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https://www.engineeringchoice.com/what-is-bulk-modulus/ | 1,713,891,094,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818732.46/warc/CC-MAIN-20240423162023-20240423192023-00782.warc.gz | 664,971,926 | 25,401 | # What Is Bulk Modulus? – Definition, and Formulas
## What is Bulk Modulus?
The bulk modulus (K or B) of a substance is a measure of how resistant to compression that substance is. It is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume.
Other moduli describe the material’s response (strain) to other kinds of stress: the shear modulus describes the response to shear stress, and Young’s modulus describes the response to normal stress.
For a fluid, only the bulk modulus is meaningful. For a complex anisotropic solid such as wood or paper, these three moduli do not contain enough information to describe its behavior, and one must use the full generalized Hooke’s law. The reciprocal of the bulk modulus at a fixed temperature is called the isothermal compressibility.
## Introduction of Bulk Modulus
Bulk modulus is a numerical constant that describes the elastic properties of a solid or fluid when it is under pressure on all surfaces. The applied pressure reduces the volume of a material, which returns to its original volume when the pressure is removed.
Sometimes referred to as the incompressibility, the bulk modulus is a measure of the ability of a substance to withstand changes in volume when under compression on all sides. It is equal to the quotient of the applied pressure divided by the relative deformation.
In this case, the relative deformation, commonly called a strain, is the change in volume divided by the original volume. Thus, if the original volume Vo of a material is reduced by an applied pressure p to a new volume Vn, the strain may be expressed as the change in volume, Vo − Vn, divided by the original volume, or (Vo − Vn)/Vo. The bulk modulus itself, which, by definition, is the pressure divided by the strain, may be expressed mathematically as
When the bulk modulus is constant (independent of pressure), this is a specific form of Hooke’s law of elasticity.
Because the denominator, strain, is a ratio without dimensions, the dimensions of the bulk modulus are those of pressure, force per unit area. In the English system the bulk modulus may be expressed in units of pounds per square inch (usually abbreviated to psi), and in the metric system, newtons per square meter (N/m2), or pascals.
The value of the bulk modulus for steel is about 2.3 × 107 psi, or 1.6 × 1011 pascals, three times the value for glass. Thus, only one-third of the pressure is needed to reduce a glass sphere the same amount as a steel sphere of the same initial size. Under equal pressure, the proportional decrease in the volume of glass is three times that of steel.
One may also say that glass is three times more compressible than steel. In fact, compressibility is defined as the reciprocal of the bulk modulus. A substance that is difficult to compress has a large bulk modulus but small compressibility. A substance that is easy to compress has high compressibility but a low bulk modulus.
## Table of Fluid Bulk Modulus (K) Values
There are bulk modulus values for solids (e.g., 160 GPa for steel; 443 GPa for diamond; 50 MPa for solid helium) and gases (e.g., 101 kPa for air at constant temperature), but the most common tables list values for liquids.
Here are representative values, in both English and metric units:
The K value varies, depending on the state of matter of a sample, and in some cases, on the temperature. In liquids, the amount of dissolved gas greatly impacts the value.
A high value of K indicates a material resists compression, while a low value indicates volume appreciably decreases under uniform pressure. The reciprocal of the bulk modulus is compressibility, so a substance with a low bulk modulus has high compressibility.
Upon reviewing the table, you can see the liquid metal mercury is very nearly incompressible. This reflects the large atomic radius of mercury atoms compared with atoms in organic compounds and also the packing of the atoms. Because of hydrogen bonding, water also resists compression.
## Bulk Modulus Formulas
The bulk modulus of a material may be measured by powder diffraction, using x-rays, neutrons, or electrons targeting a powdered or microcrystalline sample. It may be calculated using the formula:
Bulk Modulus (K) = Volumetric stress / Volumetric strain
This is the same as saying it equals the change in pressure divided by the change in volume divided by initial volume:
Bulk Modulus (K) = (p1 – p0) / [(V1 – V0) / V0]
Here, p0 and V0 are the initial pressure and volume, respectively, and p1 and V1 are the pressure and volume measured upon compression.
Bulk modulu elasticity may also be expressed in terms of pressure and density:
K = (p1 – p0) / [(ρ1 – ρ0) / ρ0]
Here, ρ0 and ρ1 are the initial and final density values.
## Example Calculation
The bulk modulus may be used to calculate the hydrostatic pressure and density of a liquid. For example, consider seawater in the deepest point of the ocean, the Mariana Trench. The base of the trench is 10994 m below sea level.
The hydrostatic pressure in the Mariana Trench may be calculated as:
p1 = ρ*g*h
Where p1 is the pressure, ρ is the density of seawater at sea level, g is the acceleration of gravity, and h is the height (or depth) of the water column.
• p1 = (1022 kg/m3)(9.81 m/s2)(10994 m)
• p1 = 110 x 106 Pa or 110 MPa
Knowing the pressure at sea level is 105 Pa, the density of the water at the bottom of the trench may be calculated:
• ρ1 = [(p1 – p)ρ + K*ρ) / K
• ρ1 = [[(110 x 106 Pa) – (1 x 105 Pa)](1022 kg/m3)] + (2.34 x 109 Pa)(1022 kg/m3)/(2.34 x 109 Pa)
• ρ1 = 1070 kg/m3
What can you see from this? Despite the immense pressure on the water at the bottom of the Mariana Trench, it isn’t compressed very much! | 1,359 | 5,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-18 | latest | en | 0.906284 |
https://www.physics.brocku.ca/PPLATO/h-flap/phys2_5.html | 1,713,090,189,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00728.warc.gz | 895,136,616 | 31,202 | # 1 Opening items
## 1.1 Module introduction
When two billiard balls collide they remain in contact for only a very short time, during which the forces between them vary rapidly. To predict the outcome of such a collision by calculating the acceleration of each ball at each instant of time we would need to know exactly how the forces vary with time, and this is not an easy matter. However, we may predict the outcome in a simpler way, using the concept of linear momentum – the subject of this module.
The total linear momentum involved in a collision is important because, under certain conditions, it has the same value both before and after the collision. In other words, it is a conserved quantity. Interestingly, when appropriately interpreted, the principle of conservation of linear momentum extends beyond the confines of classical Newtonian mechanics, into the realms of relativistic mechanics and even quantum mechanics. It is, therefore, along with the conservation of energy, one of the most wide ranging and far reaching principles in the whole of physics.
We introduce linear momentum in Section 2 and show how Newton’s second law of motion may be expressed in terms of linear momentum. The principle of conservation of linear momentum is then derived from Newton’s second and third laws of motion and it is illustrated with some simple one–dimensional applications.
In Section 3 we extend this to study collisions, and introduce the idea of energy, translational kinetic energy and the principle of conservation of energy. Collisions are classified as elastic or inelastic, according to whether kinetic energy is conserved. Elastic and inelastic collisions are then analysed in terms of the conservation principles, using examples that involve one– and two–dimensional motions. Finally, brief mention is made of a revised definition of momentum, necessary for the extension of momentum conservation into relativistic mechanics.
Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the following Fast track questions. If not, proceed directly to the Subsection 1.3Ready to study? Subsection.
## 1.2 Fast track questions
Study comment Can you answer the following Fast track questions? If you answer the questions successfully you need only glance through the module before looking at the Subsection 5.1Module summary and the Subsection 5.2Achievements. If you are sure that you can meet each of these achievements, try the Subsection 5.3Exit test. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module.
Question F1
A railway truck is moving horizontally under a stationary hopper, and is receiving coal at a rate of 200 kg s−1 from the hopper. The coal is delivered with negligible speed, and the truck and its contents are subject to no resultant force. Find the rate of change of the truck’s speed at the instant when the total mass of the truck and its contents is 4 × 103 kg and the speed of the truck is 2 m s−1.
From Newton’s second law the resultant force is given by:
$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt} = \dfrac{d(m\boldsymbol{\upsilon})}{dt} = m\dfrac{d\boldsymbol{\upsilon}}{dt} + \boldsymbol{\upsilon}\dfrac{dm}{dt}$
Therefore, if F = 0
$m\dfrac{d\boldsymbol{\upsilon}}{dt} = -\boldsymbol{\upsilon}\dfrac{dm}{dt}$
Since the direction of the motion is not changing, the rate of change of the truck’s speed is
$\dfrac{d\upsilon}{dt} = -\dfrac{\upsilon}{m} \dfrac{dm}{dt} = \rm \dfrac{2\, m\,s^{-1}\times 200\,kg\,s^{-1}}{4 \times 10^3\,kg} = -0.1\,m\,s^{-2}$
Question F2
A neutron of mass m travelling east at a speed of 4 × 104 m s−1 collides with a helium nucleus of mass 4m which is travelling north at a speed of 3 × 104 m s−1. If the neutron travels with unchanged speed after the collision, but moves in a direction 36.9° west of north, find the final speed and direction of the helium nucleus.
We will define east along the positive x–axis and north along the positive y–axis. After the collision, let the helium nucleus travel at a speed υ2 in a direction inclined at an angle θ to the positive y–axis. The pre– and post–collision situations are shown in Figure 4a and b. Using conservation of momentum in the x– and y–directions we find:
along the x–direction:
mun = 4Hesinθmunsin36.9°
so$\sin\theta = \dfrac{u_n(1 + \sin\,36.9°)}{4\upsilon_{\rm He}}$
along the y–direction:
4muHe = 4Hecosθ + muncos36.9°
so$\cos\theta = \dfrac{4u_{\rm He} - u_n\cos\,36.9°}{4\upsilon_{\rm He}}$
Dividing these two expressions to find tanθ we obtain:
$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{u_n(1 + \sin\,36.9°)}{4u_{\rm He} - u_n\cos\,36.9°} = \dfrac{4 \times 0^4 \times 160}{12 \times 10^4 -(4 \times 0.800 \times 10^4)} = 0.727$
This gives θ = 36.0°. The expression for sinθ can be rearranged to give:
$\upsilon_{\rm He} = \dfrac{u_n(1 + \sin\,36.9°)}{4\sin\theta} = \dfrac{4 \times 10^4(1+\sin\,36.9°)}{4 \sin\,36.9°}\,{\rm m\,s^{-1}} = 2.72 \times 10^4\,{\rm m\,s^{-1}}$
Therefore, after the collision the helium nucleus is travelling at a speed of 2.72 × 104 m s−1 at 36.0° east of north.
Question F3
A golf ball of mass 45 g is hit with a golf club. The mass of the club head is 0.30 kg and the mass of the shaft is negligible compared with that of the head. Immediately before the head strikes the ball, the speed of the club head is 30 m s−1. Assuming that the collision is elastic, calculate the speed with which the ball is propelled.
We assume that the club head and the ball move along the x–axis, the initial velocity of the club head is u2 and the masses and final velocities of the club head and ball are m2, υ2, and m2, υ2, respectively. From this, the conservation of momentum along the x–direction gives:
m1u1 = m1υ1 + m2υ2
and conservation of energy
½ m1u1x2 = ½ m1υ1x2 + ½ m2υ2x2
We are not interested in υ2, so let us eliminate it. The first equation gives
υ1x = u1x − (m2/m1)υ2x
and the second υ1x2 = u1x2 − (m2/m1)υ2x2
Substituting for υ1x in this last equation gives us
u1x2 − 2(m2/m1)u1xυ1x2 + (m2/m1)2υ2x2 = u1x2 − (m2/m1)υ2x2
Cancelling u1x2 on both sides, and then dividing both sides by (m2/m1)υ2x
we find−2u1x + (m2/m1)υ2x = −υ2x
so thatυ2x = 2u1x/(1 + m2/m1)
i.e.υ2x = 2 × 30 m s−1/(1 + 0.045 kg/0.30 kg) = 52 m s−1
Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module and to proceed directly to the following Ready to study? Subsection.
Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the Section 4Closing items.
Study comment In order to study this module you will need to be familiar with the following terms: acceleration, components_of_a_vectorcomponent (of a vector), force, mass, Newton’s laws of motion, scalar, speed, vector, vector addition and velocity. Also, you will need to be familiar with the use of Cartesian coordinate systems, Pythagoras’s theorem, trigonometric functions, the sine rule, and the cosine rule. Additionally, this module makes some use of the derivativederivative notation of basic differentiation to represent rates of change and assumes you are familiar with symbols such as /dt. If you are uncertain of any of these terms you can review them now by referring to the Glossary which will indicate where in FLAP they are developed. The following questions will allow you to establish whether you need to review some of the topics before embarking on this module.
Question R1
Describe the difference between scalar and vector quantities, and how the resultant or sum of two vector quantities is found diagramatically.
Scalar quantities have magnitude only whereas vector quantities have magnitude and direction and may be represented by arrows. Two vectors of similar type may be added diagrammatically using the triangle rule by placing the initial point of the second vector on the terminal point of the first vector, and then joining the initial point of the first vector to the terminal point of the second vector. (See Figure 5 for an example.)
Consult scalar and vector in the Glossary for further information.
Question R2
A triangle ABC has side AB of length 4.00 m and the angles at A and B equal to 60° and 80°, respectively. Find the lengths of the sides BC and CA.
Since the interior angles of a triangle must add up to give 180°, angle at C must be 40°. The sine rule gives:
$\dfrac{{\rm BC}}{\sin\,60°} = \dfrac{\rm CA}{\sin\,80°} = \dfrac{4.00\,{\rm m}}{\sin\,40°}$
soBC = (4.0sin60°/sin40°) m = 5.39 m
andCA = (4.0sin80°/sin40°) m = 6.13 m
Consult trigonometric functions in the Glossary for further information.
Question R3
A horizontal force of magnitude 20 N acts in a direction 40° west of north. Find the components_of_a_vectorcomponents of the force in the north and west directions.
The north component of force = (20cos40°) N = 15.3 N
the west component of force = (2sin40°) N = 12.9 N
Consult components of a vector in the Glossary for further information.
Question R4
A horizontal force of magnitude 40 N applied to a block on a frictionless horizontal surface produces an acceleration of magnitude 2 m s−2. What is the mass of the block?
Using Newton’s second law of motion F = ma, we see that
m = F/a = 40 N/2 m s−1 = 20 kg
Consult Newton’s second law of motion in the Glossary for further information.
Question R5
(a) If υ(t) represents the speed υ of a particle at time t, what is the meaning of /dt in terms of a graph of υ against t?
(b) If m is a constant, what would be the graphical interpretation of d()/dt?
(c) If m is not a constant, would you expect to find that $m\dfrac{d\upsilon}{dt} = \dfrac{d}{dt}(m\upsilon)$?
(a) At any particular time t, the value of /dt represents the gradient of the graph of υ against t at that particular time.
(b) d()/dt would represent the gradient of the graph of m × υ against t at any particular time. Since the effect of multiplying υ by a constant m is to uniformly rescale the υ–axis, the gradient of the new against t graph will be just m times the gradient of the original υ against t graph. In other words, provided m is a constant
$\dfrac{d}{dt}(m\upsilon) = m \dfrac{d\upsilon}{dt}$
(c) if m is not a constant there is no reason to expect that the above result will hold true. In fact, when m and υ both vary with time the product rule of differentiation tells us that
$\dfrac{d}{dt}(m\upsilon) = m \dfrac{d\upsilon}{dt} + \upsilon \dfrac{dm}{dt}$
Consult basic differentiation in the Glossary for further information.
# 2 Linear momentum and Newton’s second law
## 2.1 Definition of linear momentum
It is easy to imagine that a lorry travelling at 40 m s−1 is more difficult to stop than the same lorry travelling at 20 m s−1. Similarly, a lorry travelling at 40 m s−1 is more difficult to stop than a car travelling at 40 m s−1. From this it is apparent that the difficulty in stopping a body depends on both the mass and the velocity of the body. The scientific quantity identifying this property of a moving body, which we have loosely described in terms of the ‘difficulty in stopping’ is called linear momentum (more often simply momentum), p. In Newtonian mechanics, a body of mass m moving with velocity υ, has a momentum p defined by:
p = mυ(1)
As velocity is a vector quantity, linear momentum is also a vector quantity, with magnitude and a direction
which is the same as that of the velocity vector. The SI units of linear momentum are kg m s−1.
It is often convenient to express the momentum i vector p in terms of its components. If the components of the velocity vector υ are υx, υy and υz, then we write: υ = (υx, υy, υz) and p = (px, py, pz), where the components of the momentum vector are given by:
px = x py = y pz = z(1a)
In dealing with momentum we frequently encounter problems in which motion is not restricted to a single line. Bodies moving in different directions may collide and may change directions. Since momentum is a vector quantity, a change in direction represents a change in momentum, even if the magnitude of the momentum is unchanged.
Question T1
A body of mass 4 kg is moving with speed 15 m s−1 in a direction 36.9° east of north. What are the components of the momentum of the body in the north and east directions?
The magnitude of the momentum of the body is 4 kg × 15 m s−1 = 60 kg m s−1. The north and east components of the momentum are:
north:(60cos36.9°) kg m s−1 = 48 kg m s−1
east:(60sin36.9°) kg m s−1 = 36 kg m s−1
Question T2
In a game of volleyball, player 1 near the back of the court passes the ball to player 2, who is standing near the net. As it approaches player 2 the ball is moving horizontally at 3.0 m s−1. Player 2 taps the ball with a vertical force so that it acquires a 0.8 kg m s−1 component of momentum in the vertical direction. If the mass of the ball is 400 g, determine the magnitude and direction of the ball’s resultant momentum.
The x–component (horizontal) of momentum of the ball is the same before and after player 2 strikes the ball, and is given by
px = x = 0.40 kg × 3.0 m s−1 = 1.2 kg m s−1
The question tells us that the vertical component of momentum py = 0.8 kg m s−1 after player 2 strikes the ball. All we have to do now is to find the magnitude and direction of the momentum vector. The magnitude of the momentum is given by:
p2 = px2 + py2 = (1.2 kg m s−1)2 + (0.8 kg m s−1)2 = 2.08 kg2 m2 s−2
so thatp = 1.4 kg m s−1
If the angle of the ball to the horizontal is θ, then px = pcosθ, and py = psinθ. We have tanθ = px/py = 0.8/1.2, and so θ = 34°.
## 2.2 Relationship between force and rate of change of momentum
Momentum is closely associated with Newton’s second law of motion, F = ma. Indeed, if we assume that the mass m is constant, then we can express Newton’s second law in terms of momentum. To see this first note that:
$\boldsymbol{F} = m\boldsymbol{a} = m\dfrac{d\boldsymbol{\upsilon}}{dt}$
Then note that if m is constant
$m\dfrac{d\boldsymbol{\upsilon}}{dt} = \dfrac{d(m\boldsymbol{\upsilon})}{dt} = \dfrac{d\boldsymbol{p}}{dt}$ i
and so$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt}$(2)
i.e. the resultant force acting on a body is equal to the rate of change of its momentum.
Equation 2 is a vector equation and so represents the three equations, one for each of the three components of force:
$F_x = \dfrac{dp_x}{dt} \quad F_y = \dfrac{dp_y}{dt} \quad F_z = \dfrac{dp_z}{dt}$(2a)
We now have two forms of Newton’s second law when m is constant:
$\boldsymbol{F} = m\boldsymbol{a} \quad\text{and}\quad F = \dfrac{d\boldsymbol{p}}{dt}$
Faced with two ways of expressing Newton’s second law it is legitimate to ask if one is more general or more fundamental than the other. In fact, the general view is that Equation 2 is the more fundamental. It can be applied directly to systems of particles as well as to single bodies, and, in some cases, it can even be applied to bodies with changing mass, such as a rocket that accelerates by expelling the fuel that it carries.
Given experimental fact that F = dp/dt is more generally applicable throughout classical mechanics you might wonder how F = ma fails in various contexts, such as that of varying mass. In that particular case the inadequacy of F = ma is easy to demonstrate provided you are familiar with the product_rule_of_differentiationproduct rule of differentiation which enables us to write the derivative of a product as a sum of products.
If we start from Equation 2,
$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt}$(Eqn 2)
but do not assume constant mass, then:
$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt} = \dfrac{d(m\boldsymbol{\upsilon})}{dt} = m\dfrac{d\boldsymbol{\upsilon}}{dt} + \boldsymbol{\upsilon}\dfrac{dm}{dt}$(3)
From this we see that if m is not constant then we have to add a further term υdm/dt to ma to obtain the correct expression for F. There are several surprising aspects of this. First, a resultant force acting on a body does not always cause acceleration; it could be that the mass of the body is increasing with time at just the correct rate to allow constant velocity. Second, even when there is no resultant force acting on the body it may nevertheless accelerate, if its mass is changing.
✦ What force is required to push a truck across a frictionless surface at a constant velocity of 5 m s−1 in a given direction if sand is being poured into the truck at a rate of 6 kg s−1?
✧ If the velocity is constant the acceleration is zero. Therefore, from Equation 3,
$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt} = m\dfrac{ d\boldsymbol{\upsilon}}{dt} + \boldsymbol{\upsilon} \dfrac{dm}{dt}$(Eqn 3)
with the direction of motion along the x–axis we find:
$F_x = 0 + \upsilon_x \dfrac{dm}{dt}$ = (5 m s−1) × (6 kg s−1) = 30 N
The identification of ‘force’ with ‘rate of change of momentum’ sanctioned by Equation 2,
$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt}$(Eqn 2)
also has another important consequence; if F = 0 i then dp/dt = 0 thus:
If no resultant force acts on a body then its momentum does not change.
As you will see in Subsection 2.3 a generalization of this result leads to the important principle of momentum conservation.
### Impulse
There are many situations where the force on a body acts only for a very short time (e.g. bats on balls) and we are concerned only with the consequent changes in the motion of the body. If the force F is constant and acts for a short time ∆t, it follows from Equation 2 that the momentum of the body will change by an amount
p = Ft(4) i
The quantity Ft is called the impulse of the force. i
In many situations where varying forces act for short times it is only their effects (i.e. the momentum changes they cause) that are of real interest. In such cases Equation 4 may be used to find the constant force that would produce the same overall effect as the varying force that actually acted. This procedure can give surprising insights into the rapidly varying forces that arise in various impulsive interactions. For example, in American baseball the bat and ball are typically in contact for one or two thousandths of a second, during which the effective average force has a magnitude of about 6000 N. Such large forces are not unusual.
## 2.3 Conservation of momentum
The concept of momentum is particularly useful when dealing with situations where there are two or more interacting bodies which interact with each other but not with their surroundings. We call such a system an isolated system. Any forces which act on parts of an isolated system arise from the mutual interactions between bodies within the system. Each force is then one member of a Newton’s third law pair of action–reaction forces. Such forces are called internal forces, to distinguish them from the external forces which act on the system or parts of the system from outside. We will begin by considering an isolated system in which there are only two bodies and they only interact with one another.
Let p1 = m1υ1 and p2 = m2υ2 be the momenta of the two bodies in the system. These two momenta will change with time, due to the mutual interaction. The total momentum of the system, p, is given by the vector sum of the momenta of the two bodies:
p = p1 + p2
The rate of change of p with time is therefore determined by the rates of change of p1 and p2:
$\dfrac{d\boldsymbol{p}}{dt} = \dfrac{d\boldsymbol{p}_1}{dt} + \dfrac{d\boldsymbol{p}_2}{dt}$ i
At a given instant let F12 be the force on the first body due to the second and F21 be the force on the second body due to the first. Equation 2,
$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt}$(Eqn 2)
gives:$\boldsymbol{F}_{12} = \dfrac{d\boldsymbol{p}_1}{dt}\quad\text{and}\quad\boldsymbol{F}_{21} = \dfrac{d\boldsymbol{p}_2}{dt}$
Newton’s third law of motion tells us that:
F12 = −F21 or F12 + F21 = 0
Therefore$\dfrac{d\boldsymbol{p}_1}{dt} + \dfrac{d\boldsymbol{p}_2}{dt} = \dfrac{d\boldsymbol{p}}{dt} = \boldsymbol{0}$(5)
Thus p = p1 + p2 is constant with time, although both p1 and p2 may be varying. At every instant of time the total momentum is the same (since p does not depend on t). We can summarize the situation by saying that the total momentum of the isolated system is a conserved quantity.
Note that in deriving Equation 5 we have assumed that the only forces acting on the bodies are F12 and F21. If an external force F had been acting on the bodies then the system would not be isolated, Equation 5 would not be valid and the total momentum of the system would not be conserved; rather the total momentum would vary according to Equation 2,
$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt}$(Eqn 2)
Although we have considered only two bodies in deriving Equation 5, the conservation of momentum result can be extended to any number of bodies in an isolated system. The same reasoning as before may be used: the rate of change of total momentum of the isolated system is zero because the vector sum of the impulses from each action–reaction pair of forces is zero. This result constitutes one of the most important principles of physics:
The principle of conservation of momentum:
The total momentum of any isolated system is constant with time.
Alternatively, we can say that the total momentum of any system does not change through mutual interactions within the system.
✦ When a ball bounces from a wall, its momentum changes, and so it appears that momentum is not conserved. Explain why this is not so.
✧ The ball is not an isolated system when it strikes the wall. Since the wall is fixed to the Earth, the isolated system must be the ball plus the wall plus the Earth and it is the momentum of this system that is conserved. The change in momentum of the ball is equal in magnitude and opposite in direction to the change in momentum of the wall plus the Earth. However, since the Earth is so massive, you will not see any recoil of the wall when the ball strikes it; the required momentum change of the Earth corresponds to only an imperceptible change in its velocity.
Question T3
Derive the principle of conservation of momentum from Newton’s laws as applied to an isolated system consisting of three interacting particles.
The total momentum p of the three particles is given by:
p = p1 + p2 + p3
Thus$\dfrac{d\boldsymbol{p}}{dt} = \dfrac{d\boldsymbol{p}_1}{dt} + \dfrac{d\boldsymbol{p}_2}{dt} + \dfrac{d\boldsymbol{p}_3}{dt}$
However, $\dfrac{d\boldsymbol{p}_1}{dt} = \boldsymbol{F}_{12} + \boldsymbol{F}_{13}, \quad\dfrac{d\boldsymbol{p}_2}{dt} = \boldsymbol{F}_{21} + \boldsymbol{F}_{23} \quad\text{and}\quad\dfrac{d\boldsymbol{p}_3}{dt} = \boldsymbol{F}_{31} + \boldsymbol{F}_{32}$
so$\dfrac{d\boldsymbol{p}}{dt} = (\boldsymbol{F}_{12} +\boldsymbol{F}_{21}) + (\boldsymbol{F}_{23} + \boldsymbol{F}_{32}) + (\boldsymbol{F}_{13} + \boldsymbol{F}_{31})$
But from Newton’s third law of motion: F12 = −F21, F23 = −F32, F13 = −F31
so$\dfrac{d\boldsymbol{p}}{dt} = \boldsymbol{0}$ Therefore p must remain constant.
Question T4
When you walk along the road with constant velocity you have constant momentum. When you stop walking you have zero momentum. Where has your momentum gone?
When you are walking, you are not an isolated system. When you start from rest, the Earth recoils imperceptibly in the opposite direction to your motion. Similarly when you slow down and stop, you transfer momentum to the Earth in the same direction as your original motion.
## 2.4 Symmetry and conservation
If an object or a system of objects can be altered in some way without changing its overall behaviour or appearance, that object or system is said to exhibit symmetry under the alteration. For instance, a uniform sphere may be rotated through an arbitrary angle about its centre point without changing its properties or appearance in any way; such a sphere is consequently said to exhibit rotational symmetry and to be invariant (i.e. unchanging) under arbitrary rotations about its centre.
Symmetries and their related invariances are common features of physical systems and of the laws that describe them. Some of these symmetries are so readily taken for granted that it is easy to overlook them. For example, it is generally assumed amongst physicists that the outcome of a fundamental experiment does not depend on where or when it is performed. There is, of course, a good deal of evidence in support of this belief but it is impossible to test every case. Consequently, our expectation that the fundamental behaviour of physical systems will not vary from country to country or from day to day is largely based on our belief in the symmetry of nature under translations (i.e. ‘shifts’ or ‘transfers’) through space and time. In fact, much of modern science is based on the following assumptions:
• The laws of physics must be invariant under arbitrary translation through space.
• The laws of physics must be invariant under arbitrary translation through time.
If either of these fundamental invariances was violated, if momentum was only conserved on Tuesdays say, or only in Western Europe, then physics would be a much less interesting subject and even harder to master.
The reason for embarking on this discussion of symmetry and invariance is that it has been known for a long time that there is a deep link between the symmetries of nature and the existence of conserved physical quantities. It can be argued, for instance, that the invariance of physical laws under translations through time implies the existence of a conserved scalar quantity which may be identified as energy (about which more is said later in the module). Similarly, the invariance of physical laws under translation through space implies the existence of a conserved vector quantity which may be identified as momentum. Thus the principle of momentum conservation may be regarded as one of the fundamental principles of physics and you should not be surprised to learn that its validity extends beyond the confines of classical Newtonian mechanics.
The mathematical techniques needed to establish the link between symmetry and conservation are beyond the scope of FLAP but the importance of the result would be hard to overestimate. The search for fundamental symmetries, of which we have mentioned only two, has been one of the central pillars of modern physics and has fuelled much of the progress made in the last 40 years.
## 2.5 Worked examples involving conservation of momentum
In a scientific context, the term collision means a brief but powerful interaction between two particles or bodies in close proximity. To illustrate the value of the principle of conservation of momentum in solving problems, let us begin with the simple case of two bodies moving in one dimension which collide and stick together on impact. Later, we will extend our discussion to two– and three–dimensional motion and to collisions where the bodies do not stick together.
### Collisions in one dimension, where the two bodies stick together
Example 1
A body of mass 2 kg moving with a speed of 6 m s−1 along the x–axis collides with a stationary body of mass 4 kg. The bodies stick together on impact and move off as one combined body along the x–axis with speed υx. Find the value of υx, assuming this is an isolated system.
Solution
In applying the principle of conservation of momentum to the isolated system we equate the total momentum before the collision to the total momentum after the collision. Therefore, conserving momentum along the x–axis:
m1u1x + m2u2x = (m1 + m2)υx(6)
where u1x and u2x are the initial velocities of the masses m1and m2.
Thusυx = (m1u1x + m2u2x)/(m1 + m2)
i.e.υx = (2 kg × 6 m s−1 + 4 kg × 0 m s−1)/(2 kg + 4 kg)
soυx = 12 kg m s−1/6 kg = 2 m s−1
Example 2
A radioactive nucleus i at rest decays by forming an alpha–particle (an α–particle is a helium nucleus) and a daughter nucleus. The mass of the daughter nucleus is 54 times that of the α–particle. What is the velocity of the daughter nucleus if the velocity of the α–particle υα is 1 × 107 m s−1 along the positive x–axis? This can be thought of as a ‘sticking collision’, but in reverse.
Solution
The total momentum before the decay is zero and so the total momentum afterwards must also be zero. This means that the vector sum of the momenta of the two particles after the decay must be zero. To satisfy this condition the magnitudes of the two momenta must be equal and their directions must be opposite. Let m be the mass of the α–particle and υx the required velocity of the daughter nucleus. Conserving momentum along the x–axis, we have:
α + 54x = 0
soυx = −υα/54 = −1 × 107 m s−1/54 = −1.85 × 105 m s−1
This confirms that the daughter nucleus recoils in the opposite direction to the α–particle, as expected.
Question T5
Two blocks of masses 0.3 kg and 0.2 kg are moving towards one another along a frictionless, horizontal surface with speeds 1.0 m s−1 and 2.0 m s−1, respectively. If the blocks stick together on impact, find their final velocity.
Let the 0.3 kg block move in the −x–direction and the 0.2 kg block in the +x–direction. The total momentum of the two blocks before the impact is the vector sum of the momenta of the two blocks. We use Equation 6,
m1u1x + m2u2x = (m1 + m2)υx(Eqn 6)
which leads to υx = (m1u1x + m2u2x)/(m1 + m2)
Here$\upsilon_x = \dfrac{0.2\,{\rm kg} \times 2\,{\rm m\,s^{-1}} - 0.3\,{\rm kg} \times 1\,{\rm m\,s^{-1}}}{0.2\,{\rm kg} + 0.3\,{\rm kg}}$
i.e.υx = 0.1 kg m s−1/0.5 kg = 0.2 m s−1 (i.e. along the positive x–direction).
Question T6
Consider the following scenario. The population of the Earth (5.6 × 109 people in September 1994) all begin walking in the same direction at 1.0 m s−1. Estimate the effect on the Earth. (Take the mass of the Earth as 6.0 × 1024 kg.) What would happen when everybody stopped walking? i
The total momentum of the isolated system of the Earth plus the walkers is constant. On starting the walk, the walkers’ total momentum increases in the direction of the walk by an amount given by their total mass multiplied by their speed. If we estimate the average mass per person as 50 kg then this total increase is
= 5.6 × 109 × 50 kg × 1.0 m s−1 = 2.8 × 1011 kg m s−1
The Earth’s momentum in this same direction decreases by this same amount. The change in the Earth’s velocity, ∆υ, is equal to the change in the momentum, ∆p, divided by the Earth’s mass, so its speed changes by the amount:
υ = ∆p/ME = 2.8 × 1011 kg m s−1/6.0 × 1024 kg = 4.7 × 10−14 m s−1
When the walkers stop, the changes are reversed and the Earth returns to its original velocity, although the changes would have been imperceptibly small.
### Collisions in two dimensions, where the two bodies stick together
In problems in which motion is in more than one dimension we have to take care over the directions of motion of the bodies involved. This means that in applying conservation of momentum we need to remember the vector nature of momentum. In our discussions so far we have prepared the ground for this extension by stressing the vector approach throughout and by insisting on the use of component notation, even with one–dimensional problems where there was the temptation to drop the subscripts! For two– or three–dimensional problems there are two possible approaches which may be used: vector summation or use of components. We will consider these alternative approaches in the following worked examples.
Example 3
A 2000 kg van, travelling due east at 20 m s−1 collides on an icy patch of road with a 1000 kg car travelling due north at 30 m s−1. The vehicles lock together on impact and then move off as one body over the frictionless surface. Find the velocity of the vehicles after the collision.
Study comment Beware – problems which ask for the velocity cannot be fully answered until you have given the direction of motion as well as the speed of motion. This same warning applies equally for any other vector quantity, such as momentum. It is very easy to forget this!
Solution
Let the final velocity of the vehicles be υ. We need to determine both the magnitude and direction of υ in order to specify it completely. First, let us consider the vector summation approach. The magnitudes of the momenta of the van and car before the collision are:
for the van p1 = (2000 kg × 20 m s−1) = 4.0 × 104 kg m s−1
for the car p2 = (1000 kg × 30 m s−1) = 3.0 × 104 kg m s−1
These two momenta must be added vectorially, as shown in Figure 1, to determine the total momentum, p = p1 + p2 before the collision. The direction of p is at an angle θ to the original direction of the van. Using Pythagoras’s theorem or by inspection of Figure 1 (a ‘3, 4, 5 triangle’) we find the magnitude of p is 5.0 × 104 kg m s−1; also we see that tanθ = 3/4.
Since the two vehicles on a frictionless surface constitute an isolated system, conservation of momentum tells us that the total initial momentum is equal to the total final momentum and so we know that p also represents the total final momentum of the locked–together vehicles.
Sincep = (m1 + m2)υ(7)
the magnitudes of the vectors p and υ are related by
υ = p/(m1 + m2) = 5.0 × 104 kg m s−1/3000 kg = 16.7 m s−1
Furthermore, since tanθ = 3/4, it follows that θ = 36.9°.
Let us now try the alternative approach, using components. The component approach is based on the fact that if the total momentum vector, p, remains constant in an isolated system, then so too must the component of p in any given direction. In our example we need to consider, in turn, the component in the east direction (say the positive x–direction) and the component in the north direction (say the positive y–direction). In this example the van has the total x–component of momentum and the car has the total y–component of momentum. Conservation of x and y momentum then immediately gives:
px = pcosθ = p1 = m1u1x = 4.0 × 104 kg m s−1
and py = psinθ = p2 = m2u2y = 3.0 × 104 kg m s−1
withp = (px2 + py2)1/2 = 5.0 × 104 kg m s−1
andθ = arctan(py/px ) = arctan(3/4) = 36.9°
As before, υ = p/(m1 + m2) = 5.0 × 104 kg m s−1/3000 kg = 16.7 m s−1
Study comment Whilst you should practice using both the vector summation method and the component method in solving problems on momentum involving more than one dimension, the component approach is easier to use with more complicated examples. A good general strategy is to set up a clearly labelled diagram of the collision, with appropriately chosen Cartesian coordinate axes x, y, z shown. These axes are chosen so as to simplify the solution – perhaps by arranging them in such a way that one of the incoming particles is moving only along the x–axis. Then momentum conservation is applied for components along each of the axes in turn, with each direction treated entirely independently. Finally, the components are recombined to give the magnitude and direction of the required vector.
Question T7
A body of mass 6 kg moving north over a frictionless horizontal surface at a speed of 6 m s−1 collides with a stationary body of mass 4 kg. After the collision the more massive body moves in a direction 30° west of north and the less massive body in a direction 25° east of north. Find the speeds of the bodies after the collision. Use the vector summation method, then repeat the calculation using components.
First let us try the vector summation approach to this problem. Let υ1 and υ2 be the speeds of the more and less massive bodies, respectively. The principle of conservation of momentum tells us that the vector sum of the two final momenta is equal to the initial momentum of the more massive body. This initial momentum has magnitude 36 kg m s−1. The vector diagram is illustrated in Figure 5.
The sine rule is particularly useful in this example, as it enables us to determine the two unknown sides of the vector triangle. Applying the sine formula we find
$\dfrac{36\,{\rm kg\,m\,s^{-1}}}{\sin\,125°} = \dfrac{6\,{\rm kg} \times \upsilon_1}{\sin\,25°} = \dfrac{4\,{\rm kg} \times \upsilon_2}{\sin\,30°}$
soυ1 = 6 m s−1(sin25°/sin125°) = 3.10 m s−1
andυ2 = 9 m s−1(sin30°/sin125°) = 5.49 m s−1
Now let us try the component approach in solving this problem. As usual, we define east as the positive x–direction and north as the positive y–direction. First, consider components:
in the x–direction: p2sin25° − p1sin30° = 0
sop2 = p1sin30°/sin25°
in the y–direction: p = p2cos25° + p1cos30°
We can substitute for p2 from the first expression into the second expression:
p = p1sin30°cos25°/sin25° + p2cos30° = 1.938 p1
Sop1 = p/1.938 = (36/1.938) kg m s−1 = 18.57 kg m s−1
andυ1 = 18.57 kg m s−1/6 kg = 3.10 m s−1
Using the x–direction expression we can find p2 and then υ2:
p2 = p1 sin30°/sin25° = 21.97 kg m s−1
andυ2 = 21.97 kg m s−1/4 kg = 5.49 m s−1
# 3 Momentum and kinetic energy in collisional problems
In the collision problems tackled so far we have always either finished up with a single body, or started with a single body (e.g. in α–decay). The reason for this limitation is that only such problems have a unique solution, obtainable by applying momentum conservation alone. For all other collisions we need to involve other principles in addition to momentum conservation. To illustrate this we can consider a one–dimensional problem in which two colliding bodies with known masses m1 and m2, and with known initial velocities u1x and u2x collide and then separate with final velocities υ1x and υ2x. In general, when the initial conditions are given, the problem involves finding the two unknowns υ1x and υ2x. However, consideration of momentum conservation along the x–direction:
m1u1x + m2u2x = m1υ1x + m2υ2x
gives only one equation connecting these two unknowns and so the problem is insoluble without more information. i
The additional information is provided by introducing the idea of energy. i Energy can exist in several forms, but the energy that a body has due to its instantaneous linear motion is called its translational kinetic energy. In classical Newtonian mechanics the translational kinetic energy of a body of mass m and speed υ is defined as:
Etran = ½ 2(8)
Energy is a scalar quantity. In the SI system, the unit of energy is the joule (J) where 1 J = 1 kg m2 s−2.
Translational kinetic energy i is only one of several forms in which energy can appear. A body may also have potential energy due to its position or internal state, as in the case of a mass suspended above the surface of the Earth, or a compressed spring. Quite apart from these various forms of mechanical energy we can also associate energy with other physical phenomena, so we may speak of electrical energy, light energy, sound energy and thermal energy. Energy may be transformed between these different forms by physical and chemical processes but it is generally accepted that, for an isolated system, the sum of all these different forms of energy remains constant. This fundamental principle is known as the principle of conservation of energy.
Collisions may be categorized by comparing the total kinetic energy of the colliding bodies before and after the collision. If there is no change in the total kinetic energy, then the collision is an elastic collision. If the kinetic energy after the collision is less than that before the collision then the collision is an inelastic collision. In some situations (e.g. where internal potential energy is released) the total kinetic energy may increase in the collision; this type of collision is a superelastic collision.
Collisions between macroscopic objects are usually inelastic but some collisions, such as those between steel ball bearings or between billiard balls, are very nearly elastic. Collisions between subatomic particles, such as electrons, protons, commonly are elastic. The kinetic energy which is lost in an inelastic collision appears as energy in a different form, e.g. thermal energy, sound energy, light energy, so that the total energy is conserved. The collisions which we have dealt with so far, in which the bodies stick together on collision and move off together afterwards, are examples of completely inelastic collisions. In these cases the maximum amount of kinetic energy consistent with momentum conservation, is converted into other forms of energy.
## 3.1 Elastic collisions in one dimension
We begin this discussion with an example of a simple one–dimensional elastic collision between two identical masses, where one incoming projectile strikes another, initially stationary target. We wish to know the final velocities of each mass after the collision.
Example 4
A mass m moves along the x–axis with velocity u1x and collides elastically with another identical mass at rest. What are the velocities of the two masses after collision?
Solution
Let the final velocities be υ1x and υ2x. Conservation of momentum along the x–axis gives
mu1x = 1x + 2x(9)
and conservation of kinetic energy gives
½ mu1x2 = ½ 1x2 + ½ 2x2(10) i
By eliminating common factors, Equations 9 and 10 can be simplified to give
u1x = υ1x + υ2x
i.e.υ2x = u1xυ1x(9a)
andu1x2 = υ1x2 + υ2x2
i.e.υ2x2 = u1x2υ1x2 = (u1xυ1x)(u1x + υ1x)(10a)
Dividing this last equation by υ2x = u1xυ1x gives us:
υ2x = u1x + υ1x
Comparing this expression with that in Equation 9a shows us υ1x = 0 and υ2x = u1x
This result is familiar to anyone who has seen the head–on collision of two bowls on a bowling green. The moving one stops, and the stationary one moves off with the original velocity of the first.
✦ In the situation above, the two masses were equal. What would you expect to happen if the two masses were not equal? Can you predict qualitatively, i.e. without calculation, what would happen with the target mass m2 at rest if: (a) m1m2, (b) m2m1. i
✧ Experience should tell you (a) that a high mass projectile fired at a low mass target would be essentially undeflected by the collision, whereas (b) a low mass projectile fired at a massive target would bounce back with unchanged speed.
✦ In this last question we used terms such as ‘high mass’ and ‘low mass’ rather than ‘heavy’ and ‘light’. Why do you think we did this?
✧ The point is that these observations concern mass not weight. The results would be identical if the experiment were carried out well away from the Earth, where there was no gravity and therefore the objects were weightless. Terms like ‘heavy’ and ‘light’ are naturally associated with weight rather than mass, although when gravity is present these two quantities are related.
Now we consider the mathematical solution of the general one–dimensional case of an elastic collision between particles of mass m1 and m2 which move with speeds u1x and u2x before the collision and speeds υ1x and υ2x after the collision. Since the collision is elastic the kinetic energy is conserved and so we can write:
½ m1u1x2 + ½ m2u2x2 = ½ m1υ1x2 + ½ m2υ2x2(11)
This equation, together with the momentum conservation equation,
m1u1x + m2u2x = m1υ1x + m2υ2x(12)
allows υ1x and υ2x to be found, if we know the values of m1, m2 and the initial velocities u1x and u2x. The algebra is rather lengthy but it is worth the trouble because on the way we derive another useful expression relating the velocities. From Equation 11 we have:
m1(u1xυ1x)(u1x + υ1x) = m2(υ2xu2x)(υ2x + u2x)(13)
and from Equation 12:
m1(u1xυ1x) = m2(υ2xu2x)(14)
If we divide each side of Equation 13 by the corresponding side of Equation 14 we obtain:
u1x + υ1x = υ2x + u2x
i.e.u1x − u2x = −(υ1xυ2x) = υ2xυ1x(15)
Equation 15 shows that the following general result holds true:
In an elastic collision between two masses the relative velocity of approach is the negative of the relative velocity of separation.
Equation 15 can be combined with Equation 14 to solve simultaneously for υ1x and υ2x. We multiply Equation 15 by m2 to obtain:
m2u1xm2u2x = m2υ2xm2υ1x
If we subtract this equation from Equation 14 we have:
u1x(m1m2) + 2m2u2x = υ1x(m1 + m2)
soυ1x = [u1x(m1m2) + 2m2u2x]/(m1 + m2)(16)
Similarly we can derive the following expression for υ2x:
υ2x = [u2x(m2m1) + 2m1u1x]/(m1 + m2)(17)
It is interesting to examine these results for υ1x and υ2x in a few special cases, including some that have been mentioned earlier.
1
m1 = m2
If two objects of equal mass collide, Equations 16 and 17 give υ1x = u2x and υ2x = u1x. This means that the particles simply exchange velocities on collision. For example, if particle 2 is at rest initially (u2x = 0), then finally particle 1 is at rest (υ1x = 0) and particle 2 moves with the initial velocity of particle 1 (υ2x = u1x). We saw this result earlier in this subsection; it is a familiar tactic in bowls.
2
m1m2; u2x = 0
If m2 is very small compared with m1, and u2x = 0, Equations 16 and 17 give υ1xu1x and υ2x ≈ 2u1x. Thus the motion of the high mass particle is virtually unchanged by the collision but the low mass particle moves off with a velocity of twice that of the high mass particle. Tennis players serving will be familiar with this case.
3
m2m1; u2x = 0
If m1 is very small compared with m2 (which is stationary), then Equations 16 and 17 lead to υ1x ≈ −u1x and υ2x ≈ 0. i Therefore, the low mass particle rebounds with almost unchanged speed while the high mass particle remains essentially at rest. Golfers whose ball hits a tree will recognize this situation.
4
m2m1; u2x ≈ −u1x
If m1 is negligible compared with m2, and the two bodies approach head–on with equal speeds then Equations 16 and 17 lead to υ1x ≈ −3u1x and υ2xu2x. This shows that the low mass particle bounces back with three times its initial speed, while the high mass particle continues essentially unaffected by the collision. This case will be recognized by a batsman playing cricket or by a tennis player returning a serve; a less familiar example is covered in the Exit test.
The results for these four special cases accord with common experience. The results quoted above in cases 2, 3 and 4 give an upper limit to the speed that can be imparted to a ball hit by a club, bat or racquet.
Question T8
A neutron of mass m rebounds elastically in a head–on collision with a gold nucleus at rest and of mass 197m. What fraction of the neutron’s kinetic energy is transferred to the recoiling gold nucleus? Repeat this calculation when the target is a carbon nucleus at rest and of mass 12m.
We need to calculate the loss of kinetic energy for the neutron, divided by its initial kinetic energy:
$\dfrac{\Delta{\rm E_{tran}}}{{\rm E_{tran}}} = \dfrac{u_{1x}^2-\upsilon_{1x}^2}{u_{1x}^2} = 1 - \upsilon_{1x}^2/u_{1x}^2$
We use Equations 16 and 17, with u2x = 0
υ1x = [u1x(m1m2) + 2m2u2x]/(m1 + m2)(Eqn 16)
andυ2x = [u2x(m2m1) + 2m1u1x]/(m1 + m2)(Eqn 17)
soEtran/Etran = 1 − (υ1x/u1x)2 = 1 − [(m1m2)/(m1 + m2)]2
for gold:Etran/Etran = 1 − (196/198)2 = 0.020 (i.e. 2.0%)
for carbon:Etran/Etran = 1 − (11/13)2 = 0.284 (i.e. 28.4%)
We see that a low mass nucleus is much more effective than a more massive nucleus when it comes to slowing down fast neutrons by elastic collisions. As an aside, it is interesting to note that it is because of this fact that carbon is used in a nuclear reactor for just this purpose.
Question T9
A tennis player returns a service ball back in the direction of the server. The ball of mass 50 g arrives at the racquet of mass 350 g with a speed of 45 m s−1 and the racquet is travelling at 10 m s−1 at impact. Calculate the velocity of the returning ball, assuming elastic conditions.
We designate the ball as particle 1 and the racquet as particle 2, with the ball initially travelling along the positive x–direction. From Equation 16,
υ1x = [u1x(m1m2) + 2m2u2x]/(m1 + m2)(Eqn 16)
soυ1x = [45 m s−1 × (− 0.3 kg) + 2 × 0.35 kg(−10 m s−1)]/0.4 kg
υ1x = (−13.5 kg ms−1 − 7.0 kg m s−1)/0.4 kg = −51.3 m s−1
The server receives the ball back with interest!
## 3.2 Inelastic collisions in one dimension
We now extend our discussion to include cases where the total kinetic energy changes during the one–dimensional collision.
### Completely inelastic collisions
First we return to our original case, where the two particles stick together on impact; this is an example of a completely inelastic collision, which occurs with the maximum loss of kinetic energy consistent with momentum conservation. As a simple example, suppose we have two bodies of equal mass, with one initially at rest. If the initial velocity of the other is ux, the initial momentum is mux; the final momentum must be the same so, since the mass has been doubled, the final velocity is ux/2 and the final kinetic energy is 2m(ux/2)2/2 = mux2/4. Half the original kinetic energy has been transformed into other forms, mainly heat, during the collision.
For the more general case where the colliding masses are unequal, but they stick together at collision, we still have υ1x = υ2x = υx and so Equation 12 simplifies to our earlier Equation 6,
m1u1x + m2u2x = (m1 + m2)υx(Eqn 6)
and provides a full solution of the problem, without recourse to energy.
Question T10
A mass m1 travels along the x–axis with speed ux and collides with a mass m2 at rest. If the two masses stick together on impact derive an expression for the fraction of the original kinetic energy lost in the collision. What happens when m2m1?
This is a completely inelastic collision, solvable using only the conservation of momentum along the x–axis. Equation 6 gives:
m1u1x + m2u2x = (m1 + m2)υx(Eqn 6)
Withu2x = 0, υx = m1u1x/(m1 + m2)
The initial kinetic energy is m1u1x2/2 and the final kinetic energy is (m1 + m2)υx2/2 and so the fractional loss of kinetic energy is given by:
$\dfrac{\Delta{\rm E_{tran}}}{{\rm E_{tran}}} = \dfrac{m_1u_{1x}^2-(m_1+m_2)\upsilon_{x}^2}{m_1u_{1x}^2} = 1 - \left(\dfrac{m_1+m_2}{m_1}\right)\left(\dfrac{\upsilon_x}{u_{1x}}\right)^2$
Substituting for υx/u1x we find:
$\dfrac{\Delta{\rm E_{tran}}}{{\rm E_{tran}}} = 1 - \left(\dfrac{m_1+m_2}{m_1}\right)\left(\dfrac{m_1}{m_1+m_2}\right)^2 = 1 - \left(\dfrac{m_1}{m_1+m_2}\right) = \dfrac{m_2}{m_1+m_2}$
For the case where m2m1, i.e. a negligibly small mass object strikes a very massive object, the whole of the kinetic energy is lost.
### The general case of an inelastic collision in one dimension
To complete the picture, let us mention the general case where two particles collide but where the transfer of kinetic energy into other forms is less than that for the completely inelastic case. This problem has no general solution without more information, such as the fraction of kinetic energy converted. Such problems have solutions which lie between those for the two extremes of elastic and completely inelastic collisions but they must be tackled on an individual basis, using the general principles of conservation of momentum and energy.
You will see that in all these calculations we have not needed to invoke the rather complicated forces involved in the interaction of the two particles, but rather have been able to solve the problems using only the principles of conservation of momentum and energy. This is a great simplification and illustrates the power of using conservation principles whenever possible.
## 3.3 Collisions in two or three dimensions
The same principles apply when we look at collisions in two or three dimensions. We have already considered this in Subsection 2.5, for the case of completely inelastic collisions. The extension to include elastic collisions is straightforward. We simply treat the motions in each dimension as independent motions, as we did in Subsection 2.5, and apply momentum conservation separately along each Cartesian coordinate axis, with the scalar quantity of kinetic energy and its conservation providing one additional equation relating the squares of the particle speeds. Once again, since we have been careful to use vector notation throughout, this extension to two or three dimensions is easily made.
### Elastic collisions between two equal masses, with one mass at rest
Consider the elastic collision between two identical bodies of mass m, one at rest and the other approaching with velocity u1. The particles are no longer confined to move in one dimension, so our x–component equations (Equations 9 and 10), embodying momentum and kinetic energy conservation,
mu1x = 1x + 2x(Eqn 9)
½ mu1x2 = ½ 1x2 + ½ 2x2(Eqn 10)
become full vector equations:
mu1 = mυ1 + mυ2(18)
and½ mu12 = ½ 12 + ½ 22(Eqn 19)
These can be simplified to:
u1 = υ1 + υ2(20)
andu12 = υ12 + υ22(21)
Equation 20 tells us that all three velocity vectors must lie in a single plane, in order for their vector sum to be zero. These equations are most easily interpreted by a diagram. Figure 2 shows how the three vectors u1, υ1 and υ2 are related to one another. They form a triangle which must be a right–angled triangle, since the sides obey Pythagoras’s theorem. The implication of this is striking; when two equal masses collide elastically they move away at 90° to each other. This is the simplifying feature of equal–mass collisions in two or three dimensions, corresponding to the simple result of the exchange of velocities, which we found in one dimension. i
You may have noticed that this result does not tell us exactly where the bodies go after the collision. Any pair of final velocities which can be represented by Figure 2 will be equally satisfactory, and there are an infinite number of these. The reason for this is that we have said nothing about the shape or size of the bodies, and just how they collide. If we have any information about how the bodies collide, as for example in the situation shown in Figure 3, then we will usually be able to find the final velocities.
Question T11
In the example above (illustrated in Figure 2), if the moving body has an initial speed of 10 m s−1, and is deflected through 20° in the collision, find the magnitudes and directions of the velocities υ1 and υ2.
From the triangle shown in Figure 2, υ1 has a magnitude 10cos20° m s−1 = 9.40 m s−1 and is at the given angle of 20° to the x–axis; υ2 has a magnitude 10sin20° m s−1 = 3.42 m s−1 and must be at 70° to the x–axis.
Question T12
In the example above (illustrated in Question T11), if instead of the conditions in Question T11 the speed of the moving body is reduced from 10 m s−1 to 6 m s−1 by the collision, find the final velocities.
Using the triangle in Figure 2 again,
υ22 = u12υ12 = (102 − 62) m2 s−2
so that υ2 = 8 m s−1. υ1 is at an angle cos−1(6/10) = 53.1°, and υ2 at an angle cos−1(8/10) = 36.9° to the x–axis. You will observe that the two angles add up to 90°, as they should.
When the two masses are unequal the algebraic manipulations required to solve collision problems become rather complex, but no new physics is involved in the solution and we will not pursue such problems here.
## 3.4 Relativistic momentum
It is now quite easy to study collisions in which the particles are travelling at a significant fraction of c, the speed of light in vacuum. These experiments fall mainly into the realm of nuclear and particle physics and an example might be the elastic collision between a very fast electron from a particle accelerator and an electron in the material of a stationary target. From what we have said in Subsection 3.3 we would predict that the angle between the outgoing tracks of the scattered electron and the target electron ought to be 90°. The prediction proves correct at low impact speeds but as this speed is increased the angle becomes less than 90° and the reduction in the angle increases with the impact speed. At the highest attainable speeds the incident electron is barely deflected at all by its collision with the target electron while the target electron also moves off in about this same direction. The collision essentially reverts to a one–dimensional bat–ball type of collision. From our discussions in Subsection 3.1Subsections 3.1 and Subsection 3.33.3 we recognize that this is exactly what would happen if the incoming electron had a mass which increases with speed, so that its mass eventually becomes much greater than that of the target electron and it is essentially unaffected by the collision. This suggestion, that the mass of a particle is not a constant but rather increases with speed is one of the possible interpretations of this experiment. However, this experiment does not measure mass but rather momentum and so its interpretation is best made in terms of the momentum, rather than in terms of either mass or speed separately. All we can be sure of is that if momentum conservation is valid then the magnitude of the momentum at high speeds cannot simply be given by the product of mass and speed.
The magnitude of the momentum of a particle with mass m and speed υ must always be greater than . Although is a good approximation at sufficiently low speeds, it underestimates the momentum progressively as the speed increases.
In 1905 Albert Einstein (1879–1955) postulated his special theory of relativity and amongst its consequences was a new definition for the momentum of a moving particle. The new definition of the relativistic momentum of a mass m when moving with velocity υ is:
$\boldsymbol{p} = \dfrac{m\boldsymbol{\upsilon}}{\sqrt{1-\dfrac{\upsilon^2}{c^2}}}$(22) i
This definition incorporates the features required to explain our experiment.
✦ What is the expression for the relativistic momentum of a particle moving at low speeds, so that υ/c ≪ 1?
✧ The denominator in Equation 22 relativistic momentum
$\boldsymbol{p} = \dfrac{m\boldsymbol{\upsilon}}{\sqrt{1-\dfrac{\upsilon^2}{c^2}}}$(Eqn 22)
simplifies to unity so pmυ.
✦ What is the magnitude of the relativistic momentum of a particle moving at a speed of, (a) υ/c = 0.1, (b) υ/c = 0.9, and (c) what happens as υ/c approaches 1?
✧ (a) The denominator in Equation 22 relativistic momentum
$\boldsymbol{p} = \dfrac{m\boldsymbol{\upsilon}}{\sqrt{1-\dfrac{\upsilon^2}{c^2}}}$(Eqn 22)
becomes $\sqrt{0.99\os}$, so p = 1.005,
(b) p = 2.29, and
(c) p tends to infinity.
The result for part (c) (i.e. where υ/c = 1) illustrates why it is not possible to accelerate a body to c, the speed of light in vacuum; to do so would require either an infinite force or an infinite time. Note, however, that it is the speed of light in a vacuum which is the limiting speed, not the speed of light in any other material, which is always less than c. So, it is perfectly possible to have a body travelling faster than the speed of light in glass or in water or even in air! i This discussion is only the briefest mention of relativity, about which more is said elsewhere is FLAP.
# 4 Closing items
## 4.1 Module summary
1
The momentum of a body, a vector quantity, is defined as the product of its mass and its velocity in classical Newtonian mechanics
p = mυ(Eqn 1)
At high speeds this definition must be modified into that for relativistic momentum
$\boldsymbol{p} = \dfrac{m\boldsymbol{\upsilon}}{\sqrt{1-\dfrac{\upsilon^2}{c^2}}}$(Eqn 22)
2
The force acting on a body is equal to the rate of change of momentum
$\boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt}$(Eqn 2) i
This statement is a more fundamental statement of Newton’s second law of motion than that in which force is equated to the product of mass and acceleration, since the latter form is applicable only for situations in which mass is constant.
3
The principle of conservation of momentum for an isolated system can be derived from Newton’s second and third laws of motion.
4
The principle of conservation of momentum plays an essential part in the solution of collision problems.
5
Translational kinetic energy is the energy that a body has as a result of its motion. In classical Newtonian mechanics
Etran = ½ 2(Eqn 8)
6
One-dimensional problems involving elastic collisions, in which kinetic energy is conserved, may be solved completely.
7
Problems involving motion in more than one dimension can only be solved fully if some details of the collision are known. One example is the elastic collision between two equal masses, where one mass initially is at rest. Here, the departing trajectories always make an angle of 90°.
## 4.2 Achievements
Having completed this module, you should be able to:
A1
Define the terms that are emboldened and flagged in the margins of the module.
A2
Determine the momentum of a body or system of bodies, given the mass and velocity of each body.
A3
State and apply Newton’s second law in terms of momentum.
A4
Show that the force acting on a body can be expressed as the sum of two terms, one involving rate of change of velocity and the other involving the rate of change of mass.
A5
Derive the principle of conservation of momentum from Newton’s second and third laws of motion.
A6
Describe the essential features of elastic and inelastic collisions.
A7
Use the principle of conservation of momentum to solve problems involving completely inelastic collisions in one– or two–dimensional motion or ones involving the disruption of a single mass by explosion or decay.
A8
Use the principles of conservation of momentum and kinetic energy to solve elastic collision problems.
Study comment You may now wish to take the following Exit test for this module which tests these Achievements. If you prefer to study the module further before taking this test then return to the topModule contents to review some of the topics.
## 4.3 Exit test
Study comment Having completed this module, you should be able to answer the following questions, each of which tests one or more of the Achievements.
Question E1 (A2)
Find the ratio of the magnitudes of the momenta associated with a rifle bullet of mass 30 × 10−3 kg moving at 400 m s−1 and a brick of mass 1 kg moving at 10 m s−1.
The magnitudes of the momenta are as follows:
the bullet’s is:30 × 10−3 kg × 400 m s−1 = 12 kg m s−1
and the brick’s is:1 kg × 10 m s−1 = 10 kg m s−1
Therefore the ratio is 12 : 10 or 1.2 : 1.0
These two momenta are similar in size.
Question E2 (A3 and A4)
Explain quantitatively how it is possible for a body on which a resultant force acts to move with constant velocity.
Equation 3 gives:
$\displaystyle \boldsymbol{F} = \dfrac{d\boldsymbol{p}}{dt} = \dfrac{d(m\boldsymbol{\upsilon})}{dt} = m\dfrac{d\boldsymbol{\upsilon}}{dt} + \boldsymbol{\upsilon}\dfrac{dm}{dt}$(Eqn 3)
Thus if $\dfrac{dm}{dt}$ is not zero, $\dfrac{d\boldsymbol{\upsilon}}{dt}$ may be zero even if F is not zero. Thus if the mass is changing with time then a force is required to keep the body moving at constant velocity.
Question E3 (A7 and A8)
A particle of mass m moves along the x–axis with velocity ux and collides head–on with a stationary particle of mass 2m. If in the collision the first particle is brought to rest, find the velocity of the second particle. This second particle then collides head–on with a stationary third particle of mass 3m. If the second and third particles stick together on impact, find their subsequent velocity after the collision.
Let the speed of the second particle after the first collision be υ2x and the speed of the combined second and third particles after the second collision be υx. Using conservation of momentum along the x–axis:
First collision:mux + 0 = 22x and so υ2x = ux/2
Second collision:22x + 0 = (2m + 3m)υx
soυx = 2υ2x/5 = 2(ux/2)/5 = ux/5
Question E4 (A3)
Two balls of the same mass travelling at the same velocity strike a window perpendicularly. The first bounces back but the second breaks the window. Explain how this could be so.
The difference between the situations must be that one of the balls is more rigid than the other and so the time of contact during which the momentum of the ball changes on impact will then be less. The average force F required to effect this momentum change ∆p is related to the time of contact ∆t through the expression for the impulse:
p = Ft(Eqn 4)
This implies that a much larger force must be exerted by the window to bring about any given change in momentum of the more rigid ball. If this force exceeds the maximum force which the window can safely exert then the window will break.
Question E5 (A3)
A car travelling at 15 m s−1 goes out of control and hits a tree, with the car and driver coming to rest in 0.6 s. What is the initial momentum of the driver, whose mass is 70 kg? What is the magnitude of the average force exerted on the driver during the crash, and how does this force compare with the magnitude of the driver’s weight?
The magnitude of the momentum of the driver is 70 kg × 15 m s−1 = 1050 kg m s−1. Equation 4,
p = Ft(Eqn 4)
gives the magnitude of the average force as:
p/∆t = (1050 kg m s−1)/(0.6 s) = 1750 kg m s−2 = 1750 N
The driver’s weight is of magnitude 70 kg × 9.8 m s−2 = 686 N, so the average impact force is about 2.5 times the weight of the driver in terms of magnitude.
Question E6 (A8)
A spacecraft approaches the planet Jupiter, passes round behind it, and departs in precisely the opposite direction to that with which it approached. Initially, when the spacecraft is a large distance from the planet, the spacecraft is moving in the negative x–direction at a speed of 10 km s−1, and Jupiter is moving in the positive x–direction at 13 km s−1. Assuming elastic conditions, find the final speed of the spacecraft when it is once again at a large distance from the planet. i
Let the masses of the spacecraft and Jupiter be m and M, respectively, the initial velocities u1x and u2x and the final velocities υ1x and υ2x. The principle of conservation of momentum, applied along x tells us that
mu1x + Mu2x = 1x + 2x
and if there is no loss of kinetic energy during the interaction
½ mu1x2 + ½ Mu2x2 = ½ 1x2 + ½ 2x2
In these two equations there are two unknowns, υ1x and υ2x, but we are only interested in υ1x, so that we should eliminate υ2x. The neatest way to do this is to rewrite the two equations as
υ2x = u2x + (m/M)(u1xυ1x)
and(mM)(u1x2υ1x2) = υ2x2u2x2
substituting for υ2x in the second equation gives us
$\dfrac{m}{M}(u_{1x}^2-\upsilon_{1x}^2) = u_{2x}^2 + \left(\dfrac{m}{M}\right)^2\left(u_{1x}-\upsilon_{1x}\right)^2 + 2 u_{2x}\left(\dfrac{m}{M}\right)\left(u_{1x}-\upsilon_{2x}\right)^2$
cancelling the u2x2 terms, and dividing through by (m/M) × (u1xυ1x) gives:
$\displaystyle \dfrac{(u_{1x}^2-\upsilon_{1x}^2)}{u_{1x}-\upsilon_{1x})} = (u_{1x}-\upsilon_{2x}) = \left(\dfrac{m}{M}\right)(u_{1x}-\upsilon_{1x}) + 2u_{2x}$
The mass of Jupiter is so much greater than that of the spacecraft that m/M is very small and the first term on the right can be neglected compared to the second term and compared to the term on the left–hand side. In fact, this neglected term can be shown to be equal to the change in velocity of Jupiter, which is very small (even though Jupiter’s change of momentum cannot be neglected compared to that of the spacecraft). This illustrates what we saw in Equation 15 in Subsection 3.1,
u1xu2x = − (υ1xυ2x) = υ2xυ1x(Eqn 15)
that the relative velocity after collision is equal to the negative of the relative velocity before collision. So, neglecting the term, we find to a very good approximation:
u1x + υ1x ≈ 2u2x
(this is an approximation of Equation 16 with m2m1)
υ1x = [u1x(m1m2) + 2m2u2x]/(m1 + m2)(Eqn 16)
soυ1x ≈ 2u2xu1x = 2 × 13 km s−1 − (−10 km s−1) = 36 km s−1
The spacecraft has been speeded up very substantially! This is a simplified example of the so–called ‘slingshot’ method, used to give a spacecraft enough energy to reach the outermost parts of the solar system. This additional energy comes from a reduction in the planet’s kinetic energy.
Question E7 (A8)
Two steel balls travelling at the same speed collide elastically head-on. One of them is stationary after the collision. If its mass is 900 g, what is the mass of the other?
If the initial velocities of the balls are ux and −ux and the final velocity of the moving ball is υ1x, then the conservation of momentum along the x–axis gives:
m1uxm2ux = m1υ1x
and conservation of energy gives:
½ m1ux2 + ½ m2ux2 = ½ m1υ1x2
Rearranging these equations gives us
υ1x/ux = (m1m2)/m1 and (υ1x/ux)2 = (m1 + m2)/m1
so that(m1m2)2/m12 = (m1 + m2)/m1
which can be simplified to: m2 = 3m1, and so m1 = 300 g.
Question E8 (A7)
A block of stone at a quarry is blown up into three separate pieces of masses 10 tonnes, 8 tonnes, and 6 tonnes (1 tonne = 1000 kg). The 10 tonne piece moves off in the positive x–direction at a speed of 8 m s−1, and the 8 tonne piece moves in the negative y–direction at 5 m s−1. Find (a) the momentum of the third piece, and (b) the energy of the explosion, which must be equal to the extra kinetic energy after the explosion.
(a) If we take m1 = 104 kg, m2 = 8 × 103 kg, m3 = 6 × 103 kg, υ1x = 8 m s−1, and υ2y = −5 m s−1, the momentum of the third block is given by
p3x + m1υ1x = 0 and p3y + m2υ2y = 0
sop3x = −(104 kg) × (8 m s−1) = −8 × 104 kg m s−1
p3y = −(8 × 103 kg) × (−5 m s−1) = 4 × 104 kg m s−1
Alternatively
p = $\sqrt{\smash[b]{(8^2 + 4^2)}}$ × 104 kg m s−1 = 8.94 × 104 kg m s−1
at an angle to the negative x–axis of arctan(4 × 104/8 × 104) = 26.6°
(b) The speed of the third block is (8.94 × 104 kg m s−1)/(6 × 103 kg) = 14.9 m s−1
The total kinetic energy of the three blocks Etran is then
½ [(104 kg) × (8 m s−1)2 + (8 × 103 kg) × (5 m s−1)2 + (6 × 103 kg) × (14.9 m s−1)2]
So Etran = 1.09 × 106 J, equal to the total energy released in the explosion. | 19,119 | 69,854 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-18 | latest | en | 0.931164 |
https://mcqseries.com/flip-flops-3/ | 1,660,785,304,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573145.32/warc/CC-MAIN-20220818003501-20220818033501-00099.warc.gz | 353,068,207 | 24,230 | # Flip Flops – 3
3. If an active-HIGH S-R latch has a 0 on the S input and a 1 on the R input and then the R input goes to 0, the latch will be ________.
(a) SET
(b) RESET
(c) Clear
(d) Invalid
Explanation
Explanation : No answer description available for this question. Let us discuss.
General Knowledge Books
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Tags: flip, flops, flip-flops, exercise, input, electronics, engineering | 474 | 1,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-33 | latest | en | 0.622404 |
https://web2.0calc.com/questions/yay-here-we-go-again | 1,603,382,664,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879673.14/warc/CC-MAIN-20201022141106-20201022171106-00674.warc.gz | 603,745,799 | 6,976 | +0
# yay here we go again
0
576
2
+578
\(y=x^2+8x+18\)
whats the vertex, focus and directrix
Mar 12, 2018
#1
+27699
0
In vertex form:
y = x^2 +8x +16 -16 +18 ('completing the square)
y = (x+4)^2 +2 Vertex h, k = -4,2 Upward opening parabola (coefficient of (x+4)^2 is positive: +1 )
Re-arranging
y-2 = 4p (x+4)^2 4p = 1 p= 1/4 distance from vertex to directrix and focus
Directrix = y= 2- 1/4 = 1.75
Focus = y= 2+1/4 = 2 1/4 x = -4 so -4, 2 1/4
Graph:
Mar 12, 2018
#2
+111466
+1
y = x^2 + 8x + 18 complete the square on x
y = x^2 + 8x + 16 + 18 - 16
y = (x + 4)^2 + 2
(y - 2) = (x + 4)^2 (1)
In the form
4p (y - 2) = ( x + 4)^2 .....it's clear from (1) that p =1/4
This parbola turns upward
The vertex is (-4, 2)
The focus is given by : ( -4 , 2+ p) ⇒ (-4 , 2 + 1/4) ⇒ ( -4, 9/2)
The directrix is given by :
y = ( 2 - 1/4) = 7/4
Mar 12, 2018 | 456 | 950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-45 | latest | en | 0.50648 |
http://octave.org/doxygen/3.2/quadv_8m.html | 1,571,738,208,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987813307.73/warc/CC-MAIN-20191022081307-20191022104807-00217.warc.gz | 140,490,398 | 5,656 | Operators and Keywords
C++ API
## Functions
endif if (nargin< 4) tol
endif if (isa(a,"single")||isa(b,"single")) myeps
endif if (isempty(tol)) tol
endif if (isempty(trace)) trace=0
if (isinf(fa)) fa
endif if (isinf(fb)) fb
if (fcnt > 10000) warning("maximum iteration count reached")
elseif (isnan(q)||isinf(q)) warning("infinite or NaN function evaluations were returned")
elseif (hmin< (b-a)*myeps) warning("minimum step size reached -- possibly singular integral") = x != 0
endif if (trace) disp([fcnt
id q ()
endif if (fcnt==5||abs(q-q0) > tol)[q1
## Variables
function [q, fcnt]
else myeps = eps
endif c = (a + b) / 2
fa = feval (f, a, varargin{:})
fc = feval (f, c, varargin{:})
fb = feval (f, b, varargin{:})
fcnt = 3
endif h = (b - a) / 2
q = (b - a) / 6 * (fa + 4 * fc + fb)
else d = (a + c) / 2
e = (c + b) / 2
fd = feval (f, d, varargin{:})
fe = feval (f, e, varargin{:})
q1 = (c - a) / 6 * (fa + 4 * fd + fc)
q2 = (b - c) / 6 * (fc + 4 * fe + fb)
endif a
endif hmin
endif endif endfunction pi
## Function Documentation
endif endif endfunction !assert ( quadv(@sin, 0, 2 *pi) , 0 , 1e- 5 )
elseif ( ) = x != 0
elseif ( isnan(q)||isinf(q) )
endif if ( fcnt = `= 5 || abs (q - q0)`, tol )
endif if ( trace )
if ( fcnt , 10000 )
endif if ( isinf(fb) )
if ( isinf(fa) )
endif if ( isempty(trace) ) ` [pure virtual]`
endif if ( isempty(tol) )
endif if ( isa(a,"single")||isa(b,"single") )
endif if ( )
id q ( ) ` [virtual]`
Type Constraints
## Variable Documentation
endif a
endif c = (a + b) / 2
else d = (a + c) / 2
b d e = (c + b) / 2
fa = feval (f, a, varargin{:})
fb = feval (f, b, varargin{:})
fc = feval (f, c, varargin{:})
endif fcnt = 3
fd = feval (f, d, varargin{:})
fe = feval (f, e, varargin{:})
endif endfunction function[q, fcnt, hmin]
Initial value:
``` quadv (f, a, b, tol, trace, varargin)
if (nargin < 3)
print_usage ()
```
endif h = (b - a) / 2
Initial value:
``` simpsonstp (f, a, c, d, fa, fc, fd, q1, fcnt, hmin,
tol, trace, varargin{:})
```
else myeps = eps
* pi
q = (b - a) / 6 * (fa + 4 * fc + fb)
q1 = (c - a) / 6 * (fa + 4 * fd + fc)
q2 = (b - c) / 6 * (fc + 4 * fe + fb) | 833 | 2,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-43 | latest | en | 0.298714 |
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How to find the volume of 3-dimensional figures
This week in 5th grade math, we are working on finding the Volume of rectangular prisms. We have learned the formula for Volume as the V=length (l) x width (w) x height (h) or as V= area of the base (B) x the height (h).
