relatively simple: weight equals volume multiplied by density. In the case of
125 gallons of water, the volume would be 125 gallons, and the density of water is generally considered to be 8.345 pounds per gallon. Therefore, the formula for calculating the weight of 125 gallons
of water would be:
Weight = Volume x Density = 125 gallons x 8.345 pounds/gallon
Considerations for Accurate Weight Calculation
While the formula mentioned above serves as a general guideline for calculating the weight of water, there are a few factors that need to be taken into account for accurate results. One such factor
is water temperature. Water density varies slightly with temperature changes. As water temperature increases, its density decreases, and vice versa. Therefore, it is crucial to consider the
temperature of the water when calculating its weight.
To ensure accurate weight calculation, it is recommended to measure the water temperature at the time of weighing. This can be done using a reliable thermometer. Once the temperature is determined,
it is important to adjust the density value accordingly to obtain the most precise weight measurement.
Additionally, it is worth noting that the formula provided assumes standard atmospheric pressure. If the water is being weighed under non-standard atmospheric conditions, such as at a different
altitude, adjustments may need to be made to account for the variations in pressure.
By taking these considerations into account, one can accurately determine the weight of 125 gallons of water. Knowing the weight can have various practical applications and provide insights into the
impact water weight has on everyday life. In the following sections, we will explore these applications and implications in more detail.
Standard Weight of Water
Explaining the Standard Weight of Water per Gallon
In order to understand the weight of 125 gallons of water, it is important to first establish the standard weight of water per gallon. The standard weight of water is commonly accepted as 8.34 pounds
per gallon. This means that for every gallon of water, it weighs approximately 8.34 pounds.
Reasons behind the Standard Weight
The standard weight of water per gallon is determined by its density. Water is known to have a density of 1 gram per milliliter or 1 kilogram per liter. Since there are approximately 3.785 liters in
a gallon, this translates to a density of 1000 kilograms per 3785 liters or approximately 8.34 pounds per gallon.
This standard weight is important for various practical purposes. It allows for easy measurement and conversion when dealing with water in different industries and everyday situations. For example,
it helps in determining the weight of water for shipping, understanding the load-bearing capacity of structures, and managing water supplies in households and businesses.
The consistency of the standard weight also enables accurate calculations and comparisons. It provides a reliable baseline for understanding the weight of water in different contexts, allowing for
better decision-making when it comes to transportation, construction, and water usage.
Analyzing the Weight of 125 Gallons of Water
Now that we know the standard weight of water per gallon, we can calculate the weight of 125 gallons of water. Multiplying the standard weight of 8.34 pounds per gallon by 125 gallons gives us a
total weight of 1042.5 pounds.
To put this into perspective, 1042.5 pounds is roughly equivalent to the weight of a small car or a large motorbike. Visualizing this amount of weight can help us understand the immense mass of 125
gallons of water and the impact it can have on various scenarios.
Understanding the weight of 125 gallons of water is particularly important in situations where accurate weight calculations are crucial, such as shipping or construction. It allows for proper
planning and ensures that structures and transportation systems can support the weight of the water being moved or stored.
In conclusion, knowing the weight of 125 gallons of water provides valuable insights into the physical properties and practical applications of water. By understanding its weight, we can make more
informed decisions regarding water usage, transportation, and construction while appreciating the substantial nature of this essential resource.
Analyzing the Weight of 125 Gallons of Water
Present the calculation for determining the weight of 125 gallons of water
When it comes to understanding the weight of water, it is essential to consider specific quantities. One common measure used is gallons. In this section, we explore the weight of a significant volume
of water, specifically 125 gallons.
To calculate the weight of 125 gallons of water, we need to consider the density of water and its volume. The formula for calculating the weight of water is as follows:
Weight of water = Volume of water × Density of water
In this case, the volume is 125 gallons. However, we need to convert gallons to a unit that is more commonly used in weight measurements, such as pounds or kilograms. To convert gallons to pounds, we
can use the conversion factor that states 1 gallon of water is approximately 8.34 pounds.
Therefore, we can calculate the weight of 125 gallons of water as follows:
Weight of water = 125 gallons × 8.34 pounds/gallon
Weight of water = 1042.5 pounds
So, 125 gallons of water weighs approximately 1042.5 pounds.
Offer a step-by-step explanation to ensure clarity
To ensure clarity, let’s break down the calculation step by step:
Step 1: Convert the volume from gallons to pounds.
125 gallons × 8.34 pounds/gallon = 1042.5 pounds
Step 2: The result is the weight of 125 gallons of water, which is approximately 1042.5 pounds.
It is important to note that this calculation assumes the water is at a standard temperature and pressure. If the temperature deviates significantly from the standard, it can affect the density of
water and thus the accuracy of the weight calculation.
Understanding the weight of 125 gallons of water is crucial in various scenarios. It enables us to make informed decisions, such as determining the load capacity for transportation or ensuring the
structural integrity of containers and storage units. Additionally, construction projects often require accurate knowledge of water weight to determine the necessary materials and to maintain safety
In the next section, we will further explore the weight of 125 gallons of water by providing relatable examples and comparisons to common household items and other substances.
Comparison to Common Objects and Substances
Visualizing the Weight of 125 Gallons of Water
In this section, we will explore relatable examples to better visualize the weight of 125 gallons of water. By comparing it to common household items or other substances, we can gain a clearer
understanding of just how heavy this amount of water is.
To put it into perspective, imagine carrying eight full-sized washing machines. Each washing machine weighs approximately 155 pounds, which means that the combined weight of eight machines is roughly
equivalent to 125 gallons of water. This visual comparison highlights the significant weight that 125 gallons of water holds.
Another relatable example pertains to transportation. A standard car tire weighs around 20 pounds. To match the weight of 125 gallons of water, you would need to load over six car tires into your
vehicle. This comparison showcases the bulkiness and weightiness of 125 gallons of water, emphasizing that it is not a trivial amount.
Alternatively, let’s look at household beverages. An average case of 24 water bottles, with each bottle weighing half a pound, weighs a total of 12 pounds. To match the weight of 125 gallons of
water, you would need to stack more than 1,000 water bottles, which perfectly demonstrates the vastness of this volume.
By comparing the weight of 125 gallons of water to tangible objects and substances, we realize the substantial burden it presents. This understanding becomes particularly vital in scenarios that
require moving or transporting large volumes of water.
In industries like construction, where water is often used for mixing materials such as concrete, comprehending the weight of 125 gallons of water can significantly impact the planning and execution
of projects. Additionally, in transportation and shipping, knowing the weight of water plays a crucial role in ensuring weight limits are not exceeded and preventing vehicle overloading.
Overall, this comparison helps us grasp the magnitude of 125 gallons of water and its weight. It serves as a reminder that water, while essential and life-giving, can also be incredibly heavy,
necessitating careful consideration in various practical situations.
Practical Applications
Importance of Understanding Water Weight
Understanding the weight of water, particularly the weight of large quantities such as 125 gallons, has numerous practical applications in various scenarios. Recognizing the significance of water
weight can lead to more informed decision-making and better management of resources.
Transportation and Construction
One practical application of knowing the weight of water is in transportation and construction industries. Being aware of the weight of 125 gallons of water is essential for professionals involved in
the transportation of liquids, such as drivers of tanker trucks. Overloading a vehicle beyond its weight capacity can lead to serious safety risks on the road. By understanding the weight of water,
appropriate measures can be implemented to ensure that vehicles are not overloaded, preventing accidents and damage to infrastructure.
In the construction industry, knowledge of water weight is crucial for planning and executing various projects. Water is commonly used in concrete mixing, and understanding its weight helps in
determining the appropriate ratio of water to cement. Improper ratios can result in weaker concrete structures, leading to safety hazards and costly repairs. By understanding the weight of water,
construction workers can ensure the structural integrity of their projects.
Emergency Preparedness
Understanding the weight of water is also important in emergency preparedness. In cases of natural disasters or emergencies where the water supply is limited, having prior knowledge of water weight
enables individuals and organizations to store an adequate amount of water for essential needs. For example, in emergency evacuation centers, knowing the weight of 125 gallons of water helps in
planning and distributing water resources efficiently and effectively to meet the needs of the affected individuals.
Water Conservation
Knowing the weight of water can also serve as a guide for promoting water conservation practices. By understanding the weight of 125 gallons of water, individuals can visualize the amount of water
being consumed or wasted. This knowledge can encourage conscious efforts to reduce water usage, such as shorter showers or fixing leaky faucets promptly. It also highlights the significance of water
conservation on a larger scale, emphasizing the need for sustainable water management and preservation of this valuable resource.
In conclusion, understanding the weight of water, especially the weight of 125 gallons, has practical applications in various areas of life. Whether it’s in transportation, construction, emergency
preparedness, or water conservation, knowledge of water weight enables individuals and professionals to make informed decisions, ensure safety, and promote responsible water usage. By recognizing the
importance of knowing the weight of 125 gallons of water, we can contribute to a more sustainable and efficient use of this vital resource.
Implications for Everyday Life
Impact of Water Weight on Daily Activities
Understanding the weight of water, particularly 125 gallons of it, can have significant implications for everyday life. Water weight plays a crucial role in a variety of activities and decisions that
individuals make on a regular basis.
One important implication of water weight is related to transportation. Knowing the weight of 125 gallons of water is essential for determining the load capacity of vehicles, such as trucks or boats,
that transport large quantities of water. Overloading a vehicle can have serious consequences, including damage to the vehicle, increased fuel consumption, and a potential risk to public safety. By
understanding the weight of water, individuals involved in transportation can ensure that they adhere to the appropriate weight limits for their vehicles.
Water weight is also important in the construction industry. Contractors and engineers need to consider the weight of water when designing structures that incorporate water features, such as swimming
pools or water tanks. Understanding the weight of 125 gallons of water allows them to calculate the structural load and ensure that the supporting elements can handle the weight without compromising
the integrity of the construction.
Additionally, knowledge of water weight has implications for everyday decisions related to water usage. For example, individuals interested in sustainability may choose to collect rainwater for
various purposes, such as irrigation or household chores. Understanding the weight of water helps individuals determine the size and weight of the containers needed to store the collected water. This
knowledge ensures that the containers can effectively hold the amount of water without the risk of overflow or structural damage.
The weight of water also influences decisions related to physical activities. Athletes involved in water sports, such as rowing or canoeing, need to factor in the weight of the water and equipment
when determining their performance capabilities. Furthermore, understanding the weight of water in recreational environments, such as swimming pools or beaches, is essential for ensuring the safety
of individuals using these facilities.
In conclusion, understanding the weight of water, specifically 125 gallons of it, has significant implications for everyday life. It influences decisions related to transportation, construction,
water usage, and physical activities. By recognizing the impact of water weight, individuals can make informed choices that promote safety, efficiency, and sustainability in various aspects of their
daily lives.
Summarizing the Importance of Knowing the Weight of 125 Gallons of Water
In conclusion, understanding the weight of 125 gallons of water is essential due to its numerous practical applications and implications in everyday life. Throughout this article, we explored the
concept of water weight, its measurement in gallons, and the calculation of water weight. We also delved into the standard weight of water per gallon and the specific calculation for determining the
weight of 125 gallons of water. Furthermore, we provided relatable examples to help visualize the weight and discussed the practical applications of knowing water weight in various scenarios.
Knowing the weight of water is crucial in several industries and activities. In transportation, understanding water weight helps determine the maximum load a vehicle can carry, ensuring safety and
efficient transportation of goods. In construction, knowledge of water weight is essential for determining the stability and load-bearing capacity of structures, preventing accidents and ensuring the
structural integrity of buildings.
Moreover, everyday activities are influenced by water weight. Whether it is calculating the load of groceries or assessing the weight of a watering can, understanding water weight allows us to make
informed decisions and avoid straining ourselves or damaging objects.
Understanding water weight also plays a significant role in decisions related to water usage. Being aware of the weight of 125 gallons of water can help individuals and businesses estimate their
water consumption, plan storage capacities, and budget accordingly. It can also guide water conservation efforts, as knowing the weight of water emphasizes the importance of using this precious
resource wisely.
In conclusion, the weight of 125 gallons of water holds great significance in various aspects of life. By grasping the concept of water weight and understanding its calculations, we can make informed
choices, prevent accidents or damages, and contribute to sustainable water management. So next time you encounter a large quantity of water, you will have a clear understanding of its weight and the
implications it carries.
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What is inventory to sales ratio?
What is inventory to sales ratio?
The I/S ratio represents the relationship between your inventory value and your total sales. Its objective is to monitor the capital allocated to inventory, as compared to the company’s sales volume
in a given period. The lower the I/S ratio, the more efficient the company is in allocating capital to its inventory.
What percentage of sales should be inventory?
between 10-20%
Most sectors maintain inventory levels at between 10-20% of sales.
What ratio is sales divided by inventory?
inventory turnover
Sales divided by inventory levels equals inventory turnover. This ratio tells the analyst how many times the inventory sitting in stock has been moved or “turned over” during the average year.
Is 1.5 A good inventory turnover ratio?
If the cost of goods sold was $3 million, the inventory turnover ratio will be 1.5. The higher the inventory turnover ratio, the better. When the ratio is high, it means that you’re able to sell
goods quickly. A low ratio indicates weak sales.
Why is inventory to sales ratio important?
The inventory-to-sales ratio is an important metric that can help companies understand their overall profitability and sustainability. It allows them to gauge their ability to move product and
allocate resources effectively.
Is a high inventory to sales ratio good?
High or rising inventory to sales ratio indicates that the company is incurring more storage and holding cost. Low or reducing inventory to sales ratio suggests that the business is in good health
and is efficiently operating.
What is a good days in inventory ratio?
What Is a Good Days Sale of Inventory Number? In order to efficiently manage inventories and balance idle stock with being understocked, many experts agree that a good DSI is somewhere between 30 and
60 days. This, of course, will vary by industry, company size, and other factors.
What is a low inventory turnover?
What is low inventory turnover? Low inventory turnover is when stock items are slow at moving through the business e.g stock items sit on your shelves for longer than they should, affecting cashflow
and increasing carrying costs.
What is a good day sales in inventory ratio?
Is inventory to sales ratio a leading or lagging indicator?
lagging indicator
As a lagging indicator, inventories are less interesting. Instead, this study focuses on a different variant: the ratio of inventory to sales. In the United States several variants of inventory to
sales ratios are widely watched business cycle indicators.
How do you calculate inventory sales?
The formula for Days Sales of Inventory is: Days Sales of Inventory = (Average Inventory ÷ COGS), multiplied by 365.
What is average inventory formula?
Average inventory is a calculation of inventory items averaged over two or more accounting periods. To calculate the average inventory over a year, add the inventory counts at the end of each month
and then divide that by the number of months.
What is a good inventory turnover?
between 5 and 10
For most industries, the ideal inventory turnover ratio will be between 5 and 10, meaning the company will sell and restock inventory roughly every one to two months. For industries with perishable
goods, such as florists and grocers, the ideal ratio will be higher to prevent inventory losses to spoilage.