Here is a wonderful explanation about Volume. If you were out or need a refresher on how to calculate Volume click on the Math Antics link below.
Math Antics
Today's lesson will be about finding the area of the base times the height to measure the volume of two different rectangular prisms. | 153 | 623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-30 | latest | en | 0.915888 |
https://www.exampaper.com.sg/miss-loi-the-tutor/encouraging-kids-to-save-the-posb-way | 1,719,029,371,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862249.29/warc/CC-MAIN-20240622014659-20240622044659-00760.warc.gz | 660,197,148 | 33,185 | ×
# Encouraging Kids To Save – The POSB Way
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Tuition given in the topic of Miss Loi the Tutor from the desk of at 11:55 am (Singapore time)
Today’s question deals with Simple & Compound Interests, a topic that manages to keep its place within the new E-Maths 2008 syllabus.
This simple question is inspired by a true story, and Miss Loi believes that everyone (yes, even those of you who are not students anymore) should be able to solve this, though a long spreadsheet might help save some time for Part 2.
So in true Jφss Sticks fashion …
As a result of a wider effort to “instil the discipline and importance of saving” to children, a cute little boy cheerfully received a piggy bank distributed by POSB in his school.
With great diligence and discipline, he managed to fill his piggy bank to the brim with \$86.60 worth of coins from his own pocket money. Thus, it was with a great sense of achievement that he walked into his favourite POSB branch, accompanied by his proud mother, to deposit the coins that he had so painstakenly saved up.
His spirits were high as he waited in the queue, comforted by the warm, caring smile of the POSB Girl giving kids a respite from the rain, in a scene that was played over and over again on a TV screen high above.
When his turn came, he was told, to his utter disbelief, by the bank that, except for the first 200 coins, there would a charge of \$1.50 for every 100 coins deposited. As such, a total of \$4.50 was promptly deducted from his POSB account.
1. Calculate the percentage decrease in the cute little boy’s savings as a result of these charges.
2. Assuming that:
• The bank pays an attractive compound interest rate of 0.25% per annum (valid as of 6 Dec 2007), and credits this interest monthly
• This rate won’t change in the foreseeable future
• The cute little boy was too distraught to make any further transaction with the bank
Calculate how long does it take for the cute little boy to recover his \$4.50 through the bank’s interest.
After this incident, Miss Loi is confident that this cute little boy will ace his Simple & Compound Interests topic in time to come. At the same time she hopes that he has learnt a lesson that, beyond the warm, caring smiles, things in life aren’t always as straightforward as they seem.
### Revision Exercise
To show that you have understood what Miss Loi just taught you from the centre, you must:
1. Cris commented in tuition class
2007
Dec
7
Fri
12:20pm
1
Miss Loi, POSB bank where got so gd give so high interest at 0.25%? If got, please tell me leh.
That poor kid must wait 20 .7 years to recover his \$4.50 back, the bank cheat his money.
2. Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class
2007
Dec
7
Fri
12:59pm
2
Cris, you can checkout POSB's latest rates online.
Assuming you meant it literally, is 0.25% (=quarter of a percent) really that appealing to the businessman in you?
Also do be careful when using the word 'cheat'. Afterall the onus is for the cute little boy to thoroughly read and understand the carefully-worded terms in his contract fully.
3. Cris commented in tuition class
2007
Dec
7
Fri
1:05pm
3
Oops! I shall be careful with my words. Actually, 0.25% is really pathetic for me. For the poor kid, he really need to wait for 20.7 years leh. Sibei Jialat!
4. Lim Ee Hai commented in tuition class
2007
Dec
8
Sat
3:03am
4
This is a sure way of making math learning exciting - the story telling way. Real life application does allow students to appreciate math. I like the layout of your blog too. Very well done.
5. FoxTwo commented in tuition class
2007
Dec
8
Sat
1:45pm
5
Ok, seeing the pathetic rates given by POSB, I think I shall invest my money in loansharks then. At least they give better returns, and they have "Recovery Agents" to ensure the money isn't swindled.
6. Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class
2007
Dec
9
Sun
12:50am
6
Welcome back to Jφss Sticks Teacher Lim!
As we all know math mastery is about practice and practice. What better practice than to experience/feel/hear/see something for yourself in real life, as our poor cute little boy did, and remember it for life?
Anyway, this was an easy story to tell coz it's a true story as seen in our Straits Times Forum page on 6 Dec 🙂
7. Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class
2007
Dec
9
Sun
1:08am
7
FoxTwo,
Miss Loi can see that you're now comparing the rates of returns of different loansharks in the business.
And yes, they all have their own “Recovery Agents” to ensure that not a single dollar will disappear 🙂
8. Horny Ang Moh commented in tuition class
2007
Dec
9
Sun
4:14pm
8
Problem with me is how to save?? No piggy bank! I broke open the pig long long time ago!! Can milo tin can be use to store my coins??? He! He! Have a nice day!
9. Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class
2007
Dec
9
Sun
9:03pm
9
HAM,
Oh dear what did you do to the poor piggy that caused it to break prematurely?! Don't tell Miss Loi what you did began with a letter 'P'???!!!
10. FoxTwo commented in tuition class
2007
Dec
10
Mon
11:43am
10
Yeah I bet he P-unched it.
It begins with letter P... right? right?
11. eastcoastlife commented in tuition class
2007
Dec
10
Mon
4:14pm
11
The piggy also has a FaceBook acount! It got P-oked! haha....
12. Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class
2007
Dec
10
Mon
9:10pm
12
Aiyoh ECL, you actually made Miss Loi search Facebook for a profile called 'POSB Piggy'!
13. mathslover commented in tuition class
2009
Apr
21
Tue
11:52am
13
Percentage decrease = \$4.50/\$86.60*100% = 5.196% ~ 5.2%
Amount deposited after deduction = \$86.60-\$4.50=\$82.10
Interest compounded annually, but credited monthly.
1st year:
Interest credited per month = 0.25%/12*\$82.10 = \$0.0171
Total interest in 1st year = \$0.0171*12 = \$0.2052 ~ \$0.21 (really wonder if POSB will round UP.. )
Capital in 2nd year = \$82.10+\$0.21=\$82.31
*and using excel...*
End of 21 years and 5 months, the total in his bank account would be \$86.61 (finally get back the \$4.50).
I got a killer question.
'...except for the first 200 coins, there would a charge of \$1.50 for every 100 coins deposited. As such, a total of \$4.50 was promptly deducted...'
Means he deposited 500 coins. What are their denominations?
*p.s. i also not sure if it can be done...* 🙂
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* Please refer to the relevant Ten-Year Series for the questions of these suggested solutions. | 1,843 | 6,946 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-26 | latest | en | 0.976829 |
https://123dok.net/document/z1dm5ppz-default-times-arbitrage-conditions-changes-probability-measures.html | 1,627,398,191,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153392.43/warc/CC-MAIN-20210727135323-20210727165323-00346.warc.gz | 97,210,109 | 50,353 | # Default times, no-arbitrage conditions and changes of probability measures
23
## Texte intégral
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DOI 10.1007/s00780-011-0170-z
### of probability measures
Delia Coculescu· Monique Jeanblanc · Ashkan Nikeghbali
Received: 21 January 2009 / Accepted: 19 November 2010 / Published online: 10 February 2012 © Springer-Verlag 2012
Abstract In this paper, we give a financial justification, based on no-arbitrage
con-ditions, of the (H)-hypothesis in default time modeling. We also show how the (H)-hypothesis is affected by an equivalent change of probability measure. The main
technique used here is the theory of progressive enlargements of filtrations.
Keywords Default modeling· Credit risk models · Random times · Enlargements of
filtrations· Immersed filtrations · No-arbitrage conditions · Equivalent change of measure
Mathematics Subject Classification (2010) 60G07· 91G40 JEL Classification C60· G12 · G14
1 Introduction
In this paper, we study the stability of the (H)-hypothesis (or immersion property) under equivalent changes of probability measures. Given two filtrationsF ⊂ G, we
D. Coculescu (
###
)· A. Nikeghbali
Institut für Mathematik, Universität Zürich, Winterthurerstrasse 190, 8057 Zürich, Switzerland e-mail:delia.coculescu@math.uzh.ch
A. Nikeghbali
e-mail:ashkan.nikeghbali@math.uzh.ch
M. Jeanblanc
Département de Mathématiques, Université d’Evry Val d’Essonne and Institut Europlace de Finance, 23, Boulevard de France, 91037 Evry Cedex, France
e-mail:monique.jeanblanc@univ-evry.fr
A. Nikeghbali
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say thatF is immersed in G if all F-local martingales are G-local martingales. In the default risk literature, the filtrationG is obtained by the progressive enlargement of
F with a random time (the default time), and the immersion property under a
risk-neutral measure appears to be a suitable no-arbitrage condition (see [4]). Because immersion in general is not preserved under equivalent changes of probability mea-sures (see [24] and [3]), reduced-form default models are usually specified directly under a given risk-neutral measure.
However, it seems crucial to understand how the immersion property is modi-fied under an equivalent change of probability measure. This is important not only because credit markets are highly incomplete, but also because the physical default probability appears to play an important role in the presence of incomplete informa-tion. This role is emphasized by a more recent body of literature, initiated by [12] (see also [7,15,17,20], among others), which proposes to rely on accounting informa-tion, and to incorporate the imperfect information about the accounting indicators, in computing credit spreads. The default intensities are computed endogenously, using the available observations about the firm. Some of the constructions do not satisfy the immersion property [16,26]. It is therefore important to understand the role of the immersion property for pricing.
More generally, our goal in this paper is to provide efficient and precise tools from martingale theory and the general theory of stochastic processes to model default times. We wish to justify on economic grounds the default models which use the technique of progressive enlargements of filtrations, and to explain the reasons why such an approach is useful. We provide and study (necessary and) sufficient condi-tions for a market model to be arbitrage-free in the presence of default risk. More precisely, the paper is organized as follows.
In Sect.2, we describe the financial framework which uses enlargements of filtra-tions techniques and introduce the corresponding no-arbitrage condifiltra-tions. In Sect.3, we present useful tools from the theory of progressive enlargements of filtrations. Subsequently, we study how the immersion property is affected under equivalent changes of probability measures. In Sect.4, we give a simple proof of the not well-known fact (due to Jeulin and Yor [24]) that immersion is preserved under a change of probability measure whose Radon–Nikodým density isF∞-measurable. Using this result, we show that a sufficient no-arbitrage condition is that the immersion property should hold under an equivalent change of measure (not necessarily risk-neutral). Then, using a general representation property forG-martingales (Sect.5), we char-acterize the class of equivalent changes of probability measures which preserve the immersion property when the random time τ avoids theF-stopping times (Sect.6), thus extending results of Jeulin and Yor [24] to our setting. Finally, we show how the Azéma supermartingale is computed for a large class of equivalent changes of measures.
2 No-arbitrage conditions
In this section, we briefly comment on some no-arbitrage conditions appearing in de-fault models that use the progressive enlargement of a reference filtration (for further
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discussion in the case of complete default-free markets, see [4]). All notions from the theory of enlargements of filtrations used in this section are gathered in the next section. In default modeling, the technique of progressive enlargements of filtrations has been introduced by Kusuoka [26] and further developed in Elliott et al. [13]. It consists of a two-step construction of the market model, as follows.
Let (Ω,G, F = (Ft)t≥0,P) be a filtered probability space satisfying the usual
hy-potheses. For us, the probability P stands for the physical measure under which finan-cial events and prices are observed. Let τ be a random time; this is aG-measurable nonnegative random variable which usually represents the default time of a company. It is not anF-stopping time.
LetG = (Gt)t≥0be the filtration obtained by progressively enlarging the filtration
F with the random time τ . Obviously, ∀t ≥ 0, Ft⊂ Gt⊂ G.
Usually, the filtrationG plays the role of the market filtration (and is sometimes called the full market filtration), meaning that the price processes areG-adapted, and the pricing of defaultable claims is performed with respect to this filtration. On the other hand, the definition of the filtrationF (called the reference filtration) is not always clear in the literature so far, and several interpretations can be given.
Let us now suppose that the reference filtrationF contains the market price in-formation which an investor is using for evaluating some defaultable claims. Typi-cally, this is the natural filtration of a vector of semimartingales S= (St)t≥0, with
S:= (S1, . . . , Sn). The vector S is recording the prices of observable default-free (with respect to τ ) assets which are sufficiently liquid to be used for calibrating the model. Here, we may include assets without default risk, as well as assets with a dif-ferent default time than τ , typically issued by other companies than the one we are analyzing. We shall call τ -default-free assets the components of S, since these are not necessarily assets without default risk.
As usual, we let S0stand for the locally risk-free asset (i.e., the money market ac-count); the remaining assets are risky. We denote byM(F, P) the set of all equivalent local martingale measures for the numéraire S0, i.e.,
M(F, P) = Q∼ P on G S S0= S1 S0, . . . , Sn S0 is an (F, Q)-local martingale , and we suppose thatM(F, P) is not empty in order to ensure absence of arbitrage opportunities (see e.g. Delbaen and Schachermayer [8]). Notice that because we shall work with different filtrations, we prefer to always define the probability measures on the sigma-algebraG. In this way, we avoid dealing with extensions of a probability measure. When theF-market is complete, all measures belonging to M(F, P) have the sameF-restriction.
In practice, investors might use different information sets thanF, say G. In this case, they can constructG-portfolios and G-strategies. Then, from the viewpoint of arbitrage theory, one needs to understand what the relevant prices become in a differ-ent filtration.
In particular, some investors may use more than the information inF for con-structing portfolios. For instance, they might take into account the macroeconomic environment, or firm-specific accounting information which is not directly seen in
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prices. In this caseF ⊂ G. Denote M(G, P) = Q∼ P on G S S0= S1 S0, . . . , Sn S0 is a (G, Q)-local martingale . Are there (local) martingale measures forG-informed traders? One has to understand whatF-martingales become in a larger filtration. There is no general answer to this question; in general, martingales of a given filtration are not semimartingales in a larger filtration [23]. However, from a purely economic point of view, if one assumes that the information contained inG is available for all investors without cost (i.e., this is public information), then the no-arbitrage condition becomes
M(G, P) = ∅.
This is coherent with the semi-strong form of market efficiency, which says that a price process fully reflects all relevant information that is publicly available to in-vestors. This means that publicly available information cannot be used in order to obtain arbitrage profits.
Let us now come to the particular case of default models, whereF stands for the in-formation about the prices of τ -default-free assets. In general, τ is not anF-stopping time and for the purpose of pricing defaultable claims, the progressively enlarged filtrationG has to be introduced. As an illustration, let us take the filtering model introduced by Kusuoka.
Example 2.1 (Kusuoka’s filtering model [26]) Let (Bt1, Bt2)t∈[0,T ]be a 2-dimensional
Brownian motion. The default event is triggered by the process (for instance the cash flow balance of the firm, or assets’ value)
dXt:= σ1(t, Xt)dBt1+ b(t, Xt)dt, X0= x0.
Let τ:= inf{t ∈ [0, T ]|Xt= 0} be the default time. Suppose that market investors do
not observe X, but instead the process
dYt:= σ2(t, Yt)dBt2+ μ(t, Xt∧τ, Yt)dt, Y0= y0.
The process Y might be a τ -default-free asset price that is correlated with the de-faultable asset value. For instance, suppose X is the asset value of an oil company. Then the oil price is an important piece of information to take into account when estimating the default risk of the company. Then Y can be the spot price of oil. The reference filtration isFt := σ (Ys, s≤ t), and the market filtration is constructed as
Gt:= Ft∨ σ (τ ∧ s, s ≤ t).
As Kusuoka pointed out, the above example does not fulfill the immersion prop-erty. It is natural to investigate if such a model is arbitrage-free.
Let us assume that M(F, P) is not empty, i.e., the τ -default-free market is arbitrage-free, and let us introduce the following alternative no-arbitrage conditions:
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(1) There exists Q∈ M(F, P) such that every (F, Q)-(local) martingale is a (G, Q)-(local) martingale, i.e., the immersion property holds under a risk-neutral measure.
(2) There exists a measure Q∼ P such that every (F, P)-(local) martingale is a (G, Q)-(local) martingale.
The idea behind both conditions is that since default events are public information, an investor who uses this information to decide on his trading strategy should not be able to make arbitrage profits. Condition (1) says that there is (at least) one local martingale measure in common for an investor who uses information from default (filtration G) in his trading and a less informed one, who is only concerned with τ-default-free prices levels when trading (filtrationF). Condition (2) looks at first sight less restrictive, by only saying that for each such type of investor, there exists a local martingale measure (but which could a priori be different). A closer inspection tells us that the two conditions are in fact equivalent. This equivalence will be proved in Sect.4, where we also show that these conditions are equivalent to:
(3) There exists Q∼ P such that the immersion property holds under Q.
In other words, as soon as the immersion property holds under an equivalent proba-bility measure, immersion holds as well under (at least) oneF-risk-neutral measure. Furthermore,M(G, P) is not empty, i.e., absence of arbitrage holds for the default-able market. Hence the immersion property is an important no-arbitrage condition to study.
Note also that the conditions listed above are sufficient for M(G, P) to be not empty, but not necessary. One only needs that the martingale invariance property holds for the discounted price processes S/S0, not for allF-local martingales. Thus, when theF-market is incomplete, weaker conditions can be stated. We now recall some important facts from the theory of progressive enlargements of filtrations which are relevant for our study.
3 Basic facts about random times and progressive enlargements of filtrations
In this section, we recall some important facts from the general theory of stochas-tic processes which we shall need in the sequel. We assume we are given a filtered probability space (Ω,G, F = (Ft)t≥0,P) satisfying the usual assumptions. We do not
assume thatG = F.
Definition 3.1 A random time τ is defined to be a nonnegative random variable
τ: (Ω, G) → ([0, ∞], B). With a random time τ , we associate the sigma-field Fτ= σ
zτ; (zt)anyF-optional process
.
The theory of progressive enlargements of filtrations was introduced to study prop-erties of random times which are not stopping times; it originated in a paper by Bar-low [2] and was further developed by Yor and Jeulin [21–23,32]. For further details,
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the reader can also refer to Jeulin and Yor [25] which is written in French or to Man-suy and Yor [27] or Protter [31], Chap. VI, for an English text. This theory gives the decomposition of local martingales in the initial filtrationF as semimartingales in the progressively enlarged oneG. More precisely, one enlarges the initial filtration F with the one generated by the process (τ∧ t)t≥0, so that the new enlarged filtration
G = (Gt)t≥0is the smallest filtration (satisfying the usual assumptions) containingF
and making τ a stopping time, i.e.,
Gt= Kot+, whereKot = Ft∨ σ (τ ∧ t).
Let X be aG-adapted process. We denote by(o,P)X (resp.(p,P)X) the optional (resp. predictable) projection of the process X onto the filtrationF, under the mea-sure P. When there is no ambiguity about the probability meamea-sure, we simply write
oXorpX.
A few processes play a crucial role in our discussion. These are the following: – TheF-supermartingale, called Azéma supermartingale,
t = P [τ > t | Ft] (3.1)
chosen to be càdlàg, associated with τ by Azéma [1] (note that Zτ
t >0 on the set
{t < τ}).
– The F-dual optional and predictable projections of the process 1{τ≤t}, denoted, respectively, by Aτt and aτt. We recall that by definition, theF-dual optional (resp.
F-dual predictable) projection of the increasing process 1{τ≤t} is the F-optional (resp.F-predictable) increasing process Aτ (resp. aτ) that satisfies
E 0 oX sd1{τ≤s} = E 0 XsdAτs , respectively, E 0 pX sd1{τ≤s} = E 0 Xsdasτ , for any bounded and measurable process X.
– The càdlàgF-martingale
μτt = E Ft
= Aτ
t + Zτt.
– The Doob–Meyer decomposition of the supermartingale (3.1) as
Ztτ= mτt − atτ, (3.2) where mτ is anF-martingale.
In the credit risk literature, the hazard process is very often used:
Definition 3.2 Let τ be a random time such that Ztτ>0 for all t≥ 0 (in particular τ is not anF-stopping time). The nonnegative stochastic process (Γt)t≥0defined by
Γt= − log Ztτ
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The Azéma supermartingale in (3.1) is the main tool for computing the G-pre-dictable compensator of 1{τ≤t}.
Theorem 3.3 (Jeulin/Yor [23]) Let H be a boundedG-predictable process. Then 1{τ≤t}t∧τ 0 Hs sda τ s
is aG-martingale. In particular, taking H ≡ 1, we find that
Nt:= 1{τ≤t}t∧τ 0 1 Zsτda τ s is aG-martingale.
It is important to know howF-local martingales are affected under a progressive enlargement of filtrations; in general, for an arbitrary random time, anF-local martin-gale is not always aG-semimartingale (see [22,23]). However, we have the following general result.
Theorem 3.4 (Jeulin/Yor [23]) EveryF-local martingale (Mt)stopped at τ is a
G-semimartingale, with canonical decomposition
Mt∧τ= Mt+
t∧τ
0
dM, μτs
Zsτ ,
where ( Mt)is aG-local martingale.
The following assumptions are often encountered in the literature on enlargements of filtrations or on the modeling of default times:
– The (H)-hypothesis: EveryF-martingale is a G-martingale. One says that the fil-trationF is immersed in G, or that the immersion property holds.
– Assumption (A): The random time τ avoids everyF-stopping time T , in the sense that P[τ = T ] = 0.
A property weaker than the (H)-hypothesis is when everyF-martingale stopped at τ is aG-martingale. In this situation, τ is called a pseudo-stopping time.
Definition 3.5 (Nikeghbali/Yor [30]) A random time τ is a pseudo-stopping time if
E[mτ] = m0for any boundedF-martingale m.
When one assumes that the random time τ avoids F-stopping times, then one further has
Lemma 3.6 (Jeulin/Yor [23], Jeulin [22]) If τ avoidsF-stopping times (i.e.,
condi-tion (A) is satisfied), then Aτ= aτand Aτis continuous. Therefore, the compensator
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We now recall several useful equivalent characterizations of the (H)-hypothesis in the next theorem. Note that except for the last equivalence, the results are true for any filtrationsF and G such that Ft⊂ Gt. The theorem is a combination of results by
Brémaud and Yor [5], Theorem 3, and also by Dellacherie and Meyer [10], Résultat 3, in the special case when the larger filtration is obtained by progressively enlarging the smaller one with a random time.
Theorem 3.7 (Dellacherie/Meyer [10] and Brémaud/Yor [5]) The following
asser-tions are equivalent:
1. (H): EveryF-martingale is a G-martingale.
2. For all boundedF-measurable random variables F and all boundedGt
-meas-urable random variables Gt, we have
E [FGt| Ft]= E [F | Ft] E[Gt| Ft].
3. For all boundedF-measurable random variables F,
E[F | Gt] = E[F | Ft].
4. For all s≤ t,
P[τ ≤ s | Ft] = P[τ ≤ s | F].
Let us give, as a consequence of Theorem3.7, an invariance property for the Azéma supermartingale associated with τ for a particular class of equivalent changes of measure.
Proposition 3.8 Suppose that the (H)-hypothesis holds under the measure P and let Q be a probability measure which is equivalent to P onG. If dQ/dP is F -meas-urable, then
Q[τ > t | Ft] = P[τ > t | Ft] = Zτt.
Consequently, the predictable compensator of 1{τ≤t}is unchanged under such
equiv-alent changes of probability measures, i.e.,
Nt= 1{τ≤t}t∧τ 0 daτ s s
is a G-martingale under P and Q. Moreover, the (H)-hypothesis holds under the
measure Q.
Proof Let ρ= dQ/dP. We have for s ≤ t that
Q[τ > s | Ft] =
E[ρ1{τ>s}| Ft]
E[ρ | Ft]
, and from Theorem3.7(2), we have
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and hence
Q[τ > s | Ft] = P[τ > s | Ft] = P[τ > s | F] = Q[τ > s | F].
The result then follows from Theorem3.7(4).
We now indicate some consequences of the immersion property and (A).
Corollary 3.9 Suppose that the immersion property holds. Then Zτ = 1 − Aτ is a
decreasing process. If, in addition, τ avoidsF-stopping times, then Zτ is continuous.
Proof This is an immediate consequence of Theorem3.7and Lemma3.6.
Remark 3.10 (i) It is known that if τ avoidsF-stopping times, then Zτ is continuous
and decreasing if and only if τ is a pseudo-stopping time (see [30] and [6]).
(ii) When the immersion property holds and τ avoidsF-stopping times, we have from the above corollary and Theorem3.3that the compensator of 1{τ≤t}is given by logZ1τ
t∧τ = Γt∧τ.
4 Immersion property and equivalent changes of probability measures: first results
Let (Ω,G, G = (Gt)t≥0,P) be a filtered probability space satisfying the usual
as-sumptions, andF = (Ft)t≥0be such thatF ⊂ G.
Notations
– We writeF→ G if F is immersed in G under the probability measure P. Let I(P)P be the set of all probability measures Q which are equivalent to P onG and such thatF→ G.Q
– Since we deal with different probability measures, we write EP(resp. EQ) to em-phasize that the expectation is under the measure P (resp. Q). Whenever there is no ambiguity, E is used for EP.
We now want to see how the immersion property is affected by equivalent changes of probability measures. Let Q be a probability measure which is equivalent to P onG, with ρ = dQ/dP. Define dQ dP Ft = et and dQ dP Gt = Et. (4.1)
We always consider càdlàg versions of the martingales e and E.
What can one say about the (F, Q)-martingales when considered in the filtra-tionG? A simple application of Girsanov’s theorem yields
Proposition 4.1 Assume thatF→ G. Let Q be a probability measure which is equiv-P
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The decomposition of the (F, Q)-martingales in the larger filtration can be found by applying twice Girsanov’s theorem, first in the filtrationF and then in the filtra-tionG.
Theorem 4.2 (Jeulin/Yor [24]) Assume thatF→ G. With the notation introduced inP (4.1), if (Xt)is an (F, Q)-local martingale, then the stochastic process
ItX:= Xt+ t 0 EsEs 1 es− d[X, e]s− 1 Es− d[X, E]s ,
is a (G, Q)-local martingale. Moreover,
ItX= Xt+ t 0 1 ηs− d[X, η]s,
where η= e/E is a (G, Q)-martingale.
The next lemma emphasizes the fact that a change of measure using anF∞ -meas-urable Radon–Nikodým derivative preserves the (H)-hypothesis. This can be seen as a consequence of Theorem4.2, but we give below the proof, which is elementary.
Lemma 4.3 Consider a probability space (Ω,G, P) with filtrations F ⊂ G. Assume
that ˜Q∈ I(P) and Q ∼ ˜Q is such that dQ/d ˜Q is F-measurable. Then Q∈ I(P).
Proof Let (Mt)be an (F, Q)-martingale. We must show that it is also a (G,
Q)-mar-tingale, that is, Mη is a (G, ˜Q)-martingale, where ηt:= dQ/d ˜Q|Gt is by assumption
Ft-measurable. SinceF
˜Q
→ G and Mη is F-adapted, it suffices to show that Mη is an (F, ˜Q)-martingale. For this purpose, we note that η is an F-adapted (G, ˜Q)-martin-gale, hence an (F, ˜Q)-martingale. Hence the Bayes formula yields for t < s
E˜Q[Msηs| Ft] = EQ[Ms| Ft]E˜Q[ηs| Ft] = Mtηt,
as required. By localization, the proof can be extended to local martingales. Let us now state a necessary and sufficient condition forI(P) = ∅.
Proposition 4.4 Consider a probability space (Ω,G, P) with filtrations F ⊂ G. The
following conditions are equivalent:
(a) I(P) = ∅.
(b) There exists Q∼ P such that every (F, P)-martingale is a (G, Q)-martingale. We first prove a lemma which is interesting on its own.
Lemma 4.5 Consider a probability space (Ω,G, P) with filtrations F ⊂ G and such
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(i) Q= P on F; (ii) Q∈ I(P);
(iii) the Radon–Nikodým density dQ/d ˜Q isF-measurable.
Proof Suppose ˜Q∈ I(P) with ˜ρ = dP/d ˜Q. Since ˜Q ∈ I(P), the process E˜Q[ ˜ρ | Ft]
is a positive (G, ˜Q)-martingale. We define the probability measure Q ∼ P by dQ d ˜Q Gt := E˜Q[ ˜ρ | Ft], ∀t ≥ 0,
so that (iii) holds. Since dQ/d ˜Q is F-measurable, we have from Lemma4.3that
Q∈ I(P); hence (ii) is fulfilled. We now check that Q satisfies (i) as well. Indeed,
dQ dP Ft =dQ d ˜Q Ft d ˜Q dP Ft =E˜Q[ ˜ρ | Ft] E˜Q[ ˜ρ | Ft] = 1.
Proof of Proposition 4.4 (a)⇒ (b). We assume I(P) = ∅. We consider a
mea-sure Q∈ I(P) which satisfies the requirements of Lemma 4.5. It follows that all (F, P)-martingales are (F, Q)-martingales, since Q = P on F, and also (G, Q)-mar-tingales, since Q∈ I(P). We conclude that any (F, P)-martingale is a (G, Q)-mar-tingale, as required.
(b)⇒ (a). We assume there exists Q ∼ P such that every (F, P)-martingale is a (G, Q)-martingale. It suffices to show that Q = P on F, so that any (F, Q)-martingale is an (F, P)-martingale, hence, by assumption, a (G, Q)-martingale, i.e., Q ∈ I(P).
If m is any (F, P)-martingale, by statement (b), m is a (G, Q)-martingale, which isF-adapted. Therefore m is an (F, Q)-martingale. In particular, the (F, P)-martin-gale et=dQdP|Ft is an (F, Q)-martingale, which is equivalent to saying that e
2is an (F, P)-martingale. Since e and e2 are (F, P)-martingales, it follows that e ≡ 1 and
Q= P on F.
Let us now go back to the financial framework of Sect.2, where P stands for the physical measure, and let us analyze the no-arbitrage conditions introduced there. We suppose thatM(F, P) is not empty, i.e., the F-market is arbitrage-free. Now we show that if there exists an equivalent probability measure such that immersion holds, then there exists as well a risk-neutral one such that immersion holds.
Corollary 4.6 IfM(F, P) and I(P) are not empty, then M(F, P)∩I(P) = ∅.
There-fore,M(G, P) = ∅, i.e., the market model is arbitrage-free.
Proof Suppose that ˜Q∈ I(P) and P∈ M(F, P). Notice that any probability
mea-sure that has the sameF-restriction as P will also belong to M(F, P). Let us de-fine a probability measure Q as in Lemma4.5, with P replaced by P (of course, I(P) = I(P)since P∼ P). In particular, Q= P onF by (i) in Lemma4.5; hence
Q∈ M(F, P) and also Q ∈ I(P) by (ii) in Lemma4.5. Therefore, we have precisely
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The above corollary tells us that a sufficient no-arbitrage condition for the finan-cial market introduced in Sect.2isI(P) = ∅. This result is very useful. The model by Kusuoka [26] presented in Example2.1is arbitrage-free, because there exists an equivalent change of measure such that τ is independent fromFT, and hence
immer-sion holds (see [26], pp. 79–80 for details). In this setting where the filtrationG is obtained by progressively enlargingF in order to make a random time τ a stopping time, one can show that theF-measurable random times which are not stopping times do not fulfill this no-arbitrage condition.
Lemma 4.7 Let τ be a random time which isF-measurable. ThenI(P) = ∅ if and
only if τ is anF-stopping time (in this case G = F).
Proof Suppose that P∈ I(P). Then P[τ > t | Ft] = P[τ > t | F] for all t ≥ 0;
see Theorem3.7(4). Since τ isF∞-measurable, we have P[τ > t | F] = 1{τ>t}, and hence P[τ > t | Ft] = 1{τ>t}. This is possible if and only if{τ > t} ∈ Ft for
all t , that is, if and only if τ is anF-stopping time. The converse is obvious.
Remark 4.8 So-called honest times (which are ends of predictable sets) are examples
of random times which areF∞-measurable. In the financial literature, they are en-countered in models with insider information, where insiders are shown to obtain free lunches with vanishing risks (see [18]). In the default risk literature,F∞-measurable random times which are not stopping times appear in models with delayed informa-tion (see [16]). In this case, it follows from Lemma4.7thatI(P) = ∅.
In the rest of the paper, we shall only consider the particular setting where the filtrationF as well as a random time τ are initially given and G is obtained by pro-gressively enlargingF in order to make τ a stopping time, as explained in Sect.3. We should like to answer the following questions: Are there more general changes of probability measures that preserve the immersion property? More generally, how is the predictable compensator of τ modified under an equivalent change of prob-ability measure? Indeed, it is known that market-implied (i.e., risk-neutral) default intensities are very different from the ones computed using historical data from de-faults (i.e., under the physical measure). Hence for financial applications, it is crucial to understand how the predictable compensator is modified under general changes of probability measures. Note also the recent paper [14] where the particular case is studied where theF-conditional distribution of τ admits a density with respect to some non-atomic positive measure.
For the sake of completeness, we state a general result due to Jeulin and Yor [24] which is unfortunately not easy to use in practical applications.
Proposition 4.9 (Jeulin/Yor [24]) Let Q be a probability measure which is equivalent
to P onG, with ρ = dQ/dP on G. Define the processes E and e as in (4.1) and
suppose thatF→ G. Then FP → G if and only ifQ
EP[Xρ | Gt]
Et =
EP[Xρ | Ft]
et
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In particular, if ρ isF∞-measurable, then e= E and F→ G.Q
Proof Using the Bayes formula, (4.2) is equivalent to
EQ[X| Gt]= EQ[X| Ft] for all t≥ 0 and F-measurable X,
which is equivalent to the immersion property under Q by Theorem3.7.
Remark 4.10 Proposition4.9holds for general filtrations (i.e.,G need not be obtained
by progressively enlargingF with a random time). Moreover, although this is not mentioned in [24], the necessary and sufficient condition (4.2) is valid even ifF is not immersed inG under P. However, it will not directly help us to find a larger class than the change of probability measures for which the density ρ isF-measurable.
5 Some martingale representation properties
In the remainder of the paper, (Ω,G, F = (Ft)t≥0,P) is a filtered probability space
satisfying the usual assumptions, τ is a random time andG = (Gt)t≥0is the
progres-sively enlarged filtration which makes τ a stopping time. Moreover, we suppose that τis such that condition (A) holds and that the immersion property holds under P. Re-call from Sect.3that these assumptions imply that the Azéma supermartingale (Zτt) is a decreasing and continuous process. Recall also that the notation E stands for the expectation under the measure P.
Under these assumptions, we prove in this section several general martingale rep-resentation theorems for martingales of the larger filtrationG. These results will al-low us to construct in Sect.6yet larger classes of equivalent probability measures that preserve the immersion property.
We begin with a few useful lemmas.
Lemma 5.1 Assume that (A) andF→ G hold. Let H be a G-predictable processP
and let N be theG-martingale Nt= 1{τ≤t}− Γt∧τ. If E[|Hτ|] < ∞, then
E t 0 HsdNs Ft = 0.
Proof First we note that because (A) holds, any G-predictable process equals an
F-predictable process on the stochastic interval [[0, τ]] (see [22], Sect. 4.2, and [23], Lemma 1). Therefore, using the definition of N , there exists anF-predictable process
˜
H such that for all t≥ 0, we have0tHsdNs =
t
0H˜sdNs. It follows that we can
assume without loss of generality that the process H isF-predictable; this will be assumed in the rest of the proof.
Let us first show that if E[|Hτ|] < ∞, then E[
0 |Hs| dNs] < ∞, so that all quantities under consideration are well defined. It is enough to check that both
E[0∞|Hs| d1{τ≤s}] and E[ τ 0 |Hs| dAτ s
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and is hence finite. For the second quantity, using the fact that Aτ is continuous and hence predictable and using properties of predictable projections, we have
E τ 0 |H s| dAτs s = E 0 1{τ>s}|Hs| dAτs s = E 0 p 1{τ>s}|Hs| s dAτs , where p(·) denotes the (F, P)-predictable projection. Now we use the fact that p(1
τ >s)= Zτs because τ avoidsF-stopping times to conclude that
E τ 0 |Hs| dAτs s = E 0 |Hs| dA τ s = E[|Hτ|], and consequently E[0τ|Hs|dA τ s s ] is also finite.
Since N is a local martingale of finite variation, it is purely discontinuous. Now let (Mt)be any square-integrableF-martingale. Since F
P
→ G, (Mt)is also a
G-martin-gale. We also have[M, N]t= 0, because N is purely discontinuous and has a single
jump at τ which avoidsF-stopping times. Consequently, N is strongly orthogonal to all square-integrableF-martingales, and hence E[MtNt] = 0 for all t and all
square-integrableF-martingales M. This proves the lemma.
Lemma 5.2 (Brémaud/Yor [5]) Assume that F→ G. Let H be a bounded G-pre-P
dictable process and m anF-local martingale. Then
E t 0 Hsdms Ft = t 0 pH sdms,
wherepHis the (F, P)-predictable projection of the process H .
We now easily deduce from Lemma5.1the following projection formula.
Lemma 5.3 Let τ be any random time and (zt)anF-predictable process such that
E[|zτ|] < ∞.
(i) Assume that (A) holds. Then
E[zτ1{τ>t}| Ft] = E t zsdAτs Ft .
(ii) Assume further thatF→ G. ThenP
E[zτ| Ft] = E 0 zsdAτs Ft .
If moreover the hazard process Γ is defined for all t ≥ 0, that is, if Zτt >0 for
all t≥ 0, then E[zτ| Ft] = E 0 zse−ΓsdΓs Ft .
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Proof (i) This is a consequence of the projection formulae in Theorem V.25 of [9]; see also [29].
(ii) It is enough to check the result for zs = Hr1(r,u](s), with r < u and Hr an
integrableFr-measurable random variable. But in this case, the result is an immediate
consequence of Theorem3.7.
We now state and prove a first representation theorem result for some G-mar-tingales under the assumption that (Zτt)is continuous and decreasing, that is, τ is a
pseudo-stopping time that avoids stopping times (the pseudo-stopping time assump-tion is an extension of the (H)-hypothesis framework; see [30] and [6]). This result appears in [4], Proposition 3, where it is derived under the (H)-hypothesis. We give here a simpler proof which easily extends to any random time. But before, we state a lemma which we shall use in the proof.
Lemma 5.4 [4] Let τ be an arbitrary random time. Define Lt= 1{τ>t}eΓt.
Then (Lt)t≥0is aG-martingale, which is well defined for all t ≥ 0. If τ is a
pseudo-stopping time and (A) holds (or equivalently (Zτt) is continuous and decreasing),
then Lt= 1 − t 0 dNs s ,
where (Nt)is theG-martingale Nt= 1{τ≤t}− Γt∧τ.
Theorem 5.5 Let τ be a pseudo-stopping time and z anF-predictable process such
that E[|zτ|] < ∞:
(i) If (A) holds, then
E[zτ| Gt] = m0+ t∧τ 0 dms s + t 0 (zs− hs)dNs, where mt= E[ 0 zsdA τ s| Ft] and ht = (Ztτ)−1(mt− t 0zsdA τ s).
(ii) If in additionF→ G and there exists a constant c such that E[zP τ| F] = c, then
E[zτ| Gt] = c +
t
0
(zs− hs)dNs.
Proof (i) It is well known (see [11], formula XX.(75.2) or [28], Proposition 9.12)
that
E[zτ| Gt] = LtE[zτ1{τ>t}| Ft] + zτ1{τ≤t}.
Furthermore, from Lemma5.3, with the notation of Theorem5.5, we have
E[zτ1{τ>t}| Ft] = mt
t
0
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Consequently,
E[zτ| Gt] = Ltmt− Lt
t
0
zsdAτs + zτ1{τ≤t}.
Now noting that (Lt)is a purely discontinuous martingale with a single jump at τ ,
we obtain the result that (Lt)is orthogonal to anyF-martingale. An integration by
parts combined with Lemma5.4yields the desired result.
(ii) From Lemma5.3(ii), we have under the assumptionF→ G thatP mt= E 0 zsdAτs Ft = E[zτ| Ft].
Since it is assumed that E[zτ| Ft] = E[E[zτ]F| Ft] = c, the result follows at
once.
Remark 5.6 The proof of Theorem5.5(i) can be adapted so that the result holds for an
arbitrary random time that avoids stopping times. The only thing to modify is Lemma
5.4: For an arbitrary random time τ that avoids stopping times, (Zτ
t)is continuous but
not of finite variation any more, so that an extra term must be added when expressing (Lt)as a sum of stochastic integrals.
We now combine Theorem5.5(ii) with Proposition3.8to obtain a representation theorem for a larger class ofG-martingales.
Proposition 5.7 Let τ be a random time such that (A) and F→ G hold. LetP
G= F zτ, where F is an integrable, F∞-measurable random variable such that
F= 0 a.s. and z is an F-predictable process such that zτF is integrable. Then
E[G | Gt] = E[G] + t 0 E[G] + Ys− Ls mGs mF s + s 0 kudNu dmFs + t 0 LsdmGs + t 0 mFsksdNs, where mFt := E[F | Ft], mGt := E[G | Ft], Yt= t 0 Lsd mGt mFt ,
and where (kt)is anF-predictable process (which can be given explicitly).
Proof Without loss of generality, we can assume that F is strictly positive and that
E[F ] = 1 (the general case would follow by writing F = F+− F). Then we de-fine d ˜Q|G∞= F dP|G∞. By Proposition3.8, the (H)-hypothesis holds under ˜Q and
˜Q[τ > t |Ft] = P[τ > t | Ft]. We then obtain
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Using the decomposition from Theorem5.5(i), we get E˜Q[zτ| Gt] = E˜Q[zτ] + Yt+ t 0 ksdNs, where Yt= t
0Lsdms. Here, ˜m is the ˜Q-martingale defined by
˜mt:= E˜Q[zτ| Ft] = EP[zτF| Ft] mFt −1=m G t mFt and kt= zt− Ztτ−1 mtt 0 zsdAτs . Consequently, E[G | Gt] = mFt EP[G] + t 0 Lsd mGs mF s + t 0 ksdNs . Now, an integration by parts and some tedious computations lead to
E[G | Gt] = E[G]mFt + t 0 Ys− Ls mGs mF s + s 0 kudNu dmFs + t 0 LsdmGs + t 0 mFs ksdNs,
which completes the proof of our theorem.
As a corollary, we obtain the following generalization of a representation result by Kusuoka [26], which was obtained in the Brownian filtration.
Corollary 5.8 Let τ be a random time such that (A) andF→ G hold. Then anyP
G-locally square-integrable local martingale (Mt)can be written as
Mt= M0+ Vt+
t
0
hsdNs, (5.1)
where (Vt)is in the closed subspace ofG-locally square-integrable local martingales
generated by the stochastic integrals of the form0tRsdms, where (mt)is an
F-lo-cally square-integrable local martingale, (Rt)is aG-predictable process such that
t
0Rs2dm, ms is locally integrable and (ht)is anF-predictable process such that
h2τ is integrable.
Proof The result follows from Proposition5.7and the fact that anyG-measurable
random variable can be written as a limit of finite linear combinations of functions of the form Ff (τ ) where F is anF-measurable random variable and f a Borel
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Remark 5.9 Since any element V in the closed subspace ofG-locally
square-inte-grable local martingales generated by the stochastic integrals of the form0tRsdms
is strongly orthogonal to the purely discontinuous martingales of the form0thsdNs,
it follows that the decomposition (5.1) is unique.
Corollary 5.10 (Kusuoka [26]) Assume that F is the natural filtration of a
one-dimensional Brownian motion (Wt). Let τ be a random time such that (A) and
F→ G hold. Then any G-locally square-integrable local martingale M can be writ-P
ten as Mt= M0+ t 0 RsdWs+ t 0 hsdNs,
where (Rt)is aG-predictable process such that
t
0R 2
sds is locally integrable and
(ht)is anF-predictable process such that h2τ is integrable.
Remark 5.11 A result similar to the representation of Corollary5.10would hold if
the filtrationF has the predictable representation property with respect to a family of locally square-integrable local martingales.
Combining Lemma5.1and Corollary5.8, one gets
Corollary 5.12 Let τ be a random time such that (A) andF→ G hold. AssumeP
that (Mt)is a locally square-integrable localG-martingale. Then (Mt)is strongly
orthogonal to all locally square-integrable localF-martingales if and only if there
exists anF-predictable process (ht)such that h2τ is integrable and such that
Mt= M0+
t
0
hsdNs.
6 Equivalent changes of probability measures: further results
In this section, we prove two important results. We first characterize the Radon– Nikodým derivative dQ/dP of two equivalent probability measures under which the (H)-hypothesis holds. Then we generalize Proposition3.8: We compute the Azéma supermartingale Q[τ > t | Ft] for a very large class of probability measures Q which
are equivalent to P but do not necessarily preserve the immersion property.
Let us first emphasize a multiplicative decomposition of the Radon–Nikodým derivative which is very simple and general (we only needF⊂ G). We can always write onG∞that dQ/dP= F H , where F is positive F-measurable with EP[F ] = 1 and H is a positive random variable such that EP[H | F] = 1. Indeed, one has
EP[ρ | G] = EP[ρ | F]E P[ρ | G ∞] EP[ρ | F ∞]=: F H (6.1) with F= EP[ρ | F] and H =EP[ρ |G∞] EP[ρ |F
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The following theorem relates the multiplicative decomposition (6.1) to the im-mersion property.
Theorem 6.1 Let τ be a random time such that (A) and F→ G hold. Let Q beP
a probability measure which is equivalent to P with Radon–Nikodým density given
by (6.1):
(i) If H isFτ-measurable (see Definition3.1), thenF
Q → G. (ii) IfF→ G and ifQ EP e2 E <∞, (6.2)
which is equivalent to EQ[H12] < ∞, then H is Fτ-measurable.
Proof (i) Assume first that F= 1. Since H is Fτ-measurable and EP[H | F∞] = 1,
it follows from Theorem5.5(ii) that we have Et:= EP[H | Gt] = 1 +
t
0hsdNs and
EP[H | Ft] = 1. In addition, since τ avoids F-stopping times and since E is a purely
discontinuous martingale,[M, E] = 0 for any (F, P)-martingale (Mt). Hence by
Gir-sanov’s theorem, the immersion property holds under Q.
For the general case, introduce d ˜Q = F dP. This gives dQ = H d ˜Q and EQ˜[H | F
∞] = 1. From Proposition3.8, we know that the immersion property holds under ˜Q; so using the case F = 1, it follows that the immersion property also holds
under Q.
(ii) Recall that the decomposition (6.1) holds and further assume thatF→ G alsoQ holds. Assumption (6.2) is easily seen to mean that (ηt) is an L2(G, Q)-bounded
martingale. It follows from the Bayes formula that if (mt)is any L2(F, Q)-bounded
martingale, then (mtηt)is a (G, Q)-uniformly integrable martingale. Indeed, if (mt)
is an (F, Q)-martingale, then (mtet)is an (F, P)-martingale. Since F
P
→ G holds, we also find that (mtet)is a (G, P)-martingale. Now another application of the Bayes
for-mula yields that (mtEet
t), which is (by definition) (mtηt), is a (G, Q)-martingale. Put
differently, the (G, Q)-martingale η is strongly orthogonal to all (F, Q)-martingales viewed as (G, Q)-martingales (recall that by assumption F→ G). Then by Corol-Q lary5.12, ηisFτ-measurable, and so is H = (η)−1.
As we have seen, the Azéma supermartingale plays an important role in credit risk modeling, in particular for the construction of the predictable compensator of the default arrival, or the intensity process when it exists. Now we should like to display the form of the Q-Azéma supermartingale, denoted ZQ, under a large class of equivalent changes of probability measures. This will also shed new light on some of the previous results. We shall see for instance that the changes of measure appearing in Theorem6.1with F = 1 affect only the compensator of the default arrival and leave unchanged the dynamics of the τ -default-free assets in both filtrationsF and G. We shall also identify a class of changes of measure, larger that the one introduced in Proposition3.8, which do not affect the compensator.
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Before doing so, we should like to state a very useful, though somehow forgotten, result by Itô and Watanabe [19] on multiplicative decompositions of supermartin-gales. In particular, the multiplicative decomposition turns out to be useful in the study of the intensity of the default time, as we shall see.
Theorem 6.2 (Itô/Watanabe [19]) Let (Zt)be a nonnegative càdlàg
supermartin-gale, and define
T0= inf{t : Zt= 0}.
SupposeP[T0>0] = 1. Then Z admits a multiplicative decomposition as
Zt= Zt(0)Z (1)
t
with a positive local martingale (Z(t0)) and a decreasing process (Z
(1)
t ), with
Z0(1)= 1. If there are two such factorizations, then they are identical on [[0, T0[[. If Zτ is continuous, then so are (Zt(0))and (Z
(1)
t ), as well as (mτt)and (aτt)
ap-pearing in the additive (i.e., Doob–Meyer) decomposition (3.2) of Zτ. It can be easily shown by using Itô’s lemma and the Doob–Meyer decomposition (3.2) that if Zτt >0
for all t , a.s., and Zτ is continuous, then there exist a unique local martingale (mτt)
and a unique predictable increasing process (Λt)such that the multiplicative
decom-position of Zτ is t = Et · 0 dmτ s Zs e−Λt,
where the process Λ is given by Λt= t 0 1 s dasτ
andE(·) is the stochastic exponential. From Theorem3.3, we know that the process Nt:= 1{τ≤t}− Λt∧τ is aG-martingale.
Theorem 6.3 Let τ be a random time such that (A) andF→ G hold. Assume furtherP
that Ztτ>0 for all t≥ 0, so that the process Γ = − ln Zτ is well defined. Let (mt)
be an (F, P)-martingale and F a G-predictable process such that E(0·Fsdms)is
a uniformly integrableG-martingale. Let H be an F-predictable process such that
E(0·HsdNs)is a uniformly integrableG-martingale. Let
Et= Et · 0 Fsdms Et · 0 HsdNs .
Assume further that (Et) is a uniformly integrable G-martingale (this is the case
for example ifEt( · 0HsdNs)is bounded in L 2, sincet 0Fsdms and t 0HsdNs are orthogonal). Define dQ= EtdP onGt, for all t≥ 0.
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Then the Q-Azéma supermartingale associated with τ has the multiplicative decom-position ZtQ= Q[τ > t | Ft] = Et · 0 ( ˜Fs(Q,p)Fs)dms e− t 0(1+Hs)dΓs, where:
(p,Q)F is theF-predictable projection of the process F under the probability Q;
• ˜F is an F-predictable process such that 1{τ>t}Ft= 1{τ>t}˜F ; and
• mt= mt
t
0 d[m,e]s
esis an (F, Q)-martingale.
It follows that the process
NtQ:= 1{τ≤t}
t∧τ
0
(1+ Hs)dΓs
is a (G, Q)-martingale. In particular, if the process F is F-predictable, then
ZtQ= Q[τ > t | Ft] = e
t
0(1+Hs)dΓs and the immersion property holds under Q.
Remark 6.4 The process H above is taken to beF-predictable to simplify the
nota-tions. Indeed, since the martingale N is constant after τ and since aG-predictable process before τ is equal to anF-predictable process, we could as well take H to be
G-predictable.
Proof First, we need to compute et:= EP[Et| Ft]. When applying Lemma5.1to
Et= 1 + t 0 EsFsdms+ t 0 EsHsdNs,
one obtains that et= 1 + EP t 0 EsFsdms Ft = 1 + t 0 (p,P)(E sFs)dms
(see [5], Proposition 7, for the second equality above). Next we want to show that
(p,P)(E
tFt)= et(p,Q)Ft. Let T be a predictable stopping time. The martingales E
and e satisfy
EP[ET| GT] = ET− and EP[eT | FT] = eT.
Therefore, for any predictable stopping time T , we have
EP[ETFT1{T <∞}] = EP ETFT1{T <∞} = EQ (p,Q)F T1{T <∞} = EP e T(p,Q)FT1{T <∞} = EP e T(p,Q)FT1{T <∞} .
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Hence et= Et · 0 (p,Q)F sdms .
Putting this into the formula ZQt = EP[1{τ>t}Et| Ft]/etleads us to
ZtQ= e− t 0(1+Hs)dΓs Et( · 0Fsdms) Et( · 0(p,Q)Fsdms) = e−0t(1+Hs)dΓs × exp t 0 Fs(p,Q)Fs dms− 1 2 t 0 Fs2−(p,Q)Fs 2 d[m, m]s .
Using Girsanov’s theorem,mt = mt
t 0 d[m,e]s es= mt− t 0(p,Q)Fsd[m, m]s is an
(F, Q)-martingale. The result follows when replacing mt in the above expression of
ZtQbymt+
t
0(p,Q)Fsd[m, m]s.
Corollary 6.5 Suppose thatF→ G and (A) hold. Assume further that ZP τt >0 for
all t≥ 0. Define Q by et= 1 + EP t 0 EsFsdms Ft = 1 + t 0 (p,P)(E sFs)dms
with F aG-predictable process such that E is a uniformly integrable martingale.
Then, under Q, the process Nt= 1{τ≤t}− Γt∧τremains aG-martingale.
Proof It suffices to take H= 0 in Theorem6.3.
Acknowledgements The authors would like to thank an associate editor for a very careful reading and for many suggestions which helped to improve the paper. We also wish to thank two anonymous referees for very helpful comments.
D. Coculescu was supported by the National Centre of Competence in Research “Financial Valuation and Risk Management” (NCCR FINRISK) and by Credit Suisse. M. Jeanblanc benefited from the support of the “Chaire Risque de Crédit”, Fédération Bancaire Française.
References
1. Azéma, J.: Quelques applications de la théorie générale des processus I. Invent. Math. 18, 293–336 (1972)
2. Barlow, M.T.: Study of a filtration expanded to include an honest time. Z. Wahrscheinlichkeitstheor. Verw. Geb. 44, 307–324 (1978)
3. Beghdadi-Sakrani, S., Emery, M.: On certain probabilities equivalent to coin-tossing, d’après Schachermayer. In: Sém. Proba. XXIII. Lecture Notes in Mathematics, vol. 1709, pp. 240–256 (1999)
4. Blanchet-Scalliet, C., Jeanblanc, M.: Hazard rate for credit risk and hedging defaultable contingent claims. Finance Stoch. 8, 145–159 (2004)
5. Brémaud, P., Yor, M.: Changes of filtration and of probability measures. Z. Wahrscheinlichkeitstheor. Verw. Geb. 45, 269–295 (1978)
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6. Coculescu, D., Nikeghbali, A.: Hazard processes and martingale hazard processes. Math. Finance (to appear). Preprint available on doi:10.1111/j.1467-9965.2010.0047.x
7. Coculescu, D., Geman, H., Jeanblanc, M.: Valuation of default-sensitive claims under imperfect in-formation. Finance Stoch. 12, 195–218 (2008)
8. Delbaen, F., Schachermeyer, W.: A general version of the fundamental theorem of asset pricing. Math. Ann. 300, 463–520 (1994)
9. Dellacherie, C.: Capacités et Processus Stochastiques. Springer, Berlin (1972)
10. Dellacherie, C., Meyer, P.A.: A propos du travail de Yor sur les grossissements des tribus. In: Sém. Proba. XII. Lecture Notes in Mathematics, vol. 649, pp. 69–78 (1978)
11. Dellacherie, C., Maisonneuve, B., Meyer, P.A.: Probabilités et potentiel. In: Chapitres XVII–XXIV: Processus de Markov (fin), Compléments de calcul stochastique. Hermann, Paris (1992)
12. Duffie, D., Lando, D.: Term structures of credit spreads with incomplete accounting information. Econometrica 69, 633–664 (2001)
13. Elliott, R.J., Jeanblanc, M., Yor, M.: On models of default risk. Math. Finance 10, 179–196 (2000) 14. El Karoui, N., Jeanblanc, M., Jiao, Y.: What happens after a default: the conditional density approach.
Stoch. Process. Appl. 120, 1011–1032 (2010)
15. Giesecke, K., Goldberg, L.: Forecasting default in the face of uncertainty. J. Deriv. 12, 14–25 (2004)
16. Guo, X., Jarrow, R.A., Zeng, Y.: Credit risk models with incomplete information. Math. Oper. Res. 34, 320–332 (2009)
17. Frey, R., Schmidt, T.: Pricing corporate securities under noisy asset information. Math. Finance 19, 403–421 (2009)
18. Imkeller, P.: Random times at which insiders can have free lunches. Stoch. Stoch. Rep. 74, 465–487 (2002)
19. Itô, K., Watanabe, S.: Transformation of Markov processes by multiplicative functionals. Ann. Inst. Fourier Grenoble 15, 13–30 (1965)
20. Jeanblanc, M., Valchev, S.: Partial information and hazard process. Int. J. Theor. Appl. Finance 8, 807–838 (2005)
21. Jeulin, T.: Grossissement d’une filtration et applications. In: Sém. Proba. XIII. Lecture Notes in Math-ematics, vol. 721, pp. 574–609 (1979)
22. Jeulin, T.: Semi-martingales et grossissements d’une filtration. Lecture Notes in Mathematics, vol. 833. Springer, Berlin (1980)
23. Jeulin, T., Yor, M.: Grossissement d’une filtration et semimartingales: formules explicites. In: Sém. Proba. XII. Lecture Notes in Mathematics, vol. 649, pp. 78–97 (1978)
24. Jeulin, T., Yor, M.: Nouveaux résultats sur le grossissement des tribus. Ann. Sci. Èc. Norm. Super. 4(11), 429–443 (1978)
25. Jeulin, T., Yor, M. (eds.): Grossissements de Filtrations: Exemples et Applications. Lecture Notes in Mathematics, vol. 1118. Springer, Berlin (1985)
26. Kusuoka, S.: A remark on default risk models. Adv. Math. Econ. 1, 69–82 (1999)
27. Mansuy, R., Yor, M.: Random Times and Enlargements of Filtrations in a Brownian Setting. Lecture Notes in Mathematics, vol. 1873. Springer, Berlin (2006)
28. Nikeghbali, A.: An essay on the general theory of stochastic processes. Probab. Surv. 3, 345–412 (2006)
29. Nikeghbali, A.: Non stopping times and stopping theorems. Stoch. Process. Appl. 117, 457–475 (2007)
30. Nikeghbali, A., Yor, M.: A definition and some characteristic properties of pseudo-stopping times. Ann. Probab. 33, 1804–1824 (2005)
31. Protter, P.E.: Stochastic Integration and Differential Equations, 2nd edn. Springer, Berlin (2005). Ver-sion 2.1
32. Yor, M.: Grossissements d’une filtration et semi-martingales: théorèmes généraux. In: Sém. Proba. XII. Lecture Notes in Mathematics, vol. 649, pp. 61–69 (1978)
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## Références
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Polytope of Type {6,4}
Atlas Canonical Name : {6,4}*144
Also Known As : {6,4}4if this polytope has another name.
Group : SmallGroup(144,186)
Rank : 3
Schlafli Type : {6,4}
Number of vertices, edges, etc : 18, 36, 12
Order of s0s1s2 : 4
Order of s0s1s2s1 : 6
Special Properties :
Compact Hyperbolic Quotient
Locally Spherical
Orientable
Related Polytopes :
Facet
Vertex Figure
Dual
Petrial
Skewing Operation
Facet Of :
{6,4,2} of size 288
{6,4,4} of size 576
{6,4,6} of size 864
{6,4,8} of size 1152
{6,4,8} of size 1152
{6,4,4} of size 1152
{6,4,3} of size 1296
{6,4,4} of size 1296
{6,4,10} of size 1440
{6,4,12} of size 1728
Vertex Figure Of :
{2,6,4} of size 288
{3,6,4} of size 432
{4,6,4} of size 576
{6,6,4} of size 864
{6,6,4} of size 864
{8,6,4} of size 1152
{9,6,4} of size 1296
{10,6,4} of size 1440
{12,6,4} of size 1728
{12,6,4} of size 1728
Quotients (Maximal Quotients in Boldface) :
2-fold quotients : {6,4}*72
9-fold quotients : {2,4}*16
18-fold quotients : {2,2}*8
Covers (Minimal Covers in Boldface) :
2-fold covers : {6,8}*288, {12,4}*288
3-fold covers : {6,4}*432a, {6,12}*432e, {6,12}*432f, {6,4}*432b, {6,12}*432h, {6,12}*432i
4-fold covers : {6,16}*576, {12,4}*576, {12,8}*576a, {24,4}*576a, {24,4}*576b, {12,8}*576b
5-fold covers : {30,4}*720, {6,20}*720
6-fold covers : {6,8}*864a, {6,24}*864d, {6,24}*864e, {12,4}*864b, {12,12}*864f, {12,12}*864g, {12,4}*864d, {12,12}*864j, {6,8}*864b, {6,24}*864g, {6,24}*864h, {12,12}*864l
7-fold covers : {42,4}*1008, {6,28}*1008
8-fold covers : {24,4}*1152a, {12,8}*1152a, {24,8}*1152a, {24,8}*1152b, {24,8}*1152c, {24,8}*1152d, {48,4}*1152a, {12,16}*1152a, {48,4}*1152b, {12,16}*1152b, {12,4}*1152a, {12,8}*1152b, {24,4}*1152b, {6,32}*1152
9-fold covers : {18,4}*1296a, {18,4}*1296b, {6,4}*1296a, {6,12}*1296j, {6,12}*1296k, {6,12}*1296l, {6,12}*1296m, {6,12}*1296n, {6,36}*1296m, {6,12}*1296o, {6,36}*1296n, {6,36}*1296o, {6,12}*1296s, {6,12}*1296t, {6,12}*1296u
10-fold covers : {60,4}*1440, {30,8}*1440, {6,40}*1440, {12,20}*1440
11-fold covers : {66,4}*1584, {6,44}*1584
12-fold covers : {6,16}*1728a, {6,48}*1728d, {6,48}*1728e, {12,4}*1728b, {12,12}*1728f, {12,12}*1728g, {12,8}*1728a, {12,24}*1728g, {12,24}*1728h, {24,4}*1728a, {24,12}*1728i, {24,12}*1728j, {24,4}*1728c, {24,12}*1728k, {24,12}*1728l, {12,8}*1728d, {12,24}*1728m, {12,24}*1728n, {24,4}*1728f, {24,12}*1728q, {24,4}*1728g, {24,12}*1728r, {6,16}*1728b, {6,48}*1728g, {12,8}*1728g, {12,24}*1728s, {12,8}*1728h, {12,24}*1728t, {12,4}*1728c, {12,12}*1728q, {6,48}*1728h, {12,12}*1728t, {12,24}*1728u, {24,12}*1728v, {24,12}*1728w, {12,24}*1728x, {6,4}*1728, {6,12}*1728j, {12,12}*1728ab
13-fold covers : {78,4}*1872, {6,52}*1872
Permutation Representation (GAP) :
```s0 := ( 1,28)( 2,30)( 3,29)( 4,34)( 5,36)( 6,35)( 7,31)( 8,33)( 9,32)(10,19)
(11,21)(12,20)(13,25)(14,27)(15,26)(16,22)(17,24)(18,23)(37,64)(38,66)(39,65)
(40,70)(41,72)(42,71)(43,67)(44,69)(45,68)(46,55)(47,57)(48,56)(49,61)(50,63)
(51,62)(52,58)(53,60)(54,59);;
s1 := ( 1, 4)( 2, 5)( 3, 6)(10,13)(11,14)(12,15)(19,22)(20,23)(21,24)(28,31)
(29,32)(30,33)(37,49)(38,50)(39,51)(40,46)(41,47)(42,48)(43,52)(44,53)(45,54)
(55,67)(56,68)(57,69)(58,64)(59,65)(60,66)(61,70)(62,71)(63,72);;
s2 := ( 1,37)( 2,40)( 3,43)( 4,38)( 5,41)( 6,44)( 7,39)( 8,42)( 9,45)(10,46)
(11,49)(12,52)(13,47)(14,50)(15,53)(16,48)(17,51)(18,54)(19,55)(20,58)(21,61)
(22,56)(23,59)(24,62)(25,57)(26,60)(27,63)(28,64)(29,67)(30,70)(31,65)(32,68)
(33,71)(34,66)(35,69)(36,72);;
poly := Group([s0,s1,s2]);;
```
Finitely Presented Group Representation (GAP) :
```F := FreeGroup("s0","s1","s2");;
s0 := F.1;; s1 := F.2;; s2 := F.3;;
rels := [ s0*s0, s1*s1, s2*s2, s0*s2*s0*s2, s1*s2*s1*s2*s1*s2*s1*s2,
s2*s0*s1*s2*s0*s1*s2*s0*s1*s2*s0*s1,
s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1 ];;
poly := F / rels;;
```
Permutation Representation (Magma) :
```s0 := Sym(72)!( 1,28)( 2,30)( 3,29)( 4,34)( 5,36)( 6,35)( 7,31)( 8,33)( 9,32)
(10,19)(11,21)(12,20)(13,25)(14,27)(15,26)(16,22)(17,24)(18,23)(37,64)(38,66)
(39,65)(40,70)(41,72)(42,71)(43,67)(44,69)(45,68)(46,55)(47,57)(48,56)(49,61)
(50,63)(51,62)(52,58)(53,60)(54,59);
s1 := Sym(72)!( 1, 4)( 2, 5)( 3, 6)(10,13)(11,14)(12,15)(19,22)(20,23)(21,24)
(28,31)(29,32)(30,33)(37,49)(38,50)(39,51)(40,46)(41,47)(42,48)(43,52)(44,53)
(45,54)(55,67)(56,68)(57,69)(58,64)(59,65)(60,66)(61,70)(62,71)(63,72);
s2 := Sym(72)!( 1,37)( 2,40)( 3,43)( 4,38)( 5,41)( 6,44)( 7,39)( 8,42)( 9,45)
(10,46)(11,49)(12,52)(13,47)(14,50)(15,53)(16,48)(17,51)(18,54)(19,55)(20,58)
(21,61)(22,56)(23,59)(24,62)(25,57)(26,60)(27,63)(28,64)(29,67)(30,70)(31,65)
(32,68)(33,71)(34,66)(35,69)(36,72);
poly := sub<Sym(72)|s0,s1,s2>;
```
Finitely Presented Group Representation (Magma) :
```poly<s0,s1,s2> := Group< s0,s1,s2 | s0*s0, s1*s1, s2*s2,
s0*s2*s0*s2, s1*s2*s1*s2*s1*s2*s1*s2,
s2*s0*s1*s2*s0*s1*s2*s0*s1*s2*s0*s1,
s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1 >;
```
References : None.
to this polytope | 2,685 | 4,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-43 | latest | en | 0.682447 |
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May 5, 2016
# Homework Help: Math
Posted by Karen on Monday, June 17, 2013 at 8:43pm.
How do you weight probabilities to add up to 1?
Example:
Chance of rain .25
Chance of person wearing purple hat .1
Chance of person wearing red hat .05
Chance of person wearing black hat .2
Chance of person wearing yellow hat .1
Chance of person wearing white hat .3
So, You run into a person wearing a hat in the rain.
The chance of running into people in each color of hat in the rain is
Chance of person wearing purple hat .025
Chance of person wearing red hat .0125
Chance of person wearing black hat .05
Chance of person wearing yellow hat .025
Chance of person wearing white hat .075
Would I weight this to make it all add up to 1? What is what I need to do called? | 201 | 774 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2016-18 | longest | en | 0.899451 |
http://brainly.in/question/20911 | 1,477,384,148,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720000.45/warc/CC-MAIN-20161020183840-00245-ip-10-171-6-4.ec2.internal.warc.gz | 37,319,138 | 8,834 | # Free help with homework
## Why join Brainly?
• find similar questions
# What volume of 0.1M of NaOH solution is required to neutralise 100 ml of conc. aqueous sol. of sulphuric acid which contains 98% H2SO4 by mass.The density
of conc. aqueous solution is 1.84 gml-1.NaOH reaacts with H2SO4 acc. to following reaction.
2NaOH+H2SO4 = Na2SO4+2H2O
1
by rehnoorsidhu99 | 125 | 371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2016-44 | latest | en | 0.889677 |
https://de.maplesoft.com/support/help/maple/view.aspx?path=StudyGuides/MultivariateCalculus/Chapter6/Examples/Section6-3/Example6-3-4 | 1,685,595,277,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647614.56/warc/CC-MAIN-20230601042457-20230601072457-00480.warc.gz | 228,508,459 | 43,448 | Example 6-3-4 - Maple Help
Chapter 6: Applications of Double Integration
Section 6.3: Surface Area
Example 6.3.4
Calculate the surface area of the surface defined by the function whose domain is the plane region $R$, the region bounded by the graphs of $f\left(x\right)=\mathrm{arctan}\left(x+1\right)-1/2$ and $g\left(x\right)=\mathrm{sin}\left(x\right)$ on the interval $x\in \left[0,\mathrm{π}/2\right]$. See Example 6.2.4. | 146 | 440 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 74, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-23 | latest | en | 0.572841 |
https://studysoupquestions.com/questions/math/100591/explain-the-derivation-behind-the-derivative-of-sin-x-i-e-prove-f-sin-x-cos-x-how-about-cos-x-and-tan-x | 1,656,905,060,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104293758.72/warc/CC-MAIN-20220704015700-20220704045700-00419.warc.gz | 578,461,205 | 6,124 | > > Explain the derivation behind the derivative of sin(x) i.e. prove f'(sin(x)) = cos(x) How about cos(x) and tan(x)?
# Explain the derivation behind the derivative of sin(x) i.e. prove f'(sin(x)) = cos(x) How about cos(x) and tan(x)?
## Explain the derivation behind the derivative of sin(x) i.e. prove f...
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### Explain the derivation behind the derivative of sin(x) i.e. prove f'(sin(x)) = cos(x) How about cos(x) and tan(x)?
Explain the derivation behind the derivative of sin(x) i.e. prove f'(sin(x)) = cos(x) How about cos(x) and tan(x)? | 154 | 558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-27 | latest | en | 0.841088 |
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# 1
## Chapter 4, Bravais Lattice
A Bravais lattice is the collection of all (and only those) points in space reachable from the origin with
position vectors:
n
1
, n
2
, n
3
integer (+, -, or 0)
a n a n a n R
r r r
r
+ + =
a
1
, a
2
, and a
3
not all in same plane
The three primitive vectors, a
1
, a
2
, and a
3
, uniquely define a Bravais lattice. However, for one
Bravais lattice, there are many choices for the primitive vectors.