What is a good days sales in inventory ratio?
Is high or low days sales of inventory good?
Generally, a small average of days sales, or low days sales in inventory, indicates that a business is efficient, both in terms of sales performance and inventory management. Hence, it is more
favorable than reporting a high DSI.
What are 3 examples of leading indicators?
The index of consumer confidence, purchasing managers’ index, initial jobless claims, and average hours worked are examples of leading indicators.
What is the best leading indicator?
Four popular leading indicators
• The relative strength index (RSI)
• The stochastic oscillator.
• Williams %R.
• On-balance volume (OBV)
What is the average inventory?
How do you calculate inventory ratio?
What is the Inventory Ratio?
1. It can be calculated by dividing the cost of goods sold.
2. Calculation of Inventory Turnover.
3. The formula for the cost of goods sold =Opening stock + Purchases – Closing stock.
4. Secondly, average inventory can be calculated by dividing ( opening stock.
What does an inventory ratio of 5 mean?
Turnover Days in Financial Modeling
You can calculate the inventory turnover ratio by dividing the inventory days ratio by 365 and flipping the ratio. In this example, inventory turnover ratio = 1 / (73/365) = 5. This means the company
can sell and replace its stock of goods five times a year.
How do you increase days sales in inventory?
How to Improve Inventory Turnover
1. Proper forecasting.
2. Automation.
3. Effective marketing.
4. Encourage sale of old stock.
5. Efficient restocking.
6. Smart pricing strategy.
7. Negotiate price rates regularly.
8. Encourage your customers to preorder.
What is a leading KPI?
A leading KPI indicator is a measurable factor that changes before the company starts to follow a particular pattern or trend. Leading KPIs are used to predict changes in the company, but they are
not always accurate.
What are the 4 types of indicators?
The infographic differentiates between four different types, including trend, momentum, volatility, and volume indicators.
• Trend indicators. These technical indicators measure the direction and strength of a trend by comparing prices to an established baseline.
• Momentum indicators.
• Volatility Indicators.
• Volume Indicators.
What is the most accurate stock indicator?
MACD – Moving Average Convergence/Divergence
Several indicators in the stock market exist, and the Moving-Average Convergence/Divergence line or MACD is probably the most widely used technical indicator. Along with trends, it also signals the
momentum of a stock.
What is inventory formula?
The formula to calculate average inventory for an accounting period is: Average inventory = (beginning inventory + ending inventory) / 2. The inventory turnover ratio can now be calculated. The
formula is: Inventory turnover ratio = COGS / average inventory. | {"url":"https://mattstillwell.net/what-is-inventory-to-sales-ratio/","timestamp":"2024-11-03T05:53:34Z","content_type":"text/html","content_length":"41706","record_id":"<urn:uuid:d2906ac2-43b1-43c9-be21-78a7798fd93b>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00702.warc.gz"} |
Ensemble learning - Complex systems and AI
Set learning
Now suppose you have chosen the best possible model for a particular problem and are working to further improve its accuracy. In this case, you will need to apply more advanced machine learning
techniques which are collectively referred to as ensemble learning.
A set is a collection of elements that collectively contribute to a whole. A familiar example is a musical ensemble, which mixes the sounds of several musical instruments to create a beautiful
harmony, or architectural ensembles, which are a collection of buildings designed as a unit. In ensembles, the (whole) harmonious result is more important than the execution of any individual part.
Condorcet's Jury Theorem (1784) is about a set in some sense. It states that, if each member of the jury makes an independent judgment and the probability of each juror's correct decision is greater
than 0.5, then the probability of the correct decision of the entire jury increases with the total number of jurors and tends to a. On the other hand, if the probability of being right is less than
0.5 for each juror, then the probability of a correct decision by the jury as a whole decreases with the number of jurors and tends towards zero.
Consider another example of sets: an observation known as Wisdom of the Crowd. In 1906 Francis Galton visited a rural fair in Plymouth where he saw a competition held for farmers. 800 participants
tried to estimate the weight of a slaughtered bull. The actual weight of the bull was 1198 pounds. Although none of the farmers could guess the exact weight of the animal, the average of their
predictions was 1197 pounds.
A similar idea for error reduction has been adopted in the field of machine learning.
Priming (bagging and bootstrapping)
Bagging (also known as Bootstrap aggregation) is one of the earliest and most basic ensemble techniques. It was proposed by Leo Breiman in 1994. Bagging is based on the bootstrap statistical method,
which makes it possible to evaluate many statistics of complex models.
The bootstrap method proceeds as follows. Consider a sample X of size N. A new sample can be made from the original sample by drawing N elements from the latter in a random and uniform manner, with
replacement. In other words, we select a random element from the original sample of size N and do it N times. All elements are equally likely to be selected, so each element is drawn with equal
probability 1/N.
Let's say we draw balls from a bag one at a time. At each stage, the selected ball is put back in the bag so that the next selection is made in an equiprobable manner, that is to say from the same
number of balls N. Note that, since the balls are put back , there may be duplicates in the new sample. Let's call this new sample X1.
By repeating this procedure M times, we create M bootstrap samples X1, …, XM. In the end, we have a sufficient number of samples and can calculate various statistics from the original distribution.
For our example, we'll use the familiar telecom_churn dataset. Previously, when we discussed the importance of features, we saw that one of the most important features in this dataset is the number
of customer service calls. Let's visualize the data and look at the distribution of this feature.
import panda ace pd
from matplotlib import pyplot ace please
please.rcParams['figure.figsize'] = 10, 6
import sea born ace sns
%matplotlib inline
telecom_data = pd.read_csv('../../data/telecom_churn.csv')
fig = sns.kdeplot(telecom_data[telecom_data['churn'] == False]['Customer service calls'],
label = 'Loyal')
fig = sns.kdeplot(telecom_data[telecom_data['churn'] == True]['Customer service calls'],
label = 'churn')
fig.set(xlabel='Number of calls', ylabel='Density')
As you can see, loyal customers call customer service less than those who eventually left. Now, it might be a good idea to estimate the average number of customer service calls in each group. Since
our dataset is small, we wouldn't get a good estimate by just calculating the mean of the original sample. We'd better apply the bootstrap method. Let's generate 1000 new bootstrap samples from our
original population and produce an interval estimate of the mean.
import numpy ace n.p.
def get_bootstrap_samples(data, n_samples):
“””Generate bootstrap samples using the bootstrap method. » » »
clues = n.p..random.randint(0, then(data), (n_samples, then(data)))
samples = data[clues]
return samples
def stat_intervals(status, alpha):
“””Produce an interval estimate. » » »
boundaries = n.p..percentile(status, [100 * alpha / 2., 100 * (1 – alpha / 2.)])
return boundaries
# Save the data about the loyal and form customers to split the dataset
loyal_calls = telecom_data[telecom_data['churn']
== False]['Customer service calls'].values
churn_calls= telecom_data[telecom_data['churn']
== True]['Customer service calls'].values
# Set the seed for reproducibility of the results
# Generate the samples using bootstrapping and calculate the mean for each of them
loyal_mean_scores = [n.p..mean(sample)
for sample in get_bootstrap_samples(loyal_calls, 1000)]
churn_mean_scores = [n.p..mean(sample)
for sample in get_bootstrap_samples(churn_calls, 1000)]
# Print the resulting interval estimates
print(“Service calls from loyal: mean interval”,
stat_intervals(loyal_mean_scores, 0.05))
print(“Service calls from churn: mean interval”,
stat_intervals(churn_mean_scores, 0.05))
Now that you've gotten the bootstrap idea, we can move on to bagging. In a problem of regression, by averaging the individual responses, bagging reduces the root mean square error by a factor M, the
number of regressors.