A Bravais lattice is infinite. It is identical (in every aspect) when viewed from any of its lattice points.
3 3 2 2 1 1
a n a n a n R + + =
This is not a
Bravais lattice.
Honeycomb: P and Q are
equivalent. R is not.
A Bravais lattice can be defined as either the collection of lattice points, or the primitive translation
vectors which construct the lattice.
POINT OBJECT: Remember that a Bravais lattice has only points. Points, being dimensionless
and isotropic, have full spatial symmetry (invariant under any point symmetry operation).
Primitive Vectors
One sure way to find a set of primitive vectors (as described
in Problem4 8) is the following:
There are many choices for the primitive vectors of a Bravais lattice.
in Problem 4.8) is the following:
(1) a
1
is the vector to a nearest neighbor lattice point.
(2) a
2
is the vector to a lattice points closest to, but not
on, the a
1
axis.
(3) a
3
is the vector to a lattice point nearest, but not on,
the a
1
a
2
plane.
How does one prove that this is a set of primitive vectors?
Hint: there should be no lattice points inside, or on the faces
( ll l ) f h l h d ( ll l i d) f d (parallolegrams) of, the polyhedron (parallelepiped) formed
by these three vectors.
Actually, a
1
does not need to be a vector between nearest neighbors (e.g. green arrow). It just needs
to be a finite vector which is not a multiple of another vector (i.e. we cant pick the blue vector).
What happens if we choose the red vector as our a
1
?
CONCLUSION: The three primitive vectors can be chosen with considerable degree of freedom.
simul exam01
2
Bravais Lattice Examples
Simple Cubic
a
1
= [a, 0, 0];
a = [0 a 0];
Face-Centered Cubic (fcc)
a
1
= [0, 0.5, 0.5]a;
a = [0 5 0 0 5]a;
Body-Centered Cubic (bcc)
a
1
= [-0.5, 0.5, 0.5]a;
a = [0 5 -0 5 0 5]a; a
2
= [0, a, 0];
a
3
= [0, 0, a];
a
2
= [0.5, 0, 0.5]a;
a
3
= [0.5, 0.5, 0]a;
a
2
= [0.5, -0.5, 0.5]a;
a
3
= [0.5, 0.5, -0.5]a;
Lattice Points In A Cubic Cell
s.c. (0, 0, 0) b.c.c. (0, 0, 0), (0.5a, 0.5a, 0.5a)
f.c.c. (0, 0, 0), (0.5a, 0.5a, 0), (0, 0.5a, 0.5a), (0.5a, 0, 0.5a)
Verify for bcc and fcc:
1. Every lattice point is reached.
2. Every lattice point is equivalent.
3. Agree w/ prescribed method.
4. Volume of primitive cell.
Primitive Unit Cell
PRIMITIVE UNIT CELL: A volume of space that, when translated through all the vectors in
a Bravais lattice, just fills all of space without overlapping. There is an infinite number of choices
for primitive unit cell. Two common choices are the parallelepiped and the Wigner-Seitz cell.
Parallelipiped
Wigner-Seitz Cell: primitive cell with
full symmetry of the Bravais lattice
1 , , 0
3 2 1 3 3 2 2 1 1
< + + = x x x a x a x a x r
r r r r
Volume of
Primitive Cell
| ) ( |
3 2 1
a a a V
cell
r r r
=
A primitive cell contains precisely one lattice point and has a examples of valid p p y p
volume of v=1/n where n is the density of lattice points.
Given any two primitive cells of arbitrary shape, it is possible to cut
the first one into pieces, which, when translated through lattice
vectors, can be reassembled to give the second cell.
If space is divided up into subspaces belonging to each lattice point.
A primitive cell is the space associated with one lattice point.
Portions of the same unit cell dont even need to be connected.
examples of valid
primitive cell
3
Wigner-Seitz Cell
Wigner-Seitz cell about a lattice point is the region of space
that is closer to that point than to any other lattice point.
Construction of Wigner-Seitz Cell: space reached
from a lattice point without crossing any plane
bisecting lines drawn to other lattice points
What if a point in space is equidistance to two
lattice points? three lattice points? .
Generally, the larger the
facet on a Wigner-Seitz
cell, the closer is the
nearest neighbor distance
along that direction.
As we will see, all point symmetry operations of the
Bravais lattice are also symmetry operations on the
Wigner-Seitz cell, and vice versa.
Conventional Unit Cell
A non-primitive unit cell is conventionally chosen for
convenience. Typically, these unit cells have a few times
the volume of the primitive cell. They can fill space
bcc
p y p
without overlaps and gaps through translational
vectors which are sums of multiples of lattice constants.
Conventionally, lattice points are assumed to occupy
corners of the parallelepiped cells.
bcc simple cubic with two Bravais lattic points in a unit cell
lattice constant primitive vector length
bcc simple cubic with two Bravais lattic points in a unit cell
fcc simple cubic with four Bravais lattic points in a unit cell
centered tetragonal, centered monoclinic, base-centered orthorhombic, body-
centered orthorhombic two Bravais lattic points in a unit cell
face-centered orthorhombic four Bravais lattic points in a unit cell
4
Homework
Homework assignments (and hints) can be found
htt // d i b kl d / h i /t /GC745S12
Ch 4: 2, 5, 6, 8(a)
Ch 5: 1 2
Ch 6: 1, 3
Ch 7 2 5
Ch 7: 2 5
Ch 4-7 Homework Due Date: 3/2
Bravais Lattice Classification
Bravais lattices are classified according to the set of rigid symmetry operations
which take the lattice into itself. (.. meaning that the old position of every lattice
i t ill b i d b ( th ) l tti i t ft th ti ) E l f
Beginning of Chapter 7
point will be occupied by a(nother) lattice point after the operation.) Examples of
symmetry operations: translation, rotation, inversion, reflection.
The set of symmetry operations is known as a symmetry group or space group.
All translations by lattice vectors obviously belong to the space group.
The order of any space group is infinite. (Why?)
All rules of group theory apply: e.g. the identity operation, the inverse of
operation the product of any two operations all belong to the group operation, the product of any two operations all belong to the group.
A sub-group of the space group can be formed by taking those symmetry
operations which leave at least one lattice point unchanged. This is known as the
point group, which still displays all properties of a group.
The order of a point group is finite.
5
Point Symmetry Operations
Any symmetry operation of a Bravais lattice can be compounded out of a
translation T
R
through a lattice vector R and a rigid operation leaving at least
one lattice point fixed.
The full symmetry group of a Bravais lattice contains only operations of the
following form:
1. Translations through lattice vectors.
2. Operations that leave a particular point of the lattice fixed.
3. Operations that can be constructed by successive applications of (1) and (2).
Point Symmetry Operations
Proper and improper operations.
What about mirror planes that do not contain any lattice points?
6
Point Groups Crystal Systems
There are seven distinguishable point groups of
Bravais lattice. These are the seven crystal
systems systems.
What are the differences and the
similarities between Bravais lattices
belonging to the same crystal system?
The 7 Crystal Systems
The orders of the point groups can be more easily visualized by
counting the number of different ways to orient a lattice.
7
The 14 Bravais Lattices
From the full symmetries (point operations and translations) of the
Bravais lattice, 14 different space groups have been found.
Cubic(3): simple cubic, face-centered cubic, body centered cubic
Tetragonal (2): simple tetragonal, centered tetragonal
Orthorhombic (4): simple orthorhombic, body-centered
orthorhombic, face-centered orthorhombic, base-centered
orthorhombic
Monoclinic (2): simple monoclinic, centered monoclinic
Why cant we have a orthorhombic lattice which is
centered on two perpendicular faces?
Trigonal (1)
Hexagonal (1)
Triclinic (1)
NOTE: All Bravais lattices belonging to the same crystal system
have the same set of point operations which bring the lattice to
itself. For example, any point symmetry operation for a single
cubic is also a point symmetry operation for a b.c.c. or an f.c.c.
lattice.
In other words, a crystal system does not uniquely define a
Bravais lattice.
not translation operations!!
Crystal Structure: Lattice With A Basis
A Bravais lattice consists of lattice points. A crystal structure consists of
identical units (basis) located at lattice points.
Honeycomb net:
Diamond Structure
honeycomb when the word
hexagonal is mentioned.
8
Close-Packed Structures
Hexagonal Close-Packed Structure
63299 1
8
a a c 63299 . 1
3
8
= = Ideal HCP c/a ratio
two-atom basis
Close-Packed Structures
hexagonal polytypes hexagonal polytypes
twins
stacking faults
antiphase domains
fcc
9
Symmetry Operations For Real Crystal Structures
Bravais lattice constructed from translation of lattice point (point is
spherically symmetric).
Real (perfect) crystals are constructed from translation of object (unit
cell) in space.
A symmetry operation for the crystal structure is one which takes the
crystal to itself (indistinguishable from before).
Crystal symmetry depends not only on the symmetry of the Bravais
lattice of the crystal, but also on the symmetry of the unit cell.
P i t t ti (th ith iti f t l t i t Point symmetry operations (those with position of at least one point
unchanged) form a sub-group (crystal point group) of any full crystal
space symmetry group.
There are 32 different crystallographic point groups.
Cubic Point Groups
The cubic group is identical to the octahedral group.
O: no inversion
T
h
: no 4-fold, horiz. planes
T
d
: no 4-fold, diag. planes
T: no inv., no 4-fold rot.
Why is T still cubic. When does a
structure cease to be cubic?
Example: if we painted the top and the bottom of a cube black,
the rest of the faces white, to what point symmetry group does
this crystal belong?
10
32 Crystallographic Point Groups
Two point
groups are
id i l if identical if
they contain
precisely the
same
operations.
What crystallographic point group does
one get by putting trigonal objects (e.g.
NH
3
) on tetragonal Bravais lattice sites?
Moral of story: Translation vectors do not
determine crystallographic point group.
Crystallographic Space Groups
There are 230 crystallographic space groups.
New symmetry operations (not available for Bravais lattices) become possible
Example: Screw axis. non-
Bravais translation +
hcp
New sy e y ope o s ( o v b e o v s ces) beco e poss b e
for crystals.
glide plane: non-
Bravais translation +
reflection in plane
containing vector
11
Common Crystal Structures
NaCl NaCl
CsCl
You are expected to
know the details of these
structures by name.
Technologically Important Structures
zincblend Wurzite
These structures follow the stacking of layers with sequences
the same as discussed before for fcc and hcp. However, unlike
fcc and hcp, these structures are not close-packed.
12
Other Important Structures
Provskite (SrTiO
3
)
and high TC
superconductors.
Fluorite (CaF
2
)
Summary Of Crystal Symmetry
1. Bravais lattice consists of points.
2. Unit cell + Bravais lattice = crystal lattice y
3. Symmetry operations of Bravais lattice determine its point
group and space group.
4. Symmetry operations of real crystal lattice determine its
crystallographic point group and space group.
5. 14 different Bravais lattices (space groups) can be found,
falling into 7 different crystal systems (point groups). g y y (p g p )
6. 230 different crystallographic space groups can be found,
falling into 32 different point groups. | 3,257 | 11,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-26 | latest | en | 0.917233 |
http://mathhelpforum.com/algebra/39326-scientific-notation.html | 1,495,656,014,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607860.7/warc/CC-MAIN-20170524192431-20170524212431-00090.warc.gz | 221,228,618 | 10,558 | 1. ## Scientific Notation
Write the following numbers in scientific notation:
a) 201,348,761,098
b) 0.00000078
c) (2.4*10 12) 12 is expnenet * (8.6*10 11)11 is expo
d) (6.1*10 13) 13 is expo) * (9.2*10 8) 8 is the expo.
2. Originally Posted by recca
Write the following numbers in scientific notation:
a) 201,348,761,098
b) 0.00000078
c) (2.4*10 12) 12 is expnenet * (8.6*10 11)11 is expo
d) (6.1*10 13) 13 is expo) * (9.2*10 8) 8 is the expo.
a) 2.01348761098 * 10^11
b) 7.8 * 10^-7
c)2.4*10^12 * 8.6*10^11 = 2.4*8.6*10^23 = 2.064*10^24
d)6.1*10^13 * 9.2*10^8 = 6.1*9.2*10^21 = 5.612 * 10^22
3. Hi recca
to add a bit more to sean.1986
it come down to decimal places and zeros, 1 million has 6 zeros, and a total of 7 digits ie 1000000. when you are dealing with very large numbers it is easy to add or loose digit, so if you place a decimal point after the first digit, in this case 1, then count the remaining digits, in this case 6, it is then written in scientific notation as $1.0\times10^6$
The reverse of this happens on the other side of the decimal point, a millimeter is 1/1000 of a meter or 0.001m or $1.0\times10^-4$
When multiplying you add the powers, and multiply the bases
when dividing you subtract the powers and divide the bases
You can then adjust the decimal place ant the notation to suit as sean.1986 has done in the answers that he has provided
a final thing to watch for, sometimes the letter E will be substituted for x10, mainly on calculators, the trap as I found is that I thought E is the same as e, which it isn't. e is what is called a natural log.
I hope some of this makes sense
Dazza
4. ## Thank you
Thanks for your help, it does make sense
,
,
### scientific notation of 0.00000078
Click on a term to search for related topics. | 617 | 1,777 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-22 | longest | en | 0.856449 |
https://communities.sas.com/t5/Base-SAS-Programming/Dataset-structure-for-regression/td-p/74405?nobounce | 1,531,992,054,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590794.69/warc/CC-MAIN-20180719090301-20180719110301-00487.warc.gz | 625,610,474 | 28,487 | ## Dataset structure for regression
Occasional Contributor
Posts: 15
# Dataset structure for regression
I have a dataset block, I call it A.
A is the actual data, which is large.
Then I have A_1, A_2, ... A_3,....A_n which are simulated, smaller blocks of data that have the same # of columns as A but a lot less rows than A. (n=1000, say).
I want to regress using the dataset that has A_1 appended to A.
Then, repeat with A_2 appended to A, until A_n appended to A.
Is there any good way to do this as efficiently as possible, without appending the data A_1,...., A_n to A through a loop data step and performing regression?
Thank you.
Contributor
Posts: 22
## Re: Dataset structure for regression
Not that I know of (using SAS for 27 years). I would just write a macro loop to append the data for each case and do any analysis. Something like:
[pre]
%macro loopy;
%do i=1 %to 1000;
data Aplus;
set A A_&i;
run;
proc reg data=Aplus;
*--- etc. ----;
run;
%end;
%mend;
[/pre]
%loopy
Posts: 3,852
## Re: Dataset structure for regression
I would try to avoid 1000 data steps followed by 1000 calls to PROC REG.
A data step view seems like a good choice and with a bit of help from the macro language to write it you can run one data step and one call to PROC REG.
[pre]
data A;
set sashelp.class;
run;
*** You don't need this as you already have 1000 sim datas;
%macro simdata(arg);
data
%do i = 1 %to &arg;
A&i
%end;
;
set sashelp.class(obs=5);
run;
%mend simdata;
options mprint=1;
%simdata(1000);
*** Combine each SIM data with a copy of A;
%macro combine(arg);
data all / view=all;
set
%do i = 1 %to &arg;
A&i A
%end;
indsname=indsname open=defer;
retain simgroup;
from = indsname;
if indsname ne 'WORK.A' then simgroup = indsname;
run;
%mend combine;
%combine(1000);
proc print data=all(obs=100);
run;
proc reg data=all noprint outest=est;
by NOTSORTED simgroup;
model weight = age height;
run;
quit;
[/pre]
Contributor
Posts: 22
## Re: Dataset structure for regression
I absolutely agree with data _null_ - I would want to avoid 1000 datasetps & 1000 proc regs - there is almost always a better solution by restructuring the problem - one way being the one suggested - however, there are rare times when you just gotta bite the bullet and muscle through the 1000 or more steps.
Posts: 3,852
## Re: Dataset structure for regression
I agree too.
I think the program will scale ok using OPEN=DEFER should allow the data step to concatenate the data sets efficiently.
For my similated 1000 data sets I was surprised that I could create 1000 data sets in a single step. If they had more variables it might be a problem.
Even if you could not concatenate 1000 data sets in one step the problem could probably be broken up to smaller say 100 data set groups.
I'd like to see the OP's program that created the 1000 simulated data sets.
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## ASTROPHYSCL PROCESS
by: Solon Leuschke
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# ASTROPHYSCL PROCESS ASTRO 405
Solon Leuschke
ISU
GPA 3.81
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## Popular in Astronomy and Astrophysics
This 2 page Class Notes was uploaded by Solon Leuschke on Sunday September 27, 2015. The Class Notes belongs to ASTRO 405 at Iowa State University taught by Staff in Fall. Since its upload, it has received 24 views. For similar materials see /class/214521/astro-405-iowa-state-university in Astronomy and Astrophysics at Iowa State University.
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Date Created: 09/27/15
Astro 405505 fall semester 2005 Additional problems Problem 1 plasma You intend to study a gas cloud that collapses to form a star Suppose the cloud has a radius R 1015 cm a temperature T 1000 K a density of neutral atoms 72 1012 cm g and a density of ionized atoms and electrons n6 108 cm g Explain with what technique you would describe the collapse Problem 2 emission spectra Suppose an emission region is homogeneous and the radiation coef cients have the following form j AV 1 04 B V73 0 0 A B const Calculate the angular distribution of the intensity ID that would be seen by an observer at a xed position at a distance D from the emission region and the total flux a Assume the emission region has the form of a cube with sidelength R that is observed exactly along the normal of one of the sides b Assume the emission region has the form of a sphere with radius R Problem 3 stellar emission The conditions in stellar photospheres are often well represented by a Local Thermodynamic Equilibrium LTE Neglecting scattering please derive a quantitative estimate for the angular distribution of the intensity ID that would be seen by an outside observer a Assume the temperature T is constant independent of the vertical optical depth which itself is assumed independent of frequency b Assume the temperature is constant in each of three zones with values To OSDSH TTz 2To 73913 739 S 7392 4T0 TQSTZ Problem 4 pinhole camera A pinhole camera consists of a small circular hole of diameter d on the front side of a box which is at a distance L from the lm plane on the back side of the box Ifd denotes the angle to the optical axis perpendicular to the front and back planes of the camera show that the flux at the lm plane depends on the intensity eld 167 g5 as 7r cos4 F11 ll67 Problem 5 radiation transport Consider a cloud of gas in LTE with temperature Ty and diameter and thickness D box geometry Assume emission and absorption processes for continuum and lines to operate in the cloud Derive the solution of this radiation transport problem and discuss it for the limiting cases of very large and very small optical depth Suppose a background star with Te gtgt Tg was located behind the gas cloud What would the spectrum of the star look like Problem 6 Review questions a What conservation laws do the hydrodynamical equations describe b What is the di erence between intensity and flux c How would you estimate the level occupation density of hydrogen in the solar photosphere and in a diffuse interstellar gas cloud Can the two be treated the same way d Why do accretion disks form and how do they help the accretion of matter on compact objects e What is the color of blackbody radiation f What can you say about the directionality of the radiation eld if the scattering optical depth 7390 5 d3 03 is very large and absorption is inef cient
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posted by .
If 720.0 grams of steam at 400.0 degrees Celsius absorbs 800.0 kJ of energy, what will be its increase in temperature?
• Chemistry I HN -
See your other post. Same formula.
• Chemistry I HN -
758
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Fixed Gear Bicycle Wheels
# Fixed Gear Bicycle Wheels
Posted in the Marion Forum
Since: Apr 12
#1 Oct 10, 2012
Single speed bicycle should be focus on the ratio that front disc and after fixed gear http://www.chinacarbonwheels.com/carbon-fixed... to their actual situation choice and go to appropriate refit it.My present empirical data is that:tire specification for 26 * 1 3/8 (one and three 8) of the single speed bicycle, fixed gear ratio should be 2.5 as a benchmark. If someone Habit high step frequency lighter power by 2.35 2.5, strength pedal frequency slightly slow can consider 2.5 2.6. This tooth advantage of care shook the bike ride long gentle slope, and even shorter steep slope can also deal with. http://www.chinacarbonwheels.com/carbon-fixed... best step frequency is recreational ride between 75-90, sports ride between 90-100. Wave long gentle slope stepped frequency can also keep in between 50-75. Wave steep slope (such as street overpass) sometimes even step frequency dropped to about 30.
Here give a step frequency:
Formula: speed /(grail tooth/pannikin tooth * wheel circumference * 60 seconds)= per minute (RPM) step frequency
Such as the use of 26 * 1 3/8 tires, shroud 36 t, t fly and 26 normal car, its step frequency calculated as follows: speed of 25000 meters per hour ÷(36 ÷ 14 x 2.086 x 60 seconds)= 77.68 RPM this is 25 per hour of step frequency, then the speed of 30 of step frequency is how much, speed of 30000 meters per hour ÷(36 ÷ 14 x 2.086 x 60 seconds)= 93.22 RPM
Commonly used wheel circumference value (unit: m)
26 * 1 2.086 (26 ordinary car tyre) http://www.chinacarbonwheels.com/carbon-fixed...
26 * 1 1.973 (mountain slicks)
26 * 1.5 2.026 (mountain slicks or fine lines tire)
26 * 1.75 2.070 (mountain tooth twins or fine lines tire) http://www.chinacarbonwheels.com/carbon-fixed...
In the mountainous single speed car ride to physical demand is higher, larry technology also need good (the correct posture, the reasonable rhythm and systemic compatibility), on muscle endurance has certain value of exercise. From strengthening fitness, practice technology point of view, with variable speed shop across the even get also has its values, believe can also help to improve the ability to ride a variable speed car.http://www.chinacarbonwhee ls.com/carbon-fixed-gear-wheel s-clincher-88mm.html
Maintenance key
In addition to various parts of daily inspection and maintenance outside, and single speed car maintenance is focused on axial and pedal. http://www.chinacarbonwheels.com/700c-carbon-... time out of town to key check its working conditions so that give you flow the road, cry cry not to come out, all; Before and after the brake is key to take care of the object, always ensure the have enough braking dynamics, brake line to year to change in time.
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### Marion Discussions
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# What equal 33?
Wiki User
2016-11-04 15:24:56
1 x 33, 3 x 11 = 33
Wiki User
2016-11-04 15:24:56
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
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Wiki User
2012-12-04 22:52:50
11x3=33
Princess Pony girl
Lvl 1
2022-10-14 14:28:59
hmm Ok your correct but i was more looking for addtion | 163 | 417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-06 | latest | en | 0.801662 |
https://fr.mathworks.com/matlabcentral/answers/2025017-how-to-solve-this-equation?s_tid=prof_contriblnk | 1,716,643,803,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058829.24/warc/CC-MAIN-20240525131056-20240525161056-00754.warc.gz | 218,459,066 | 30,715 | # How to solve this equation?
3 vues (au cours des 30 derniers jours)
Ali Almakhmari le 24 Sep 2023
Modifié(e) : John D'Errico le 25 Sep 2023
I have this equation:
s is a complex number. I want to use MATLAB to figure out what critical value of τ that causes this equation to have the solution, in other words, s, to have a positive number for the real part of s.
##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
Torsten le 24 Sep 2023
Modifié(e) : Torsten le 24 Sep 2023
Which equation ? What you write is a mathematical expression, not an equation.
I'm surprised "fmincon" cannot do better:
format long
s0 = [1 1 1];
sol = fmincon(@f,s0,[],[],[],[],[0 -Inf -Inf],[Inf Inf Inf],[],optimset('MaxFunEvals',10000,'MaxIter',10000));
Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
sol
sol = 1×3
1.0e+03 * 0.000000000000109 -0.000000731599559 4.292766616773154
f(sol)
ans =
1.000003668571633
function obj = f(s)
sr = s(1);
si = s(2);
tau = s(3);
obj = norm((sr+1i*si)^2+4*(sr+1i*si)+3+2*exp(-tau*(sr+1i*si)))^2;
end
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### Réponse acceptée
Sam Chak le 24 Sep 2023
It seems that there are no real solutions for τ. From the graph, we can see that we need to lower the quadratic by more than 3 units in order to obtain one positive real root for s. In other words, there are no real solutions for because . You can check when τ is an infinitely large value.
s = linspace(-4, 3, 7001);
y1 = s.^2 + 4*s + 3; % 2nd-degree polynomial (poly2)
y2 = 2*exp(-(+1)*s);
y3 = 2*exp(-(-1)*s);
plot(s, [y1; y2; y3]), ylim([-5 20]), grid on
xline(0, '--')
yline(3, '--', 'y = 3')
xlabel('s')
legend('poly2', '\tau = 1', '\tau = -1', '', '', 'fontsize', 12, 'location', 'best')
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### Plus de réponses (1)
John D'Errico le 24 Sep 2023
Modifié(e) : John D'Errico le 25 Sep 2023
As I see it, from your question, I think people have been missing the point. As has been said, there are no real solutions for s, for any (real) value of tau. But since you are willing to accept complex solutions, as long as the real part of s is positive, then what can you do? You asked for the critical value of tau, that yields a solution with positive real part. That is different from a real solution for s.
And of course, solve cannot handle this probem. I'll arbitrarily choose tau = 1 first, just to get a feeling for what is happening. A plot will always make things easy to understand.
tau = 1;
Rs = @(sr,sc) real((sr+i*sc).^2 + 4*(sr+i*sc) + 3 + 2*exp(-tau*(sr+i*sc)));
Is = @(sr,sc) imag((sr+i*sc).^2 + 4*(sr+i*sc) + 3 + 2*exp(-tau*(sr+i*sc)));
Since I care about solutions with a positive real part, look at the right half plane, even though I also plotted the left half plane too.
fimplicit(Rs,'r',[-10,10,-10,10])
hold on
fimplicit(Is,'b',[-10,10,-10,10])
hold off
grid on
xlabel 'Real part of s'
ylabel 'Imaginary part of s'
Remember, we need to look for where the red and blue lines cross. And they do not cross in the right half plane. Different values of tau, as long as tau is a positive number, never seem to show a crossing in the RIGHT half plane. If tau is negative, then solutions do exist.
tau = -1;
Rs = @(sr,sc) real((sr+i*sc).^2 + 4*(sr+i*sc) + 3 + 2*exp(-tau*(sr+i*sc)));
Is = @(sr,sc) imag((sr+i*sc).^2 + 4*(sr+i*sc) + 3 + 2*exp(-tau*(sr+i*sc)));
Again, since I care about solutions with a positive real part, look at the right half plane.
fimplicit(Rs,'r',[-10,10,-10,10])
hold on
fimplicit(Is,'b',[-10,10,-10,10])
hold off
grid on
xlabel 'Real part of s'
ylabel 'Imaginary part of s'
And there we see multiple solutions in the right half plane. Given the figure, I'd expect to see a solution at roughly s=3+4.5i, when tau=-1.
syms S
vpasolve(S.^2 + 4*S + 3 + 2*exp(-tau*S),S,3+4.5i)
ans =
I might hazard a guess that for any negative value for tau, a solution does exist with positive real part. In fact, for negative values of tau, I might conjecture there will be infinitely many solutions with positive real part. Conversely, for positive values of tau, we might be able to prove there exists no solution with positive real part. That would take some thought of course. Anyway, What happens at tau==0?
solve(S.^2 + 4*S + 3 + 2*exp(-0*S))
ans =
Ah. When tau is zero, then there is no positive real part solution, yet we found a solution for at least one negative value of tau. And that suggests there could possibly be a "critical" negative value of tau where the solutions begin to arise. Or it might suggest that a solution exists for only negative tau, and that as tau approoaches zero from belo, the solutions go to infinity.
tau = -0.25
tau = -0.2500
Rs = @(sr,sc) real((sr+i*sc).^2 + 4*(sr+i*sc) + 3 + 2*exp(-tau*(sr+i*sc)));
Is = @(sr,sc) imag((sr+i*sc).^2 + 4*(sr+i*sc) + 3 + 2*exp(-tau*(sr+i*sc)));
fimplicit(Rs,'r',[0,50,-50,50])
hold on
fimplicit(Is,'b',[0,50,-50,50])
hold off
grid on
xlabel 'Real part of s'
ylabel 'Imaginary part of s'
And there we do find solutions.
So, are negative values of tau acceptable? Is there a valid reason to chase down this rabbit hole to actually identify IF some critical value does exist, and what it is for negative tau?
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Note:
This worksheet draws on the results of customer data gathered using Analysis 2
and Analysis 66
. Please review these analyses for further information and context for the steps in this worksheet.
Step 1:
Interview a few of the best sales and marketing people in the company to define Win and Fail benefits over each of the last three years. In this step, you would use the Benefit Code Packet you developed in Worksheet #3: Defining the Purpose of Roles. Use the benefits stated in each category of the Customer Buying Hierarchy to define Win and Fail benefits.
Benefit Code Packets: Commercial Printing Industry or Less Than Truckload Trucking
Step 2:
Apply these Win/Fail definitions to each instance of volatility in the random customer samples to determine the volatile sales volumes connected with Wins and Failures.
Step 3:
Separate the "Weak Wins" from "Wins" in all instances of Positive Volatility.
Step 4:
Add the total volume of sales from all occurrences of Positive Volatility. For each reason for Positive Volatility, total the volume of sales for the reason and divide that total by the total volume of sales in all instances of Positive Volatility. Convert this fraction into a percentage. Rank order these reasons by percentages. These percentage rankings will give you guidance on the relative importance of the reasons for Positive Volatility.
Step 5:
Separate the total of percentages for "Wins" and for "Weak Wins." The total percentage of "Wins" plus the total percentage of "Weak Wins" should equal 100% of the Positive Volatility. The total percentage of "Wins" equals the percentage of "Wins" in the market (see Analysis 21 and Basic Strategy Guide 7). The total percentage of "Weak Wins" equals the total percentage of "Failures."
Step 6:
Add the total volume of sales for all instances of Negative Volatility. Then total the sales for each reason for Negative Volatility.
Step 7:
Total the sales for each reason for Negative Volatility and divide each reason's sales total by the total sales for all instances of Negative Volatility. Separate the reasons that result in "Failures" from the reasons that result in a loss after a competitor's "Win." Rank order the reasons for "Failures". These percentage rankings will give you guidance on the relative importance of the reasons for "Failures".
Step 8:
Confirm the conclusions the company draws from its random 100 customer sample with the results of fifteen to twenty minute second round random interviews of 25 of the largest two segments of customers that the company might seek. See the introduction to Analysis 66. In undertaking this step, you would use the Code Packet you developed in Worksheet #3: Defining the Purpose of Roles, and the attendant Access or Excel databases, in order to summarize your results.
Benefit Code Packets: Commercial Printing Industry or Less Than Truckload Trucking
• Once the company has created minimum size breaks for customer size segments based on the 100 results in the random customer sample, the company can define the customers the company has the capacity to serve. Ideally, the company would like to serve the Very Large customer set. However, if it does not have the capability to do so, then the largest customers the company should seek would be Large customers.
If the company is able to serve Very Large customers, then it should define a random sample of 25 Very Large and Large customers. The company conducts an individual interview, lasting 15 to 20 minutes, to determine how these customers buy and to explain volatility. If the company is unable to serve Very Large customers, then the sample of 25 customers for external interviews would include a total of 25 Large and Medium customers.
• The information to obtain from each of these interviews is included as Analysis 2 and Analysis 66 in the Tools/Analyses section of StrategyStreet.
• It is important for the quality of these interviews that the customer interviewed does not know which company is seeking the information. In order to accomplish this, the company may use temporary employees, such as college or graduate students working over the summer, or external resources, such as consulting or market research firms. | 863 | 4,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-10 | longest | en | 0.89509 |
https://www.jiskha.com/questions/1732128/why-do-the-shiites-believe-that-uyyades-should-not-be-in-power | 1,632,693,967,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057973.90/warc/CC-MAIN-20210926205414-20210926235414-00322.warc.gz | 819,377,423 | 5,027 | # social studies
why do the shiites believe that uyyades should not be in power
1. 👍
2. 👎
3. 👁
1. Are you sure of the spelling of uyyades?
1. 👍
2. 👎
👤
Ms. Sue
1. 👍
2. 👎
1. 👍
2. 👎
👤
Ms. Sue
4. this didnt really help snything else?
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1. 👍
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Ms. Sue
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Question 10 of 25 Modeling/Multiple Representation
The distance between parallel lines p and m is 12 units. Point A is on line p. How many points are equidistant from lines p and m and 8 units from point A?
1. 1 2. 2 3. 3 4. 4 | 88 | 281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2013-20 | latest | en | 0.83789 |
https://community.qlik.com/thread/47770 | 1,532,230,763,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593004.92/warc/CC-MAIN-20180722022235-20180722042235-00562.warc.gz | 607,825,688 | 22,042 | 2 Replies Latest reply: Feb 22, 2012 3:21 AM by Thomas Ljungström
# replace parameter in set analysis
Hi!
I'm loading a fair amount of expressions to calculate stuff. However, the expressions are quite the same with just a parameter different. How to create a 'standard' expression and then in my chart supply the parameter?
This is what I have:
```vBelopp_a=sum({\$<Region_own={[ON-US]},PosEntryDesc={Annat}>}Transbelopp)
vBelopp_c=sum({\$<Region_own={[ON-US]},PosEntryDesc={Chip}>}Transbelopp)
```
I would like to write:
```vBelopp=sum({\$<Region_own={[ON-US]},PosEntryDesc={\$1}>}Transbelopp)
```
and then supply the parameter in chart like:
```\$(vBelopp('{Annat}'))
\$(vBelopp('{Chip}'))
```
I can't get the syntax right (if it's possible to do like this...).
Thanks
• ###### replace parameter in set analysis
Hi,
Try with this
`vBelopp=sum({\$<Region_own={[ON-US]},PosEntryDesc=\$1>} Transbelopp)`
Then use
`\$(vBelopp('{Annat}'))\$(vBelopp('{Chip}'))`
Celambarasan
• ###### Re: replace parameter in set analysis
Thanks Celambarasan, but no. But it got me continue and this is how you do it:
```The variable:
sum({\$<Region_own={[ON-US]},PosEntryDesc={\$1}>}Transbelopp)
The expression:
\$(vBeloppRegion_annat('Annat'))
```
Extend with more parameters:
```The variable:
sum({\$<Region_own={\$2},PosEntryDesc={\$1}>}Transbelopp)
The expression:
\$(vBeloppRegion_annat('Annat','ON-US'))
``` | 419 | 1,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-30 | latest | en | 0.476651 |
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# Answer to Question #26296 in C++ for red rose
Question #26296
An angle is considered acute if it is less than 90 degrees, obtuse if it is greater than 90 degrees, and a right angle if it is equal to 90 degrees. Using this information, write a C++ program that accepts an angle, in degrees, and displays the type of angle corresponding to the degrees entered.
A student’s letter grade is calculated according to the following schedule:
Greater than or equal to 90 A
Less than 90 but greater than or equal to 80 B
Less than 80 but greater than or equal to 70 C
Less than 70 but greater than or equal to 60 D
Less than 60 F
Using this information, write a C++ program that accepts a student’s numerical grade, converts the numerical grade to an equivalent letter grade, and displays the letter grade.
1
2013-03-14T13:05:37-0400
#include <iostream>
using namespace std;
int main()
{
/* Ask the user to enter an angle in degrees */
double angle;
cout << "Enter an angle in degrees: ";
cin >> angle;
/* Normalize the angle if it's bigger than 360 degrees */
while (angle > 360)
angle -= 360;
/* Print the type of the angle */
if (angle == 90)
cout << "The angle is right." << endl;
else if (angle < 90)
cout << "The angle is acute." << endl;
else
cout << "The angle is obtuse." << endl;
return 0;
}
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for any assignment or question with DETAILED EXPLANATIONS! | 397 | 1,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-21 | latest | en | 0.802341 |
www.hunggartorino.it | 1,695,913,309,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00347.warc.gz | 60,244,701 | 14,975 | # The sorting numbers routines
A classic of programming required in the industrial technical schools is the realization of the so-called bubble sort : comparing two numbers so that the smallest “leaps” in the upper position. It can be realized with a pair of cycles and a temporary variable (as support ) that allows the exchange of the array values, thus avoiding the loss of one of the two numbers.
With SuperBasic to sort in ascending order we’ll proceed in this way:
240 FOR I = A-1 TO 1 STEP -1
250 FOR J = 0 TO I – 1
260 IF matrice\$(J)>matrice\$(J+1) THEN
270 Temp\$ = matrice\$(J)
280 matrice\$(J) = matrice\$(J+1)
290 matrice\$(J+1) = Temp\$
300 END IF
310 NEXT J
320 NEXT I
while the reverse order (highest to lowest ) is so made:
340 FOR I = A-1 TO 1 STEP -1
350 FOR J = 0 TO I – 1
360 IF matrice\$(J)<matrice\$(J+1) THEN
370 Temp\$ = matrice\$(J)
380 matrice\$(J) = matrice\$(J+1)
390 matrice\$(J+1) = Temp\$
400 END IF
410 NEXT J
420 NEXT I
So, the implementation on the QL is similar to what also happens with the other Basic dialects except from the END IF statement which, for example, is missing in ZX Spectrum Basic and therefore is necessary to use GOSUBs or set all on the same line. As well is missing the ELSE keyword, so we have to utilize jumps thus making the program less clear.
## One thought on “The sorting numbers routines”
1. J.J. van de Molengraaf secretary SIN_QL_AIR says:
Keep up the good work! | 409 | 1,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-40 | longest | en | 0.813947 |
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06-09-2015, 08:41 AM #1
Class 3B Laser Join Date: Jan 2014 Location: Iraq Posts: 4,287 Rep Power: 2132
Class 3B Laser
Join Date: Jan 2014
Location: Iraq
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OMG Laser Shotgun
__________________
http://www.lexellaser.com/techinfo_wavelengths.htm
Laser Safety:
http://www.laserpointersafety.com/index.html
Angular Size Calculator; use with diode angle of radiation specification to calculate the needed lens diameter for a given FL:
http://www.1728.org/angsize.htm
Divergence to spot size calculator: http://tinyurl.com/spotsize - 1 mRad is about .057 degrees which from the earth would expand to be ~10% the diameters of the moon or sun at their distances.
Divergence Calculator: http://www.pseudonomen.com/lasers/ca...alculator.html - Measure your lasers beamwidth at 1 foot & then at a further distance to calculate the divergence.
Online calculator to determine spot intensity at different mRad's & powers @ distances: http://tinyurl.com/divergence-calculator
Laser Power Density Calculator: http://www.ophiropt.com/laser-measur...ity-calculator
How to build a laser beam expander to reduce divergence: http://tinyurl.com/BeamExpander
RHD's Relative Perceived Brightness Calculator. Compare brightness @nm: http://lsrtools.1apps.com/relativebrightness
YAG Power Calculator, i.e. convert ns @ mJ to peak power in watts http://tinyurl.com/YAG-Pulse
The forum costs more to run than donations received, if you wish to help click this link: http://laserpointerforums.com/donations.htm
_______
Last edited by Alaskan; 06-09-2015 at 08:44 AM.
06-09-2015, 08:43 AM #2
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Constandinos97
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Re: OMG Laser Shotgun
Quote:
To put the 40 watts of laser power in perspective: the most powerful laser you can buy off the shelf is 0.005 watts
lol
So they think...
__________________
*Insert intellectual quote here*
06-09-2015, 09:07 AM #3
Class 3B Laser Join Date: May 2013 Location: Snohomish County Washington, US Posts: 3,800 Rep Power: 1086
Pi R Squared
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Re: OMG Laser Shotgun
Alan
__________________
Keychain 650nm <5mW
Quartet 4-in-1 630-680nm <1mW
AtlasNova 635nm <5mW
MillionAccessories 532nm <5mW broken
M462 462nm one of a kind (in progress)
PLTB450B 450nm 1913mW G2 lens
PL520 520nm 82mW acrylic lens
S06J 12X 405nm 590mW G2 lens
C6 M140 445nm 1.5W 3 element lens
9mm 445nm with G2 lens in a stainless steel host
Radiant Electronics X4 3.7W Laser Power Meter
___________________________________________
The light that shines twice as bright burns half as long.
06-09-2015, 09:11 AM #4
Class 3B Laser Join Date: Nov 2008 Location: Germany Posts: 4,239 Rep Power: 292
Elektrotechniker
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06-09-2015, 03:07 PM #5
Class 3B Laser Join Date: Jan 2014 Location: Iraq Posts: 4,287 Rep Power: 2132
Class 3B Laser
Join Date: Jan 2014
Location: Iraq
Posts: 4,287
Rep Power: 2132
Re: OMG Laser Shotgun
Hell dude, it's the first time I saw it and I'm in this forum daily.
__________________
http://www.lexellaser.com/techinfo_wavelengths.htm
Laser Safety:
http://www.laserpointersafety.com/index.html
Angular Size Calculator; use with diode angle of radiation specification to calculate the needed lens diameter for a given FL:
http://www.1728.org/angsize.htm
Divergence to spot size calculator: http://tinyurl.com/spotsize - 1 mRad is about .057 degrees which from the earth would expand to be ~10% the diameters of the moon or sun at their distances.
Divergence Calculator: http://www.pseudonomen.com/lasers/ca...alculator.html - Measure your lasers beamwidth at 1 foot & then at a further distance to calculate the divergence.
Online calculator to determine spot intensity at different mRad's & powers @ distances: http://tinyurl.com/divergence-calculator
Laser Power Density Calculator: http://www.ophiropt.com/laser-measur...ity-calculator
How to build a laser beam expander to reduce divergence: http://tinyurl.com/BeamExpander
RHD's Relative Perceived Brightness Calculator. Compare brightness @nm: http://lsrtools.1apps.com/relativebrightness
YAG Power Calculator, i.e. convert ns @ mJ to peak power in watts http://tinyurl.com/YAG-Pulse
The forum costs more to run than donations received, if you wish to help click this link: http://laserpointerforums.com/donations.htm
_______
06-09-2015, 03:36 PM #6
Class 3R Laser Join Date: Jan 2015 Location: USA Posts: 1,244 Rep Power: 352
Laser Chick
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Re: OMG Laser Shotgun
LOL, they are right next to other. No worries, as we are all human!!
06-09-2015, 04:30 PM #7
Class 3B Laser Join Date: Jan 2014 Location: Iraq Posts: 4,287 Rep Power: 2132
Class 3B Laser
Join Date: Jan 2014
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Re: OMG Laser Shotgun
Yea i know each member owns the forum, some more than others... Reminds me why i quit a job in Germany, not that you wont find disagreeables everywhere, just their culture has their noses up everyone elses back side far more than most.... that said, America has their own idiocies too, my problem is disliking any critism and saying so!
Last edited by Alaskan; 06-09-2015 at 04:32 PM.
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http://bootmath.com/dense-in-mathbbr-and-dense-in-modulus-1.html | 1,529,918,811,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867644.88/warc/CC-MAIN-20180625092128-20180625112128-00118.warc.gz | 45,355,769 | 5,838 | # dense in $\mathbb{R}$ and dense in $$modulus 1 I want to solve this problem given here. The problem is: (2) The exercise is useful to help us show the following lemma. \{ar +b: a,b \in \mathbb Z\} where r \in \mathbb Q^c is dense in \mathbb R. Its equivalent to: \{ar : a \in \mathbb Z\}, where r \in \mathbb Q^c is dense in [0,1] modulus 1. But, I don’t know the meaning of dense in \mathbb{R} and dense in [0,1] modulus 1 as well as I don’t know what does the “modulus” mean here. Please help. Thank you. #### Solutions Collecting From Web of "dense in \mathbb{R} and dense in$$ modulus $1$"
A set $A\subseteq\Bbb R$ is dense in $\Bbb R$ if $(a,b)\cap A\ne\varnothing$ whenever $a,b\in\Bbb R$ and $a<b$; in other words, $A$ is dense in $\Bbb R$ if every non-empty open interval in $\Bbb R$ contains some member of $A$. An example of a dense subset of $\Bbb R$ is $\Bbb Q$, the set of rational numbers.
A set $D\subseteq[0,1]$ is dense in $[0,1]$ if $D\cap(a,b)\ne\varnothing$ whenever $(a,b)$ is an open interval in $\Bbb R$ such that $(a,b)\cap[0,1]\ne\varnothing$.
Finally, $A\subseteq\Bbb R$ is dense in $[0,1]$ modulus $1$ if $\{x-\lfloor x\rfloor:x\in A\}$ is dense in $[0,1]$, where $\lfloor x\rfloor$ is the greatest integer in $x$, i.e., the largest integer $n$ such that $n\le x$; $x-\lfloor x\rfloor$ is the fractional part of $x$.
A set $S$ is dense in $\mathbb{R}$ if, for any $r \in \mathbb{R}$ and any $\delta > 0$, there is a point $s\in S$ such that
$$r – \delta < s < r+\delta$$
“Modulus 1” means that you take the fractional part of a number. ie.
$$s\,\text{mod}(1) = s – [s]$$
where $[s]$ denotes the integer part of $s$.
In this case, you want to show that, for any $x\in [0,1]$ and any $\delta > 0$ there is a number $s \in S$ such that
$$x-\delta < s-[s] < x+\delta$$ | 642 | 1,799 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-26 | latest | en | 0.786522 |
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Save the Tram donations
Looks like the Kemmerer family really, really wants to go with option A. They're just 20 million short.
http://www.jacksonholenews.com/
I guess you can send donations to the "Tram Team."
Thanks for the link. Never been to JH. What kind of terrain will the route of proposed Tbar/poma traverse? Something tells me they are likely to sustain more injuries from a steep tbar than they would from running the old tram a few more years until a proper replacement is built.
Quote:
Originally Posted by Jamesj Thanks for the link. Never been to JH. What kind of terrain will the route of proposed Tbar/poma traverse? Something tells me they are likely to sustain more injuries from a steep tbar than they would from running the old tram a few more years until a proper replacement is built.
You might be right in a strict statistical sense but if the tram were to fail, the consequences would be devastating for those involved. Granted, tram failure (even at its current age) may be a very low probability event but any probablility above virtual zero makes it devil's calculus to specify an acceptable limit.
Quote:
Originally Posted by lifer You might be right in a strict statistical sense but if the tram were to fail, the consequences would be devastating for those involved. Granted, tram failure (even at its current age) may be a very low probability event but any probablility above virtual zero makes it devil's calculus to specify an acceptable limit.
I've been on plenty of lifts in Europe that are far older (St. Anton, Lech Garmisch), but are still operational. Just like many airplanes (DC-3s, B-24s, B-52s, etc) that are still operational beyond their service life years, each unit must be taken in its individual circumstance. That tram is still in excellent shape could be used for years beyond the original 40 year life span. They could spend that \$5 million on repairs/mods and make the tram last even longer.
Gondola
Looks like Option B is a gondola. Am I wrong, or would a gondola have greater hourly capacity than the tram would have? I understand that 20% of the time, it wouldn't operate. But, to tell you the truth, that's the 20% you least want to be at the top of Rendezvous mountain anyways. And, also being honest, a gondola is a bit more comfortable than a tram (my wife, for instance, hates trams, and you can sit down on a gondola, which you can't on a tram). What are the pluses and minuses here?
"Tram Option A, the preferred option, would replace the existing lift with an 80-passenger, top-to-bottom tram that would run along the existing tram line. The new tram would have an enclosed top terminal with a restaurant and would carry 520 skiers per hour, up from the current capacity of 400 skiers per hour."
A base-to-summit gondola would move over 1300 skiers per hour and cost \$9 million less than a tram.
Trams are very high capital cost/high maintenance cost in terms of skiers per hour. Asking federal and Wyoming taxpayers to help fund this is a ridiculous extravagance that benefits a very few. Unless the Kemmerer family wants to pay it back under terms of a low-interest loan, for example.
Except for access to backcountry, Corbet's and a few other chutes, Rendezvous Bowl is one of the least desirable steeps at JH.
People who really want to get above Sublette lift should hike for their turns.
I don't know much about the Kemmerer family but they sound like weasels. I would love to take a look at their financials. I don't buy this annualized \$27K return. If the market demands a tram, the market will pay for the tram through increased ticket prices and season pass prices which, if I'm not mistaken, are above the industry averages already. No way should taxpayer funding subsidize the construction of a new tram.
Jeez, my \$70 lift ticket isn't enough, now I need to my a donation to keep the tram too?
Quote:
Originally Posted by Jamesj Thanks for the link. Never been to JH. What kind of terrain will the route of proposed Tbar/poma traverse? Something tells me they are likely to sustain more injuries from a steep tbar than they would from running the old tram a few more years until a proper replacement is built.
It looks like it will take the same route (it will not traverse any terrian, it will climb terrain) as the previous Poma, along skier's right of Rendezvous Bowl. No higher incident rate was witnessed when the previous Poma was in.
No, the Kemmerer family is not looking for donations or partners. They are looking for low interest loans or grants, such that they can maintain control of the mountain.
HB
I'll donate money towards the poma. I know it was not popular with the locals, but as the only way to the top, it would add to the mystique of Corbetts and S&S, and make Ten Sleep, Expert Chutes a little harder to get to, and R'vous Bowl itself will get less traffic. I would rather see it be a hike if we can't raise the funds for a poma or rope tow.
Quote:
Originally Posted by HarkinBanks It looks like it will take the same route (it will not traverse any terrian, it will climb terrain) as the previous Poma, along skier's right of Rendezvous Bowl. No higher incident rate was witnessed when the previous Poma was in. No, the Kemmerer family is not looking for donations or partners. They are looking for low interest loans or grants, such that they can maintain control of the mountain. HB
Harkin:
I was in a day-long meeting today and the Pres of the Ski Corp was one of the presenters. According to him, the "new" surface lift would go up the LOOKER'S right side of Rendezvous Bowl (probably near the current tree line along the north edge of R-Bowl proper). The old poma went up looker's left and then jogged right along the ridgeline to the top.
This new route would allow for a straight line access to the top of the mountain. What's sort of being de-emphasized right at the moment is how really, REALLY difficult a route like that would be for snowboarders. Going up any surface lift seems to be difficult for boarders, and this one would be VERY steep in comparison.
Also, he was adamant about the fact that this new surface lift would be temporary until the eventual tram/gondola is constructed.
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# Ch15 transforms
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### Ch15 transforms
1. 1. Transforms
2. 2. A sine wave 8 5*sin (2π4t) 6 Amplitude = 5 4 Frequency = 4 Hz 2 0 -2 -4 -6 -8 0 0.1 0.2 0.3 0.4 0.5 seconds 0.6 0.7 0.8 0.9 1
3. 3. A sine wave signal 8 5*sin(2π4t) 6 Amplitude = 5 4 Frequency = 4 Hz 2 Sampling rate = 256 samples/second 0 -2 Sampling duration = 1 second -4 -6 -8 0 0.1 0.2 0.3 0.4 0.5 seconds 0.6 0.7 0.8 0.9 1
4. 4. An undersampled signal sin(2π8t), SR = 8.5 Hz 2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
5. 5. The Nyquist Frequency • The Nyquist frequency is equal to one-half of the sampling frequency. • The Nyquist frequency is the highest frequency that can be measured in a signal.
6. 6. Fourier series • Periodic functions and signals may be expanded into a series of sine and cosine functions
7. 7. The Fourier Transform • A transform takes one function (or signal) and turns it into another function (or signal)
8. 8. The Fourier Transform • A transform takes one function (or signal) and turns it into another function (or signal) • Continuous Fourier Transform: close your eyes if you don’t like integrals
9. 9. The Fourier Transform • A transform takes one function (or signal) and turns it into another function (or signal) • Continuous Fourier Transform: ∞ H ( f ) = ∫ h ( t ) e 2πift dt −∞ ∞ h ( t ) = ∫ H ( f ) e −2πift df −∞
10. 10. The Fourier Transform • A transform takes one function (or signal) and turns it into another function (or signal) • The Discrete Fourier Transform: N −1 H n = ∑ hk e 2πikn N k =0 1 N −1 hk = ∑ H n e −2πikn N N n =0
11. 11. Fast Fourier Transform • The Fast Fourier Transform (FFT) is a very efficient algorithm for performing a discrete Fourier transform • FFT principle first used by Gauss in 18?? • FFT algorithm published by Cooley & Tukey in 1965 • In 1969, the 2048 point analysis of a seismic trace took 13 ½ hours. Using the FFT, the same task on the same machine took 2.4 seconds!
12. 12. Famous Fourier Transforms 2 1 Sine wave 0 -1 -2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 300 250 200 Delta function 150 100 50 0 0 20 40 60 80 100 120
13. 13. Famous Fourier Transforms 0.5 0.4 Gaussian 0.3 0.2 0.1 0 0 5 10 15 20 25 30 35 40 45 50 6 5 4 Gaussian 3 2 1 0 0 50 100 150 200 250
14. 14. Famous Fourier Transforms 1.5 1 Sinc function 0.5 0 -0.5 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 6 5 4 Square wave 3 2 1 0 -100 -50 0 50 100
15. 15. Famous Fourier Transforms 1.5 1 Sinc function 0.5 0 -0.5 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 6 5 4 Square wave 3 2 1 0 -100 -50 0 50 100
16. 16. Famous Fourier Transforms 1 0.8 Exponential 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 30 25 20 Lorentzian 15 10 5 0 0 50 100 150 200 250
17. 17. FFT of FID 2 1 0 f = 8 Hz SR = 256 Hz T2 = 0.5 s -1 -2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 70 60 50 40 30 20 10 0 0 20 40 60 80 100 ( t ) = sin( 2πft ) exp − t F T 2 120
18. 18. FFT of FID 2 f = 8 Hz SR = 256 Hz T2 = 0.1 s 1 0 -1 -2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 14 12 10 8 6 4 2 0 0 20 40 60 80 100 120
19. 19. FFT of FID 2 1 0 -1 -2 f = 8 Hz SR = 256 Hz T2 = 2 s 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 200 150 100 50 0 0 20 40 60 80 100 120
20. 20. Effect of changing sample rate 2 1 0 -1 -2 f = 8 Hz T2 = 0.5 s 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 70 35 60 30 50 25 40 20 30 15 20 10 10 5 0 0 10 20 30 40 50 60 0
21. 21. Effect of changing sample rate 2 SR = 256 Hz SR = 128 Hz 1 0 -1 -2 f = 8 Hz T2 = 0.5 s 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 70 35 60 30 50 25 40 20 30 15 20 10 10 5 0 0 10 20 30 40 50 60 0
22. 22. Effect of changing sample rate • Lowering the sample rate: – Reduces the Nyquist frequency, which – Reduces the maximum measurable frequency – Does not affect the frequency resolution
23. 23. Effect of changing sampling duration 2 1 0 -1 -2 f = 8 Hz T2 = .5 s 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 2 4 6 8 10 12 14 16 18 20 70 60 50 40 30 20 10 0
24. 24. Effect of changing sampling duration 2 1 ST = 2.0 s ST = 1.0 s 0 -1 -2 f = 8 Hz T2 = .5 s 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 2 4 6 8 10 12 14 16 18 20 70 60 50 40 30 20 10 0
25. 25. Effect of changing sampling duration • Reducing the sampling duration: – Lowers the frequency resolution – Does not affect the range of frequencies you can measure
26. 26. Effect of changing sampling duration 2 1 0 -1 -2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 200 150 100 50 0 f = 8 Hz T2 = 2.0 s 0 2 4 6 8 10 12 14 16 18 20
27. 27. Effect of changing sampling duration 2 ST = 2.0 s ST = 1.0 s 1 0 f = 8 Hz T2 = 0.1 s -1 -2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 2 4 6 8 10 12 14 16 18 20 14 12 10 8 6 4 2 0
28. 28. Measuring multiple frequencies 3 f = 80 Hz, T2 = 1 s 1 2 1 f = 90 Hz, T2 = .5 s 2 2 f = 100 Hz, T2 = 0.25 s 1 3 3 0 -1 -2 -3 SR = 256 Hz 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 120 100 80 60 40 20 0 0 20 40 60 80 100 120
29. 29. Measuring multiple frequencies 3 f = 80 Hz, T2 = 1 s 1 2 1 f = 90 Hz, T2 = .5 s 2 2 f = 200 Hz, T2 = 0.25 s 1 3 3 0 -1 -2 -3 SR = 256 Hz 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 120 100 80 60 40 20 0 0 20 40 60 80 100 120
30. 30. L: period; u and v are the number of cycles fitting into one horizontal and vertical period, respectively of f(x,y).
31. 31. Discrete Fourier Transform 1 w−1 h −1 2π (ux + vy ) 2π (ux + vy ) F (u , v) = ∑∑ f ( x, y) cos wh ÷+ j sin wh ÷ wh x =0 y =0 1 w−1 h −1 − j 2π ( ux + vy ) / wh F (u , v) = ∑∑ f ( x, y)e wh x =0 y =0
32. 32. Discrete Fourier Transform (DFT). • When applying the procedure to images, we must deal explicitly with the fact that an image is: – Two-dimensional – Sampled – Of finite extent • These consideration give rise to the The DFT of an NxN image can be written:
33. 33. Discrete Fourier Transform • For any particular spatial frequency specified by u and v, evaluating equation 8.5 tell us how much of that particular frequency is present in the image. • There also exist an inverse Fourier Transform that convert a set of Fourier coefficients into an image. 1 f ( x, y ) = N N −1 N −1 F (u , v)e j 2π ( ux +vy ) / N ∑∑ x =0 y =0
34. 34. 2 P (u , v) = F (u , v) = R (u , v) + I (u , v) 2 2 PSD • The magnitudes correspond to the amplitudes of the basic images in our Fourier representation. • The array of magnitudes is termed the amplitude spectrum (or sometime ‘spectrum’). • The array of phases is termed the phase spectrum. • The power spectrum is simply the square of its amplitude spectrum:
35. 35. FFT • The Fast Fourier Transform is one of the most important algorithms ever developed – Developed by Cooley and Tukey in mid 60s. – Is a recursive procedure that uses some cool math tricks to combine sub-problem results into the overall solution.
36. 36. DFT vs FFT
37. 37. DFT vs FFT
38. 38. DFT vs FFT
39. 39. Periodicity assumption • The DFT assumes that an image is part of an infinitely repeated set of “tiles” in every direction. This is the same effect as “circular indexing”.
40. 40. Periodicity and Windowing • Since “tiling” an image causes “fake” discontinuities, the spectrum includes “fake” highfrequency components Spatial discontinuities
41. 41. Discrete Cosine Transform Real-valued π( 2i + 1) m π( 2k + 1) n Gc ( m, n) = α ( m)α ( n)∑ ∑ g( i, k ) cos cos 2 N 2N i= 0 k = 0 with an inverse N −1 N −1 π( 2i + 1) m π ( 2k + 1) n gc ( i, k ) = ∑ ∑ α ( m)α ( n)Gc ( m, n) cos cos 2 N 2N m= 0 n = 0 where N −1 N −1 α ( 0) = 1 N and α ( m) = 2 N for 1≤ m ≤ N
42. 42. DCT in Matrix Form G c = CgC where the kernel elements are Ci , m π( 2i + 1) m = α( m) cos 2N
43. 43. Discrete Sine Transform Most Convenient when N=2 p - 1 2 N −1 N −1 π( i + 1) ( m + 1) π( k + 1) ( n + 1) Gsin ( m, n) = g( i, k ) sin ∑∑ sin N + 1 i=0 k =0 N +1 N +1 with an inverse 2 N −1 N −1 π( i + 1) ( m + 1) π( k + 1) ( n + 1) gsin ( i, k ) = ∑0 ∑ Gs ( m, n) sin N + 1 sin N + 1 N + 1 m= n = 0
44. 44. DST in Matrix Form G c = TgT where the kernel elements are Ti , k = 2 π( i + 1) ( k + 1) sin N +1 N +1
45. 45. DCT Basis Functions*
46. 46. (Log Magnitude) DCT Example*
47. 47. Hartley Transform • Alternative to Fourier • Produces N Real Numbers • Use Cosine Shifted 45o to the Right cas( θ) = cos( θ) + sin( θ) π = 2 cos θ − 4
48. 48. Square Hartley Transform 2π ( im + kn ) 1 N −1 N −1 GHartley ( m, n ) = ∑0 ∑= 0 g H ( i, k ) cas N N * N i= k with an inverse 2π ( im + kn ) g Hartley ( i, k ) = ∑ ∑ GH ( m, n ) cas N m= 0 n= 0 N −1 N −1
49. 49. Rectangular Hartley Transform 1 2π mx 2π ny GHartley ( m, n ) = ∑ ∑ g H ( x, y ) cas + wh y = 0 x = 0 w h h − 1 w− 1 m ∈ [ 0..h ] , n ∈ [ 0..w] with an inverse 2π mx 2π ny g Hartley ( m, n ) = ∑ ∑ GH ( x, y ) cas + w h y= 0 x= 0 h − 1 w− 1
50. 50. Hartley in Matrix Form G Hartley = TgT where the kernel elements are Ti , k 1 2πik = cas N N
51. 51. What is an even function? • the function f is even if the following equation holds for all x in the domain of f:
52. 52. Hartley Convolution Theorem • Computational Alternative to Fourier Transform • If One Function is Even, Convolution in one Domain is Multiplication in Hartley Domain g( x ) = f ( x )* h( x ) ⇔ G( ν) = F ( ν) H even ( ν) + F ( − ν) H odd ( ν)
53. 53. Rectangular Wave Transforms • Binary Valued {1, -1} • Fast to Compute • Examples – – – – Hadamard Walsh Slant Haar
54. 54. Hadamard Transform • Consists of elements of +/- 1 • A Normalized N x N Hadamard matrix satisfies the relation H Ht = I 1 1 1 H2 = 2 1 −1 Walsh Tx can be constructed as H 2N 1 H N = 2 H N HN −H N
55. 55. Walsh Transform, N=4 Gonzalez, Wintz *
56. 56. Non-ordered Hadamard Transform H8 +1 +1 +1 +1 H8 = +1 +1 +1 +1 +1 −1 +1 −1 +1 −1 +1 −1 +1 +1 −1 −1 +1 +1 −1 −1 +1 −1 −1 +1 +1 −1 −1 +1 +1 +1 +1 +1 −1 −1 −1 −1 +1 −1 +1 −1 −1 +1 −1 +1 +1 +1 −1 −1 −1 −1 +1 +1 +1 −1 −1 +1 −1 +1 +1 −1
57. 57. Sequency • In a Hadamard Transform, the Number of Sign Changes in a Row Divided by Two • It is Possible to Construct an H matrix with Increasing Sequency per row
58. 58. Ordered Hadamard Transform 1 F ( u, v ) = N N −1 N −1 ∑ ∑ F ( j, k )( −1) q ( j , k , u, v ) j =0 k =0 where N −1 q( j, k , u, v ) = ∑ [ g i ( u) ji + g i ( v ) k i ] i=0 and g 0 (u ) ≡ un −1 g1 (u) ≡ un −1 + un − 2 g 2 ( u ) ≡ un − 2 + un − 3 g n-1 (u ) ≡ u1 + u0
59. 59. Ordered Hadamard Transform* Gonzalez, Wintz *
60. 60. Haar Transform • Derived from Haar Matrix • Sampling Process in which Subsequent Rows Sample the Input Data with Increasing Resolution • Different Types of Differential Energy Concentrated in Different Regions – Power taken two at a time – Power taken a power of two at a time, etc.
61. 61. 1 1 2 0 H4 = * 1 1 2 0 Castleman Haar Transform*, H4 1 1 −1 −1 0 0 2 − 2
62. 62. Karhunen-Loeve Transform • Variously called the K-L, Hotelling, or Eignevector • Continuous Form Developed by K-L • Discrete Version Credited to Hotelling • Transforms a Signal into a Set of Uncorrelated Representational Coefficients • Keep Largest Coefficients for Image Compression
63. 63. Discrete K-L N −1 N −1 F (u, v ) = ∑ ∑ F ( j, k ) A( j, k ; u, v ) j =0 k =0 where the kernel satisfies N −1 N −1 λ( u, v ) A( j, k ; u, v ) = ∑ ∑ K F ( j, k ; j ′, k ′) A( j ′, k ′; u, v ) j =0 k =0 where K F ( j, k ; j ′, k ′ ) is the image covariance function λ( u, v ) is a constant for a fixed ( u, v ) , the eigenvalues of K F
64. 64. Singular Value Decomposition An NxN matrix A can be expressed as A = UΛV t where Columns of U are Eigenvectors of AA t Columns of V are Eigenvectors of A t A L is the NxN diagonal matrix of singular values Λ = U t AV
65. 65. Singular Value Decomposition • If A is symmetric, then U=V • Kernel Depends on Image Being Transformed • Need to Compute AAt and AtA and Find the Eigenvalues • Small Values can be Ignored to Yield Compression
66. 66. Transform Domain Filtering • Similar to Fourier Domain Filtering • Applicable to Images in which Noise is More Easily Represented in Domain other than Fourier – Vertical and horizontal line detection: Haar transform produces non-zero entries in first row and/or first column | 5,594 | 12,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-22 | latest | en | 0.712993 |
https://phys.libretexts.org/Courses/Joliet_Junior_College/Physics_201_-_Fall_2019v2/Book%3A_Custom_Physics_textbook_for_JJC/03%3A_Vectors/3.11%3A_Vectors_(Exercises) | 1,722,736,136,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00863.warc.gz | 380,638,852 | 38,898 | # 3.11: Vectors (Exercises)
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## Conceptual Questions
### 2.1 Scalars and Vectors
1. A weather forecast states the temperature is predicted to be −5 °C the following day. Is this temperature a vector or a scalar quantity? Explain.
2. Which of the following is a vector: a person’s height, the altitude on Mt. Everest, the velocity of a fly, the age of Earth, the boiling point of water, the cost of a book, Earth’s population, or the acceleration of gravity?
3. Give a specific example of a vector, stating its magnitude, units, and direction.
4. What do vectors and scalars have in common? How do they differ?
5. Suppose you add two vectors $$\vec{A}$$ and $$\vec{B}$$. What relative direction between them produces the resultant with the greatest magnitude? What is the maximum magnitude? What relative direction between them produces the resultant with the smallest magnitude? What is the minimum magnitude?
6. Is it possible to add a scalar quantity to a vector quantity?
7. Is it possible for two vectors of different magnitudes to add to zero? Is it possible for three vectors of different magnitudes to add to zero? Explain.
8. Does the odometer in an automobile indicate a scalar or a vector quantity?
9. When a 10,000-m runner competing on a 400-m track crosses the finish line, what is the runner’s net displacement? Can this displacement be zero? Explain.
10. A vector has zero magnitude. Is it necessary to specify its direction? Explain.
11. Can a magnitude of a vector be negative?
12. Can the magnitude of a particle’s displacement be greater that the distance traveled?
13. If two vectors are equal, what can you say about their components? What can you say about their magnitudes? What can you say about their directions?
14. If three vectors sum up to zero, what geometric condition do they satisfy?
### 2.2 Coordinate Systems and Components of a Vector
1. Give an example of a nonzero vector that has a component of zero.
2. Explain why a vector cannot have a component greater than its own magnitude.
3. If two vectors are equal, what can you say about their components?
4. If vectors $$\vec{A}$$ and $$\vec{B}$$ are orthogonal, what is the component of $$\vec{B}$$ along the direction of $$\vec{A}$$? What is the component of $$\vec{A}$$ along the direction of $$\vec{B}$$?
5. If one of the two components of a vector is not zero, can the magnitude of the other vector component of this vector be zero?
6. If two vectors have the same magnitude, do their components have to be the same?
### 2.4 Products of Vectors
1. What is wrong with the following expressions? How can you correct them?
1. $$C = \vec{A} \vec{B}$$,
2. $$\vec{C} = \vec{A} \vec{B}$$,
3. $$C = \vec{A} \times \vec{B}$$,
4. $$C = A \vec{B}$$,
5. $$C + 2 \vec{A} = B$$,
6. $$\vec{C} = A \times \vec{B}$$,
7. $$\vec{A} \cdotp \vec{B} = \vec{A} \times \vec{B}$$,
8. $$\vec{C} = 2 \vec{A} \cdotp \vec{B}$$,
9. $$C = \vec{A} / \vec{B}$$, and
10. $$C = \vec{A} /B$$ .
2. If the cross product of two vectors vanishes, what can you say about their directions?
3. If the dot product of two vectors vanishes, what can you say about their directions?
4. What is the dot product of a vector with the cross product that this vector has with another vector?
## Problems
### 2.1 Scalars and Vectors
1. A scuba diver makes a slow descent into the depths of the ocean. His vertical position with respect to a boat on the surface changes several times. He makes the first stop 9.0 m from the boat but has a problem with equalizing the pressure, so he ascends 3.0 m and then continues descending for another 12.0 m to the second stop. From there, he ascends 4 m and then descends for 18.0 m, ascends again for 7 m and descends again for 24.0 m, where he makes a stop, waiting for his buddy. Assuming the positive direction up to the surface, express his net vertical displacement vector in terms of the unit vector. What is his distance to the boat?
2. In a tug-of-war game on one campus, 15 students pull on a rope at both ends in an effort to displace the central knot to one side or the other. Two students pull with force 196 N each to the right, four students pull with force 98 N each to the left, five students pull with force 62 N each to the left, three students pull with force 150 N each to the right, and one student pulls with force 250 N to the left. Assuming the positive direction to the right, express the net pull on the knot in terms of the unit vector. How big is the net pull on the knot? In what direction?
3. Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point and what is the compass direction of a line connecting your starting point to your final position? Use a graphical method.
4. For the vectors given in the following figure, use a graphical method to find the following resultants:
1. $$\vec{A} + \vec{B}$$,
2. $$\vec{C} + \vec{B}$$,
3. $$\vec{D} + \vec{F}$$,
4. $$\vec{A} − \vec{B}$$,
5. $$\vec{D} − \vec{F}$$,
6. $$\vec{A} + 2 \vec{F}$$,
7. $$\vec{A} − 4 \vec{D} + 2 \vec{F}$$.
1. A delivery man starts at the post office, drives 40 km north, then 20 km west, then 60 km northeast, and finally 50 km north to stop for lunch. Use a graphical method to find his net displacement vector.
2. An adventurous dog strays from home, runs three blocks east, two blocks north, one block east, one block north, and two blocks west. Assuming that each block is about 100 m, how far from home and in what direction is the dog? Use a graphical method.
3. In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind shifts a great deal during the day and he is blown along the following directions: 2.50 km and 45.0° north of west, then 4.70 km and 60.0° south of east, then 1.30 km and 25.0° south of west, then 5.10 km straight east, then 1.70 km and 5.00° east of north, then 7.20 km and 55.0° south of west, and finally 2.80 km and 10.0° north of east. Use a graphical method to find the castaway’s final position relative to the island.
4. A small plane flies 40.0 km in a direction 60° north of east and then flies 30.0 km in a direction 15° north of east. Use a graphical method to find the total distance the plane covers from the starting point and the direction of the path to the final position.
5. A trapper walks a 5.0-km straight-line distance from his cabin to the lake, as shown in the following figure. Use a graphical method (the parallelogram rule) to determine the trapper’s displacement directly to the east and displacement directly to the north that sum up to his resultant displacement vector. If the trapper walked only in directions east and north, zigzagging his way to the lake, how many kilometers would he have to walk to get to the lake?
1. A surveyor measures the distance across a river that flows straight north by the following method. Starting directly across from a tree on the opposite bank, the surveyor walks 100 m along the river to establish a baseline. She then sights across to the tree and reads that the angle from the baseline to the tree is 35°. How wide is the river?
2. A pedestrian walks 6.0 km east and then 13.0 km north. Use a graphical method to find the pedestrian’s resultant displacement and geographic direction.
3. The magnitudes of two displacement vectors are A = 20 m and B = 6 m. What are the largest and the smallest values of the magnitude of the resultant $$\vec{R} = \vec{A} + \vec{B}$$?
### 2.2 Coordinate Systems and Components of a Vector
1. Assuming the +x-axis is horizontal and points to the right, resolve the vectors given in the following figure to their scalar components and express them in vector component form.
1. Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point? What is your displacement vector? What is the direction of your displacement? Assume the +x-axis is to the east.
2. You drive 7.50 km in a straight line in a direction 15° east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (b) Show that you still arrive at the same point if the east and north legs are reversed in order. Assume the +xaxis is to the east.