From our previous lesson, let's recall the components that make up the total out-of-sample error:
Bagging reduces the variance of a classifier by decreasing the error difference when we train the model on different datasets. In other words, bagging prevents overfitting. The effectiveness of
bagging comes from the fact that the individual models are quite different due to different training data and their errors cancel each other out when voting. Also, outliers are likely omitted in some
of the training starter samples.
The scikit-learn library supports bagging with the BaggingRegressor and BaggingClassifier meta-estimators. You can use most algorithms as a base.
Let's take a look at how bagging works in practice and compare it with the decision tree. For this we will use an example from the sklearn documentation.
The error for the decision tree:
0.0255 = 0.0003 (bias²)+ 0.0152 (variance) + 0.0098 (σ²)
The error when using bagging:
0.0196 = 0.0004 (bias²) + 0.0092 (variance) + 0.0098 (σ²)
As you can see from the graph above, the error variance is much lower for bagging. Remember that we have already proven this theoretically.
Bagging is efficient on small datasets. Removing even a small part of the training data leads to the construction of significantly different base classifiers. If you have a large data set, you will
generate bootstrap samples of a much smaller size.
The example above is unlikely to apply to real work. This is because we made the strong assumption that our individual errors are uncorrelated. More often than not, this is far too optimistic for
real-world applications. When this assumption is false, the reduction in error will not be as great. In subsequent lectures, we will discuss some more sophisticated ensemble methods, which allow for
more accurate predictions in real-world problems.
In the future, in the case of random forest, it is not necessary to use cross-validation or exclusion samples to obtain an unbiased error estimate. Why? Because, in ensemble techniques, error
estimation takes place internally.
Random trees are constructed using different bootstrap samples from the original dataset. About 37 % of the inputs are excluded from a particular bootstrap sample and are not used in the construction
of the K-th tree.
Let's see how Out-of-Bag (or OOBE) error estimation works:
The upper part of the figure above represents our original dataset. We've split it into practice (left) and test (right) sets. In the image on the left, we draw a grid that neatly divides our dataset
by classes. Now we use the same grid to estimate the share of correct answers on our test set. We can see that our classifier gave incorrect answers in these 4 cases which were not used during
training (left). Therefore, the precision of our classifier is 11/15*100 % = 73.33 %.
To sum up, each algorithm base is trained on ~63 % of the original examples. It can be validated on the remaining ~37%. The Out-of-Bag estimate is nothing more than the average estimate of the base
algorithms over the ~37 % of inputs that have not been trained.
Leo Breiman succeeded in applying the bootstrap not only in statistics but also in machine learning. Together with Adel Cutler, he extended and improved the Random Forest algorithm proposed by Tin
Kam Ho. They combined the construction of uncorrelated trees using CART, bagging and the random subspace method.
Decision trees are a good choice for the basic classifier in bagging because they are quite sophisticated and can achieve zero classification errors on any sample. The random subspace method reduces
the correlation between the shafts and thus avoids overfitting. With bagging, the core algorithms are trained on different random subsets of the original feature set.
The following algorithm builds a set of models using the random subspace method:
1. Suppose the number of instances is equal to n and the number of dimensions of the entity is equal to d.
2. Choose M as the number of individual models in the set.
3. For each model m, choose the number of features dm < d. As a general rule, the same dm value is used for all models.
4. For each model m, create a training set by selecting dm features at random from the feature set d.
5. Train each model.
6. Apply the resulting ensemble model to a new input by combining the results of all models of M. You can use either majority voting or posterior probability aggregation.
The algorithm is as follows:
The final classifier is the mean of the trees.
For classification problems, it is advisable to set m equal to the square root of d. For regression problems, we usually take m = d/3, where d is the number of features. It is recommended to build
each tree until all its leaves contain only 1 instance for classification and 5 instances for regression.
You can think of Random Forest as a clustering of decision trees with the changing selection of a random subset of features with each split.
Here are the results for the three algorithms:
As we can see from our graphs and the MSE values above, a Random Forest of 10 trees achieves a better result than a single decision tree and is comparable to bagging with 10 trees. The main
difference between Random Forests and bagging is that, in a Random Forest, the best feature for a split is selected from a random subset of the available features while, in bagging, all features are
considered for the next best split.
We can also look at the benefits of random forests and classification problems.
The figures above show that the decision boundary of the decision tree is quite irregular and has many sharp angles that suggest overfitting and poor ability to generalize. We would struggle to make
reliable predictions on new test data. In contrast, the bagging algorithm has a rather smooth boundary and shows no obvious signs of overfitting.
Now let's look at some parameters that can help us increase the accuracy of the model.
Parameters to increase accuracy
The scikit-learn library implements random forests by providing two estimators: RandomForestClassifier and RandomForestRegressor.
Below are the parameters we need to pay attention to when building a new model:
• n_estimators is the number of trees in the forest;
• criterion is the function used to measure the quality of a split;
• max_features is the number of features to consider when finding the best distribution;
• min_samples_leaf is the minimum number of samples required to be at a leaf node;
• max_depth is the maximum depth of the tree.
The most important fact about random forests is that its accuracy does not decrease when we add trees, so the number of trees is not a hyperparameter of complexity unlike max_depth and
min_samples_leaf. This means you can tune the hyperparameters with, say, 10 trees, then increase the number of trees up to 500 and be sure that the accuracy will only improve.
Extremely random trees use a greater degree of randomization to the choice of cutpoint when splitting a tree node. As in Random Forests, a random subset of features is used. But, instead of searching
for optimal thresholds, their values are randomly selected for each possible feature, and the best among these randomly generated thresholds is used as the best rule to split the node. This usually
compensates for a slight reduction in model variance with a slight increase in bias.
In the scikit-learn library, there are 2 extremely random tree implementations: ExtraTreesClassifier and ExtraTreesRegressor.
This method should be used if you have overdone it with random forests or gradient boosting.
Conclusion on the random forest
• High prediction accuracy; will perform better than linear algorithms in most problems; the precision is comparable to that of boosting;
• Robust to outliers, thanks to random sampling;
• Insensitive to feature scaling as well as any other monotonic transformation due to random subspace selection;
• Doesn't require fine-tuning of settings, works quite well out of the box. With tuning, an accuracy gain of 0.5 to 3 % can be achieved, depending on the problem tuning and data;
• Effective for datasets with a large number of features and classes;
• Handles both continuous and discrete variables;
• Rarely oversized. In practice, an increase in the number of trees almost always improves the composition. But, after reaching a certain number of trees, the learning curve is very close to the
• There are methods developed to estimate the importance of features;
• Works well with missing data and maintains good levels of precision even when a large portion of the data is missing;
• Provides means to weight classes over the dataset as well as for each sample tree;
• Under the hood, calculates proximities between pairs of instances which can then be used in clustering, outlier detection, or interesting data representations;
• The above functionality and properties can be extended to unlabeled data to allow unsupervised grouping, data visualization and detection of outliers;
• Easily parallelized and highly scalable.
The inconvenients:
• Compared to a single decision tree, the output of Random Forest is more difficult to interpret.
• There are no formal p-values for estimating feature importance.