3. A sledge is being pulled by two horses on a flat terrain. The net force on the sledge can be expressed in the Cartesian coordinate system as vector $$\vec{F}$$= (−2980.0 $$\hat{i}$$ + 8200.0 $$\hat{j}$$)N , where $$\hat{i}$$ and $$\hat{j}$$ denote directions to the east and north, respectively. Find the magnitude and direction of the pull.
4. A trapper walks a 5.0-km straight-line distance from her cabin to the lake, as shown in the following figure. Determine the east and north components of her displacement vector. How many more kilometers would she have to walk if she walked along the component displacements? What is her displacement vector?
1. The polar coordinates of a point are $$\frac{4 \pi}{3}$$ and 5.50 m. What are its Cartesian coordinates?
2. Two points in a plane have polar coordinates P1 (2.500 m, $$\frac{\pi}{6}$$) and P2 (3.800 m, $$\frac{2 \pi}{3}$$). Determine their Cartesian coordinates and the distance between them in the Cartesian coordinate system. Round the distance to a nearest centimeter.
3. A chameleon is resting quietly on a lanai screen, waiting for an insect to come by. Assume the origin of a Cartesian coordinate system at the lower left-hand corner of the screen and the horizontal direction to the right as the +x-direction. If its coordinates are (2.000 m, 1.000 m), (a) how far is it from the corner of the screen? (b) What is its location in polar coordinates?
4. Two points in the Cartesian plane are A(2.00 m, −4.00 m) and B(−3.00 m, 3.00 m). Find the distance between them and their polar coordinates.
5. A fly enters through an open window and zooms around the room. In a Cartesian coordinate system with three axes along three edges of the room, the fly changes its position from point b(4.0 m, 1.5 m, 2.5 m) to point e(1.0 m, 4.5 m, 0.5 m). Find the scalar components of the fly’s displacement vector and express its displacement vector in vector component form. What is its magnitude?
### 2.3 Algebra of Vectors
1. For vectors $$\vec{B} = − \hat{i} − 4 \hat{j}$$ and $$\vec{A} = −3 \hat{i} − 2 \hat{j}$$, calculate (a) $$\vec{A} + \vec{B}$$ and its magnitude and direction angle, and (b) $$\vec{A} − \vec{B}$$ and its magnitude and direction angle.
2. A particle undergoes three consecutive displacements given by vectors $$\vec{D}_{1}$$ = (3.0 $$\hat{i}$$ − 4.0 $$\hat{j}$$ − 2.0 $$\hat{k}$$)mm, $$\vec{D}_{2}$$ = (1.0 $$\hat{i}$$ − 7.0 $$\hat{j}$$ + 4.0 $$\hat{i}$$)mm, and $$\vec{D}_{3}$$ = (−7.0 $$\hat{i}$$ + 4.0 $$\hat{j}$$ + 1.0 $$\hat{k}$$)mm. (a) Find the resultant displacement vector of the particle. (b) What is the magnitude of the resultant displacement? (c) If all displacements were along one line, how far would the particle travel?
3. Given two displacement vectors $$\vec{A}$$ = (3.00 $$\hat{i}$$ − 4.00 $$\hat{j}$$ + 4.00 $$\hat{k}$$)m and $$\vec{B}$$ = (2.00 $$\hat{i}$$ + 3.00 $$\hat{j}$$ − 7.00 $$\hat{k}$$)m, find the displacements and their magnitudes for (a) $$\vec{C} = \vec{A} + \vec{B}$$ and (b) $$\vec{D} = 2 \vec{A} − \vec{B}$$.
4. A small plane flies 40.0 km in a direction 60° north of east and then flies 30.0 km in a direction 15° north of east. Use the analytical method to find the total distance the plane covers from the starting point, and the geographic direction of its displacement vector. What is its displacement vector?
5. . In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind shifts a great deal during the day, and she is blown along the following straight lines: 2.50 km and 45.0° north of west, then 4.70 km and 60.0° south of east, then 1.30 km and 25.0° south of west, then 5.10 km due east, then 1.70 km and 5.00° east of north, then 7.20 km and 55.0° south of west, and finally 2.80 km and 10.0° north of east. Use the analytical method to find the resultant vector of all her displacement vectors. What is its magnitude and direction?
6. Assuming the +x-axis is horizontal to the right for the vectors given in the following figure, use the analytical method to find the following resultants:
1. $$\vec{A} + \vec{B}$$,
2. $$\vec{C} + \vec{B}$$,
3. $$\vec{D} + \vec{F}$$,
4. $$\vec{A} - \vec{B}$$,
5. $$\vec{D} - \vec{F}$$,
6. $$\vec{A} + 2 \vec{F}$$,
7. $$\vec{C} - 2 \vec{B} + 3 \vec{F}$$, and
8. $$\vec{A} - 4 \vec{D} + 2 \vec{F}$$.
7. Given the vectors in the preceding figure, find vector $$\vec{R}$$ that solves equations (a) $$\vec{D} + \vec{R} = \vec{F}$$ and (b) $$\vec{C} - 2 \vec{D} + 5 \vec{R} = 3 \vec{F}$$. Assume the +x-axis is horizontal to the right.
8. A delivery man starts at the post office, drives 40 km north, then 20 km west, then 60 km northeast, and finally 50 km north to stop for lunch. Use the analytical method to determine the following: (a) Find his net displacement vector. (b) How far is the restaurant from the post office? (c) If he returns directly from the restaurant to the post office, what is his displacement vector on the return trip? (d) What is his compass heading on the return trip? Assume the +x-axis is to the east.
9. An adventurous dog strays from home, runs three blocks east, two blocks north, and one block east, one block north, and two blocks west. Assuming that each block is about a 100 yd, use the analytical method to find the dog’s net displacement vector, its magnitude, and its direction. Assume the +x-axis is to the east. How would your answer be affected if each block was about 100 m?
10. If $$\vec{D}$$ = (6.00 $$\hat{i}$$ − 8.00 $$\hat{j}$$m, $$\vec{B}$$ = (−8.00 $$\hat{i}$$ + 3.00 $$\hat{j}$$)m , and $$\vec{A}$$ = (26.0 $$\hat{i}$$ + 19.0 $$\hat{j}$$)m, find the unknown constants a and b such that a $$\vec{D} + b \vec{B} + \vec{A} = \vec{0}$$.
11. Given the displacement vector $$\vec{D}$$ = (3 $$\hat{i}$$ − 4 $$\hat{j}$$)m, find the displacement vector $$\vec{R}$$ so that $$\vec{D}$$ + $$\vec{R}$$ = −4D $$\hat{j}$$.
12. Find the unit vector of direction for the following vector quantities: (a) Force $$\vec{F}$$ = (3.0 $$\hat{i}$$ − 2.0 $$\hat{j}$$)N, (b) displacement $$\vec{D}$$ = (−3.0 $$\hat{i}$$ − 4.0 $$\hat{j}$$)m, and (c) velocity $$\vec{v}$$ = (−5.00 $$\hat{i}$$ + 4.00 $$\hat{j}$$)m/s.
13. At one point in space, the direction of the electric field vector is given in the Cartesian system by the unit vector $$\hat{E} = \frac{1}{\sqrt{5}} \hat{i} - \frac{2}{\sqrt{5}} \hat{j}$$. If the magnitude of the electric field vector is E = 400.0 V/m, what are the scalar components Ex, Ey, and Ez of the electric field vector $$\vec{E}$$ at this point? What is the direction angle $$\theta_{E}$$ of the electric field vector at this point?
14. A barge is pulled by the two tugboats shown in the following figure. One tugboat pulls on the barge with a force of magnitude 4000 units of force at 15° above the line AB (see the figure and the other tugboat pulls on the barge with a force of magnitude 5000 units of force at 12° below the line AB. Resolve the pulling forces to their scalar components and find the components of the resultant force pulling on the barge. What is the magnitude of the resultant pull? What is its direction relative to the line AB?
15. In the control tower at a regional airport, an air traffic controller monitors two aircraft as their positions change with respect to the control tower. One plane is a cargo carrier Boeing 747 and the other plane is a Douglas DC-3. The Boeing is at an altitude of 2500 m, climbing at 10° above the horizontal, and moving 30° north of west. The DC-3 is at an altitude of 3000 m, climbing at 5° above the horizontal, and cruising directly west. (a) Find the position vectors of the planes relative to the control tower. (b) What is the distance between the planes at the moment the air traffic controller makes a note about their positions?
### 2.4 Products of Vectors
1. Assuming the +x-axis is horizontal to the right for the vectors in the following figure, find the following scalar products:
1. $$\vec{A} \cdotp \vec{C}$$,
2. $$\vec{A} \cdotp \vec{F}$$,
3. $$\vec{D} \cdotp \vec{C}$$,
4. $$\vec{A} \cdotp ( \vec{F} + 2 \vec{C})$$,
5. $$\hat{i} \cdotp \vec{B}$$,
6. $$\hat{j} \cdotp \vec{B}$$,
7. $$(3 \hat{i} - \hat{j}) \cdotp \vec{B}$$ and
8. $$\hat{B} \cdotp \vec{B}$$.
1. Assuming the +x-axis is horizontal to the right for the vectors in the preceding figure, find (a) the component of vector $$\vec{A}$$ along vector $$\vec{C}$$, (b) the component of vector $$\vec{C}$$ along vector $$\vec{A}$$, (c) the component of vector $$\hat{i}$$ along vector $$\vec{F}$$, and (d) the component of vector $$\vec{F}$$ along vector $$\hat{i}$$.
2. Find the angle between vectors for
1. $$\vec{D}$$ = (−3.0 $$\hat{i}$$ − 4.0 $$\hat{j}$$)m and $$\vec{A}$$ = (−3.0 $$\hat{i}$$ + 4.0 $$\hat{j}$$)m and
2. $$\vec{D}$$ = (2.0 $$\hat{i}$$ − 4.0 $$\hat{j}$$ + $$\hat{k}$$)m and $$\vec{B}$$ = (−2.0 $$\hat{i}$$ + 3.0 $$\hat{j}$$ + 2.0 $$\hat{k}$$)m.
3. Find the angles that vector $$\vec{D}$$ = (2.0 $$\hat{i}$$ − 4.0 $$\hat{j}$$ + $$\hat{k}$$)m makes with the x-, y-, and z-axes.
4. Show that the force vector $$\vec{D}$$ = (2.0 $$\hat{i}$$ − 4.0 $$\hat{j}$$ + $$\hat{k}$$)N is orthogonal to the force vector $$\vec{G}$$ = (3.0 $$\hat{i}$$ + 4.0 $$\hat{j}$$+ 10.0 $$\hat{k}$$)N.
5. Assuming the +x-axis is horizontal to the right for the vectors in the following figure, find the following vector products:
1. $$\vec{A} \times \vec{C}$$,
2. $$\vec{A} \times \vec{F}$$,
3. $$\vec{D} \times \vec{C}$$
4. $$\vec{A} \times (\vec{F} + 2 \vec{C})$$,
5. $$\hat{i} \times \vec{B}$$,
6. $$\hat{j} \times \vec{B}$$,
7. $$(3 \hat{i} - \hat{j}) \times \vec{B}$$ and
8. $$\hat{B} \times \vec{B}$$.
1. Find the cross product $$\vec{A} \times \vec{C}$$ for
1. $$\vec{A}$$ = 2.0 $$\hat{i}$$ − 4.0 $$\hat{j}$$ + $$\hat{k}$$ and $$\vec{C}$$ = 3.0 $$\hat{i}$$ + 4.0 $$\hat{j}$$ + 10.0$$\hat{k}$$,
2. $$\vec{A}$$ = 3.0 $$\hat{i}$$ + 4.0 $$\hat{j}$$ + 10.0 $$\hat{k}$$ and $$\vec{C}$$ = 2.0 $$\hat{i}$$ − 4.0 $$\hat{j}$$ + $$\hat{k}$$,
3. $$\vec{A}$$ = −3.0 $$\hat{i}$$ − 4.0 $$\hat{j}$$ and $$\vec{C}$$ = −3.0 $$\hat{i}$$ + 4.0 $$\hat{j}$$ , and
4. $$\vec{C}$$ = −2.0 $$\hat{i}$$ + 3.0 $$\hat{j}$$ + 2.0 $$\hat{k}$$ and $$\vec{A}$$ = −9.0 $$\hat{j}$$ .
2. For the vectors in the following figure, find (a) ($$\vec{A} \times \vec{F} \cdotp \vec{D}$$), (b) ($$\vec{A} \times \vec{F}) \cdotp (\vec{A} \times \vec{C}$$), and (c) ($$\vec{A} \cdotp \vec{F})(\vec{D} \times \vec{B}$$).
1. (a) If $$\vec{A} \times \vec{F} = \vec{B} \times \vec{F}$$, can we conclude $$\vec{A}$$ = $$\vec{B}$$? (b) If $$\vec{A} \cdotp \vec{F}$$ = $$\vec{B} \cdotp \vec{F}$$, can we conclude $$\vec{A}$$ = $$\vec{B}$$? (c) If $$F \vec{A}$$ = $$\vec{B} F$$, can we conclude $$\vec{A}$$ = $$\vec{B}$$? Why or why not?
1. You fly 32.0 km in a straight line in still air in the direction 35.0° south of west. (a) Find the distances you would have to fly due south and then due west to arrive at the same point. (b) Find the distances you would have to fly first in a direction 45.0° south of west and then in a direction 45.0° west of north. Note these are the components of the displacement along a different set of axes—namely, the one rotated by 45° with respect to the axes in (a).
2. Rectangular coordinates of a point are given by (2, y) and its polar coordinates are given by (r, $$\frac{\pi}{6}$$). Find y and r.
3. If the polar coordinates of a point are (r,$$\varphi$$) and its rectangular coordinates are (x, y), determine the polar coordinates of the following points: (a) (−x, y), (b) (−2x, −2y), and (c) (3x, −3y).
4. Vectors $$\vec{A}$$ and $$\vec{B}$$ have identical magnitudes of 5.0 units. Find the angle between them if $$\vec{A}$$ + $$\vec{B}$$ = 5 2 $$\hat{j}$$.
5. Starting at the island of Moi in an unknown archipelago, a fishing boat makes a round trip with two stops at the islands of Noi and Poi. It sails from Moi for 4.76 nautical miles (nmi) in a direction 37° north of east to Noi. From Noi, it sails 69° west of north to Poi. On its return leg from Poi, it sails 28° east of south. What distance does the boat sail between Noi and Poi? What distance does it sail between Moi and Poi? Express your answer both in nautical miles and in kilometers. Note: 1 nmi = 1852 m.
6. An air traffic controller notices two signals from two planes on the radar monitor. One plane is at altitude 800 m and in a 19.2-km horizontal distance to the tower in a direction 25° south of west. The second plane is at altitude 1100 m and its horizontal distance is 17.6 km and 20° south of west. What is the distance between these planes?
7. Show that when $$\vec{A}$$ + $$\vec{B}$$ = $$\vec{C}$$, then C2 = A2 + B2 + 2AB cos $$\varphi$$, where $$\varphi$$ is the angle between vectors $$\vec{A}$$ and $$\vec{B}$$.
8. Four force vectors each have the same magnitude f. What is the largest magnitude the resultant force vector may have when these forces are added? What is the smallest magnitude of the resultant? Make a graph of both situations.
9. A skater glides along a circular path of radius 5.00 m in clockwise direction. When he coasts around one-half of the circle, starting from the west point, find (a) the magnitude of his displacement vector and (b) how far he actually skated. (c) What is the magnitude of his displacement vector when he skates all the way around the circle and comes back to the west point?
10. A stubborn dog is being walked on a leash by its owner. At one point, the dog encounters an interesting scent at some spot on the ground and wants to explore it in detail, but the owner gets impatient and pulls on the leash with force $$\vec{F}$$ = (98.0 $$\hat{i}$$ + 132.0 $$\hat{j}$$ + 32.0 $$\hat{j}$$)N along the leash. (a) What is the magnitude of the pulling force? (b) What angle does the leash make with the vertical?
11. If the velocity vector of a polar bear is $$\vec{u}$$ = (−18.0 $$\hat{i}$$ − 13.0 $$\hat{j}$$)km/h , how fast and in what geographic direction is it heading? Here, $$\hat{i}$$ and $$\hat{j}$$ are directions to geographic east and north, respectively.
12. Find the scalar components of three-dimensional vectors $$\vec{G}$$ and $$\vec{H}$$ in the following figure and write the vectors in vector component form in terms of the unit vectors of the axes.
1. A diver explores a shallow reef off the coast of Belize. She initially swims 90.0 m north, makes a turn to the east and continues for 200.0 m, then follows a big grouper for 80.0 m in the direction 30° north of east. In the meantime, a local current displaces her by 150.0 m south. Assuming the current is no longer present, in what direction and how far should she now swim to come back to the point where she started?
2. A force vector $$\vec{A}$$ has x- and y-components, respectively, of −8.80 units of force and 15.00 units of force. The x- and y-components of force vector $$\vec{B}$$ are, respectively, 13.20 units of force and −6.60 units of force. Find the components of force vector $$\vec{C}$$ that satisfies the vector equation $$\vec{A}$$− $$\vec{B}$$ + 3 $$\vec{C}$$ = 0.
3. Vectors $$\vec{A}$$ and $$\vec{B}$$ are two orthogonal vectors in the xy-plane and they have identical magnitudes. If $$\vec{A}$$ = 3.0 $$\hat{i}$$ + 4.0 $$\hat{j}$$, find $$\vec{B}$$.
4. For the three-dimensional vectors in the following figure, find (a) $$\vec{G} \times \vec{H}$$, (b) |$$\vec{G} \times \vec{H}$$| , and (c) $$\vec{G} \cdotp \vec{H}$$.
1. Show that $$(\vec{B} \times \vec{C}) \cdotp \vec{A}$$ is the volume of the parallelepiped, with edges formed by the three vectors in the following figure.
## Challenge Problems
1. Vector $$\vec{B}$$ is 5.0 cm long and vector $$\vec{A}$$ is 4.0 cm long. Find the angle between these two vectors when |$$\vec{A} + \vec{B}$$| = 3.0 cm and |$$\vec{A}$$ − $$\vec{B}$$| = 3.0 cm.
2. What is the component of the force vector $$\vec{G}$$ = (3.0 $$\hat{i}$$ + 4.0 $$\hat{j}$$ + 10.0 $$\hat{k}$$)N along the force vector $$\vec{H}$$ = (1.0 $$\hat{i}$$ + 4.0 $$\hat{j}$$)N?
3. The following figure shows a triangle formed by the three vectors $$\vec{A}$$, $$\vec{B}$$ and $$\vec{C}$$. If vector $$\vec{C}\; '$$ is drawn between the midpoints of vectors $$\vec{A}$$ and $$\vec{B}$$, show that $$\vec{C}\; '$$ = $$\frac{\vec{C}}{2}$$.
1. Distances between points in a plane do not change when a coordinate system is rotated. In other words, the magnitude of a vector is invariant under rotations of the coordinate system. Suppose a coordinate system S is rotated about its origin by angle $$\varphi$$ to become a new coordinate system S′, as shown in the following figure. A point in a plane has coordinates (x, y) in S and coordinates (x′, y′) in S′.
1. Show that, during the transformation of rotation, the coordinates in S′ are expressed in terms of the coordinates in S by the following relations: $$\begin{cases} x' = x \cos \varphi + y \sin \varphi \\ y' = -x \sin \varphi + y \cos \varphi \end{cases} \ldotp$$
2. Show that the distance of point P to the origin is invariant under rotations of the coordinate system. Here, you have to show that $$\sqrt{x^{2} + y^{2}} = \sqrt{x'^{2} + y'^{2}} \ldotp$$
3. Show that the distance between points P and Q is invariant under rotations of the coordinate system. Here, you have to show that $$\sqrt{(x_{P} - x_{Q})^{2} + (y_{P} - y_{Q})^{2}} = \sqrt{(x'_{P} - x'_{Q})^{2} + (y'_{P} - y'_{Q})^{2}} \ldotp$$
## Contributors
Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).
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# Closed interval
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Title: Closed interval Author: World Heritage Encyclopedia Language: English Subject: Closure (mathematics) Collection: Publisher: World Heritage Encyclopedia Publication Date:
### Closed interval
In mathematics, a (real) interval is a set of real numbers with the property that any number that lies between two numbers in the set is also included in the set. For example, the set of all numbers Template:Mvar satisfying 0 ≤ x ≤ 1 is an interval which contains 0 and 1, as well as all numbers between them. Other examples of intervals are the set of all real numbers $\R$, the set of all negative real numbers, and the empty set.
Real intervals play an important role in the theory of integration, because they are the simplest sets whose "size" or "measure" or "length" is easy to define. The concept of measure can then be extended to more complicated sets of real numbers, leading to the Borel measure and eventually to the Lebesgue measure.
Intervals are central to interval arithmetic, a general numerical computing technique that automatically provides guaranteed enclosures for arbitrary formulas, even in the presence of uncertainties, mathematical approximations, and arithmetic roundoff.
Intervals are likewise defined on an arbitrary totally ordered set, such as integers or rational numbers. The notation of integer intervals is considered in the special section below.
## Notations for intervals
The interval of numbers between Template:Mvar and Template:Mvar, including Template:Mvar and Template:Mvar, is often denoted Template:Closed-closed. The two numbers are called the endpoints of the interval. In countries where numbers are written with a decimal comma, a semicolon may be used as a separator, to avoid ambiguity.
### Excluding the endpoints
To indicate that one of the endpoints is to be excluded from the set, the corresponding square bracket can be either replaced with a parenthesis, or reversed. Both notations are described in International standard ISO 31-11. Thus, in set builder notation,
\begin\left\{align\right\}
(a,b) = \mathopen{]}a,b\mathclosea,b\mathclose{]} &= \{x\in\R\,|\,a Note that Template:Open-open, Template:Closed-open, and Template:Open-closed represent the empty set, whereas Template:Closed-closed denotes the set {a} . When a > b, all four notations are usually assumed to represent the empty set.
Both notations may overlap with other uses of parentheses and brackets in mathematics. For instance, the notation $\left(a,b\right)$ is often used to denote an ordered pair in set theory, the coordinates of a point or vector in analytic geometry and linear algebra, or (sometimes) a complex number in algebra. The notation $\left[a,b\right]$ too is occasionally used for ordered pairs, especially in computer science.
Some authors use $\right]a,b\left[$ to denote the complement of the interval Template:Open-open; namely, the set of all real numbers that are either less than or equal to Template:Mvar, or greater than or equal to Template:Mvar.
### Infinite endpoints
In both styles of notation, one may use an infinite endpoint to indicate that there is no bound in that direction. Specifically, one may use $a=-\infty$ or $b=+\infty$ (or both). For example, Template:Open-open is the set of all positive real numbers, and Template:Open-open is the set of real numbers.
The notations Template:Closed-closed , Template:Closed-open , Template:Closed-closed , and Template:Open-closed are ambiguous. For authors who define intervals as subsets of the real numbers, those notations are either meaningless, or equivalent to the open variants. In the latter case, the interval comprising all real numbers is both open and closed, Template:Open-open = Template:Closed-closed = Template:Closed-open = Template:Open-closed .
On the extended real number line the intervals are all different as this includes −∞ and +∞ elements. For example Template:Open-closed means the extended real numbers excluding only −∞.
### Integer intervals
The notation Template:Closed-closed when Template:Mvar and Template:Mvar are integers, or {a .. b} , or just a .. b is sometimes used to indicate the interval of all integers between Template:Mvar and Template:Mvar, including both. This notation is used in some programming languages; in Pascal, for example, it is used to define the set of valid indices of a vector.
An integer interval that has a finite lower or upper endpoint always includes that endpoint. Therefore, the exclusion of endpoints can be explicitly denoted by writing a .. b − 1 , a + 1 .. b , or a + 1 .. b − 1. Alternate-bracket notations like Template:Closed-open or [a .. b[ are rarely used for integer intervals.
## Terminology
An open interval does not include its endpoints, and is indicated with parentheses. For example Template:Open-open means greater than 0 and less than 1. A closed interval includes its endpoints, and is denoted with square brackets. For example Template:Closed-closed means greater than or equal to 0 and less than or equal to 1.
A degenerate interval is any set consisting of a single real number. Some authors include the empty set in this definition. A real interval that is neither empty nor degenerate is said to be proper, and has infinitely many elements.
An interval is said to be left-bounded or right-bounded if there is some real number that is, respectively, smaller than or larger than all its elements. An interval is said to be bounded if it is both left- and right-bounded; and is said to be unbounded otherwise. Intervals that are bounded at only one end are said to be half-bounded. The empty set is bounded, and the set of all reals is the only interval that is unbounded at both ends. Bounded intervals are also commonly known as finite intervals.
Bounded intervals are bounded sets, in the sense that their diameter (which is equal to the absolute difference between the endpoints) is finite. The diameter may be called the length, width, measure, or size of the interval. The size of unbounded intervals is usually defined as +∞, and the size of the empty interval may be defined as 0 or left undefined.
The centre (midpoint) of bounded interval with endpoints Template:Mvar and Template:Mvar is (a + b)/2, and its radius is the half-length Template:Mabs/2. These concepts are undefined for empty or unbounded intervals.
An interval is said to be left-open if and only if it has no minimum (an element that is smaller than all other elements); right-open if it has no maximum; and open if it has both properties. The interval Template:Closed-open = Template:Mset, for example, is left-closed and right-open. The empty set and the set of all reals are open intervals, while the set of non-negative reals, for example, is a right-open but not left-open interval. The open intervals coincide with the open sets of the real line in its standard topology.
An interval is said to be left-closed if it has a minimum element, right-closed if it has a maximum, and simply closed if it has both. These definitions are usually extended to include the empty set and to the (left- or right-) unbounded intervals, so that the closed intervals coincide with closed sets in that topology.
The interior of an interval Template:Mvar is the largest open interval that is contained in Template:Mvar; it is also the set of points in Template:Mvar which are not endpoints of Template:Mvar. The closure of Template:Mvar is the smallest closed interval that contains Template:Mvar; which is also the set Template:Mvar augmented with its finite endpoints.
For any set Template:Mvar of real numbers, the interval enclosure or interval span of Template:Mvar is the unique interval that contains Template:Mvar and does not properly contain any other interval that also contains Template:Mvar.
## Classification of intervals
The intervals of real numbers can be classified into eleven different types, listed below; where Template:Mvar and Template:Mvar are real numbers, with $a < b$:
empty: $\left[b,a\right] = \left(a,a\right) = \left[a,a\right) = \left(a,a\right] = \\left\{ \\right\} = \emptyset$
degenerate: $\left[a,a\right] = \\left\{a\\right\}$
proper and bounded:
open:
closed: $\left[a,b\right]=\\left\{x\,|\,a\leq x\leq b\\right\}$
left-closed, right-open:
left-open, right-closed:
left-bounded and right-unbounded:
left-open: $\left(a,\infty\right)=\\left\{x\,|\,x>a\\right\}$
left-closed: $\left[a,\infty\right)=\\left\{x\,|\,x\geq a\\right\}$
left-unbounded and right-bounded:
right-open:
right-closed: $\left(-\infty,b\right]=\\left\{x\,|\,x\leq b\\right\}$
unbounded at both ends: $\left(-\infty,+\infty\right)=\R$
### Intervals of the extended real line
In some contexts, an interval may be defined as a subset of the extended real numbers, the set of all real numbers augmented with −∞ and +∞.
In this interpretation, the notations Template:Closed-closed , Template:Closed-open , Template:Closed-closed , and Template:Open-closed are all meaningful and distinct. In particular, Template:Open-open denotes the set of all ordinary real numbers, while Template:Closed-closed denotes the extended reals.
This choice affects some of the above definitions and terminology. For instance, the interval Template:Open-open = $\R$ is closed in the realm of ordinary reals, but not in the realm of the extended reals.
## Properties of intervals
The intervals are precisely the connected subsets of $\R$. It follows that the image of an interval by any continuous function is also an interval. This is one formulation of the intermediate value theorem.
The intervals are also the convex subsets of $\R$. The interval enclosure of a subset $X\subseteq \R$ is also the convex hull of $X$.
The intersection of any collection of intervals is always an interval. The union of two intervals is an interval if and only if they have a non-empty intersection or an open end-point of one interval is a closed end-point of the other (e.g., $\left(a,b\right) \cup \left[b,c\right] = \left(a,c\right]$).
If $\R$ is viewed as a metric space, its open balls are the open bounded sets Template:Open-open, and its closed balls are the closed bounded sets Template:Closed-closed.
Any element Template:Mvar of an interval Template:Mvar defines a partition of Template:Mvar into three disjoint intervals Template:Mvar1, Template:Mvar2, Template:Mvar3: respectively, the elements of Template:Mvar that are less than Template:Mvar, the singleton $\left[x,x\right] = \\left\{x\\right\}$, and the elements that are greater than Template:Mvar. The parts Template:Mvar1 and Template:Mvar3 are both non-empty (and have non-empty interiors) if and only if Template:Mvar is in the interior of Template:Mvar. This is an interval version of the trichotomy principle.
A dyadic interval is a bounded real interval whose endpoints are $\frac\left\{j\right\}\left\{2^n\right\}$ and $\frac\left\{j+1\right\}\left\{2^n\right\}$, where $j$ and $n$ are integers. Depending on the context, either endpoint may or may not be included in the interval.
Dyadic intervals have some nice properties, such as the following:
• The length of a dyadic interval is always an integer power of two.
• Every dyadic interval is contained in exactly one "parent" dyadic interval of twice the length.
• Every dyadic interval is spanned by two "child" dyadic intervals of half the length.
• If two open dyadic intervals overlap, then one of them must be a subset of the other.
The dyadic intervals thus have a structure very similar to an infinite binary tree.
Dyadic intervals are relevant to several areas of numerical analysis, including adaptive mesh refinement, multigrid methods and wavelet analysis. Another way to represent such a structure is p-adic analysis (for Template:Mvar=2).[1]
## Generalizations
### Multi-dimensional intervals
In many contexts, an $n$-dimensional interval is defined as a subset of $\R^n$ that is the Cartesian product of $n$ intervals, $I = I_1\times I_2 \times \cdots \times I_n$, one on each coordinate axis.
For $n=2$, this generally defines a rectangle whose sides are parallel to the coordinate axes; for $n=3$, it defines an axis-aligned rectangular box.
A facet of such an interval $I$ is the result of replacing any non-degenerate interval factor $I_k$ by a degenerate interval consisting of a finite endpoint of $I_k$. The faces of $I$ comprise $I$ itself and all faces of its facets. The corners of $I$ are the faces that consist of a single point of $\R^n$.
### Complex intervals
Intervals of complex numbers can be defined as regions of the complex plane, either rectangular or circular.[2]
## Topological algebra
Intervals can be associated with points of the plane and hence regions of intervals can be associated with regions of the plane. Generally, an interval in mathematics corresponds to an ordered pair (x,y) taken from the direct product R × R of real numbers with itself. Often it is assumed that y > x. For purposes of mathematical structure, this restriction is discarded,[3] and "reversed intervals" where yx < 0 are allowed. Then the collection of all intervals [x,y] can be identified with the topological ring formed by the direct sum of R with itself where addition and multiplication are defined component-wise.
The direct sum algebra $\left( R \oplus R, +, \times\right)$ has two ideals, { [x,0] : x ∈ R } and { [0,y] : y ∈ R }. The identity element of this algebra is the condensed interval [1,1]. If interval [x,y] is not in one of the ideals, then it has multiplicative inverse [1/x, 1/y]. Endowed with the usual topology, the algebra of intervals forms a topological ring. The group of units of this ring consists of four quadrants determined by the axes, or ideals in this case. The identity component of this group is quadrant I.
Every interval can be considered a symmetric interval around its midpoint. In a reconfiguration published in 1956 by M Warmus, the axis of "balanced intervals" [x, −x] is used along with the axis of intervals [x,x] that reduce to a point. Instead of the direct sum $R \oplus R$, the ring of intervals has been identified[4] with the split-complex number plane by M. Warmus and D. H. Lehmer through the identification
z = (x + y)/2 + j (xy)/2.
This linear mapping of the plane, which amounts of a ring isomorphism, provides the plane with a multiplicative structure having some analogies to ordinary complex arithmetic, such as polar decomposition. | 3,399 | 14,567 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 42, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2020-24 | latest | en | 0.906115 |
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In The Name of God, The All Gracious, The All Merciful
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Introduction to the Prime-Index Mathematics Analysis Generating the Quran
The Mathematics of Prime-Index, Generator of the Quran, (please click on the “Prime-Index Mathematic...” button) provides a mathematical basis by which the Quran is generated. Ultimate Mathematics is a 19-based blind analysis approach and it generates the “Hafs” (please click on the “Quran verse structure” button) version of the Quran in circulation, however, it only generates 127 verses for chapter 9. The analysis methodology, based on primes, twin primes, twin prime companions, composites, semi-primes, tri-primes, quad-primes, quint-primes … and their respective positional ranks or indices as well as inter-indices relations is beyond human knowledge of number theory and therefore it is a mathematical problem with no known solution.
Quran 42:13 - He decreed for you the same religion that He decreed for Noah, and what We inspired to you is the same as what We decreed to Abraham, Moses, and Jesus; you shall uphold this religion, and do not be divided therein. Your call is heavy to bear for those who set up partners for god. God chooses for Himself whomever He wills and guides to Himself those who repent.
A few salient features of the Prime-Index Mathematics
Index Prime Composite
19 67 30
• Chapter 67 has 30 verses.
• 19 x 67 = 1273 which is the frequency of the word God from 1:1 to 9:127.
• 19 x 30 = 570 and 570th Gaussian Prime is 9127
• Chapter 9 has 127 verses.
• 570 is the frequency of the initials in Chapter 32.
• The exponent of the 19th Mersenne Prime is 4253
• Chapter 42 has 53 verses.
• The exponent of the 19th Gaussian Mersenne Prime is 353.
• 353 is the frequency of HaMeem in Chapter 42. Permutation Group Properties of 4253
• The 19th group element is 5234.
• The index of composite 5234 is 4537.
• Chapter 45 has 37 verses.
• The 22nd group element is 5342.
• The index of 5342 is 4635.
• Chapter 46 has 35 verses.
• The 9th group element is 3425.
• 3425 is the index of composite 3975.
• Chapter 39 has 75 verses.
• The 10th group element is 3452.
• The index of composite 3452 is 2969.
• Chapter 29 has 69 verses.
• The 19th semi-prime is 55.
• 5519 is the index of composite 6346.
• 6346 is the total number of verses in the Quran.
• 1955 is the 1657th composite.
• 1657 in base 19 is 9127 in base 10.
Leviticus 24 : 17 “‘Anyone who takes the life of a human being is to be put to death. 18 Anyone who takes the life of someone’s animal must make restitution—life for life. 19 Anyone who injures their neighbor is to be injured in the same manner: 20 fracture for fracture, eye for eye, tooth for tooth. The one who has inflicted the injury must suffer the same injury. 21 Whoever kills an animal must make restitution, but whoever kills a human being is to be put to death. 22 You are to have the same law for the foreigner and the native-born. I am the Lord your God.’” | 829 | 3,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-38 | latest | en | 0.851185 |
http://www.nasdaq.com/article/basics-of-etf-risk-part-ii-beta-and-alpha-cm101638 | 1,435,711,458,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375094629.80/warc/CC-MAIN-20150627031814-00297-ip-10-179-60-89.ec2.internal.warc.gz | 547,845,539 | 39,091 | # Basics Of ETF Risk, Part II: Beta And Alpha
Â
While ETF performance descriptions might bring to mind Mark Twainâs phrase âlies, damn lies, and statistics,â risk metrics are indispensible when evaluating a fund.
When analyzing ETFs, we often evaluate pairs of data. For example, we compare a fundâs market price against its net asset value ( NAV ). Or we might look at a fundâs NAV versus the index it tracks. I described these fundamental relationships in a previous piece looking at what I consider to be crucial terminology.
Basic performance statistics that compare data sets arenât complicated, but the terms themselves often carry baggage that obscures their meaning.
For example, alpha is often associated with risk takers, and beta with the follow-the-herd crowd. I donât buy these characterizations. Moreover, I think the mystique around these terms just gets in the way.
Beta and alpha come from regressions. Hereâs the basic idea:
Take two sets of numbers, such as daily returns. Plot all the returns on a simple grid, with one set on the horizontal axis and the other on the vertical axis. The regression is the best estimate of a straight line that comes closest to fitting these points. Beta is simply the slope of this line and alpha is the intercept.
Beta
Beta is typically used to compare a fund to a broad index. Letâs say youâre looking at an equal-weight fund like the Rydex S'P Equal Weight ETF (NYSEArca:RSP). You want to know how the fund stacks up against a comparable cap-weighted fund like the SPDR S'P 500 ETF (NYSEArca:SPY).
Running the regression on 60 months of daily NAV data, we get a beta of 1.10.
Hereâs why it matters. Think of beta as a performance multiple. The regression estimates that when SPY is up 1 percent, RSP is up 1.10 percent. When SPY is down 1 percent, the fund is down 1.10 percent. RSPâs 1.1 beta tells us that itâs a bit riskier than SPY, so you should expect more return in compensation.
Bottom line:Beta provides a measure of comparative risk. Beta is not confined to measuring market risk, though thatâs often the case. You can use it to compare any two sets of returns. The key is to understand whatâs being compared.
Â
Â
Goodness Of Fit
The beta estimate comes from a regression. But how do we know whether to trust the regression itself? R 2 , pronounced âR-squared,â describes the overall âgoodness of fitâ of the regression to actual data. R 2 ranges from 0 to 1, with 1 as a perfect fit.
Correlation and R 2 are kissing cousins. In fact, R 2 is simply the square of correlation in simple regressions. In the graph above, the data points cling tightly to the regression line. Thatâs high R 2 (0.976 in this case). A low R 2 regression would have data points farther away from the regression line.
Bottom line:You can only trust the beta number when R 2 is high. When R 2 is low, beta doesnât tell us much.
Alpha
Alpha is a measure of outperformance. From a visual standpoint, alpha shows where the regression line crosses the vertical axis, or the Y-intercept. In the example above, alpha is simply the estimate of RSPâs return on days when SPYâs return is zero.
Alpha has its own measure of accuracy. I wonât bore you with too many details, but in short, alpha typically lacks statistical significance. Thatâs the case for the regression above. Oftentimes, marketing materials claiming a particular investment generates alpha wonât refer to statistical significance at all, in which case youâre probably better off ignoring it.
Bottom line:All regressions spit out an alpha number, but most of the time itâs meaningless. Real alpha is rare. And itâs quite possibly negative to boot. Be skeptical.
Itâs All Relative
Beta, R 2 and alpha come in handy when comparing funds side by side. Looking at fund Aâs beta to the S'P 500 compared with fund Bâs provides insight into the relative market risk between the two funds.
Like all performance measures though, these metrics come with major limitations. Theyâre backward looking, they change over time and theyâre sensitive to the time period used.
Hereâs another takeaway:Beta and alpha are relative to whatever baseline is used.
Consider a growth ETF that uses a fancy stock-selection method. Youâll be interested in its beta to a vanilla growth ETF and maybe its beta to the broad market as well.
Some folks have a hard time thinking about the presence of alpha in the context of a passive vehicle like an ETF. But statistically, significant alpha can occasionally be found even comparing a fundâs NAV to its index.
In short, a clear picture of what beta and alpha mean can help you cut through the veiling mists that often cloud the ETF landscape.
Â
The views and opinions expressed herein are the views and opinions of the author and do not necessarily reflect those of The NASDAQ OMX Group, Inc.
Referenced Stocks: NAV , RSP ,
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# Are 14% of M&M Candies Yellow? Mars, Inc. claims that 14%
ISBN: 9780321836960 18
## Solution for problem 12BSC Chapter 5.4
Elementary Statistics | 12th Edition
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Problem 12BSC
Are 14% of M&M Candies Yellow? Mars, Inc. claims that 14% of its M&M plain candies are yellow, and a sample of 100 such candies is randomly selected.
a. Find the mean and standard deviation for the number of yellow candies in such groups of 100.
b. Data Set 20 in Appendix B consists of a random sample of 100 M&Ms, including 8 that are yellow. Is this result unusually low? Does it seem that the claimed rate of 14% is wrong?
Step-by-Step Solution:
Step 1:
Given that, Mars, Inc. claims that 14% of its M&M plain candies are yellow, and a sample of 100 such candies is randomly selected.
Here, n = 100,p = 14% = 0.14, q = 1- p = 1- 0.14 = 0.86.
a). The mean and standard deviation for the number of yellow candies in such groups of 100.
For binomial distribution the mean and variance is given by the following formula.
Here ‘x’ follows binomial distribution with sample size “n” and proportion “p”.
i.e, X
Probability mass function of binomial distribution is given by
P(x) = , x = 0,1,2,...,n.
Mean μ = np = 1000.14 = 14.
Variance = npq = 100(0.14)(0.86) = 12.04.
And standard deviation = = =
Step 2 of 2
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# Spectrum Analysis using Java, Frequency Resolution versus Data Length
August 10, 2004
Java Programming, Notes # 1483
## Preface
The how and the why of spectral analysis
A previous lesson entitled Fun with Java, How and Why Spectral Analysis Works explained some of the fundamentals regarding spectral analysis. An understanding of that lesson is a prerequisite to an understanding of this lesson.
Another previous lesson entitled Spectrum Analysis using Java, Sampling Frequency, Folding Frequency, and the FFT Algorithm presented and explained several Java programs for doing spectral analysis. In that lesson, I used a DFT program to illustrate several aspects of spectral analysis that center around the sampling frequency and the Nyquist folding frequency.
I also used and briefly explained two different plotting programs that were originally explained in the earlier lesson entitled Plotting Engineering and Scientific Data using Java.
An understanding of the lesson entitled Spectrum Analysis using Java, Sampling Frequency, Folding Frequency, and the FFT Algorithm is also a prerequisite to an understanding of this lesson.
Frequency resolution versus data length
In this lesson I will use similar programs to explain and illustrate the manner in which spectral frequency resolution behaves with respect to data length.
A hypothetical situation
Consider a hypothetical situation in which you are performing spectral analysis on underwater acoustic signals in an attempt to identify enemy submarines.
You are aware that the enemy submarine contains a device that operates occasionally in short bursts. You are also aware that this device contains two rotating machines that rotate at almost but not quite the same speed.
During an operating burst of the device, each of the two machines contained in the device will emit acoustic energy that may appear as a peak in your spectral analysis output. (Note that I said, "may appear" and did not say, "will appear.") If you can identify the two peaks, you can conclusively identify the acoustic source as an enemy submarine.
The big question
How long must the operating bursts of this device be in order for you to resolve the peaks and identify the enemy submarine under ideal conditions? That is the question that I will attempt to answer in this lesson by teaching you about the relationship between frequency resolution and data length.
Viewing tip
You may find it useful to open another copy of this lesson in a separate browser window. That will make it easier for you to scroll back and forth among the different figures and listings while you are reading about them.
Supplementary material
I recommend that you also study the other lessons in my extensive collection of online Java tutorials. You will find those lessons published at Gamelan.com. However, as of the date of this writing, Gamelan doesn't maintain a consolidated index of my Java tutorial lessons, and sometimes they are difficult to locate there. You will find a consolidated index at www.DickBaldwin.com.
## Preview
Before I get into the technical details, here is a preview of the programs and their purposes that I will present and explain in this lesson:
• Dsp031 - Illustrates frequency resolution versus pulse length for pulses consisting of a truncated single sinusoid.
• Dsp031a - Displays the pulses analyzed by Dsp031.
• Dsp032 - Illustrates frequency resolution versus pulse length for pulses consisting of the sum of two truncated sinusoids with closely spaced frequencies.
• Dsp032a - Displays the pulses analyzed by Dsp032.
• Dsp033 - Illustrates frequency resolution versus pulse length for pulses consisting of the sum of two truncated sinusoids whose frequencies are barely resolvable.
• Dsp033a - Displays the pulses analyzed by Dsp033.
In addition, I will use the following programs that I explained in the lesson entitled Spectrum Analysis using Java, Sampling Frequency, Folding Frequency, and the FFT Algorithm.
• ForwardRealToComplex01 - Class that implements the DFT algorithm for spectral analysis.
• Graph03 - Used to display various types of data. (The concepts were explained in an earlier lesson.)
• Graph06 - Also used to display various types of data in a somewhat different format. (The concepts were also explained in an earlier lesson.)
## Discussion and Sample Code
Let's begin by looking at the time series data that will be used as input to the first spectral analysis experiment. Figure 1 shows five pulses in the time domain. Figure 2 and Figure 3 show the result of performing a spectral analysis on each of these pulses.
(The display in Figure 1 was produced by the program named Dsp031a, which I will explain later.)
``` Figure 1 Five pulses in the time domain. ```
The length of the pulses
If you examine Figure 1 carefully, you will see that each pulse is twice as long as the pulse above it. (There is a tick mark on the horizontal axes every twenty-five samples.) The bottom pulse is 400 samples long while the top pulse is 25 samples long.
Truncated sinusoids
Each pulse consists of a cosine wave that has been truncated at a different length. The frequency of the cosine wave is the same for every pulse. As you will see when we examine the code, the frequency of the cosine wave is 0.0625 times the sampling frequency. If you do the arithmetic, you will conclude that this results in 16 samples per cycle of the cosine wave.
In all five cases, the length of the time series upon which spectral analysis will be performed is 400 samples. For those four cases where the length of the pulse is less than 400 samples, the remaining samples in the time series have a value of zero.
Will compute at 400 frequencies
When the spectral analysis is performed later, the number of individual frequencies at which the amplitude of the spectral energy will be computed will be equal to the total data length. Therefore, the amplitude of the spectral energy will be computed at the same 400 frequencies for each of the five time series. That makes it convenient for us to stack the spectral plots up vertically and compare them (as in Figure 2). This makes it easy for us to compare the distribution of energy across the frequency spectrum for pulses of different lengths.
Graph03 and Graph06
The plots in Figure 1 were produced using the program named Graph03. Other plots in this lesson will be produced using the program named Graph06. I explained those programs in earlier lessons, and I provided the source code for both programs in the previous lesson entitled Spectrum Analysis using Java, Sampling Frequency, Folding Frequency, and the FFT Algorithm. Therefore, I won't repeat those explanations or provide the source code for those programs in this lesson.
The program named Dsp031a
A complete listing of the program named Dsp031a is provided in Listing 9 near the end of the lesson.
This program displays sinusoidal pulses identical to those processed by the program named Dsp031, which will be discussed later.
Time series containing sinusoidal pulses
The program named Dsp031a creates and displays five separate time series, each 400 samples in length. Each time series contains a pulse and the pulses are different lengths.
Each pulse consists of a truncated sinusoid. The frequency of the sinusoid for each of the pulses is the same.
Frequency values are specified as type double as a fraction of the sampling frequency. The frequency of each sinusoid is 0.0625 times the sampling frequency.
The pulse lengths
The lengths of the five pulses are:
• 25 samples
• 50 samples
• 100 samples
• 200 samples
• 400 samples
Will discuss in fragments
This program is very similar to programs that I explained in previous lessons in this series, so my explanation will be very brief. As usual, I will discuss the program in fragments.
The beginning of the class, along with the declaration and initialization of several variables is shown in Listing 1. The names of the variables along with the embedded comments should make the code self explanatory.
```class Dsp031a implements GraphIntfc01{ final double pi = Math.PI; int len = 400;//data length int numberPulses = 5; //Frequency of the sinusoids double freq = 0.0625; //Amplitude of the sinusoids double amp = 160; //Following arrays will contain sinusoidal data double[] data1 = new double[len]; double[] data2 = new double[len]; double[] data3 = new double[len]; double[] data4 = new double[len]; double[] data5 = new double[len]; Listing 1 ```
The constructor
Listing 2 shows the constructor, which creates the raw sinusoidal data and stores that data in the array objects created in Listing 1.
(Recall that all element values in the array objects are initialized with a value of zero. Therefore, the code in Listing 2 only needs to store the non-zero values in the array objects.)
``` public Dsp031a(){//constructor //Create the raw data for(int x = 0;x < len/16;x++){ data1[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len/8;x++){ data2[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len/4;x++){ data3[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len/2;x++){ data4[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len;x++){ data5[x] = amp*Math.cos(2*pi*x*freq); }//end for loop }//end constructor Listing 2 ```
The code in the conditional clause of each of the for loops in Listing 2 controls the length of each of the sinusoidal pulses.
The interface methods
As you can see in Listing 1, the class implements the interface named GraphIntfc01. I introduced this interface in the earlier lesson entitled Plotting Engineering and Scientific Data using Java and also discussed it in the previous lesson entitled Spectrum Analysis using Java, Sampling Frequency, Folding Frequency, and the FFT Algorithm.
The remaining code for the class named Dsp031a consists of the methods necessary to satisfy the interface. These methods are invoked by the plotting programs named Graph03 and Graph06 to obtain and plot the data returned by the methods. As implemented in Dsp031a, these interface methods return the values stored in the array objects referred to by data1 through data5. Thus, the values stored in those array objects are plotted in Figure 1.
Spectral analysis results
Figure 2 shows the result of using the program named Dsp031 to perform a spectral analysis on each of the five pulses shown in Figure 1. These results were plotted using the program named Graph06. With this plotting program, each data value is plotted as a vertical bar. However, in this case, the sides of each of the bars are so close together that the area under the spectral curve appears to be solid black.
(When you run this program, you can expand the display to full screen and see the individual vertical bars. However, I can't to that and maintain the narrow publication format required for this lesson.)
``` Figure 2 Spectral analyses of five pulses. ```
Interpretation of the results
Before I get into the interpretation, I need to point out that I normalized the data plotted in Figure 2 to cause each spectral peak to have approximately the same value. Otherwise, the spectral analysis result values for the short pulses would have been too small to be visible in this plotting format.
Therefore, the fact that the area under the curve in the top plot is greater than the area under the curve in the bottom plot doesn't indicate that the first pulse contains more energy than the last pulse. It simply means that I normalized the data for best results in plotting.
Spectrum of an ideal sinusoid
That having been said, different people will probably interpret these results in different ways. Let's begin by stating that the theoretical spectrum for a sinusoid of infinite length in the absence of noise is a single vertical line having zero width and infinite height.
In the real world of measurements, however, there is no such thing as a sinusoid of infinite length. Rather, every measurement that we make must truncate the sinusoid at some point in time. For a theoretical signal of infinite length, every spectral analysis that we can perform is an imperfect estimate of the spectrum.
Two viewpoints
There are at least two ways to think of the pulses shown in Figure 1.
1. Each pulse is a truncated section of an ideal sinusoid of infinite length.
2. Each pulse is a signal having a definite planned start and stop time.
The way that you interpret the results shown in Figure 2 depends on your viewpoint regarding the pulses.
The first viewpoint
If your viewpoint is that each pulse is a truncated section of an ideal sinusoid of infinite length, then the width of each of the peaks (beyond zero width) is the result of measurement error introduced by the truncation process.
The second viewpoint
If your viewpoint is that each pulse is a signal having a definite planned start and stop time, then the widths and the shape of each of the peaks describes the full range of frequency components required to physically generate such a pulse. This is the viewpoint that is consistent with the hypothetical situation involving a device on a submarine that I described earlier in this lesson.
A simplified hypothetical situation
Assume for the moment that the hypothetical device on the submarine contains only one rotating machine and that this device is turned on and off occasionally in short bursts. Because of the rotating machine, when the device is turned on, it will emit acoustic energy whose frequency matches the rotating speed of the machine.
(In reality, it will probably also emit acoustic energy at other frequencies as well, but we will consider it to be a very ideal machine. We will also assume the complete absence of any other acoustic noise in the environment.)
Assume that you have a recording window of 400 samples, and that you are able to record five such bursts within each of five separate recording windows. Further assume that the lengths of the individual bursts match the time periods indicated by the pulses in Figure 1.
The spectrum of the bursts
If you perform spectral analysis on each of the five individual 400-sample windows containing the bursts, and if you normalize the peak values for plotting purposes, you should get results similar to those shown in Figure 2.
The spectral bandwidth of the signal
The frequency range over which energy is distributed is referred to as the bandwidth of the signal. As you can see in Figure 2, shorter pulses require wider bandwidth.
For example, considerably more bandwidth is required of a communication system that is required to reliably transmit a series of short truncated sinusoids than one that is only required to reliably transmit a continuous tone at a single frequency.
At the same time, it is very difficult to convey very much information with a signal consisting of a continuous tone at a single frequency (other than the fact that the tone exists). Communication systems designed to convey information usually encode that information by either turning the tone on and off or by causing it to shift among a set of previously defined frequencies. The tone is often referred to as the carrier and the encoding of the information is often referred to as modulating the carrier.
The relationship between pulse length and bandwidth
So far, we can draw one important conclusion from our experiment.
Shorter pulses require greater bandwidth.
This leads to an important question. What is the numerical relationship between pulse length and bandwidth? Although we can draw the above general conclusion from Figure 2, it is hard to draw any quantitative conclusions from Figure 2. That brings us to the expanded plot of the spectral data shown in Figure 3.
An expanded plot of the spectral results
Figure 3 shown the left one-fourth of the spectral results from Figure 2 plotted in the same horizontal space. In other words, Figure 3 discards the upper three-fourths of the spectral results from Figure 2 and shows only the lower one-fourth of the spectral results on an expanded scale. Figure 3 also provides tick marks that make it convenient to perform measurements on the plots.
Also, whereas Figure 2 was plotted using the program named Graph06, Figure 3 was plotted using the program named Graph03. Thus, Figure 3 uses a different plotting format than Figure 2
``` Figure 3 Expanded spectral analyses of five pulses. ```
Picking numeric values
The curves in Figure 3 are spread out to the point that we can pick some approximate numeric values off the plot, and from this, we can draw a very significant conclusion.
For purposes of our approximation, consider the bandwidth to be the distance along the frequency axis between the points where the curves hit zero on either side of the peak. Using this approximation, the bandwidth indicated by the spectral analyses in Figure 3 shows the bandwidth of each spectrum to be twice as wide as the one below it.
Referring back to Figure 1, recall that the length of each pulse was half that of the one below it. The conclusion is:
The bandwidth of a truncated sinusoidal pulse is inversely proportional to the length of the pulse.
If you reduce the length of the pulse by a factor of two, you must double the bandwidth of a transmission system designed to reliably transmit a pulse of that length.
This will also be an important conclusion regarding our ability to separate and identify the two spectral peaks in the burst of acoustic energy described in our original hypothetical situation.
Let's see some code
The generation of the signals and the spectral analysis for the results presented in Figure 2 and Figure 3 were performed using the program named Dsp031. A complete listing of the program is shown in Listing 10 near the end of the lesson.
Description of the program named Dsp031
This program performs spectral analyses on five separate time series, each 400 samples in length.
Each time series contains a pulse and the pulses are different lengths. (The lengths of the individual pulses match that shown in Figure 1.) Each pulse consists of a truncated sinusoid. The frequency of the sinusoid for each pulse is the same.
All frequency values are specified as type double as a fraction of the sampling frequency. The frequency of all five sinusoids is 0.0625 times the sampling frequency.
The lengths of the pulses are:
• 25 samples
• 50 samples
• 100 samples
• 200 samples
• 400 samples
If this sounds familiar, it is because the pulses are identical to those displayed in Figure 1 and discussed under Dsp031a above.
Uses a DFT algorithm
The spectral analysis process uses a DFT algorithm and computes the amplitude of the spectral energy at 400 equally spaced frequ4encies between zero and the folding frequency.
(Recall from the previous lesson entitled Spectrum Analysis using Java, Sampling Frequency, Folding Frequency, and the FFT Algorithm that the folding frequency is one-half the sampling frequency.)
This program computes and displays the amplitude spectrum at frequency intervals that are one-half of the frequency intervals for a typical FFT algorithm.
Normalize the results
The results of the spectral analysis are multiplied by the reciprocal of the lengths of the individual pulses to normalize all five plots to the same peak value. Otherwise, the results for the short pulses would be too small to see on the plots.
Will discuss in fragments
Once again, this program is very similar to programs explained in the previous lesson entitled Spectrum Analysis using Java, Sampling Frequency, Folding Frequency, and the FFT Algorithm. Therefore, this discussion will be very brief.
The code in Listing 3 declares and initializes some variables and creates the array objects that will contain the sinusoidal pulses.
In addition, the code in Listing 3 declares reference variables that will be used to refer to array objects containing results of the spectral analysis process that are not used in this program.
Finally, Listing 3 declares reference variables that will be used to refer to array objects containing the results plotted in Figure 2 and Figure 3.
Given the names of the variables, the comments, and what you learned in the previous lesson, the code in Listing 3 should be self explanatory.
```class Dsp031 implements GraphIntfc01{ final double pi = Math.PI; int len = 400;//data length //Sample that represents zero time. int zeroTime = 0; //Low and high frequency limits for the // spectral analysis. double lowF = 0.0; double highF = 0.5; int numberSpectra = 5; //Frequency of the sinusoids double freq = 0.0625; //Amplitude of the sinusoids double amp = 160; //Following arrays will contain data that is // input to the spectral analysis process. double[] data1 = new double[len]; double[] data2 = new double[len]; double[] data3 = new double[len]; double[] data4 = new double[len]; double[] data5 = new double[len]; //Following arrays receive information back // from the spectral analysis that is not used // in this program. double[] real; double[] imag; double[] angle; //Following arrays receive the magnitude // spectral information back from the spectral // analysis process. double[] mag1; double[] mag2; double[] mag3; double[] mag4; double[] mag5; Listing 3 ```
The constructor
The constructor begins in Listing 4. The code in Listing 4 is identical to that shown earlier in Listing 2. This code generates the five sinusoidal pulses and stores the data representing those pulses in the arrays referred to by data1 through data5. So far, except for the declaration of some extra variables, this program isn't much different from the program named Dsp031a discussed earlier in this lesson.
``` public Dsp031(){//constructor //Create the raw data for(int x = 0;x < len/16;x++){ data1[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len/8;x++){ data2[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len/4;x++){ data3[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len/2;x++){ data4[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len;x++){ data5[x] = amp*Math.cos(2*pi*x*freq); }//end for loop Listing 4 ```
Perform the spectral analysis
The remainder of the constructor is shown in Listing 5. This code invokes the transform method of the ForwardRealToComplex01 class five times in succession to perform the spectral analysis on each of the five pulses shown in Figure 1.
(I explained the transform method in detail in the previous lesson entitled Spectrum Analysis using Java, Sampling Frequency, Folding Frequency, and the FFT Algorithm.)
``` mag1 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data1,real, imag,angle,mag1,zeroTime,lowF,highF); mag2 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data2,real, imag,angle,mag2,zeroTime,lowF,highF); mag3 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data3,real, imag,angle,mag3,zeroTime,lowF,highF); mag4 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data4,real, imag,angle,mag4,zeroTime,lowF,highF); mag5 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data5,real, imag,angle,mag5,zeroTime,lowF,highF); }//end constructor Listing 5 ```
Each time the transform method is invoked, it computes the magnitude spectra for the incoming data and saves it in the output array.
(Note that the real, imag, and angle arrays are not used later, so they are discarded each time a new spectral analysis is performed.)
The interface methods
The Dsp031 class also implements the interface named GraphIntfc01. The remaining code in the program consists of the methods required to satisfy that interface. Except for the identification of the arrays from which the methods extract data to be returned for plotting, these methods are identical to those defined in the earlier class named Dsp031a. Therefore, I won't discuss them further.
What we have learned so far
So far, the main things that we have learned is that shorter pulses require greater bandwidth, and the bandwidth required to faithfully represent a truncated sinusoidal pulse is the reciprocal of the length of the pulse.
Where do we go from here?
Now we will look at the issues involved in using spectral analysis to separate and identify the frequencies of two closely-spaced spectral peaks for a pulse composed of the sum of two sinusoids. Once again, we will begin by looking at some results and then discuss the code that produced those results.
The five pulses
The five pulses that we will be working with in this example are shown in Figure 4. As you can see, these pulses are a little more ugly than the pulses shown in Figure 1. As you can also see, as was the case in Figure 1, each pulse appears to be a shorter or longer version of the other pulses in terms of its waveform.
``` Figure 4 Five pulses with two sinusoids each. ```
Produced by Dsp032a
The plots in Figure 4 were produced by the program named Dsp032a, which I will briefly discuss later. (A complete listing of the program is shown in Listing 11 near the end of the lesson.) This program creates and displays pulses identical to those processed by the program named Dsp032, which I will also briefly discuss later. (A complete listing of the program named Dsp32 is presented in Listing 12.)
Five time series containing pulses
The program creates and displays five separate time series, each 400 samples in length. Each time series contains a pulse and the pulses are different lengths. As before, each of the pulses shown in Figure 4 is half the length below the pulse below it.
The sum of two sinusoids
Each pulse consists of the sum of two sinusoids at closely spaced frequencies. The frequencies of the two sinusoids for all pulses are the same.
All frequency values are specified as type double as a fraction of the sampling frequency. The frequencies of the two sinusoids are equidistant from a center frequency of 0.0625 times the sampling frequency.
(Recall that 0.0625 was the frequency of the only sinusoid contained in the pulses shown in Figure 1 and processed by the program named Dsp031.)
The frequencies and pulse lengths
The frequency of one sinusoid is (0.0625 - 2.0/len) times the sampling frequency, where len is the length of the time series containing the pulse. (The value for len is 400 samples in this program.) The frequency of the other sinusoid is (0.0625 + 2.0/len) times the sampling frequency.
The lengths of the five pulses are:
• 25 samples
• 50 samples
• 100 samples
• 200 samples
• 400 samples
(Note that Figure 4 has tick marks every 25 samples.)
The program named Dsp032a
The only new code in this program is the code in the constructor that creates the pulses and stores them in the data arrays. This code consists of five separate for loops, one for each pulse. The code for the first for loop, which is typical of the five, is shown in Listing 6.
``` public Dsp032a(){//constructor //Create the raw data for(int x = 0;x < len/16;x++){ data1[x] = amp*Math.cos(2*pi*x*freq1) + amp*Math.cos(2*pi*x*freq2); }//end for loop //... code removed for brevity }//end constructor Listing 6 ```
As you can see from Listing 6, the values that make up the pulse are produced by adding together the values of two different cosine functions having different frequencies. The values for freq1 and freq2 are as described above.
You can view the remainder of this program in Listing 11.
Spectral analysis output
The results of running the program named Dsp032 and displaying the results with the program named Graph03 are shown in Figure 5.
``` Figure 5 Spectral analyses of five pulses. ```
Each of the peaks in the third, fourth, and fifth plots in Figure 5 corresponds to the frequency of one of the two sinusoids that were added together to produce the pulses shown in Figure 4.
Can we answer the original question now?
The question posed in the original hypothetical situation was "how long must the operating bursts of this device be in order for you to resolve the peaks and identify the enemy submarine under ideal conditions?"
We are looking at very ideal conditions in Figure 4 and Figure 5. In particular, the pulses exist completely in the absence of noise.
(The existence of wide-band noise added to the pulses in Figure 4 would cause a change in the spectral results in Figure 5. That change might be described as having the appearance of grass and weeds growing on the baseline across the entire spectrum. The stronger the wide-band noise, the taller would be the weeds.)
Cannot resolve two peaks for first two pulses
Clearly for the ideal condition of recording the bursts in the total absence of noise, you cannot resolve the peaks from the top two plots in Figure 5. For those two pulses, the spectral peaks simply merge together to form a single broad peak. Therefore, for this amount of separation between the frequencies of the two sinusoids, the lengths of the first two pulses in Figure 4 are insufficient to allow for separation and identification of the separate peaks.
We must be in luck
We seem to have the problem bracketed. (Were we really lucky, or did I plan it this way?) Under the ideal conditions of this experiment, the peaks are separable in the middle plot of Figure 5. Thus, for the amount of separation between the frequencies of the two sinusoids, the length of the third pulse is Figure 4 is sufficient to allow for separation and identification of the separate peaks.
A qualified answer to the question
The peaks are even better separated in the bottom two plots in Figure 5. For the five pulses used in this experiment and the amount of separation between the frequencies of the two sinusoids, any pulse as long or longer than the length of the third pulse is Figure 4 is sufficient to allow for separation and identification of the separate peaks.
What about the effects of noise?
If you were to add a nominal amount of wide-band noise to the mix, it would become more difficult to resolve the peaks for the bottom three plots in Figure 5 because the peaks would be growing out of a bed of weeds.
(If you add enough wide-band noise, you couldn't resolve the peaks using any of the plots, because the peaks would be completely "lost in the noise.")
What can we learn from this?
Since we have concluded that the middle pulse in Figure 4 is sufficiently long to allow us to resolve the two peaks, let's see what we can learn from the parameters that describe that pulse.
Pulse length and frequency separation
To begin with, the length of the pulse is 100 samples.
What about the frequency separation of the two sinusoids? Recall that the frequency of one sinusoid is (0.0625 - 2.0/len) times the sampling frequency, where len is the length of the time series containing the pulse. The frequency of the other sinusoid is (0.0625 + 2.0/len) times the sampling frequency.
Thus, total separation between the two frequencies is 4/len, or 4/400. Dividing through by 4 we see that the separation between the two frequencies is 1/100.
Eureka, we have found it
For the third pulse, the frequency separation is the reciprocal of the length of the pulse. Also, the length of the third pulse is barely sufficient to allow for separation and identification of the two peaks in the spectrum. Thus, the two spectral peaks are separable in the absence of noise if the frequency separation is the reciprocal of the pulse length.
(That is too good to be a coincidence. I must have planned that way.)
Thus, we have reached another conclusion.
Under ideal conditions, the two peaks in the spectrum can be resolved when the separation between the frequencies of the two sinusoids is equal to the reciprocal of the pulse length.
There is no single answer to the question "how long must the operating bursts of this device be in order for you to resolve the peaks and identify the enemy submarine under ideal conditions?"
The answer depends on the frequency separation. The general answer is that the length of the bursts must be at least as long as the reciprocal of the frequency separation for the two sinusoids. If the separation is large, the pulse length may be short. If the separation is small, the pulse length must be long.
The program named Dsp032
As I indicated earlier, the plots shown in Figure 5 were the result of running the program named Dsp032 and displaying the data with the program named Graph03.
The only thing that is new in this program is the code that generates the five pulses and saves them in their respective data arrays. Even that code is not really new, because it is identical to the code shown in Listing 6. Therefore, I won't discuss this program further in this lesson.
One more experiment
As you can surmise from the conclusions reached above, in order to be able to resolve the two peaks in the spectrum, you can either keep the pulse length the same and increase the frequency separation, or you can keep the frequency separation the same and increase the pulse length.
Let's examine an example where we keep the pulse lengths the same as before and adjust the frequency separation between the two sinusoids to make it barely possible to resolve the peaks for each of the five pulses.
We will need to increase the frequency separation for the first two pulses, and we can decrease the frequency separation for the fourth and fifth pulses. We will leave the frequency separation the same as before for the third pulse since it already seems to have the optimum relationship between pulse length and frequency separation.
The five pulses
The five pulses used in this experiment are shown in Figure 6.
``` Figure 6 Five pulses with additive sinusoids. ```
Unlike in the previous two cases shown in Figure 1 and Figure 4, each of these pulses has a different shape from the others. In other words, in the previous two cases, each pulse simply looked like a longer or shorter version of the other pulses. That is not the case in this example.
(Note however that the third pulse in Figure 6 looks just like the third pulse in Figure 4. They were created using the same parameters. However, none of the other pulses in Figure 6 look like the corresponding pulses in Figure 4, and none of the pulses in Figure 6 look like the pulses in Figure 1.)
Spectral analysis results
Figure 7 shows the result of performing spectral analysis on the five time series containing the pulses shown in Figure 6.
``` Figure 7 Spectral analyses of five pulses. ```
Peaks for first two pulses are now resolvable
When we examine the code, you will see that the frequency separation for the first two pulses has been increased to the reciprocal of the pulse length in each case. This results in the two peaks in the spectrum for each of the first two pulses being resolvable in Figure 7.
Third pulse hasn't changed
The spectrum for the third pulse shown in Figure 7 is almost identical to the spectrum for the third pulse shown in Figure 5. The only difference is that I had to decrease the vertical scaling on all of the plots in Figure 5 to keep the peak in the top plot within the bounds of the plot.
Spectral peaks for last two pulses are closer
When we examine the code, you will also see that the frequency separation for the last two pulses has been decreased to the reciprocal of the pulse length in each case. This results in the two peaks in the spectrum for each of the last two pulses being closer than before in Figure 7.
The peaks in the bottom two plots in Figure 7 appear to be resolvable, but we can't be absolutely certain because they are so close together, particularly for the last plot.
(If you expand the Frame to full screen when you run this program, you will see that the two peaks are resolvable, but I can't do that and stay within this narrow publication format.)
Expand the horizontal plotting scale
Figure 8 adjusts the plotting parameters to cause the left-most one-fourth of the data in Figure 7 to be plotted in the full width of the Frame in Figure 8.
``` Figure 8 Expanded spectral analyses of five pulses. ```
The peaks are barely resolvable
Figure 8 shows that the two peaks are barely resolvable for all five of the pulses shown in Figure 6.
(There is no space between the peaks at the baseline in Figure 8, but the plots do go almost down to the baseline half way between the two peaks.)
The program named Dsp033
The plots in Figure 7 and Figure 8 were produced by running the program named Dsp033 and plotting the results with the program named Graph03.
A complete listing of the program named Dsp033 is shown in Listing 14 near the end of the lesson.
This program is the same as Dsp032 except that the separation between the frequencies of the two sinusoids is the reciprocal of the length of the pulse in each case.
The program performs spectral analysis on five separate time series, each 400 samples in length. Each time series contains a pulse and the pulses are different lengths.
Each pulse consists of the sum of two sinusoids at closely spaced frequencies. The frequencies of the two sinusoids are equidistant from a center frequency of 0.0625 times the sampling frequency. The total separation between the frequencies of the two sinusoids is the reciprocal of the length of the pulse.
All frequency values are specified as type double as a fraction of the sampling frequency.
The lengths of the pulses are:
• 25 samples
• 50 samples
• 100 samples
• 200 samples
• 400 samples
The spectral analysis
The spectral analysis computes the spectra at 400 equally spaced frequencies between zero and the folding frequency (one-half the sampling frequency).
The results of the spectral analysis are multiplied by the reciprocal of the lengths of the individual pulses to normalize the five plots. Otherwise, the results for the short pulses would be too small to see on the plots.
Because of the similarity of this program to the previous programs, my discussion of the code will be very brief.
Computation of the frequencies
The code in Listing 7 shows the computation of the frequencies of the sinusoids that will be added together to form each of the five pulses.
``` //Frequencies of the sinusoids double freq1a = 0.0625 - 8.0/len; double freq2a = 0.0625 + 8.0/len; double freq1b = 0.0625 - 4.0/len; double freq2b = 0.0625 + 4.0/len; double freq1c = 0.0625 - 2.0/len; double freq2c = 0.0625 + 2.0/len; double freq1d = 0.0625 - 1.0/len; double freq2d = 0.0625 + 1.0/len; double freq1e = 0.0625 - 0.5/len; double freq2e = 0.0625 + 0.5/len; Listing 7 ```
Create the pulses
The code in Listing 8 uses those frequency values to create the data for the pulses and to store that data in the arrays used to hold the pulses.
``` //Create the raw data for(int x = 0;x < len/16;x++){ data1[x] = amp*Math.cos(2*pi*x*freq1a) + amp*Math.cos(2*pi*x*freq2a); }//end for loop for(int x = 0;x < len/8;x++){ data2[x] = amp*Math.cos(2*pi*x*freq1b) + amp*Math.cos(2*pi*x*freq2b); }//end for loop for(int x = 0;x < len/4;x++){ data3[x] = amp*Math.cos(2*pi*x*freq1c) + amp*Math.cos(2*pi*x*freq2c); }//end for loop for(int x = 0;x < len/2;x++){ data4[x] = amp*Math.cos(2*pi*x*freq1d) + amp*Math.cos(2*pi*x*freq2d); }//end for loop for(int x = 0;x < len;x++){ data5[x] = amp*Math.cos(2*pi*x*freq1e) + amp*Math.cos(2*pi*x*freq2e); }//end for loop Listing 8 ```
Other than the code shown in Listing 7 and Listing 8, the program named Dsp033 is the same as the programs that were previously explained, and I won't discuss it further.
## Run the Programs
I encourage you to copy, compile, and run the programs provided in this lesson. Experiment with them, making changes and observing the results of your changes.
Create more complex experiments. For example, you could create pulses containing three or more sinusoids at closely spaced frequencies, and you could cause the amplitudes of the sinusoids to be different. See what it takes to cause the peaks in the spectra of those pulses to be separable and identifiable.