• Performs less well than linear methods in the case of sparse data: text entries, bag of words, etc. ;
• Unlike linear regression, Random Forest is unable to extrapolate. But this can also be seen as an advantage because outliers do not cause outliers in random forests;
• Prone to overfitting in some problems, especially when dealing with noisy data;
• In the case of categorical variables with varying numbers of levels, random forests favor variables with a larger number of levels. The tree structure will adapt more to multi-level
functionality, as it becomes more precise;
• If a dataset contains groups of correlated features with similar importance for the predicted classes, preference will be given to smaller groups;
• The resulting model is large and requires a lot of RAM. | {"url":"https://complex-systems-ai.com/en/analyse-des-donnees/learning-sets/","timestamp":"2024-11-05T06:32:24Z","content_type":"text/html","content_length":"175775","record_id":"<urn:uuid:2a2c0a27-ec56-4350-b75a-6df5b7c5151c>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00143.warc.gz"} |
How Many Seconds In 100 Years - How Many Sumo
How Many Seconds In 100 Years
The measurement of time plays a fundamental role in our understanding of the world and the events that occur within it. The concept of time allows us to organize our lives, schedule our activities,
and make sense of the passage of days, months, and years.
While we often think of time in terms of larger units such as years and centuries, it is also possible to break it down into smaller, more precise increments, such as seconds. In this article, we
will explore the precise calculation of how many seconds are contained within a span of 100 years, using an analytical and objective approach.
To determine the number of seconds in 100 years, we must first establish the number of seconds in a single year. A year consists of 365 days, each of which contains 24 hours, 60 minutes, and 60
seconds. By multiplying these values together (365 x 24 x 60 x 60), we find that there are 31,536,000 seconds in one year.
With this information, we can then calculate the number of seconds in 100 years by multiplying the number of seconds in a single year by 100. Thus, the total number of seconds in 100 years would be
By breaking down time into its smallest increments, we are able to precisely analyze and calculate the number of seconds contained within a century.
The Concept of Time Measurement
The concept of time measurement involves quantifying the duration of events and intervals, such as determining the number of seconds in a given time span, such as 100 years.
Time perception refers to the subjective experience of the passage of time, which can vary among individuals and cultures. Cultural differences have a profound influence on how time is perceived and
measured. Different cultures have distinct ways of understanding and organizing time, which can be influenced by factors such as historical events, religious beliefs, and societal norms.
In some cultures, time is seen as a linear progression from the past to the future, with a focus on punctuality and efficiency. These cultures tend to view time as a finite resource that should be
used wisely and not wasted.
On the other hand, some cultures have a more cyclical view of time, where events are seen as recurring in a continuous cycle. In these cultures, the emphasis is on the present moment, and there may
be less importance placed on punctuality and strict adherence to schedules.
Cultural differences in time perception can also be seen in the concept of ‘clock time’ versus ‘event time.’ Clock time is based on the ticking of a clock or the measurement of seconds, minutes, and
hours. It is a standardized and objective way of measuring time that is commonly used in Western societies.
In contrast, event time is more flexible and subjective, where time is perceived in relation to specific events or activities. This can be seen in cultures where the focus is on the completion of a
task or the achievement of a goal, rather than strict adherence to a specific schedule.
The concept of time measurement involves quantifying the duration of events and intervals. Time perception varies among individuals and cultures, influenced by factors such as historical events,
religious beliefs, and societal norms. Cultural differences can be observed in the ways time is understood and organized, with some cultures emphasizing linear progression and punctuality, while
others have a more cyclical view and prioritize the present moment.
Understanding these cultural differences is important in various fields, such as cross-cultural communication, international business, and psychology.
Calculating the Number of Seconds in a Year
Calculating the duration of 100 years results in a significant accumulation of seconds. In order to determine the number of seconds in a year, it is necessary to convert the time unit of years into
seconds. This can be achieved by considering the conversion factors between years, months, days, hours, minutes, and seconds.
To illustrate this, a table can be used to showcase the conversion process. In the first column of the table, the various time units are listed, starting with years and ending with seconds. The
second column represents the conversion factor for each time unit, indicating the number of that unit in relation to the preceding unit. For example, there are 12 months in a year, 365 days in a
year, 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute. By multiplying these conversion factors together, we can determine the number of seconds in a year.
Time Unit Conversion Factor
Year 1
Month 12
Day 365
Hour 24
Minute 60
Second 60
Starting with 100 years, we can multiply the conversion factors together to calculate the number of seconds. By multiplying 100 years by 12 months per year, then by 365 days per year, then by 24
hours per day, then by 60 minutes per hour, and finally by 60 seconds per minute, we obtain the total number of seconds in 100 years. This calculation can be easily performed using a calculator or a
spreadsheet software, resulting in a substantial accumulation of seconds over the course of a century.
Determining the Number of Seconds in a Century
Determining the number of seconds in a century involves multiplying the number of seconds in a year by the number of years in a century.
This calculation accounts for the fact that a century consists of 100 years.
Additionally, it is important to consider leap years, which occur every four years, except for years that are divisible by 100 but not by 400.
By incorporating these factors, one can accurately determine the total number of seconds in a century.
Multiplying seconds by years
Multiplying the number of seconds by the number of years allows for the determination of the total duration in seconds. When calculating the number of seconds in a given number of years, it is
essential to consider the conversion factor between years and seconds.
This conversion factor is derived from the fact that there are 60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, and 365 days in a year. By multiplying these conversion factors, one
can determine the total number of seconds in a year, and subsequently, in a given number of years.
To calculate the total duration in seconds, one needs to multiply the number of years by the number of seconds in a year. For example, if we multiply 100 years by the number of seconds in a year
(which is 60 seconds 60 minutes 24 hours * 365 days), we can determine the total duration in seconds for 100 years.
This multiplication calculation allows for a precise and logical determination of the number of seconds in a given number of years.
By following the conversion factors and performing the multiplication calculations, one can accurately determine the number of seconds in a specific number of years. This objective and analytical
approach to calculating time conversions provides a reliable method for understanding the duration of time in seconds.
Accounting for leap years
Accounting for leap years necessitates considering the additional day added to the calendar every four years, which affects the precise calculation of the duration in seconds for a given number of
Leap year calculations involve accounting for the fact that a standard year consists of 365 days, while a leap year has 366 days. This extra day is added to the month of February, extending its
duration from 28 to 29 days. Consequently, when determining the number of seconds in a given number of years, it is crucial to adjust for the leap years within that time frame.
To incorporate leap year adjustments into the calculation, one must first determine the number of leap years within the given period. This can be achieved by dividing the total number of years by
four, as every fourth year is a leap year. However, it is important to note that centuries not divisible by 400 are not leap years. For instance, the years 1700, 1800, and 1900 were not leap years
despite being divisible by four. This exception ensures that the calendar remains aligned with the Earth’s revolutions around the Sun.
Once the number of leap years is determined, it can be multiplied by 24 (as each day consists of 24 hours), 60 (as each hour consists of 60 minutes), and 60 (as each minute consists of 60 seconds) to
calculate the additional seconds accounted for by the leap years.
By incorporating these leap year adjustments, one can accurately calculate the precise duration in seconds for a given number of years.
Reflecting on the Significance of Time
The fleeting nature of seconds invites us to reflect on the significance of time.
Each second that passes is a reminder of the transience of life and the limited time we have to accomplish our goals.
This realization has implications for both our personal and societal perspectives on time.
On an individual level, it highlights the importance of making the most out of every moment and prioritizing meaningful experiences.
From a societal standpoint, it emphasizes the need for efficient resource allocation and long-term planning to ensure the well-being of future generations.
Ultimately, understanding the fleeting nature of seconds can lead to a more conscious and intentional approach to time management.
Considering the fleeting nature of seconds
Considering the transitory essence of seconds, it is intriguing to ponder the cumulative number of seconds in a span of 100 years.