If you really want to get fancy, you could create a pulse consisting of a sinusoid whose frequency changes with time from the beginning to the end of the pulse. (A pulse of this type is often referred to as a frequency modulated sweep signal.) See what you can conclude from doing spectral analysis on a pulse of this type.
Try using the random number generator of the Math class to add some random noise to every value in the 400-sample time series. See what this does to your spectral analysis results.
Move the center frequency up and down the frequency axis. See if you can explain what happens as the center frequency approaches zero and as the center frequency approaches the folding frequency.
Most of all, enjoy yourself and learn something in the process.
## Summary
This program provides the code for three spectral analysis experiments of increasing complexity.
Bandwidth versus pulse length
The first experiment performs spectral analyses on five simple pulses consisting of truncated sinusoids. This experiment shows:
• Shorter pulses require greater bandwidth.
• The bandwidth of a truncated sinusoidal pulse is inversely proportional to the length of the pulse.
Peak resolution versus pulse length and frequency separation
The second experiment performs spectral analyses on five more complex pulses consisting of the sum of two truncated sinusoids having closely spaced frequencies. The purpose is to determine the required length of the pulse in order to use spectral analysis to resolve spectral peaks attributable to the two sinusoids. The experiment shows that the peaks are barely resolvable when the length of the pulse is the reciprocal of the frequency separation between the two sinusoids.
Five pulses with barely resolvable spectral peaks
The third experiment also performs spectral analyses on five pulses consisting of the sum of two truncated sinusoids having closely spaced frequencies. In this case, the frequency separation for each pulse is the reciprocal of the length of the pulse. The results of the spectral analysis reinforce the conclusions drawn in the second experiment.
## What's Next?
So far, the lessons in this series have ignored the complex nature of the results of spectral analysis. The complex results have been converted into real results by computing the square root of the sum of the squares of the real and imaginary parts.
The next lesson in the series will meet the issue of complex spectral results head on and will explain the concept of phase angle. In addition, the lesson will explain the behavior of the phase angle with respect to time shifts in the input time series.
## Complete Program Listings
Complete listings of all the programs discussed in this lesson are provided in this section.
```/* File Dsp031a.java Copyright 2004, R.G.Baldwin Revised 5/17/2004 Displays sinusoidal pulses identical to those processed by Dsp031. Creates and displays five separate time series, each 400 samples in length. Each time series contains a pulse and the pulses are different lengths. Each pulse consists of a truncated sinusoid. The frequency of the sinusoid for all pulses is the same. All frequency values are specified as type double as a fraction of the sampling frequency. The frequency of all sinusoids is 0.0625 times the sampling frequency. The lengths of the pulses are: 25 samples 50 samples 100 samples 200 samples 400 samples Tested using J2SEE 1.4.2 under WinXP. ************************************************/ import java.util.*; class Dsp031a implements GraphIntfc01{ final double pi = Math.PI; int len = 400;//data length int numberPulses = 5; //Frequency of the sinusoids double freq = 0.0625; //Amplitude of the sinusoids double amp = 160; //Following arrays will contain sinusoidal data double[] data1 = new double[len]; double[] data2 = new double[len]; double[] data3 = new double[len]; double[] data4 = new double[len]; double[] data5 = new double[len]; public Dsp031a(){//constructor //Create the raw data for(int x = 0;x < len/16;x++){ data1[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len/8;x++){ data2[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len/4;x++){ data3[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len/2;x++){ data4[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len;x++){ data5[x] = amp*Math.cos(2*pi*x*freq); }//end for loop }//end constructor //-------------------------------------------// //The following six methods are required by the // interface named GraphIntfc01. public int getNmbr(){ //Return number of functions to process. // Must not exceed 5. return 5; }//end getNmbr //-------------------------------------------// public double f1(double x){ int index = (int)Math.round(x); if(index < 0 || index > data1.length-1){ return 0; }else{ //Scale the amplitude of the pulses to make // them compatible with the default // plotting amplitude of 100.0. return data1[index]*90.0/amp; }//end else }//end function //-------------------------------------------// public double f2(double x){ int index = (int)Math.round(x); if(index < 0 || index > data2.length-1){ return 0; }else{ return data2[index]*90.0/amp; }//end else }//end function //-------------------------------------------// public double f3(double x){ int index = (int)Math.round(x); if(index < 0 || index > data3.length-1){ return 0; }else{ return data3[index]*90.0/amp; }//end else }//end function //-------------------------------------------// public double f4(double x){ int index = (int)Math.round(x); if(index < 0 || index > data4.length-1){ return 0; }else{ return data4[index]*90.0/amp; }//end else }//end function //-------------------------------------------// public double f5(double x){ int index = (int)Math.round(x); if(index < 0 || index > data5.length-1){ return 0; }else{ return data5[index]*90.0/amp; }//end else }//end function }//end sample class Dsp031a Listing 9 ```
```/* File Dsp031.java Copyright 2004, R.G.Baldwin Revised 5/17/2004 Performs spectral analysis on five separate time series, each 400 samples in length. Each time series contains a pulse and the pulses are different lengths. Each pulse consists of a truncated sinusoid. The frequency of the sinusoid for all pulses is the same. All frequency values are specified as type double as a fraction of the sampling frequency. The frequency of all sinusoids is 0.0625 times the sampling frequency. The lengths of the pulses are: 25 samples 50 samples 100 samples 200 samples 400 samples The spectral analyis computes the spectra at 400 equally spaced points between zero and the folding frequency (one-half the sampling frequency). The results of the spectral analysis are multiplied by the reciprocal of the lengths of the individual pulses to normalize all five plots to the same peak value. Otherwise, the results for the short pulses would be too small to see on the plots. Tested using J2SEE 1.4.2 under WinXP. ************************************************/ import java.util.*; class Dsp031 implements GraphIntfc01{ final double pi = Math.PI; int len = 400;//data length //Sample that represents zero time. int zeroTime = 0; //Low and high frequency limits for the // spectral analysis. double lowF = 0.0; double highF = 0.5; int numberSpectra = 5; //Frequency of the sinusoids double freq = 0.0625; //Amplitude of the sinusoids double amp = 160; //Following arrays will contain data that is // input to the spectral analysis process. double[] data1 = new double[len]; double[] data2 = new double[len]; double[] data3 = new double[len]; double[] data4 = new double[len]; double[] data5 = new double[len]; //Following arrays receive information back // from the spectral analysis that is not used // in this program. double[] real; double[] imag; double[] angle; //Following arrays receive the magnitude // spectral information back from the spectral // analysis process. double[] mag1; double[] mag2; double[] mag3; double[] mag4; double[] mag5; public Dsp031(){//constructor //Create the raw data for(int x = 0;x < len/16;x++){ data1[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len/8;x++){ data2[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len/4;x++){ data3[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len/2;x++){ data4[x] = amp*Math.cos(2*pi*x*freq); }//end for loop for(int x = 0;x < len;x++){ data5[x] = amp*Math.cos(2*pi*x*freq); }//end for loop //Compute magnitude spectra of the raw data // and save it in output arrays. Note that // the real, imag, and angle arrays are not // used later, so they are discarded each // time a new spectral analysis is performed. mag1 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data1,real, imag,angle,mag1,zeroTime,lowF,highF); mag2 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data2,real, imag,angle,mag2,zeroTime,lowF,highF); mag3 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data3,real, imag,angle,mag3,zeroTime,lowF,highF); mag4 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data4,real, imag,angle,mag4,zeroTime,lowF,highF); mag5 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data5,real, imag,angle,mag5,zeroTime,lowF,highF); }//end constructor //-------------------------------------------// //The following six methods are required by the // interface named GraphIntfc01. public int getNmbr(){ //Return number of functions to process. // Must not exceed 5. return 5; }//end getNmbr //-------------------------------------------// public double f1(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag1.length-1){ return 0; }else{ //Scale the magnitude data by the // reciprocal of the length of the sinusoid // to normalize the five plots to the same // peak value. return mag1[index]*16.0; }//end else }//end function //-------------------------------------------// public double f2(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag2.length-1){ return 0; }else{ return mag2[index]*8.0; }//end else }//end function //-------------------------------------------// public double f3(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag3.length-1){ return 0; }else{ return mag3[index]*4.0; }//end else }//end function //-------------------------------------------// public double f4(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag4.length-1){ return 0; }else{ return mag4[index]*2.0; }//end else }//end function //-------------------------------------------// public double f5(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag5.length-1){ return 0; }else{ return mag5[index]*1.0; }//end else }//end function }//end sample class Dsp031 Listing 10 ```
```/* File Dsp032a.java Copyright 2004, R.G.Baldwin Revised 5/17/2004 Displays sinusoidal pulses identical to those processed by Dsp032. Creates and displays five separate time series, each 400 samples in length. Each time series contains a pulse and the pulses are different lengths. Each pulse consists of the sum of two sinusoids at closely spaced frequencies. The frequencies of the two sinusoids for all pulses are the same. All frequency values are specified as type double as a fraction of the sampling frequency. The frequencies of the two sinusoids are equidistant from 0.0625 times the sampling frequency. The frequency of one sinusoid is (0.0625 - 2.0/len) times the sampling frequency. The frequency of the other sinusoid is (0.0625 + 2.0/len) times the sampling frequency. The lengths of the pulses are: 25 samples 50 samples 100 samples 200 samples 400 samples Tested using J2SEE 1.4.2 under WinXP. ************************************************/ import java.util.*; class Dsp032a implements GraphIntfc01{ final double pi = Math.PI; int len = 400;//data length int numberPulses = 5; //Frequencies of the sinusoids double freq1 = 0.0625 - 2.0/len; double freq2 = 0.0625 + 2.0/len; //Amplitude of the sinusoids double amp = 160; //Following arrays will contain sinusoidal data double[] data1 = new double[len]; double[] data2 = new double[len]; double[] data3 = new double[len]; double[] data4 = new double[len]; double[] data5 = new double[len]; public Dsp032a(){//constructor //Create the raw data for(int x = 0;x < len/16;x++){ data1[x] = amp*Math.cos(2*pi*x*freq1) + amp*Math.cos(2*pi*x*freq2); }//end for loop for(int x = 0;x < len/8;x++){ data2[x] = amp*Math.cos(2*pi*x*freq1) + amp*Math.cos(2*pi*x*freq2); }//end for loop for(int x = 0;x < len/4;x++){ data3[x] = amp*Math.cos(2*pi*x*freq1) + amp*Math.cos(2*pi*x*freq2); }//end for loop for(int x = 0;x < len/2;x++){ data4[x] = amp*Math.cos(2*pi*x*freq1) + amp*Math.cos(2*pi*x*freq2); }//end for loop for(int x = 0;x < len;x++){ data5[x] = amp*Math.cos(2*pi*x*freq1) + amp*Math.cos(2*pi*x*freq2); }//end for loop }//end constructor //-------------------------------------------// //The following six methods are required by the // interface named GraphIntfc01. public int getNmbr(){ //Return number of functions to process. // Must not exceed 5. return 5; }//end getNmbr //-------------------------------------------// public double f1(double x){ int index = (int)Math.round(x); if(index < 0 || index > data1.length-1){ return 0; }else{ //Scale the amplitude of the pulses to make // them compatible with the default // plotting amplitude of 100.0. return data1[index]*40.0/amp; }//end else }//end function //-------------------------------------------// public double f2(double x){ int index = (int)Math.round(x); if(index < 0 || index > data2.length-1){ return 0; }else{ return data2[index]*40.0/amp; }//end else }//end function //-------------------------------------------// public double f3(double x){ int index = (int)Math.round(x); if(index < 0 || index > data3.length-1){ return 0; }else{ return data3[index]*40.0/amp; }//end else }//end function //-------------------------------------------// public double f4(double x){ int index = (int)Math.round(x); if(index < 0 || index > data4.length-1){ return 0; }else{ return data4[index]*40.0/amp; }//end else }//end function //-------------------------------------------// public double f5(double x){ int index = (int)Math.round(x); if(index < 0 || index > data5.length-1){ return 0; }else{ return data5[index]*40.0/amp; }//end else }//end function }//end sample class Dsp032a Listing 11 ```
```/* File Dsp032.java Copyright 2004, R.G.Baldwin Revised 5/17/2004 Performs spectral analysis on five separate time series, each 400 samples in length. Each time series contains a pulse and the pulses are different lengths. Each pulse consists of the sum of two sinusoids at closely spaced frequencies. The frequencies of the two sinusoids for all pulses are the same. All frequency values are specified as type double as a fraction of the sampling frequency. The frequencies of the two sinusoids are equidistant from 0.0625 times the sampling frequency. The frequency of one sinusoid is (0.0625 - 2.0/len) times the sampling frequency. The frequency of the other sinusoid is (0.0625 + 2.0/len) times the sampling frequency. The lengths of the pulses are: 25 samples 50 samples 100 samples 200 samples 400 samples The spectral analyis computes the spectra at 400 equally spaced points between zero and the folding frequency (one-half the sampling frequency). The results of the spectral analysis are multiplied by the reciprocal of the lengths of the individual pulses to normalize the five plots. Otherwise, the results for the short pulses would be too small to see on the plots. Tested using J2SEE 1.4.2 under WinXP. ************************************************/ import java.util.*; class Dsp032 implements GraphIntfc01{ final double pi = Math.PI; int len = 400;//data length //Sample that represents zero time. int zeroTime = 0; //Low and high frequency limits for the // spectral analysis. double lowF = 0.0; double highF = 0.5; int numberSpectra = 5; //Frequencies of the sinusoids double freq1 = 0.0625 - 2.0/len; double freq2 = 0.0625 + 2.0/len; //Amplitude of the sinusoids double amp = 160; //Following arrays will contain data that is // input to the spectral analysis process. double[] data1 = new double[len]; double[] data2 = new double[len]; double[] data3 = new double[len]; double[] data4 = new double[len]; double[] data5 = new double[len]; //Following arrays receive information back // from the spectral analysis that is not used // in this program. double[] real; double[] imag; double[] angle; //Following arrays receive the magnitude // spectral information back from the spectral // analysis process. double[] mag1; double[] mag2; double[] mag3; double[] mag4; double[] mag5; public Dsp032(){//constructor //Create the raw data for(int x = 0;x < len/16;x++){ data1[x] = amp*Math.cos(2*pi*x*freq1) + amp*Math.cos(2*pi*x*freq2); }//end for loop for(int x = 0;x < len/8;x++){ data2[x] = amp*Math.cos(2*pi*x*freq1) + amp*Math.cos(2*pi*x*freq2); }//end for loop for(int x = 0;x < len/4;x++){ data3[x] = amp*Math.cos(2*pi*x*freq1) + amp*Math.cos(2*pi*x*freq2); }//end for loop for(int x = 0;x < len/2;x++){ data4[x] = amp*Math.cos(2*pi*x*freq1) + amp*Math.cos(2*pi*x*freq2); }//end for loop for(int x = 0;x < len;x++){ data5[x] = amp*Math.cos(2*pi*x*freq1) + amp*Math.cos(2*pi*x*freq2); }//end for loop //Compute magnitude spectra of the raw data // and save it in output arrays. Note that // the real, imag, and angle arrays are not // used later, so they are discarded each // time a new spectral analysis is performed. mag1 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data1,real, imag,angle,mag1,zeroTime,lowF,highF); mag2 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data2,real, imag,angle,mag2,zeroTime,lowF,highF); mag3 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data3,real, imag,angle,mag3,zeroTime,lowF,highF); mag4 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data4,real, imag,angle,mag4,zeroTime,lowF,highF); mag5 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data5,real, imag,angle,mag5,zeroTime,lowF,highF); }//end constructor //-------------------------------------------// //The following six methods are required by the // interface named GraphIntfc01. public int getNmbr(){ //Return number of functions to process. // Must not exceed 5. return 5; }//end getNmbr //-------------------------------------------// public double f1(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag1.length-1){ return 0; }else{ //Scale the magnitude data by the // reciprocal of the length of the sinusoid // to normalize the five plots to the same // peak value. return mag1[index]*16.0; }//end else }//end function //-------------------------------------------// public double f2(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag2.length-1){ return 0; }else{ return mag2[index]*8.0; }//end else }//end function //-------------------------------------------// public double f3(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag3.length-1){ return 0; }else{ return mag3[index]*4.0; }//end else }//end function //-------------------------------------------// public double f4(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag4.length-1){ return 0; }else{ return mag4[index]*2.0; }//end else }//end function //-------------------------------------------// public double f5(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag5.length-1){ return 0; }else{ return mag5[index]*1.0; }//end else }//end function }//end sample class Dsp032 Listing 12 ```
```/* File Dsp033a.java Copyright 2004, R.G.Baldwin Revised 5/17/2004 Displays sinusoidal pulses identical to those processed by Dsp033. Creates and displays five separate time series, each 400 samples in length. Each time series contains a pulse and the pulses are different lengths. Each pulse consists of the sum of two sinusoids at closely spaced frequencies. The frequencies of the two sinusoids are equidistant from 0.0625 times the sampling frequency. The total separation between the frequencies of the two sinusoids is the reciprocal of the length of the pulse. All frequency values are specified as type double as a fraction of the sampling frequency. The lengths of the pulses are: 25 samples 50 samples 100 samples 200 samples 400 samples Tested using J2SEE 1.4.2 under WinXP. ************************************************/ import java.util.*; class Dsp033a implements GraphIntfc01{ final double pi = Math.PI; int len = 400;//data length int numberPulses = 5; //Frequencies of the sinusoids double freq1a = 0.0625 - 8.0/len; double freq2a = 0.0625 + 8.0/len; double freq1b = 0.0625 - 4.0/len; double freq2b = 0.0625 + 4.0/len; double freq1c = 0.0625 - 2.0/len; double freq2c = 0.0625 + 2.0/len; double freq1d = 0.0625 - 1.0/len; double freq2d = 0.0625 + 1.0/len; double freq1e = 0.0625 - 0.5/len; double freq2e = 0.0625 + 0.5/len; //Amplitude of the sinusoids double amp = 160; //Following arrays will contain sinusoidal data double[] data1 = new double[len]; double[] data2 = new double[len]; double[] data3 = new double[len]; double[] data4 = new double[len]; double[] data5 = new double[len]; public Dsp033a(){//constructor //Create the raw data for(int x = 0;x < len/16;x++){ data1[x] = amp*Math.cos(2*pi*x*freq1a) + amp*Math.cos(2*pi*x*freq2a); }//end for loop for(int x = 0;x < len/8;x++){ data2[x] = amp*Math.cos(2*pi*x*freq1b) + amp*Math.cos(2*pi*x*freq2b); }//end for loop for(int x = 0;x < len/4;x++){ data3[x] = amp*Math.cos(2*pi*x*freq1c) + amp*Math.cos(2*pi*x*freq2c); }//end for loop for(int x = 0;x < len/2;x++){ data4[x] = amp*Math.cos(2*pi*x*freq1d) + amp*Math.cos(2*pi*x*freq2d); }//end for loop for(int x = 0;x < len;x++){ data5[x] = amp*Math.cos(2*pi*x*freq1e) + amp*Math.cos(2*pi*x*freq2e); }//end for loop }//end constructor //-------------------------------------------// //The following six methods are required by the // interface named GraphIntfc01. public int getNmbr(){ //Return number of functions to process. // Must not exceed 5. return 5; }//end getNmbr //-------------------------------------------// public double f1(double x){ int index = (int)Math.round(x); if(index < 0 || index > data1.length-1){ return 0; }else{ //Scale the amplitude of the pulses to make // them compatible with the default // plotting amplitude of 100.0. return data1[index]*40.0/amp; }//end else }//end function //-------------------------------------------// public double f2(double x){ int index = (int)Math.round(x); if(index < 0 || index > data2.length-1){ return 0; }else{ return data2[index]*40.0/amp; }//end else }//end function //-------------------------------------------// public double f3(double x){ int index = (int)Math.round(x); if(index < 0 || index > data3.length-1){ return 0; }else{ return data3[index]*40.0/amp; }//end else }//end function //-------------------------------------------// public double f4(double x){ int index = (int)Math.round(x); if(index < 0 || index > data4.length-1){ return 0; }else{ return data4[index]*40.0/amp; }//end else }//end function //-------------------------------------------// public double f5(double x){ int index = (int)Math.round(x); if(index < 0 || index > data5.length-1){ return 0; }else{ return data5[index]*40.0/amp; }//end else }//end function }//end sample class Dsp033a Listing 13 ```
```/* File Dsp033.java Copyright 2004, R.G.Baldwin Revised 5/17/2004 Same as Dsp032 except that the separation between the frequencies of the two sinusoids is the reciprocal of the length of the pulse. Performs spectral analysis on five separate time series, each 400 samples in length. Each time series contains a pulse and the pulses are different lengths. Each pulse consists of the sum of two sinusoids at closely spaced frequencies. The frequencies of the two sinusoids are equidistant from 0.0625 times the sampling frequency. The total separation between the frequencies of the two sinusoids is the reciprocal of the length of the pulse. All frequency values are specified as type double as a fraction of the sampling frequency. The lengths of the pulses are: 25 samples 50 samples 100 samples 200 samples 400 samples The spectral analyis computes the spectra at 400 equally spaced points between zero and the folding frequency (one-half the sampling frequency). The results of the spectral analysis are multiplied by the reciprocal of the lengths of the individual pulses to normalize the five plots. Otherwise, the results for the short pulses would be too small to see on the plots. Tested using J2SEE 1.4.2 under WinXP. ************************************************/ import java.util.*; class Dsp033 implements GraphIntfc01{ final double pi = Math.PI; int len = 400;//data length //Sample that represents zero time. int zeroTime = 0; //Low and high frequency limits for the // spectral analysis. double lowF = 0.0; double highF = 0.5; int numberSpectra = 5; //Frequencies of the sinusoids double freq1a = 0.0625 - 8.0/len; double freq2a = 0.0625 + 8.0/len; double freq1b = 0.0625 - 4.0/len; double freq2b = 0.0625 + 4.0/len; double freq1c = 0.0625 - 2.0/len; double freq2c = 0.0625 + 2.0/len; double freq1d = 0.0625 - 1.0/len; double freq2d = 0.0625 + 1.0/len; double freq1e = 0.0625 - 0.5/len; double freq2e = 0.0625 + 0.5/len; //Amplitude of the sinusoids double amp = 160; //Following arrays will contain data that is // input to the spectral analysis process. double[] data1 = new double[len]; double[] data2 = new double[len]; double[] data3 = new double[len]; double[] data4 = new double[len]; double[] data5 = new double[len]; //Following arrays receive information back // from the spectral analysis that is not used // in this program. double[] real; double[] imag; double[] angle; //Following arrays receive the magnitude // spectral information back from the spectral // analysis process. double[] mag1; double[] mag2; double[] mag3; double[] mag4; double[] mag5; public Dsp033(){//constructor //Create the raw data for(int x = 0;x < len/16;x++){ data1[x] = amp*Math.cos(2*pi*x*freq1a) + amp*Math.cos(2*pi*x*freq2a); }//end for loop for(int x = 0;x < len/8;x++){ data2[x] = amp*Math.cos(2*pi*x*freq1b) + amp*Math.cos(2*pi*x*freq2b); }//end for loop for(int x = 0;x < len/4;x++){ data3[x] = amp*Math.cos(2*pi*x*freq1c) + amp*Math.cos(2*pi*x*freq2c); }//end for loop for(int x = 0;x < len/2;x++){ data4[x] = amp*Math.cos(2*pi*x*freq1d) + amp*Math.cos(2*pi*x*freq2d); }//end for loop for(int x = 0;x < len;x++){ data5[x] = amp*Math.cos(2*pi*x*freq1e) + amp*Math.cos(2*pi*x*freq2e); }//end for loop //Compute magnitude spectra of the raw data // and save it in output arrays. Note that // the real, imag, and angle arrays are not // used later, so they are discarded each // time a new spectral analysis is performed. mag1 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data1,real, imag,angle,mag1,zeroTime,lowF,highF); mag2 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data2,real, imag,angle,mag2,zeroTime,lowF,highF); mag3 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data3,real, imag,angle,mag3,zeroTime,lowF,highF); mag4 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data4,real, imag,angle,mag4,zeroTime,lowF,highF); mag5 = new double[len]; real = new double[len]; imag = new double[len]; angle = new double[len]; ForwardRealToComplex01.transform(data5,real, imag,angle,mag5,zeroTime,lowF,highF); }//end constructor //-------------------------------------------// //The following six methods are required by the // interface named GraphIntfc01. public int getNmbr(){ //Return number of functions to process. // Must not exceed 5. return 5; }//end getNmbr //-------------------------------------------// public double f1(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag1.length-1){ return 0; }else{ //Scale the magnitude data by the // reciprocal of the length of the sinusoid // to normalize the five plots to the same // peak value. return mag1[index]*16.0; }//end else }//end function //-------------------------------------------// public double f2(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag2.length-1){ return 0; }else{ return mag2[index]*8.0; }//end else }//end function //-------------------------------------------// public double f3(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag3.length-1){ return 0; }else{ return mag3[index]*4.0; }//end else }//end function //-------------------------------------------// public double f4(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag4.length-1){ return 0; }else{ return mag4[index]*2.0; }//end else }//end function //-------------------------------------------// public double f5(double x){ int index = (int)Math.round(x); if(index < 0 || index > mag5.length-1){ return 0; }else{ return mag5[index]*1.0; }//end else }//end function }//end sample class Dsp033 Listing 14 ```
Copyright 2004, Richard G. Baldwin. Reproduction in whole or in part in any form or medium without express written permission from Richard Baldwin is prohibited. | 16,706 | 70,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2013-48 | longest | en | 0.907653 |
https://p.pdfkul.com/chapter-2-github_5acb71827f8b9a056b8b4567.html | 1,718,365,991,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861546.27/warc/CC-MAIN-20240614110447-20240614140447-00329.warc.gz | 412,706,095 | 12,320 | Chapter 2 DJM 30 January 2018 What is this chapter about? Problems with regression, and in particular, linear regression A quick overview: 1. 2. 3. 4. 5.
The truth is almost never linear. Collinearity can cause difficulties for numerics and interpretation. The estimator depends strongly on the marginal distribution of X. Leaving out important variables is bad. Noisy measurements of variables can be bad, but it may not matter.
Asymptotic notation • The Taylor series expansion of the mean function µ(x) at some point u ∂µ(x) |x=u + O(kx − uk2 ) ∂x • The notation f (x) = O(g(x)) means that for any x there exists a constant C such that f (x)/g(x) < C. µ(x) = µ(u) + (x − u)>
• More intuitively, this notation means that the remainder (all the higher order terms) are about the size of the distance between x and u or smaller. • So as long as we are looking at points u near by x, a linear approximation to µ(x) = E [Y | X = x] is reasonably accurate.
What is bias? • We need to be more specific about what we mean when we say bias. • Bias is neither good nor bad in and of itself. • A very simple example: let Z1 , . . . , Zn ∼ N (µ, 1). • We don’t know µ, so we try to use the data (the Zi ’s) to estimate it. • I propose 3 estimators: 1. µ b1 = 12, 2. µ b2 = Z6 , 3. µ b3 = Z. • The bias (by definition) of my estimator is E [b µ] − µ. • Calculate the bias and variance of each estimator.
1
Regression in general • If I want to predict Y from X, it is almost always the case that µ(x) = E [Y | X = x] 6= x> β • There are always those errors O(kx − uk)2 , so the bias is not zero. • We can include as many predictors as we like, but this doesn’t change the fact that the world is non-linear.
Covariance between the prediction error and the predictors • In theory, we have (if we know things about the state of nature) −1 β ∗ = arg min E kY − Xβk2 = Cov [X, X] Cov [X, Y ] β
−1
• Define v −1 = Cov [X, X]
.
• Using this optimal value β , what is Cov [Y − Xβ ∗ , X]? Cov [Y − Xβ ∗ , X] = Cov [Y, X] − Cov [Xβ ∗ , X] = Cov [Y, X] − Cov X(v −1 Cov [X, Y ]), X = Cov [Y, X] − Cov [X, X] v −1 Cov [X, Y ]
(Cov is linear) (substitute the def. of β ∗ ) (Cov is linear in the first arg)
= Cov [Y, X] − Cov [X, Y ] = 0.
Bias and Collinearity • • • • •
Adding or dropping variables may impact the bias of a model Suppose µ(x) = β0 + β1 x1 . It is linear. What is our estimator of β0 ? If we instead estimate the model yi = β0 , our estimator of β0 will be biased. How biased? But now suppose that x1 = 12 always. Then we don’t need to include x1 in the model. Why not? Form the matrix [1 x1 ]. Are the columns collinear? What does this actually mean?
When two variables are collinear, a few things happen. 1. We cannot numerically calculate (X> X)−1 . It is rank deficient. 2. We cannot intellectually separate the contributions of the two variables. 3. We can (and should) drop one of them. This will not change the bias of our estimator, but it will alter our interpretations. 4. Collinearity appears most frequently with many categorical variables. 5. In these cases, software automatically drops one of the levels resulting in the baseline case being in the intercept. Alternately, we could drop the intercept! 6. High-dimensional problems (where we have more predictors than observations) also lead to rank deficiencies. 7. There are methods (regularizing) which attempt to handle this issue (both the numerics and the interpretability). We may have time to cover them slightly.
2
White noise White noise is a stronger assumption than Gaussian. Consider a random vector . 1. ∼ N(0, Σ). 2. i ∼ N(0, σ 2 (xi )). 3. ∼ N(0, σ 2 I). The third is white noise. The are normal, their variance is constant for all i and independent of xi , and they are independent.
Asymptotic efficiency This and MLE are covered in 420. There are many properties one can ask of estimators θb of parameters θ h i 1. Unbiased: E θb − θ = 0 n→∞ 2. Consistent: hθb − i−−−→ θ b 3. Efficient: V θ is the smallest of all unbiased estimators 4. Asymptotically efficient: Maybe not efficient for every n, but in the limit, the variance is the smallest of all unbiased estimators. 5. Minimax: over all possible estimators in some class, this one has the smallest MSE for the worst problem. 6. . . .
Problems with R-squared SSE =1− 2 i=1 (Yi − Y )
R2 = 1 − Pn
1 n
M SE SSE =1− 2 SST (Y − Y ) i=1 i
Pn
• • • •
This gets spit out by software X and Y are both normal with (empirical) correlation r, then R2 = r2 In this nice case, it measures how tightly grouped the data are about the regression line Data that are tightly grouped about the regression line can be predicted accurately by the regression line. • Unfortunately, the implication does not go both ways. • High R2 can be achieved in many ways, same with low R2 • You should just ignore it completely (and the adjusted version), and encourage your friends to do the same
High R-squared with non-linear relationship genY <- function(X, sig) Y = sqrt(X)+sig*rnorm(length(X)) sig=0.05; n=100 X1 = runif(n,0,1) X2 = runif(n,1,2) X3 = runif(n,10,11) df = data.frame(x=c(X1,X2,X3), grp = rep(letters[1:3],each=n)) df\$y = genY(df\$x,sig) ggplot(df, aes(x,y,color=grp)) + geom_point() +
3
geom_smooth(method = 'lm', fullrange=TRUE,se = FALSE) + ylim(0,4) + stat_function(fun=sqrt,color='black')
4
3
grp y
a 2
b c
1
0 0
3
6
x df %>% group_by(grp) %>% summarise(rsq = summary(lm(y~x))\$r.sq) ## ## ## ## ## ##
# A tibble: 3 x 2 grp rsq 1 a 0.924 2 b 0.845 3 c 0.424
4
9
## Chapter 2 - GitHub
Jan 30, 2018 - More intuitively, this notation means that the remainder (all the higher order terms) are about the size of the distance between ... We don't know µ, so we try to use the data (the Zi's) to estimate it. ⢠I propose 3 ... Asymptotically efficient: Maybe not efficient for every n, but in the limit, the variance is the smallest.
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In StarOffice Writer, to make text bold, click. A) B) C) D). 05. In StarOffice Writer, to make text italic, click. A) B) C) D). 06. In StarOffice Writer, to make text underline ... | 3,276 | 11,243 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-26 | latest | en | 0.888949 |
https://www.trackopedia.info/encyclopedia/infrastructure/turnouts/concepts-for-turnouts/ | 1,610,743,979,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703496947.2/warc/CC-MAIN-20210115194851-20210115224851-00168.warc.gz | 1,040,738,040 | 11,483 | # Concepts for turnouts
The term turnouts, includes classical turnouts, crossings and diamond crossings with slips. In addition to the traditional geometric parameters of main line track, for the turnout, gauging parameters, the radius of the diverging line and the turnout inclination are of importance.
## Turnout inclination
Turnouts can be distinguished by their basic shape and are clearly limited in their extent by the start and end of the turnout. The start of the turnout is the tangent point of the diverging track on the main line. The end of the turnout is described as that point at which both rails (the main line and the diverging line) are so far apart that welding or fish-plating is possible. This point also allows the use of separate rail fastening systems and requires a space of at least 200 to 220 mm. The tangent at the turnout end points makes with the track axis of the main line the angle of the turnout. The tangent of this angle is also called the turnout inclination. The turnout inclination is given as the simplest possible break. Such a simplification is sometimes only possible if the constructive turnout is pushed further down the diverging line. This geometric turnout end is a fictional point and is designated as the designed end of the turnout. The distance between the main and the diverging line is designated as the spreading dimension.
Running edge diagrams are often used to provide a simplified representation of the different types of turnouts. This gives a "Representation of a turnout in ground plan, where - by leaving out the constructive details - only the running edge of the running and guiding rail is drawn."
## Track guidance measurement
For safe operation the track guidance measurement in the turnout must be continually monitored and maintained. Essential for this are, among other things, the gauge, the flange groove, the distance apart of the guide rails and the check rail gauge. All values are measured with a track gauge normal to the track axis between 0 and 14 mm below the upper edge of the rail.
Track gauge: The distance between the two rail heads.
Flange groove: The distance between the guide rail and the running rail.
Check rail distance: Distance between the check rail and the associated wing rail.
Check rail gauge: Distance between the check rail and nose of the crossing.
Detailed descriptions can be seen in the European Standard EN 13232.
To choose the right radius of curvature for the branch, and thus the curvature of the diverging line, it is necessary to consider different points of view. On one hand, it is advisable from an operating point of view to choose a large radius in order to raise the permitted speed through the turnout.
On the other hand, larger radii lead to increased space requirements and increasing costs both in investment and also in maintenance. Depending on the permissible centrifugal acceleration, the required radius is:
R=apermissible*3.6²/Vpermissible²
or
Vpermissible=√(3.6×Rpermissible×apermissible
If it is assumed that the diverging branch is built without superelevation and a maximum lateral acceleration apermissible of 0.65 m/s² is authorised, the maximum speed is:
Vpermissible=2.91×√(R0)
The branch curve begins at the start of the turnout. Circular curves, basket arches or clothoids[1] are used for the shape. The latter are primarily used on high-speed lines, in order to reduce the resulting acceleration difference.
In addition to the turnout sleepers, the associated rail fastening systems and control devices and the running surface of the turnout forms the centre of the turnout. This consists of the tongue, the closure rails and the crossings with check rails and running rails. In conventional turnouts the wheelset has no guidance in the area of the gap at the nose of the crossing (between closure rail and crossing nose). Check rails in the main line and branch line, however, guarantee safe wheel guidance.
[1] The term clothoid refers to a curve that begins at zero curvature at the straight section (the tangent) and increases linearly with its curve length.
## Parts of the turnout
A simple turnout contains the components shown below.
Turnouts in the main line should be arranged in such away that mutual interference can be avoided as far as possible. For the installation of safety facilities (e.g. axle counters, insulating joints, etc.) there should be a minimum distance between turnouts > 7 m.
From the beginning to the end of the turnout the individual components of the running surface of the turnout are fixed to the crossing timbers. In this section the turnout assembly forms a fixed construction. For turnouts primarily reinforced concrete sleepers are used with and without elastic coating and hardwood sleepers. Turnouts are arranged differently :
• perpendicular to the main line
• perpendicular to the bisector
• fan-shaped
The resulting partial rotation of the turnout places particular challenges on the correction of the position of the turnouts.
crossings: basics of crossings (Kr)
curve crossing (BKr)
crossing turnout (KW): single crossing turnout (EKW)
outer curve crossing turnout (EABKW)
double crossing turnout (DKW)
single curve crossing turnout: inner curve crossing turnout (EIBKW)
double curve crossing turnout (DBKW)
single turnout (EW)
single turnout (EW) as straight turnout (Ewl) left
double turnout (DW) single-sided double turnout (EinsDW)
double-sided double turnout (DW)
inner curve turnout (IBW)
outer curve turnout (ABW)
## You can find suitable specialist literature to this topic here:
### The Basic Principles of Mechanised Track Maintenance
This book is dedicated to the many people involved in the day to day planning and performance of track maintenance activities. Providing a practical approach to everyday challenges in mechanised track maintenance, it is not just intended as a theoretical approach to the track system.
Railways aim at transporting people and freight safely, rapidly, regularly, comfortably and on time from one place to another. This book is directed to track infrastructure departments contributing to the above objective by ensuring the track infrastructure’s reliability, availability, maintainability and safety – denoted by the acronym RAMS. Regular, effective and affordable track maintenance enable RAMS to be achieved.
1. [1] Fendrich, L.; Fengler, W.: Handbuch Eisenbahninfrastruktur. Springer, Berlin, 2013.
2. [2] Berg, G.; Henker, H.: Weichen. Eisenbahnbau. transpress Verl. f. Verkehrswesen, Berlin, 1978.
3. [3] Freystein, H.; Muncke, M.; Schollmeier, P.: Handbuch Entwerfen von Bahnanlagen. Regelwerke, Planfeststellung, Bau, Betrieb, Instandhaltung; Eurailpress, Hamburg, 2015. | 1,419 | 6,820 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-04 | latest | en | 0.931793 |
https://gitlab.science.ru.nl/benoit/tweetnacl/-/commit/82f2b86ec696408c6022fe9afe60c2bd151bfa45 | 1,660,620,361,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572220.19/warc/CC-MAIN-20220816030218-20220816060218-00005.warc.gz | 277,616,969 | 51,581 | Commit 82f2b86e by Benoit Viguier
### 9.75 pages
parent f920cbbe
... ... @@ -296,7 +296,7 @@ in Coq (for the sake of simplicity we do not display the conversion in the theor For all $n \in \N, n < 2^{255}$ and where the bits 1, 2, 5 248, 249, 250 are cleared and bit 6 is set, for all $P \in E(\F{p^2})$, for all $p \in \F{p}$ such that $P.x = p$, there exists $Q \in E(\F{p^2})$ such that $Q = nP$, $Q.x = q$ and $q$ = \VSTe{CSM} $n$ $p$. there exists $Q \in E(\F{p^2})$ such that $Q = nP$ where $Q.x = q$ and $q$ = \VSTe{CSM} $n$ $p$. \end{theorem} A more complete description in Coq of Theorem \ref{CSM-correct} with the associated conversions is as follow: ... ...
\section{Linking C and Coq} \section{Linking C and Reflections} In this section we describe techniques used to prove the equivalence between the Clight description of TweetNaCl and Coq functions producing similar behaviors. ... ... @@ -8,27 +8,38 @@ Clight description of TweetNaCl and Coq functions producing similar behaviors. As described in Section \ref{sec:impl}, numbers in \TNaCle{gf} are represented in base $2^{16}$ and we can use a direct mapping to represent that array as a list integers in Coq. However in order to show the correctness of the basic operations, we need to convert this number as a full integer. For this reason we define \Coqe{ZofList : Z -> list Z -> Z}: \begin{Coq} we need to convert this number as a full integer. \begin{definition} Let \Coqe{ZofList} : $\Z \rightarrow \texttt{list} \Z \rightarrow \Z$, a parametrized map by $n$ betwen a list $l$ and its it's little endian representation with a base $2^n$. \end{definition} We define it in Coq as: \begin{lstlisting}[language=Coq] Fixpoint ZofList {n:Z} (a:list Z) : Z := match a with | [] => 0 | h :: q => h + (pow 2 n) * ZofList q end. \end{lstlisting} We define a notation where $n$ is $16$. \begin{lstlisting}[language=Coq] Notation "Z16.lst A" := (ZofList 16 A). \end{Coq} It converts any list of integer from a base $2^n$ to an single integer. In our case we define a notation where $n$ is $16$. The following Coq notation is defined to convert any integer to its representant in \Zfield. \begin{Coq} \end{lstlisting} We also define a notation to do the modulo, projecting any numbers in $\Zfield$. \begin{lstlisting}[language=Coq] Notation "A :GF" := (A mod (2^255-19)). \end{Coq} \end{lstlisting} Remark that this representation is different from \Coqe{Zmodp}. However the equivalence between operations over $\Zfield$ and $\F{p}$ is easily proven. Using these two definitions, we proved lemmas such as the correctness of the multiplication \Coqe{M}: Using these two definitions, we proved intermediates lemmas such as the correctness of the multiplication \Coqe{M} where \Coqe{M} replicate the computations and steps done in C. \begin{lemma} For all list of integers \texttt{a} and \texttt{b} of length 16 representing $A$ and $B$ in $\Zfield$, the number represented in $\Zfield$ by the list \Coqe{(M a b)} is equal to $A \times B \bmod \p$. \end{lemma} And seen in Coq as follows: \begin{Coq} Lemma mult_GF_Zlength : forall (a:list Z) (b:list Z), ... ... @@ -38,14 +49,10 @@ Lemma mult_GF_Zlength : (Z16.lst a * Z16.lst b) :GF. \end{Coq} For all list of integers \texttt{a} and \texttt{b} of length 16 representing $A$ and $B$ in \Zfield, the number represented in \Zfield by the list \Coqe{(M a b)} is equal to $A \times B \bmod 2^{255}-19$. \subsection{Inversions in \Zfield} In a similar fashion we can define a Coq version of the inversion mimicking the behavior of \TNaCle{inv25519} over \Coqe{list Z}. \Coqe{step_pow} contains the body of the for loop. the behavior of \TNaCle{inv25519} over \Coqe{list Z}. \begin{lstlisting}[language=Ctweetnacl] sv inv25519(gf o,const gf a) { ... ... @@ -59,18 +66,15 @@ sv inv25519(gf o,const gf a) FOR(i,16) o[i]=c[i]; } \end{lstlisting} \begin{Coq} We specify this with 2 functions: a recursive \Coqe{pow_fn_rev} to to simulate the for loop and a simple \Coqe{step_pow} containing the body. Note the off by one for the loop. \begin{lstlisting}[language=Coq] Definition step_pow (a:Z) (c g:list Z) : list Z := let c := Sq c in if a <>? 1 && a <>? 3 then M c g else c. \end{Coq} \Coqe{pow_fn_rev} is responsible of the iteration of the loop by making recursive calls. \begin{Coq} Function pow_fn_rev (a:Z) (b:Z) (c g: list Z) {measure Z.to_nat a} : (list Z) := if a <=? 0 ... ... @@ -78,23 +82,46 @@ Function pow_fn_rev (a:Z) (b:Z) (c g: list Z) else let prev := pow_fn_rev (a - 1) b c g in step_pow (b - 1 - a) prev g. \end{Coq} % This \Coqe{Function} requires a proof of termination. It is done by proving the % Well-foundness of the decreasing argument: \Coqe{measure Z.to_nat a}. \end{lstlisting} Calling \Coqe{pow_fn_rev} 254 times allows us to reproduce the same behavior as the Clight definition. \begin{Coq} This \Coqe{Function} requires a proof of termination. It is done by proving the Well-foundness of the decreasing argument: \Coqe{measure Z.to_nat a}. Calling \Coqe{pow_fn_rev} 254 times allows us to reproduce the same behavior as the \texttt{Clight} definition. \begin{lstlisting}[language=Coq] Definition Inv25519 (x:list Z) : list Z := pow_fn_rev 254 254 x x. \end{Coq} \end{lstlisting} Similarily we define the same function over $\Z$. \begin{lstlisting}[language=Coq] Definition step_pow_Z (a:Z) (c:Z) (g:Z) : Z := let c := c * c in if a <>? 1 && a <>? 3 then c * g else c. \subsection{Inversions and Reflections} Function pow_fn_rev_Z (a:Z) (b:Z) (c:Z) (g: Z) {measure Z.to_nat a} : Z := if (a <=? 0) then c else let prev := pow_fn_rev_Z (a - 1) b c g in step_pow_Z (b - 1 - a) prev g. In TweetNaCl, \TNaCle{inv25519} computes an inverse in \Zfield. It uses the Fermat's little theorem by the exponentiation to $2^{255}-21$. To prove the correctness of the result we can use multiple strategy such as: Definition Inv25519_Z (x:Z) : Z := pow_fn_rev_Z 254 254 x x. \end{lstlisting} And prove their equivalence in $\Zfield$. \begin{lstlisting}[language=Coq] Lemma Inv25519_Z_GF : forall (g:list Z), length g = 16 -> (Z16.lst (Inv25519 g)) :GF = (Inv25519_Z (Z16.lst g)) :GF. \end{lstlisting} In TweetNaCl, \TNaCle{inv25519} computes an inverse in $\Zfield$. It uses the Fermat's little theorem by doing an exponentiation to $2^{255}-21$. This is done by applying a square-and-multiply algorithm. The binary representation of $p-2$ implies to always do a multiplications aside for bit 2 and 4, thus the if case. To prove the correctness of the result we can use multiple strategies such as: \begin{itemize} \item Proving it is special case of square-and-multiply algorithm applied to a specific number and then show that this number is indeed $2^{255}-21$. ... ... @@ -102,94 +129,44 @@ correctness of the result we can use multiple strategy such as: $x^a \times x^b = x^{(a+b)}$ and $(x^a)^2 = x^{(2 \times a)}$. We can prove that the resulting exponent is $2^{255}-21$. \end{itemize} We choose the second method. The only drawback is that it requires to apply the unrolling and exponentiation lemmas 255 times. This can be automated in Coq with tacticals such as \Coqe{repeat}, but it generates a big proof object which We use the second method for the benefits of simplicity. However it requires to apply the unrolling and exponentiation formulas 255 times. This can be automated in Coq with tacticals such as \Coqe{repeat}, but it generates a proof object which will take a long time to verify. \subsubsection{Speeding up with Reflections} \subsection{Speeding up with Reflections} In order to speed up the verification, we use a technique called reflection. It provides us with flexibility such as we don't need to know the number of times the lemmas needs to be applied. It provides us with flexibility such as we don't need to know the number of times nor the order in which the lemmas needs to be applied (chapter 15 in \cite{CpdtJFR}). The idea is to \textit{reflect} the goal into a decidable environment. In our case, we show that for a property $P$, we can define a decidable boolean property $P_{bool}$ such that if $P_{bool}$ is \Coqe{true} then $P$ holds. $$P_{bool} = true \implies P$$ With VST we proved that \TNaCle{inv25519} in \texttt{Clight} is equivalent to its Coq functional definition \Coqe{Inv25519}. We also proved that \Coqe{Inv25519} over \Coqe{list Z} is equivalent to applying \Coqe{Inv25519_Z} over \Coqe{Z} % With VST we proved that \TNaCle{inv25519} in \texttt{Clight} is equivalent to its Coq % functional definition \Coqe{Inv25519}. We also proved that \Coqe{pow_fn_rev} % over \Coqe{list Z} is equivalent to applying % \Coqe{pow_fn_rev_Z} over \Coqe{Z} \begin{Coq} Lemma Inv25519_Z_GF : forall (g:list Z), length g = 16 -> (Z16.lst (Inv25519 g)) :GF = (Inv25519_Z (Z16.lst g)) :GF. \end{Coq} % \begin{Coq} % Lemma pow_fn_rev_Z_GF : % forall (a:Z) (b:Z) (c:list Z) (g:list Z), % Zlength c = 16 -> % Zlength g = 16 -> % (Z16.lst (pow_fn_rev a b c g)) :GF = % (pow_fn_rev_Z a b (Z16.lst c) (Z16.lst g)) :GF . % \end{Coq} where \Coqe{Inv25519_Z} reproduces the same pattern as \Coqe{Inv25519}. \begin{Coq} Definition Inv25519_Z (x:Z) : Z := pow_fn_rev_Z 254 254 x x. \end{Coq} \Coqe{pow_fn_rev_Z} mimicks \Coqe{pow_fn_rev} over \Coqe{Z} instead of \Coqe{list Z}. The structure is the same: the application \Coqe{step_pow_Z} and the loop itself \Coqe{pow_fn_rev_Z} \begin{Coq} Definition step_pow_Z (a:Z) (c:Z) (g:Z) : Z := let c := c * c in if a <>? 1 && a <>? 3 then c * g else c. Function pow_fn_rev_Z (a:Z) (b:Z) (c:Z) (g: Z) {measure Z.to_nat a} : Z := if (a <=? 0) then c else let prev := pow_fn_rev_Z (a - 1) b c g in step_pow_Z (b - 1 - a) prev g. \end{Coq} % % In line with the definition of \Coqe{Inv25519}, we define the inversion modulo % $2^{255}-19$ as an instance of \Coqe{pow_fn_rev_Z} where \Coqe{a} and \Coqe{b} are 254. \subsubsection{A Simple Domain Specific Language} We show that for a property $P$, we can define a decidable boolean property $P_{bool}$ such that if $P_{bool}$ is \Coqe{true} then $P$ holds. $$reify\_P : P_{bool} = true \implies P$$ By applying $reify\_P$ on $P$ our goal become $P_{bool} = true$. We can then compute the result of $P_{bool}$. If the decision goes well we are left with the tautology $true = true$. To prove that the \Coqe{Inv25519_Z} is computing $x^{2^{255}-21}$, we define a Domain Specific Language: \begin{Coq} we define a Domain Specific Language. \begin{definition} Let \Coqe{expr_inv} denote an expression which can be either a term; a multiplication of expressions; a squaring of an expression or a power of an expression. And Let \Coqe{formula_inv} denote an equality between two expressions. \end{definition} \begin{lstlisting}[language=Coq] Inductive expr_inv := | R_inv : expr_inv | M_inv : expr_inv -> expr_inv -> expr_inv | S_inv : expr_inv -> expr_inv | P_inv : expr_inv -> positive -> expr_inv. \end{Coq} An expression can be either a term, a multiplication, a squaring or a power. This is denoted as follow: \begin{Coq} Inductive formula_inv := | Eq_inv : expr_inv -> expr_inv -> formula_inv. \end{lstlisting} The denote functions are defined as follows: \begin{lstlisting}[language=Coq] Fixpoint e_inv_denote (m:expr_inv) : Z := match m with | R_inv => ... ... @@ -201,37 +178,13 @@ Fixpoint e_inv_denote (m:expr_inv) : Z := | P_inv x p => pow (e_inv_denote x) (Z.pos p) end. \end{Coq} We defined \Coqe{step_inv} and \Coqe{pow_inv} to mirror the behavior of \Coqe{step_pow_Z} and respectively \Coqe{pow_fn_rev_Z} over our new domain. \begin{Coq} Lemma step_inv_step_pow_eq : forall (a:Z) (c:expr_inv) (g:expr_inv), e_inv_denote (step_inv a c g) = step_pow_Z a (e_inv_denote c) (e_inv_denote g). \end{Coq} \begin{Coq} Lemma pow_inv_pow_fn_rev_eq : forall (a:Z) (b:Z) (c:expr_inv) (g:expr_inv), e_inv_denote (pow_inv a b c g) = pow_fn_rev_Z a b (e_inv_denote c) (e_inv_denote g). \end{Coq} We also define what is a formula in our language: a simple equality. \begin{Coq} Inductive formula_inv := | Eq_inv : expr_inv -> expr_inv -> formula_inv. \end{Coq} This is denoted as follow: \begin{Coq} Definition f_inv_denote (t : formula_inv) : Prop := match t with | Eq_inv x y => e_inv_denote x = e_inv_denote y end. \end{Coq} With such Domain Specific Language we have the equivalence between: \end{lstlisting} With such Domain Specific Language we have the equality between: \begin{lstlisting}[backgroundcolor=\color{white}] f_inv_denote (Eq_inv (M_inv R_inv (S_inv R_inv)) ... ... @@ -239,22 +192,48 @@ f_inv_denote = (x * x^2 = x^3) \end{lstlisting} \subsubsection{Deciding formulas} We define \Coqe{step_inv} and \Coqe{pow_inv} to mirror the behavior of \Coqe{step_pow_Z} and respectively \Coqe{pow_fn_rev_Z} over our DSL and we prove their equality. \begin{lstlisting}[language=Coq] Lemma step_inv_step_pow_eq : forall (a:Z) (c:expr_inv) (g:expr_inv), e_inv_denote (step_inv a c g) = step_pow_Z a (e_inv_denote c) (e_inv_denote g). Lemma pow_inv_pow_fn_rev_eq : forall (a:Z) (b:Z) (c:expr_inv) (g:expr_inv), e_inv_denote (pow_inv a b c g) = pow_fn_rev_Z a b (e_inv_denote c) (e_inv_denote g). \end{lstlisting} We can then derive the following lemma. \begin{lemma} \label{reify} With an appropriate choice of variables, \Coqe{pow_inv} denotes \Coqe{Inv25519_Z}. \end{lemma} In order to prove formulas in \Coqe{formula_inv}, we define a decidable procedure. We first compute the power of each side of the formula and then check their equality. we have the following a decidable procedure. We define \Coqe{pow_expr_inv}, a function which returns the power of an expression. We can then compare the two values and decide over their equality. \begin{Coq} Fixpoint pow_expr_inv (t:expr_inv) : Z := match t with | R_inv => 1 (* power of a term is 1. *) | M_inv x y => (pow_expr_inv x) + (pow_expr_inv y) (* power of a multiplication is the sum of the exponents. *) | S_inv x => 2 * (pow_expr_inv x) (* power of a squaring is the double of the exponent. *) | P_inv x p => (Z.pos p) * (pow_expr_inv x) (* power of a power is the multiplication of the exponents. *) end. Definition decide_e_inv (l1 l2:expr_inv) : bool := ... ... @@ -265,20 +244,24 @@ Definition decide_f_inv (f:formula_inv) : bool := | Eq_inv x y => decide_e_inv x y end. \end{Coq} We proved that our procedure is correct: for all formulas in \Coqe{formula_inv}, if \Coqe{decide_f_inv} returns \Coqe{true}, then the denoted equality is correct. We prove our decision procedure correct. \begin{lemma} \label{decide} For all formulas $f$, if the decision over $f$ returns \Coqe{true}, then the denoted equality by $f$ is true. \end{lemma} Which can be formalized as: \begin{Coq} Lemma decide_formula_inv_impl : forall (f:formula_inv), decide_f_inv f = true -> f_inv_denote f. \end{Coq} By reification to our Domain Specific Language and then by applying \texttt{decide\_formula\_inv\_impl} and computing \texttt{decide\_f\_inv}, we proved that \Coqe{Inv25519} is indeed computing an inverse in modulo $2^{255}-19$. By reification to over DSL (lemma \ref{reify}) and by applying our decision (lemma \ref{decide}). we proved the following theorem. \begin{theorem} \Coqe{Inv25519_Z} computes an inverse in \Zfield. \end{theorem} \begin{Coq} Theorem Inv25519_Z_correct : forall (x:Z), ... ... @@ -287,7 +270,9 @@ Theorem Inv25519_Z_correct : From \Coqe{Inv25519_Z_correct} and \Coqe{Inv25519_Z_GF}, we conclude the functionnal correctness of the inversion over \Zfield. \begin{corollary} \Coqe{Inv25519} computes an inverse in \Zfield. \end{corollary} \begin{Coq} Corollary Inv25519_Zpow_GF : forall (g:list Z), ... ... @@ -296,18 +281,19 @@ Corollary Inv25519_Zpow_GF : (pow (Z16.lst g) (2^255-21)) :GF. \end{Coq} \subsubsection{Packing and other applications} \subsection{Packing and other applications of reflection} Reflection can also be used where proofs requires computing and a small and We prove the functional correctness of \Coqe{Inv25519} with reflections. This technique can also be used where proofs requires some computing or a small and finite domain of variable to test e.g. for all $i$ such that $0 \le i < 16$. Using reflections we proved that we can split the for loop in \TNaCle{pack25519} into two parts. Using reflection we prove that we can split the for loop in \TNaCle{pack25519} into two parts. \begin{lstlisting}[language=Ctweetnacl] for(i=1;i<15;i++) { m[i]=t[i]-0xffff-((m[i-1]>>16)&1); m[i-1]&=0xffff; } \end{lstlisting} The first is computing the substraction while the second is applying the carrying. The first loop is computing the substraction while the second is applying the carrying. \begin{lstlisting}[language=Ctweetnacl] for(i=1;i<15;i++) { m[i]=t[i]-0xffff ... ... @@ -318,8 +304,8 @@ for(i=1;i<15;i++) { } \end{lstlisting} This loop separation allows simpler proofs. The first loop is seen as the substraction of a number in \Zfield. We then proved than with the iteration of the second loop, the number represented in \Zfield stays the same. This lead to the proof that \TNaCle{pack25519} is effectively reducing mod $2^{255}-19$ and returning a number in base $2^8$. We then prove that with the iteration of the second loop, the number represented in \Zfield stays the same. This leads to the proof that \TNaCle{pack25519} is effectively reducing mod $\p$ and returning a number in base $2^8$. \begin{Coq} Lemma Pack25519_mod_25519 : ... ...
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The mean marks obtained by 300 students in a subject are 60. The mean of top 100 students was found to be 80 and the mean of last 100 students was found to be 50. The mean marks of the remaining 100 students are? A) 70 B) 65 C) 60 D) 50 Correct Answer :50 Explanation :Total marks of remaining(middle 100) students = 300*60 - 100*80 - 100*50 = 18000 - 8000 - 5000 = 18000 - 13000 = 5000 Hence, Mean marks of remaining 100 students = 5000/100 = 50 Post/View Answer Post comment Cancel Thanks for your comment.! Write a comment(Click here) ... | 173 | 585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2020-24 | latest | en | 0.862697 |
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# Lecture 4
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### Lecture 4
1. 1. LOCAL NEIGHBORHOOD PROCESSING. CONCEPT OF SPATIAL DOMAIN FILTERING CS-467 Digital Image Processing 1
2. 2. Spatial Domain • The Spatial Domain is a domain (the plane) where a digital image is defined by spatial coordinates of its pixels • (Another domain considered in image processing is the frequency domain where a digital image is defined by braking down into the spatial frequencies participating in its formation – we will consider it later) 2
3. 3. Local Neighborhood in Image Processing • Many image processing operations in the spatial domain (particularly, spatial domain filtering) are reduced to local neighborhood processing 3
4. 4. Local Neighborhood in Image Processing • Let be the set of coordinates of a neighborhood centered on an arbitrary pixel (x, y) in an image f. • Neighborhood processing generates a corresponding pixel at the same coordinates in an output image g, such that the intensity value in that that pixel is determined by a specific operation involving the pixels in the input image with coordinates in 4 xyS xyS
5. 5. Local Neighborhood in Image Processing 5
6. 6. Local Neighborhood in Spatial Domain Image Processing • The spatial domain processing can be represented by the following expression where f(x,y) is the input image, g(x,y) is the output image and T is an operator defined over a local neighborhood of pixel with the coordinates (x,y) 6 ( ) ( )( ), ,Tg x y f x y=
7. 7. Local Contrast Enhancement • Local contrast enhancement is utilized through local histogram processing • The local histogram in an n x m window of a digital image with intensity levels {0,1,…,L-1} is a discrete function 7 xyS ; 0,1,..., 1xy xy S k k S h r n k L = = −
8. 8. Local Contrast Enhancement • Local contrast enhancement leads to better visibility of local details and, as a result, to the extraction of those details (not necessarily small details) that are hidden in the areas with a poor local contrast • Local contrast enhancement can be utilized, for example, through the same methods that are used for the global contrast enhancement 8
9. 9. Local Contrast Enhancement 9 © 1992-2008 R.C. Gonzalez & R.E. Woods
10. 10. Filtering • Filtering in signal processing is the process of accepting (passing) or rejecting certain frequency components • Lowpass filters preserve low frequencies, rejecting the high ones • Highpass filters preserve high frequencies, rejecting the low ones 10
11. 11. Filtering • Lowpass filters are used in image processing for removal or reduction of noise • Highpass filters are used in image processing for edge detection and distinguishing of small details 11
12. 12. Models of Noise • Models of noise • Additive noise • Multiplicative (speckle) noise • Impulse noise where is the probability of distortion ( in percents is called the corruption rate) 12 ( ) ( ), , ( , )g x y f x y x yη= + ( ) ( ), , ( , )g x y f x y x yη= × ( ) ( ) , ( , ) , 1 , , n n p x y g x y p f x y η = − np np 100%np ⋅
13. 13. The Goal of Noise Filtering • The goal of noise filtering is to obtain the best approximation of the “ideal” image in terms of the mean square error (root mean square error) or, which is the same, in terms of PSNR • Filtering is utilized through a filtering operator applied to a noisy image: 13 ( ),f x y( )ˆ ,f x y ( ) ( )( )ˆ , ,f x y F g x y=
14. 14. “Ideal” Filtering • If there are k realizations of the same image scene corrupted by the same kind of additive noise, and , then the original image can be restored by taking component-wise mean of its realizations: • The larger is k, the closer is to 14 k → ∞ ( ) ( ) ( ) ( ) 1 , , ( , ); 1,2,..., ,... 1ˆ , , i i k i i g x y f x y x y i k f x y g x y k η = = + = = ∑ ( )ˆ ,f x y ( ),f x y
15. 15. Linear Filters • Linear filtering is utilized through a linear operator • The operator F is called linear, if for arbitrary constants a and b the following property holds: 15 ( )( ) ( )( ) ( ) ( )( ) ( ) , ( , ) , ( , ) , ( , ) F af x y b x y F af x y F b x y aF f x y bF x y η η η + = + = +
16. 16. Linear Filters • A filter, which is represented by a linear operator is called a linear filter • Ideally, a linear filter may separate an image from additive noise. If F is a linear filter, then and an image can be restored as follows 16 ( )( ) ( )( ) ( )( ) ( ), , ( , ) , ( , )F g x y F f x y x y F f x y F x yη η= + = + ( ) ( )( )( ) ( )( ) ( )( )1 1 , , , ( , )f x y F F f x y F F g x y F x yη− − = −
17. 17. Linear and Nonlinear Filters • Since can be estimated only in the frequency domain and this estimation is possible only for some particular kinds of noise, the “ideal” method has a limited applicability • Speckle and impulse noise cannot be separated from an image using a linear filter. Thus, for their filtering nonlinear filters shall be used 17 ( )( , )F x yη
18. 18. Linear and Nonlinear Filters • A filter, which is represented by an operator, which is not linear, is called a nonlinear filter • Linear filters can be implemented in the frequency domain (where they originally were applied by Norbert Wiener in 1940s ) and in the spatial domain • Most of nonlinear filters can be implemented only in the spatial domain (except some specific nonlinear transformations of spectra that are applicable only in the frequency domain) 18
19. 19. Filter Design • When any new filter is designed, then to test its efficiency, the following model should be used: A clean image shall be artificially corrupted using a certain kind of noise Then an artificially corrupted image shall be filtered and the result shall be compared to the “ideal” clean image in terms of PSNR A filter showing stable good results, being applied to different images (when compared to other filters), should be considered good 19
20. 20. Modeling of Additive Noise • To add noise to a clean image, it is necessary to generate a matrix of noise with a desired distribution (Gaussian is most frequently used) using a random numbers generator, with exactly the same sizes as the ones of the clean image, the same mean as the one of the clean image and σn (standard deviation) equal to of the clean image. • Then this noise shall be added to the clean image component-wise and their common mean shall be component-wise subtracted from the sum 20 0.2 0.5σ σ− ( ) ( ), , ( , )g x y f x y x y meanη= + −
21. 21. Spatial Domain Filtering • Any spatial domain filter is reduced to processing of a local neighborhood of every pixel • Filtering creates a new pixel with coordinates equal to the coordinates of the center of the neighborhood and whose value is the result of the filtering operator applied to the neighborhood of the processed pixel • Filtering of different pixels is typically independent and in such a case can be done in parallel 21
22. 22. Spatial Domain Filtering • Any lowpass spatial domain filter averages intensities (linearly or non-linearly) over an n x m local neighborhood of each pixel and smooths the image due to this averaging. This means that some image details whose size is about 0.5n x 0.5m may become undistinguishable. • Lowpass spatial domain filters are used for noise reduction. Averaging over each n x m local neighborhood , they “dissolve” a noise in an image 22
23. 23. Spatial Domain Filtering 23
24. 24. Linear Spatial Domain Filtering • Linear spatial filtering of an image of size M x N with a filter of size m x n is defined by the expression where g(x,y) is an image to be processed, a=(m-1)/2, b=(n-1)/2, and the weights w(s,t) create a filter kernel 24 ( ) ( ) ( )ˆ , , , a b s a t b f x y w s t g x s y t =− =− = + +∑ ∑
25. 25. Linear Spatial Domain Filtering 25 © 1992-2008 R.C. Gonzalez & R.E. Woods
26. 26. Border Effects in Spatial Domain Filtering • To process image borders, it is necessary to extend an image in all directions, otherwise it will not be possible to build local neighborhoods for the border pixels • The simplest way of such an extension is zero- padding . However, this method always creates a “black” frame along the image borders. 26
27. 27. Border Effects in Spatial Domain Filtering • To take care of border effects, mirroring shall be used. An N x M image, before it is processed by a spatial domain filter with an n x m kernel, shall be extended to an image as follows 27 2 2 n m N M + × + ( ) ( ) ( ) ( ) ( ) ( ) , ; 0,1,..., 1, 0,1,..., 1 , ; 1,..., / 2 , 0,1,..., 1 ( 1) ( 1), ; ,..., / 2 1, 0,1,..., 1 , ; 0,1,..., 1, 1,..., / 2, ,( 1) ( 1) ; 0,1,..., 1, ,... f x y x N y M f x y x n y M f N x N y x N N n y M f x y x N y mf x y f x N y N x N y N = −= − − =− − = − − − − + = + −= − − = − =− − = − − − + = −= ( ) ( ) , / 2 1 , ; , 1, 2,... ( 1) ( 1),( 1) ( 1) ; , , 1,... N m f x y x y f N x N N y N x y N N + − − − =− − − − − + − − − + = +
28. 28. Border Effects in Spatial Domain Filtering 28 0th row M-1th row 0th col. 1st col. -1th row Mth row -1st col. Mirroring for a 3x3 local neighborhood window
29. 29. Arithmetic Mean Filter – the Simplest Spatial Domain Linear Filter • Arithmetic mean filter replaces the intensity value in each pixel by the local arithmetic mean taken over a local n x m processing window: • This corresponds to the filter kernel (mask) 29 ( ) ( ) 1ˆ , , a b s a t b f x y g x s y t nm =− =− = + +∑ ∑ 1 1 1 ... 1 1 1 ... ... ... ... ... 1 1 1 ... nm nm nm nm nm nm nm nm nm
30. 30. Useful Filtering Mask • Vertical and horizontal adjacent pixels in a 3x3 window are located “closer” to the center of the window than the diagonal pixels. Evidently, the distance from a center of a square to any of its sides is shorter than the one to its vertices: 30
31. 31. Useful Filtering Mask • So it can be reasonable to emphasize contribution of the central pixel and its closest neighbors to the resulting intensity value. This can be done by the following filter mask (kernel) 31 1 2 1 1 2 4 2 16 1 2 1 | 2,974 | 10,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-34 | longest | en | 0.851651 |
https://www.math-english.com/reasoning-beginner/clocks/ | 1,679,844,193,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00550.warc.gz | 992,573,496 | 8,177 | # Clocks
## Basics of Clocks
### Definition of Clock
A Clock is an electronic instrument used for indicating and maintaining the time. It represents time in hours, minutes and seconds.
### Hands of Clock
A clock has three hands:
• Hour hand - clock is divided along the circumference into 12 equal spaces denoting 12 hours.
• Minute hand - clock is divided along the circumference into 60 equal spaces denoting 60 minutes.
• Second hand
One hour = 60 minutes = 60 × 60 seconds = 3600 seconds
#### Different types of questions asked:
• Find the angle made by the two hands of a clock at a particular time.
• Find the time at which two hands make a particular angle with each other.
• Questions on faulty watches
Share on: | 165 | 731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-14 | latest | en | 0.900629 |
https://www.opengl.org/discussion_boards/archive/index.php/t-137065.html | 1,521,768,962,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648113.87/warc/CC-MAIN-20180323004957-20180323024957-00024.warc.gz | 832,591,700 | 5,221 | PDA
View Full Version : Trying to do the impossible.??.
dabeav
06-05-2002, 05:37 AM
Here I am again, attempting the impossible. Why would one not be able to aply multiple normals per vertex, just like you could apply multiple textures per vertex. I understand, i could simply create another identical vertex, and give it another normal, but then thats 2 verts that have be transformed each frame. Isnt there ANY way to simply assign multiple normals to ONE vertex?? This would make LOW poly models, able to look VERY VERY HIGH poly. I think Q3 did something like this, but im not aware how. Any clues??
Bob
06-05-2002, 05:56 AM
A vertex can only have one normal. What would be the point in using several normals for a vertex? OpenGL doesn't use more than one anyway.
Multiple sets of texture coordinates has a function, since you can apply several textures to a primitive, and therefore need a set of texture coordinates for each texture. Multiple normals have no function in OpenGL.
06-05-2002, 12:26 PM
Wow, what exactly are you trying to achieve?
I actually thought you needed a whole triangle to have a normal. The normal is perpindicular to the triangle. You can't have 2 normals per triangle because there is only 1 thats perpindicular to the triangle.. right? Maybe I'm thinking of something else.
That seems right though.
What were you going to do with 2 normals? maybe theres a different way to get the same effect.
chmod
06-05-2002, 12:48 PM
You can always try and average the normals for a common vertex.
Jambolo
06-05-2002, 07:11 PM
Multiple normals would make the data very complicated and there really isn't much use for them.
Also, while a vertex can have multiple texture coordinates, it can't have one set of coords for one poly and a different set for the other poly. That's what you want to do with the normals, right?
Roadmaster: if you want to do smooth shading, you need to specify a normal for each vertex. If you are doing flat shading, you can specify a normal for each poly but OpenGL still sets a normal for each vertex (using the same one on all three vertices).
[This message has been edited by Jambolo (edited 06-05-2002).]
T2k
06-06-2002, 01:24 AM
why multiple normals, if you want to have smooth edges you simply calc: vertex_normal= normalized(sum of triangle_normals[using this vertex]). and if you want to have sharp edges you have to use a bit more data...
T2k
Arsenal
06-11-2002, 11:05 AM
I think what dabeav is trying to do is implement a DOOM3-similar way of modeling, just that he's going by the approach all wrong.
DOOM3 uses both ultra-high poly models and lower "good for games" poly models for the games. The high-poly model is used to generate a "normal map" which is basically a bumpmap over the entire model. This normal map is rendered onto the game poly model to make it look much more detailed than it actually is.
I assume the DOOM3 model is bumpmapped using object-space vectors, not tangent-space vectors. Most bumpmapping tutorials use tangent-space vectors.
jmg
06-11-2002, 04:14 PM
dabeav,
In OpenGL, there is only the concept of "current normal". Whatever the current normal is set to, that's what it is for subsequent vertices.
The meaning of OpenGL's "normal" not *quite* the same as it is in math class. http://www.opengl.org/discussion_boards/ubb/smile.gif | 816 | 3,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-13 | longest | en | 0.937284 |
https://cutlogpro.com/what-length-chainsaw-chain/ | 1,679,818,419,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00660.warc.gz | 229,058,192 | 23,514 | # What Length Chainsaw Chain?
## How long of a chainsaw chain do I need?
Measure the length of the chainsaw bar from its front tip to the cutter closest to the saw’s body. To get the closest measurement, round it up to the nearest inch. A bar with a called length of 20 will be eighteen feet long.
## How do I know what chain I need for my chainsaw?
The distances between the links on the chain are described by the pitch measurement. To calculate the pitch of your chain, divide the distance between the three rivets by two. The pitches for chainsaw chains are between 0.25 and 0.406 inches.
## How do I know my chain size?
If you only own the sprocket, use the calipers to measure between the teeth. Take a look at where the chain roller will set between the teeth. You can determine what chain size is needed by knowing the pitch.
## What size bar on a chainsaw should I get?
The length of a chainsaw bar is related to the thickness of the wood being cut. It is better to have a smaller chainsaw bar that you occasionally use for larger jobs than it is to go the largest chainsaw bar and use it on the smaller jobs.
## What size chainsaw do I need for a 36 inch bar?
The diameter of the trees you cut should not be more than twice the length of the bar on the chainsaw. That means a 20-inch chainsaw can cut a 36-inch tree.
## Does chainsaw bar length matter?
It’s important to consider the nature of your work and the size of your work area when choosing a blade length for your chainsaw. Most of the time, a chainsaw with a saw bar or blade that can cut through a tree limb is enough.
## Are all 14 inch chainsaw chains the same?
The guide bar size is not always the same as the chain saw’s chain size. Determining the correct size of chain for a chain saw with a 14-inch guide bar involves a number of factors.
## What size chain is on my Stihl chainsaw?
The number on the side of the chain can be used to determine pitch, gauge, and other characteristics of the chain. You have a chain with the number 33RM2 stamped on it. The chain’s characteristics are marked by the last three numbers.
## What is the difference between .325 and 3/8 chain?
It is the. It may not be the best choice for your daily needs. The three-eighths inch chain lasts longer than its smaller cousin. One of the most popular switches for chainsaw users is this one.
## Does Oregon chain fit Stihl?
Oregon R55 AdvanceCut Chainsaw Chain for 16-Inch Bar -55 Drive Links is a low kickback chain that can be used on several models. The product can only be purchased through retailers or dealers. It is designed to fit most chainsaw models that have a low profile pitch.
## What is a 420 chain?
USA Roller Chain has a common non-standard roller chain called the 420 roller chain. The chain is made with solid rollers and uses heat-treated components for better performance.
## What do Oregon chain numbers mean?
It’s up to the person to pitch. The distance between any three consecutive rivets divided by two is referred to as the chain pitch. The smallest, most popular, and largest pitches are used to make the chain.
## What do Stihl chain numbers mean?
The gauge of the chain can be found by looking at the number on the drive link. A 1 is equivalent to a 0.043-inch gauge, 3 is equivalent to a 0.050-inch gauge, 5 is equivalent to a 0.058-inch gauge, and 6 is equivalent to a 0.063-inch gauge. A 0 means the chain has a small gauge, but it’s not found on the Harvester chain.
## Can you put a different size bar on a chainsaw?
Depending on the bar size, most chainsaws can operate up to three different bars. The guide bars range in size from 14 to over 40 inches. You can check the specifications of your model’s operator’s manual to find out what size bars you can run on your saw.
## Is a 14 inch chainsaw big enough?
If you only need it for light-duty work like trimming small branches, a less powerful chainsaw with a shorter bar is all you need. Something over 14” is enough for this type of work.
## What size tree can a 16 chainsaw cut?
How large is a tree for a chainsaw cut? When choosing a bar length, most professionals recommend that you size it up by a couple of inches. It is possible to cut a trunk with a width of up to 14 inches if you use a 16 chainsaw.
## What is a good size chainsaw for a homeowner?
A good length for a chainsaw is 14 to 16 inches. It’s easy to control the saw because it’s long enough to limb up trees, but also short enough to cut down small trees.
## Can you put an 18 inch bar on a Stihl ms180?
The guides for the homeowner models should be between 12 and 16 inches. The larger 45cc engines in theMS 250 andMS 251 models allow them to drive a bar of 18 inches.
Fine sawdust can be created by cross cutting against the grain of the wood. The saw is making smoke even if your chain is oiled. An even cut can be produced by the chainsaw. It’s possible that you have dull cutting teeth on one side.
## What are the different sizes of chainsaw chains?
You can find chainsaw chains in the following sizes. There is a low-profile of 325″, 3/8″, and. A number of times, “A number of times”.
## What size chain is on a STIHL Farm Boss?
The standard model of the farm boss chainsaw has a 20 inch guide bar and chain.
## Are .325 and 3/8 chains interchangeable?
There is a way to replace a power match nose. There is a nose on the Oregon Power Match bar. Then when you’re done. The chain had to be put back on.
## What size file do you use on a 3/8 chainsaw chain?
The pitch of the chain can be found on the packaging of the chainsaw blade. There is a 5/32 file needed for these pitches.
## Does it matter how many drive links on a chainsaw chain?
A specific chainsaw chain link count is used in the design of the chainsaw. It might be better to have a longer chain with more links for jobs like felling trees or bucking logs that are heavier.
## What is a low profile chainsaw chain?
The cutters on a low profile chain are not as tall as the ones on a standard chainsaw chain. A shallow cut is made by the low profile chain. It’s a good fit for low-horsepower saws and electric saws because it weighs less than standard chain.
## How many links does a 14 chainsaw have?
Oregon S52 Advance Cut Chainsaw Chain for 14-Inch Bar 52 Drive Links is a low kickback chain that fits many people. The product can only be purchased through retailers or dealers. It is designed to fit all chainsaw brands, with models that need a small pitch.
## Can I use a different brand chain on my Stihl chainsaw?
If you want to use your own chains, after-market manufacturers can produce similar chains that you can use on a chain saw. If you want to buy a chain from a brand, you need to know how the chain is described.
## Can I use a Husqvarna chain on a Stihl?
A general answer is that chains from Husqvarna can be used on the chainsaw. The two chainsaw brands use different pitch sizes, so they can’t fit on the same chainsaw.
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Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A306995 Number of primes q < prime(n), such that 2*prime(n) + q is prime. 2
%I #12 Apr 10 2019 22:31:55
%S 0,0,1,2,1,3,3,2,2,2,3,3,3,5,6,4,3,5,7,3,6,5,6,6,7,5,5,8,6,7,8,5,9,9,
%T 4,8,8,12,8,10,8,9,11,10,10,10,8,10,12,9,10,10,11,11,9,10,11,13,15,11,
%U 12,13,14,11,16,15,11,13,10,15,14,12,13,16,16,13,17,15,15
%N Number of primes q < prime(n), such that 2*prime(n) + q is prime.
%C Conjecture: a(n) > 0 for all n >= 3. - _Dmitry Kamenetsky_, Mar 18 2019
%F a(n) = A071127(prime(n)). - _Michel Marcus_, Mar 19 2019
%e 2*7 + 3 = 17 and 2*7 + 5 = 19 are prime. Hence a(4) = 2.
%o (PARI) a(n) = my(s=0, p=prime(n)); forprime(i=1, p-1, s += isprime(2*p+i)); s; \\ _Michel Marcus_, Mar 19 2019
%Y Cf. A071127, A306996, A306997.
%K nonn
%O 1,4
%A _Dmitry Kamenetsky_, Mar 18 2019
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Last modified July 15 09:15 EDT 2024. Contains 374324 sequences. (Running on oeis4.) | 549 | 1,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.58643 |
https://nrich.maths.org/public/leg.php?code=-56&cl=4&cldcmpid=1928 | 1,519,133,278,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812959.48/warc/CC-MAIN-20180220125836-20180220145836-00572.warc.gz | 762,760,065 | 6,483 | # Search by Topic
#### Resources tagged with Golden ratio similar to Golden Ratio:
Filter by: Content type:
Stage:
Challenge level:
### There are 20 results
Broad Topics > Fractions, Decimals, Percentages, Ratio and Proportion > Golden ratio
### Golden Ratio
##### Stage: 5 Challenge Level:
Solve an equation involving the Golden Ratio phi where the unknown occurs as a power of phi.
### Pentakite
##### Stage: 4 and 5 Challenge Level:
ABCDE is a regular pentagon of side length one unit. BC produced meets ED produced at F. Show that triangle CDF is congruent to triangle EDB. Find the length of BE.
### Darts and Kites
##### Stage: 4 Challenge Level:
Explore the geometry of these dart and kite shapes!
### Golden Mathematics
##### Stage: 5
A voyage of discovery through a sequence of challenges exploring properties of the Golden Ratio and Fibonacci numbers.
### Golden Thoughts
##### Stage: 4 Challenge Level:
Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.
### Golden Fibs
##### Stage: 5 Challenge Level:
When is a Fibonacci sequence also a geometric sequence? When the ratio of successive terms is the golden ratio!
### Golden Fractions
##### Stage: 5 Challenge Level:
Find the link between a sequence of continued fractions and the ratio of succesive Fibonacci numbers.
### Gold Yet Again
##### Stage: 5 Challenge Level:
Nick Lord says "This problem encapsulates for me the best features of the NRICH collection."
### Leonardo of Pisa and the Golden Rectangle
##### Stage: 2, 3 and 4
Leonardo who?! Well, Leonardo is better known as Fibonacci and this article will tell you some of fascinating things about his famous sequence.
### About Pythagorean Golden Means
##### Stage: 5
What is the relationship between the arithmetic, geometric and harmonic means of two numbers, the sides of a right angled triangle and the Golden Ratio?
### Pentabuild
##### Stage: 5 Challenge Level:
Explain how to construct a regular pentagon accurately using a straight edge and compass.
### Gold Again
##### Stage: 5 Challenge Level:
Without using a calculator, computer or tables find the exact values of cos36cos72 and also cos36 - cos72.
### Golden Eggs
##### Stage: 5 Challenge Level:
Find a connection between the shape of a special ellipse and an infinite string of nested square roots.
### Golden Construction
##### Stage: 5 Challenge Level:
Draw a square and an arc of a circle and construct the Golden rectangle. Find the value of the Golden Ratio.
### Golden Powers
##### Stage: 5 Challenge Level:
You add 1 to the golden ratio to get its square. How do you find higher powers?
### Pent
##### Stage: 4 and 5 Challenge Level:
The diagram shows a regular pentagon with sides of unit length. Find all the angles in the diagram. Prove that the quadrilateral shown in red is a rhombus.
### Golden Triangle
##### Stage: 5 Challenge Level:
Three triangles ABC, CBD and ABD (where D is a point on AC) are all isosceles. Find all the angles. Prove that the ratio of AB to BC is equal to the golden ratio.
### Whirling Fibonacci Squares
##### Stage: 3 and 4
Draw whirling squares and see how Fibonacci sequences and golden rectangles are connected.
### The Golden Ratio, Fibonacci Numbers and Continued Fractions.
##### Stage: 4
An iterative method for finding the value of the Golden Ratio with explanations of how this involves the ratios of Fibonacci numbers and continued fractions.
### Pythagorean Golden Means
##### Stage: 5 Challenge Level:
Show that the arithmetic mean, geometric mean and harmonic mean of a and b can be the lengths of the sides of a right-angles triangle if and only if a = bx^3, where x is the Golden Ratio. | 842 | 3,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-09 | longest | en | 0.858627 |
https://stats.stackexchange.com/posts/324337/revisions | 1,568,821,671,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573309.22/warc/CC-MAIN-20190918151927-20190918173927-00231.warc.gz | 658,640,743 | 18,013 | For a project, I am looking for idea to model for the distribution of corners in football matches. I know that the number of goals can be model by a Poisson distribution, but for the number of corners, the distribution has larger tails. I think it is partly due to the fact that after a corner, the ball is put back in the game on one extrema of the field and this makes the probability of a second corner more likely. Would any of you have an idea of a model I could use?
EDITPrecisions on the model I plan to use I realized reading the comment that my first message was not precise enough. I do not have a specific distribution in mind, myMy initial idea is to adapt the Maher modelMaher Model for the number of goals in soccer matches. In this model the number of corners of scored by team $$T$$ when playing vs team $$T'$$ is assumed to followedfollow a Poisson distribution with Parameter $$a_T d_{T'}$$ where $$a_T$$ and $$d_T$$ are the attack and defense scores of team $$T$$ which are estimated from the data using macimummaximum likelihood.
To answer one question from the comment; from there, I am looking for a distribution different from the Poisson distribution (fatter tails) which generalizes suchthe Maher model to the distribution of corners. A distribution for which estimating the parameters via maximal likelihood is possible would be perfect.
For a project, I am looking for idea to model for the distribution of corners in football matches. I know that the number of goals can be model by a Poisson distribution, but for the number of corners, the distribution has larger tails. I think it is partly due to the fact that after a corner, the ball is put back in the game on one extrema of the field and this makes the probability of a second corner more likely. Would any of you have an idea of a model I could use?
EDIT I realized reading the comment that my first message was not precise enough. I do not have a specific distribution in mind, my initial idea is to adapt the Maher model for the number of goals in soccer matches. In this model the number of corners of team $$T$$ vs team $$T'$$ is assumed to followed a Poisson distribution with Parameter $$a_T d_{T'}$$ where $$a_T$$ and $$d_T$$ are the attack and defense scores of team $$T$$ which are estimated from the data using macimum likelihood.
To answer one question from the comment; from there, I am looking for a distribution different from the Poisson distribution which generalizes such model to the distribution of corners. A distribution for which estimating the parameters via maximal likelihood is possible would be perfect.
For a project, I am looking for idea to model for the distribution of corners in football matches. I know that the number of goals can be model by a Poisson distribution, but for the number of corners, the distribution has larger tails. I think it is partly due to the fact that after a corner, the ball is put back in the game on one extrema of the field and this makes the probability of a second corner more likely. Would any of you have an idea of a model I could use?
Precisions on the model I plan to use My initial idea is to adapt the Maher Model for the number of goals in soccer matches. In this model the number of corners of scored by team $$T$$ when playing vs team $$T'$$ is assumed to follow a Poisson distribution with Parameter $$a_T d_{T'}$$ where $$a_T$$ and $$d_T$$ are the attack and defense scores of team $$T$$ which are estimated from the data using maximum likelihood.
I am looking for a distribution different from the Poisson distribution (fatter tails) which generalizes the Maher model to the distribution of corners. A distribution for which estimating the parameters via maximal likelihood is possible would be perfect.
2 added 779 characters in body
For a project, I am looking for idea to model for the distribution of corners in football matches. I know that the number of goals can be model by a Poisson distribution, but for the number of corners, the distribution has larger tails. I think it is partly due to the fact that after a corner, the ball is put back in the game on one extrema of the field and this makes the probability of a second corner more likely. Would any of you have an idea of a model I could use?
EDIT I realized reading the comment that my first message was not precise enough. I do not have a specific distribution in mind, my initial idea is to adapt the Maher model for the number of goals in soccer matches. In this model the number of corners of team $$T$$ vs team $$T'$$ is assumed to followed a Poisson distribution with Parameter $$a_T d_{T'}$$ where $$a_T$$ and $$d_T$$ are the attack and defense scores of team $$T$$ which are estimated from the data using macimum likelihood.
To answer one question from the comment; from there, I am looking for a distribution different from the Poisson distribution which generalizes such model to the distribution of corners. A distribution for which estimating the parameters via maximal likelihood is possible would be perfect.
For a project, I am looking for idea to model for the distribution of corners in football matches. I know that the number of goals can be model by a Poisson distribution, but for the number of corners, the distribution has larger tails. I think it is partly due to the fact that after a corner, the ball is put back in the game on one extrema of the field and this makes the probability of a second corner more likely. Would any of you have an idea of a model I could use?
For a project, I am looking for idea to model for the distribution of corners in football matches. I know that the number of goals can be model by a Poisson distribution, but for the number of corners, the distribution has larger tails. I think it is partly due to the fact that after a corner, the ball is put back in the game on one extrema of the field and this makes the probability of a second corner more likely. Would any of you have an idea of a model I could use?
EDIT I realized reading the comment that my first message was not precise enough. I do not have a specific distribution in mind, my initial idea is to adapt the Maher model for the number of goals in soccer matches. In this model the number of corners of team $$T$$ vs team $$T'$$ is assumed to followed a Poisson distribution with Parameter $$a_T d_{T'}$$ where $$a_T$$ and $$d_T$$ are the attack and defense scores of team $$T$$ which are estimated from the data using macimum likelihood.
To answer one question from the comment; from there, I am looking for a distribution different from the Poisson distribution which generalizes such model to the distribution of corners. A distribution for which estimating the parameters via maximal likelihood is possible would be perfect.
1
# Modeling the number of corners in soccer
For a project, I am looking for idea to model for the distribution of corners in football matches. I know that the number of goals can be model by a Poisson distribution, but for the number of corners, the distribution has larger tails. I think it is partly due to the fact that after a corner, the ball is put back in the game on one extrema of the field and this makes the probability of a second corner more likely. Would any of you have an idea of a model I could use? | 1,561 | 7,317 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 30, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-39 | latest | en | 0.95863 |
https://freecourseweb.com/tutorials3/teaching-academics/electrical-schematics-industrial-controls/ | 1,606,911,135,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141708017.73/warc/CC-MAIN-20201202113815-20201202143815-00072.warc.gz | 269,383,590 | 26,864 | # Electrical Schematics (Industrial Controls)
Learn to read, understand and design electrical schematics / electrical wiring drawings!!
Electrical Schematics (Industrial Controls) is a basic course to start learning schematics or electrical drawing (power and control circuits). We have covered various concepts that are required to design schematics. We start with very basics and eventually build on it.
What you’ll learn
• How to read, understand & design schematics.
• Concept of power and control circuits.
• Designing a circuit from scratch : Starter Circuits.
• Indication circuits.
• Local Remote & Auto Manual Operation.
• Basics of electrical interlocking.
• Metering circuits.
• Learn certain key switchgears.
• Electrical symbols.
• Cross-referencing and Designations in Schematics.
• Starter Circuit.
• Concept of NO & NC contacts.
• Contactor’s Contact Numbering.
Course Content
• Introduction and Concept Building –> 9 lectures • 26min.
• Section 2 : Key Switchgears –> 7 lectures • 25min.
• Section 3 : Designing of Schematics –> 18 lectures • 1hr 19min.
• Section 4 : Basic of Interlocking –> 2 lectures • 11min.
• Section 5: Metering Circuits –> 3 lectures • 11min.
• Section 6 : Application of Schematics and Quick course recap –> 2 lectures • 5min.
Requirements
• Should have a basic degree or knowledge in Electrical Engineering – B tech, B.E., Diploma or any similar relevant educational background.
Electrical Schematics (Industrial Controls) is a basic course to start learning schematics or electrical drawing (power and control circuits). We have covered various concepts that are required to design schematics. We start with very basics and eventually build on it.
Some of the topics that you will come across are:
1) Basic details on schematics – What is schematic ? Where is it used ? Why is it needed?
2) Reading & Understanding Schematic – Title block, Disclaimer, Template, Designations, Cross-referencing, Legends etc.
3) Symbols & Switchgears – Electrical Symbols along with definition & images. Key technical details about switchgear like circuit breakers, Contactors Isolators, Meters , Indicators & Protection devices
a) Learn Off load & On Load devices
b) Difference between isolators & circuit breakers
c) NO & NC contacts
d) Power & Auxiliary Contacts – contact numbering etc.
e) Instrument Transformers
4) Designing circuit – in a step by step method – Control of motors, indication circuits, Local remote and auto manual operations & metering circuits: Starter Circuit
5) Basics of electrical interlocking
6) Basics of Metering Circuits
7) Application in Switchboards and practice questions.
Get Tutorial | 597 | 2,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-50 | latest | en | 0.815627 |
https://theredseamarket.com/qa/quick-answer-how-do-you-write-an-algorithm.html | 1,606,386,825,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141187753.32/warc/CC-MAIN-20201126084625-20201126114625-00154.warc.gz | 529,569,929 | 7,556 | # Quick Answer: How Do You Write An Algorithm?
## How do you use algorithm in a sentence?
Algorithms sentence examplesSelf-teaching algorithms will get better and better at making suggestions.
The speed and quality of those algorithms will get ever better.
Algorithms get smarter, computers faster.More items….
## How algorithms are written?
An algorithm is a set of steps designed to solve a problem or accomplish a task. Algorithms are usually written in pseudocode, or a combination of your speaking language and one or more programming languages, in advance of writing a program.
## What are the types of algorithm?
Algorithm types we will consider include:Simple recursive algorithms.Backtracking algorithms.Divide and conquer algorithms.Dynamic programming algorithms.Greedy algorithms.Branch and bound algorithms.Brute force algorithms.Randomized algorithms.
## What are the four characteristics of algorithms?
Algorithm and its characteristicsFiniteness. An algorithm must always terminate after a finite number of steps.Definiteness. Each step of an algorithm must be precisely defined; the actions to be carried out must be rigorously and unambiguously specified for each case.Input. … Output. … Effectiveness.
## How do you write a simple algorithm?
There are many ways to write an algorithm….An Algorithm Development ProcessStep 1: Obtain a description of the problem. This step is much more difficult than it appears. … Step 2: Analyze the problem. … Step 3: Develop a high-level algorithm. … Step 4: Refine the algorithm by adding more detail. … Step 5: Review the algorithm.
## How do you write algorithms with two numbers?
Examples Of Algorithms In ProgrammingAlgorithm to add two numbers entered by the user Step 1: Start Step 2: Declare variables num1, num2 and sum. … Find the largest number among three different numbers Step 1: Start Step 2: Declare variables a,b and c. … Factorial of a number entered by the user.More items…
## What is algorithm and its types?
An algorithm is a set of self contained sequence of instructions or actions that contains finite space or sequence and that will give us a result to a specific problem in a finite amount of time. It is a logical and mathematical approach to solve or crack a problem using any possible method.
## What is an algorithm and an example?
An algorithm is a step by step procedure to solve logical and mathematical problems. A recipe is a good example of an algorithm because it says what must be done, step by step. It takes inputs (ingredients) and produces an output (the completed dish). … Informally, an algorithm can be called a “list of steps”.
## How do you write an algorithm in C?
Let’s try to learn algorithm-writing by using an example.Problem − Design an algorithm to add two numbers and display the result.Step 1 − START.Step 2 − declare three integers a, b & c.Step 3 − define values of a & b.Step 4 − add values of a & b.Step 5 − store output of step 4 to c.Step 6 − print c.Step 7 − STOP.More items…
## What is a good algorithm?
Input: a good algorithm must be able to accept a set of defined input. Output: a good algorithm should be able to produce results as output, preferably solutions. Finiteness: the algorithm should have a stop after a certain number of instructions. Generality: the algorithm must apply to a set of defined inputs.
## What is algorithm example?
One of the most obvious examples of an algorithm is a recipe. It’s a finite list of instructions used to perform a task. For example, if you were to follow the algorithm to create brownies from a box mix, you would follow the three to five step process written on the back of the box.
## What are basic algorithms?
Algorithm is a step-by-step procedure, which defines a set of instructions to be executed in a certain order to get the desired output. Algorithms are generally created independent of underlying languages, i.e. an algorithm can be implemented in more than one programming language.
## What is another word for algorithm?
What is another word for algorithm?processprogrammeUKtaskbatchcodescriptbinaryfunctionsmechanicsprocedures2 more rows
## What is an example of an algorithm in math?
A step-by-step solution. Each step has clear instructions. Like a recipe. Long Division is another example of an algorithm: when you follow the steps you get the answer.
## What are five things algorithms must have?
An algorithm must have five properties:Input specified.Output specified.Definiteness.Effectiveness.Finiteness. | 938 | 4,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-50 | latest | en | 0.860101 |
https://proficientwritershub.com/wk-1-discussion-statistics-due-thurs/ | 1,719,332,718,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00646.warc.gz | 413,268,425 | 18,482 | # wk 1 discussion statistics due thurs
## Post a total of 3 substantive responses over 2 separate days for full participation. This includes your initial post and 2 replies to other students. You have two due dates for your responses. The first response is due by Thursday and the other responses are due by Monday. Note that you can always post more than 3 responses to the weekly discussion and your responses can be posted prior to the due dates.
Due by Thursday
Respond to the following in a minimum of 175 words:
Review Figure 1.1 and Figure 1.2 from Chapter 1 and note the difference between a parameter and a statistic. Figure 1.3 describes the levels of measurement.
Consider how the type of variable (qualitative and quantitative) and the level of measurement used influence the presentation of the data collected for the variable and the statistics that can be calculated for the variable. Provide your own examples.
Due by Monday
Review the concepts of measures of central tendency and variation from Chapter 3 and consider the following situation: You are standing on the bank of a river.
Assume the river current is not a factor and you only have the mean as the information for the depth of the river posted on the river bank. Is this sufficient information to allow a person who is 5 ft 7in tall to safely cross the river?
Reply to at least 2 of your classmates in 100 word count. Be constructive and professional in your responses.
BY Brianna
Consider how the type of variable (qualitative and quantitative) and the level of measurement used influence the presentation of the data collected for the variable and the statistics that can be calculated for the variable. Provide your own examples.
In Chapter 1 we review the statistical concepts and the levels of measurement. A statistic and a parameter are different yet similar. The huge difference is that a parameter describes an entire population and a statistic describes a sample. In Figure 1.1 the key elements are graphically displayed of the statistics: collect data, analyze data, interpret data and present findings. Figure 1.2 shows the inferential process which is when a researcher gathers data from a sample and uses the statistics generated to reach conclusions about the population from which the sample was taken. Then we have Figure 1.3 which shows the highest level of data measurement to the lowest level of data measurement (ratio, interval, ordinal, and nominal).
An example would be if production artists have made a new reality tv show and they want to find out how many people are watching, in what states are the number of people watching higher, what type of people are watching, etc. The researchers would take these findings and feedback from the viewers into consideration and would provide the mean, median, average and range and other numbers to gather more data on their study. The info used will help to see if the producers are going in the right direction with the show and if it’s successful then they may start to work on a new season. They do what is done in Figure 1.1 collect data, analyze the data, interpret the data and present their results that they found from the data.
Reference
Black, K. (2017). Business Statistics: For Contemporary Decision Making, (9th Edition). Hoboken, NJ:
By Lakesha
### a-Kesha Morrison
17 hours ago, at 4:43 PM
NEW
The differences in parameters and statistics are that parameters look at a populations descriptive measures and statistics are an analysis of gathered data that is used as a numerical representation of scientific data (Black, 2017). Statistics are subdivided and a parameter is one of the subdivisions of a statistic. In Figures 1.1 and 1.2, figure 1.1 is a representation of the components that researchers use to gather data to be analyzed and to make a hypothesis and then draw a fact based conclusion based on the research that was conducted and data gathered. The end results are then used to present to an audience of a target population. For example, when researchers are looking to gather information on the effects associated with a new medication, the population is usually an population of people with a particular condition or illness in which the medication is targeted. The researchers monitor these case studies over a particular amount of time and they rely heavily on information from these individuals to provide them with feedback, researchers will make notations based on reports and observations to complete a final analysis of their findings. In Figure 1.2, the diagram is an example of a parameter and sample (both are subdivisions of statistics). A sample is derived from a particular population of interests. The parameters look specifically at variances within a sample to provide descriptive measures of the population. So as previously mentioned in the example, the researchers would provide the mean, median, average and range, standard deviations and so on and so forth based on the numerical data gathered during the study. The information is used to report to the FDA which is usually included in the medication pamphlet for patients to review the medications compositions, as well as case findings during research.
Black, K. (2017). Business Statistics: For Contemporary Decision Making (9th Edition). Hoboken, N.J.: Wiley | 1,061 | 5,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-26 | latest | en | 0.916201 |
http://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition/chapter-4-inverse-exponential-and-logarithmic-functions-chapter-4-test-prep-review-exercises-page-492/56 | 1,524,256,147,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944682.35/warc/CC-MAIN-20180420194306-20180420214306-00257.warc.gz | 420,681,049 | 13,373 | # Chapter 4 - Inverse, Exponential, and Logarithmic Functions - Chapter 4 Test Prep - Review Exercises: 56
$e^{3}$
#### Work Step by Step
$\ln x+\ln x^{3}=12\Rightarrow \ln x+3\ln x=12\Rightarrow 4\ln x=12\Rightarrow \ln x=3\Rightarrow x=e^{3}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 121 | 408 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-17 | latest | en | 0.661144 |
https://vustudents.ning.com/group/cs401computerarchitectureandassemblylanguageprogra/forum/topics/cs401-assignment-1-dated-9-may-2019-to-16-may-2019-1?commentId=3783342%3AComment%3A6224350&groupId=3783342%3AGroup%3A59348 | 1,571,019,418,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986648481.7/warc/CC-MAIN-20191014003258-20191014030258-00416.warc.gz | 837,866,165 | 20,913 | We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>
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# CS401 Assignment No 01 Spring 2019 Solution & Discussion Due Date: 16-05-2019
CS401 Assignment No 01 Spring 2019 Solution & Discussion Due Date: 16-05-2019
Q1. Calculate the Effective Address and Physical Address for the following set of values. (10)
Segment: 0xFFFF
Offset =0x0800
Base: 0x0111
Index=0x0001
Q2. Calculate the Segment Address for the given set of Physical and Effective addresses. (10)
Physical address = 0x119B, Effective address = 0x10AB
Physical address = 0xFF9B, Effective address = 0x10FF
Spring%202019_CS401_1.docx
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### Replies to This Discussion
physical address = segment + offset (handouts pg no 14)
well effect add has more formullas in the book one them is this
effective add = base + offset + index ( handouts pg no 28)
Please Discuss here about this assignment.Thanks
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dear this is wrong solution please delete it. i will provide correct solution soon.
Ali73 ya sure, if u have corrected solution then u can upload there .thanks
Ali bhai, waiting...
CS401 assignment solution
part 1
part 2
part 3
ye correct hai solution?
according to me yes. read handout page no 14 and 28
ThankYou!
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© 2019 Created by + M.Tariq Malik. Powered by | 1,154 | 3,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-43 | latest | en | 0.776474 |
http://www.kosmologia.gr/intro_en/final_in_physics_8b_en.htm | 1,480,702,192,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540409.8/warc/CC-MAIN-20161202170900-00178-ip-10-31-129-80.ec2.internal.warc.gz | 553,024,889 | 14,367 | THE COMPLETE COSMOS, THE MATTER & THE WAVE PHENOMENA
"If I could know in the beginning that I learn now !" popular Greek saying
THE RELATIVITY OF TIME AND ENERGY --- THE MICROSCOPIC EXPLOSIONS OF "EMPTY" SPACE
??????????????????????????????????????????????????????????????????
??????????????????????????????????????????????????????????????????
c =2,997924 x108 m/s
h = 6,62606 x 10-34 kg m2 / sec
π = 3,141592...
2π = 6,283185
c2 =8,98755 x1016
μο = 4π x 10-7 H/m
εο=1/36π 109 F/m
√ μο/εο =zo =μο c
c = 1 / √ μο εο
c2 = 1 / μο εο
Ε = h f = hc / λ
Εo = m c2 = F λ
Fη = k q1 q2 / r2
c=120π /4π x10-7
zo=120π
c / h =fmax
c = fmax λmin
c = amax tmin
c = zo / μo
fmax=Vmax /λmin
Fmax=Emax /λmin
λmin = c2 /amax
amax = c2 / λmin
amax=λmin fmax2
Emax=Fmax λmin
Emax = fmax h
e=1,602176 x10-19
√2 = 1,4142135
0,452444 x1042
1,35639 x1050
1,23559 x1020
2,26873 x1023
h / c λmin = Mmax
Vmax = Mmax V / M
h c / λ = Mpl G / λ
λe / re 2π = c / Vκ
1rad / s ~ 0,1591 Hz
.
Relativity of time or energy? - The big traps of experience - the small explosions of indivisible space…
##### THE BIG EXPLOSION IN THE FIELD OF SCIENCE
The natural laws, which regulate the existence and the structure of material elements are not messages that come from their outside, from far and insubstantial way. These laws begin from the common " contact " that the all structural elements have with one and the same " empty " space, with one and common completed reality, with the same " communal " quantity of energy, finally from the common (invisible) their substance. The total quantity of energy is the same for the all things and the all structural elements takes shape with the same fluctuations of energy in one and same dynamic space.
The first publication in the world
#### THE MATHEMATIC MEETING OF PHILOSOPHY WITH THE PHYSICS AND THE SCIENCE FOREVER
• If we meet so much thoughtlessness, lack of ideas and initiative, unconcern for the clues, long-lasting tolerance in the biases, cowardice for the expression of our opinion, discriminatory application of knowledge with psychological and economically motives and bureaucratic mentality, if we meet all these in persons that teach the sciences and in researchers: Then, we should think more beyond limits of this question (about the role and reliability of the Science), how many more ineffectual and unreliable is the human thought for the problems in life and society!
If an opinion of a madman is correct, then it is correct. We don't downgrades and invalidates, in order to we select hears from a celebrated person and with some ceremonious preparations or because we want a frame with narrations for the curious and historical memoirs
WHY THE CENTRAL IDEA OF THE "BIG BANG" THEORY, IT IS THE BIGGEST FOOLISHNESS IN THE SCIENCE ! READ HERE
Interpretation & Math investigation
SHORT ANSWERS - BIG SURPRISES !
κοσμολογία, cosmology, in German: Kosmologie, in Italian and Portuguese: cosmologia, in France: cosmologie, in Spanish: cosmología, in Russian: космология, in Chinese: 宇宙论, in Hindi: ब्रह्मांड विज्ञान (), in Turkish: kozmoloji, in Swedish and Norwegian: kosmologi, in Arabic: الكوزمولوجيا علم الكونيات , in Hebrew: קוסמולוגיה
φιλοσοφία, philosophy, in German: Philosophie, in Italian and Portuguese: filosofia, in France: philosophie, in Spanish: filosofía, in Russian: философию, in Chinese: 哲学, in Hindi: दर्शन (tattvadnyaan), in Turkish: felsefe, in Swedish and Norwegian: filosofi, in Arabic: الفلسفة , in Hebrew: פילוסופיה
on 2016-05-23.
G = 6,6725 x10-11 m3 kg-1 s-2
M=E/c2 = h / f λ2
M = h f μο εο
V = S / t
p = M V = E / V
aκ = V2 / r
S = 1/2 a t2
g = G M / r2
F = M a
a = V2 /λ = λ f2
V =√GM / r
Fg = G M1 M2 / r2
G=c2 / Smax
zo=50G /μο
Mmax=Fmax / amin
c2 / λmin = Gmax
c2 = amax λmin
ω = 2 π f = V/ r
2π = T Vκ / r
f =ω / 2π =V/2π r
λmax=amin Tmax2
λmax = c2 /amin
amin = c / Tmax
amax = c / Tmin
c = h / M λ
M=m c / √c2- V2
L = M V r
0,73725 x10-50
cos 90° = 0
c G = 0,02
0,44929 x1019
9,10938 x10-31
1,67262 x10-27
Mpl2 =h c / G
F r2 = M1M2 G
G = Fλ2 / Mpl2
DH / 1Mpc = c / VH
parsec~3,086x1016 m
. | 1,405 | 4,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2016-50 | latest | en | 0.741586 |
https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_18 | 1,709,265,946,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00547.warc.gz | 105,156,348 | 10,855 | # 1982 AHSME Problems/Problem 18
## Problem 18
In the adjoining figure of a rectangular solid, $\angle DHG=45^\circ$ and $\angle FHB=60^\circ$. Find the cosine of $\angle BHD$.
$[asy] import three;defaultpen(linewidth(0.7)+fontsize(10)); currentprojection=orthographic(1/3+1/10,1-1/10,1/3); real r=sqrt(3); triple A=(0,0,r), B=(0,r,r), C=(1,r,r), D=(1,0,r), E=O, F=(0,r,0), G=(1,0,0), H=(1,r,0); draw(D--G--H--D--A--B--C--D--B--F--H--B^^C--H); draw(A--E^^G--E^^F--E, linetype("4 4")); label("A", A, N); label("B", B, dir(0)); label("C", C, N); label("D", D, W); label("E", E, NW); label("F", F, S); label("G", G, W); label("H", H, S); triple H45=(1,r-0.15,0.1), H60=(1-0.05, r, 0.07); label("45^\circ", H45, dir(125), fontsize(8)); label("60^\circ", H60, dir(25), fontsize(8));[/asy]$
$\text {(A)} \frac{\sqrt{3}}{6} \qquad \text {(B)} \frac{\sqrt{2}}{6} \qquad \text {(C)} \frac{\sqrt{6}}{3} \qquad \text{(D)}\frac{\sqrt{6}}{4}\qquad \text{(E)}\frac{\sqrt{6}-\sqrt{2}}{4}$
## Solution
WLOG, let $CD=1$.
Looking at square GHDC, we see that $\angle DHC=45$, which implies that $DC=CH=1$ and $DH=\sqrt{2}$
Taking each cross-section one at a time, we look at square DHFB. We obviously know that CHB is a $30$ degree angle, giving $BH=\frac{2\sqrt{3}}{3}$, and $BC=\frac{\sqrt{3}}{3}$.
Looking at square ABCD, we see that $BD^2=1+\frac{1}{3}\implies BD^2=\frac{4}{3}=\frac{2\sqrt{3}}{3}$.
We now look at triangle DBH, which has side lengths $BD=\frac{2\sqrt{3}}{3}$, $BH=\frac{2\sqrt{3}}{3}$ and $DH=\sqrt{2}$. Because $BD$ is opposite angle DHB, by the law of cosines, $\frac{4}{3}=\frac{4}{3}+2-2(\frac{2\sqrt{3}}{3})(\sqrt{2})\cos(\theta)\implies 2=\frac{4\sqrt{6}\cos(\theta)}{3} \implies \frac{2}{\frac{4\sqrt{6}}{3}}\implies \frac{6}{4\sqrt{6}} \implies \frac{\sqrt{6}}{4}$
Thus, the answer is $(\textbf{D})$ | 781 | 1,821 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-10 | latest | en | 0.449107 |
https://phys.libretexts.org/Bookshelves/University_Physics/Mechanics_and_Relativity_(Idema)/04%3A_Momentum/4.07%3A_Totally_Elastic_Collisions | 1,720,809,103,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00277.warc.gz | 371,812,734 | 30,019 | Skip to main content
# 4.7: Totally Elastic Collisions
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For a totally elastic collision, we can invoke both conservation of momentum and (by definition of a totally elastic collision) of kinetic energy. We also have an additional variable, as compared to the totally inelastic case, because in this case the objects do not stick together and thus get different end speeds. The two equations governing a totally elastic collision are:
$m_{1} v_{1, \mathrm{i}}+m_{2} v_{2, \mathrm{i}}=m_{1} v_{1, \mathrm{f}}+m_{2} v_{2, \mathrm{f}} \label{momentumcons}$
for momentum conservation, and
$\frac{1}{2} m_{1} v_{1, \mathrm{i}}^{2}+\frac{1}{2} m_{2} v_{2, \mathrm{i}}^{2}=\frac{1}{2} m_{1} v_{1, \mathrm{f}}^{2}+\frac{1}{2} m_{2} v_{2, \mathrm{f}}^{2} \label{kineticecons}$
for kinetic energy conservation.
When the collision occurs in one dimension, we can combine equations (\ref{momentumcons}) and (\ref{kineticecons}) to calculate the final velocities as functions of the initial ones. We first rewrite the two equations so that everything associated with particle 1 is on the left, and the terms for particle 2 are on the right:
$m_{1}\left(v_{1, \mathrm{i}}-v_{1, \mathrm{f}}\right)=m_{2}\left(v_{2, \mathrm{f}}-v_{2, \mathrm{i}}\right) \label{firstorder}$
and
$m_{1}\left(v_{1, \mathrm{i}}^{2}-v_{1, \mathrm{f}}^{2}\right)=m_{2}\left(v_{2, \mathrm{f}}^{2}-v_{2, \mathrm{i}}^{2}\right) \label{squared}$
We can expand the terms in parentheses in Equation (\ref{squared}), which gives:
$m_{1}\left(v_{1, \mathrm{i}}-v_{1, \mathrm{f}}\right)\left(v_{1, \mathrm{i}}+v_{1, \mathrm{f}}\right)=m_{2}\left(v_{2, \mathrm{f}}-v_{2, \mathrm{i}}\right)\left(v_{2, \mathrm{f}}+v_{2, \mathrm{i}}\right) \label{expanded}$
Dividing Equation (\ref{expanded}) by Equation (\ref{firstorder}), we get a relation between the velocities alone:
$v_{1, \mathrm{i}}+v_{1, \mathrm{f}}=v_{2, \mathrm{i}}+v_{2, \mathrm{f}} \label{pretty}$
From Equation (\ref{pretty}) we can isolate $$v_{2,f}$$ and substitute back in (\ref{firstorder}) to find $$v_{1,f}$$ in terms of the initial velocities:
$v_{1, \mathrm{f}}=\frac{m_{1}-m_{2}}{m_{1}+m_{2}} v_{1, \mathrm{i}}+2 \frac{m_{2}}{m_{1}+m_{2}} v_{2, \mathrm{i}} \label{v1final}$
Naturally, we could just as well have calculated $$v_{2,f}$$, the equation for which is just (\ref{v1final}) with the 1’s and 2’s swapped:
$v_{2, \mathrm{f}}=2 \frac{m_{1}}{m_{1}+m_{2}} v_{1, \mathrm{i}}+\frac{m_{2}-m_{1}}{m_{1}+m_{2}} v_{2, \mathrm{i}}$
We note that in the limit case that $$m_{2}>>m_{1}, v_{2}$$ is hardly affected, and $$v_{1, \mathrm{f}} \simeq-v_{1, \mathrm{i}}+2 v_{2, \mathrm{f}}$$.
This page titled 4.7: Totally Elastic Collisions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Timon Idema (TU Delft Open) via source content that was edited to the style and standards of the LibreTexts platform.
• Was this article helpful? | 2,755 | 7,257 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-30 | latest | en | 0.202846 |
https://uk.mathworks.com/matlabcentral/cody/problems/1305-creation-of-2d-sinc-surface/solutions/1202072 | 1,607,049,792,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141733120.84/warc/CC-MAIN-20201204010410-20201204040410-00122.warc.gz | 516,555,669 | 18,367 | Cody
Problem 1305. Creation of 2D Sinc Surface
Solution 1202072
Submitted on 1 Jun 2017 by George Pantazis
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Test Suite
Test Status Code Input and Output
1 Pass
nrc=65; xymax=3; freq=1; m=sinx_div_x(xymax,nrc,freq); %figure(3);imagesc(m) %figure(4);surf(m) xv=repmat(-xymax:2*xymax/(nrc-1):xymax,nrc,1); yv=xv'; mexp=zeros(nrc); for r=1:nrc for c=1:nrc Rv=sqrt(xv(r,c)^2+yv(r,c)^2); if Rv>eps mexp(r,c)=sin(Rv*pi*freq)/(Rv*pi*freq); else mexp(r,c)=1; end end % c end %r %figure(1);imagesc(mexp) %figure(2);surf(mexp) assert(~any(any(isnan(m)))) assert(max(max(abs(m-mexp)))<.01)
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2 Pass
nrc=127; xymax=3; freq=4; m=sinx_div_x(xymax,nrc,freq); %figure(3);imagesc(m) %figure(4);surf(m) xv=repmat(-xymax:2*xymax/(nrc-1):xymax,nrc,1); yv=xv'; mexp=zeros(nrc); for r=1:nrc for c=1:nrc Rv=sqrt(xv(r,c)^2+yv(r,c)^2); if Rv>eps mexp(r,c)=sin(Rv*pi*freq)/(Rv*pi*freq); else mexp(r,c)=1; end end % c end %r %figure(1);imagesc(mexp) %figure(2);surf(mexp) assert(~any(any(isnan(m)))) assert(max(max(abs(m-mexp)))<.01)
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3 Pass
nrc=96; xymax=16; freq=0.5; m=sinx_div_x(xymax,nrc,freq); %figure(3);imagesc(m) %figure(4);surf(m) xv=repmat(-xymax:2*xymax/(nrc-1):xymax,nrc,1); yv=xv'; mexp=zeros(nrc); for r=1:nrc for c=1:nrc Rv=sqrt(xv(r,c)^2+yv(r,c)^2); if Rv>eps mexp(r,c)=sin(Rv*pi*freq)/(Rv*pi*freq); else mexp(r,c)=1; end end % c end %r %figure(1);imagesc(mexp) %figure(2);surf(mexp) assert(~any(any(isnan(m)))) assert(max(max(abs(m-mexp)))<.01)
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Start Hunting! | 15,548 | 26,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-50 | latest | en | 0.378411 |
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11-7.
Howie and Steve are making cookies for themselves and some friends. The recipe they are using will make $48$ cookies, but they only want to make $16$ cookies. They have no trouble reducing the amounts of flour and sugar, but the original recipe calls for $1\frac{3}{4}$ cups of butter. Help Howie and Steve determine how much butter they will need.
Per $48$ cookies there needs to be $1\frac{3}{4}$ cups of butter.
Set up a proportion.
$\frac{48 \text{ cookies}}{1\frac{3}{4}\text{ cups of butter}}=\frac{16\text{ cookies}}{x\text{ cups of butter}}$
Solve for $x$.
$\frac{7}{12}$ of a cup | 199 | 655 | {"found_math": true, "script_math_tex": 8, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-33 | latest | en | 0.844536 |
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Spectrometer- Determination of Cauchy's constants
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## Aim:
To determine the Cauchy’s constant of the given prism.
## Apparatus:
Spectrometer, prism, prism clamp, Magnifying glass, mercury vapor lamp, etc.
## Theory:
Cauchy's equation is an empirical relationship between the refractive index and wavelength of light for a particular transparent material. It is named for the mathematician Augustin-Louis Cauchy, who defined it in 1836.
Augustin-Louis Cauchy
(1789-1857)
The most general form of Cauchy's equation is
. . . . .. . . (1)
where n is the refractive index, λ is the wavelength, B, C, D, etc., are coefficients that can be determined for a material by fitting the equation to measured refractive indices at known wavelengths.
The refractive index n of the material of the prism for a wavelength λ is given by.
. . . . .. .. ..(2)
Where A and B are called Cauchy’s constants for the prism.
If the refractive indices n1 and n2 for any two known wavelength λ1 and λ2 are determined by a spectrometer, the Cauchy’s constants A and B can be calculated from the above equation.
*The theory of light-matter interaction on which Cauchy based this equation was later found to be incorrect. In particular, the equation is only valid for regions of normal dispersion in the visible wavelength region. In the infrared, the equation becomes inaccurate, and it cannot represent regions of anomalous dispersion.
Cite this Simulator:
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Copyright @ 2024 Under the NME ICT initiative of MHRD | 447 | 1,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-33 | latest | en | 0.875024 |
http://www.instructables.com/id/William-Jones-Pi-Pie-mathematically-infused/ | 1,419,079,190,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802769844.62/warc/CC-MAIN-20141217075249-00131-ip-10-231-17-201.ec2.internal.warc.gz | 603,008,928 | 35,040 | # William Jones' Pi Pie (mathematically infused)
Here i show to properly make a William Jones' Pi pie. All of the steps and materials are provided in order to make this truly mathematical pie. All of the bold digits are the first 30 numbers, in chronological order, of pi. Enjoy this instructable more for the entertainment purposes rather than the practicality of what we are creating.
3.14159265358979323846264383279
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## Step 1: Materials
(3 point)s before we begin: The name of our Pi Pie comes from the name of the man who proposed the use of the greek symbol for pi as a representation of the ratio of the circumference of a circle to its diameter. Secondly, this pie is much more pi related than pie related. That leads to our third point- please DON'T actually attempt to make William Jones' Pi Pie because its probably dangerous, so enjoy this instructable for its mathematical goodness rather than its baked goodness.
On that note, it is time to gather your materials:
(1) Pie tin
(4) Tablespoons of sugar
(5) Oreos
(9) Slices/pieces of a certain fruit (we used apple slices)
(2) Graphing calculators
(6) Drops of math extract (purchased from our local office store)
(5) Pencils
(3) Protractors
(5) Tablespoons of butter
## Step 2: Prepping the PI
Place the pie crust so that it fits the pan that you are using. You don't necessarily have to use a circular pie tin for this because the mathematics infused in the recipe will alter the tin so that is circular because nature says you can't have pi without a nice circle. If you do choose to use your own pie crust then roll it NO MORE than (8) or (9) times.
mistyp2 years ago
Cute :-)
2 years ago
Also, good luck on the contest!
w3-m4k3-stuff (author) mistyp2 years ago
Thank you!
mamalaoshi2 years ago | 476 | 1,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2014-52 | longest | en | 0.837269 |
https://www.riddles.com/logic-puzzles?page=10 | 1,660,492,460,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572043.2/warc/CC-MAIN-20220814143522-20220814173522-00756.warc.gz | 845,488,120 | 42,510 | Riddle: If it is 1,800 kilometers to America, 1,200 kilometers to Japan, 2,400 kilometers to New Zealand, and 1,400 kilometers to Brazil- How far is Morocco?
Answer: The answer is 1,700 kilometers, as vowels in the countries' names are worth 300 kilometers and the consonats are worth 200 kilometers.
Riddle: I like indigo but not blue, I like onions but not turnips, I like forms but not shapes. According to the same rule, do I like tomatoes or avocados?
Riddle: There are three guards and three prisoners who need to cross a river. Their boat only holds two people at a time, and the number of prisoners must NEVER be allowed to outnumber the number of guards on either side of the river; otherwise, the prisoners will overpower the guards and, well, the story will come to an abrupt end. List each of the trips that need to be made and who is in the boat, and who is on each of the riverbanks during each trip. How many trips it will take to safely transport all of the guards and prisoners across the river? | 239 | 1,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-33 | latest | en | 0.948304 |
https://whatisconvert.com/387-grams-in-long-tons | 1,680,426,827,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950422.77/warc/CC-MAIN-20230402074255-20230402104255-00316.warc.gz | 682,395,591 | 7,303 | ## Convert 387 Grams to Long Tons
To calculate 387 Grams to the corresponding value in Long Tons, multiply the quantity in Grams by 9.8420652761106E-7 (conversion factor). In this case we should multiply 387 Grams by 9.8420652761106E-7 to get the equivalent result in Long Tons:
387 Grams x 9.8420652761106E-7 = 0.00038088792618548 Long Tons
387 Grams is equivalent to 0.00038088792618548 Long Tons.
## How to convert from Grams to Long Tons
The conversion factor from Grams to Long Tons is 9.8420652761106E-7. To find out how many Grams in Long Tons, multiply by the conversion factor or use the Mass converter above. Three hundred eighty-seven Grams is equivalent to zero point zero zero zero three eight one Long Tons.
## Definition of Gram
The gram (alternative spelling: gramme; SI unit symbol: g) is a metric system unit of mass. A gram is defined as one one-thousandth of the SI base unit, the kilogram, or 1×10−3 kg, which itself is now defined, not in terms of grams, but as being equal to the mass of a physical prototype of a specific alloy kept locked up and preserved by the International Bureau of Weights and Measures.
## Definition of Long Ton
A long ton is defined as exactly 2,240 pounds. The long ton arises from the traditional British measurement system: A long ton is 20 cwt, each of which is 8 stone (1 stone = 14 pounds). Thus a long ton is 20 × 8 × 14 lb = 2,240 lb. Long ton, also known as the imperial ton or displacement ton is the name for the unit called the "ton" in the avoirdupois or Imperial system of measurements standardised in the thirteenth century that is used in the United Kingdom
## Using the Grams to Long Tons converter you can get answers to questions like the following:
• How many Long Tons are in 387 Grams?
• 387 Grams is equal to how many Long Tons?
• How to convert 387 Grams to Long Tons?
• How many is 387 Grams in Long Tons?
• What is 387 Grams in Long Tons?
• How much is 387 Grams in Long Tons?
• How many uk ton are in 387 g?
• 387 g is equal to how many uk ton?
• How to convert 387 g to uk ton?
• How many is 387 g in uk ton?
• What is 387 g in uk ton?
• How much is 387 g in uk ton? | 598 | 2,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-14 | latest | en | 0.882514 |
https://www.wyzant.com/resources/blogs/234736/how_to_explain_why_division_by_zero_is_undefined | 1,444,150,028,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736678861.8/warc/CC-MAIN-20151001215758-00006-ip-10-137-6-227.ec2.internal.warc.gz | 1,240,138,958 | 12,219 | Search 81,550 tutors
# How to explain why division by zero is undefined.
If we try to divide 2 by 0, why isn't that just infinity? Well, infinity isn't really a number. It's just a concept and it represents a place that you can't ever really get to. Math deals with numbers, and it just can't handle this.
Here's a way to think about it. If I divide 20 by 5, I get 4 as the answer. Turning that around, it says that I would need 4 groups of size 5 to make 20. So what would 20 divided by zero mean? How many groups of size zero would I need to make 20? There is no number of groups that would ever add up to 20. Even if you could add up an infinite number of zeros, the total would still be zero. You just can't get to 20! The answer is therefore undefined.
This is by no means mathematically rigorous, but it gives one an intuitive picture of why you just can't do that.
I like that explanation! I have a middle-school math student who is often curious about things and comes up with questions about things like that. I'll have to remember that example in case he ever asks me about that sometime.
Thanks for the comment! I've seen this question addressed once in a while, and the answer was always couched in deep mathematical terms that didn't give you anything that you could intuitively see. I came up with this when tutoring a student about fractions. It was a simple twist to make one particular explanation about fractions apply to division by zero. (I saved it for more advanced students. She didn't need to be worried about this while trying to understand fractions.)
Hi Gene. Here is another point of view:
Let's assume, that 20 divide by "0" is some, different from zero, number "n"
20 / 0 = n
then
n * 0 = 20
but we know that n * 0 = 0
Hi Nataliya,
I've seen a lot of your work in Answers. You're one of a select few who I really admire for your math skills. For those of us for whom a mathematical proof clearly demonstrates a point, your method is great, but I've long looked for something that would make sense to a youngster who hasn't yet gotten to the point where a valid proof really "feels" like a proof. I was well into college math before I realized that seeing a derivation was really a proof to me at the "gut feel" level. I think what I came up with is something that accomplishes what I wanted. Thanks for your comment!
\$30p/h
Gene G.
You can do it! I'll show you how.
400+ hours | 599 | 2,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2015-40 | longest | en | 0.983824 |
https://math.stackexchange.com/questions/859334/how-to-find-the-solution-of-a-quadratic-equation-with-complex-coefficients | 1,576,158,851,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540543850.90/warc/CC-MAIN-20191212130009-20191212154009-00077.warc.gz | 455,237,654 | 31,427 | # How to find the solution of a quadratic equation with complex coefficients?
I know how to find the solution for a quadratic equation with real coefficients. But if the coefficient changes to complex numbers then what is the change in the solution? Want an example of such equation with solution.
• Same as usual formula. Just need to be careful with the branches of square root. – copper.hat Jul 7 '14 at 18:57
• But how to find the two square roots of a complex number ?? – Bob Jul 7 '14 at 19:00
• "How to find the two square roots of a complex number" -- see the link in my answer below. – mweiss Jul 7 '14 at 19:01
• Write the complex number in exponential form (for example, $i = e^{i {\pi \over 2}}$, hence $\sqrt{i} = \pm e^{i {\pi \over 4}} = \pm {1+i \over \sqrt{2}}$). – copper.hat Jul 7 '14 at 19:06
It's no different. The quadratic formula works regardless of whether the coefficients are real or complex.
Consider the example $$(3+i)x^2 + (2-i)x + (5+2i) = 0$$
$$x = \frac{-(2-i) \pm \sqrt{ (2-i)^2-4(3+i)(5+2i) } }{2(3+i)}$$
Simplifying this is kind of a pain, of course. Under the radical you have to multiply everything out and combine terms. Eventually you get the radical into the form $\sqrt{M+Ni}$ where $M$ and $N$ are some constants -- in this example, they will be integers. Then the question is, how do you simplify the square root of a complex number? (For that, see How do I get the square root of a complex number?). | 414 | 1,450 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2019-51 | latest | en | 0.846167 |
https://www.bartleby.com/questions-and-answers/let-a-b-be-coprime-integers.-prove-that-every-integer-x-greater-ab-a-b-can-be-written-as-na-mb-where/4f193b51-af44-406e-ac59-af40f0678f21 | 1,579,966,313,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251672537.90/warc/CC-MAIN-20200125131641-20200125160641-00188.warc.gz | 772,231,809 | 23,029 | # Let a, b be coprime integers. Prove that every integer x > ab- a- b canbe written as na mb where n, m are non-negative integers. Prove thatab -a - b cannot be expressed in this form
Question
12 views
Abstract Algebra
check_circle
Step 1
To find non-negative representations under the given conditions
Step 2
In the following, a and b are coprime positive integers; Note that , by Euclidean algorithm, ANY integer x is an integral linear combination na+mb. The point is we seek sufficient condtions on the integer x ( namely x>ab-a-b) na+nb=x with both n and m non-negative,For instance , consider a =10, b =3, clearly relatively prime positive integers . Now x=10+3=13 is a non-negative integral linear combination of a and b , but 14,16, 17 are not. The theorem asserts that whenver x > 10*3-10-3=17,x can be expressed as 10n+3m , with n and m non-negative integers.
Step 3
General solution of (1) is derived , using the ...
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,194 | 3,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-47 | latest | en | 0.830409 |
http://energia.nu/reference/en/language/variables/data-types/float/ | 1,537,729,348,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159570.46/warc/CC-MAIN-20180923173457-20180923193857-00333.warc.gz | 82,243,068 | 3,129 | ### Description
Datatype for floating-point numbers, a number that has a decimal point. Floating-point numbers are often used to approximate analog and continuous values because they have greater resolution than integers. Floating-point numbers can be as large as 3.4028235E+38 and as low as -3.4028235E+38. They are stored as 32 bits (4 bytes) of information.
Floats have only 6-7 decimal digits of precision. That means the total number of digits, not the number to the right of the decimal point. Unlike other platforms, where you can get more precision by using a double (e.g. up to 15 digits), on the Arduino, double is the same size as float.
Floating point numbers are not exact, and may yield strange results when compared. For example 6.0 / 3.0 may not equal 2.0. You should instead check that the absolute value of the difference between the numbers is less than some small number.
Floating point math is also much slower than integer math in performing calculations, so should be avoided if, for example, a loop has to run at top speed for a critical timing function. Programmers often go to some lengths to convert floating point calculations to integer math to increase speed.
If doing math with floats, you need to add a decimal point, otherwise it will be treated as an int. See the Floating point constants page for details.
### Syntax
`float var=val;`
`var` - your float variable name `val` - the value you assign to that variable
### Example Code
`````` float myfloat;
float sensorCalbrate = 1.117;
int x;
int y;
float z;
x = 1;
y = x / 2; // y now contains 0, ints can't hold fractions
z = (float)x / 2.0; // z now contains .5 (you have to use 2.0, not 2)`````` | 416 | 1,706 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-39 | longest | en | 0.890454 |
https://www.o.vg/unit/fuel-efficiency-volume/joule-milliliter-to-calorie-IT-liter.php | 1,725,788,866,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650976.41/warc/CC-MAIN-20240908083737-20240908113737-00869.warc.gz | 896,901,861 | 11,053 | Convert Joule/Milliliter to Calorie (IT)/Liter | cal/L to J/mL
document.write(document.title);
J/mL to cal/L Converter
From J/mL to cal/L: 1 J/mL = 238.846 cal/L;
From cal/L to J/mL: 1 cal/L = 0.0041867981879537 J/mL;
How to Convert Joule/Milliliter to Calorie (IT)/Liter?
As we know One J/mL is equal to 238.846 cal/L (1 J/mL = 238.846 cal/L).
To convert Joule/Milliliter to Calorie (IT)/Liter, multiply your J/mL figure by 238.846.
Example : convert 25 J/mL to cal/L:
25 J/mL = 25 × 238.846 cal/L = cal/L
To convert Calorie (IT)/Liter to Joule/Milliliter, divide your cal/L figure by 238.846.
Example : convert 25 cal/L to J/mL:
25 cal/L = 25 ÷ 238.846 J/mL = J/mL
How to Convert Calorie (IT)/Liter to Joule/Milliliter?
As we know One cal/L is equal to 0.0041867981879537 J/mL (1 cal/L = 0.0041867981879537 J/mL).
To convert Calorie (IT)/Liter to Joule/Milliliter, multiply your cal/L figure by 0.0041867981879537.
Example : convert 45 cal/L to J/mL:
45 cal/L = 45 × 0.0041867981879537 J/mL = J/mL
To convert Joule/Milliliter to Calorie (IT)/Liter, divide your J/mL figure by 0.0041867981879537.
Example : convert 45 J/mL to cal/L:
45 J/mL = 45 ÷ 0.0041867981879537 cal/L = cal/L
Convert Joule/Milliliter or Calorie (IT)/Liter to Other Fuel Efficiency - Volume Units
Joule/Milliliter Conversion Table
J/mL to J/m³ 1 J/mL = 1000000 J/m³ J/mL to J/km³ 1 J/mL = 1.0E+15 J/km³ J/mL to J/dm³ 1 J/mL = 1000 J/dm³ J/mL to J/cm³ 1 J/mL = 1 J/cm³ J/mL to J/mm³ 1 J/mL = 0.001 J/mm³ J/mL to J/L 1 J/mL = 1000 J/L J/mL to J/µL 1 J/mL = 0.001 J/µL J/mL to kJ/m³ 1 J/mL = 1000 kJ/m³ J/mL to kJ/km³ 1 J/mL = 1000000000000 kJ/km³ J/mL to kJ/dm³ 1 J/mL = 1 kJ/dm³ J/mL to kJ/cm³ 1 J/mL = 0.001 kJ/cm³ J/mL to kJ/L 1 J/mL = 1 kJ/L J/mL to mJ/mL 1 J/mL = 1000 mJ/mL J/mL to mJ/µL 1 J/mL = 1 mJ/µL J/mL to µJ/mL 1 J/mL = 1000000 µJ/mL J/mL to µJ/µL 1 J/mL = 1000 µJ/µL J/mL to nJ/µL 1 J/mL = 1000000 nJ/µL J/mL to erg/mL 1 J/mL = 10000000 erg/mL J/mL to erg/µL 1 J/mL = 10000 erg/µL J/mL to kW·h/m³ 1 J/mL = 16.666666666667 kW·h/m³ J/mL to kW·h/L 1 J/mL = 0.016666666666667 kW·h/L J/mL to kW·s/m³ 1 J/mL = 1000 kW·s/m³ J/mL to kW·s/L 1 J/mL = 1 kW·s/L J/mL to W·h/m³ 1 J/mL = 16666.666666667 W·h/m³ J/mL to W·h/km³ 1 J/mL = 16666666.666667 W·h/km³ J/mL to W·h/L 1 J/mL = 16.666666666667 W·h/L J/mL to W·s/m³ 1 J/mL = 1000000 W·s/m³ J/mL to W·s/L 1 J/mL = 1000 W·s/L J/mL to N·m/m³ 1 J/mL = 1000000 N·m/m³ J/mL to N·m/L 1 J/mL = 1000 N·m/L J/mL to N·m/mL 1 J/mL = 1 N·m/mL J/mL to W·s/in³ 1 J/mL = 16.4 W·s/in³ J/mL to N·m/yd³ 1 J/mL = 764554.858 N·m/yd³ J/mL to N·m/ft³ 1 J/mL = 28316.847 N·m/ft³ J/mL to J/yd³ 1 J/mL = 765158.4 J/yd³ J/mL to J/ft³ 1 J/mL = 28339.2 J/ft³ J/mL to J/in³ 1 J/mL = 16.4 J/in³ J/mL to kJ/yd³ 1 J/mL = 765.1584 kJ/yd³ J/mL to kJ/ft³ 1 J/mL = 28.3392 kJ/ft³ J/mL to kJ/in³ 1 J/mL = 0.0164 kJ/in³ J/mL to kW·h/yd³ 1 J/mL = 12.75264 kW·h/yd³
Joule/Milliliter Conversion Table
J/mL to kW·h/ft³ 1 J/mL = 0.47232 kW·h/ft³ J/mL to kW·h/in³ 1 J/mL = 0.00027333333333333 kW·h/in³ J/mL to kW·s/yd³ 1 J/mL = 765.1584 kW·s/yd³ J/mL to kW·s/ft³ 1 J/mL = 28.3392 kW·s/ft³ J/mL to kW·s/in³ 1 J/mL = 0.0164 kW·s/in³ J/mL to W·h/yd³ 1 J/mL = 12752.64 W·h/yd³ J/mL to W·h/ft³ 1 J/mL = 472.32 W·h/ft³ J/mL to W·h/in³ 1 J/mL = 0.27333333333333 W·h/in³ J/mL to W·s/yd³ 1 J/mL = 765158.4 W·s/yd³ J/mL to W·s/ft³ 1 J/mL = 28339.2 W·s/ft³ J/mL to cal (th)/yd³ 1 J/mL = 182732.997 cal (th)/yd³ J/mL to kcal (th)/yd³ 1 J/mL = 182.733 kcal (th)/yd³ J/mL to hp·h/m³ 1 J/mL = 0.373 hp·h/m³ J/mL to hp·h/L 1 J/mL = 0.000373 hp·h/L J/mL to hp·h/yd³ 1 J/mL = 0.285 hp·h/yd³ J/mL to hp·h/ft³ 1 J/mL = 0.0106 hp·h/ft³ J/mL to kcal/L 1 J/mL = 0.239 kcal/L J/mL to kcal/yd³ 1 J/mL = 182.611 kcal/yd³ J/mL to kcal (th)/L 1 J/mL = 0.23923444976077 kcal (th)/L J/mL to cal/L 1 J/mL = 238.846 cal/L J/mL to cal/ft³ 1 J/mL = 6763.363 cal/ft³ J/mL to cal (th)/m³ 1 J/mL = 239005.73613767 cal (th)/m³ J/mL to cal (th)/L 1 J/mL = 239.00573613767 cal (th)/L J/mL to Btu/L 1 J/mL = 0.948 Btu/L J/mL to Btu/yd³ 1 J/mL = 724.658 Btu/yd³ J/mL to Btu (th)/L 1 J/mL = 0.948 Btu (th)/L J/mL to Btu (th)/yd³ 1 J/mL = 725.144 Btu (th)/yd³ J/mL to J/gal (US) 1 J/mL = 3785.4120857494 J/gal (US) J/mL to J/gal (UK) 1 J/mL = 4546.0899648474 J/gal (UK) J/mL to W·h/gal (US) 1 J/mL = 63.090201421197 W·h/gal (US) J/mL to W·h/gal (UK) 1 J/mL = 75.768166080789 W·h/gal (UK) J/mL to kcal/gal (US) 1 J/mL = 0.90413014343669 kcal/gal (US) J/mL to kcal (th)/gal (US) 1 J/mL = 0.90473520210557 kcal (th)/gal (US) J/mL to kcal/gal (UK) 1 J/mL = 1.0858149338032 kcal/gal (UK) J/mL to kcal (th)/gal (UK) 1 J/mL = 1.0865415785964 kcal (th)/gal (UK) J/mL to hp·h/gal (US) 1 J/mL = 0.0014100892296977 hp·h/gal (US) J/mL to hp·h/gal (UK) 1 J/mL = 0.0016934464073308 hp·h/gal (UK) Created @ o.vg Free Unit Converters
FAQ
What is 9 Joule/Milliliter in Calorie (IT)/Liter?
cal/L. Since one J/mL equals 238.846 cal/L, 9 J/mL in cal/L will be cal/L.
How many Calorie (IT)/Liter are in a Joule/Milliliter?
There are 238.846 cal/L in one J/mL. In turn, one cal/L is equal to 0.0041867981879537 J/mL.
How many J/mL is equal to 1 cal/L?
1 cal/L is approximately equal to 0.0041867981879537 J/mL.
What is the J/mL value of 8 cal/L?
The Joule/Milliliter value of 8 cal/L is J/mL. (i.e.,) 8 x 0.0041867981879537 = J/mL.
J/mL to cal/L converter in batch
Cite this Converter, Content or Page as:
"" at https://www.o.vg/unit/fuel-efficiency-volume/joule-milliliter-to-calorie-IT-liter.php from www.o.vg Inc,09/08/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters | 2,506 | 5,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-38 | latest | en | 0.270813 |
http://tasks.illustrativemathematics.org/content-standards/K/CC/B/tasks/1143 | 1,695,865,236,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510334.9/warc/CC-MAIN-20230927235044-20230928025044-00383.warc.gz | 45,308,947 | 9,503 | # Number Rods
Alignments to Content Standards: K.CC.B
#### Materials
• Unifix cubes or snap cubes, composed into rods with 1-10 cubes (or any counting sequence the class is currently working on within 20)
If the chosen number range includes numbers greater then 10, then make the rods using two colors so that the set of 10 in easily identified (see the image in the solution).
• A number line for the chosen number range
#### Action
The students count the number of unifix cubes then match the rod up to the correct number of the number line. Students can work in pairs or on their own.
## IM Commentary
This activity gives students practice counting for meaning. This task also allows students to see the size of the rod grow as the number gets larger.
One variation of this game is to give students a tub of unifix cubes and have them build the rods. However, this version is for students who are proficient counters. Students whose counting skill are still emerging will do better with the number rods already made for them.
As an extension, students can be asked to figure out why 2 colors were used to represent the numbers larger than 10. Also students can be asked what they notice about the cubes as you go from one number to the next (up or down). This could lead to a good discussion on one more and one less.
Materials Note: Very long rods of unifix cubes can break easily. Check to make sure each rod is intact before students start the game. Also instruct students to handle the rods carefully if the rods are longer than ten cubes. If the teacher does not have access to unifix cubes or snap cubes, or colored number strips could be made on heavy-duty paper.
To see an annotated version of this and other Illustrative Mathematics tasks as well as other Common Core aligned resources, visit Achieve the Core.
## Solution
Students will match the rods to the number line so that each rod is above its number. | 406 | 1,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2023-40 | latest | en | 0.947634 |
bookwormhomeschool.com | 1,656,880,953,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104249664.70/warc/CC-MAIN-20220703195118-20220703225118-00182.warc.gz | 184,899,967 | 55,764 | # Mathematics Education Month: What We’re Using for Math
March had National Pi Day, but April is Mathematics Education Month. I’ve done several write-ups about math including this post on math manipulatives and a Pi Day post for Dandelion Moms.
Around here, we try to do math every day, although we do mix it up a bit. For example, last week we did one or two worksheets out of our Saxon Math 1st grade workbooks, he finished the last few problems in his 2nd grade Lakeshore Common Core Workbooks, he started his 3rd grade Common Core workbook, and we spent a day practicing counting money (I’d give him different denominations and he had to see how much it all added up to.) We got the common core workbooks because we wanted to see how it was different from what we were already doing, and I chose different grades because I wanted to see what he already knows and what we need to start working on. He actually really enjoys doing these workbooks, minus the in-depth explanation of how he found his answer.
Since we started adding Usborne books to our collection, we have found several new books and tools that make math fun to learn and practice. We love the First Illustrated Math Dictionary. The illustrations are so playful that my 3-year old loves to sit down and read it. She practices counting and names all the shapes she finds. The book covers everything in a fun and kid-friendly way from counting from 1-100, how to use a calculator, lines of symmetry, number lines, counting in groups and so much more. We love it. It’s perfect for pre-k and up.
The next step up is the Illustrated Elementary Math Dictionary. This book has the same fun illustrations but adds more difficult mathematics to the mix. This is a great resource for students, but is also super-helpful for parents! This one gives you the tools you need to understand the math your kids are bringing home, and how to help them work through the problems.
We also love the Wipe-Clean activity books. The repetition is great for practice and the kids use their books over and over again. Two of our wipe-clean books are two years old and we still use them on a regular basis. | 465 | 2,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-27 | longest | en | 0.975493 |
https://us.metamath.org/mpeuni/tgrpbase.html | 1,716,850,821,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00450.warc.gz | 521,257,686 | 6,118 | Mathbox for Norm Megill < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > tgrpbase Structured version Visualization version GIF version
Theorem tgrpbase 37947
Description: The base set of the translation group is the set of all translations (for a fiducial co-atom 𝑊). (Contributed by NM, 5-Jun-2013.)