Seconds, as the smallest unit of time measurement, hold a fleeting nature that often goes unnoticed in our daily lives. However, when we reflect on the significance of time, it becomes apparent that
seconds play a crucial role in shaping our perception of time.
The fleeting nature of seconds can be experienced in various ways, such as in the rapid passing of moments or the brevity of a single heartbeat. Each second that ticks away is a reminder of the
impermanence of time and the ever-changing nature of our existence.
Time perception is deeply intertwined with the concept of seconds. Our perception of time can vary depending on the context and our state of mind. When engaged in an enjoyable activity, seconds may
seem to fly by, while in moments of boredom or discomfort, they can stretch out endlessly. This subjective experience of time highlights the significance of seconds as a unit of measurement.
By measuring time in seconds, we can objectively track the passing of moments and analyze our experiences with precision. This analytical approach allows us to understand the fleeting nature of time
and its impact on our perception.
The fleeting nature of seconds underscores their significance in shaping our perception of time. By considering the cumulative number of seconds in a span of 100 years, we are reminded of the
transitory essence of time and the impermanence of our existence.
The subjective experience of time perception further highlights the importance of seconds as a precise and analytical unit of measurement. As we delve deeper into the concept of time, it becomes
evident that understanding the nature of seconds is essential in comprehending our own place in the ever-changing fabric of time.
Implications for personal and societal perspectives on time
Implications for personal and societal perspectives on time can be observed through the reflection on the fleeting nature of seconds and the subjective experience of time perception. The realization
that seconds pass by quickly and cannot be retrieved creates a sense of urgency and highlights the importance of making the most of every moment. This personal reflection on the ephemeral nature of
time can lead individuals to prioritize their goals, make better use of their time, and strive for personal growth and fulfillment.
Moreover, cultural perceptions of time also play a significant role in shaping personal and societal perspectives. Different cultures may have varying attitudes towards time, with some emphasizing
punctuality and productivity, while others prioritize a more relaxed and leisurely approach. These cultural differences can influence individuals’ perception of time, their work-life balance, and
their overall well-being.
To evoke an emotional response in the audience, consider the following list:
1. Regret: The fleeting nature of seconds can evoke feelings of regret when individuals realize they have not made the most of their time or missed out on important opportunities.
2. Motivation: The realization that seconds are passing quickly can serve as a powerful motivator, inspiring individuals to seize the moment, set goals, and take action towards their aspirations.
3. Gratitude: Reflecting on the fleeting nature of seconds can instill a sense of gratitude for the time we have, encouraging individuals to appreciate and savor each moment.
4. Existential contemplation: The brevity of seconds can lead individuals to ponder the larger questions of life’s purpose and meaning, prompting introspection and personal growth.
By considering personal reflection and cultural perceptions, it becomes apparent that the fleeting nature of seconds can have profound implications for individuals and society as a whole. It
underscores the importance of being mindful of time, making conscious choices, and embracing a balanced approach to work, leisure, and personal growth. Recognizing the value of each passing second
can lead to a more purposeful and fulfilled life.
In conclusion, the concept of time measurement is fundamental in our understanding of the world. By calculating the number of seconds in a year, we can determine the number of seconds in a century.
This analytical approach allows us to accurately quantify time and its significance in our lives.
Time measurement is a precise and logical process that helps us organize our activities and make sense of the world around us. By breaking down time into smaller units such as seconds, we can measure
and compare different durations. Through this analytical approach, we can calculate the number of seconds in a year, which in turn allows us to determine the number of seconds in a century. This
logical calculation demonstrates the importance of time in our lives and how it shapes our understanding of the world.
The significance of time cannot be overstated. It is a universal concept that affects every aspect of our lives. By accurately measuring and quantifying time, we can better plan and manage our
The precise calculation of the number of seconds in a century helps us appreciate the vastness of time and how it unfolds over long periods. This analytical approach to time measurement allows us to
gain a deeper understanding of the world and our place in it.
In conclusion, the study of time is crucial for our understanding of the world and our ability to navigate through it.
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On minimization problems which approximate Hardy L<sup>p</sup> inequality
Let Ω be a smooth bounded domain in ℝ^N with 0 ∈ Ω and let p ∈ (1,∞)\{N}. By a classical inequality of Hardy we have ∫Ω|∇v|^p>c[p,N]^*∫ Ω|v|^p/|x|^p, for all 0≠v ∈ W[0]^1,p(Ω\{0}), with c[p,N]^*=|
(N-p)/p|^p being the best constant in this inequality. More generally, for η ∈ C(Ω̄) such that η≥0,η≠0 and η(0)=0 we have, for certain values of λ, that ∫Ω|∇v|^p-λη|v|^p/|x|^p >c[p,N]^*∫Ω|v|^p/|x|^p ,
for all 0≠v ∈ W[0]^1,p(Ω\{0}). In particular, it follows that there is no minimizer for this inequality. We consider then a family of approximating problems, namelyinf0≠v ∈ W[0]^1,p(Ω\{0})∫Ω|∇v|^p-
λη|v|^p/|x|^p∫Ω|v|^p-ε/| x|^pfor ε>0, and study the asymptotic behavior, as ε→0, of the positive minimizers {u[ε]} which are normalized by ∫Ωu[ε]^p=1. We prove the convergence u[ε]→u[*] in ∩[1<q<p]W
[0]^1,q(Ω\{0}), where u[*] is the unique positive solution (up to a multiplicative factor) of the equation -Δ[p]u=(u^p-1/|x|^p)(c[p,N]^ *+λη(x)) in Ω\{0}, with u=0 on ∂Ω.
• Hardy's inequality
• P-Laplacian
• Singular elliptic problem
ASJC Scopus subject areas
• Analysis
• Applied Mathematics
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Compound Interest MCQ Quiz [PDF] Questions Answers | Compound Interest MCQs App Download & e-Book
Class 8 Math Online Tests
Compound Interest MCQ (Multiple Choice Questions) PDF Download
The Compound interest Multiple Choice Questions (MCQ Quiz) with Answers PDF (Compound Interest MCQ PDF e-Book) download to practice Grade 8 Math Tests. Learn Math Applications Multiple Choice
Questions and Answers (MCQs), Compound Interest quiz answers PDF to study online homeschool courses. The Compound Interest MCQ App Download: Free learning app for taxation, percentage calculations,
hire purchase test prep for online elementary school courses.
The MCQ: John invested $8000 at 6.5% per annum compound interest which is compounded daily. The amount at the end of fifth year is; "Compound Interest" App Download (Free) with answers: $13,960.70;
$10,960.70; $11,960.70; $12,960.70; to study online homeschool courses. Solve Math Applications Quiz Questions, download Google eBook (Free Sample) for online teaching certification programs.
Compound Interest MCQs PDF: Questions Answers Download
MCQ 1:
The compound interest on $2500 for 5 years at 4% per annum (compounded annually) is
1. $432
2. $542
3. $642
4. $452
MCQ 2:
John invested $8000 at 6.5% per annum compound interest which is compounded daily. The amount at the end of fifth year is
1. $13,960.70
2. $10,960.70
3. $11,960.70
4. $12,960.70
MCQ 3:
The compound interest on $7500 for 1 year at 8.5% per annum which is compounded monthly is
1. $363
2. $636
3. $663
4. $366
MCQ 4:
Jane invests $15000 at 5% per annum compound interest which is compounded daily. If a year is equal to 365 days then the amount at the end of fifth day is
1. $15,010.28
2. $18,010.28
3. $16,010
4. $14,010
Class 8 Math Practice Tests
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θ - (Computational Complexity Theory) - Vocab, Definition, Explanations | Fiveable
from class:
Computational Complexity Theory
In computational complexity theory, the notation θ (theta) represents a tight bound on the asymptotic growth of a function. Specifically, a function f(n) is said to be in θ(g(n)) if there exist
positive constants c1, c2, and n0 such that for all n ≥ n0, c1*g(n) ≤ f(n) ≤ c2*g(n). This means that f(n) grows at the same rate as g(n) as n approaches infinity, allowing for a precise
characterization of an algorithm's performance.
congrats on reading the definition of θ. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. The θ notation is significant because it encapsulates both the upper and lower bounds of a function, making it useful for giving an exact asymptotic behavior.