Hypotheses
Ref Expression
tgrpset.h 𝐻 = (LHyp‘𝐾)
tgrpset.t 𝑇 = ((LTrn‘𝐾)‘𝑊)
tgrpset.g 𝐺 = ((TGrp‘𝐾)‘𝑊)
tgrp.c 𝐶 = (Base‘𝐺)
Assertion
Ref Expression
tgrpbase ((𝐾𝑉𝑊𝐻) → 𝐶 = 𝑇)
Proof of Theorem tgrpbase
Dummy variables 𝑓 𝑔 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 tgrpset.h . . . 4 𝐻 = (LHyp‘𝐾)
2 tgrpset.t . . . 4 𝑇 = ((LTrn‘𝐾)‘𝑊)
3 tgrpset.g . . . 4 𝐺 = ((TGrp‘𝐾)‘𝑊)
41, 2, 3tgrpset 37946 . . 3 ((𝐾𝑉𝑊𝐻) → 𝐺 = {⟨(Base‘ndx), 𝑇⟩, ⟨(+g‘ndx), (𝑓𝑇, 𝑔𝑇 ↦ (𝑓𝑔))⟩})
54fveq2d 6657 . 2 ((𝐾𝑉𝑊𝐻) → (Base‘𝐺) = (Base‘{⟨(Base‘ndx), 𝑇⟩, ⟨(+g‘ndx), (𝑓𝑇, 𝑔𝑇 ↦ (𝑓𝑔))⟩}))
6 tgrp.c . 2 𝐶 = (Base‘𝐺)
72fvexi 6667 . . 3 𝑇 ∈ V
8 eqid 2824 . . . 4 {⟨(Base‘ndx), 𝑇⟩, ⟨(+g‘ndx), (𝑓𝑇, 𝑔𝑇 ↦ (𝑓𝑔))⟩} = {⟨(Base‘ndx), 𝑇⟩, ⟨(+g‘ndx), (𝑓𝑇, 𝑔𝑇 ↦ (𝑓𝑔))⟩}
98grpbase 16601 . . 3 (𝑇 ∈ V → 𝑇 = (Base‘{⟨(Base‘ndx), 𝑇⟩, ⟨(+g‘ndx), (𝑓𝑇, 𝑔𝑇 ↦ (𝑓𝑔))⟩}))
107, 9ax-mp 5 . 2 𝑇 = (Base‘{⟨(Base‘ndx), 𝑇⟩, ⟨(+g‘ndx), (𝑓𝑇, 𝑔𝑇 ↦ (𝑓𝑔))⟩})
115, 6, 103eqtr4g 2884 1 ((𝐾𝑉𝑊𝐻) → 𝐶 = 𝑇)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 399 = wceq 1538 ∈ wcel 2115 Vcvv 3479 {cpr 4550 ⟨cop 4554 ∘ ccom 5542 ‘cfv 6338 ∈ cmpo 7142 ndxcnx 16471 Basecbs 16474 +gcplusg 16556 LHypclh 37185 LTrncltrn 37302 TGrpctgrp 37943 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1912 ax-6 1971 ax-7 2016 ax-8 2117 ax-9 2125 ax-10 2146 ax-11 2162 ax-12 2179 ax-ext 2796 ax-rep 5173 ax-sep 5186 ax-nul 5193 ax-pow 5249 ax-pr 5313 ax-un 7446 ax-cnex 10580 ax-resscn 10581 ax-1cn 10582 ax-icn 10583 ax-addcl 10584 ax-addrcl 10585 ax-mulcl 10586 ax-mulrcl 10587 ax-mulcom 10588 ax-addass 10589 ax-mulass 10590 ax-distr 10591 ax-i2m1 10592 ax-1ne0 10593 ax-1rid 10594 ax-rnegex 10595 ax-rrecex 10596 ax-cnre 10597 ax-pre-lttri 10598 ax-pre-lttrn 10599 ax-pre-ltadd 10600 ax-pre-mulgt0 10601 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2071 df-mo 2624 df-eu 2655 df-clab 2803 df-cleq 2817 df-clel 2896 df-nfc 2964 df-ne 3014 df-nel 3118 df-ral 3137 df-rex 3138 df-reu 3139 df-rab 3141 df-v 3481 df-sbc 3758 df-csb 3866 df-dif 3921 df-un 3923 df-in 3925 df-ss 3935 df-pss 3937 df-nul 4275 df-if 4449 df-pw 4522 df-sn 4549 df-pr 4551 df-tp 4553 df-op 4555 df-uni 4822 df-int 4860 df-iun 4904 df-br 5050 df-opab 5112 df-mpt 5130 df-tr 5156 df-id 5443 df-eprel 5448 df-po 5457 df-so 5458 df-fr 5497 df-we 5499 df-xp 5544 df-rel 5545 df-cnv 5546 df-co 5547 df-dm 5548 df-rn 5549 df-res 5550 df-ima 5551 df-pred 6131 df-ord 6177 df-on 6178 df-lim 6179 df-suc 6180 df-iota 6297 df-fun 6340 df-fn 6341 df-f 6342 df-f1 6343 df-fo 6344 df-f1o 6345 df-fv 6346 df-riota 7098 df-ov 7143 df-oprab 7144 df-mpo 7145 df-om 7566 df-1st 7674 df-2nd 7675 df-wrecs 7932 df-recs 7993 df-rdg 8031 df-1o 8087 df-oadd 8091 df-er 8274 df-en 8495 df-dom 8496 df-sdom 8497 df-fin 8498 df-pnf 10664 df-mnf 10665 df-xr 10666 df-ltxr 10667 df-le 10668 df-sub 10859 df-neg 10860 df-nn 11626 df-2 11688 df-n0 11886 df-z 11970 df-uz 12232 df-fz 12886 df-struct 16476 df-ndx 16477 df-slot 16478 df-base 16480 df-plusg 16569 df-tgrp 37944 This theorem is referenced by: tgrpgrplem 37950 tgrpabl 37952
Copyright terms: Public domain W3C validator | 2,177 | 3,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-22 | latest | en | 0.110131 |
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# how to position my world around my player
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I am currently making a 2D Zelda type game and I have come quite far with it. I am trying to add a feature to my map editor that allows the user to set the start position of the player on the map. I am having problems trying to set the world position so that black edges don''t appear around the screen when the map is loaded and the player is put on the screen. Example A 36x36 (tiles each 32x32 pixels in size) map with a start position at tile 35,35. How can I make it so the bottom tiles are drawn at the bottom of the screen and not at the top with a bunch of blackness below? I am not neccessarily looking for some source code to do this(would be helpful) but some insight on the algorithm involved in making this happen. My scrolling function does bounds checking to prevent scrolling beyond the edges of the map. I thought I may use this to scroll the map to the start position when the map is loaded but I thought there could be a better way. Does anyone have any ideas that may help me?
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think of your screen as a camera looking down on the map through a window centered at the player position as he moves around the world.
So, the world will be 1152 x 1152 pixels.
The screen is a window made of (x, y)-(x+640, y+480), or whatever your display resolution is. So, to avoid borders, x >= 0, y >= 0, and x < 1152-640, y < 1152-480.
(x, y) will be the coordinates on the map, and you move (x, y) so that (x+320, y+240) points to your player position on the map.
the player is at pos xplayer, yplayer on the map.
so if the player is centered on the display, the window start position is x = xplayer-320, y = yplayer-240
and also, you have the constrains on x and y.
x >= 0 && x < (1152-640)
y >= 0 && y < (1152-480)
so there you go
int player_x, player_y;GetPlayerCoords(player_x, player_y);//----------------------------------------------------------// Move the dispaly window on the map so that the player // is at the centre of the screen//----------------------------------------------------------display_x = player_x - (display_resolution_width /2);display_y = player_y - (display_resolution_height/2);//----------------------------------------------------------// constrain the display window to the map boundaries//----------------------------------------------------------int min_x = 0;int min_y = 0;int max_x = (map_num_tiles_x * map_tile_pixels_width) - (display_resolution_width /2) - 1;int max_y = (map_num_tiles_y * map_tile_pixels_height) - (display_resolution_height/2) - 1;if (display_x < min_x) display_x = min_x; else if (display_x > max_x) display_x = max_x; if (display_y < min_y) display_y = min_y; else if (display_y > max_y) display_y = max_y;
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Thanks, I got it to work in my engine. I had something similar to this before but the start position would get messed up if it started in the bottom right corner. Oh well it works now! Now on to the fun stuff!
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# Linear programming Problems using excel
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Sunco Oil Co. manufactures three types of gasoline: Gas 1, Gas 2 and Gas 3. Each type is produced by blending three type of crude oil: Crude 1, Crude 2 and Crude 3. The sales price per barrel of gasoline and the purchase price per barrel of crude oil is given in the following table:
Gasoline Type Gas Selling Price Per Barrel Crude Oil Type Crude Oil Purchase Price Per Barrel
1 \$70 1 \$45
2 \$60 2 \$35
3 \$50 3 \$25
Sunco can purchase up to 5,000 barrels of each type of crude oil each day.
The three types of gasoline differ in their octane rating and sulfur content.
The crude oil blended to create Gas 1 must have an average octane rating of at least 10 and contain at most 1% sulfur. The crude oil blended to create Gas 2 must have an average octane rating of at least 8 and contain at most 2% sulfur. The crude oil blended to create Gas 3 must have an average octane rating of at least 6 and contain at most 1% sulfur.
The octane rating and sulfur content of the three types of crude oil is given in the following table:
Crude Oil Type Octane Rating Sulfur Content (%)
1 12 0.5
2 6 2.0
3 8 3.0
It costs \$4 to transform one barrel of crude oil into one barrel of gasoline, and Sunco's refinery can produce up to 14,000 barrels of gasoline per day. Sunco has the following contractual obligations to its customers that must be met:
Gasoline Type Barrels / Day
1 3,000
2 2,000
3 1,000
Sunco also has the option of advertising to stimulate demand for its gasoline products. Each dollar spent daily on advertising a particular type of gasoline increases the daily demand for that type of gasoline by 10 barrels.
Create and solve a linear program which maximizes Sunco's daily profits. What are the optimum decisions, i.e. the barrels of crude oil used to create the gasoline and the advertising dollars spent on stimulating the demand for gasoline?
---
(See attached file for full problem description) | 527 | 2,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-49 | latest | en | 0.92164 |
https://tutorbin.com/questions-and-answers/b-assuming-that-it-is-true-for-k-you-get-that-k-3k-1-so-expandingrepla | 1,718,352,886,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00309.warc.gz | 546,790,946 | 12,651 | Question
2. Prove that n > 3n – 1 for "-3,4,5,6,[2), using Mathematical Induction. a) Verify that it is true for n = 3 b) Assuming that it is true for
"= k, you get that k > 3k – 1. So expandingreplacing k with 3k –1 leads to(k +1)and (k +1) = k² + 2k +1> 3k – 1+ 2k +1 = 5k = 3k + 2k >= 3k +______
Fig: 1
Fig: 2
Fig: 3
Fig: 4 | 153 | 332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-26 | latest | en | 0.77559 |
https://pastebin.com/U0qNNTg3 | 1,600,632,851,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198652.6/warc/CC-MAIN-20200920192131-20200920222131-00098.warc.gz | 568,056,857 | 7,511 | # determinant
May 7th, 2019
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1. #include <stdio.h>
2. #include <stdlib.h>
3. #include <math.h>
4.
5. #define ERROR_TEXT "ALLOCATING MEMORY PROBLEMS "
6.
7. double **allocate_matrix(int matrix_size);
8. void fill_array(double **matrix, int matrix_size);
9. void show_array(double **matrix, int matrix_size1);
10. double calculate_determinant(double **matrix, double *result, int matrix_size);
11. void free_matrix(double **matrix, int matrix_size);
12.
13. void free_matrix(double **matrix, int matrix_size){ //free allocate memory
14. int current_row;
15. for(current_row=0;current_row<matrix_size;current_row++) free(matrix[current_row]);
16. free(matrix);
17. }
18.
19. double **allocate_matrix(int matrix_size){
20. int given_row, current_row;
21. double **matrix=malloc((matrix_size)*sizeof(double*)); //allocates space for number of rows
22.
23. if(matrix==NULL) printf(ERROR_TEXT);
24. else{
25. for(given_row=0;given_row<matrix_size;given_row++){
26. matrix[given_row]=malloc(sizeof(double)*(matrix_size)); //fullfils rows with space for data
27. if(matrix[given_row]==NULL){
28. printf(ERROR_TEXT);
29. for(current_row=0;current_row<given_row;current_row++) //empties space allocated prieviously
30. {free(matrix[current_row]);}
31. free(matrix); //empites space allocated for number of rows
32. }
33. }
34. }
35. return matrix;
36. }
37.
38. void fill_array(double **matrix, int matrix_size){
39. int current_row, current_column;
40. printf("\n\n");
41. for(current_row=0;current_row<matrix_size;current_row++){
42. for(current_column=0;current_column<matrix_size;current_column++){
43. scanf("%lf",&matrix[current_row][current_column]);
44. }
45. }
46. }
47.
48. void show_array(double **matrix, int matrix_size){
49. int current_row, current_column;
50. for(current_row=0;current_row<matrix_size;current_row++){
51. for(current_column=0;current_column<matrix_size;current_column++){
52. printf("%lf \t",matrix[current_row][current_column]);
53. }
54. printf("\n");
55. }
56. }
57.
58. double calculate_determinant(double **matrix,double *result,int matrix_size){
59. double determinant=0;
60. double **minor=0;
61. int det_column, current_row, current_column;
62.
63. if(matrix_size==1){
64. *result=matrix[0][0];
65. }
66. else if(matrix_size == 2){
67. *result=matrix[0][0]*matrix[1][1]-matrix[0][1]*matrix[1][0];
68. }
69. else{
70. minor=allocate_matrix(matrix_size-1);
71. if(minor==NULL) return-1;
72.
73. for(det_column=0;det_column<matrix_size;det_column++){
74. for(current_row=1;current_row<matrix_size;current_row++){
75.
76. int minor_column=0;
77. for(current_column=0;current_column<matrix_size;current_column++){
78. if(current_column!=det_column){
79. minor[current_row-1][minor_column]=matrix[current_row][current_column];
80. minor_column++;
81. }
82. }
83.
84. }
85. if(calculate_determinant(minor, &determinant, matrix_size-1)==-1){ //&determinat to adres *result
86. free_matrix(minor,matrix_size-1);
87. return-1;
88. }
89. else{
90. *result+=matrix[0][det_column]*pow((-1),(2+det_column))*determinant;
91. determinant=0;
92. }
93. }
94. free_matrix(minor, matrix_size-1);
95. }
96.
97. return 1;
98. }
99.
100. int main(){
101. int matrix_size;
102. double det=0;
103. double **matrix=NULL;
104.
105. printf("Please insert size of the matrix");
106. scanf("%d", &matrix_size);
107.
108. matrix=allocate_matrix(matrix_size);
109. if(matrix!=NULL){
110. fill_array(matrix,matrix_size);
111. show_array(matrix,matrix_size);
112.
113. if(calculate_determinant(matrix, &det, matrix_size)){
114. printf("\nDeterminant is equal = %lf\n\n", det);
115. }
116.
117. free_matrix(matrix, matrix_size);
118. }
119.
120. return 0;
121. }
RAW Paste Data | 1,197 | 4,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-40 | latest | en | 0.181768 |
https://us.metamath.org/nfeuni/reueq1.html | 1,723,239,830,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782236.55/warc/CC-MAIN-20240809204446-20240809234446-00502.warc.gz | 470,632,197 | 3,311 | New Foundations Explorer < Previous Next > Nearby theorems Mirrors > Home > NFE Home > Th. List > reueq1 GIF version
Theorem reueq1 2809
Description: Equality theorem for restricted uniqueness quantifier. (Contributed by NM, 5-Apr-2004.)
Assertion
Ref Expression
reueq1 (A = B → (∃!x A φ∃!x B φ))
Distinct variable groups: x,A x,B
Allowed substitution hint: φ(x)
Proof of Theorem reueq1
StepHypRef Expression
1 nfcv 2489 . 2 xA
2 nfcv 2489 . 2 xB
31, 2reueq1f 2805 1 (A = B → (∃!x A φ∃!x B φ))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 176 = wceq 1642 ∃!wreu 2616 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1546 ax-5 1557 ax-17 1616 ax-9 1654 ax-8 1675 ax-6 1729 ax-7 1734 ax-11 1746 ax-12 1925 ax-ext 2334 This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-tru 1319 df-ex 1542 df-nf 1545 df-sb 1649 df-eu 2208 df-cleq 2346 df-clel 2349 df-nfc 2478 df-reu 2621 This theorem is referenced by: reueqd 2817
Copyright terms: Public domain W3C validator | 473 | 1,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.293767 |
http://www.abs.gov.au/websitedbs/D3310114.nsf/home/CURF:+Relative+Standard+Error | 1,477,680,560,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988725451.13/warc/CC-MAIN-20161020183845-00201-ip-10-171-6-4.ec2.internal.warc.gz | 294,196,879 | 7,010 | Relative Standard Error
Sources of Variability
There are two sources of uncertainty or variability associated with survey estimates that are released by Survey TableBuilder. The first source of variability is due to sampling and the second is due to random adjustment of cell values.
Variability due to Sampling
Since the estimates from surveys are based on information obtained from a sub-sample of usual residents of a sample of dwellings, they are subject to sampling variability; that is, they may differ from those that would have been produced if all usual residents of all dwellings had been included in the survey. This component of variability is measured by the group Jackknife method.
The random adjustment of totals and subtotals introduces another source of variability into the estimates. As these adjustments are generated in a predictable way the impact they have on estimates can be measured directly.
Standard Errors
The variability due to sampling and random adjustment is combined into a single measure called the Standard Error (SE). The SE indicates the extent to which an estimate might have varied by chance, because only a sample of dwellings was included, and by random adjustment.
There are about two chances in three that a sample estimate will differ by less than one SE from the number that would have been obtained if all dwellings had been included and there was no random adjustment. There are about 19 chances in 20 that the difference will be less than two SEs. Another measure of the likely difference is the relative standard error (RSE), which is obtained by expressing the SE as a percentage of the estimate.
RSEs of proportions and percentages
Proportions and percentages formed from the ratio of two estimates are also subject to sampling errors. The size of the error depends of the accuracy of both the numerator and denominator. A formula to approximate the RSE of a proportion is given below:
Note - this formula only holds when the x is a subset of y. It should not be used if this is not the case i.e. estimates of 'rates' as opposed to proportions.
SEs may also be used to calculate SEs for the difference between two survey estimates (numbers or percentages). The sampling error of the difference between the two estimates depends on their individual SEs and the relationship (correlation) between them. An approximate SE of the difference between two estimates (x-y) may be calculated by the following formula:
SE(x-y) =sqrt[SE(x)]2 +[SE(y)]2
While this formula will only be exact for differences between separate and uncorrelated characteristics of subpopulations, it is expected to provide a reasonable approximation for most differences likely to be of interest in relation to this survey.
Note that it is the RSE of a percentage that is displayed, from which the SE may be calculated. For example if the estimated proportion is 30% with an RSE of 20%, then the SE for the proportion is 6%.
In some cases, the formula for the approximation of the RSE of a proportion may be unsuitable to use because the RSE of the numerator is very close to, or below, the RSE of the denominator. In this case the RSE will be suppressed. It is recommended to use the RSE of the numerator as a proxy for the RSE of the proportion in the event of this occurrence.
Standard Errors of Means and Sums
The estimates of means and sums of continuous variables are subject to sampling variability and random adjustment. As for population estimates, the variability due to sampling and random adjustment is combined into the calculated Standard Error, and the Relative Standard Error is reported. The component of variability arising from sampling is calculated using the Jackknife method.
Standard Errors of Quantiles
The estimates of quantiles such as medians, quartiles, quintiles and deciles are subject to sampling variability and random adjustment. As for population estimates, the variability due to sampling and random adjustment is combined into the calculated Standard Error, and the Relative Standard Error is reported. The component of variability arising from sampling is calculated using the Woodruff method. This is also true for Equal Distribution Quantiles.
Reliability of Estimates
Estimates with RSEs of 25% or more are not considered reliable for most purposes. Estimates with RSEs greater than 25% but less than or equal to 50% are annotated by an asterisk to indicate they are subject to high SEs and should be used with caution. Estimates with RSEs greater than 50% have their RSE suppressed in order to prevent the release of confidential data, and are annotated by a double asterisk. These estimates are considered too unreliable for general use.
Non-Sampling Error
The imprecision due to sampling variability and random adjustment should not be confused with inaccuracies that may occur because of imperfections in reporting by respondents and recording by interviewers, and errors made in coding and processing data. Inaccuracies of this kind are referred to as non-sampling error, and they may occur in any enumeration, whether it be a full count or a sample. Every effort is made to reduce non-sampling error to a minimum by careful design of questionnaires, intensive training and supervision of interviewers, and efficient operating procedures. | 1,049 | 5,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2016-44 | latest | en | 0.969841 |
https://plainmath.net/17022/linear-equations-second-order-constant-coefficients-solutions-equal | 1,627,848,813,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154219.62/warc/CC-MAIN-20210801190212-20210801220212-00478.warc.gz | 468,892,259 | 7,904 | Question
# Linear equations of second order with constant coefficients. Find all solutions on (−∞,+∞). y''−4y=0
Linear equations and graphs
Linear equations of second order with constant coefficients. Find all solutions on $$\displaystyle{\left(−∞,+∞\right)}$$. y''−4y=0
The characteristic equation is given by $$\displaystyle{\left({r}^{{2}}\right)}-{4}={0}$$ and has the roots $$r_1=2 \text{ and } r_2=-2$$. The general solution is given by
$$\displaystyle{y}={c}_{1}{e}^{{2}}{x}+{c}_{2}{e}^{{-{{2}}}}{x}$$ where e1 and e2 are arbitrary costant. | 186 | 549 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-31 | latest | en | 0.764145 |
http://www.drdobbs.com/database/median-cut-color-quantization/184409309?pgno=24 | 1,408,510,908,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500800168.29/warc/CC-MAIN-20140820021320-00014-ip-10-180-136-8.ec2.internal.warc.gz | 340,500,847 | 34,419 | # Median-Cut Color Quantization
SEP94: Median-Cut Color Quantization
# Median-Cut Color Quantization
## Fitting true-color images onto VGA displays
### Anton Kruger
In many instances, the number of colors in an image exceeds the number of displayable colors. For example, the output from 24-bit color scanners and ray-tracing software produces what's known as true-color images, which simply means that the red, green, and blue (RGB) components are each eight bits wide. It is often said that a true-color image can have up to 224, or 16 million colors, but the image dimensions normally determine the upper limit; a 512x512 true-color image, for instance, can't have more than 5122, or 262,000 colors.
While true-color displays are common on high-end workstations and some true-color video cards are becoming available for the PC, most of us still have to be satisfied with VGA cards that can display no more than 256 colors at a time. Another reason for wanting to convert a true-color image to a 256-color image is the savings in disk space. A 512x512 true-color image requires 512x512x3786 Kbytes of disk space, while a 256-color version of the image requires one byte per pixel, or 512x512262 Kbytes of disk space. The problem then is: Given a true-color image, what 256 colors should we use to display the image?
Note that the RGB components of a color can be viewed as lying along the axes of a color space, or RGB cube; see Figure 1. For a 24-bit image, the color space is (practically speaking) continuous, since the smallest difference between colors is imperceptible. This continuous color space must be mapped to 256 discrete colors, which are then used to represent the color space. Mapping a continuous variable to a discrete set of values is called quantization.
A simple approach to color quantization is to take the 256 most-common colors in the true-color image. For each color in the input image, we then search for the closest color in the set of 256 colors, and display that color instead. Actually, the 256 colors are loaded in the display system's color map or video look-up table (LUT), and the pixels' LUT indexes are written out to the display hardware. The algorithm that picks the 256 most-common colors to load into the LUT is the popularity algorithm. Its simplicity comes at a price, because it gives poor results for many images. Other color-quantization algorithms may give excellent results, but may be very time consuming. For example, as Wan et al. report, it "may take more than 20 hours on a VAX 780 computer to produce 256 clusters for a full-color image" using an algorithm known as the K-means algorithm. Color quantization is a post-processing procedure on an image, and should normally take only a fraction of the time to generate the image. Thus, such run times are in most cases unacceptable.
Some algorithms quantize an image without regard to its content; their dominant feature is speed. They use a fixed number of standard colors. For example, 256 colors require eight bits of color per pixel, and the bits are divided among red, green, and blue. Since the human eye is less sensitive to blue, red and green each get three bits, and blue two bits. This gives eight shades of red, eight of green, and four of blue. Other colors are mixtures of these primaries. The problem is that for a particular image, most standard colors may never be used. For example, if an image contains a lot of red and very little blue, all the entries assigned to blue are effectively wasted. This is a fundamental flaw of image-independent quantizers--if the image and the quantizer don't match, the results are poor.
### The Median-cut Algorithm
Heckbert's median-cut algorithm is an image-based, color-quantization algorithm that gives good results for many images. If properly implemented, it's also quite fast. The aim of the median-cut algorithm is to have each of the 256 output colors represent the same number of pixels in the input image. The starting point is the RGB cube that corresponds to the whole image, around which a tight-fitting cube is placed. The cube is then split at the median of the longest axis. This ensures that about the same number of colors is assigned to each of the new cubes. The procedure is recursively applied to the two new cubes until 256 cubes are generated. The centroids (average values) of the cubes become the 256 output colors.
For example, Table 1 is a histogram for 14 pixels that have six unique colors; for clarity, the colors have no blue component. The task is to apply the median-cut algorithm to this histogram and find four output colors.
The initial rectangle is in Figure 2(a); crosses are used to indicate the pixels' colors. The distance from maximum to minimum along the R-axis is 80--5=75; along the G-axis, 80--20=60. Thus, the longest distance from maximum to minimum is along the R-axis. The median along this axis is R=30, so this is where the rectangle is split, and two tight-fitting rectangles are placed around the two new regions; see Figure 2(b).
When applying the procedure to the left rectangle, the axis with the longest distance is the G-axis. The median along this axis is G=50--this is where the axis is split. For the right rectangle, the axis with the longest distance from maximum to minimum is also the G-axis. The median along this axis is G=40, and this is where the axis is split. The final rectangles are shown in Figure 2(c), where each rectangle has about the same number of pixels. Following this, the centroids of the colors in each rectangle are computed. The resulting values are the output colors; see Figure 2(d).
There are two approaches to remapping an input image: fast remapping and best remapping. With fast remaps, the centroid of a cube represents all the colors enclosed by the cube. This is often good enough, but does not give the best results because the centroid of the cube in which a color falls may not be the closest centroid. Best remap searches the list of output colors for the closest color, but because searching is involved, it's slower.
### Implementing the Median-cut Algorithm
The first division of the RGB cube results in two smaller cubes called CubeA1 and CubeB1. Then CubeA1 is split into CubeA2 and CubeB2. Next CubeA2 is split into CubeA3 and CubeB3, rendering four cubes. See Figure 3(b), where CubeB1 is never split, generating a depth-first spanning tree. To solve this problem, associate with each cube a level where the initial RGB cube has level 0, CubeA1 and CubeB1 have level 1, and so on. To end up with four cubes, we now require that the maximum level for any cube be log24=2; see Figure 3(c). Smaller cubes are also split on each recursive call. However, when the proper level is reached, the algorithm returns and splits the larger regions.
Another problem an implementation should address is the special case in which a cube encloses only one color, typically repeated many times. Obviously, such a cube cannot be split. One method of dealing with this is to split one of the other cubes; otherwise, you end up with one less cube than you want. To do this, you need a list of all the current cubes. However, this isn't available with the algorithms in Figure 3(a) and Figure 3(c), since you implicitly let the algorithm save the cubes on the program stack during the recursive calls. You can deal with this by increasing the maximum allowable level by 1 when encountering a cube with a single color. Unfortunately, this again opens the door for the problem of dividing smaller cubes while a larger cube is still waiting on the stack. Furthermore, the situation where the desired number of final cubes is not a power of 2, further complicates the process.
To address these problems, I used the following nonrecursive method. A list of the cubes computed so far is maintained in an array. During each iteration, the list of cubes is scanned for the cube with the smallest level, but cubes with single colors are ignored. Several cubes are normally candidates for splitting, and you could devise a rule for picking the best one to split, but I simply used the first one found. This cube is then split at the median, perpendicular to the longest axis. This increases the number of cubes by one. One of the new cubes is saved in the slot of the cube just split, and the other is added to the bottom of the list of cubes. When the desired number of cubes is generated, or all of the cubes enclose a single color, the splitting phase terminates. This algorithm is summarized in Figure 3(d).
At this point you have a list of cubes, and the next step is to compute their centroids. These colors are the output colors, or output color map. The histogram is no longer required, and you can use its space for a look-up table that holds the output colors--an inverse color map. The inverse color map works just like the histogram, where 24-bit colors are mapped to 15-bit colors that serve as indexes into the histogram. However, the inverse color map contains the input colors' indexes in the color map, instead of their count.
I've seen implementations of the median-cut algorithm that compute the inverse color map for the whole RGB cube. However, this is unnecessary because you know which colors each cube encloses, and they're the only ones required. Initialization of the inverse color map depends on the kind of remapping desired. With fast remap, initialization is accomplished by replacing the counts for all the colors in a cube with the centroid of the cube. With best remap, initialization is accomplished by replacing the counts for all the colors in a cube by the closest centroid, found by searching the list of centroids.
### Finding the Median
The time-consuming part of quantization is finding the median along an axis of a cube. Algorithms exist for finding the median of a set of numbers with a run-time complexity of O(N), but because of its simplicity, I used the following method. First sort the numbers in ascending order and compute their sum. Then start at the smallest number and compute a running sum. When the running sum is equal to half the total, you've found the median.
I've used the C standard-library function qsort (based on the Quicksort algorithm with its average run-time complexity of O(N log2N)) to do the sort. However, the algorithm has a potentially serious pitfall. When it's called to sort data that is already in order, the run-time complexity is O(N2). The difference between an algorithm with O(N log2N) complexity and one with O(N2) is dramatic when N is large. For example, with 32K=215 colors, the O(N2) algorithm takes more than 2000 times longer. Because of the way the histogram is constructed, with five bits of each color packed into a 15-bit color, one color is always sorted. The macros in Listing One are used to pack the 15-bit colors, which results in the image histogram being initially sorted on the blue component. A worst-case scenario is when a predominantly blue image is encountered--qsort is then called several times to sort a large set of colors already in order.
Another problem with Quicksort is that it is often implemented as a recursive routine, and can rapidly deplete the program stack when a large number of data points must be sorted. A nonrecursive implementation needs much less auxiliary storage. Thus, a good implementation of Quicksort is normally nonrecursive, and it randomizes the input data somewhat to provide for the case were the data is already in sorted order. Many implementations, however, don't do this. For example, I examined the Microsoft C 5.1 run-time library source, and its qsort is a fairly simple recursive implementation of Quicksort. For a production version of the median-cut algorithm, you might want to replace the qsort routine with a sort routine that's less efficient on the average, but has a better worst-case performance. Heapsort has both an average and worst-case run-time complexity of O(N log2N), and it needs no (or very little) auxiliary storage.
### Data Structures
The histogram, which I call Hist, is accessed indirectly via the array HistPtr that functions as an index into the histogram. HistPtr contains the position of the colors in the actual histogram. When the histogram (or parts of it) must be sorted, HistPtr (or parts of it) is sorted. A structure is used to represent a cube. In Listing One, this structure (cube_t) has several members, but the essential members are lower and upper, and they point to the opposite corners of a cube in HistPtr.
To see how this works, return to the data in Table 1, and rework the example in terms of the Hist and HistPtr. Also, keep an eye on Figure 2 to see the correspondence between the data structures and the geometric interpretation of the median-cut algorithm. Assume that after the histogram is constructed, it looks like Hist; see Figure 4(a). Note the "holes" in the histogram--this is typical. Also, this example assumes that the colors are initially sorted on green. C1 has the smallest green component, C4 has the second smallest, and so on. Just after Hist is constructed, HistPtr is filled as in Figure 4(a). The initial cube is the whole RGB cube, so that Cube.lower is set to 0, and Cube.upper to 5. This completes the initialization.
Now the cube must be split. The red axis is the longest, so we sort the colors along this axis. To do this, sort the array HistPtr instead of Hist. To find the median along the axis, start at Cube.lower; its value is 0. We look in HistPtr[0]; its value is 5, and this corresponds to C2 in Hist. C2's count is 4, and this is the running sum. Now we look at Cube.lower+1's corresponding color, C0, and the running sum becomes 3+4=7. This is half the total for the cube, so it is split to form two cubes, CubeA1, and CubeB1. The result is in Figure 4(b).
Next we split CubeB1. The colors in CubeB1 are sorted along the longest axis, the green axis. This is shown in Figure 4(c), while Figure 4(d) depicts the situation after CubeB1 is split. Finally, we split CubeA1 to get the desired four cubes; see Table 2.
To compute the centroid for, say, CubeB2, start at CubeB2.lower, whose value is 4. Look in HistPtr[4] to find 4. The color at Hist[4] is C5, and its count is 2. Next find the color that corresponds to CubeB2.lower+1, which is also the last color in CubeB2. This is color C3, which has a count of 2. The centroid for this cube is shown in Example 1. All centroids are computed this way.
### The MedianCut Program
MedianCut, the program in Listing One was developed on and for DOS PCs. To ease porting, I've tried to adhere to ANSI C. I also used typedefs for some variables when the number of bits was important. When moving to a different platform, you might want to change these so that they reflect the sizes of the target system. The code compiles and executes cleanly with Microsoft's 5.1 compiler (large memory model). If the /W4 compiler switch is used with version 6.0 of this compiler, some warning messages are issued, but they can safely be ignored.
MedianCut takes three arguments: the image histogram Hist, which contains the 15-bit colors; maxcubes, the desired number of output colors (this can be any number between 1 and 256, but the upper limit can be changed by altering the #define MAXCOLORS preprocessor directive); and a maxcubes x 3 array ColMap that MedianCut fills with the output colors, where the red component of color i is r=ColMap[i][0]; that of green, g=ColMap[i][1]; and that of blue, b=ColMap[i][2]. MedianCut returns the number of actual colors, which may be less than maxcubes if the input image already contains less colors than the requested number.
The #define FAST_REMAP preprocessor directive controls whether the inverse color map is initialized with the fast-remap or the best-remap method. If this directive is deleted or commented out, initialization is done according to the best-remap method.
Additionally, I've written a typical driver routine for MedianCut that, in this case, converts a Targa type-2 image file to a Targa type 1. Truevision's Targa type-2 image format is a popular format for true-color images, and Targa type-1 images are color-mapped, so to convert from type 1 to type 2, we must quantize the true-color image. This program is available electronically; see page 3.
Instead of dynamically allocating the space for HistPtr and the array for the list of cubes, I've chosen to have these static variables. These arrays occupy about 70 Kbytes. The histogram occupies another 64 Kbytes, and depending on your compiler's qsort routine, several Kbytes of stack may be required. Add to this about another 20 Kbytes for the file buffers, as well as the space required for the program code, and a total of about 200 Kbytes of memory are required.
Figures 5 and 6 each show two images before and after application of the median-cut algorithm (fast remap). Both were generated with the POV-Ray (Persistence of Vision) ray-tracing program (see "Ray Tracing and the POV-Ray Toolkit," by Craig A. Lindley, DDJ, July 1994). The 256-color version of "grid" is indistinguishable from the true-color original, which shows how effective the median-cut algorithm can be. However, with the "lamp" image, there are some false contours in areas of low contrast, but this may not show up in the photographs, since it is quite difficult to reproduce a video display faithfully in print.
This implementation of the median-cut algorithm is quite fast. With the fast-remap method, it takes about eight seconds to quantize 640x480 true-color versions of the images in Figures 5 and 6 on a 20-MHz AT clone. Much of this time is I/O, and the actual color quantization takes less than three seconds.
### References
Wan, S.J., S.K.M. Wong, and P. Prusinkiewicz. "An Algorithm for Multidimensional Data Clustering." ACM Transactions on Mathematical Software (June 1988).
Heckbert, P. "Color Image Quantization for Frame Buffer Display." Computer Graphics (July 1982).
Figure 1 The RGB cube.
Figure 2 The median-cut algorithm: (a) original rectangle; (b) after splitting once; (c) after splitting three times; (d) output colors.
#### Figure 3: (a) A flawed recursive implementation of the median-cut algorithm; (b) depth-first division as a result of the algorithm; (c) another recursive implementation; (d) pseudocode for the nonrecursive, cube-splitting algorithm.
```(a)
Split(Cube){
if (ncubes == 4) return;
find longest axis of Cube;
cut Cube at median to form CubeA, CubeB;
Split(CubeA);
Split(CubeB);
}
(b)
<a href="/showArticle.jhtml?documentID=ddj9409e&pgno=20">Figure</A>
(c)
maxlevel = 2;
Split(Cube,level){
if (ncubes == 4) return;
if (Cube's level == maxlevel) return;
find longest axis of Cube;
cut Cube at median to form CubeA, CubeB;
Split(CubeA, level+1);
Split(CubeB, level+1);
}
(d)
build initial cube from histogram;
set initial cube's level to 0;
insert initial cube in list of cubes;
ncubes = 1;
while (ncubes < maxcubes){
search for Cube with smallest level;
find the longest axis of Cube;
find the median along this axis;
cut Cube at median to form CubeA, CubeB;
set CubeA's level = Cube's level + 1;
set CubeB's level = Cube's level + 1;
insert CubeA in Cube's slot;
add CubeB to end of list of cubes;
ncubes = ncubes + 1;
}```
Figure 4 (a) Hist and HistPtr after initialization; (b) sorted on red and split at the median; (c) CubeB1, sorted on green; (d) CubeB1, split at the median.
Figure 5 Grid: (a) original, true-color image; (b) 256-color version (image-file courtesy of Dan Farmer).
Figure 6 Lamp: (a) original, true-color image; (b) 256-color version (anonymous image-file description).
#### Table 1: Histogram for median-cut example.
```Color (r,g)-coordinates Count
C0 (20,40) 3
C1 (40,20) 2
C2 (5,60) 4
C3 (50,80) 2
C4 (60,30) 1
C5 (80,50) 2```
#### Table 2: Contents of the data structures after splitting three times.
```Cube HistPtr.lower HistPtr.upper Colors Enclosed Cube Centroid
A3 0 0 C0 (20,40)
B3 1 1 C2 (5,60)
A2 2 3 C1,C4 (46.7,23.3)
B2 4 5 C5,C3 (65,65)
```
Example 1 Computing cube centroids.
#### Listing One
```/* median.c -- Anton Kruger, Copyright (c) Truda Software, 215 Marengo Rd,
** #2, Oxford, IA 52322-9383
** Description: Contains an implementation of Heckbert's median-cut color
** quantization algorithm.
** Compilers: MSC 5.1, 6.0.
** Note: 1) Compile in large memory model. 2) Delete "#define FAST_REMAP"
** statement below in order to deactivate fast remapping.
*/
#define FAST_REMAP
#include <stdio.h>
#include <stddef.h> /* for NULL */
#include <stdlib.h> /* for "qsort" */
#include <float.h> /* for FLT_MAX, FLT_MIN */
#define MAXCOLORS 256 /* maximum # of output colors */
#define HSIZE 32768 /* size of image histogram */
typedef unsigned char byte; /* range 0-255 */
typedef unsigned short word; /* range 0-65,535 */
typedef unsigned long dword; /* range 0-4,294,967,295 */
/* Macros for converting between (r,g,b)-colors and 15-bit */
/* colors follow. */
#define RGB(r,g,b) (word)(((b)&~7)<<7)|(((g)&~7)<<2)|((r)>>3)
#define RED(x) (byte)(((x)&31)<<3)
#define GREEN(x) (byte)((((x)>>5)&255)<< 3)
#define BLUE(x) (byte)((((x)>>10)&255)<< 3)
typedef struct { /* structure for a cube in color space */
word lower; /* one corner's index in histogram */
word upper; /* another corner's index in histogram */
dword count; /* cube's histogram count */
int level; /* cube's level */
byte rmin,rmax;
byte gmin,gmax;
byte bmin,bmax;
} cube_t;
static cube_t list[MAXCOLORS]; /* list of cubes */
static int longdim; /* longest dimension of cube */
static word HistPtr[HSIZE]; /* points to colors in "Hist" */
void Shrink(cube_t * Cube);
void InvMap(word * Hist, byte ColMap[][3],word ncubes);
int compare(const void * a1, const void * a2);
word MedianCut(word Hist[],byte ColMap[][3], int maxcubes)
{
/* Accepts "Hist", a 32,768-element array that contains 15-bit color counts
** of input image. Uses Heckbert's median-cut algorithm to divide color
** space into "maxcubes" cubes, and returns centroid (average value) of each
** cube in ColMap. Hist is also updated so that it functions as an inverse
** color map. MedianCut returns the actual number of cubes, which may be
** less than "maxcubes". */
byte lr,lg,lb;
word i,median,color;
dword count;
int k,level,ncubes,splitpos;
void *base;
size_t num,width;
cube_t Cube,CubeA,CubeB;
/* Create the initial cube, which is the whole RGB-cube. */
ncubes = 0;
Cube.count = 0;
for (i=0,color=0;i<=HSIZE-1;i++){
if (Hist[i] != 0){
HistPtr[color++] = i;
Cube.count = Cube.count + Hist[i];
}
}
Cube.lower = 0; Cube.upper = color-1;
Cube.level = 0;
Shrink(&Cube);
list[ncubes++] = Cube;
/* Main loop follows. Search the list of cubes for next cube to split, which
** is the lowest level cube. A special case is when a cube has only one
** color, so that it cannot be split. */
while (ncubes < maxcubes){
level = 255; splitpos = -1;
for (k=0;k<=ncubes-1;k++){
if (list[k].lower == list[k].upper)
; /* single color */
else if (list[k].level < level){
level = list[k].level;
splitpos = k;
}
}
if (splitpos == -1) /* no more cubes to split */
break;
/* Must split the cube "splitpos" in list of cubes. Next, find longest
** dimension of cube, and update external variable "longdim" which is
** used by sort routine so that it knows along which axis to sort. */
Cube = list[splitpos];
lr = Cube.rmax - Cube.rmin;
lg = Cube.gmax - Cube.gmin;
lb = Cube.bmax - Cube.bmin;
if (lr >= lg && lr >= lb) longdim = 0;
if (lg >= lr && lg >= lb) longdim = 1;
if (lb >= lr && lb >= lg) longdim = 2;
/* Sort along "longdim". This prepares for the next step, namely finding
** median. Use standard lib's "qsort". */
base = (void *)&HistPtr[Cube.lower];
num = (size_t)(Cube.upper - Cube.lower + 1);
width = (size_t)sizeof(HistPtr[0]);
qsort(base,num,width,compare);
/* Find median by scanning through cube, computing a running sum. When
** running sum equals half the total for cube, median has been found. */
count = 0;
for (i=Cube.lower;i<=Cube.upper-1;i++){
if (count >= Cube.count/2) break;
color = HistPtr[i];
count = count + Hist[color];
}
median = i;
/* Now split "Cube" at median. Then add two new cubes to list of cubes.*/
CubeA = Cube; CubeA.upper = median-1;
CubeA.count = count;
CubeA.level = Cube.level + 1;
Shrink(&CubeA);
list[splitpos] = CubeA; /* add in old slot */
CubeB = Cube; CubeB.lower = median;
CubeB.count = Cube.count - count;
CubeB.level = Cube.level + 1;
Shrink(&CubeB);
list[ncubes++] = CubeB; /* add in new slot */
if ((ncubes % 10) == 0)
fprintf(stderr,"."); /* pacifier */
}
/* We have enough cubes, or we have split all we can. Now compute the color
** map, inverse color map, and return number of colors in color map. */
InvMap(Hist, ColMap,ncubes);
return((word)ncubes);
}
void Shrink(cube_t * Cube)
{
/* Encloses "Cube" with a tight-fitting cube by updating (rmin,gmin,bmin)
** and (rmax,gmax,bmax) members of "Cube". */
byte r,g,b;
word i,color;
Cube->rmin = 255; Cube->rmax = 0;
Cube->gmin = 255; Cube->gmax = 0;
Cube->bmin = 255; Cube->bmax = 0;
for (i=Cube->lower;i<=Cube->upper;i++){
color = HistPtr[i];
r = RED(color);
if (r > Cube->rmax) Cube->rmax = r;
if (r < Cube->rmin) Cube->rmin = r;
g = GREEN(color);
if (g > Cube->gmax) Cube->gmax = g;
if (g < Cube->gmin) Cube->gmin = g;
b = BLUE(color);
if (b > Cube->bmax) Cube->bmax = b;
if (b < Cube->bmin) Cube->bmin = b;
}
}
void InvMap(word * Hist, byte ColMap[][3],word ncubes)
{
/* For each cube in list of cubes, computes centroid (average value) of
** colors enclosed by that cube, and loads centroids in the color map. Next
** loads histogram with indices into the color map. A preprocessor directive
** #define FAST_REMAP controls whether cube centroids become output color
** for all the colors in a cube, or whether a "best remap" is followed. */
byte r,g,b;
word i,j,k,index,color;
float rsum,gsum,bsum;
float dr,dg,db,d,dmin;
cube_t Cube;
for (k=0;k<=ncubes-1;k++){
Cube = list[k];
rsum = gsum = bsum = (float)0.0;
for (i=Cube.lower;i<=Cube.upper;i++){
color = HistPtr[i];
r = RED(color);
rsum += (float)r*(float)Hist[color];
g = GREEN(color);
gsum += (float)g*(float)Hist[color];
b = BLUE(color);
bsum += (float)b*(float)Hist[color];
}
/* Update the color map */
ColMap[k][0] = (byte)(rsum/(float)Cube.count);
ColMap[k][1] = (byte)(gsum/(float)Cube.count);
ColMap[k][2] = (byte)(bsum/(float)Cube.count);
}
#ifdef FAST_REMAP
/* Fast remap: for each color in each cube, load the corresponding slot
** in "Hist" with the centroid of the cube. */
for (k=0;k<=ncubes-1;k++){
Cube = list[k];
for (i=Cube.lower;i<=Cube.upper;i++){
color = HistPtr[i];
Hist[color] = k;
}
if ((k % 10) == 0) fprintf(stderr,"."); /* pacifier */
}
#else
/* Best remap: for each color in each cube, find entry in ColMap that has
** smallest Euclidian distance from color. Record this in "Hist". */
for (k=0;k<=ncubes-1;k++){
Cube = list[k];
for (i=Cube.lower;i<=Cube.upper;i++){
color = HistPtr[i];
r = RED(color); g = GREEN(color); b = BLUE(color);
/* Search for closest entry in "ColMap" */
dmin = (float)FLT_MAX;
for (j=0;j<=ncubes-1;j++){
dr = (float)ColMap[j][0] - (float)r;
dg = (float)ColMap[j][1] - (float)g;
db = (float)ColMap[j][2] - (float)b;
d = dr*dr + dg*dg + db*db;
if (d == (float)0.0){
index = j; break;
}
else if (d < dmin){
dmin = d; index = j;
}
}
Hist[color] = index;
}
if ((k % 10) == 0) fprintf(stderr,"."); /* pacifier */
}
#endif
return;
}
int compare(const void * a1, const void * a2)
{
/* Called by the sort routine in "MedianCut". Compares two
** colors based on the external variable "longdim". */
word color1,color2;
byte C<SUB>1</SUB>,C<SUB>2</SUB>;
color1 = (word)*(word *)a1;
color2 = (word)*(word *)a2;
switch (longdim){
case 0:
C<SUB>1</SUB> = RED(color1), C<SUB>2</SUB> = RED(color2);
break;
case 1:
C<SUB>1</SUB> = GREEN(color1), C<SUB>2</SUB> = GREEN(color2);
break;
case 2:
C<SUB>1</SUB> = BLUE(color2), C<SUB>2</SUB> = BLUE(color2);
break;
}
return ((int)(C<SUB>1</SUB>-C<SUB>2</SUB>));
}
```
### More Insights
To upload an avatar photo, first complete your Disqus profile. | View the list of supported HTML tags you can use to style comments. | Please read our commenting policy. | 7,322 | 29,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2014-35 | longest | en | 0.879286 |
https://www.giantbomb.com/everyday-genius-squarelogic/3030-29096/ | 1,632,500,885,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057558.23/warc/CC-MAIN-20210924140738-20210924170738-00120.warc.gz | 816,050,740 | 60,842 | ## Gameplay / Concept
Squarelogic is, at its core, a clone of KenKen-style games. The game takes place on a square grid, where no two numbers may share a row or column. Unlike Sudoku style games, there are no numbers on the grid to begin with. Instead, the numbers are derived through the process of elimination. Segments of the grid have part of a math problem on them, such as "4x." In that case, the numbers within the cage would have to multiply to result in 4, i.e. 4x1 or 2x2. In normal Kenken games, there are also addition, subtraction and division cages.
Here are the rules used:
Normal KenKen rules
2. Subtraction
3. Multiplication
4. Division
Special Squarelogic rules
1. Odd: All squares in this cage must have odd values (1, 3, 5, 7, or 9).
2. Even: All squares in this cage must have odd values (2, 4, 6, or 8).
3. Greater Than, Lesser Than: The value of the squares must be greater or lesser than the one next to it according to the brackets
4. Straight: All squares in this cage must form a "straight" (like poker), IN ANY ORDER (for example: 4,3,5)
5. Hidden: Blank squares that do not show their rule but must be colored in.
6. None: No rule at all.
Some puzzles are Double Board puzzles; for these puzzles, both boards have the same answers, and you solve them both simultaneously.
## Regions
The game is separated into six regions of which only five are visible from the start. Each new region represents a more difficult size of grid, and this features more difficult puzzles. These regions are separated further into locations. Each location contains a specific type of puzzle. There are 12 practice puzzles, followed by a challenge puzzle. Completing the challenge puzzle unlocks the next location, as well as puzzles of varying difficulty in the current location.
The regions and locations within the game are as follows:
Ocean - 4x4Canyon - 5x5Forest 6x6
• Beginnings
• Ocean Floor
• Shipwreck
• Kelp
• Shelf
• Arch
• Shallows
• Shore
• Grotto
• Tunnel
• Cave
• Canyon
• Rise
• Path
• Stream
• Bridge
• Trail
• Grove
• Glen
• Hilltop
• View
Mountain - 7x7Sky - 8x8Beyond - 9x9
• Pass
• Waterfall
• Crest
• Vista
• Precipice
• Spire
• Summit
• Horizon
• Cloud
• Vapor
• Rain
• Storm
• Sky
• Twilight
• Secret
• Enigma
• Tranquility
• Serenity
• Self-Awareness
• Self-Reliance
• Beyond
sizepositionchange
sizepositionchange
positionchange
positionchange
positionchange | 642 | 2,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-39 | latest | en | 0.924525 |
http://www.spicysms.in/smslist/lang/A/cid/16/page/16 | 1,513,183,573,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948529738.38/warc/CC-MAIN-20171213162804-20171213182804-00200.warc.gz | 449,270,498 | 5,818 | # Brain Teasers
You have a jug that holds five gallons, and a jug that holds three gallons. You have no other containers, and there are no markings on the jugs. You need to obtain exactly seven gallons of water from a faucet. How can you do it?
Second Problem: You need exactly four gallons. How do you do it?
Ans. Fill 3 gallon jug , pour in 5 gallon jug, fill 3 gallon, pour 2 gallon in 5 gallon, you have 1 gallon left in the 3 gallon jug. Empty the 5 gallon. Pour the 1 gal into the 5 gallon, fill the 3 gallon and pour it in the 5 gallon (at this step you have 4 gallon). Fill the 3 gallon jug one last time and you now have 7 gallons.
I Like SMS - Like: 40 - SMS Length: 640
A farmer is taking a fox, a chicken, and a bag of grain home. To get there, he must cross a river, but he's only allowed to take one item across the bridge with them at a time. If the fox is left alone with the chicken, the fox will eat the chicken. If the chicken is left alone with the grain, the chicken will eat the grain. How can the farmer cross the river without any of his possessions being eaten?
Ans. He'll carry chicken to other side. Then back and he'll carry the grain. On other bank he'll put the grain and take the chicken back. And on dis side , he'll put the chicken and he'll carry the fox, and return. And finaly he'll carry the chicken frm dis end.
I Like SMS - Like: 56 - SMS Length: 680
I want you think about yourself in a squared room. There are no windows or doors. No exits at all. You can't break down the walls. It is empty with only you in it. How do you get out?
Ans. "Stop thinking" :P
I Like SMS - Like: 37 - SMS Length: 221
A man who lives on the tenth floor takes the elevator down to the first floor every morning and goes to work. In the evening, when he comes back; on a rainy day, or if there are other people in the elevator, he goes to his floor directly. Otherwise, he goes to the seventh floor and walks up three flights of stairs to his apartment.
Can you explain why?
Ans. The man is of short stature. He can't reach the upper elevator buttons, but he can ask people to push them for him. He can also push them with his umbrella.
I Like SMS - Like: 33 - SMS Length: 528
A Petri dish hosts a healthy colony of bacteria. Once a minute every bacterium divides into two. The colony was founded by a single cell at noon. At exactly 12:43 (43 minutes later) the Petri dish was half full.
At what time will the dish be full?
Ans. The dish will be full at 12:44.
I Like SMS - Like: 36 - SMS Length: 298
The day before yesterday I was 25 and the next year I will be 28. This is true only one day in a year. What day is my birthday?
Ans. He was born on December 31st and spoke about it on January 1st. [The day before yesterday (December 30) he was 25. He turned 26 on December 31. He talked about it on January 1. That year he would turn 27, so "next year" he would turn 28.]
I Like SMS - Like: 39 - SMS Length: 389
Three brothers share a family sport:
A non-stop marathon
The oldest one is fat and short
And trudges slowly on
The middle brother's tall and slim
The youngest runs just like the wind,
Speeding through the race
He's young in years, we let him run,
The other brothers say
Though he's surely number one,
He's second, in a way.
Who r d three brothers??
Ans. The hands on a clock (hour, minute, and second)!!!
I Like SMS - Like: 39 - SMS Length: 460
« Previous 11 12 13 14 15 16 17 18 19 20 Next » | 912 | 3,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-51 | latest | en | 0.933539 |
https://www.jiskha.com/display.cgi?id=1321520088 | 1,503,414,401,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110792.29/warc/CC-MAIN-20170822143101-20170822163101-00629.warc.gz | 897,532,848 | 3,375 | # math
posted by .
what is the GCF of 300 and 225 using prime factorization
• math -
75
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Write the prime factorization using exponents of each number .Then find the greatest common factor (GCF)of the numbers. 1\24.............. 36.............. 2\21.............. 56.............. Please,,help me.😞 I tried to solve …
More Similar Questions | 318 | 1,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-34 | latest | en | 0.805169 |
https://forum.arduino.cc/t/about-port-manipulation/313453 | 1,670,505,836,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711336.41/warc/CC-MAIN-20221208114402-20221208144402-00346.warc.gz | 261,333,577 | 7,604 | I am making a 16-bit decimal to binary converter with 8 leds using port manipulation. User enters the
decimal then program converts it to binary then writes the leds. But I have a problem with port manipulation. I keep binary number as a type of String so I can't write number to pins. Code like this;
void setup() {
DDRC = B11111111;
Serial.begin(9600);
}
void loop() {
// put your main code here, to run repeatedly:
if(Serial.available()>0){
long input = Serial.parseInt();
if (input>=2){
y = input/2;
while (y>0){
r = input % 2;
out = String(r) + out;
y = input/2;
input = y ;
}}
else
out = String(input);
int digits = out.length();
for(int i=0; i<(16-digits); i++){
out = "0" + out;
}
String firstEight = "B" + out.substring(0,8);
String lastEight = "B" + out.substring(8,16);
Then I write end of this code;
PORTC = firstEight;
and I get this error message : cannot convert 'String' to 'volatile uint8_t {aka volatile unsigned char}' in assignment
Sorry if this is actually common knowledge, but I couldn't find it.
Alper.
Hello and welcome,
Why use String? You want numbers so use numbers.
``````long input = Serial.parseInt();
PORTC = input & 0xFF; // first 8 bits
PORTX = (input >> 8) & 0xFF; // next 8 bits
``````
Most here would recommend you not use String as there are problems with using it.
I make it a habit of not looking at sketches that are not commented.
Actually this is my homework so i have to convert decimal to binary using my own code. That is why i used the String. It keeps the bits of binary number.
Modifying your original post makes no sense at at all.
Here is a similar function that converts Decimal to Seven-Segment. To do discrete LEDs, just replace the PortB values. e.g.:
case 2: PORTB = 0b00000010; break; // Segments for "2"
`````` // ****FUNCTION******* SevenSeg ********************************
// Code to drive 7-segment display, process one digit at a time
void SevenSeg(int TDisplay)
{ switch (TDisplay)
{// gfedcba. segments to light
case 0: PORTB = 0b01111110; break; // Segments for "0"
case 1: PORTB = 0b00001100; break; // Segments for "1"
case 2: PORTB = 0b10110110; break; // Segments for "2"
case 3: PORTB = 0b10011110; break; // Segments for "3"
case 4: PORTB = 0b11001100; break; // Segments for "4"
case 5: PORTB = 0b11011010; break; // Segments for "5"
case 6: PORTB = 0b11111010; break; // Segments for "6"
case 7: PORTB = 0b00001110; break; // Segments for "7"
case 8: PORTB = 0b11111110; break; // Segments for "8"
case 9: PORTB = 0b11011110; break; // Segments for "9"
}//switch
}// End FUNCTION SevenSeg ---------------------------
``````
Quick different example, as this is your homework.
You don't want to use port manipulation, as that automatically
does the conversion. (Output decimal to port, you see binary.)
I doubt, if I understand the assignment correctly, that the
educator would accept you using such a method, unless it's
only breaking the 2 byte integer into it's low and high byte,
in which case I see no purpose to the assignment. I mean you
have way more code than that would require, right?
So, as an example only, say you set the following variables.
(I'm NOT giving full real code, as that would be cheating.)
Integer PortBase // Set this to the lowest Bit pin number.
(The above assumes bit 0 is the lowest pin and 7 the highest.)
(If the pin numbers are reversed, remember to take that into
account. Use Highest Pin # and Subtract ( 7 - I ) as below.)
Do your input whatever way into a Long Integer "Input".
Use the "Old Way" from Boolean algebra. Here Goes.
(Note: I am a "C" noob. This is a translate from assembly
code to "C" so it may not be optimal, but demonstrates OK.)
for (int I=7; I>=0; I--) { // Setup the bit loop
if ((Input - (2 Pwr I)) >= 0) { // Decode to Binary
// (By Pwr - I Mean exponent - 2 to the power of I)
// (Remember, example, not finished code)
// (Suggest left shift rather than "call" pwr function.)
// (Right shift from max to calc on the fly, etc.)
// Now do these bits individually
digitalWrite (I + Portbase, HIGH); // Binary 1
Input -= (2 Pwr I); // Decrease Input
(Save bit to array here, If Req'd)
// The decrement removes the bit from the value
// so you can just continue on to the next one.
}
else {
DigitalWrite (I + Portbase, LOW); // Binary 0
(Save bit to array here, If Req'd)
}
} // This is the end of the "For" loop - All Done
Finally, for 16 bit conversion, start I at 15, break decode
as desired. If port numbers are reverse order, use the
Portbase - ( 7 - I ) or ( 15 - I ) as needed to get the
correct pin number on the port. I really can't write
more without actually doing most of the assignment.
Of course, the loop may be broken into 2 half processes, so
you could display the first half, wait for any serial input,
then display the second half, but that part is up to you.
This was a common method in old days but I have no idea
if this helps as displaying 16 bits with 8 leds has already
got me slightly confused. As MY instructors would have said,
"It's cheating to let the compiler do the work." With this
method, you make no "calls" and store the bits directly in
the port, bit by bit, so no save is really req'd. If it IS
required, you can store the bits in a boolean array, etc.
Also, You have done all the work, not the compiler and it
clearly does a decimal to binary conversion, even though, in
reality, the parseint function already did one. (You didn't
think the CPU ever stored in decimal, did you?) If the real
object is to convert the string input to a decimal value, and
then convert to binary, follow standard build methods, count
the chars, add (Char Value - 48) * 10 Pwr (Char Position)
where Char position is Max - 0, loop through all, you already
know all this part and I'm just rambling now.
I hope this is of use to you. Thanks for reading.
The only purpose of the assignment is converting decimal to binary(with your own code) and writing the leds. I have already did converting part, I had a problem with writing part but I changed something on my code and I solve the problem. Actually your answer helped me Arthur_Clark. Thanks for your answer. | 1,686 | 6,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-49 | latest | en | 0.738916 |
https://numberworld.info/452 | 1,653,321,559,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558030.43/warc/CC-MAIN-20220523132100-20220523162100-00453.warc.gz | 475,377,726 | 3,774 | # Number 452
### Properties of number 452
Cross Sum:
Factorization:
2 * 2 * 113
Divisors:
1, 2, 4, 113, 226, 452
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
1c4
Base 32:
e4
sin(452)
-0.3795798508808
cos(452)
0.9251589792059
tan(452)
-0.41028607991959
ln(452)
6.1136821798322
lg(452)
2.6551384348114
sqrt(452)
21.260291625469
Square(452)
### Number Look Up
Look Up
452 which is pronounced (four hundred fifty-two) is a very great number. The cross sum of 452 is 11. If you factorisate 452 you will get these result 2 * 2 * 113. 452 has 6 divisors ( 1, 2, 4, 113, 226, 452 ) whith a sum of 798. The number 452 is not a prime number. 452 is not a fibonacci number. The number 452 is not a Bell Number. The figure 452 is not a Catalan Number. The convertion of 452 to base 2 (Binary) is 111000100. The convertion of 452 to base 3 (Ternary) is 121202. The convertion of 452 to base 4 (Quaternary) is 13010. The convertion of 452 to base 5 (Quintal) is 3302. The convertion of 452 to base 8 (Octal) is 704. The convertion of 452 to base 16 (Hexadecimal) is 1c4. The convertion of 452 to base 32 is e4. The sine of the figure 452 is -0.3795798508808. The cosine of 452 is 0.9251589792059. The tangent of the number 452 is -0.41028607991959. The square root of 452 is 21.260291625469.
If you square 452 you will get the following result 204304. The natural logarithm of 452 is 6.1136821798322 and the decimal logarithm is 2.6551384348114. I hope that you now know that 452 is amazing number! | 587 | 1,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-21 | latest | en | 0.78078 |
http://mathwaycalculus.com/page/2/ | 1,670,284,829,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711064.71/warc/CC-MAIN-20221205232822-20221206022822-00081.warc.gz | 35,517,885 | 13,514 | ## How to solve 1 + 1/2 + 1/3 + ?= 12
Our mission is to systematically share mathematics information to people around the world and to make it universally accessible and … Read more
## How to solve 5/2+7/2+(0.5)^2+4=x
Welcome to my article How to solve 5/2+7/2+(0.5)^2+4=xThis question is taken from the simplification lesson.The solution of this question has … Read more
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Welcome to my article How to solve 12(4-2)+1/2-5/6=x+2. This question is taken from the simplification lesson.The solution of this question … Read more
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Welcome to my article Fraction to Decimal: 1/4. This question is taken from the simplification lesson.The solution of this question … Read more
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Our mission is to systematically share mathematics (What are Types of Triangles) information to people around the world and to … Read more
## How to solve 1/2 as a fraction ?
Welcome to my article How to solve 1/2 as a fraction ?. This question is taken from the simplification lesson.The … Read more | 362 | 1,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2022-49 | latest | en | 0.894595 |
http://web2.0calc.com/questions/1-cos-2x-cos-6x-cos-8x | 1,516,219,190,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886964.22/warc/CC-MAIN-20180117193009-20180117213009-00734.warc.gz | 362,263,012 | 6,719 | +0
# 1 - cos(2x) + cos(6x) - cos(8x)
0
685
8
1 - cos(2x) + cos(6x) - cos(8x)
Guest May 7, 2015
#1
+18827
+15
1 - cos(2x) + cos(6x) - cos(8x) = 0 x =?
$$\small{\text{ \begin{array}{rcl} 1-\cos{(2x)}+\cos{(6x)}-\cos{(8x)}&=&0\\ 1-\cos{(2x)} &=&\cos{(8x)}-\cos{(6x)}\\ &&Formula:\\ && ~\boxed{\mathbf{ \cos{2x}=1-2\sin^2{(x)} \quad or \quad 1-\cos{2x}=2\sin^2{(x)} }}\\ 2\sin^2{(x)} &=& \cos{(8x)}-\cos{(6x)}\\ &&Formula:\\ && ~\boxed{\mathbf{ \cos{(a)}-\cos{(b)}=2\sin{ \left( \dfrac{a+b}{2} \right)}\sin{ \left( \dfrac{b-a}{2} \right)} }}\\ && \cos{(8x)}-\cos{(6x)}=2\sin{ \left( \dfrac{14x}{2} \right)}\sin{ \left( \dfrac{-2x}{2} \right)}\\ && \cos{(8x)}-\cos{(6x)}= -2\sin{ (7x) }\sin{(x)}\\ 2\sin^2{(x)} &=& -2\sin{ (7x) }\sin{(x)}\\ \sin^2{(x)} &=& -\sin{ (7x) }\sin{(x)}\\ \sin^2{(x)} +\sin{ (7x) }\sin{(x)} &=& 0\\ \sin{(x)} \left[ \sin{(x)} + \sin{ (7x) } \right] &=& 0\\ &&Formula:\\ && ~\boxed{\mathbf{ \sin{(a)}+\sin{(b)}=2\sin{ \left( \dfrac{a+b}{2} \right)}\cos{ \left( \dfrac{b-a}{2} \right)} }}\\ && \sin{(x)}+\sin{(7x)}=2\sin{ \left( \dfrac{8x}{2} \right)}\cos{ \left( \dfrac{6x}{2} \right)}\\ && \sin{(x)}+\sin{(7x)}=2\sin{ (4x) }\cos{ (3x) }\\ \sin{(x)}\cdot 2 \cdot\sin{ (4x) }\cos{ (3x) } &=& 0\\ \end{array} }}$$
The solution set of the given equation is:
$$\\\boxed{\mathbf{ \sin{(x)}=0 \qquad {x = k\cdot\pi \qquad k \in \mathrm{Z} } }}\\ \boxed{\mathbf{ \sin{(4x)}=0 \qquad 4x = k\cdot \pi \qquad {x = k\cdot \dfrac{\pi}{4} \qquad k \in \mathrm{Z} } }}\\ \boxed{\mathbf{ \cos{(3x)}=0 \qquad 3x = \pm \dfrac{\pi}{2}+ k\cdot 2\cdot \pi \qquad {x = \pm\dfrac{\pi}{6} + k\cdot \dfrac{2}{3}~\pi \qquad k \in \mathrm{Z} } }}$$
heureka May 8, 2015
Sort:
#1
+18827
+15
1 - cos(2x) + cos(6x) - cos(8x) = 0 x =?
$$\small{\text{ \begin{array}{rcl} 1-\cos{(2x)}+\cos{(6x)}-\cos{(8x)}&=&0\\ 1-\cos{(2x)} &=&\cos{(8x)}-\cos{(6x)}\\ &&Formula:\\ && ~\boxed{\mathbf{ \cos{2x}=1-2\sin^2{(x)} \quad or \quad 1-\cos{2x}=2\sin^2{(x)} }}\\ 2\sin^2{(x)} &=& \cos{(8x)}-\cos{(6x)}\\ &&Formula:\\ && ~\boxed{\mathbf{ \cos{(a)}-\cos{(b)}=2\sin{ \left( \dfrac{a+b}{2} \right)}\sin{ \left( \dfrac{b-a}{2} \right)} }}\\ && \cos{(8x)}-\cos{(6x)}=2\sin{ \left( \dfrac{14x}{2} \right)}\sin{ \left( \dfrac{-2x}{2} \right)}\\ && \cos{(8x)}-\cos{(6x)}= -2\sin{ (7x) }\sin{(x)}\\ 2\sin^2{(x)} &=& -2\sin{ (7x) }\sin{(x)}\\ \sin^2{(x)} &=& -\sin{ (7x) }\sin{(x)}\\ \sin^2{(x)} +\sin{ (7x) }\sin{(x)} &=& 0\\ \sin{(x)} \left[ \sin{(x)} + \sin{ (7x) } \right] &=& 0\\ &&Formula:\\ && ~\boxed{\mathbf{ \sin{(a)}+\sin{(b)}=2\sin{ \left( \dfrac{a+b}{2} \right)}\cos{ \left( \dfrac{b-a}{2} \right)} }}\\ && \sin{(x)}+\sin{(7x)}=2\sin{ \left( \dfrac{8x}{2} \right)}\cos{ \left( \dfrac{6x}{2} \right)}\\ && \sin{(x)}+\sin{(7x)}=2\sin{ (4x) }\cos{ (3x) }\\ \sin{(x)}\cdot 2 \cdot\sin{ (4x) }\cos{ (3x) } &=& 0\\ \end{array} }}$$
The solution set of the given equation is:
$$\\\boxed{\mathbf{ \sin{(x)}=0 \qquad {x = k\cdot\pi \qquad k \in \mathrm{Z} } }}\\ \boxed{\mathbf{ \sin{(4x)}=0 \qquad 4x = k\cdot \pi \qquad {x = k\cdot \dfrac{\pi}{4} \qquad k \in \mathrm{Z} } }}\\ \boxed{\mathbf{ \cos{(3x)}=0 \qquad 3x = \pm \dfrac{\pi}{2}+ k\cdot 2\cdot \pi \qquad {x = \pm\dfrac{\pi}{6} + k\cdot \dfrac{2}{3}~\pi \qquad k \in \mathrm{Z} } }}$$
heureka May 8, 2015
#2
+91436
+5
Like WOW Heureka,
MathsGod1 is learning LaTex. He is only 12 but he is learning very quickly.
I should send him to you :))
I'm quite happy to keep teaching him but you can if you would like to.
You are Lord of LaTex in Camelot. :))
Melody May 9, 2015
#3
+80875
0
Very nice, heureka....!!!!
CPhill May 9, 2015
#4
+4664
+5
aha Melody, I've noticed Heureka's LaTeX ever sinced i layed my eyes on this website and LaTeX.
The LaTeX is amazing but i think i should take it slow before i get to your level...Really slow :D
MathsGod1 May 9, 2015
#5
+80875
0
Your LaTex abilities are improving all the time, MG1....!!!!
Maybe....one day......you can teach me....LOL !!!!!
CPhill May 9, 2015
#6
+4664
+5
Hahah!
The first time i teach a grown up.
Lol.
Ok, we'll trade in exchange for LaTeX give me 99% of your points!
Just joking...We'll need someone like you as Moderator I'm not sure I'll do great.
LaTeX looks hard but once leart it actually isn't.
Just a whole bunch of numbers, letters and symbols!
MathsGod1 May 9, 2015
#7
+80875
0
.......Just a whole bunch of numbers, letters and symbols!.....
Hey......that kinda' sounds like.......MATH....!!!!
CPhill May 9, 2015
#8
+4664
0
That is pure logic.
MathsGod1 May 10, 2015
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See the FAQ
or other info.
# Polytope of Type {6,3,2}
Atlas Canonical Name : {6,3,2}*648
if this polytope has a name.