2. Using θ helps in comparing different algorithms by providing a more refined understanding of their time or space complexities rather than just upper or lower bounds.
3. When using θ, one must find constants c1, c2, and n0 that satisfy the definition; this is often done through limits or inequalities.
4. θ notation is essential in proving that certain problems belong to specific complexity classes, such as P or NP.
5. Many common algorithms have known θ notations, which help in classifying their efficiency; for example, the merge sort algorithm has a time complexity of θ(n log n).
Review Questions
• How does θ notation differ from Big O and Omega notations in terms of describing algorithmic complexity?
□ θ notation differs from Big O and Omega notations by providing a tight bound on a function's growth rate. While Big O only gives an upper limit on how fast a function can grow and Omega gives
a lower limit, θ specifies both bounds, meaning it tightly characterizes an algorithm's efficiency. This allows for more precise comparisons between algorithms, as knowing both upper and
lower bounds can provide better insights into performance.
• Explain why understanding θ notation is important when analyzing algorithms in computational complexity theory.
□ Understanding θ notation is crucial because it allows researchers and practitioners to accurately evaluate and compare the efficiency of different algorithms. By establishing a function's
growth rate through θ notation, one can make informed decisions about which algorithm to use based on performance expectations. Additionally, it aids in classifying problems within specific
complexity classes, providing insights into whether problems can be solved efficiently.
• Evaluate the impact of using θ notation on the development of efficient algorithms within the field of computational complexity theory.
□ Using θ notation significantly impacts the development of efficient algorithms by enabling clear communication about algorithm performance. When researchers know the exact growth rate of an
algorithm's time or space complexity, they can identify potential improvements or optimizations more easily. This understanding also encourages the design of algorithms that are optimal for
their specific applications by allowing developers to analyze trade-offs based on precise performance metrics. Ultimately, it fosters innovation and refinement in algorithm design by focusing
on achieving efficient solutions.
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Generate Correlated Data Using Rank Correlation
This example shows how to use a copula and rank correlation to generate correlated data from probability distributions that do not have an inverse cdf function available, such as the Pearson flexible
distribution family.
Step 1. Generate Pearson random numbers.
Generate 1000 random numbers from two different Pearson distributions, using the pearsrnd function. The first distribution has the parameter values mu equal to 0, sigma equal to 1, skew equal to 1,
and kurtosis equal to 4. The second distribution has the parameter values mu equal to 0, sigma equal to 1, skew equal to 0.75, and kurtosis equal to 3.
rng default % For reproducibility
p1 = pearsrnd(0,1,-1,4,1000,1);
p2 = pearsrnd(0,1,0.75,3,1000,1);
At this stage, p1 and p2 are independent samples from their respective Pearson distributions, and are uncorrelated.
Step 2. Plot the Pearson random numbers.
Create a scatterhist plot to visualize the Pearson random numbers.
The histograms show the marginal distributions for p1 and p2. The scatterplot shows the joint distribution for p1 and p2. The lack of pattern to the scatterplot shows that p1 and p2 are independent.
Step 3. Generate random numbers using a Gaussian copula.
Use copularnd to generate 1000 correlated random numbers with a correlation coefficient equal to –0.8, using a Gaussian copula. Create a scatterhist plot to visualize the random numbers generated
from the copula.
u = copularnd('Gaussian',-0.8,1000);
The histograms show that the data in each column of the copula have a marginal uniform distribution. The scatterplot shows that the data in the two columns are negatively correlated.
Step 4. Sort the copula random numbers.
Using Spearman's rank correlation, transform the two independent Pearson samples into correlated data.
Use the sort function to sort the copula random numbers from smallest to largest, and to return a vector of indices describing the rearranged order of the numbers.
[s1,i1] = sort(u(:,1));
[s2,i2] = sort(u(:,2));
s1 and s2 contain the numbers from the first and second columns of the copula, u, sorted in order from smallest to largest. i1 and i2 are index vectors that describe the rearranged order of the
elements into s1 and s2. For example, if the first value in the sorted vector s1 is the third value in the original unsorted vector, then the first value in the index vector i1 is 3.
Step 5. Transform the Pearson samples using Spearman's rank correlation.
Create two vectors of zeros, x1 and x2, that are the same size as the sorted copula vectors, s1 and s2. Sort the values in p1 and p2 from smallest to largest. Place the values into x1 and x2, in the
same order as the indices i1 and i2 generated by sorting the copula random numbers.
x1 = zeros(size(s1));
x2 = zeros(size(s2));
x1(i1) = sort(p1);
x2(i2) = sort(p2);
Step 6. Plot the correlated Pearson random numbers.
Create a scatterhist plot to visualize the correlated Pearson data.
The histograms show the marginal Pearson distributions for each column of data. The scatterplot shows the joint distribution of p1 and p2, and indicates that the data are now negatively correlated.
Step 7. Confirm Spearman rank correlation coefficient values.
Confirm that the Spearman rank correlation coefficient is the same for the copula random numbers and the correlated Pearson random numbers.
copula_corr = corr(u,'Type','spearman')
copula_corr = 2×2
1.0000 -0.7858
-0.7858 1.0000
pearson_corr = corr([x1,x2],'Type','spearman')
pearson_corr = 2×2
1.0000 -0.7858
-0.7858 1.0000
The Spearman rank correlation is the same for the copula and the Pearson random numbers.
See Also
copularnd | corr | sort
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What comes after quarter? - Answers
The Arizona State Quarter comes out next. New Mexico was the last one and after Arizona, Alaska comes out.
Quarter = cuarto; the r comes before the t Four = cuatro; the r comes after the t
The phase right before the last quarter is the third quarter.
The First Quarter moon comes after the Waxing Crescent moon.
The phase of the moon that comes after the third quarter is the waning crescent. It appears as a small sliver of light on the left side of the moon.
After the first quarter moon comes the waxing gibbous moon. The waxing gibbous moon appears as the illuminated portion of the moon grows larger each night until it reaches a full moon. | {"url":"https://math.answers.com/math-and-arithmetic/What_comes_after_quarter","timestamp":"2024-11-07T12:20:39Z","content_type":"text/html","content_length":"157409","record_id":"<urn:uuid:9560ad96-7ad6-4277-9a71-e3b8e26e733d>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00665.warc.gz"} |
M^2 - N^2 = 12...Prove there is only one solution in positive integers and much more
(1) See the visualization for the difference of squares posted on 1-9-08.
(2) Read the comments in this post for considerable clarification and instructor guidelines and suggestions. Mathmom's and Eric's comments are particularly insightful.
This post can be developed into an activity for prealgebra through first-year algebra students (or even 2nd year algebra). The last part is more challenging.
The focus here is on developing a method/strategy that can be used to solve similar Diophantine equations. The other objective is to introduce the ideas and methods of proof. This problem may later
be used to solve a recent math contest problem for which I obtained permission to discuss on this blog. I am fully aware that many students will 'solve' these equations by Guess-Test methods, but
they need to go further.
(a) Prove there is only one solution in positive integers for the equation:
M^2 - N^2 = 12
Note: If we omit the word positive, what would the solution(s) be?
(b) Determine all positive integer solutions:
M^2 - N^2 = 15
(c) Determine all positive integer solutions:
M^2 - N^2 = 36
(d) Let's investigate for what positive integer values of P, M^2 - N^2 = P has NO solutions in positive integers.
(i) Determine at least 5 positive integer values of P for which the above equation has no positive integer solutions.
(ii) (More challenging) Describe all values of P for which the above equation has no solutions. Justify your result.
Note: All students should have success with (i), although some may struggle to find 5 values. Part(ii) should challenge the student who has finished the other parts in rapid order and sits there
Additional Comment: If P is itself a perfect square, our equation is obviously related to the most famous equation in geometry. Thus, if P = 9 or P = 16, for example, students should recognize
something! For this reason you may want to have students consider these values when doing this investigation. More to come...
8 comments:
mathmom said...
I can tell you for sure none of my pre-algebra students can "prove" any of this. They might be able to guess/check to get solutions, but I'm skeptical of even that.
Diophantine equations were college algebra for me, and I remember nothing. I can see that there is only one answer to (a) by "brute force" (looking at the increasing pattern of differences
between the squares) but I see no easy way to prove it; certainly nothing accessible to pre-algebra students, even if they were comfortable with the notion of a "proof" (which they are not). So,
am I missing something, or is this not really accessible to a pre-algebra level.
By the way, what was the contest?
With such problems, it's always best to find a multiplicative formulation. For example, the Goldbach conjecture asks whether every even integer ≥ 4 is the sum of two primes. The answer is
unknown. But every positive integer is the product of primes in exactly one way.
So, write (a) as:
12 = M²−N² = (M+N) (M−N).
But the two factors differ by 2N; so they are both even or both odd. Since the product is even, they're both even. Factorizing 12 gives
M+N = 6
M−N = 2.
M = 4, N = 2.
(b) By the same argument,
M+N = 15, M−N = 1, or
M+N = 5, M−N = 3.
M=8, N=7, or
M=4, N=1.
(c) Again:
M+N = 18, M−N = 2, or
M+N = 6, M−N = 6.
M = 10, N = 8, or
M = 6, N = 0.
(d) What sort of number is a square? Do squares have special characteristics? I do doubt whether any high schooler would realize without prompting that squares ≡ 0 or 1 (mod 4). If they do, then
they can find that M²−N² ≡ −1, 0, or 1 (mod 4).
So, numbers ≡2 (mod 4) satisfy d.
Going to higher mathematics, one of the first ways to factorize numbers came from this equation. It was invented by Blaise Pascal. I'll write again about it later.
Mathmom, Eric--
This problem is certainly ambitious for the younger student. Eric's in-depth analysis is certainly beyond the expectations of middle schoolers, however, take a look at today's post which uses an
area diagram to demonstrate the factorizaton of the difference of squares. This is appropriate for prealgebra students IMO.
Certainly, most students would begin by guessing values to solve these equations. The instructor should have them organize their findings in a table (see below) from which conjectures could be
more easily made. This might be more than enough for the prealgebra students. The justifications would be for the older student -- I should have made that clearer.
Eric, in addition to P=1, P=2 and P=4, which are exceptional cases, I believe some students would be able to see the generalization that non-representable values are of the form 2⋅(an odd
integer) =
2(2n+1) = 4n+2, which is equivalent to your modular representation.
Here's why:
P = (M+N(M-N)
Case I:If P is odd (and greater than 1), we CAN find a representation as follows:
Therefore, M= (P+1)/2 and N = (P-1)/2 would be a solution.
Case II: The problem arises when P is even. If P is divisible by 4 (but not equal to 4), then we can find two DIFFERENT EVEN factors of P and proceed as before. If P is even but NOT divisible by
4, we CANNOT find a pair of factors that are BOTH EVEN, so there is no solution. This is precisely the case that P equals 2 times an odd integer or of the form 4N+2 as you suggested.
Note that 2 itself is of the form 2 times an odd!
Does this make sense or am I missing something? I believe some students would discover this if they proceed in a systematic fashion.
I was referring to the last question from the high school math league question from the December contest. I will post it soon.
also, Eric, Mathmom--
The restriction to positive integer solutions was critical since it excluded 1 and 4 from being representable.
High school students should consider the significance of using perfect square values for P and its connection to Pythagorean Triples.
Obvoiusly M^2 - N^2 = P has a soution if
P is odd or a square number itself (then
let N be 0 and M is sqrt(P))
If P is even there can only be a solution
for M and M if P contains square factors.
P = q^2 * P'
M^2 - N^2 = q^2 * P'
this leads to
(M/q)^2 - (N^q)^2 = P'
Recursivly this equation only has a
solution for the properties described above.
The values of M and N can then be obtained by back
substitution. And we can conclude that M^2 - N^2 = P
has no solution if P is even or square factor free.
How about 18? 18 is not square-factor free but I don't believe M^2 - N^2 = 18 has a solution.
Also, I restricted M,N to be positive so M^2 - N^2 = 4 has no solution either! Read my comment above in which I described the even values of P for which there is no solution. Someone needs to
verify this!
Another way to approach this is to consider all numbers of the form n(n+e) where e is any even number. If any number satisfies this equation, then the two squares can be easily found. Take, for
instance, 5(5+6) = 55. Divide e by two, in this case 6/2 = 3, square the result, and add it to the original number. Thus we have: 55 + (6/2)^2 = 55 + 9 = 64, then 64 - 9 = 55. Hence if you know
the factors of n, you can determine whether or not n can be expressed as the difference of two squares.
Take 18 as another example. There is no way to arrange the factors of 18 such that n(n+e) = 18: 18=6*3, 18=9*2, 18=18*1. Therefore 18 cannot be expressed as the difference of two squares.
Incidentally this shows that all odd numbers have at least one such expression (including primes) -- o*1 = o (example: 7*1 = 7). So 1(1+6) = 7 thus 7 + 9 = 16 and 16 - 9 = 7. That also follows
from the figurate representation of squares:
16 = 1 + 3 + 5 + 7
9 = 1 + 3 + 5
Therefore 16 - 9 = 7, and it is trivial to see that all odd numbers can be expressed as the difference of successive squares.
Interesting analysis...
The idea of completing the square is inspired. In effect, a number can be represented as a difference of squares if one can add a square to the number and produce another square. I followed
through your argument algebraically:
n(n+e) n^2 + en.
Adding (e/2)^2, the standard term when completing the square, produces
n^2 + en + (e/2)^2 = (n+e/2)^2.
Thus, n(n+e) = (n+e/2)^2 - (e/2)^2.
Applying this approach to odd integers of the form 2k+1 produces:
2k+1 = (2k+1)((2k+1)+(-2k)) =
((2k+1)+(-k))^2 - (-k)^2 =
(k+1)^2 - k^2, a nice way to represent any odd number! For example, 11 = 2x5 + 1 = (5+1)^2 - (5^2) or 11 = 6^2-5^2.
The only numbers that are NOT representable as a difference of squares are those of the form 2*odd, i.e., even numbers NOT divisible by 4.
Even numbers that are MULTIPLES OF 4 are easily shown to be representable using your approach:
This fits into your form since 4k = 2(2k) = 2(2 + (2k-2)), so n = 2 and e = 2k-2. Thus, 4k = (2+(k-1))^2 - (k-1)^2 = (k+1)^2 - (k-1)^2.
For example, 24 = 4x6 = (6+1)^2 - (6-1)^2, i.e., 24 = 7^2 - 5^2. Cool...
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