Group : SmallGroup(648,301)
Rank : 4
Schlafli Type : {6,3,2}
Number of vertices, edges, etc : 54, 81, 27, 2
Order of s0s1s2s3 : 18
Order of s0s1s2s3s2s1 : 2
Special Properties :
Degenerate
Universal
Orientable
Flat
Related Polytopes :
Facet
Vertex Figure
Dual
Facet Of :
{6,3,2,2} of size 1296
{6,3,2,3} of size 1944
Vertex Figure Of :
{2,6,3,2} of size 1296
Quotients (Maximal Quotients in Boldface) :
3-fold quotients : {6,3,2}*216
9-fold quotients : {6,3,2}*72
27-fold quotients : {2,3,2}*24
Covers (Minimal Covers in Boldface) :
2-fold covers : {6,6,2}*1296c
3-fold covers : {18,3,2}*1944a, {6,9,2}*1944c, {6,9,2}*1944e, {6,3,2}*1944, {6,3,6}*1944e
Permutation Representation (GAP) :
```s0 := ( 4, 8)( 5, 9)( 6, 7)(10,19)(11,20)(12,21)(13,26)(14,27)(15,25)(16,24)
(17,22)(18,23)(31,35)(32,36)(33,34)(37,46)(38,47)(39,48)(40,53)(41,54)(42,52)
(43,51)(44,49)(45,50)(58,62)(59,63)(60,61)(64,73)(65,74)(66,75)(67,80)(68,81)
(69,79)(70,78)(71,76)(72,77);;
s1 := ( 1,10)( 2,12)( 3,11)( 4,13)( 5,15)( 6,14)( 7,16)( 8,18)( 9,17)(20,21)
(23,24)(26,27)(28,66)(29,65)(30,64)(31,69)(32,68)(33,67)(34,72)(35,71)(36,70)
(37,57)(38,56)(39,55)(40,60)(41,59)(42,58)(43,63)(44,62)(45,61)(46,75)(47,74)
(48,73)(49,78)(50,77)(51,76)(52,81)(53,80)(54,79);;
s2 := ( 1,28)( 2,30)( 3,29)( 4,33)( 5,32)( 6,31)( 7,35)( 8,34)( 9,36)(10,53)
(11,52)(12,54)(13,46)(14,48)(15,47)(16,51)(17,50)(18,49)(19,40)(20,42)(21,41)
(22,45)(23,44)(24,43)(25,38)(26,37)(27,39)(55,57)(58,59)(62,63)(64,79)(65,81)
(66,80)(67,75)(68,74)(69,73)(70,77)(71,76)(72,78);;
s3 := (82,83);;
poly := Group([s0,s1,s2,s3]);;
```
Finitely Presented Group Representation (GAP) :
```F := FreeGroup("s0","s1","s2","s3");;
s0 := F.1;; s1 := F.2;; s2 := F.3;; s3 := F.4;;
rels := [ s0*s0, s1*s1, s2*s2, s3*s3, s0*s2*s0*s2,
s0*s3*s0*s3, s1*s3*s1*s3, s2*s3*s2*s3,
s1*s2*s1*s2*s1*s2, s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1,
s2*s0*s1*s0*s1*s2*s0*s1*s0*s1*s2*s0*s1*s0*s1*s2*s0*s1*s0*s1*s2*s0*s1*s0*s1*s2*s0*s1*s0*s1 ];;
poly := F / rels;;
```
Permutation Representation (Magma) :
```s0 := Sym(83)!( 4, 8)( 5, 9)( 6, 7)(10,19)(11,20)(12,21)(13,26)(14,27)(15,25)
(16,24)(17,22)(18,23)(31,35)(32,36)(33,34)(37,46)(38,47)(39,48)(40,53)(41,54)
(42,52)(43,51)(44,49)(45,50)(58,62)(59,63)(60,61)(64,73)(65,74)(66,75)(67,80)
(68,81)(69,79)(70,78)(71,76)(72,77);
s1 := Sym(83)!( 1,10)( 2,12)( 3,11)( 4,13)( 5,15)( 6,14)( 7,16)( 8,18)( 9,17)
(20,21)(23,24)(26,27)(28,66)(29,65)(30,64)(31,69)(32,68)(33,67)(34,72)(35,71)
(36,70)(37,57)(38,56)(39,55)(40,60)(41,59)(42,58)(43,63)(44,62)(45,61)(46,75)
(47,74)(48,73)(49,78)(50,77)(51,76)(52,81)(53,80)(54,79);
s2 := Sym(83)!( 1,28)( 2,30)( 3,29)( 4,33)( 5,32)( 6,31)( 7,35)( 8,34)( 9,36)
(10,53)(11,52)(12,54)(13,46)(14,48)(15,47)(16,51)(17,50)(18,49)(19,40)(20,42)
(21,41)(22,45)(23,44)(24,43)(25,38)(26,37)(27,39)(55,57)(58,59)(62,63)(64,79)
(65,81)(66,80)(67,75)(68,74)(69,73)(70,77)(71,76)(72,78);
s3 := Sym(83)!(82,83);
poly := sub<Sym(83)|s0,s1,s2,s3>;
```
Finitely Presented Group Representation (Magma) :
```poly<s0,s1,s2,s3> := Group< s0,s1,s2,s3 | s0*s0, s1*s1, s2*s2,
s3*s3, s0*s2*s0*s2, s0*s3*s0*s3, s1*s3*s1*s3,
s2*s3*s2*s3, s1*s2*s1*s2*s1*s2, s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1,
s2*s0*s1*s0*s1*s2*s0*s1*s0*s1*s2*s0*s1*s0*s1*s2*s0*s1*s0*s1*s2*s0*s1*s0*s1*s2*s0*s1*s0*s1 >;
```
to this polytope | 1,855 | 3,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-51 | latest | en | 0.449396 |
https://differencess.com/one-dimensional-1d-array-vs-two-dimensional-2d-array/ | 1,719,128,846,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00740.warc.gz | 173,357,277 | 19,062 | # One Dimensional 1d Array Vs Two Dimensional 2d Array
In this article, we will be discussing the difference between 1d arrays and 2d arrays. We will start by looking at what 1d arrays are and how they can be used, and then move on to 2d arrays.
## What is a One Dimensional Array?
A one-dimensional array is a data structure that stores information in a linear fashion. This means that the data is organized in rows and columns, like a table.
One-dimensional arrays are useful for storing small amounts of data. They are also simple to use and can be stored in memory or on disk.
One-dimensional arrays are sometimes called row-major arrays because the first dimension (the number of rows) is important. Other terms used to describe one-dimensional arrays include row vector and column vector.
## What is a Two Dimensional Array?
A two-dimensional array is a data structure that stores data in rows and columns. It is often used in computer science and mathematics, especially in the fields of artificial intelligence and machine learning.
One-dimensional arrays are similar to two-dimensional arrays, but they only store data in one direction. This makes them useful for certain types of tasks, but they are not as versatile as two-dimensional arrays.
## The Differences Between One Dimensional and Two Dimensional Arrays
1. One-dimensional arrays are simpler to understand and use than two-dimensional arrays.
2. Two-dimensional arrays are more efficient when it comes to storage space and processing time.
One-dimensional arrays are simpler to understand and use than two-dimensional arrays. They consist of only a single row and column, whereas two-dimensional arrays have multiple rows and columns.
Two-dimensional arrays are more efficient when it comes to storage space and processing time. The layout of data in two-dimensional arrays is accordance with the rules of mathematics, which makes it faster to process the data.
## What is a 1d Array?
A 1d array is a data structure that stores data in row-major order. This means that the first dimension of the array is always the largest dimension and the last dimension is always the smallest dimension.
When you create a 1d array, you specify its size in terms of length (number of elements). The first element in the array will have a length of 1, the second element will have a length of 2, and so on.
A 2d array is also known as a d array. A d array stores data in column-major order. This means that the first dimension of the array is always the smallest dimension and the last dimension is always the largest dimension.
When you create a 2d array, you specify its size in terms of width (number of columns) and height (number of rows). The first element in the array will have a width of 1 and a height of 1, the second element will have a width of 2 and a height of 2, and so on.
## What is a 2d Array?
A 2d array is a data structure that stores data in two dimensions. It is similar to a one-dimensional array, but it has additional capabilities.
One important difference between a one-dimensional and a two-dimensional array is that a two-dimensional array can hold more data than a one-dimensional array.
Another advantage of using a 2d array is that you can access data in any direction. This allows you to more easily search for specific pieces of data.
2d arrays are commonly used in computer science and engineering applications. They are also used in video gaming and other graphical applications.
## How does the 1d Array compare to the 2d Array?
One-dimensional arrays are often used when the data is small and the array will not be accessed often. Two-dimensional arrays allow for more efficient storage of data as each row and column can be stored in a separate memory location. This is important when large arrays are being processed.
One major advantage of two-dimensional arrays is that they can be processed more quickly. For example, if you have an array of 1000 data points and you want to find the median value, you can do this in a two-dimensional array in just a few seconds rather than minutes or hours with a one-dimensional array.
When comparing one-dimensional arrays to two-dimensional arrays, it is important to consider the data type, size, usage, and complexity of the problem.
## How does an 1d Array differ from a 2d Array?
An 1d Array is a one-dimensional array that contains data in a single row and column. A 2d Array is a two-dimensional array that contains data in two rows and two columns.
1d Arrays are faster to access than 2d Arrays because they require less memory. However, 1d Arrays do not allow you to search across the entire array. 2d Arrays, on the other hand, allow you to search across both the rows and the columns of the array.
1d Arrays are most commonly used when you need to store data that is organized in a grid or matrix format. For example, you might use an 1d Array to store data about a football team.
2d Arrays are more often used when you need to store data that is not organized in a grid or matrix format. For example, you might use a 2d Array to store pictures of people.
## How to create an 1d Array and a 2d Array?
In this blog post, we will be discussing how to create an 1d Array and a 2d Array.
To create an 1d Array, you will need to use the following code:
var arr1 = new Array(); arr1.push(“a”); arr1.push(“b”); arr1.push(“c”);
arr1.push(“d”);
To create a 2d Array, you will need to use the following code:
var arr2 = new Array(); arr2.push({x: 5, y: 8}); arr2.push({x: -5, y: 9});
arr2.push({x: 10, y: 11});
arr2.push({x: 12, y: 13});
arr2.push({x: 14, y: 15});
## How to use an 1d Array and a 2d Array?
In this blog post, we will be discussing how to use an 1d Array and a 2d Array. An 1d Array is a type of data structure that stores data in a single row, while a 2d Array is a type of data structure that stores data in two rows.
An 1d Array is useful when you need to store small amounts of data, while a 2d Array is more suitable when you need to store large amounts of data.
Let’s take a look at an example to see how these arrays work. Suppose you have the following list of numbers:
1, 2, 3, 4
You can create an 1d Array using the following code:
var numArray = new Array(); numArray[0] = 1; numArray[1] = 2; numArray[2] = 3; //etc…
The above code creates an array containing the values 1, 2, 3, 4. You can access any element in the array using the index number. For example, suppose you want to get the value stored at position 3 in the array:
numArray[3]
## Summary
1. In this article, we will compare and contrast the two most popular d-array data storage formats: one-dimensional (1D) and two-dimensional (2D).
2. 1D arrays are simple to understand and manage, but they can only store a finite amount of data. Two-dimensional arrays, on the other hand, are more complex to understand and manage, but they can hold a virtually unlimited amount of data.
3. The main advantage of 2D arrays is that they can be more efficient when it comes to storing data. This is because they can use more space per row and per column, which means that your data will be stored faster and will take up less space on your hard drive.
4. Ultimately, the decision whether or not to use a 1D or 2D array depends on the specific needs of your application. If you need to store a relatively small amount of data, then a 1D array is likely the best option for you. However, if you need to store a large amount of data, then a 2D array may be a better option for you. | 1,763 | 7,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-26 | latest | en | 0.935756 |
https://www.doorsteptutor.com/Exams/IMO/Class-7/Questions/Part-19.html | 1,591,330,295,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348492427.71/warc/CC-MAIN-20200605014501-20200605044501-00532.warc.gz | 678,025,513 | 11,249 | # IMO Level 1- Mathematics Olympiad (SOF) Class 7: Questions 131 - 137 of 1022
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## Question number: 131
Edit
MCQ▾
### Question
If Find the value of a, b and c respectively.
### Choices
Choice (4)Response
a.
b.
9,
c.
d.
## Question number: 132
Edit
MCQ▾
### Question
Tom and Shubh have Rs. 41 altogether. of Tom’s money is Rs. 2 more than of Shubh’s money. How much does Tom have?
### Choices
Choice (4)Response
a.
Rs. 30.50
b.
Rs. 29
c.
Rs. 27.5
d.
Rs. 20
## Question number: 133
Edit
MCQ▾
### Question
A bowler took 20 wickets for 642 runs, and then his average score per wicket is ________.
### Choices
Choice (4)Response
a.
12.3 runs/wicket
b.
32.1 runs/wicket
c.
21.4 runs/wicket
d.
22 runs/wicket
## Question number: 134
Edit
MCQ▾
### Question
Kavya runs miles every Tuesday, Wednesday and Thursday. How many miles will she run in a month in which there are 4 Tuesday, 4 Wednesday and 4 Thursday?
### Choices
Choice (4)Response
a.
44 miles
b.
18 miles
c.
52 miles
d.
36 miles
## Question number: 135
Edit
MCQ▾
### Question
A jar is full of mango juice. This amount is equal to 10 full glasses. When 1 full glass is drunk, what fraction of the jar is still left with orange juice?
### Choices
Choice (4)Response
a.
b.
c.
d.
## Question number: 136
Edit
MCQ▾
### Question
What fraction of the figure is shaded?
### Choices
Choice (4)Response
a.
b.
c.
d.
## Question number: 137
Edit
MCQ▾
### Question
What fraction of the figure is shaded?
### Choices
Choice (4)Response
a.
b.
c.
d.
Developed by: | 556 | 1,861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-24 | latest | en | 0.805363 |
https://www.unitconverters.net/flow-mass/pound-minute-to-gigagram-second.htm | 1,726,854,252,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00245.warc.gz | 946,637,081 | 3,073 | Home / Flow - Mass Conversion / Convert Pound/minute to Gigagram/second
Convert Pound/minute to Gigagram/second
Please provide values below to convert pound/minute [lb/min] to gigagram/second [Gg/s], or vice versa.
From: pound/minute To: gigagram/second
Pound/minute to Gigagram/second Conversion Table
Pound/minute [lb/min]Gigagram/second [Gg/s]
0.01 lb/min7.5598728333333E-11 Gg/s
0.1 lb/min7.5598728333333E-10 Gg/s
1 lb/min7.5598728333333E-9 Gg/s
2 lb/min1.5119745666667E-8 Gg/s
3 lb/min2.26796185E-8 Gg/s
5 lb/min3.7799364166667E-8 Gg/s
10 lb/min7.5598728333333E-8 Gg/s
20 lb/min1.5119745666667E-7 Gg/s
50 lb/min3.7799364166667E-7 Gg/s
100 lb/min7.5598728333333E-7 Gg/s
1000 lb/min7.5598728333333E-6 Gg/s
How to Convert Pound/minute to Gigagram/second
1 lb/min = 7.5598728333333E-9 Gg/s
1 Gg/s = 132277357.31093 lb/min
Example: convert 15 lb/min to Gg/s:
15 lb/min = 15 × 7.5598728333333E-9 Gg/s = 1.133980925E-7 Gg/s | 379 | 930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-38 | latest | en | 0.373883 |
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Lecture17
Lecture17 - Advanced Operations Research Techniques IE316...
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Advanced Operations Research Techniques IE316 Lecture 17 Dr. Ted Ralphs
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IE316 Lecture 17 1 Reading for This Lecture Bertsimas 7.1-7.3
IE316 Lecture 17 2 Network Flow Problems Networks are used to model systems in which a commodity or commodities must be transported from one or supply points to one or more demand points along defined pathways. These models occur naturally in many contexts. Transportation Logistics Telecommunications Network flow problems are defined on graphs that define the structure of the pathways in the network.
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IE316 Lecture 17 3 Undirected Graphs An undirected graph G = ( N, E ) consists of A set of nodes N representing the supply and demand points. A set E of unordered pairs of nodes called edges representing the pathways joining pairs of nodes. We say that the edge { i, j } is incident to nodes i and j and i and j are its endpoints . The degree of a node is the number of edges incident to it. The degree of a graph is the maximum of the degrees of its nodes.
IE316 Lecture 17 4 Basic Definitions (Undirected) A walk is a finite sequence of nodes i 1 , . . . , i t such that { i k , i k +1 } ∈ E k = 1 , 2 , . . . , t - 1 . A walk is called a path if it has no repeated nodes. A cycle is a path with i 1 = i t , t > 2 . An undirected graph is said to be connected if for every pair of nodes i and j , there is a path from i to j . For undirected graphs, our convention will be to denote | N | = n and | E | = m .
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IE316 Lecture 17 5 Directed Graphs A directed graph G = ( N, A ) consists of a set of N nodes and a set A of ordered pairs of nodes called arcs .
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Lecture17 - Advanced Operations Research Techniques IE316...
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Ask a homework question - tutors are online | 576 | 2,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-13 | latest | en | 0.90404 |
https://braingenie.ck12.org/skills/102492/learn | 1,603,340,080,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878879.33/warc/CC-MAIN-20201022024236-20201022054236-00628.warc.gz | 243,059,487 | 1,645 | ### Sample Problem
David weighed 92 lb. 9 oz. last year. He gained 9 lb. 9 oz. since last year. How much does he weigh now?
lb. oz.
#### Solution
Sum: 92 lb. 9 oz. + 9 lb. 9 oz. = 102 lb. 2 oz. | 74 | 197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2020-45 | latest | en | 0.840409 |
https://www.exceldemy.com/ranking-in-excel-based-on-multiple-criteria/ | 1,716,828,878,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059044.17/warc/CC-MAIN-20240527144335-20240527174335-00495.warc.gz | 669,261,486 | 76,766 | # Ranking Based on Multiple Criteria in Excel (4 Cases)
Often you may need to rank items from a dataset based on multiple criteria. More specifically, you have to accomplish this task when there are ties in one column. In this instructive session, I’ll demonstrate 4 cases with the proper explanation of ranking in Excel based on multiple criteria.
## Ranking in Excel Based on Multiple Criteria: 4 Practical Cases
Let’s introduce today’s dataset where Scores of the Student in Math and Psychology are given according to their corresponding Group. Here, D6 and D7 cells are tied in column D. So, let’s apply the ranking with consideration of column E.
### 1. Using RANK.EQ and COUNTIFS Functions
In the beginning method, I’ll show you the combined use of the RANK.EQ function and the COUNTIFS function. To rank based on the two Scores, insert the following formula.
`=RANK.EQ(\$C5,\$C\$5:\$C\$15)+COUNTIFS(\$C\$5:\$C\$15,\$C5,\$D\$5:\$D\$15,">"&\$D5)`
Here, C5 and D5 are the starting cell of Score (Math) i.e. column C, and Score (Psychology) i.e. column D respectively.
⧬ Formula Explanation:
• The RANK.EQ function returns the rank number from the C5:C15 cell range based on the C5 cell. Unfortunately, it provides the same rank for the duplicate scores (e.g. rank number is 7 for C6, C7, and C12 cells).
• So, the COUNTIFS function is assigned in descending order (“>”&\$D5) to count duplicate scores. For example, the function returns 1 for the C7 cell and 2 for the C12 cell.
• However, when you sum the two outputs i.e. the output RANK.EQ function and the output of the COUNTIFS function, you’ll get the unique rank number for all students.
After pressing ENTER and using the Fill Handle tool, you’ll get the following output.
If you look closely at the above image, you’ll get that the Rank for Robert Smith is 7 (look at the B6:E6 cells) whereas it is 8 for Jim Brown (look at the B7:E7 cells).
### 2. Ranking Based on Multiple Criteria Using COUNTIF and COUNTIFS Functions
Similarly, you may use the COUNTIF function instead of the RANK.EQ function.
`=COUNTIF(\$C\$5:\$C\$15,"<"&\$C5)+COUNTIFS(\$C\$5:\$C\$15,\$C5,\$D\$5:\$D\$15,"<"&\$D5)+1`
Here, I want to rank the scores in ascending order (“<“&\$D5).
⧬ Formula Explanation:
• The COUNTIF function counts the number of cells having values greater than the corresponding cell (like C5 for James Smith, C6 for Jim Brown, and so on).
• Lastly, you have to add 1 with the output as the COUNTIF function returns 0 for the smallest values i.e. for the C13 cell.
So, the output will be as follows.
### 3. Applying RANK and SUMPRODUCT FunctionsÂ
Also, you can utilize both the RANK function and SUMPRODUCT function for ranking the items based on multiple criteria.
Now, look at the following dataset from where you need to rank based on the GRE Score (Quant) and Financial Aid. But the cell values of C10 and C11 are tied.
So, insert the following combined formula.
`=RANK(C5,\$C\$5:\$C\$15)+SUMPRODUCT(--(\$C\$5:\$C\$15=\$C5),--(D5<\$D\$5:\$D\$15))`
⧬ Formula Explanation:
• The RANK function returns the rank number from the \$C\$5:\$C\$15 cell range based on the C5 cell with the duplicates value in the C10 and C11 cells (the rank number is 2).
• And, the SUMPRODUCT function finds 0 in case of no tied values. But it returns 1 for the C10 cell.
• Notably, the () operator is used to return 1 instead of getting TRUE and 0 for FALSE.
• Thus, you can easily avoid the duplicate rank number using this formula.
Finally, the output will look as follows.
Instead of utilizing the RANK function, you can use the COUNTIF function. But you have to add 1 in that case.
`=COUNTIF(\$C\$5:\$C\$15,">"&\$C5)+SUMPRODUCT(--(\$C\$5:\$C\$15=\$C5),--(D5<\$D\$5:\$D\$15))+1`
Certainly, you’ll get the same output.
### 4. Ranking with Multiple Criteria by Group
What if you have some common Groups in your dataset? For example, the Science group covers C5:C6 and C11:C12 cells.
Fortunately, you can get the unique rank number dealing with both Group and Score. We have functions that can help us rank in Excel based on multiple criteria on groups.
#### 4.1. Using COUNTIFS Function
Using the COUNTIFS function, you can easily rank the Score by the assigned Group in descending order (“>”&D5).
`=COUNTIFS(\$C\$5:\$C\$15,C5,\$D\$5:\$D\$15,">"&D5)+1`
⧬ Formula Explanation:
• The COUNTIFS(\$C\$5:\$C\$15,C5) returns 4 as there are 4 strings available namely Science.
• And, the COUNTIFS(\$C\$5:\$C\$15,C5,\$D\$5:\$D\$15,”>”&D5) syntax returns 0 for the highest scores (e.g. for the E6 cell). That’s why you need to add 1.
Here, the scores are ranked based on the Group separately. For example, Jim Brown (B6 cell) is ranked 1st though the score of Mary Smith (B13 cell) is greeted than him.
Read More: How to Rank Within Group in Excel
#### 4.2. Using SUMPRODUCT Function
Similarly, you can use the following formula where the SUMPRODUCT function is used (ranking in ascending order).
`=SUMPRODUCT((C5=\$C\$5:\$C\$15)*(\$D5<\$D\$5:\$D\$15))+1`
⧬ Formula Explanation:
• The SUMPRODUCT((C5=\$C\$5:\$C\$15) returns 0.
• Besides, the SUMPRODUCT((C5=\$C\$5:\$C\$15)*(\$D5<\$D\$5:\$D\$15)) finds 2. But the SUMPRODUCT function returns for E7 cell is 0 as it is the smallest score. So, you need to 1 to avoid such type of error.
## Conclusion
That’s the end of today’s session. This is how you may accomplish raking in Excel based on multiple criteria. Anyway, if you have any queries or recommendations, please share them in the comments section.
## Related Articles
<< Go Back to Excel RANK Function | Excel Functions | Learn Excel
Get FREE Advanced Excel Exercises with Solutions!
MD. ABDUL KADER is an engineer with a talent for Excel and a passion for VBA programming. To him, programming is like a time-saving wizard, making data manipulation, file handling, and internet interactions a breeze. His skill set extends to SWM, GIS, RS, and SPSS. He holds a B.Sc in Urban & Regional Planning from Chittagong University of Engineering and Technology and has shifted to become a content developer. In this role, he crafts technical content centred around... Read Full Bio
1. I need your support to find a formula for aging stock to put a value to a range of time based on the remaining quantity and aging from invoiced quantity
2. Hi NGÂN,
The solution you want will require a combination of some functions like TODAY, COUNTIF, VLOOKUP, etc. Here is a post on our website that will help you.
https://www.exceldemy.com/stock-ageing-analysis-formula-in-excel/
We have several posts related to this topic too.
https://www.exceldemy.com/make-inventory-aging-report-in-excel/
https://www.exceldemy.com/excel-ageing-formula-for-30-60-90-days/
https://www.exceldemy.com/aging-of-accounts-receivable-in-excel/
I hope these articles will help get your job done. If not, please remember that we are just a text away!!
Thank you. Have a good day.
3. Thanks for providing the guide. How do you rank without duplicates in the case of 4. Ranking with Multiple Criteria by Group? So instead of having duplicate ranks, I want to avoid them without skipping any number. Thanks in advance
Rubayed Razib Suprov May 28, 2023 at 5:12 PM
Greetings Edward,
Thanks a lot for your Question in our blog post. Now the issue you have is a little bit unclear to me. Can you provide a sample output manually which will contain your desired result? In that way, your problem be more clear to us and in turn it will help us to resolve your problem.
4. How would you rank with multiple criteria and duplicates? In the initial example above, you see Jim Brown and Henry James with Science and 65 scores. How would you rank and not repeat numbers?
Thank you.
Mahfuza Anika Era Feb 29, 2024 at 11:51 AM
Hello GUILLERMO ALCALA,
Thank you for your comment. In the initial example, Jim Brown, Robert Smith, and Henry James scored 65 in Science. So if you rank them according to the score of Science, you will get repeated ranks. But, as you can see, these 3 students got different scores in Psychology. So, we have ranked them according to the E column (Psychology). Hence, the rank is not repeated, and they got different ranks according to Psychology score.
Regards
Mahfuza Anika Era
Exceldemy
5. I am looking for a ranking formula that will rank salesmen by region based on number of units sold and then (in the case of ties) by total sales amount. I need unique rankings so there are no duplicates. Any help you can provide is greatly appreciated. I have tried this so many ways and never getting the desired outcome.
Lutfor Rahman Shimanto Mar 24, 2024 at 12:34 PM
Hello Denise Sano
Thanks for visiting our blog and leaving an exciting comment. You want to rank salespeople by region based on number of units sold and then (in the case of ties) by total sales amount. However, developing such a formula using Excel’s built-in function would be time-consuming.
1. Press Alt+F11.
2. Click on Insert followed by Module.
3. Paste the following code into the Module and run it:
``````Sub RankSalesRepByRegion()
Dim ws As Worksheet
Dim lastRow As Long
Dim salesRepCol As Range, regionCol As Range, unitsSoldCol As Range, salesAmountCol As Range, rankCol As Range
Dim region As Range, uniqueRegions As New Collection
Dim i As Long
Set ws = ThisWorkbook.Sheets("Sheet1")
lastRow = ws.Cells(ws.Rows.Count, "D").End(xlUp).Row
Set salesRepCol = ws.Range("A2:A" & lastRow)
Set regionCol = ws.Range("B2:B" & lastRow)
Set unitsSoldCol = ws.Range("C2:C" & lastRow)
Set salesAmountCol = ws.Range("D2:D" & lastRow)
Set rankCol = ws.Range("E2:E" & lastRow)
For Each region In regionCol
On Error Resume Next
On Error GoTo 0
Next region
For i = 1 To uniqueRegions.Count
Dim regionName As String
regionName = uniqueRegions(i)
Dim salesReps() As String
Dim unitsSold() As Long
Dim salesAmount() As Double
Dim ranks() As Integer
Dim j As Long
Dim uniqueCount As Long
uniqueCount = 0
For j = 1 To lastRow - 1
If regionCol.Cells(j, 1).Value = regionName Then
uniqueCount = uniqueCount + 1
ReDim Preserve salesReps(1 To uniqueCount)
ReDim Preserve unitsSold(1 To uniqueCount)
ReDim Preserve salesAmount(1 To uniqueCount)
salesReps(uniqueCount) = salesRepCol.Cells(j, 1).Value
unitsSold(uniqueCount) = unitsSoldCol.Cells(j, 1).Value
salesAmount(uniqueCount) = salesAmountCol.Cells(j, 1).Value
End If
Next j
For j = 1 To uniqueCount - 1
For k = j + 1 To uniqueCount
If unitsSold(j) < unitsSold(k) Or (unitsSold(j) = unitsSold(k) And salesAmount(j) < salesAmount(k)) Then
Dim tempSalesRep As String
tempSalesRep = salesReps(j)
salesReps(j) = salesReps(k)
salesReps(k) = tempSalesRep
Dim tempUnitsSold As Long
tempUnitsSold = unitsSold(j)
unitsSold(j) = unitsSold(k)
unitsSold(k) = tempUnitsSold
Dim tempSalesAmount As Double
tempSalesAmount = salesAmount(j)
salesAmount(j) = salesAmount(k)
salesAmount(k) = tempSalesAmount
End If
Next k
Next j
ReDim ranks(1 To uniqueCount)
ranks(1) = 1
For j = 2 To uniqueCount
If unitsSold(j - 1) = unitsSold(j) And salesAmount(j - 1) = salesAmount(j) Then
ranks(j) = ranks(j - 1)
Else
ranks(j) = j
End If
Next j
For j = 1 To uniqueCount
For k = 1 To lastRow
If salesRepCol.Cells(k, 1).Value = salesReps(j) Then
rankCol.Cells(k, 1).Value = ranks(j)
Exit For
End If
Next k
Next j
Next i
End Sub``````
As a result, you get the intended rank like the following GIF.
I have attached the solution workbook for better understanding. Hopefully, the idea will help; good luck.
Regards
Lutfor Rahman Shimanto
Excel & VBA Developer
ExcelDemy
Advanced Excel Exercises with Solutions PDF | 3,150 | 11,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-22 | longest | en | 0.848291 |
https://calculator-online.net/salestax-calculator/ | 1,720,975,908,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00057.warc.gz | 127,575,764 | 54,286 | Pages
Category
# Sales Tax Calculator
The free sales tax calculator readily calculates the sales tax for your business revenues against the amounts and tax percentage you provide.
\$
%
How to calculate sales tax is the most common question that frequently asked by many people. The experts of calculator-online provided an efficient and tested sales tax calculator to get the ease of calculating sales tax. In this post, we are decided to tell you about the sale tax calculator and all you need to know about sales tax!
## What Is Sales Tax?
Sales tax is a type of consumption tax which you pay to the seller when you buy anything. It is imposed on everything. There is a fixed percentage of sales tax which retailers charge from the and then pay it to the government. It is imposed on daily life usage things as household items, fruits, vegetables and on luxury items. Different or related types of sales tax are:
• Wholesale sales tax
• Manufacturers sales tax
• Retail sales tax
• Gross receipts tax
• Excise tax
• Use tax
• Fair tax
• Turnover tax
• Securities turnover excise tax
Sales tax is different from value-added tax in a way that sales tax is only collected at the final point of sales from the consumer of the product rather than at every stage of product processing as a vat. It is an added amount based on the fixed percentage of the selling price of the goods.
### If you are buying any item costing Rs.2000 and the sales tax is 17%.
The total amount which you’ll have to pay is Rs.2340.
Some sales tax is not to be kept with the retailer. They are collecting sales tax from buyers because of the agreed with the government that they’ll pay tax value for each item if they allow them to work in a particular area.
## Sales Tax Calculator:
To calculate sales tax, you have to use a tool which can do it quickly. We have sales tax calculator that helps to unfold the question of how to find sales tax. You can use this sale tax calculator anytime sitting anywhere just by entering the amount and sales tax rate. Then hitting the calculate button will give you the total sum after sales tax. Moreover, when you know the sales tax rate, the sellers can’t cheat you in charging the wrong amount. If you want to figure out how much amount you have to pay sales tax in the mentioned prices if they are exclusive of tax you can use our sales tax, calculator. We know many of you people are not sharp enough in calculations and can’t add or subtract a percentage in mind quickly.
So, for those people, we have a fantastic tool for sales tax calculator. A sales tax is imposed on most items you need to calculate it many times of the day. Even if you went out shopping or you go out to have a cup of tea or eating dinner, you have to pay the sales tax on your total bill. So, to cross-check the bill amount you need to have a tool which can do it quickly. Here you need to go to calculator-online.net and enter into the sales tax calc to know how much is sales tax you have to pay.
### How to Calculate Sales Tax With Our Sale Tax Calculator:
Swipe down to unfold the question of how to figure sales tax with this efficient tool.
• In the above sales tax rate calculator, you ought to enter the net price of a product in the field of 'Net Price.'
• Right after, you have to enter the tax amount in the field of 'Sale Tax %.'
• Finally, hit the calculate button to find out the gross price and sale tax value
Well, there is no need to waste your time in the lengthy calculation, you just have to stick with the above steps to get sale tax value within a couple of seconds. And if the seller is charging you more amount you can confidently argue with him on the tax rate and total amount. We know the need for fastest performing tools in your busy life. No one wants to spend extra time on any calculations which is not much needed. In this life, many rich people don’t bother about the value of tax they are paying. But for limited-income individuals and for those who can’t bear to spend extra money on lavish things the cost of the tax is significant as they had to plan their whole month in a limited amount of money. Our site is giving you a tool to which you can benefit in giving the right sum of money while purchasing anything. The sales tax calculator is allowing you to know the proper amount of money to give to the seller with the added amount of sales tax in the selling price.
Once you know that you have paid the right amount you’ll have the mental satisfaction that you have not wasted your money. It is possible only if you calculate the sales tax at the moment when you are buying anything. Otherwise regretting after purchase and coming to the home is of no use if the seller has charged you more money with the added sales tax. Use our best free online available sales tax rate calculator whenever you are buying anything or enjoying a dinner at a restaurant to know the amount you have to pay. And the sellers can’t cheat you in the sales tax amount.
## Importance of Sales Tax:
The money of sales tax is used by the government for the country’s development and economic growth. It is used to provide free public services to the community. This money is used to make more public parks, schools, roads, bridges and recreational places. The government also uses these funds of the tax to maintain already established public places. It is used to pay the salaries of various government workers such as police, army or to pay those people who came to collect garbage at your doorstep daily. It is therefore not wasted as the government is using for the betterment of a country.
## How to Figure Out Sales Tax and Gross Price?
If you want to figure out sales tax rate manually, then don’t here we are decided to explain it with example. Let us make it 7%.
• First of all, you ought to find out the net price of a product, let’s use here 75
• Very next, you ought to multiply your net price by 7%, so by 0.07, to determine the tax amount: means, 75 * 0.07 = 5.25
• Finally, you ought to add the tax amount into the net price to figure out the gross price: means, 75+5.25 = 80.25
With the above example:
Gross Price = 80.25
However, you can get the same result with the ease of the above sale tax calculator. Well, stay with us to know about the United States Sales Tax Rate!
## United States Sales Tax Rate:
In the US or United States, the sales tax rate is an amount that is charged to consumers depends on the purchase price of certain goods and services. Generally, sales taxes are collected on all sales of tangible goods (and sometimes on services) in the state. Keep in mind, sale tax rates are varied in different states of the US; you can utilize our advanced ca sales tax calculator to get the sale tax value in CA. Additionally, the sale tax rate ranges from 0% to 16% concerning the type of good or service that is given in the state, and remember that all states differ in their enforcement of sales tax. If you are in California and wants to calculate sale tax ca, then you ought to stick with the above sales tax calculator California!
## Benefit of Using Sales Tax Calculator
Our simple sales tax calculator can be used by anyone whether one is good or poor in mathematics. It can benefit both users equally. We have not designed it complexly; we kept it simple so that even a layperson can use it to calculate the value of sales tax. Although calculating sales tax doesn’t involve complicated formulas, but still, it’s hard for some people to calculate it manually or in mind. They have to use a calculator for knowing the sales tax amount. For these people, calculator-online.net is the site to go. Remember or bookmark this site and use our free sale tax calculator whenever you want to calculate any amount on which sales tax is imposed. | 1,662 | 7,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-30 | latest | en | 0.972327 |
https://uk.mathworks.com/matlabcentral/cody/problems/1855-usage-of-java-math-n-choose-k-with-unlimited-precision/solutions/2911972 | 1,603,522,784,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882102.31/warc/CC-MAIN-20201024051926-20201024081926-00257.warc.gz | 584,258,464 | 17,233 | Cody
# Problem 1855. Usage of java.math : N Choose K with unlimited precision
Solution 2911972
Submitted on 4 Sep 2020 by pign_an
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
tic N=5;K=2; NK=nchoosekJava(N,K); toc assert(strcmp(NK,num2str(nchoosek(N,K))))
Elapsed time is 0.022205 seconds.
2 Pass
tic N=randi(10); K=randi(N); NK=nchoosekJava(N,K); toc assert(strcmp(NK,num2str(nchoosek(N,K))))
Elapsed time is 0.004225 seconds.
3 Pass
tic N=100; K=50; NK=nchoosekJava(N,K); toc assert(strcmp(NK,'100891344545564193334812497256'))
Elapsed time is 0.017855 seconds.
4 Pass
tic N=200; K=75; NK=nchoosekJava(N,K); toc assert(strcmp(NK,'168849997346404286704489530268603459022868706883102845056'))
Elapsed time is 0.023110 seconds.
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
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# Transmission Line Basics
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### Transmission Line Basics
1. 1. Transmission Line Basics II - Class 6Prerequisite Reading assignment: CH2TARA SAIKUMAR
2. 2. 2Real Computer Issues data Dev a Dev b Signal Measured Clk here Switch ThresholdAn engineer tells you the measured clock is non-monotonicand because of this the flip flop internally may double clockthe data. The goal for this class is to by inspectiondetermine the cause and suggest whether this is a problemor not. Transmission Lines Class 6
3. 3. 3Agenda The Transmission Line Concept Transmission line equivalent circuits and relevant equations Reflection diagram & equation Loading Termination methods and comparison Propagation delay Simple return path ( circuit theory, network theory come later) Transmission Lines Class 6
4. 4. 4Two Transmission Line Viewpoints Steady state ( most historical view) Frequency domain Transient Time domain Not circuit element Why? We mix metaphors all the time Why convenience and history Transmission Lines Class 6
5. 5. 5 Transmission Line ConceptPower Frequency (f) is @ 60 Hz Power Wavelength (λ ) is 5× 10 m 6 Plant ( Over 3,100 Miles) Consumer Home Transmission Lines Class 6
6. 6. PC Transmission Lines 6Signal Frequency (f) is approaching 10 GHz Integrated Circuit Stripline Wavelength (λ ) is 1.5 cm ( 0.6 inches) Microstrip T PCB substrate Cross section view taken here Stripline W Via Micro- FR4 Dielectric Copper Trace Cross Section of Above PCB Strip Signal (microstrip) Ground/Power T Copper Plane Signal (stripline) Signal (stripline) Ground/Power Signal (microstrip) W Transmission Lines Class 6
7. 7. 7Key point about transmission line operation Voltage and current on a transmission line is a function of both time and position. I2 V = f ( z, t ) I1 I = f ( z, t ) V1 V2 dz The major deviation from circuit theory with transmission line, distributed networks is this positional dependence of voltage and current! Must think in terms of position and time to understand transmission line behavior This positional dependence is added when the assumption of the size of the circuit being small compared to the signaling wavelength Transmission Lines Class 6
8. 8. 8Examples of Transmission LineStructures- I Cables and wires (a) Coax cable (b) Wire over ground (c) Tri-lead wire (d) Twisted pair (two-wire line) Long distance interconnects + + - (a) (b) - - + - + - (c) (d) Transmission Lines Class 6
9. 9. 9Segment 2: Transmission line equivalentcircuits and relevant equations Physics of transmission line structures Basic transmission line equivalent circuit ?Equations for transmission line propagation Transmission Lines Class 6
10. 10. 10E & H Fields – Microstrip Case How does the signal move Signal path from source to load? Y Z (into the page) X Electric field Remember fields are setup given field Magnetic an applied forcing function. (Source) Ground return path The signal is really the wave propagating between the conductors Transmission Lines Class 6
11. 11. Transmission Line “Definition” 11 General transmission line: a closed system in which power is transmitted from a source to a destination Our class: only TEM mode transmission lines A two conductor wire system with the wires in close proximity, providing relative impedance, velocity and closed current return path to the source. Characteristic impedance is the ratio of the voltage and current waves at any one position on the transmission line V Z0 = I Propagation velocity is the speed with which signals are transmitted through the transmission line in its surrounding medium. c v= εr Transmission Lines Class 6
12. 12. 12Presence of Electric and Magnetic Fields H I I + ∆I I I + ∆I + + + + E V V + ∆V V V + ∆V H I I + ∆I I I + ∆I - - - - Both Electric and Magnetic fields are present in the transmission lines These fields are perpendicular to each other and to the direction of wave propagation for TEM mode waves, which is the simplest mode, and assumed for most simulators(except for microstrip lines which assume “quasi-TEM”, which is an approximated equivalent for transient response calculations). Electric field is established by a potential difference between two conductors. Implies equivalent circuit model must contain capacitor. Magnetic field induced by current flowing on the line Implies equivalent circuit model must contain inductor. Transmission Lines Class 6
13. 13. 13T-Line Equivalent Circuit General Characteristics of Transmission Line Propagation delay per unit length (T0) { time/distance} [ps/in] Or Velocity (v0) {distance/ time} [in/ps] Characteristic Impedance (Z0) Per-unit-length Capacitance (C0) [pf/in] Per-unit-length Inductance (L0) [nf/in] Per-unit-length (Series) Resistance (R0) [Ω/in] Per-unit-length (Parallel) Conductance (G0) [S/in] lR0 lL0 lG0 lC0 Transmission Lines Class 6
14. 14. 14Ideal T Line Ideal (lossless) Characteristics of Transmission Line lL0 Ideal TL assumes: Uniform line lC0 Perfect (lossless) conductor (R0→ 0) Perfect (lossless) dielectric (G0→ 0) We only consider T0, Z0 , C0, and L0. A transmission line can be represented by a cascaded network (subsections) of these equivalent models. The smaller the subsection the more accurate the model The delay for each subsection should be no larger than 1/10th the signal rise time. Transmission Lines Class 6
15. 15. Signal Frequency and Edge Rate 15 vs. Lumped or Tline ModelsIn theory, all circuits that deliver transient power fromone point to another are transmission lines, but if thesignal frequency(s) is low compared to the size of thecircuit (small), a reasonable approximation can beused to simplify the circuit for calculation of the circuittransient (time vs. voltage or time vs. current)response. Transmission Lines Class 6
16. 16. 16T Line Rules of Thumb So, what are the rules of thumb to use? May treat as lumped Capacitance Use this 10:1 ratio for accurate modeling of transmission lines Td < .1 Tx May treat as RC on-chip, and treat as LC for PC board interconnect Td < .4 Tx Transmission Lines Class 6
17. 17. Other “Rules of Thumb” 17 Frequency knee (Fknee) = 0.35/Tr (so if Tr is 1nS, Fknee is 350MHz) This is the frequency at which most energy is below Tr is the 10-90% edge rate of the signal Assignment: At what frequency can your thumb be used to determine which elements are lumped? Assume 150 ps/in Transmission Lines Class 6
18. 18. When does a T-line become a T-Line? 18 Whether it is a bump or a mountain depends on the ratio of its When do we need to size (tline) to the use transmission line size of the vehicle analysis techniques vs. (signal lumped circuit wavelength) analysis? Similarly, whether or not a line is to be considered as a transmission line depends on the ratio of length of the line (delay) to the wavelength of Wavelength/edge rate Tline the applied frequency or the rise/fall edge of the Transmission Lines Class 6 signal
19. 19. Equations & Formulas How to model & explain transmission line behavior
20. 20. 20Relevant Transmission Line Equations Propagation equation γ = ( R + jωL)(G + jωC ) = α + jβ α is the attenuation (loss) factor β is the phase (velocity) factor Characteristic Impedance equation ( R + j ωL ) Z0 = (G + jωC ) In class problem: Derive the high frequency, lossless approximation for Z0 Transmission Lines Class 6
21. 21. 21Ideal Transmission Line Parameters Knowing any two out of Z0, Td, C0, and L0, the other two L0 can be calculated. Z0 = ; T d = L0 C0 ; C0 and L0 are reciprocal C0 functions of the line cross- T0 sectional dimensions and C0 = ; L0 = Z 0 T 0 ; Z0 are related by constant me. 1 ε is electric permittivity v0 = ; C0 L0 = µε; ε 0= 8.85 X 10-12 F/m (free space) µε ε ri s relative dielectric constant µ = µr µ0 ; ε = εr ε0 . µ is magnetic permeability µ 0= 4p X 10-7 H/m (free space) µ r is relative permeability Don’t forget these relationships and what they mean! Transmission Lines Class 6
22. 22. Parallel Plate Approximation 22 Assumptions TC TEM conditions ε TD Uniform dielectric (ε ) between conductors WC TC<< TD; WC>> TD ε * PlateArea Base T-line characteristics are C= function of: d equation Material electric and WC F WC pF magnetic properties C0 ε⋅ ⋅ 8.85 ⋅ε r ⋅ ⋅ TD m TD m Dielectric Thickness (TD) TD F T D µH Width of conductor (WC) L0 µ⋅ ⋅ 0.4 ⋅π ⋅µ r ⋅ ⋅ Trade-off WC m WC m TD ; C0 , L0 , Z0 TD µr Z0 377 ⋅ ⋅ ⋅Ω WC ; C0 , L0 , Z0 WC εrTo a first order, t-line capacitance and inductance canbe approximated using the parallel plate approximation. Transmission Lines Class 6
23. 23. 23Improved Microstrip Formula Parallel Plate Assumptions + WC Large ground plane with TC zero thickness ε TD To accurately predict microstrip impedance, you must calculate the effective dielectric constant. From Hall, Hall & McCall: 87 5.98TD Z0 ≈ ln Valid when: εr + 1.41 0.8WC + TC 0.1 < WC/TD < 2.0 and 1 < r < 15 εr + 1 εr − 1 TC εe = + + F − 0.217( εr − 1) 2 12TD WCTD 2 1+ WC 2 You can’t beat WC 0.02(εr −1)1 − a field solver WC for TD <1 F= TD 0 for WC >1 TD Transmission Lines Class 6
24. 24. 24Improved Stripline Formulas Same assumptions as WC TD1 used for microstrip ε TC apply here TD2 From Hall, Hall & McCall:Symmetric (balanced) Stripline Case TD1 = TD2 60 4(TD1 + TD1) Z 0 sym ≈ ln 0.67π (0.8WC + TC ) εr Valid when WC/(TD1+TD2) < 0.35 and TC/(TD1+TD2) < 0.25 You can’t beat aOffset (unbalanced) Stripline Case TD1 > TD2 field solver Z 0 sym(2 A, WC , TC , εr ) ⋅ Z 0 sym(2 B, WC , TC , εr ) Z 0offset ≈ 2 Z 0 sym(2 A,WC , TC , εr ) + Z 0 sym(2 B,WC , TC , εr ) Transmission Lines Class 6
25. 25. 25Refection coefficient Signal on a transmission line can be analyzed by keeping track of and adding reflections and transmissions from the “bumps” (discontinuities) Refection coefficient Amount of signal reflected from the “bump” Frequency domain ρ=sign(S11)*|S11| If at load or source the reflection may be called gamma (ΓL or Γs) Time domain ρ is only defined a location The “bump” Time domain analysis is causal. Frequency domain is for all time. We use similar terms – be careful Reflection diagrams – more later Transmission Lines Class 6
26. 26. Reflection and Transmission 26 Incident 1+ρ Transmitted ρ ReflectedReflection Coeficient Transmission Coeffiecent Zt − Z0 ρ Zt − Z0 τ (1 + ρ) "" → "" τ 1+ Zt + Z0 Zt + Z0 2⋅ Zt τ Zt + Z0 Transmission Lines Class 6
27. 27. 27Special Cases to Remember A: Terminated in Zo Zs − Zo Zo ρ = Zo Zo = 0 Vs Zo + Zo B: Short Circuit Zs − Zo ρ = 0 Zo = −1 Vs 0 + Zo C: Open Circuit Zs ∞ − Zo Zo ρ= =1 Vs ∞ + Zo Transmission Lines Class 6
28. 28. 28Assignment – Building the SI Tool Box Compare the parallel plate approximation to the improved microstrip and stripline formulas for the following cases: Microstrip: WC = 6 mils, TD = 4 mils, TC = 1 mil, εr = 4 Symmetric Stripline: WC = 6 mils, TD1 = TD2 = 4 mils, TC = 1 mil, εr = 4 Write Math Cad Program to calculate Z0, Td, L & C for each case. What factors cause the errors with the parallel plate approximation? Transmission Lines Class 6
29. 29. 29Transmission line equivalent circuits andrelevant equations Basic pulse launching onto transmission lines Calculation of near and far end waveforms for classic load conditions Transmission Lines Class 6
30. 30. Review: Voltage Divider Circuit 30 Consider the simple circuit that RS contains source voltage VS, source RL VS VL resistance RS, and resistive load RL. The output voltage, VL is RL easily calculated VL = VS from the source RL + R S amplitude and the values of the two series resistors. Why do we care for? Next page…. Transmission Lines Class 6
31. 31. 31Solving Transmission Line ProblemsThe next slides will establish a procedure that will allow you to solve transmission line problems without the aid of a simulator. Here are the steps that will be presented: Determination of launch voltage & final “DC” or “t =0” voltage Calculation of load reflection coefficient and voltage delivered to the load Calculation of source reflection coefficient and resultant source voltage These are the steps for solving all t-line problems. Transmission Lines Class 6
32. 32. Determining Launch Voltage 32 TD Rs A B Vs Zo 0 Vs Rt (initial voltage) t=0, V=Vi Z0 Rt Vi = VS Vf = VS Z 0 + RS Rt + RS Step 1 in calculating transmission line waveforms is to determine the launch voltage in the circuit. The behavior of transmission lines makes it easy to calculate the launch & final voltages – it is simply a voltage divider! Transmission Lines Class 6
33. 33. 33Voltage Delivered to the Load TD Vs Rs A B Vs Zo Rt 0 (initial voltage) t=0, V=Vi (signal is reflected) t=2TD, ρΒ = Rt − Zo V=Vi + ρB(Vi) +ρA(ρB)(Vi ) t=TD, V=Vi +ρB(Vi ) Vreflected = ρ Β (Vincident) Rt + Zo VB = Vincident + Vreflected Step 2: Determine VB in the circuit at time t = TD The transient behavior of transmission line delays the arrival of launched voltage until time t = TD. VB at time 0 < t < TD is at quiescent voltage (0 in this case) Voltage wavefront will be reflected at the end of the t-line VB = Vincident + Vreflected at time t = TD Transmission Lines Class 6
34. 34. 34Voltage Reflected Back to the Source Rs A B Vs Zo 0 Vs ρA ρB Rt TD (initial voltage) t=0, V=Vi (signal is reflected) t=2TD, V=Vi + ρB (Vi) + ρA B )(Vi ) (ρ t=TD, V=Vi + ρB (Vi ) Transmission Lines Class 6
35. 35. Voltage Reflected Back to the Source 35 − Zo Vreflected = ρ Α (Vincident) ρ Α = Rs Rs + Zo VA = Vlaunch + Vincident + VreflectedStep 3: Determine VA in the circuit at time t = 2TD The transient behavior of transmission line delays the arrival of voltage reflected from the load until time t = 2TD. VA at time 0 < t < 2TD is at launch voltage Voltage wavefront will be reflected at the source VA = Vlaunch + Vincident + Vreflected at time t = 2TDIn the steady state, the solution converges to VB = VS[Rt / (Rt + Rs)] Transmission Lines Class 6
36. 36. 36Problems Solved Homework Consider the circuit shown to the right with a resistive load, assume propagation RS I1 I2 Z0 ,Τ 0 delay = T, RS= Z0 . VS V1 l V2 RL Calculate and show the wave forms of V1(t),I1(t),V2(t), and I2(t) for (a) RL= ∞ and (b) RL= 3Z0 Transmission Lines Class 6
37. 37. 37Step-Function into T-Line: Relationships Source matched case: RS= Z0 V1(0) = 0.5VA, I1(0) = 0.5IA Γ S = 0, V(x,∞ ) = 0.5VA(1+ Γ L) Uncharged line V2(0) = 0, I2(0) = 0 Open circuit means RL= ∞ Γ L = ∞ /∞ = 1 V1(∞) = V2(∞) = 0.5VA(1+1) = VA I1(∞) = I2 (∞) = 0.5IA(1-1) = 0 Solution Transmission Lines Class 6
38. 38. 38Step-Function into T-Line with Open Ckt At t = T, the voltage wave reaches load end and doubled wave travels back to source end V1(T) = 0.5VA, I1(T) = 0.5VA/Z0 V2(T) = VA, I2 (T) = 0 At t = 2T, the doubled wave reaches the source end and is not reflected V1(2T) = VA, I1(2T) = 0 V2(2T) = VA, I2(2T) = 0 Solution Transmission Lines Class 6
39. 39. 39Waveshape:Step-Function into T-Line with Open Ckt I1 IA I2 RS I1 I2Current (A) 0.75IA Z0 ,Τ 0 l 0.5I A VS V1 V2 Open 0.25IA 0 Τ 2Τ 3Τ 4Τ Time (ns) VA V1 This is called V2 “reflected waveVoltage (V) 0.75VA switching” 0.5VA 0.25VA Solution 0 Τ 2Τ 3Τ 4Τ Time (ns) Transmission Lines Class 6
40. 40. 40Problem 1b: Relationships Source matched case: RS= Z0 V1(0) = 0.5VA, I1(0) = 0.5IA Γ S = 0, V(x,∞ ) = 0.5VA(1+ Γ L) Uncharged line V2(0) = 0, I2(0) = 0 RL= 3Z0 Γ L = (3Z0 -Z0) / (3Z0 +Z0) = 0.5 V1(∞) = V2(∞) = 0.5VA(1+0.5) = 0.75VA I1(∞) = I2(∞) = 0.5IA(1-0.5) = 0.25IA Solution Transmission Lines Class 6
41. 41. 41Problem 1b: Solution At t = T, the voltage wave reaches load end and positive wave travels back to the source V1(T) = 0.5VA, I1(T) = 0.5IA V2(T) = 0.75VA , I2(T) = 0.25IA At t = 2T, the reflected wave reaches the source end and absorbed V1(2T) = 0.75VA , I1(2T) = 0.25IA V2(2T) = 0.75VA , I2(2T) = 0.25IA Solution Transmission Lines Class 6
42. 42. Waveshapes for Problem 1b 42 I1 IA I2 I1 I2 RS 0.75IA Z0 ,Τ 0 Current (A) l 0.5IA VS V1 V2 RL 0.25IA 0 Τ 2Τ 3Τ 4Τ Time (ns) I1 VA I2 Note that a 0.75VA properly terminated Voltage (V) 0.5VA wave settle out at 0.5 V 0.25VA Solution 0 Τ 2Τ 3Τ 4Τ Time (ns) Solution Transmission Lines Class 6
43. 43. Transmission line step response 43 Introduction to lattice diagram analysis Calculation of near and far end waveforms for classic load impedances Solving multiple reflection problems Complex signal reflections at different types of transmission line “discontinuities” will be analyzed in this chapter. Lattice diagrams will be introduced as a solution tool. Transmission Lines Class 6
44. 44. 44Lattice Diagram Analysis – Key Concepts V(source) Zo V(load) Vs Rs The lattice diagram is a 0 TD = N ps Vs Rt tool/technique to simplify the accounting of ρsource ρload reflections and waveforms V(load) Time V(source) Diagram shows the boundaries 0 a (x =0 and x=l) and the reflection A’ coefficients (GL and GL ) N ps A Time (in T) axis shown b vertically Slope of the line should 2N ps c B’ indicate flight time of signal Particularly important for multiple reflection problems using both 3N ps B microstrip and stripline mediums. d Calculate voltage amplitude C’ for each successive reflected 4N ps e wave Total voltage at any point is the 5N ps sum of all the waves that have reached that point Transmission Lines Class 6 | 5,496 | 17,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-17 | latest | en | 0.766667 |
https://www.jiskha.com/display.cgi?id=1267507460 | 1,511,576,011,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809229.69/warc/CC-MAIN-20171125013040-20171125033040-00651.warc.gz | 795,398,501 | 3,733 | # geography
posted by .
what city is anabijd scrambled?
• geography -
Sra (aka Mme)
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https://origin.geeksforgeeks.org/calculator-using-pysimplegui-python/ | 1,660,678,170,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572515.15/warc/CC-MAIN-20220816181215-20220816211215-00174.warc.gz | 415,078,792 | 30,061 | # Calculator using PySimpleGUI – Python
• Difficulty Level : Hard
• Last Updated : 24 Feb, 2021
Prerequisite: PySimpleGUI, eval
PySimpleGUI is a Python package that enables Python programmers of all levels to create GUIs. You specify your GUI window using a “layout” that contains widgets (they’re called “Elements” in PySimpleGUI). In this article, we will learn, how to make a calculator using PySimpleGUI in Python.
Before starting we need to install this package:
`pip install PySimpleGUI`
Approach:
• Import the PySimpleGUI Module
• Create GUI layout and Window
• Add any number of widgets to the main window
• Apply the event Trigger on the widgets.
Below is what the GUI looks like:
Let’s create a GUI-based simple calculator using the Python PySimpleGUI module, which can perform basic arithmetic operation’s addition, subtraction, multiplication, and division.
Let’s Understand step by step implementation:-
Step1: Create Text Box, Buttons
For creating a text box, we will use Txt() method.
Syntax: Txt(Enter Text, *attr)
Step 2: Create an infinite loop, read the button value and perform the operation.
## Python3
`# Result Value ` `Result ``=` `'' ` ` ` `# Make Infinite Loop ` `while` `True``: ` ` ``# Button Values ` ` ``button, value ``=` `form.Read() ` ` ` ` ``# Check Press Button Values ` ` ``if` `button``=``=``'c'``: ` ` ``Result ``=` `'' ` ` ``form.FindElement(``'input'``).Update(Result) ` ` ``elif` `button``=``=``'«'``: ` ` ``Result ``=` `Result[:``-``1``] ` ` ``form.FindElement(``'input'``).Update(Result) ` ` ``elif` `len``(Result) ``=``=` `16` `: ` ` ``pass` ` ` ` ``# Results ` ` ``elif` `button``=``=``'='``: ` ` ``Answer ``=` `eval``(Result) ` ` ``Answer ``=` `str``(``round``(``float``(Answer),``3``)) ` ` ``form.FindElement(``'input'``).Update(Answer) ` ` ``Result ``=` `Answer ` ` ``# close the window ` ` ``elif` `button``=``=``'Quit'` `or` `button``=``=``None``: ` ` ``break` ` ``else``: ` ` ``Result ``+``=` `button ` ` ``form.FindElement(``'input'``).Update(Result)`
Below is the full implementation :
## Python3
`# Import Module ` `from` `PySimpleGUI ``import` `*` ` ` `# GUI Layout ` `layout ``=` `[[Txt('' ``*` `10``)], ` ` ``[Text('``', size = (15, 1), font = ('``Helvetica', ``18``), ` ` ``text_color ``=` `'black'``, key ``=` `'input'``)], ` ` ``[Txt('' ``*` `10``)], ` ` ``[ReadFormButton(``'c'``), ReadFormButton(``'«'``)], ` ` ``[ReadFormButton(``'7'``), ReadFormButton(``'8'``), ReadFormButton(``'9'``), ReadFormButton(``'/'``)], ` ` ``[ReadFormButton(``'4'``), ReadFormButton(``'5'``), ReadFormButton(``'6'``), ReadFormButton(``'*'``)], ` ` ``[ReadFormButton(``'1'``), ReadFormButton(``'2'``), ReadFormButton(``'3'``), ReadFormButton(``'-'``)], ` ` ``[ReadFormButton(``'.'``), ReadFormButton(``'0'``), ReadFormButton(``'='``), ReadFormButton(``'+'``)], ` ` ``] ` ` ` `# Set PySimpleGUI ` `form ``=` `FlexForm(``'CALCULATOR'``, default_button_element_size ``=` `(``5``, ``2``), ` ` ``auto_size_buttons ``=` `False``, grab_anywhere ``=` `False``) ` `form.Layout(layout) ` ` ` `# Result Value ` `Result ``=` `'' ` ` ` `# Make Infinite Loop ` `while` `True``: ` ` ``# Button Values ` ` ``button, value ``=` `form.Read() ` ` ` ` ``# Check Press Button Values ` ` ``if` `button ``=``=` `'c'``: ` ` ``Result ``=` `'' ` ` ``form.FindElement(``'input'``).Update(Result) ` ` ``elif` `button``=``=``'«'``: ` ` ``Result ``=` `Result[:``-``1``] ` ` ``form.FindElement(``'input'``).Update(Result) ` ` ``elif` `len``(Result) ``=``=` `16` `: ` ` ``pass` ` ` ` ``# Results ` ` ``elif` `button ``=``=` `'='``: ` ` ``Answer ``=` `eval``(Result) ` ` ``Answer ``=` `str``(``round``(``float``(Answer),``3``)) ` ` ``form.FindElement(``'input'``).Update(Answer) ` ` ``Result ``=` `Answer ` ` ` ` ``# close the window ` ` ``elif` `button ``=``=` `'Quit'` `or` `button ``=``=` `None``: ` ` ``break` ` ``else``: ` ` ``Result ``+``=` `button ` ` ``form.FindElement(``'input'``).Update(Result)`
Output:
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Applied Math Applied Math Forum
March 11th, 2019, 04:51 PM #1 Newbie Joined: Feb 2019 From: Watertown NY USA Posts: 5 Thanks: 0 3 = [f_p][ln(b)][b^(f_p)]/ [b^(f_p) - 1]. On Wednesday, February 8, 2017 at 12:20:05 PM UTC-5, Simon Roberts wrote: > Result (version 5.4): (all version prior to feb 08 2017 may be wrong) > > (14->15) 3 = [f_p][ln(b)][b^(f_p)]/ [b^(f_p) - 1]. > > where, > > (6) b = e^[48 x 10^(-12)/ T](Kelvins*sec). > > and, > f_p is the frequency of peak intensity > > Planck's Law for a given temperature, T. > > Proof. > > Givens. > > (1) B(f,T) = c_0*(f^3/(b^f - 1)) where, > > (2) b = e^(h/(k_b*T)) > > (3) k_b = 1.38 x 10^(-23) Joules / Kelvin. > > (4) h = 6.626×10-34 Joule.sec > > (5) h/(k_b*T) = [48 x 10^(-12)/ T](Kelvins*sec). > > (6) b = e^[48 x 10^(-12)/ T](Kelvins*sec). > > (7) ln(b) = [48 x 10^(-12)/ T](Kelvins*sec). > > ( c_0 = 2*h / c^2. > > (9) c = 299 792 458 m / s. > > (10) c_0 = 2(6.626×10-34 Joule.sec/[299.792458x10^(6) m / s]^2. > > (10.1) c_0 = .147448 x 10^(-49) joules*sec^3/meter^2 > > (10.2) c_0 = 14.7448 x 10^(-51) joules*sec^3/meter^2 > > End Givens. > > Main section of proof. > > Taking the natural logarithm of B(f), > > (11) ln(B(f)) = ln(c_0) + 3ln(f) - ln(b^f - 1), > > Taking derivative of (11) w.r.t. f yields, > > (12) B'(f)/B(f) = 3/(f_p) - [ln(b)b^(f_p)]/ [b^(f_p) - 1]. > > Assuming B(f) =/= 0 and f = f_p: B(f_p) is a maxima; > > (d/df)B(f) = B'(f_p)= 0. > > (12->13) 0 = 3/(f_p) - [ln(b)b^(f_p)]/ [b^(f_p) - 1]. > > (13 -> 14) 3/(f_p) = [ln(b)b^(f_p)]/ [b^(f_p) - 1]. > > (14->15) 3 = [f_p][ln(b)][b^(f_p)]/ [b^(f_p) - 1]. > > where, > > (6) b = e^[48 x 10^(-12)/ T](Kelvins*sec). > > and, > f_p is the frequency of peak intensity of Planck's Law, for a given temperature T. Simon C. Robert
March 17th, 2019, 05:01 PM #2 Senior Member Joined: Feb 2016 From: Australia Posts: 1,801 Thanks: 636 Math Focus: Yet to find out. ?? Thanks from Greens
March 17th, 2019, 06:06 PM #3
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Originally Posted by Joppy ??
???
March 17th, 2019, 06:08 PM #4
Senior Member
Joined: Feb 2016
From: Australia
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Math Focus: Yet to find out.
Quote:
Originally Posted by Denis ???
(??)^2
March 17th, 2019, 06:29 PM #5
Math Team
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From: The Astral plane
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Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by retenshun On Wednesday, February 8, 2017 at 12:20:05 PM UTC-5, Simon Roberts wrote: > Result (version 5.4): (all version prior to feb 08 2017 may be wrong) > > (14->15) 3 = [f_p][ln(b)][b^(f_p)]/ [b^(f_p) - 1]. > > where, > > (6) b = e^[48 x 10^(-12)/ T](Kelvins*sec). > > and, > f_p is the frequency of peak intensity > > Planck's Law for a given temperature, T. > > Proof. > > Givens. > > (1) B(f,T) = c_0*(f^3/(b^f - 1)) where, > > (2) b = e^(h/(k_b*T)) > > (3) k_b = 1.38 x 10^(-23) Joules / Kelvin. > > (4) h = 6.626×10-34 Joule.sec > > (5) h/(k_b*T) = [48 x 10^(-12)/ T](Kelvins*sec). > > (6) b = e^[48 x 10^(-12)/ T](Kelvins*sec). > > (7) ln(b) = [48 x 10^(-12)/ T](Kelvins*sec). > > ( c_0 = 2*h / c^2. > > (9) c = 299 792 458 m / s. > > (10) c_0 = 2(6.626×10-34 Joule.sec/[299.792458x10^(6) m / s]^2. > > (10.1) c_0 = .147448 x 10^(-49) joules*sec^3/meter^2 > > (10.2) c_0 = 14.7448 x 10^(-51) joules*sec^3/meter^2 > > End Givens. > > Main section of proof. > > Taking the natural logarithm of B(f), > > (11) ln(B(f)) = ln(c_0) + 3ln(f) - ln(b^f - 1), > > Taking derivative of (11) w.r.t. f yields, > > (12) B'(f)/B(f) = 3/(f_p) - [ln(b)b^(f_p)]/ [b^(f_p) - 1]. > > Assuming B(f) =/= 0 and f = f_p: B(f_p) is a maxima; > > (d/df)B(f) = B'(f_p)= 0. > > (12->13) 0 = 3/(f_p) - [ln(b)b^(f_p)]/ [b^(f_p) - 1]. > > (13 -> 14) 3/(f_p) = [ln(b)b^(f_p)]/ [b^(f_p) - 1]. > > (14->15) 3 = [f_p][ln(b)][b^(f_p)]/ [b^(f_p) - 1]. > > where, > > (6) b = e^[48 x 10^(-12)/ T](Kelvins*sec). > > and, > f_p is the frequency of peak intensity of Planck's Law, for a given temperature T. Simon C. Robert
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http://web2.0calc.com/questions/in-the-diagram-below-we-have-overline-mn-parallel-overline-pq-om-9-oq-6-and-pn-24-find-op-asy-pair-o-m-nn-p-q-m-0-4-0-5-nn | 1,511,461,376,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806856.86/warc/CC-MAIN-20171123180631-20171123200631-00365.warc.gz | 318,019,415 | 8,605 | +0
# In the diagram below, we have $\overline{MN}\parallel\overline{PQ}$, $OM = 9$, $OQ = 6$, and $PN = 24$. Find $OP$. [asy] pair O,M,NN,P,Q; M = (-0.4,0.5); NN =
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$$In the diagram below, we have \overline{MN}\parallel\overline{PQ}, OM = 9, OQ = 6, and PN = 24. Find OP. [asy] pair O,M,NN,P,Q; M = (-0.4,0.5); NN = (0.3,0.5); P = -0.9*NN; Q = -0.9*M; label("O",O,E); label("M", M,N); label("N", NN,N); label("P",P,S); label("Q",Q,S); draw(M--NN--P--Q--M); [/asy]$$In the diagram below, we have , , , and . Find .
Mellie Oct 31, 2015
#9
+1035
+10
LancelotLink turns into a really POd chimp when I hire away his de/coders–especially if I don’t cross his palm with a few bananas. He’s almost as POed when he catches Sir LaTex gold-bricking –he’s very skilled at it. Lancelot throws banana peels at him after catching him perusing his stamp collection on company time.
Along with roasted peanuts, Sir La-Tex likes the occasional mouse, but I am uncertain if he likes shrimp/prawns. I’ve not had BBQed prawns before, but I often have BBQed crawdads –or eat them raw – they are a troll staple and easily found near troll toll bridges. I may try some BBQed prawns the next time I BBQ a steak for my dogs. We dogs (and trolls) really like our steaks.
Nauseated Nov 2, 2015
Sort:
#1
+91049
+10
Sorry Mellie,
Melody Oct 31, 2015
#2
+1035
+10
The embedded (La)tex compiler for this forum does not render Asy(mptote) code.
In the diagram below, we have $\overline{MN}\parallel\overline{PQ}$, $OM = 9$, $OQ = 6$, and $PN = 24$. Find $OP$. [asy] pair O,M,NN,P,Q; M = (-0.4,0.5); NN = (0.3,0.5); P = -0.9*NN; Q = -0.9*M; label("$O$",O,E); label("$M$", M,N); label("$N$", NN,N); label("$P$",P,S); label("$Q$",Q,S); draw(M--NN--P--Q--M); [/asy]
Here’s image of the rendered code:
Nauseated Oct 31, 2015
#3
+91049
+10
Thanks Nauseated.
MN is parallel to PQ
Consider triangle MNO and triangle QPO
< MOP = < QOM Vertiacally opposite angles are congruent.
<MNO = < QPO Alternate angles on parallel lines are congruent
<NMO = < PQO Alternate angles on parallel lines are congruent
Therefore triangle MNO and triangle QPO (3 equal angles.)
Therefore
$$\frac{0M}{ON}=\frac{OQ}{OP}\\ \frac{9}{ON}=\frac{6}{OP}\\ \frac{9}{6}=\frac{ON}{OP}\\ \frac{3}{2}=\frac{ON}{OP}\\ But \;NP=24\\ \mbox{We need to divide 24 in the ratio 3:2}\\ 3/5*24 = 14.4\\ 2/5*24 = 9.6\\ So\\ ON=14.4\;units\;\;\; and\;\;\; OP=9.6\;units$$
Mellie, if you do not understand you need to ask questions :)
Melody Oct 31, 2015
#4
+78750
+10
Triangle MON ≈ Triangle QOP by AA congruency....therefore.....
OM / ON = OQ / OP Let ON = x.....then OP = 24 - x.......and we have
9 / x = 6 / [ 24 - x] cross-multiply
9 [ 24 - x ] = 6x simplify
216 - 9x = 6x add 9x to both sides
216 = 15 x
x = 216/15 = 14.4 = ON
And OP = 24 - 14.4 = 9.6
CPhill Oct 31, 2015
#5
+78750
+10
Don't worry, Nauseated.....I think I have found the perfect "LaTex" detective amongst your brood of monkeys.....Sir "LA-TeX"......that name is not a coincidence......he actually has legal residences in both states.......he's a whiz at deciphering the indecipherable.......he's working on Mellie's codes as we speak........having some experience with cracking the Japanese "Purple Codes" and the German "Enigma Codes" in WWII.....this should be cinch for him.....!!!
CPhill Oct 31, 2015
edited by CPhill Oct 31, 2015
#6
+91049
+10
WOW Chris, this is great. We now have Sir La-TeX working with us.
I know he works for bananas too!!
Just as well we have no greenbacks or Aussie plastic to pay him with!
We could probably throw in a few peanuts though. I know chimps are partial to peanuts.
Welcome to the team Sir La-TeX
Melody Oct 31, 2015
#7
+78750
+5
Sir LA-TeX isn't partial to "normal" chimp food........he's more of a steak and seafood type........maybe you can throw a shrimp on the barbie for him.....????
CPhill Oct 31, 2015
edited by CPhill Oct 31, 2015
#8
+91049
+10
Australians do not have shrimps. We have prawns and we rarely BBQ them.
You have been watching too much of Paul Hogan's silly ads!! Shame on you!!
Anyway, as much as La-TeX may like prawns, they are not good for him.
If we allow him to eat rubbish he will get sick and then we will not get our peanuts and bananas worth out of him.
NO La-TeX will not be fed prawns on my watch!
Melody Oct 31, 2015
#9
+1035
+10
LancelotLink turns into a really POd chimp when I hire away his de/coders–especially if I don’t cross his palm with a few bananas. He’s almost as POed when he catches Sir LaTex gold-bricking –he’s very skilled at it. Lancelot throws banana peels at him after catching him perusing his stamp collection on company time.
Along with roasted peanuts, Sir La-Tex likes the occasional mouse, but I am uncertain if he likes shrimp/prawns. I’ve not had BBQed prawns before, but I often have BBQed crawdads –or eat them raw – they are a troll staple and easily found near troll toll bridges. I may try some BBQed prawns the next time I BBQ a steak for my dogs. We dogs (and trolls) really like our steaks.
Nauseated Nov 2, 2015
#10
+91049
+5
I knew Sir La-TeX is a very generous philanthropist, He has agreeed to help us I believe. If he is a gold bricker sometimes that is ok, Sir Lancelot should not come down on him to hard. He is working for bananas after all! I suppose when a chimp has the name like Sir Phil LaTeX it is very obvious that he would be a philatelist! I have some old stamps he may be interested in seeing! Give him a little space. Let him study his stamps. He may find some great code in there somewhere.
I think any Troll who is worth his salt would like sushi. He probably would like it best when it has ben naturally putrified by the resident croc or gator. Crocs and gators like to stow there dinner in underwater crevices and come back for it in a few weeks when it is soft and ripe for they have no teeth to chew it with. I think any self respecting troll would like it that way too! Steak can be prepared in this way too
Melody Nov 2, 2015
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